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in-animals-there-are-special-glands-secreting-hormones-whereas-there-are-no-glands-in-plants-where-are-plant-hormones-formed-how-are-the-hormones-tran

in-animals-there-are-special-glands-secreting-hormones-whereas-there-are-no-glands-in-plants-where-are-plant-hormones-formed-how-are-the-hormones-tran-68496

<div class="question">In animals, there are special glands secreting hormones, whereas there are no glands in plants. Where are plant hormones formed? How are the hormones translocated to the site of activity?</div>

['Biology', 'Plant - Growth and Development']

None

None

<div class="solution">The plant hormones are synthesised by the plant cells needed. Few hormones are specifically synthesised at a particular part of the plant like auxin synthesised in growing shoot apices and ethylene is secretes by ripened fruits. Cytokinin is found in dividing cells. Unlike plants animal being more advanced, and organised they have proper hormone secreting glands and organs. These are transported through the transport system of their body in both plant and animals. In plants, hormone are translocated via xylem and phloem to the site of activity.</div>

MarksBatch2_P1.db

in-dicot-stem-the-xylem-is

in-dicot-stem-the-xylem-is-58762

<div class="question">In dicot stem, the xylem is</div>

['Biology', 'Anatomy of Flowering Plants', 'NEET']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">Exarch</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">Mesarch</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">Centarch</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">Endarch</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value">Endarch</span> </div>

<div class="solution">There are four types of primary xylem development: <strong>exarch, endarch, centrarch, and mesarch.</strong><br/>In <strong>endarch </strong>development, the protoxylem begins its development from the innermost procambial cells located adjacent to the pith and development progresses outward. Therefore, the protoxylem is found toward the inside and metaxylem toward the outside of the stem. Endarch development is considered the most highly advanced type of primary xylem development.</div>

MarksBatch2_P1.db

in-the-cyanide-extraction-process-of-silver-from-argentite-ore-the-oxidising-and-reducing-agents-used-are

in-the-cyanide-extraction-process-of-silver-from-argentite-ore-the-oxidising-and-reducing-agents-used-are-72603

<div class="question">In the cyanide extraction process of silver from argentite ore, the oxidising and reducing agents used are</div>

['Chemistry', 'General Principles and Processes of Isolation of Metals', 'JEE Advanced', 'JEE Advanced 2012 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$\mathrm{O}_{2}$ and $\mathrm{CO}$ respectively</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$\mathrm{O}_{2}$ and $\mathrm{Zn}$ dust respectively</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$\mathrm{HNO}_{3}$ and $\mathrm{Zn}$ dust respectively</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$\mathrm{HNO}_{3}$ and $\mathrm{CO}$ respectively</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value">$\mathrm{O}_{2}$ and $\mathrm{Zn}$ dust respectively</span> </div>

<div class="solution">The reactions involved in cyanide extraction process are: <br/> <br/>$\underset{\text { (argentite) }}{\mathrm{Ag}_{2} \mathrm{~S}}+4 \mathrm{NaCN} \rightarrow 2 \mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]+\mathrm{Na}_{2} \mathrm{~S}$ <br/> <br/>$4 \mathrm{Na}_{2} \mathrm{~S}+\underset{\text { Oxiding agent }}{5 \mathrm{O}_{2}}+2 \mathrm{H}_{2} \mathrm{O} \rightarrow 2 \mathrm{Na}_{2} \mathrm{SO}_{4}+4 \mathrm{NaOH}+2 \mathrm{~S}$ <br/> <br/>$2 \mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]+\underset{\begin{array}{c}\text { (reducing } \\ \text { agent) }\end{array}}{\mathrm{Zn}} \rightarrow \mathrm{Na}_{2}\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]+2 \mathrm{Ag} \downarrow$</div>

MarksBatch2_P1.db

in-the-determination-of-youngs-modulus-y-l-d-2-4-m-lg-by-using-searles-method-a-wire-of-length-l-2-m-and-diameter-d-05-mm-is-used-for-a-load-m-25-kg-a

in-the-determination-of-youngs-modulus-y-l-d-2-4-m-lg-by-using-searles-method-a-wire-of-length-l-2-m-and-diameter-d-05-mm-is-used-for-a-load-m-25-kg-a-99762

<div class="question">In the determination of Young's modulus $\left(Y=\frac{4 M L g}{\pi l d^{2}}\right)$ by using Searle's method, a wire of length $L=2 \mathrm{~m}$ and diameter $d=0.5 \mathrm{~mm}$ is used. For a load $M=2.5 \mathrm{~kg}$, an extension $l=0.25 \mathrm{~mm}$ in the length of the wire is observed. Quantities $d$ and $l$ are measured using a screw gauge and a micrometer, respectively. They have the same pitch of $0.5 \mathrm{~mm}$. The number of divisions on their circular scale is 100 . The contributions to the maximum probable error of the $Y$ measurement</div>

['Physics', 'Mathematics in Physics', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">due to the errors in the measurements of $d$ and $l$ are the same.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">due to the error in the measurement of $d$ is twice that due to the error in the measurement of $l$.</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">due to the error in the measurement of $l$ is twice that due to the error in the measurement of $d$.</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">due to the error in the measurement of $d$ is four times that due to the error in the measurement of $l$.</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value">due to the errors in the measurements of $d$ and $l$ are the same.</span> </div>

<div class="solution">The maximum possible error in $Y$ due to $l$ and $d$ $\frac{\Delta Y}{Y}=\frac{\Delta l}{l}+\frac{2 \Delta d}{d}$ Least count $=\frac{\text { Pitch }}{\text { No. of division on circular scale }}$ $=\frac{0.5}{100} \mathrm{~mm}=0.005 \mathrm{~mm}$ <br/> <br/>Here, $\Delta d=\Delta l=0.005 \mathrm{~mm}$ <br/> <br/>Error contribution of $l=\frac{\Delta l}{l}=\frac{0.005 \mathrm{~mm}}{0.25 \mathrm{~mm}}=\frac{1}{50}$ <br/> <br/>Error contribution of <br/> <br/>$d=\frac{2 \Delta d}{d}=\frac{2 \times 0.005 \mathrm{~mm}}{0.5 \mathrm{~mm}}=\frac{1}{50}$ <br/> <br/>Hence contribution to the maximum possible error in the measurement of $y$ due to $l$ and $d$ is the same.</div>

MarksBatch2_P1.db

in-the-experiment-to-determine-the-speed-of-sound-using-a-resonance-column

in-the-experiment-to-determine-the-speed-of-sound-using-a-resonance-column-53529

<div class="question">In the experiment to determine the speed of sound using a resonance column:</div>

['Physics', 'Waves and Sound', 'JEE Advanced', 'JEE Advanced 2007 (Paper 2)']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>Prongs of the tuning fork are kept in a vertical plane<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>Prongs of the tuning fork are kept in a horizontal plane<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>In one of the two resonances observed, the length of the resonating air column is close to the wavelength of sound in air<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>In one of the two resonances observed, the length of the resonating air column is close to half of the wavelength of sound in air</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>Prongs of the tuning fork are kept in a vertical plane<br/></span> </div>

<div class="solution">Length of air column in resonance is odd integer multiple of $\frac{\lambda}{4}$. Correct option is (a).</div>

MarksBatch2_P1.db

in-the-following-carbocation-h-ch-3-that-is-most-likely-to-migrate-to-the-positively-charged-carbon-is

in-the-following-carbocation-h-ch-3-that-is-most-likely-to-migrate-to-the-positively-charged-carbon-is-49327

<div class="question">In the following carbocation, $\mathrm{H} / \mathrm{CH}_3$ that is most likely to migrate to the positively charged carbon is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/SL8Rsu5u8piwlrnZ6cwJsXNsIB9cai7C5FH3HNiqyBU.original.fullsize.png"/><br/></div>

['Chemistry', 'General Organic Chemistry', 'JEE Advanced', 'JEE Advanced 2009 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\mathrm{CH}_3$ at $\mathrm{C}-4$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\mathrm{H}$ at $\mathrm{C}-4$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\mathrm{CH}_3$ at $\mathrm{C}-2$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$\mathrm{H}$ at $\mathrm{C}-2$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$\mathrm{H}$ at $\mathrm{C}-2$</span> </div>

<div class="solution">Hydride shift from $\mathrm{C}-2$ will give most stable resonance stabilised carbocation as Fig.</div>

MarksBatch2_P1.db

in-the-following-reaction-sequence-the-compound-j-is-an-intermediate-j-c-9-h-8-o-2-gives-effervescence-on-treatment-with-nahco-3-and-a-positive-baeyer-1

in-the-following-reaction-sequence-the-compound-j-is-an-intermediate-j-c-9-h-8-o-2-gives-effervescence-on-treatment-with-nahco-3-and-a-positive-baeyer-1-22814

<div class="question">In the following reaction sequence, the compound $\mathrm{J}$ is an intermediate. <br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/wEQdUNBgA0a1fYiZUUKvCk8wMg-9NCH2X7b_nCIMm3Q.original.fullsize.png"/><br/> <br/> <br/>$\mathbf{J}\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{2}\right)$ gives effervescence on treatment with $\mathrm{NaHCO}_{3}$ and a positive Baeyer's test.<strong> Question:</strong> The compound $\mathbf{K}$ is</div>

['Chemistry', 'Aldehydes and Ketones', 'JEE Advanced', 'JEE Advanced 2012 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/GJpVx3Z5Kjrw3UYhem7itdrTO4TaiMoigUAMz-h4PHE.original.fullsize.png"/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/KolWgJGmAxktnW3C3T3iB8xUtV0SZCNdmz5WSzM20ME.original.fullsize.png"/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/pCWVN9JiSgXxc_KWdW_rXvmW3HAfvAZbJpBNL_cfHCg.original.fullsize.png"/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/u_T7Cf0hYaMEGcVrJRUWQxTblFcUF2Qaq3ZuiQIhFDY.original.fullsize.png"/></span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/KolWgJGmAxktnW3C3T3iB8xUtV0SZCNdmz5WSzM20ME.original.fullsize.png"/></span> </div>

<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/shdddhtgg76t676g3gh.png"/></div>

MarksBatch2_P1.db

in-the-following-reaction-sequence-the-compound-j-is-an-intermediate-j-c-9-h-8-o-2-gives-effervescence-on-treatment-with-nahco-3-and-a-positive-baeyer-2

in-the-following-reaction-sequence-the-compound-j-is-an-intermediate-j-c-9-h-8-o-2-gives-effervescence-on-treatment-with-nahco-3-and-a-positive-baeyer-2-16444

<div class="question">In the following reaction sequence, the compound $\mathrm{J}$ is an intermediate. <br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/wEQdUNBgA0a1fYiZUUKvCk8wMg-9NCH2X7b_nCIMm3Q.original.fullsize.png"/><br/> <br/> <br/>$\mathbf{J}\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{2}\right)$ gives effervescence on treatment with $\mathrm{NaHCO}_{3}$ and a positive Baeyer's test.<strong> Question:</strong> The compound $\mathbf{I}$ is</div>

['Chemistry', 'Aldehydes and Ketones', 'JEE Main']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/DFnR6NI_l1GHvwTdefAymtSlz3fe_-1fJk_APNOF6HI.original.fullsize.png"/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/cD1NtWzOfi1ZhqVr1ncyJhBdA1gsgHcYOjb94jRxclA.original.fullsize.png"/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/LtE_4vaHfn-ocsFI-ffFslfDRKVE7tcPR1WMWf9dv28.original.fullsize.png"/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/o7Am4486gKoykHYLsk2LpWs2noPCnZUcna8-qsZp6PE.original.fullsize.png"/></span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/DFnR6NI_l1GHvwTdefAymtSlz3fe_-1fJk_APNOF6HI.original.fullsize.png"/></span> </div>

<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/shdddhtgg76t676g3gh.png"/></div>

MarksBatch2_P1.db

in-the-following-reaction-sequence-the-compound-j-is-an-intermediate-j-c-9-h-8-o-2-gives-effervescence-on-treatment-with-nahco-3-and-a-positive-baeyer

in-the-following-reaction-sequence-the-compound-j-is-an-intermediate-j-c-9-h-8-o-2-gives-effervescence-on-treatment-with-nahco-3-and-a-positive-baeyer-26143

<div class="question">In the following reaction sequence, the compound $\mathrm{J}$ is an intermediate. <br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/wEQdUNBgA0a1fYiZUUKvCk8wMg-9NCH2X7b_nCIMm3Q.original.fullsize.png"/><br/> <br/> <br/>$\mathbf{J}\left(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{2}\right)$ gives effervescence on treatment with $\mathrm{NaHCO}_{3}$ and a positive Baeyer's test.<strong> Question:</strong> The compound $\mathbf{I}$ is</div>

['Chemistry', 'Aldehydes and Ketones', 'JEE Advanced', 'JEE Advanced 2012 (Paper 2)']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/DFnR6NI_l1GHvwTdefAymtSlz3fe_-1fJk_APNOF6HI.original.fullsize.png"/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/cD1NtWzOfi1ZhqVr1ncyJhBdA1gsgHcYOjb94jRxclA.original.fullsize.png"/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/LtE_4vaHfn-ocsFI-ffFslfDRKVE7tcPR1WMWf9dv28.original.fullsize.png"/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/o7Am4486gKoykHYLsk2LpWs2noPCnZUcna8-qsZp6PE.original.fullsize.png"/></span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/DFnR6NI_l1GHvwTdefAymtSlz3fe_-1fJk_APNOF6HI.original.fullsize.png"/></span> </div>

<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/shdddhtgg76t676g3gh.png"/></div>

MarksBatch2_P1.db

in-the-following-reaction-sequence-the-correct-structures-of-e-f-and-g-are-1

in-the-following-reaction-sequence-the-correct-structures-of-e-f-and-g-are-1-34398

<div class="question">In the following reaction sequence, the correct structures of $E, F$ and $G$ are<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/vRmeGTCFSFL0D1alGxSueCfK-zdY5zwtUD9rhX6j-Mg.original.fullsize.png"/><br/></div>

['Chemistry', 'Alcohols Phenols and Ethers', 'JEE Main']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/k3F3-7je2haDq7_12n1P1m5iFD6n82eA0DYWjE1HQbs.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/nhD7m9Clo_dCMTHogT3HcNGdzB6jIs1g_8t6MetAbwA.original.fullsize.png"/><br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/VnlxbctDCuLlZbYpQySukWy9tIGd9deajMeEqbrS6bQ.original.fullsize.png"/><br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/OL4wL-dWZlA7CibKKqumxbpIDNibFSKMC_iHMcv5R2A.original.fullsize.png"/><br/></span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/VnlxbctDCuLlZbYpQySukWy9tIGd9deajMeEqbrS6bQ.original.fullsize.png"/><br/></span> </div>

<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/M_hmHQ0sm0mLvJzSz32L6boXFj6Tv5MPfvvVGHq4HVU.original.fullsize.png"/><br/></div>

MarksBatch2_P1.db

in-the-following-reaction-sequence-the-correct-structures-of-e-f-and-g-are

in-the-following-reaction-sequence-the-correct-structures-of-e-f-and-g-are-47720

<div class="question">In the following reaction sequence, the correct structures of $E, F$ and $G$ are<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/vRmeGTCFSFL0D1alGxSueCfK-zdY5zwtUD9rhX6j-Mg.original.fullsize.png"/><br/></div>

['Chemistry', 'Alcohols Phenols and Ethers', 'JEE Advanced', 'JEE Advanced 2008 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/k3F3-7je2haDq7_12n1P1m5iFD6n82eA0DYWjE1HQbs.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/nhD7m9Clo_dCMTHogT3HcNGdzB6jIs1g_8t6MetAbwA.original.fullsize.png"/><br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/VnlxbctDCuLlZbYpQySukWy9tIGd9deajMeEqbrS6bQ.original.fullsize.png"/><br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/OL4wL-dWZlA7CibKKqumxbpIDNibFSKMC_iHMcv5R2A.original.fullsize.png"/><br/></span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/VnlxbctDCuLlZbYpQySukWy9tIGd9deajMeEqbrS6bQ.original.fullsize.png"/><br/></span> </div>

<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/M_hmHQ0sm0mLvJzSz32L6boXFj6Tv5MPfvvVGHq4HVU.original.fullsize.png"/><br/></div>

MarksBatch2_P1.db

in-the-following-reaction-the-structure-of-the-major-product-x-is-1

in-the-following-reaction-the-structure-of-the-major-product-x-is-1-89851

<div class="question">In the following reaction,<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/BeYLjJnusd9Xzxl0vznAVx5d0UUSSlQE221UaLXUoJc.original.fullsize.png"/><br/><br/>$$<br/>\text { The structure of the major product } X \text { is }<br/>$$</div>

['Chemistry', 'Amines', 'JEE Advanced', 'JEE Advanced 2007 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/26YC4HDKD_i5rzE2S69CDLXPFvNtlToWRf3zX9dT5rs.original.fullsize.png"/><br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/RMr84-hG7z5Kz2mC9SK5vlioGK4LCoSiGYozBfbXyUE.original.fullsize.png"/><br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/0GMRPjdKfsKZ3p2JjvHqXxJU5b7ymc8nbnUpmhTnxG4.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/> &lt; smiles&gt;O=C(Nc1cccc([N+](=O)[O-])c1)c1ccccc1 &lt; /smiles&gt;</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/RMr84-hG7z5Kz2mC9SK5vlioGK4LCoSiGYozBfbXyUE.original.fullsize.png"/><br/></span> </div>

<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/H9cdbc58aJUxa1ExyAOXQsSe72xfyiieY5zntgW4w3g.original.fullsize.png"/><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/wgJoK_F7PbJi-ma8hY9layaKmAHf498O9xvDMW8tGzU.original.fullsize.png"/><br/><br/>Ring 1 is more active, electrophilic substitution takes place over ring 1. - $\mathrm{NH}-\mathrm{C}-\mathrm{Ph}$ is ortho-para directing, para product is predominating.</div>

MarksBatch2_P1.db

in-the-following-reaction-the-structure-of-the-major-product-x-is-2

in-the-following-reaction-the-structure-of-the-major-product-x-is-2-43697

<div class="question">In the following reaction,<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/BeYLjJnusd9Xzxl0vznAVx5d0UUSSlQE221UaLXUoJc.original.fullsize.png"/><br/><br/>$$<br/>\text { The structure of the major product } X \text { is }<br/>$$</div>

['Chemistry', 'Amines', 'JEE Main']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/26YC4HDKD_i5rzE2S69CDLXPFvNtlToWRf3zX9dT5rs.original.fullsize.png"/><br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/RMr84-hG7z5Kz2mC9SK5vlioGK4LCoSiGYozBfbXyUE.original.fullsize.png"/><br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/0GMRPjdKfsKZ3p2JjvHqXxJU5b7ymc8nbnUpmhTnxG4.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/> &lt; smiles&gt;O=C(Nc1cccc([N+](=O)[O-])c1)c1ccccc1 &lt; /smiles&gt;</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/RMr84-hG7z5Kz2mC9SK5vlioGK4LCoSiGYozBfbXyUE.original.fullsize.png"/><br/></span> </div>

<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/H9cdbc58aJUxa1ExyAOXQsSe72xfyiieY5zntgW4w3g.original.fullsize.png"/><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/wgJoK_F7PbJi-ma8hY9layaKmAHf498O9xvDMW8tGzU.original.fullsize.png"/><br/><br/>Ring 1 is more active, electrophilic substitution takes place over ring 1. - $\mathrm{NH}-\mathrm{C}-\mathrm{Ph}$ is ortho-para directing, para product is predominating.</div>

MarksBatch2_P1.db

in-the-following-x-denotes-the-greatest-integer-less-than-or-equal-to-x-match-the-functions-in-column-i-with-the-properties-column-ii-1

in-the-following-x-denotes-the-greatest-integer-less-than-or-equal-to-x-match-the-functions-in-column-i-with-the-properties-column-ii-1-59536

<div class="question">In the following $[x]$ denotes the greatest integer less than or equal to $x$. Match the functions in Column I with the properties Column II.<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/ZOJzgUXgnR79Tqmkejx4TZv8JzPFeZlpF6VMfpG-z9A.original.fullsize.png"/><br/></div>

['Mathematics', 'Continuity and Differentiability', 'JEE Main']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>A-p, r; B-p, q, s; C-p; D-q, r<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>A-p, s; B-p, r, s; C-p, s; D-q, r<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>A-p, q, r; B-p, s; C-r, s; D-p, q<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>A-r, s; B-q; C-p; D-p, q, s</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>A-p, q, r; B-p, s; C-r, s; D-p, q<br/></span> </div>

<div class="solution">(A) $x|x|$ is continuous, differentiable and strictly increasing in $(-1,1)$.<br/>(B) $\sqrt{|x|}$ is continuous in $(-1,1)$ and not differentiable at $x=0$.<br/>(C) $x+[x]$ is strictly increasing in $(-1,1)$ and discontinuous at $x=0$ $\Rightarrow$ not differentiable at $x=0$.<br/>(D) $|x-1|+|x+1|=2$ in $(-1,1)$<br/>The function is continuous and differentiable in $(-1,1)$.</div>

MarksBatch2_P1.db

in-the-following-x-denotes-the-greatest-integer-less-than-or-equal-to-x-match-the-functions-in-column-i-with-the-properties-column-ii

in-the-following-x-denotes-the-greatest-integer-less-than-or-equal-to-x-match-the-functions-in-column-i-with-the-properties-column-ii-51645

<div class="question">In the following $[x]$ denotes the greatest integer less than or equal to $x$. Match the functions in Column I with the properties Column II.<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/ZOJzgUXgnR79Tqmkejx4TZv8JzPFeZlpF6VMfpG-z9A.original.fullsize.png"/><br/></div>

['Mathematics', 'Continuity and Differentiability', 'JEE Advanced', 'JEE Advanced 2007 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>A-p, r; B-p, q, s; C-p; D-q, r<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>A-p, s; B-p, r, s; C-p, s; D-q, r<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>A-p, q, r; B-p, s; C-r, s; D-p, q<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>A-r, s; B-q; C-p; D-p, q, s</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>A-p, q, r; B-p, s; C-r, s; D-p, q<br/></span> </div>

<div class="solution">(A) $x|x|$ is continuous, differentiable and strictly increasing in $(-1,1)$.<br/>(B) $\sqrt{|x|}$ is continuous in $(-1,1)$ and not differentiable at $x=0$.<br/>(C) $x+[x]$ is strictly increasing in $(-1,1)$ and discontinuous at $x=0$ $\Rightarrow$ not differentiable at $x=0$.<br/>(D) $|x-1|+|x+1|=2$ in $(-1,1)$<br/>The function is continuous and differentiable in $(-1,1)$.</div>

MarksBatch2_P1.db

in-the-given-circuit-a-charge-of-80-c-is-given-to-the-upper-plate-of-the-4-f-capacitor-then-in-the-steady-state-the-charge-on-the-upper-plate-of-the-3-1

in-the-given-circuit-a-charge-of-80-c-is-given-to-the-upper-plate-of-the-4-f-capacitor-then-in-the-steady-state-the-charge-on-the-upper-plate-of-the-3-1-50593

<div class="question">In the given circuit, a charge of $+80 \mu C$ is given to the upper plate of the $4 \mu F$ capacitor. Then in the steady state, the charge on the upper plate of the $3 \mu F$ capacitor is <br/> <br/><img src=" data:image/png;base64,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"/></div>

['Physics', 'Capacitance', 'JEE Main']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$+32 \mu \mathrm{C}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$+40 \mu \mathrm{C}$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$+48 \mu \mathrm{C}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$+80 \mu \mathrm{C}$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value">$+48 \mu \mathrm{C}$</span> </div>

<div class="solution">The total charge on plate $A$ will be $80 \mu \mathrm{C}$. <br/> <br/>$2 \mu F$ and $3 \mu F$ capacitors are in parallel. Therefore, $C_{e q}=2+3=5 \mathrm{HF}$ <br/> <br/>Charge on capacitor of $3 \mu \mathrm{F}$ capacitance $q=\frac{3}{5} \times 80=48 \mu C$ <br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/hg2rttPavpB446uZyDbOiN4HL3hyjh-OD7VgG6EegZM.original.fullsize.png"/></div>

MarksBatch2_P1.db

in-the-given-circuit-a-charge-of-80-c-is-given-to-the-upper-plate-of-the-4-f-capacitor-then-in-the-steady-state-the-charge-on-the-upper-plate-of-the-3

in-the-given-circuit-a-charge-of-80-c-is-given-to-the-upper-plate-of-the-4-f-capacitor-then-in-the-steady-state-the-charge-on-the-upper-plate-of-the-3-24198

<div class="question">In the given circuit, a charge of $+80 \mu C$ is given to the upper plate of the $4 \mu F$ capacitor. Then in the steady state, the charge on the upper plate of the $3 \mu F$ capacitor is <br/> <br/><img src=" data:image/png;base64,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"/></div>

['Physics', 'Capacitance', 'JEE Advanced', 'JEE Advanced 2012 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$+32 \mu \mathrm{C}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$+40 \mu \mathrm{C}$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$+48 \mu \mathrm{C}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$+80 \mu \mathrm{C}$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value">$+48 \mu \mathrm{C}$</span> </div>

<div class="solution">The total charge on plate $A$ will be $80 \mu \mathrm{C}$. <br/> <br/>$2 \mu F$ and $3 \mu F$ capacitors are in parallel. Therefore, $C_{e q}=2+3=5 \mathrm{HF}$ <br/> <br/>Charge on capacitor of $3 \mu \mathrm{F}$ capacitance $q=\frac{3}{5} \times 80=48 \mu C$ <br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/hg2rttPavpB446uZyDbOiN4HL3hyjh-OD7VgG6EegZM.original.fullsize.png"/></div>

MarksBatch2_P1.db

in-the-given-circuit-the-ac-source-has-100-rad-s-considering-the-inductor-and-capacitor-to-be-ideal-the-correct-choices-is-are

in-the-given-circuit-the-ac-source-has-100-rad-s-considering-the-inductor-and-capacitor-to-be-ideal-the-correct-choices-is-are-78356

<div class="question">In the given circuit, the $\mathrm{AC}$ source has $\omega=100 \mathrm{rad} / \mathrm{s}$. Considering the inductor and capacitor to be ideal, the correct choice(s) is (are) <br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/OyGpPSLZMw0huKfOCnZEjjxWU3IFQF8cPowE_JIDs6E.original.fullsize.png"/><br/></div>

['Physics', 'Alternating Current', 'JEE Advanced', 'JEE Advanced 2012 (Paper 2)']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">The current through the circuit, $I$ is $0.3 \mathrm{~A}$.</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">The current through the circuit, $I$ is $0.3 \sqrt{2} A$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">The voltage across $100 \Omega$ resistor $=10 \sqrt{2} \mathrm{~V}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">The voltage across $50 \Omega$ resistor $=10 \mathrm{~V}$</span> </li> </ul>

<div class="correct-answer"> The correct answers are: <span class="option-value">The current through the circuit, $I$ is $0.3 \mathrm{~A}$., The voltage across $100 \Omega$ resistor $=10 \sqrt{2} \mathrm{~V}$</span> </div>

<div class="solution">Impedance across $A B, R C$ part of the circuit <br/> <br/>$\begin{aligned} \mathrm{Z}_{1}=&amp; \sqrt{X_{c}^{2}+R_{1}^{2}}=\sqrt{\left(\frac{1}{\omega C}\right)^{2}+R_{1}^{2}} \\=&amp; \sqrt{(100)^{2}+(100)^{2}}=100 \sqrt{2} \end{aligned}$ <br/> <br/>$\therefore \quad I_{1}=\frac{V}{Z_{1}}=\frac{20}{100 \sqrt{2}}$ <br/> <br/>$\quad\left[\right.$ leads emf by $\left.\phi_{1}\right]$ <br/> <br/>where $\cos \phi_{1}=\frac{R}{Z_{1}}=\frac{100}{100 \sqrt{2}}=\frac{1}{\sqrt{2}} \Rightarrow \theta=45^{\circ}$ <br/> <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/ed64m2hdoG2W80ioJw89XZcyqG4WBGJtaNWRyTdBbrs.original.fullsize.png"/><br/> <br/> <br/>Impedance across $C D, L R$ part of the circuit. <br/> <br/>$\mathrm{Z}_{2}=\sqrt{X_{L}^{2}+R_{2}^{2}}=\sqrt{(\omega L)^{2}+R_{2}^{2}}$ <br/> <br/>$=\sqrt{(0.5 \times 100)^{2}+(50)^{2}}=50 \sqrt{2} \Omega$ <br/> <br/>$\therefore \quad I_{2}=\frac{V}{Z_{2}}=\frac{20}{50 \sqrt{2}}$<img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/srQHHHyf8YGl8W5QJp-lTf3vffyvgV-w4nhnlGsmfq0.original.fullsize.png"/><br/> <br/> <br/>where $\cos \phi_{2}=\frac{R}{Z_{2}}=\frac{50}{50 \sqrt{2}}=\frac{1}{\sqrt{2}} \Rightarrow \phi_{2}=45^{\circ}$ $\therefore \quad$ Current $I$ from the circuit <br/> <br/>$I=\frac{20}{100 \sqrt{2}}+\frac{20}{50 \sqrt{2}}=\mathrm{I}_{1}+\mathrm{I}_{2} \simeq 0.3 \mathrm{~A}$</div>

MarksBatch2_P1.db

in-the-newman-projection-for-22dimethylbutane-x-and-y-can-respectively-be

in-the-newman-projection-for-22dimethylbutane-x-and-y-can-respectively-be-16190

<div class="question">In the Newman projection for 2,2-dimethylbutane<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/jyUNg_6mD-YRdCFk2RndTgNILG1dMpSe27gLekPA9TA.original.fullsize.png"/><br/><br/>$X$ and $Y$ can respectively be</div>

['Chemistry', 'General Organic Chemistry', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\mathrm{H}$ and $\mathrm{H}$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$\mathrm{H}$ and $\mathrm{C}_2 \mathrm{H}_5$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\mathrm{C}_2 \mathrm{H}_5$ and $\mathrm{H}$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$\mathrm{CH}_3$ and $\mathrm{CH}_3$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

<div class="correct-answer"> The correct answers are: <span class="option-value"><br/>$\mathrm{H}$ and $\mathrm{C}_2 \mathrm{H}_5$<br/>, <br/>$\mathrm{CH}_3$ and $\mathrm{CH}_3$</span> </div>

<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/LRcYxFJ8pgtTu7gJBL1lLnFmiLfSD0N9djoHPP9yG8U.original.fullsize.png"/><br/><br/><br/>Conformation projection along $\mathrm{C}_1-\mathrm{C}_2$<br/>$\mathrm{C}_1$ contains all three $\mathrm{Hs}$<br/>So, $\quad X=\mathrm{H}$<img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/UmKWGE4IzP_J8gN7hc4PzbG2cHLO0s06uQ6HUgtSrR8.original.fullsize.png"/><br/><br/><br/>$\mathrm{C}_2$ contains two methyl and one ethyl group<br/>So, $\quad Y=\mathrm{C}_2 \mathrm{H}_5$<br/>Conformational projection along<br/>$$<br/>\mathrm{C}_2-\mathrm{C}_3<br/>$$<br/>$\mathrm{C}_2$ contains three methyl groups<br/>$\left(\mathrm{C}_2\right.$ form back carbon in the given structure)<br/>So, $\quad Y=\mathrm{CH}_3$<img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/99XNPwbz6laYONQD43P2IwseZwQ1EWIPEfxCtXPMZA8.original.fullsize.png"/><br/><br/>$\mathrm{C}_3$ contains two Hs and one methyl group<br/>$\left(\mathrm{C}_3\right.$ form front carbon in the given structure)<br/>So, $\quad X=\mathrm{CH}_3$<br/>$\mathrm{C}_3$ contains two Hs and one methyl group<br/>$\left(\mathrm{C}_3\right.$ form front carbon in the given structure)<br/>So, $\quad X=\mathrm{CH}_3$<br/>Isomerism (Stereochemistry)<br/>Conceptual (Structural visualisation) III</div>

MarksBatch2_P1.db

in-the-options-given-below-let-e-denote-the-rest-mass-energy-of-a-nucleus-and-n-a-neutron-the-correct-option-is-2

in-the-options-given-below-let-e-denote-the-rest-mass-energy-of-a-nucleus-and-n-a-neutron-the-correct-option-is-2-49560

<div class="question">In the options given below, let $E$ denote the rest mass energy of a nucleus and $n$ a neutron. The correct option is</div>

['Physics', 'Nuclear Physics', 'JEE Advanced', 'JEE Advanced 2007 (Paper 1)']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$E\left({ }_{92}^{236} \mathrm{U}\right)&gt;E\left({ }_{53}^{137} \mathrm{I}\right)+E\left({ }_{36}^{97} Y\right)+2 E(n)$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\left.E{ }_{92}^{236} \mathrm{U}\right) &lt; E\left({ }_{53}^{137} \mathrm{I}\right)+E\left({ }_{39}^{97} Y\right)+2 E(n)$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$E\left({ }_{92}^{236} \mathrm{U}\right) &lt; E\left(5_{56}^{140} \mathrm{Ba}\right)+E\left({ }_{36}^{94} \mathrm{Kr}\right)+2 E(n)$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$E\left(9_{92}^{236} \mathrm{U}\right) &lt; E\left(5_{56}^{140} \mathrm{Ba}\right)+E\left({ }_{36}^{94} \mathrm{Kr}\right)+2 E(n)$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$E\left({ }_{92}^{236} \mathrm{U}\right)&gt;E\left({ }_{53}^{137} \mathrm{I}\right)+E\left({ }_{36}^{97} Y\right)+2 E(n)$<br/></span> </div>

<div class="solution">Rest mass of parent nucleus should be greater than the rest mass of daughter nuclei. Therefore, option (a) will be correct.</div>

MarksBatch2_P1.db

in-the-reaction-2-x-b-2-h-6-bh-2-x-2-bh-4-t-h-e-amin-e-s-x-is-are

in-the-reaction-2-x-b-2-h-6-bh-2-x-2-bh-4-t-h-e-amin-e-s-x-is-are-10748

<div class="question">In the reaction,<br/>$$<br/>2 \mathrm{X}+\mathrm{B}_2 \mathrm{H}_6 \longrightarrow\left[\mathrm{BH}_2(\mathrm{X})_2\right]^{+}\left[\mathrm{BH}_4\right]^{-}<br/>$$<br/>The amine (s) $X$, is (are)</div>

['Chemistry', 'p Block Elements (Group 13 & 14)', 'JEE Advanced', 'JEE Advanced 2009 (Paper 2)']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\mathrm{NH}_3$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$\mathrm{CH}_3 \mathrm{NH}_2$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$\left(\mathrm{CH}_3\right)_2 \mathrm{NH}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\left(\mathrm{CH}_3\right)_3 \mathrm{~N}$</span> </li> </ul>

<div class="correct-answer"> The correct answers are: <span class="option-value"><br/>$\mathrm{NH}_3$<br/>, <br/>$\mathrm{CH}_3 \mathrm{NH}_2$<br/>, <br/>$\left(\mathrm{CH}_3\right)_2 \mathrm{NH}$<br/></span> </div>

<div class="solution">Small amines such as $\mathrm{NH}_3 \cdot \mathrm{CH}_3 \mathrm{NH}_2$ and $\left(\mathrm{CH}_3\right)_2 \mathrm{NH}$ give unsymmetrical cleavage of diborane according to following reaction,<br/>$$<br/>\begin{aligned}<br/>&amp; \mathrm{B}_2 \mathrm{H}_6+2 \mathrm{NH}_3 \longrightarrow \\<br/>&amp; {\left[\mathrm{H}_2 \mathrm{~B}\left(\mathrm{NH}_3\right)_2\right]^{+}\left[\mathrm{BH}_4\right]^{-}} \\<br/>&amp;<br/>\end{aligned}<br/>$$<br/>Large amines, such as $\left(\mathrm{CH}_3\right)_3 \mathrm{~N}$ give symmetrical cleavage of diborane according to following reaction,<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/YhfSvFekusS1ekfppnWA9ksQ_OT9BCLkCB9xM2Mq8yE.rendered.fullsize.png"/></div>

MarksBatch2_P1.db

in-the-reaction-hbr-the-products-are

in-the-reaction-hbr-the-products-are-56247

<div class="question">In the reaction,<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/WpcSmf9mSruKQidMpbn1ZkSwknUTJW66ijOz6VU1J7s.original.fullsize.png"/><br/><br/>$\stackrel{\mathrm{HBr}}{\longrightarrow}$<br/>the products are,</div>

['Chemistry', 'Alcohols Phenols and Ethers', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/FCC6s1TCUwlVXhDdrao6sdSUvSPNT92S_AwtN33AbTQ.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/eWQApNuBIFyrGtkKrexTVAXUOi-DpUX5TlEBJOE1sXo.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/fCSpZTT3bifF7LTiLb9bcf1RWs2S4KRJkAhqBldx4kE.original.fullsize.png"/><br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/pRzZgsoOl-UImBOA0IhkHwBv8rKdO35XQk13rMQb8hk.original.fullsize.png"/><br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/pRzZgsoOl-UImBOA0IhkHwBv8rKdO35XQk13rMQb8hk.original.fullsize.png"/><br/></span> </div>

<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/02XAyKlCumqJiH_vIJ0TSA_ZowYSW4yJVsuZjF_Svbg.original.fullsize.png"/><br/><br/>Alcohol-phenol-ether<br/>Conceptual, understanding of mechanism<br/>II (Given options raised the difficulty level)</div>

MarksBatch2_P1.db

in-the-reaction-the-intermediates-is-are

in-the-reaction-the-intermediates-is-are-26288

<div class="question">In the reaction, the intermediate(s) is (are)<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/7Ig40VgdgL1GnJfo-MIrPzx591YZTtZidoA2jAweZ0I.original.fullsize.png"/><br/></div>

['Chemistry', 'Alcohols Phenols and Ethers', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/QKOeqjn-5kWFb3dgBVLOMMNvejmzW0S0PkCly4CI_vk.original.fullsize.png"/><br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/Csqt09PIiI4kqtYRh-Ggeo0-p636NxhX1iwx2q4muV0.original.fullsize.png"/><br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/4i8s67LE1UC5KfnYcyWN7MgpC6A7Y5k8bP4LYSjLhuc.original.fullsize.png"/><br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/neorBxWHY9M_V6zZV-hua3ZZnPVLPSkNKDN6jDjcGgE.original.fullsize.png"/><br/></span> </li> </ul>

<div class="correct-answer"> The correct answers are: <span class="option-value"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/QKOeqjn-5kWFb3dgBVLOMMNvejmzW0S0PkCly4CI_vk.original.fullsize.png"/><br/>, <br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/4i8s67LE1UC5KfnYcyWN7MgpC6A7Y5k8bP4LYSjLhuc.original.fullsize.png"/><br/></span> </div>

<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/5RdSSwS0JrRF8Uf9VPicMs07VYx1jh9ckx_ouXuPQGw.original.fullsize.png"/><br/><br/>- $\mathrm{OH}$ and in aqueous in alkaline medium $-\overline{\mathrm{O}}$ are ortho and para directing. We expect the main intermediate products as mono ortho/para substituted product. Remember! if we consider (b), which could be there in small amount then we need to consider even (d) because $-\overline{\mathrm{O}}$ is strongest activating group that facilitates electrophilic substitution reaction overall on the ring. $-\overline{\mathrm{O}}$<br/>activates ortho and para position more but doesn't deactivate the meta position. We get a small amount of even the meta substituted product. But, as the question appears to be directly concerned answer should be only (a) and (c).<br/>Reaction mechanism (Electrophilic substitution)<br/>Conceptual<br/>II</div>

MarksBatch2_P1.db

in-the-reaction-the-structure-of-the-product-t-is-1

in-the-reaction-the-structure-of-the-product-t-is-1-89697

<div class="question">In the reaction the structure of the product $T$ is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/ty1ZkhKwuyldG1DQNcjd0AT3HPZOA43ocCLBqaMwhlg.original.fullsize.png"/><br/></div>

['Chemistry', 'Carboxylic Acid Derivatives', 'JEE Main']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/TSu69oPn4xTuC50RwgJPSGnSOC6XJVJZ-mowin5UJg0.original.fullsize.png"/><br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/QKC3ihcIr9XuXXsAK3kUGuAJy7UWULhCP56OhPlrx5A.original.fullsize.png"/><br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/JSMfI_RJhvcjmYjkePSDh0TGHLSxAyjrjbspJDmZuLc.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/WGmILMBmvsnyTghZRpx_lM0tY_NIiaNyxEG-TAi0vTM.original.fullsize.png"/><br/></span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/QKC3ihcIr9XuXXsAK3kUGuAJy7UWULhCP56OhPlrx5A.original.fullsize.png"/><br/></span> </div>

<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/eUfPSVZUxXHhPXPRNB41PEarJhi5jLOvwDpPGp4wf0U.original.fullsize.png"/><br/><br/>Acids and their derivatives Conceptual (Reaction understanding) II</div>

MarksBatch2_P1.db

in-the-reaction-the-structure-of-the-product-t-is

in-the-reaction-the-structure-of-the-product-t-is-63199

<div class="question">In the reaction the structure of the product $T$ is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/ty1ZkhKwuyldG1DQNcjd0AT3HPZOA43ocCLBqaMwhlg.original.fullsize.png"/><br/></div>

['Chemistry', 'Carboxylic Acid Derivatives', 'JEE Advanced', 'JEE Advanced 2010 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/TSu69oPn4xTuC50RwgJPSGnSOC6XJVJZ-mowin5UJg0.original.fullsize.png"/><br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/QKC3ihcIr9XuXXsAK3kUGuAJy7UWULhCP56OhPlrx5A.original.fullsize.png"/><br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/JSMfI_RJhvcjmYjkePSDh0TGHLSxAyjrjbspJDmZuLc.original.fullsize.png"/><br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/WGmILMBmvsnyTghZRpx_lM0tY_NIiaNyxEG-TAi0vTM.original.fullsize.png"/><br/></span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/QKC3ihcIr9XuXXsAK3kUGuAJy7UWULhCP56OhPlrx5A.original.fullsize.png"/><br/></span> </div>

<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/eUfPSVZUxXHhPXPRNB41PEarJhi5jLOvwDpPGp4wf0U.original.fullsize.png"/><br/><br/>Acids and their derivatives Conceptual (Reaction understanding) II</div>

MarksBatch2_P1.db

in-the-scheme-given-below-the-total-number-of-intramolecular-aldol-condensation-products-formed-from-y-is-1

in-the-scheme-given-below-the-total-number-of-intramolecular-aldol-condensation-products-formed-from-y-is-1-41519

<div class="question">In the scheme given below, the total number of intramolecular aldol condensation products formed from ' $Y$ ' is<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/5H7_nrPU-8mhsPUhRpuQ421CGQImWgh3nDByt7s54gI.original.fullsize.png"/><br/></div>

['Chemistry', 'Alcohols Phenols and Ethers', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)']

None

<div class="correct-answer"> The correct answer is: <span class="option-value">1</span> </div>

<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/2Zrp04_0tXQ4SVhpaVe-xuXBOlFDWBEbgSoc8krhlZg.original.fullsize.png"/><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/-OIeOWmBWVaTrc0z1CbgH4SQuxniqCdxlYbyqMYVSwA.original.fullsize.png"/><br/><br/>For aldol condensation $\mathrm{C}-5$ and $\mathrm{C}-7$ can attack to C-1 similarly C-2 and C-10 can attack to C-6 but all give same product.<br/>Carbonyl compounds<br/>Conceptual<br/>III</div>

MarksBatch2_P1.db

incandescent-bulbs-are-designed-by-keeping-in-mind-that-the-resistance-of-their-filament-increases-with-the-increase-in-temperature-if-at-room-tempera

incandescent-bulbs-are-designed-by-keeping-in-mind-that-the-resistance-of-their-filament-increases-with-the-increase-in-temperature-if-at-room-tempera-80357

<div class="question">Incandescent bulbs are designed by keeping in mind that the resistance of their filament increases with the increase in temperature. If at room temperature, $100 \mathrm{~W}, 60 \mathrm{~W}$ and $40 \mathrm{~W}$ bulbs have filament resistances $R_{100}, R_{60}$ and $R_{40}$, respectively, the relation between these resistances is</div>

['Physics', 'Current Electricity', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{1}{R_{100}}=\frac{1}{R_{40}}+\frac{1}{R_{60}}$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$R_{100}=R_{40}+R_{60}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$R_{100}&gt;R_{60}&gt;R_{40}$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{1}{R_{100}}&gt;\frac{1}{R_{60}}&gt;\frac{1}{R_{40}}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$\frac{1}{R_{100}}&gt;\frac{1}{R_{60}}&gt;\frac{1}{R_{40}}$</span> </div>

<div class="solution">$R=\frac{V^2}{P}$ or $R \propto \frac{1}{P}$ $\therefore \quad \frac{1}{R_{100}}&gt;\frac{1}{R_{60}}&gt;\frac{1}{R_{40}}$<br/>Hence, the correct option is (d).</div>

MarksBatch2_P1.db

inradius-of-a-circle-which-is-inscribed-in-an-isosceles-triangle-one-of-whose-angle-is-2-3-is-3-then-area-of-triangle-is

inradius-of-a-circle-which-is-inscribed-in-an-isosceles-triangle-one-of-whose-angle-is-2-3-is-3-then-area-of-triangle-is-76117

<div class="question">Inradius of a circle which is inscribed in an isosceles triangle one of whose angle is $2 \pi / 3$, is $\sqrt{3}$, then area of triangle is</div>

['Mathematics', 'Properties of Triangles', 'JEE Advanced', 'JEE Advanced 2006']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$4 \sqrt{3}$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$12-7 \sqrt{3}$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$12+7 \sqrt{3}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>None of these</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$12+7 \sqrt{3}$<br/></span> </div>

<div class="solution">Let $A B=A C$ and $\angle A=120^{\circ}$<br/>$\therefore$ Area of $\Delta=\frac{1}{2} a^2 \sin 120^{\circ}$<br/>where, $a=A D+B D$<br/>$=\sqrt{3} \tan 30^{\circ}+\sqrt{3} \cot 15^{\circ}$<br/>$=1+\sqrt{3}\left(\frac{1+\tan 45^{\circ} \tan 30^{\circ}}{\tan 45^{\circ}-\tan 30^{\circ}}\right)$<br/>$=1+\sqrt{3}\left(\frac{\sqrt{3}+1}{\sqrt{3}-1}\right)$<br/>$\therefore \quad a=4+2 \sqrt{3}$<br/>$\therefore$ Area of $\Delta=\frac{1}{2}(4+2 \sqrt{3})^2\left(\frac{\sqrt{3}}{2}\right)=12+7 \sqrt{3}$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/PXl03TAYLnGL7W1m-BtgA8nH4CI9IIu9JAkQGn0XaRE.original.fullsize.png"/><br/></div>

MarksBatch2_P1.db

internal-bisector-of-a-of-a-bc-meets-side-bc-at-d-a-line-drawn-through-d-perpendicular-to-a-d-intersects-the-side-a-c-at-e-and-side-a-b-at-f-if-a-b-an

internal-bisector-of-a-of-a-bc-meets-side-bc-at-d-a-line-drawn-through-d-perpendicular-to-a-d-intersects-the-side-a-c-at-e-and-side-a-b-at-f-if-a-b-an-37619

<div class="question">Internal bisector of $\angle A$ of $\triangle A B C$ meets side $B C$ at $D$. A line drawn through $D$ perpendicular to $A D$ intersects the side $A C$ at $E$ and side $A B$ at $F$. If $a, b$ and $c$ represent sides of $\triangle A B C$, then</div>

['Mathematics', 'Properties of Triangles', 'JEE Advanced', 'JEE Advanced 2006']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$A E$ is $\mathrm{HM}$ of $b$ and $c$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$A D=\frac{2 b c}{b+c} \cos \frac{A}{2}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$E F=\frac{4 b c}{b+c} \sin \frac{A}{2}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>the $\triangle A E F$ is isosceles</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

<div class="correct-answer"> The correct answers are: <span class="option-value"><br/>$A E$ is $\mathrm{HM}$ of $b$ and $c$<br/>, <br/>$A D=\frac{2 b c}{b+c} \cos \frac{A}{2}$<br/>, <br/>$E F=\frac{4 b c}{b+c} \sin \frac{A}{2}$<br/>, <br/>the $\triangle A E F$ is isosceles</span> </div>

<div class="solution">We have, $\triangle A B C=\triangle A B D+\triangle A C D$<br/>$$<br/>\begin{aligned}<br/>&amp; \Rightarrow \frac{1}{2} b c \sin A=\frac{1}{2} c A D \sin \frac{A}{2}+\frac{1}{2} b A D \sin \frac{A}{2} \\<br/>&amp; \Rightarrow \quad A D=\frac{2 b c}{b+c} \cos \frac{A}{2}<br/>\end{aligned}<br/>$$<br/>Again, $\quad A E=A D \sec \frac{A}{2}=\frac{2 b c}{b+c}$<br/>$\therefore A E$ is $\mathrm{HM}$ of $b$ and $c$.<br/>$$<br/>\begin{aligned}<br/>E F &amp; =E D+D F=2 D E=2 A D \tan \frac{A}{2} \\<br/>&amp; =2 \frac{2 b c}{b+c} \cos \frac{A}{2} \tan \frac{A}{2}=\frac{4 b c}{b+c} \sin \frac{A}{2}<br/>\end{aligned}<br/>$$<br/>As $A D \perp E F$ and $D E=D F$ and $A D$ is bisector.<br/>$\Rightarrow \triangle A E F$ is isosceles.<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/_Xl9hjzWwQRk4E-I6GNC2M5Xwg65OHYLXnkwwQZg4pM.original.fullsize.png"/><br/></div>

MarksBatch2_P1.db

let-0-2-be-such-that-2-cos-1-sin-sin-2-tan-2-cot-2-cos-1-tan-2-0-and-1-sin-2-3-then-cannot-satisfy

let-0-2-be-such-that-2-cos-1-sin-sin-2-tan-2-cot-2-cos-1-tan-2-0-and-1-sin-2-3-then-cannot-satisfy-27871

<div class="question">Let $\theta, \varphi \in[0,2 \pi]$ be such that $2 \cos \theta(1-\sin \varphi)=\sin ^{2} \theta \times$ $\left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \varphi-1, \tan (2 \pi-\theta)&gt;0$ and <br/> <br/>$-1 &lt; \sin \theta &lt; -\frac{\sqrt{3}}{2}$, then $\varphi$ cannot satisfy</div>

['Mathematics', 'Trigonometric Equations', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$0 &lt; \varphi &lt; \frac{\pi}{2}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$\frac{\pi}{2} &lt; \varphi &lt; \frac{4 \pi}{3}$</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$\frac{4 \pi}{3} &lt; \varphi &lt; \frac{3 \pi}{2}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$\frac{3 \pi}{2} &lt; \varphi &lt; 2 \pi$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

<div class="correct-answer"> The correct answers are: <span class="option-value">$0 &lt; \varphi &lt; \frac{\pi}{2}$, $\frac{4 \pi}{3} &lt; \varphi &lt; \frac{3 \pi}{2}$, $\frac{3 \pi}{2} &lt; \varphi &lt; 2 \pi$</span> </div>

<div class="solution">As $\tan (2 \pi-\theta)&gt;0$ and $-1 &lt; \sin \theta &lt; -\frac{\sqrt{3}}{2}, \theta \in[0,2 \pi]$ <br/> <br/>Hence $\frac{3 \pi}{2} &lt; \theta &lt; \frac{5 \pi}{3}$ <br/> <br/>Now $2 \cos \theta(1-\sin \varphi)=\sin ^{2} \theta\left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \varphi-1$ <br/> <br/>$\Rightarrow 2 \cos \theta(1-\sin \varphi)=2 \sin \theta \cos \varphi-1$ <br/> <br/>$\Rightarrow 2 \cos \theta+1=2 \sin (\theta+\varphi)$ <br/> <br/>As $\quad \theta \in\left(\frac{3 \pi}{2}, \frac{5 \pi}{3}\right), 1 &lt; 2 \sin (\theta+\varphi) &lt; 2$ <br/> <br/>As $\theta+\varphi \in\left(\frac{\pi}{6}, \frac{5 \pi}{6}\right)$ or $(\theta+\varphi) \in\left(\frac{13 \pi}{6}, \frac{17 \pi}{6}\right)$ <br/> <br/>We have $\varphi \in\left(-\frac{3 \pi}{2},-\frac{2 \pi}{3}\right) \cup\left(\frac{2 \pi}{3}, \frac{7 \pi}{6}\right)$</div>

MarksBatch2_P1.db

let-0-4-and-t-1-tan-t-a-n-t-2-tan-c-o-t-and-t-4-cot-t-a-n-then

let-0-4-and-t-1-tan-t-a-n-t-2-tan-c-o-t-and-t-4-cot-t-a-n-then-71133

<div class="question">Let $\theta \in\left(0, \frac{\pi}{4}\right)$ and $t_1=(\tan \theta)^{\tan \theta}, t_2=(\tan \theta)^{\cot \theta}$ and $t_4=(\cot \theta)^{\tan \theta}$, then</div>

['Mathematics', 'Trigonometric Ratios & Identities', 'JEE Advanced', 'JEE Advanced 2006']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$t_1&gt;t_2&gt;t_3&gt;t_4$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$t_4&gt;t_3&gt;t_1&gt;t_2$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$t_3&gt;t_1&gt;t_2&gt;t_4$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$t_2&gt;t_3&gt;t_1&gt;t_4$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$t_4&gt;t_3&gt;t_1&gt;t_2$<br/></span> </div>

<div class="solution">As when $\theta \in\left(0, \frac{\pi}{4}\right)$<br/>$\tan \theta &lt; \cot \theta$<br/>Since, $\tan \theta &lt; 1$ and $\cot \theta&gt;1$<br/>$\therefore(\tan \theta)^{\cot \theta} &lt; 1$ and $(\cot \theta)^{\tan \theta}&gt;1$<br/>$\therefore t_4&gt;t_1$ which only holds in option (b).</div>

MarksBatch2_P1.db

let-1-be-a-cube-root-of-unity-and-s-be-the-set-of-all-nonsingular-matrices-of-the-form-1-2-a-1-b-c-1-where-each-of-a-b-and-c-is-either-or-2-then-the-n

let-1-be-a-cube-root-of-unity-and-s-be-the-set-of-all-nonsingular-matrices-of-the-form-1-2-a-1-b-c-1-where-each-of-a-b-and-c-is-either-or-2-then-the-n-71815

<div class="question">Let $\omega \neq 1$ be a cube root of unity and $S$ be the set of all non-singular matrices of the form $\left[\begin{array}{ccc}1 &amp; a &amp; b \\ \omega &amp; 1 &amp; c \\ \omega^2 &amp; \omega &amp; 1\end{array}\right]$, where each of $a, b$ and $c$ is either $\omega$ or $\omega^2$. Then, the number of distinct matrices in the set $S$ is</div>

['Mathematics', 'Matrices', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>2<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>6<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>4<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>8</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>2<br/></span> </div>

<div class="solution">$|A| \neq 0$, as non-singular.<br/>$$<br/>\begin{aligned}<br/>&amp; \therefore \quad\left|\begin{array}{ccc}<br/>1 &amp; a &amp; b \\<br/>\omega &amp; 1 &amp; c \\<br/>\omega^2 &amp; \omega &amp; 1<br/>\end{array}\right| \neq 0 \\<br/>&amp; \Rightarrow \quad 1(1-c \omega)-a\left(\omega-c \omega^2\right) \\<br/>&amp; +b\left(\omega^2-\omega^2\right) \neq 0 \\<br/>&amp; \Rightarrow \quad 1-c \omega-a \omega+a c \omega^2 \neq 0 \\<br/>&amp;<br/>\end{aligned}<br/>$$<br/><br/>$$<br/>\begin{aligned}<br/>&amp; \Rightarrow \quad(1-c \omega)(1-a \omega) \neq 0 \\<br/>&amp; \Rightarrow \quad a \neq \frac{1}{\omega}, c \neq \frac{1}{\omega} \Rightarrow a=\omega, c=\omega<br/>\end{aligned}<br/>$$<br/>and $b \in\left\{\omega, \omega^2\right\} \Rightarrow 2$ solutions</div>

MarksBatch2_P1.db

let-a-1-a-2-a-3-a-11-be-real-numbers-satisfying-a-1-15-27-2-a-2-0-and-a-k-2-a-k-1-a-k-2-for-k-3-4-11-if-11-a-1-2-a-2-2-a-11-2-90-then-the-value-of-11-

let-a-1-a-2-a-3-a-11-be-real-numbers-satisfying-a-1-15-27-2-a-2-0-and-a-k-2-a-k-1-a-k-2-for-k-3-4-11-if-11-a-1-2-a-2-2-a-11-2-90-then-the-value-of-11-66431

<div class="question">Let $a_1, a_2, a_3, \ldots, a_{11}$ be real numbers satisfying $a_1=15,27-2 a_2&gt;0$ and $a_k=2 a_{k-1}-a_{k-2}$ for $k=3,4, \ldots, 11$.<br/>If $\frac{a_1^2+a_2^2+\ldots+a_{11}^2}{11}=90$, then the value of $\frac{a_1+a_2+\ldots+a_{11}}{11}$ is equal to</div>

['Mathematics', 'Sequences and Series', 'JEE Advanced', 'JEE Advanced 2010 (Paper 2)']

None

<div class="correct-answer"> The correct answer is: <span class="option-value">0</span> </div>

<div class="solution">$$<br/>\text { } \begin{aligned}<br/>&amp; a_k=2 a_{k-1}-a_{k-2} \\<br/>&amp; \Rightarrow \quad a_1, a_2, \ldots, a_{11} \text { are in AP } \\<br/>&amp; \therefore \frac{a_1^2+a_2^2+\ldots+a_{11}^2}{11} \\<br/>&amp;=\frac{11 a^2+35 \times 11 d^2+10 a d}{11}=90 \\<br/>&amp; \Rightarrow \quad 225+35 d^2+150 d=90 \\<br/>&amp; 35 d^2+150 d+135=0 \\<br/>&amp; \Rightarrow \quad d=-3,-\frac{9}{7}<br/>\end{aligned}<br/>$$<br/>Given, $a_2 &lt; \frac{27}{2} \therefore d=-3$ and $d \neq-\frac{9}{7}$<br/>$$<br/>\Rightarrow \quad \frac{a_1+a_2+\ldots+a_{11}}{11}<br/>$$<br/>$$<br/>=\frac{11}{2}[30-10 \times 3]=0<br/>$$</div>

MarksBatch2_P1.db

let-a-1-a-2-a-3-a-100-be-an-arithmetic-progression-with-a-1-3-and-s-p-i-1-p-a-i-1-p-100-for-any-integer-n-with-1-n-20-let-m-5-n-if-s-n-s-m-does-not-de

let-a-1-a-2-a-3-a-100-be-an-arithmetic-progression-with-a-1-3-and-s-p-i-1-p-a-i-1-p-100-for-any-integer-n-with-1-n-20-let-m-5-n-if-s-n-s-m-does-not-de-24201

<div class="question">Let $a_1, a_2, a_3, \ldots ., a_{100}$ be an arithmetic progression with $a_1=3$ and $S_p=\sum_{i=1}^p a_i$, $1 \leq p \leq 100$. For any integer $n$ with $1 \leq n \leq 20$, let $m=5 n$. If $\frac{S_m}{S_n}$ does not depend on $n$, then $a_2$ is</div>

['Mathematics', 'Sequences and Series', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)']

None

<div class="correct-answer"> The correct answer is: <span class="option-value">3</span> </div>

<div class="solution">Given, $a_1=3, m=5 n$ and $a_1, a_2, \ldots$ are in AP. $\therefore \frac{S_m}{S_n}=\frac{S_{5 n}}{S_n}$ is independent of $n$. Now, $\quad \frac{\frac{5 n}{2}[2 \times 3+(5 n-1) d]}{\frac{n}{2}[2 \times 3+(n-1) d]}$ $\Rightarrow \quad \frac{f\{(6-d)+5 n\}}{(6-d)+n}$ independent of $n$, if $\begin{aligned} \quad 6-d= &amp; 0 \Rightarrow d=6 \\ \therefore \quad a_2= &amp; a_1+d=3+6=9 \\ &amp; \text { Or }\end{aligned}$<br/>If $d=0$ $\frac{S_m}{S_n}$ is independent of $n$.<br/>$$<br/>\therefore \quad a_2=3<br/>$$</div>

MarksBatch2_P1.db

let-a-1-a-2-a-3-be-in-harmonic-progression-with-a-1-5-and-a-20-25-the-least-positive-integer-n-for-which-a-n-0-is

let-a-1-a-2-a-3-be-in-harmonic-progression-with-a-1-5-and-a-20-25-the-least-positive-integer-n-for-which-a-n-0-is-70554

<div class="question">Let $a_{1}, a_{2}, a_{3}, \ldots . .$ be in harmonic progression with $a_{1}=5$ and $a_{20}=25$. The least positive integer $n$ for which $a_{n} &lt; 0$ is</div>

['Mathematics', 'Sequences and Series', 'JEE Advanced', 'JEE Advanced 2012 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">22</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">23</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">24</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">25</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value">25</span> </div>

<div class="solution">$\because a_{1}, a_{2}, a_{3}, \ldots \ldots$ are in H.P. <br/> <br/>$\begin{array}{l} <br/> <br/>\therefore \frac{1}{a_{1}}, \frac{1}{a_{2}}, \frac{1}{a_{3}} \ldots \ldots \text { are in A.P. } \\ <br/> <br/>\therefore \quad \frac{1}{a_{1}}=\frac{1}{5} \text { and } \frac{1}{a_{20}}=\frac{1}{25} \\ <br/> <br/>\frac{1}{a_{1}}+19 d=\frac{1}{a_{20}} \Rightarrow \frac{1}{5}+19 d=\frac{1}{25} \Rightarrow d=\frac{-4}{475} \\ <br/> <br/>\text { Now } \frac{1}{a_{n}}=\frac{1}{5}+(n-1)\left(\frac{-4}{475}\right) \\ <br/> <br/>\text { Clearly } a_{n} &lt; 0, \text { if } \frac{1}{a_{n}} &lt; 0 \Rightarrow \frac{1}{5}-\frac{4 n}{475}+\frac{4}{475} &lt; 0 \\ <br/> <br/>\Rightarrow \quad-4 n &lt; -99 \text { or } n&gt;\frac{99}{4}=24 \frac{3}{4} \quad \therefore n \geq 25 \\ <br/> <br/>\therefore \quad \text { Least value of } n \text { is } 25 . <br/> <br/>\end{array}$</div>

MarksBatch2_P1.db

let-a-and-a-be-the-roots-of-the-equation-3-1-a-1-x-2-1-a-1-x-6-1-a-1-0-where-a-1-then-lim-a-0-a-and-lim-x-0-a-are

let-a-and-a-be-the-roots-of-the-equation-3-1-a-1-x-2-1-a-1-x-6-1-a-1-0-where-a-1-then-lim-a-0-a-and-lim-x-0-a-are-32087

<div class="question">Let $\alpha(a)$ and $\beta(a)$ be the roots of the equation $(\sqrt[3]{1+a}-1) x^{2}+(\sqrt{1+a}-1) x+(\sqrt[6]{1+a}-1)=0$ where $a$ $&gt;-1$. Then $\lim _{a \rightarrow 0^{+}} \alpha(a)$ and $\lim _{x \rightarrow 0^{+}} \beta(a)$ are</div>

['Mathematics', 'Limits', 'JEE Advanced', 'JEE Advanced 2012 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$-\frac{5}{2}$ and 1</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$-\frac{1}{2}$ and $-1$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$-\frac{7}{2}$ and 2</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$-\frac{9}{2}$ and 3</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value">$-\frac{1}{2}$ and $-1$</span> </div>

<div class="solution">$(\sqrt[3]{1+a}-1) x^{2}+(\sqrt{1+a}-1) x+(\sqrt[6]{1+a}-1)=0$ <br/> <br/>Let $a+1=y$, then equation reduces to $\left(y^{1 / 3}-1\right) x^{2}+\left(y^{1 / 2}-1\right) x+\left(y^{1 / 6}-1\right)=0$ <br/> <br/>On dividing both sides by $y-1$, we get $\left(\frac{y^{1 / 3}-1}{y-1}\right) x^{2}+\left(\frac{y^{1 / 2}-1}{y-1}\right) x+\left(\frac{y^{1 / 6}-1}{y-1}\right)=0$ <br/> <br/>On taking limit as $y \rightarrow 1$ i.e. $a \rightarrow 0$ on both sides, we get $\frac{1}{3} x^{2}+\frac{1}{2} x+\frac{1}{6}=0 \Rightarrow 2 x^{2}+3 x+1=0$ <br/> <br/>$\Rightarrow x=-1,-\frac{1}{2}$ (roots of the equation) <br/> <br/>$\therefore \quad \lim _{a \rightarrow 0^{+}} \alpha(a)=-1, \lim _{a \rightarrow 0^{+}} \beta(a)=-\frac{1}{2}$</div>

MarksBatch2_P1.db

let-a-and-b-be-nonzero-and-real-numbers-then-the-equation-a-x-2-b-y-2-c-x-2-5-x-y-6-y-2-0-represents

let-a-and-b-be-nonzero-and-real-numbers-then-the-equation-a-x-2-b-y-2-c-x-2-5-x-y-6-y-2-0-represents-53393

<div class="question">Let $a$ and $b$ be non-zero and real numbers. Then, the equation $\left(a x^2+b y^2+c\right)\left(x^2-5 x y+6 y^2\right)=0$ represents</div>

['Mathematics', 'Pair of Lines', 'JEE Advanced', 'JEE Advanced 2008 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>Four straight lines, when $c=0$ and $a, b$ are of the same sign<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>Two straight lines and a circle, when $a=b$ and $c$ is of sign opposite to that of $a$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>Two straight lines and a hyperbola, when $a$ and $b$ are of the same sign and $c$ is of sign opposite to that of $a$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>A circle and an ellipse, when $a$ and $b$ are of the same sign and $c$ is of sign opposite to that of $a$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>Two straight lines and a circle, when $a=b$ and $c$ is of sign opposite to that of $a$<br/></span> </div>

<div class="solution">Let $a$ and $b$ be non- zero real numbers.<br/>Therefore, the given equation $\left(a x^2+b y^2+c\right)\left(x^2-5 x y+6 y^2\right)=0$ implies either<br/>$$<br/>\begin{array}{rc} <br/>&amp; x^2-5 x y+6 y^2=0 \\<br/>\Rightarrow &amp; (x-2 y)(x-3 y)=0 \\<br/>\Rightarrow &amp; x=2 y \text { and } x=3 y<br/>\end{array}<br/>$$<br/>represent two straight lines passing through origin.<br/>Or $\quad a x^2+b y^2+c=0$<br/>When $c=0$ and $a$ and $b$ are of same signs, then<br/>$$<br/>\begin{aligned}<br/>&amp; a x^2+b y^2+c=0 \\<br/>\Rightarrow &amp; x=0 \text { and } y=0<br/>\end{aligned}<br/>$$<br/>which is a point specified as the origin.<br/>When $a=b$ and $c$ is of sign opposite to that of $a, a x^2+b y^2+c=0$ represents a circle.<br/>Hence, the given equation $\left(a x^2+b y^2+c\right)\left(x^2-5 x y+6 y^2\right)=0$<br/>may represent two straight lines and a circle.</div>

MarksBatch2_P1.db

let-a-and-b-be-two-distinct-points-on-the-parabola-y-2-4-x-if-the-axis-of-the-parabola-touches-a-circle-of-radius-r-having-a-b-as-its-diameter-then-th

let-a-and-b-be-two-distinct-points-on-the-parabola-y-2-4-x-if-the-axis-of-the-parabola-touches-a-circle-of-radius-r-having-a-b-as-its-diameter-then-th-50269

<div class="question">Let $A$ and $B$ be two distinct points on the parabola $y^2=4 x$. If the axis of the parabola touches a circle of radius $r$ having $A B$ as its diameter, then the slope of the line joining $A$ and $B$ can be</div>

['Mathematics', 'Parabola', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$-\frac{1}{r}$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{1}{r}$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{2}{r}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$-\frac{2}{r}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

<div class="correct-answer"> The correct answers are: <span class="option-value"><br/>$\frac{2}{r}$<br/>, <br/>$-\frac{2}{r}$</span> </div>

<div class="solution">Here, coordinate $M=\left(\frac{t_1^2+t_2^2}{2}, t_1+t_2\right) i e$, mid point of chord $A B$.<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/5OP5-giT0HF2WPAeV49UhRy4TxSM3EtXoEcLRC-CfFc.original.fullsize.png"/><br/><br/>$M P=t_1+t_2=r$ Also, $\quad m_{A B}=\frac{2 t_2-2 t_1}{t_2^2-t_1^2}=\frac{2^{\ldots(i)}}{t_2+t_1}$ (when $A B$ is chord)<br/>$$<br/>\Rightarrow \quad m_{A B}=\frac{2}{r}<br/>$$<br/>[from Eq. (i)]<br/>Also, $\quad m_{A^{\prime} B^{\prime}}=-\frac{2}{r}$ (when $A^{\prime} B^{\prime}$ is chord)<br/>Hence, (c, d) is the correct option.</div>

MarksBatch2_P1.db

let-a-b-c-be-unit-vectors-such-that-a-b-c-0-which-one-of-the-following-is-correct

let-a-b-c-be-unit-vectors-such-that-a-b-c-0-which-one-of-the-following-is-correct-97227

<div class="question">Let $\mathbf{a}, \mathbf{b}, \mathbf{c}$ be unit vectors such that $\mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0}$. Which one of the following is correct?</div>

['Mathematics', 'Vector Algebra', 'JEE Advanced', 'JEE Advanced 2007 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\mathbf{a} \times \mathbf{b}=\mathbf{b} \times \mathbf{c}=\mathbf{c} \times \mathbf{a}=\mathbf{0}$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$\mathbf{a} \times \mathbf{b}=\mathbf{b} \times \mathbf{c}=\mathbf{c} \times \mathbf{a} \neq \mathbf{0}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\mathbf{a} \times \mathbf{b}=\mathbf{b} \times \mathbf{c}=\mathbf{a} \times \mathbf{c}=\mathbf{0}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\mathbf{a} \times \mathbf{b}, \mathbf{b} \times \mathbf{c}, \mathbf{c} \times \mathbf{a}$ are mutually perpendicular</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$\mathbf{a} \times \mathbf{b}=\mathbf{b} \times \mathbf{c}=\mathbf{c} \times \mathbf{a} \neq \mathbf{0}$<br/></span> </div>

<div class="solution">Since $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are unit vectors and $\mathbf{a}+\mathbf{b}+\mathbf{c}=\mathbf{0}$ $\mathbf{a}, \mathbf{b}$ and $\mathbf{c}$ represent an equilateral triangle.<br/>$$<br/>\therefore \quad \mathbf{a} \times \mathbf{b}=\mathbf{b} \times \mathbf{c}=\mathbf{c} \times \mathbf{a} \neq \mathbf{0} .<br/>$$</div>

MarksBatch2_P1.db

let-a-b-c-p-q-be-real-numbers-suppose-are-the-roots-of-the-equation-x-2-2-p-x-q-0-and-1-are-the-roots-of-the-equation-a-x-2-2-b-x-c-0-where-2-1-0-1-st

let-a-b-c-p-q-be-real-numbers-suppose-are-the-roots-of-the-equation-x-2-2-p-x-q-0-and-1-are-the-roots-of-the-equation-a-x-2-2-b-x-c-0-where-2-1-0-1-st-10949

<div class="question">Let $a, b, c, p, q$ be real numbers.<br/>Suppose, $\alpha, \beta$ are the roots of the equation $x^2+2 p x+q=0$ and $\alpha, \frac{1}{\beta}$ are the roots of the equation $a x^2+2 b x+c=0$, where $\beta^2 \notin\{-1,0,1\}$.<br/>Statement $1\left(p^2-q\right)\left(b^2-a c\right) \geq 0$.<br/>Statement $2 b \neq p a$ or $c \neq q a$.</div>

['Mathematics', 'Quadratic Equation', 'JEE Advanced', 'JEE Advanced 2008 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>Statement 1 is true, Statement 2 is false.<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>Statement 1 is false, Statement 2 is true</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.<br/></span> </div>

<div class="solution">Given, $\alpha$ and $\beta$ are the roots of $x^2+2 p x+q=0$.<br/>$$<br/>\begin{array}{rlrl} <br/>&amp; \therefore &amp; \alpha+\beta &amp; =-2 p \\<br/>\text { and } &amp; \alpha \beta &amp; =q<br/>\end{array}<br/>$$<br/>$\alpha$ and $\frac{1}{\beta}$ are the roots of $a x^2+2 b x+c=0$<br/>and<br/>$$<br/>\alpha+\frac{1}{\beta}=\frac{-2 b}{a}<br/>$$<br/><br/>and<br/>$$<br/>\frac{\alpha}{\beta}=\frac{c}{a}<br/>$$<br/>$$<br/>\text { Now, } \begin{aligned}<br/>\left(p^2-q\right)\left(b^2-a c\right) &amp; =\left[\left(\frac{\alpha+\beta}{-2}\right)^2-\alpha \beta\right]\left[\left(\frac{\alpha+\frac{1}{\beta}}{2}\right)^2-\frac{\alpha}{\beta}\right] a^2 \\<br/>&amp; =\frac{(\alpha-\beta)^2}{16}\left(\alpha-\frac{1}{\beta}\right)^2 \cdot a^2 \geq 0<br/>\end{aligned}<br/>$$<br/>Statement 1 is true.<br/>$$<br/>\begin{array}{ll}<br/>\text { Again now, } &amp; p a=-\left(\frac{\alpha+\beta}{2}\right) a=-\frac{a}{2}(\alpha+\beta) \\<br/>\text { and } &amp; b=-\frac{a}{2}\left(\alpha+\frac{1}{\beta}\right) \\<br/>&amp; p a \neq b \Rightarrow \alpha+\frac{1}{\beta} \neq \alpha+\beta \\<br/>\Rightarrow &amp; \beta \neq 1 \\<br/>\because &amp; \beta^2 \neq\{-1,0,1\}, \text { correct. } \\<br/>\text { Similarly, if } &amp; c \neq q x \\<br/>\Rightarrow &amp; a \frac{\alpha}{\beta} \neq a \alpha \beta \Rightarrow \alpha\left(\beta-\frac{1}{\beta}\right) \neq 0<br/>\end{array}<br/>$$<br/><br/>$$<br/>\begin{array}{ll}<br/>\Rightarrow &amp; \alpha \neq 0 \text { and } \beta-\frac{1}{\beta} \neq 0 \\<br/>\Rightarrow &amp; \beta \neq\{-1,0,1\}<br/>\end{array}<br/>$$<br/>Statement 2 is true.<br/>Both Statement 1 and Statement 2 are true. But Statement 2 does not explain Statement 1.</div>

MarksBatch2_P1.db

let-a-bc-and-a-b-c-be-two-noncongruent-triangles-with-sides-a-b-4-a-c-a-c-2-2-and-angle-b-3-0-the-absolute-value-of-the-difference-between-the-areas-o

let-a-bc-and-a-b-c-be-two-noncongruent-triangles-with-sides-a-b-4-a-c-a-c-2-2-and-angle-b-3-0-the-absolute-value-of-the-difference-between-the-areas-o-15439

<div class="question">Let $A B C$ and $A B C^{\prime}$ be two non-congruent triangles with sides $A B=4$, $A C=A C^{\prime}=2 \sqrt{2}$ and angle $B=30^{\circ}$. The absolute value of the difference between the areas of these triangles is</div>

['Mathematics', 'Properties of Triangles', 'JEE Advanced', 'JEE Advanced 2009 (Paper 2)']

None

<div class="correct-answer"> The correct answer is: <span class="option-value">4</span> </div>

<div class="solution">In $\triangle A B C$, by sine rule,<br/>$$<br/>\begin{aligned}<br/>&amp; \frac{a}{\sin A}=\frac{2 \sqrt{2}}{\sin 30^{\circ}}=\frac{4}{\sin C} \\<br/>\Rightarrow \quad &amp; C=45^{\circ}, C^{\prime}=135^{\circ} \\<br/>\Rightarrow &amp; A=180^{\circ}-\left(45^{\circ}+30^{\circ}\right)=105^{\circ}<br/>\end{aligned}<br/>$$<br/><br/><br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/BUgaugr3-2iN2vBBsw8lq7vfyVWLNvDQoBTiS3mnVkM.original.fullsize.png"/><br/><br/><br/><br/>When $\quad C^{\prime}=135^{\circ}$, then<br/>$$<br/>\begin{aligned}<br/>&amp; \quad A=180^{\circ}-\left(135^{\circ}+30^{\circ}\right)=15^{\circ} \\<br/>&amp; \text { Area of } \triangle A B C=\frac{1}{2} A B \times A C \sin A \\<br/>&amp; =\frac{1}{2} \times 4 \times 2 \sqrt{2} \sin \left(105^{\circ}\right) \\<br/>&amp; =4 \sqrt{2} \times \frac{\sqrt{3}+1}{2 \sqrt{2}}=2(\sqrt{3}+1) \\<br/>&amp; \text { Area of } \triangle A B C^{\prime}=\frac{1}{2} A B \times A C \sin A \\<br/>&amp; =\frac{1}{2} \times 4 \times 2 \sqrt{2} \sin \left(15^{\circ}\right)=2(\sqrt{3}-1)<br/>\end{aligned}<br/>$$<br/>Difference of areas of triangles<br/>$$<br/>=|2(\sqrt{3}+1)-2(\sqrt{3}-1)|=4<br/>$$<img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/7lJ98Kv3OvF_T4cGscG-lVOXIIQusN3nwnSa4WPdPhg.original.fullsize.png"/><br/><br/>$$<br/>A D=2, D C=2<br/>$$<br/>Difference of areas of $\triangle A B C$ and $\triangle A B C^{\prime}$<br/><br/>$$<br/>\begin{aligned}<br/>&amp; =\text { Area of } \triangle A C C^{\prime} \\<br/>&amp; =\frac{1}{2} A D \times C C^{\prime}=\frac{1}{2} \times 2 \times 4=4<br/>\end{aligned}<br/>$$</div>

MarksBatch2_P1.db

let-a-bc-be-a-triangle-such-that-a-cb-6-and-let-a-b-and-c-denote-the-lengths-of-the-sides-opposite-to-a-b-and-c-respectively-the-values-of-x-for-which

let-a-bc-be-a-triangle-such-that-a-cb-6-and-let-a-b-and-c-denote-the-lengths-of-the-sides-opposite-to-a-b-and-c-respectively-the-values-of-x-for-which-66914

<div class="question">Let $A B C$ be a triangle such that $\angle A C B=\frac{\pi}{6}$ and let $a, b$ and $c$ denote the lengths of the sides opposite to $A, B$ and $C$ respectively. The value(s) of $x$ for which $a=x^2+x+1, b=x^2-1$ and $c=2 x+1$ is (are)</div>

['Mathematics', 'Properties of Triangles', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$-(2+\sqrt{3})$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$1+\sqrt{3}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$2+\sqrt{3}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$4 \sqrt{3}$</span> </li> </ul>

<div class="correct-answer"> The correct answers are: <span class="option-value"><br/>$1+\sqrt{3}$<br/></span> </div>

<div class="solution">Using, $\cos C=\frac{a^2+b^2-c^2}{2 a b}$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/89PQ-836ICof7ENzJeD4KV6Ji82KVNt2qE7sAheUO2c.original.fullsize.png"/><br/><br/>$$<br/>\begin{gathered}<br/>\Rightarrow \frac{\sqrt{3}}{2}=\frac{\left(x^2+x+1\right)^2+\left(x^2-1\right)^2-(2 x+1)^2}{2\left(x^2+x+1\right)\left(x^2-1\right)} \\<br/>\Rightarrow(x+2)(x+1)(x-1) x+\left(x^2-1\right)^2 \\<br/>=\sqrt{3}\left(x^2+x+1\right)\left(x^2-1\right) \\<br/>\Rightarrow x^2+2 x+\left(x^2-1\right)=\sqrt{3}\left(x^2+x+1\right) \\<br/>\Rightarrow(2-\sqrt{3}) x^2+(2-\sqrt{3}) x-(\sqrt{3}+1)=0<br/>\end{gathered}<br/>$$<br/><br/>$\Rightarrow x=-(2+\sqrt{3})$ and $x=1+\sqrt{3}$<br/>But, $x=-(2+\sqrt{3}) \Rightarrow c$ is negative. $\therefore \quad x=1+\sqrt{3}$ is the only solution.<br/>Hence, (b) is the correct option.</div>

MarksBatch2_P1.db

let-a-bc-d-be-a-quadrilateral-with-area-18-with-side-a-b-parallel-to-the-side-c-d-and-a-b-2-c-d-let-a-d-be-perpendicular-to-a-b-and-c-d-if-a-circle-is-1

let-a-bc-d-be-a-quadrilateral-with-area-18-with-side-a-b-parallel-to-the-side-c-d-and-a-b-2-c-d-let-a-d-be-perpendicular-to-a-b-and-c-d-if-a-circle-is-1-51680

<div class="question">Let $A B C D$ be a quadrilateral with area 18 , with side $A B$ parallel to the side $C D$ and $A B=2 C D$. Let $A D$ be perpendicular to $A B$ and $C D$. If a circle is drawn inside the quadrilateral $A B C D$ touching all the sides, then its radius is</div>

['Mathematics', 'Circle', 'JEE Main']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>3<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>2<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{3}{2}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>1</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>2<br/></span> </div>

<div class="solution">$$<br/>\text { } 18=\frac{1}{2}(3 \alpha)(2 r) \Rightarrow \alpha r=6<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/t_pMgp4UiyFCXUuzb7KvmMfhX6yIJXigJy_YvZXXXZU.original.fullsize.png"/><br/><br/>Line, $y=-\frac{2 r}{\alpha}(x-2 \alpha)$ is tangent to circle<br/>$$<br/>\begin{aligned}<br/>(x-r)^2+(y-r)^2 &amp; =r^2 \\<br/>2 \alpha &amp; =3 r \text { and } \alpha=6 \\<br/>r &amp; =2<br/>\end{aligned}<br/>$$<br/><br/>$$<br/>\begin{aligned}<br/>\frac{1}{2}(x+2 x) \times 2 r &amp; =18 \\<br/>x r &amp; =6 \\<br/>\tan \theta &amp; =\frac{x-r}{r} \\<br/>\tan \left(90^{\circ}-\theta\right) &amp; =\frac{2 x-r}{r} \\<br/>\frac{x-r}{r} &amp; =\frac{r}{2 x-r} \times(2 x-3 r)=0<br/>\end{aligned}<br/>$$<img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/RYKRVuCPEMy3tbKorSAXw9Jpm6M4NNkl3z1Zw8MBA1Y.original.fullsize.png"/><br/><br/>$$<br/>x=\frac{3 r}{2}<br/>$$<br/>From Eqs. (i) and (ii), we get<br/>$$<br/>r=2<br/>$$</div>

MarksBatch2_P1.db

let-a-bc-d-be-a-quadrilateral-with-area-18-with-side-a-b-parallel-to-the-side-c-d-and-a-b-2-c-d-let-a-d-be-perpendicular-to-a-b-and-c-d-if-a-circle-is

let-a-bc-d-be-a-quadrilateral-with-area-18-with-side-a-b-parallel-to-the-side-c-d-and-a-b-2-c-d-let-a-d-be-perpendicular-to-a-b-and-c-d-if-a-circle-is-49338

<div class="question">Let $A B C D$ be a quadrilateral with area 18 , with side $A B$ parallel to the side $C D$ and $A B=2 C D$. Let $A D$ be perpendicular to $A B$ and $C D$. If a circle is drawn inside the quadrilateral $A B C D$ touching all the sides, then its radius is</div>

['Mathematics', 'Circle', 'JEE Advanced', 'JEE Advanced 2007 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>3<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>2<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{3}{2}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>1</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>2<br/></span> </div>

<div class="solution">$$<br/>\text { } 18=\frac{1}{2}(3 \alpha)(2 r) \Rightarrow \alpha r=6<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/t_pMgp4UiyFCXUuzb7KvmMfhX6yIJXigJy_YvZXXXZU.original.fullsize.png"/><br/><br/>Line, $y=-\frac{2 r}{\alpha}(x-2 \alpha)$ is tangent to circle<br/>$$<br/>\begin{aligned}<br/>(x-r)^2+(y-r)^2 &amp; =r^2 \\<br/>2 \alpha &amp; =3 r \text { and } \alpha=6 \\<br/>r &amp; =2<br/>\end{aligned}<br/>$$<br/><br/>$$<br/>\begin{aligned}<br/>\frac{1}{2}(x+2 x) \times 2 r &amp; =18 \\<br/>x r &amp; =6 \\<br/>\tan \theta &amp; =\frac{x-r}{r} \\<br/>\tan \left(90^{\circ}-\theta\right) &amp; =\frac{2 x-r}{r} \\<br/>\frac{x-r}{r} &amp; =\frac{r}{2 x-r} \times(2 x-3 r)=0<br/>\end{aligned}<br/>$$<img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/RYKRVuCPEMy3tbKorSAXw9Jpm6M4NNkl3z1Zw8MBA1Y.original.fullsize.png"/><br/><br/>$$<br/>x=\frac{3 r}{2}<br/>$$<br/>From Eqs. (i) and (ii), we get<br/>$$<br/>r=2<br/>$$</div>

MarksBatch2_P1.db

let-a-be-vector-parallel-to-line-of-intersection-of-planes-p-1-and-p-2-through-origin-p-1-is-parallel-to-the-vectors-2-j-3-k-and-4-j-3-k-and-p-2-is-pa

let-a-be-vector-parallel-to-line-of-intersection-of-planes-p-1-and-p-2-through-origin-p-1-is-parallel-to-the-vectors-2-j-3-k-and-4-j-3-k-and-p-2-is-pa-96897

<div class="question">Let $\mathbf{A}$ be vector parallel to line of intersection of planes $P_1$ and $P_2$ through origin. $P_1$ is parallel to the vectors $2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ and $P_2$ is parallel to $\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$, then the angle between vector $\mathbf{A}$ and $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ is</div>

['Mathematics', 'Vector Algebra', 'JEE Advanced', 'JEE Advanced 2006']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{\pi}{2}$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{\pi}{4}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{\pi}{6}$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{3 \pi}{4}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

<div class="correct-answer"> The correct answers are: <span class="option-value"><br/>$\frac{\pi}{4}$<br/>, <br/>$\frac{3 \pi}{4}$</span> </div>

<div class="solution">Let vector $\mathbf{A O}$ be parallel to line of intersection of planes $P_1$ and $P_2$ through, i.e.<br/>$$<br/>\begin{aligned}<br/>&amp; {[(2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \times(4 \hat{\mathbf{j}}-3 \hat{\mathbf{k}})] \times[(\hat{\mathbf{j}}-\hat{\mathbf{k}}) \times(3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}})]=54(\hat{\mathbf{j}}-\hat{\mathbf{k}}) .} \\<br/>&amp; \therefore \text { Angle between } 54(\hat{\mathbf{j}}-\hat{\mathbf{k}}) \text { and }(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \\<br/>&amp; \Rightarrow \cos \theta=\pm\left(\frac{54+108}{3.54 \cdot \sqrt{2}}\right)=\pm \frac{1}{\sqrt{2}} \\<br/>&amp; \therefore \quad \theta=\frac{\pi}{4}, \frac{3 \pi}{4}<br/>\end{aligned}<br/>$$</div>

MarksBatch2_P1.db

let-a-i-2-j-k-b-i-j-k-c-i-j-k-a-vector-coplanar-to-a-and-b-has-a-projection-along-c-of-magnitude-3-1-then-the-vector-is

let-a-i-2-j-k-b-i-j-k-c-i-j-k-a-vector-coplanar-to-a-and-b-has-a-projection-along-c-of-magnitude-3-1-then-the-vector-is-34659

<div class="question">Let $\mathbf{a}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{c}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$. A vector coplanar to $\mathbf{a}$ and $\mathbf{b}$ has a projection along c of magnitude $\frac{1}{\sqrt{3}}$, then the vector is</div>

['Mathematics', 'Vector Algebra', 'JEE Advanced', 'JEE Advanced 2006']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$4 \hat{\mathbf{i}}+\hat{\mathbf{j}}-4 \hat{\mathbf{k}}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>None of these</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}$<br/></span> </div>

<div class="solution">Let vector $\mathbf{r}$ be coplanar to $\mathbf{a}$ and $\mathbf{b}$.<br/>$$<br/>\begin{array}{ll}<br/>\therefore &amp; \mathbf{r}=\mathbf{a}+t \mathbf{b} \\<br/>\Rightarrow &amp; \mathbf{r}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})+t(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=\hat{\mathbf{i}}(1+t)+\hat{\mathbf{j}}(2-t)+\hat{\mathbf{k}}(1+t)<br/>\end{array}<br/>$$<br/>The projection of $\mathbf{r}$ on $\mathbf{c}=\frac{1}{\sqrt{3}} \Rightarrow \frac{\mathbf{r} \cdot \mathbf{c}}{|\mathbf{c}|}=\frac{1}{\sqrt{3}}$ $\Rightarrow \frac{|1 \cdot(1+t)+1 \cdot(2-t)-1 \cdot(1+t)|}{\sqrt{3}}=\frac{1}{\sqrt{3}}$ $\Rightarrow|2-t|=\pm 1 \Rightarrow t=1$ or 3<br/>When, $t=1$ we have $\mathbf{r}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$<br/>When, $t=3$ we have $\mathbf{r}=4 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}$</div>

MarksBatch2_P1.db

let-a-i-j-k-b-i-j-k-and-c-i-j-k-be-three-vectors-a-vector-v-in-the-plane-of-a-and-b-whose-projection-of-c-is-3-1-is-given-by

let-a-i-j-k-b-i-j-k-and-c-i-j-k-be-three-vectors-a-vector-v-in-the-plane-of-a-and-b-whose-projection-of-c-is-3-1-is-given-by-15759

<div class="question">Let $\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{c}=\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}$ be three vectors. A vector $\mathbf{v}$ in the plane of $\mathbf{a}$ and $\mathbf{b}$ whose projection of $\mathbf{c}$ is $\frac{1}{\sqrt{3}}$, is given by</div>

['Mathematics', 'Vector Algebra', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$-3 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}-\hat{\mathbf{k}}$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\hat{\mathbf{i}}+3 \hat{\mathbf{k}}-3 \hat{\mathbf{k}}$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}$<br/></span> </div>

<div class="solution">Let $\mathbf{v}=\mathbf{a}+\lambda \mathbf{b}$<br/>$$<br/>\mathbf{v}=(1+\lambda) \hat{\mathbf{i}}+(1-\lambda) \hat{\mathbf{j}}+(1+\lambda) \hat{\mathbf{k}}<br/>$$<br/>Projection of $\mathbf{v}$ on $\mathbf{c}=\frac{1}{\sqrt{3}}$<br/>$$<br/>\begin{array}{lc}<br/>\Rightarrow &amp; \frac{\mathbf{v} \cdot \mathbf{c}}{|\mathbf{c}|}=\frac{1}{\sqrt{3}} \\<br/>\Rightarrow &amp; \frac{(1+\lambda)-(1-\lambda)-(1+\lambda)}{\sqrt{3}}=\frac{1}{\sqrt{3}} \\<br/>\Rightarrow &amp; 1+\lambda-1+\lambda-1-\lambda=1 \\<br/>\Rightarrow &amp; \lambda-1=1 \Rightarrow \lambda=2 \\<br/>\therefore &amp; \mathbf{v}=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+3 \hat{\mathbf{k}}<br/>\end{array}<br/>$$</div>

MarksBatch2_P1.db

let-a-i-k-b-i-j-and-c-i-2-j-3-k-be-three-given-vectors-if-r-is-a-vector-such-that-r-b-c-b-and-r-a-0-then-the-value-of-r-b-is

let-a-i-k-b-i-j-and-c-i-2-j-3-k-be-three-given-vectors-if-r-is-a-vector-such-that-r-b-c-b-and-r-a-0-then-the-value-of-r-b-is-88519

<div class="question">Let $\mathbf{a}=-\hat{\mathbf{i}}-\hat{\mathbf{k}}, \mathbf{b}=-\hat{\mathbf{i}}+\hat{\mathbf{j}}$ and $\mathbf{c}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ be three given vectors. If $\mathbf{r}$ is a vector such that $\mathbf{r} \times \mathbf{b}=\mathbf{c} \times \mathbf{b}$ and $\mathbf{r} \cdot \mathbf{a}=0$, then the value of $\mathbf{r} \cdot \mathbf{b}$ is</div>

['Mathematics', 'Vector Algebra', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)']

None

<div class="correct-answer"> The correct answer is: <span class="option-value">9</span> </div>

<div class="solution">$$<br/>\begin{aligned}<br/>&amp; \mathbf{r} \times \mathbf{b}=\mathbf{c} \times \mathbf{b} \\<br/>&amp; \Rightarrow \quad(\mathbf{r}-\mathbf{c}) \times \mathbf{b}=0 \Rightarrow \mathbf{r}-\mathbf{c}+\lambda \mathbf{b} \\<br/>&amp; \text { or } \quad \mathbf{r}=\mathbf{c}+\lambda \mathbf{b} \\<br/>&amp;<br/>\end{aligned}<br/>$$<br/>Given, $\mathbf{r} \cdot \mathbf{a}=0$, taking dot product with a for Eq. (i).<br/>Now, $\mathbf{r} \cdot \mathbf{a}=\mathbf{a} \cdot \mathbf{c}+\lambda \mathbf{a} \cdot \mathbf{b}$<br/>$\therefore \quad \lambda=\frac{-\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}} \quad[\because \vec{r} \cdot \vec{a}=0] \ldots$ (i)<br/>From Eqs. (i) and (ii), we get<br/>$$<br/>\mathbf{r}=\mathbf{c}-\frac{\mathbf{a} \cdot \mathbf{c}}{\mathbf{a} \cdot \mathbf{b}} \mathbf{b}<br/>$$<br/>Taking dot with $\mathbf{b}$, we get<br/>$$<br/>\mathbf{r} \cdot \mathbf{b}=\mathbf{c} \cdot \mathbf{b}-\frac{\mathbf{a} \cdot \mathbf{c}}{\mathbf{a} \cdot \mathbf{b}}(\mathbf{b} \cdot \mathbf{b})<br/>$$<br/><br/>where, $\left[\begin{array}{l}\mathbf{a}=-\hat{\mathbf{i}}-\hat{\mathbf{k}} \\ \mathbf{b}=-\hat{\mathbf{i}}+\hat{\mathbf{j}} \\ \mathbf{c}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\end{array}\right]$<br/>$$<br/>=(-1+2)-\frac{(-1-3)}{(1)}(1+1)=1+8=9<br/>$$</div>

MarksBatch2_P1.db

let-a-n-4-3-4-3-2-4-3-3-1-n-1-4-3-n-and-b-n-1-a-n-find-a-least-odd-natural-number-n-0-so-that-b-n-a-n-n-n-0

let-a-n-4-3-4-3-2-4-3-3-1-n-1-4-3-n-and-b-n-1-a-n-find-a-least-odd-natural-number-n-0-so-that-b-n-a-n-n-n-0-69545

<div class="question">Let $A_n=\left(\frac{3}{4}\right)-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3+\ldots+(-1)^{n-1}\left(\frac{3}{4}\right)^n$ and $B_n=1-A_n$. Find a least odd natural number $n_0$, so that $B_n&gt;A_n, \forall n \geq n_0$.</div>

['Mathematics', 'Sequences and Series', 'JEE Advanced', 'JEE Advanced 2006']

None

<div class="correct-answer"> The correct answer is: <span class="option-value">7</span> </div>

<div class="solution">$B_n=1-A_n&gt;A_n \Rightarrow A_n &lt; \frac{1}{2}$ Now, $A_n=\frac{\frac{3}{4}\left(1-\left(-\frac{3}{4}\right)^n\right)}{1+\frac{3}{4}} &lt; \frac{1}{2} \Rightarrow\left(-\frac{3}{4}\right)&gt;-\frac{1}{6}$<br/>Obviously, it is true for all even values of $n$. But, for<br/>$$<br/>\begin{aligned}<br/>n &amp; =1, &amp; \frac{3}{4} &amp; &lt; \frac{1}{6} \\<br/>n &amp; =3, &amp; \frac{27}{64} &amp; &lt; \frac{1}{6} \\<br/>n &amp; =5, &amp; \frac{243}{1024} &amp; &gt;\frac{1}{6}<br/>\end{aligned}<br/>$$<br/>which is true for $n=7$.<br/>Obviously, $n_0=7$</div>

MarksBatch2_P1.db

let-a-n-denote-the-number-of-all-n-digit-positive-integers-formed-by-the-digits-01-or-both-such-that-no-consecutive-digits-in-them-are-0-let-b-n-the-n-1

let-a-n-denote-the-number-of-all-n-digit-positive-integers-formed-by-the-digits-01-or-both-such-that-no-consecutive-digits-in-them-are-0-let-b-n-the-n-1-55664

<div class="question">Let $a_{n}$ denote the number of all $n$-digit positive integers formed by the digits 0,1 or both such that no consecutive digits in them are 0 . Let $b_{n}=$ the number of such $n$-digit integers ending with digit 1 and $c_{n}=$ the number of such $n$-digit integers ending with digit 0 . <br/> <br/><strong> Question:</strong> Which of the following is correct?</div>

['Mathematics', 'Permutation Combination', 'JEE Advanced', 'JEE Advanced 2012 (Paper 2)']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$a_{17}=a_{16}+a_{15}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$c_{17} \neq c_{16}+c_{15}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$b_{17} \neq b_{16}+c_{16}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$a_{17}=c_{17}+b_{16}$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value">$a_{17}=a_{16}+a_{15}$</span> </div>

<div class="solution">By recurring formula, $a_{17}=a_{16}+a_{15}$ is correct <br/> <br/>Also $c_{17} \neq c_{16}+c_{15}$ <br/> <br/>$\Rightarrow a_{15} \neq a_{14}+a_{13}\left(\because c_{n}=a_{n-2}\right)$ <br/> <br/>$\therefore$ Incorrect <br/> <br/>Similarly, other parts are also incorrect.</div>

MarksBatch2_P1.db

let-a-n-denote-the-number-of-all-n-digit-positive-integers-formed-by-the-digits-01-or-both-such-that-no-consecutive-digits-in-them-are-0-let-b-n-the-n

let-a-n-denote-the-number-of-all-n-digit-positive-integers-formed-by-the-digits-01-or-both-such-that-no-consecutive-digits-in-them-are-0-let-b-n-the-n-10346

<div class="question">Let $a_{n}$ denote the number of all $n$-digit positive integers formed by the digits 0,1 or both such that no consecutive digits in them are 0 . Let $b_{n}=$ the number of such $n$-digit integers ending with digit 1 and $c_{n}=$ the number of such $n$-digit integers ending with digit 0 . <br/> <br/><strong> Question:</strong> The value of $b_{6}$ is</div>

['Mathematics', 'Permutation Combination', 'JEE Advanced', 'JEE Advanced 2012 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">7</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">8</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">9</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">11</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value">8</span> </div>

<div class="solution">$\because a_{n}=$ number of all $n$ digit +ve integers formed by the digits 0,1 or both such that no consecutive digits in them are 0 . <br/> <br/>and $b_{n}=$ number of such $n$ digit integers ending with 1 $c_{n}=$ number of such $n$ digit integers ending with 0 <br/> <br/>Clearly, $a_{n}=b_{n}+c_{n}\left(\because a_{n}\right.$ can end with 0 or 1) <br/> <br/>Also $b_{n}=a_{n-1}$ and $c_{n}=a_{n-2}[\because$ if last digit is 0 , second last has to be 1] <br/> <br/>$\therefore$ We get $a_{n}=a_{n-1}+a_{n-2}, n \geq 3$ <br/> <br/>Also $a_{1}=1, a_{2}=2$, <br/> <br/>Now by this recurring formula, we get <br/> <br/>$\begin{aligned} a_{3} &amp;=a_{2}+a_{1}=3 \\ a_{4} &amp;=a_{3}+a_{2}=3+2=5 \\ a_{5} &amp;=a_{4}+a_{3}=5+3=8 \\ \text { Also } b_{6} &amp;=a_{5}=8 \end{aligned}$</div>

MarksBatch2_P1.db

let-a-solution-y-y-x-of-the-differential-equation-x-x-2-1-d-y-y-y-2-1-d-x-0-satisfy-y-2-3-2-statement-1-y-x-sec-sec-1-x-6-statement-2-y-x-is-given-by-

let-a-solution-y-y-x-of-the-differential-equation-x-x-2-1-d-y-y-y-2-1-d-x-0-satisfy-y-2-3-2-statement-1-y-x-sec-sec-1-x-6-statement-2-y-x-is-given-by-74693

<div class="question">Let a solution $y=y(x)$ of the differential equation $x \sqrt{x^2-1} d y-y \sqrt{y^2-1} d x=0$ satisfy $y(2)=\frac{2}{\sqrt{3}}$<br/>Statement $1 y(x)=\sec \left(\sec ^{-1} x-\frac{\pi}{6}\right)$<br/>Statement $2 y(x)$ is given by $\frac{1}{y}=\frac{2 \sqrt{3}}{x}-\sqrt{1-\frac{1}{x^2}}$</div>

['Mathematics', 'Differential Equations', 'JEE Advanced', 'JEE Advanced 2008 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>Statement 1 is true, Statement 2 is false.<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>Statement 1 is false, Statement 2 is true</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>Statement 1 is true, Statement 2 is false.<br/></span> </div>

<div class="solution">$\because \frac{d y}{d x}=\frac{y \sqrt{y^2-1}}{x \sqrt{x^2-1}} \Rightarrow \int \frac{d y}{y \sqrt{y^2-1}}=\int \frac{d x}{x \sqrt{x^2-1}}$ $\Rightarrow \quad \sec ^{-1} y=\sec ^{-1} x+C$ At $\quad x=2, y=\frac{2}{\sqrt{3}}$ $$ \frac{\pi}{6}=\frac{\pi}{3}+C \Rightarrow C=-\frac{\pi}{6} $$<br/>Now, $y=\sec \left(\sec ^{-1} x-\frac{\pi}{6}\right)$<br/><br/>$$<br/>\begin{aligned}<br/>&amp; =\cos \left[\cos ^{-1} \frac{1}{x}-\cos ^{-1} \frac{\sqrt{3}}{2}\right] \\<br/>&amp; =\cos \left[\cos ^{-1}\left(\frac{\sqrt{3}}{2 x}+\sqrt{1-\frac{1}{x^2}} \sqrt{1-\frac{3}{4}}\right)\right] \\<br/>\Rightarrow \quad \frac{1}{y} &amp; =\frac{\sqrt{3}}{2 x}+\frac{1}{2} \sqrt{1-\frac{1}{x^2}} \\</div>

MarksBatch2_P1.db

let-and-be-the-roots-of-x-2-6-x-2-0-with-if-a-n-n-n-for-n-1-then-the-value-of-2-a-9-a-10-2-a-8-is

let-and-be-the-roots-of-x-2-6-x-2-0-with-if-a-n-n-n-for-n-1-then-the-value-of-2-a-9-a-10-2-a-8-is-46905

<div class="question">Let $\alpha$ and $\beta$ be the roots of $x^2-6 x-2=0$ with $\alpha&gt;\beta$. If $a_n=\alpha^n-\beta^n$ for $n \geq 1$, then the value of $\frac{a_{10}-2 a_8}{2 a_9}$ is</div>

['Mathematics', 'Quadratic Equation', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>1<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>2<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>3<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>4</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>3<br/></span> </div>

<div class="solution">$\frac{a_{10}-2 a_8}{2 a_9}=\frac{\left(\alpha^{10}-\beta^{10}\right)-2\left(\alpha^8-\beta^8\right)}{2\left(\alpha^9-\beta^9\right)}$ $=\frac{\alpha^8\left(\alpha^2-2\right)-\beta^8\left(\beta^2-2\right)}{2\left(\alpha^9-\beta^9\right)}$ $\left[\because \alpha\right.$ is root of $x^2-6 x-2=0$ $\left.\alpha^2-2=6 \alpha\right]$<br/>Also, $\beta$ is root of $x^2-6 x-2=0$<br/>$$<br/>\begin{aligned}<br/>&amp; \Rightarrow \quad \beta^2-2=6 \beta \\<br/>&amp; =\frac{\alpha^8(6 \alpha)-\beta^8(6 \beta)}{2\left(\alpha^9-\beta^9\right)}=\frac{6\left(\alpha^9-\beta^9\right)}{2\left(\alpha^9-\beta^9\right)}=3<br/>\end{aligned}<br/>$$</div>

MarksBatch2_P1.db

let-be-a-complex-cube-root-of-unity-with-1-a-fair-die-is-thrown-three-times-if-r-1-r-2-and-r-3-are-the-numbers-obtained-on-the-die-then-the-probabilit

let-be-a-complex-cube-root-of-unity-with-1-a-fair-die-is-thrown-three-times-if-r-1-r-2-and-r-3-are-the-numbers-obtained-on-the-die-then-the-probabilit-32393

<div class="question">Let $\omega$ be a complex cube root of unity with $\omega \neq 1$. A fair die is thrown three times. If $r_1, r_2$ and $r_3$ are the numbers obtained on the die, then the probability that $\omega^{r_1}+\omega^{r_2}+\omega^{r_3}=0$ is</div>

['Mathematics', 'Probability', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{1}{18}$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{1}{9}$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{2}{9}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{1}{36}$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$\frac{2}{9}$<br/></span> </div>

<div class="solution">Sample space A dice is thrown thrice, $n(s)=6 \times 6 \times 6$.<br/>Favorable events $\omega^{r_1}+\omega^{r_2}+\omega^{r_3}=0$ ie, $\left(r_1, r_2, r_3\right)$ are ordered 3-triples which can take values, $\left.\begin{array}{llll}(1,2,3), &amp; (1,5,3), &amp; (4,2,3), &amp; (4,5,3) \\ (1,2,6), &amp; (1,5,6), &amp; (4,2,6), &amp; (4,5,6)\end{array}\right\}$ ie, 8 ordered pairs and each can be arranged in 3 ! ways $=6$<br/>$$<br/>\begin{array}{ll}<br/>\therefore &amp; n(E)=8 \times 6 \\<br/>\Rightarrow &amp; P(E)=\frac{8 \times 6}{6 \times 6 \times 6}=\frac{2}{9}<br/>\end{array}<br/>$$</div>

MarksBatch2_P1.db

let-be-the-complex-number-cos-3-2-i-sin-3-2-then-the-number-of-distinct-complex-number-z-satisfying-z-1-2-z-2-1-2-1-z-0-is-equal-to

let-be-the-complex-number-cos-3-2-i-sin-3-2-then-the-number-of-distinct-complex-number-z-satisfying-z-1-2-z-2-1-2-1-z-0-is-equal-to-72006

<div class="question">Let $\omega$ be the complex number $\cos \frac{2 \pi}{3}+i \sin \frac{2 \pi}{3}$. Then the number of distinct complex number $z$ satisfying $\left|\begin{array}{ccc}z+1 &amp; \omega &amp; \omega^2 \\ \omega &amp; z+\omega^2 &amp; 1 \\ \omega^2 &amp; 1 &amp; z+\omega\end{array}\right|=0$ is equal to</div>

['Mathematics', 'Complex Number', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)']

None

<div class="correct-answer"> The correct answer is: <span class="option-value">1</span> </div>

<div class="solution">Let $A=\left[\begin{array}{ccc}1 &amp; \omega &amp; \omega^2 \\ \omega &amp; \omega^2 &amp; 1 \\ \omega^2 &amp; 1 &amp; \omega\end{array}\right]$<br/>Now, $\quad A^2=\left[\begin{array}{lll}0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0 \\ 0 &amp; 0 &amp; 0\end{array}\right]$<br/>and $\operatorname{Tr}(A)=0,|A|=0$<br/>$\therefore \quad A^3=0$<br/><br/>$\Rightarrow\left|\begin{array}{ccc}z+1 &amp; \omega &amp; \omega^2 \\ \omega &amp; z+\omega^2 &amp; 1 \\ \omega^2 &amp; 1 &amp; z+\omega\end{array}\right|=|A+z I|=0$<br/>$\Rightarrow \quad z^3=0$<br/>$\Rightarrow z=0$, the number of $z$ satisfying the given equation is 1 .</div>

MarksBatch2_P1.db

let-be-the-roots-of-the-equation-x-2-p-x-r-0-and-2-2-be-the-roots-of-the-equation-x-2-q-x-r-0-then-the-value-of-r-is

let-be-the-roots-of-the-equation-x-2-p-x-r-0-and-2-2-be-the-roots-of-the-equation-x-2-q-x-r-0-then-the-value-of-r-is-80075

<div class="question">Let $\alpha, \beta$ be the roots of the equation $x^2-p x+r=0$ and $\frac{\alpha}{2}, 2 \beta$ be the roots of the equation $x^2-q x+r=0$. Then, the value of $r$ is</div>

['Mathematics', 'Quadratic Equation', 'JEE Advanced', 'JEE Advanced 2007 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{2}{9}(p-q)(2 q-p)$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{2}{9}(q-p)(2 p-q)$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$$<br/>\frac{2}{9}(q-2 p)(2 q-p)<br/>$$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{2}{9}(2 p-q)(2 q-p)$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$\frac{2}{9}(2 p-q)(2 q-p)$</span> </div>

<div class="solution">The equation $x^2-p x+r=0$ has roots $(\alpha, \beta)$ and the equation $x^2-q x+r=0$ has roots<br/>$$<br/>\begin{array}{ll}<br/>\left(\frac{\alpha}{2}, 2 \beta\right) . &amp; \\<br/>\Rightarrow &amp; r=\alpha \beta \text { and } \alpha+\beta=p \text { and } \frac{\alpha}{2}+2 \beta=q \\<br/>\Rightarrow &amp; \beta=\frac{2 q-p}{3} \text { and } \alpha=\frac{2(2 p-q)}{3} \\<br/>\Rightarrow &amp; \alpha \beta=r=\frac{2}{9}(2 q-p)(2 p-q)<br/>\end{array}<br/>$$</div>

MarksBatch2_P1.db

let-e-and-f-be-two-independent-events-the-probability-that-exactly-one-of-them-occurs-is-1125-and-the-probability-of-none-of-them-occurring-is-225-if-

let-e-and-f-be-two-independent-events-the-probability-that-exactly-one-of-them-occurs-is-1125-and-the-probability-of-none-of-them-occurring-is-225-if-64101

<div class="question">Let $E$ and $F$ be two independent events. The probability that exactly one of them occurs is $11 / 25$ and the probability of none of them occurring is $2 / 25$. If $P(T)$ denotes the probability of occurrence of the event $T$, then</div>

['Mathematics', 'Probability', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$P(E)=\frac{4}{5}, P(F)=\frac{3}{5}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$P(E)=\frac{1}{5}, P(F)=\frac{2}{5}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$P(E)=\frac{2}{5}, P(F)=\frac{1}{5}$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$P(E)=\frac{3}{5}, P(F)=\frac{4}{5}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

<div class="correct-answer"> The correct answers are: <span class="option-value"><br/>$P(E)=\frac{4}{5}, P(F)=\frac{3}{5}$<br/>, <br/>$P(E)=\frac{3}{5}, P(F)=\frac{4}{5}$</span> </div>

<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/npte3oGDly9GUN3k8gBrVXSvn3jwoEwWiCLTtMUQ5Ng.original.fullsize.png"/><br/><br/><br/>$$<br/>P(E \cup F)-P(E \cap F)=\frac{11}{25}<br/>$$<br/>(i.e. only $E$ or only $F$ )<img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/mBHGp2I9tOtkfP60SSq3iGSoDXC_ovlYuNpy4rtKhKE.original.fullsize.png"/><br/><br/>Neither of them occurs $=\frac{2}{25}$ $\Rightarrow \quad P(\bar{E} \cap \bar{F})=\frac{2}{25}$<br/>From Eq. (i), we get<br/>$$<br/>P(E)+P(F)-2 P(E \cap F)=\frac{11}{25}<br/>$$<br/>From Eq. (ii), we get<br/>$$<br/>\begin{gathered}<br/>(1-P(E))(1-P(F))=\frac{2}{25} \\<br/>\Rightarrow 1-P(E)-P(F)+P(E) \cdot P(F)=\frac{2}{25}<br/>\end{gathered}<br/>$$<br/>(iv)<br/>From Eqs. (iii) and (iv), we get<br/>$$<br/>\begin{aligned}<br/>&amp; P(E)+P(F)=\frac{7}{5} \text { and } P(E) \cdot P(F)=\frac{12}{25} \\<br/>&amp; \therefore \quad P(E) \cdot\left\{\frac{7}{5}-P(E)\right\}=\frac{12}{25}<br/>\end{aligned}<br/>$$<br/><br/>$$<br/>\begin{aligned}<br/>&amp; \Rightarrow \quad(P(E))^2-\frac{7}{5} P(E)+\frac{12}{25}=0 \\<br/>&amp; \Rightarrow \quad\left(P(E)-\frac{3}{5}\right)\left(P(E)-\frac{4}{5}\right)=0 \\<br/>&amp; \therefore \quad P(E)=\frac{3}{4} \text { or } \frac{4}{5} \Rightarrow P(F)=\frac{4}{5} \text { or } \frac{3}{5}<br/>\end{aligned}<br/>$$</div>

MarksBatch2_P1.db

let-e-c-denotes-the-complement-of-an-event-e-let-e-f-g-be-pairwise-independent-events-with-p-g-0-and-p-e-f-g-0-then-p-e-c-f-c-g-equals

let-e-c-denotes-the-complement-of-an-event-e-let-e-f-g-be-pairwise-independent-events-with-p-g-0-and-p-e-f-g-0-then-p-e-c-f-c-g-equals-50316

<div class="question">Let $E^c$ denotes the complement of an event $E$. Let $E, F, G$ be pairwise independent events with $P(G)&gt;0$ and $P(E \cap F \cap G)=0$. Then, $P\left(E^c \cap F^c \mid G\right)$ equals</div>

['Mathematics', 'Probability', 'JEE Advanced', 'JEE Advanced 2007 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$P\left(E^c\right)+P\left(F^c\right)$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$P\left(E^c\right)-P\left(F^c\right)$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$P\left(E^c\right)-P(F)$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$P(E)-P\left(F^c\right)$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$P\left(E^c\right)-P(F)$<br/></span> </div>

<div class="solution">$$<br/>\text { } \begin{aligned}<br/>P\left(\frac{E^c \cap F^c}{G}\right) &amp; =\frac{P\left(E^c \cap F^c \cap G\right)}{P(G)} \\<br/>&amp; =\frac{P(G)-P(E \cap G)-P(G \cap F)}{P(G)} \\<br/>&amp; =\frac{P(G)[1-P(E)-P(F)]}{P(G)} \\<br/>&amp; =1-P(E)-P(F) \\<br/>&amp; =P\left(E^c\right)-P(F) .<br/>\end{aligned}<br/>$$</div>

MarksBatch2_P1.db

let-e-i-3-and-a-b-c-x-y-z-be-nonzero-complex-numbers-such-that-a-b-c-x-a-b-c-2-y-a-b-2-c-z-then-the-value-of-a-2-b-2-c-2-x-2-y-2-z-2-is

let-e-i-3-and-a-b-c-x-y-z-be-nonzero-complex-numbers-such-that-a-b-c-x-a-b-c-2-y-a-b-2-c-z-then-the-value-of-a-2-b-2-c-2-x-2-y-2-z-2-is-35077

<div class="question">Let $\omega=e^{i \pi / 3}$ and $a, b, c, x, y, z$ be non-zero complex numbers such that $a+b+c=x, a+b \omega+c \omega^2=y, a+b \omega^2+c \omega=z$.<br/>Then, the value of $\frac{|x|^2+|y|^2+|z|^2}{|a|^2+|b|^2+|c|^2}$ is</div>

['Mathematics', 'Complex Number', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)']

None

<div class="correct-answer"> The correct answer is: <span class="option-value">3</span> </div>

<div class="solution">Here, $\omega=e^{i 2 \pi / 3}$, then only integer solution exists.<br/>Then, $\frac{\left|x^2\right|+\left|y^2\right|+\left|z^2\right|}{\left|a^2\right|+\left|b^2\right|+\left|c^2\right|}=3$</div>

MarksBatch2_P1.db

let-f-0-1-r-be-defined-by-f-x-1-b-x-b-x-where-b-is-a-constant-such-that-0-b-1-then

let-f-0-1-r-be-defined-by-f-x-1-b-x-b-x-where-b-is-a-constant-such-that-0-b-1-then-66708

<div class="question">Let $f:(0,1) \rightarrow R$ be defined by $f(x)=\frac{b-x}{1-b x}$, where $b$ is a constant such that $0 &lt; b &lt; 1$. Then,</div>

['Mathematics', 'Functions', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$f$ if not invertible on $(0,1)$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$f \neq f^{-1}$ on $(0,1)$ and $f^{\prime}(b)=\frac{1}{f^{\prime}(0)}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$f=f^{-1}$ on $(0,1)$ and $f^{\prime}(b)=\frac{1}{f^{\prime}(0)}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$f^{-1}$ is differentiable on $(0,1)$</span> </li> </ul>

<div class="correct-answer"> The correct answers are: <span class="option-value"><br/>$f$ if not invertible on $(0,1)$<br/></span> </div>

<div class="solution">Here, $f(x)=\frac{b-x}{1-b x}$ where, $0 &lt; b &lt; 1,0 &lt; x &lt; 1$ For function to be invertible it should be one-one onto.<br/>$\therefore$ Check range :<br/>Let $\quad f(x)=y \Rightarrow y=\frac{b-x}{1-b x}$<br/>$$<br/>\begin{aligned}<br/>&amp; \Rightarrow y-b x y=b-x \Rightarrow x(1-b y)=b-y \\<br/>&amp; \Rightarrow \quad x=\frac{b-y}{1-b y}<br/>\end{aligned}<br/>$$<br/><br/>where, $0 &lt; x &lt; 1$<br/>$$<br/>\begin{gathered}<br/>\therefore \quad 0 &lt; \frac{b-y}{1-b y} &lt; 1 \\<br/>\frac{b-y}{1-b y}&gt;0 \text { and } \frac{b-y}{1-b y} &lt; 1 \\<br/>\quad+\quad-\quad+ \\<br/>\Rightarrow \quad b \\<br/>y &lt; b \text { or } y&gt;\frac{1}{b} \\<br/>\frac{(b-1)(y+1)}{1-b y} &lt; -1 &lt; y &lt; \frac{1}{b}<br/>\end{gathered}<br/>$$<br/>From Eqs. (i) and (ii), we get $y \in\left(-1, \frac{1}{b}\right) \subset$ Codomain<br/>Thus, $f(x)$ is not invertible.</div>

MarksBatch2_P1.db

let-f-1-1-i-r-be-such-that-f-cos-4-2-s-e-c-2-2-for-0-4-4-2-then-the-value-s-of-f-3-1-is-are

let-f-1-1-i-r-be-such-that-f-cos-4-2-s-e-c-2-2-for-0-4-4-2-then-the-value-s-of-f-3-1-is-are-66058

<div class="question">Let $f:(-1,1) \rightarrow I R$ be such that $f(\cos 4 \theta)=\frac{2}{2-\sec ^{2} \theta}$ for $\theta \in\left(0, \frac{\pi}{4}\right) \cup\left(\frac{\pi}{4}, \frac{\pi}{2}\right)$. Then the value (s) of $f\left(\frac{1}{3}\right)$ is (are)</div>

['Mathematics', 'Trigonometric Ratios & Identities', 'JEE Advanced', 'JEE Advanced 2012 (Paper 2)']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data">$1-\sqrt{\frac{3}{2}}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$1+\sqrt{\frac{3}{2}}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$1-\sqrt{\frac{2}{3}}$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$1+\sqrt{\frac{2}{3}}$</span> </li> </ul>

<div class="correct-answer"> The correct answers are: <span class="option-value">$1-\sqrt{\frac{3}{2}}$, $1+\sqrt{\frac{3}{2}}$</span> </div>

<div class="solution">Given : $f(\cos 4 \theta)=\frac{2}{2-\sec ^{2} \theta}=\frac{2 \cos ^{2} \theta}{2 \cos ^{2} \theta-1}$ <br/> <br/>$=\frac{1+\cos 2 \theta}{\cos 2 \theta}=1+\frac{1}{\cos 2 \theta}$ <br/> <br/>Let $\cos 4 \theta=\frac{1}{3} \Rightarrow 2 \cos ^{2} 2 \theta-1=\frac{1}{3} \Rightarrow \cos 2 \theta=\pm \sqrt{\frac{2}{3}}$ <br/> <br/>$\therefore f(\cos 4 \theta)=1+\frac{1}{\cos 2 \theta}=1 \pm \sqrt{\frac{3}{2}}$ or $f\left(\frac{1}{3}\right)=1 \pm \sqrt{\frac{3}{2}}$</div>

MarksBatch2_P1.db

let-f-1-2-0-be-a-continuous-function-such-that-f-x-f-1-x-for-all-x-1-2-let-r-1-1-2-x-f-x-d-x-and-r-2-be-the-area-of-the-region-bounded-by-y-f-x-x-1-x-

let-f-1-2-0-be-a-continuous-function-such-that-f-x-f-1-x-for-all-x-1-2-let-r-1-1-2-x-f-x-d-x-and-r-2-be-the-area-of-the-region-bounded-by-y-f-x-x-1-x-68472

<div class="question">Let $f:[-1,2] \rightarrow[0, \infty)$ be a continuous function such that $f(x)=f(1-x)$ for all $x \in[-1,2]$. Let $R_1=\int_{-1}^2 x f(x) d x$ and $R_2$ be the area of the region bounded by $y=f(x), x=-1, x=2$ and the $X$-axis. Then,</div>

['Mathematics', 'Definite Integration', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$R_1=2 R_2$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$R_1=3 R_2$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$2 R_1=R_2$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$3 R_1=R_2$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$2 R_1=R_2$<br/></span> </div>

<div class="solution">Here, $R_1=\int_{-1}^2 x f(x) d x$<br/>Using, $\quad \int_a^b f(x) d x=\int_a^b f(a+b-x) d x$<br/>$$<br/>\begin{aligned}<br/>&amp; R_1=\int_{-1}^2(1-x) f(1-x) d x, \\<br/>\therefore \quad &amp; \left.R_1=\int_{-1}^2(1-x) f(x) d x \quad \ldots(1-x)\right]<br/>\end{aligned}<br/>$$<br/>Given, $R_2$ is area bounded by<br/>$$<br/>\begin{aligned}<br/>&amp; f(x), x-1 \text { and } x=2 \\<br/>\therefore &amp; R_2=\int_{-1}^2 f(x) d x<br/>\end{aligned}<br/>$$<br/>Adding Eqs. (i) and (ii), we get<br/>$$<br/>2 R_1=\int_{-1}^2 f(x) d x<br/>$$<br/><br/>$\therefore$ From Eqs. (iii) and (iv), we get<br/>$$<br/>2 R_1=R_2<br/>$$</div>

MarksBatch2_P1.db

let-f-1-2-be-a-differentiable-function-such-that-f-1-2-if-6-1-x-f-t-d-t-3-x-f-x-x-3-for-all-x-1-then-the-value-of-f-2-is

let-f-1-2-be-a-differentiable-function-such-that-f-1-2-if-6-1-x-f-t-d-t-3-x-f-x-x-3-for-all-x-1-then-the-value-of-f-2-is-18706

<div class="question">Let $f:[1, \infty) \rightarrow[2, \infty)$ be a differentiable function such that $f(1)=2$. If $6 \int_1^x f(t) d t=3 x f(x)-x^3$ for all $x \geq 1$, then the value of $f(2)$ is</div>

['Mathematics', 'Definite Integration', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)']

None

<div class="correct-answer"> The correct answer is: <span class="option-value">2.667</span> </div>

<div class="solution">Given, $f(1)=\frac{1}{3}$ and $6 \int_1^x f(t) d t$ $=3 x f(x)-x^3$, for all $x \geq 1$<br/>Using (Newton-Leibnitz formula),<br/>On Differentiating both sides,<br/>$$<br/>\begin{aligned}<br/>&amp; 6 f(x) \cdot 1-0-3 f(x)+3 x f^{\prime}(x)-3 x^2 \\<br/>\Rightarrow &amp; 3 x f^{\prime}(x)-3 f(x)=3 x^2 \\<br/>\Rightarrow &amp; f^{\prime}(x)-\frac{1}{x} f(x)=x<br/>\end{aligned}<br/>$$<br/><br/>$$<br/>\Rightarrow \frac{x f^{\prime}(x)-f(x)}{x^2}=1 \Rightarrow \frac{d}{d x}\left\{\frac{f(x)}{x}\right\}=1<br/>$$<br/>On integrating both sides,<br/>$$<br/>\begin{array}{ll} <br/>&amp; \frac{f(x)}{x}=x+C \quad\left[\because f(1)=\frac{1}{3}\right] \\<br/>\Rightarrow \quad &amp; \frac{1}{3}=1+C \Rightarrow \quad C=-\frac{2}{3} \\<br/>&amp; \text { Now, } \quad f(x)=x^2-\frac{2}{3} x \\<br/>\Rightarrow \quad &amp; f(2)=4-\frac{4}{3}=\frac{8}{3}<br/>\end{array}<br/>$$<br/>Note Here, $f(1)=2$ does not satisfy given function.<br/>$$<br/>\therefore \quad f(1)=\frac{1}{3}<br/>$$<br/>For that, $f(x)=x^2-\frac{2}{3} x$ and $\quad f(2)=4-\frac{4}{3}=\frac{8}{3}$</div>

MarksBatch2_P1.db

let-f-and-g-be-real-valued-functions-defined-on-interval-1-1-such-that-g-x-is-continuous-g-0-0-g-0-0-g-0-0-and-f-x-g-x-sin-x-statement-1-lim-x-0-g-x-c

let-f-and-g-be-real-valued-functions-defined-on-interval-1-1-such-that-g-x-is-continuous-g-0-0-g-0-0-g-0-0-and-f-x-g-x-sin-x-statement-1-lim-x-0-g-x-c-95023

<div class="question">Let $f$ and $g$ be real valued functions defined on interval $(-1,1)$ such that $g^{\prime \prime}(x)$ is continuous, $g(0) \neq 0, g^{\prime}(0)=0, g^{\prime \prime}(0) \neq 0$ and $f(x)=g(x) \sin x$.<br/>Statement $1 \lim _{x \rightarrow 0}[g(x) \cos x-g(0) \operatorname{cosec} x]=f^{\prime \prime}(0)$.<br/>Statement $2 f^{\prime}(0)=g(0)$.</div>

['Mathematics', 'Continuity and Differentiability', 'JEE Advanced', 'JEE Advanced 2008 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1.<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>Statement 1 is true, Statement 2 is false.<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>Statement 1 is false, Statement 2 is true</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 1.<br/></span> </div>

<div class="solution">We have,<br/>$$<br/>\begin{aligned}<br/>&amp; \lim _{x \rightarrow 0} \frac{g(x) \cos x-g(0)}{\sin x} \\<br/>&amp; =\lim _{x \rightarrow 0} \frac{g^{\prime}(x) \cos x-g(x) \sin x}{\cos x}=0<br/>\end{aligned}<br/>$$<br/>Since, $f(x)=g(x) \sin x$<br/>$$<br/>\begin{aligned}<br/>&amp; \Rightarrow \quad f^{\prime}(x)=g^{\prime}(x) \sin x+g(x) \cos x \\<br/>&amp; \Rightarrow \quad f^{\prime \prime}(x)=g^{\prime \prime}(x) \sin x+2 g^{\prime}(x) \cos x-g(x) \sin x \\<br/>&amp; \Rightarrow \quad f^{\prime \prime}(0)=0<br/>\end{aligned}<br/>$$<br/>Thus, $\lim _{x \rightarrow 0}[g(x) \cos x-g(0) \operatorname{cosec} x]=0=f^{\prime \prime}(0)$<br/>$\Rightarrow$ Statement 1 is true.<br/>Statement 2. $f^{\prime}(x)=g^{\prime}(x) \sin x+g(x) \cos x$<br/>$$<br/>\Rightarrow \quad f^{\prime}(0)=g(0)<br/>$$<br/>Statement 2 is not a correct explanation of Statement 1.</div>

MarksBatch2_P1.db

let-f-be-a-function-defined-on-r-the-set-of-all-real-numbers-such-that-f-x-2010-x-2009-x-2010-2-x-2011-3-x-2012-4-for-all-x-r-if-g-is-a-function-defin

let-f-be-a-function-defined-on-r-the-set-of-all-real-numbers-such-that-f-x-2010-x-2009-x-2010-2-x-2011-3-x-2012-4-for-all-x-r-if-g-is-a-function-defin-19026

<div class="question">Let $f$ be a function defined on $R$ (the set of all real numbers) such that $f^{\prime}(x)=2010(x-2009)$ $(x-2010)^2(x-2011)^3(x-2012)^4$, for all $x \in R$. If $g$ is a function defined on $R$ with values in the interval $(0, \infty)$ such that $f(x)=\ln (g(x))$, for all $x \in R$, then the number of points in $R$ at which $g$ has a local maximum is</div>

['Mathematics', 'Application of Derivatives', 'JEE Advanced', 'JEE Advanced 2010 (Paper 2)']

None

<div class="correct-answer"> The correct answer is: <span class="option-value">1</span> </div>

<div class="solution">Let $g(x)=e^{f(x)}, \forall x \in R$<br/>$$<br/>\Rightarrow g^{\prime}(x)=e^{f(x)} \cdot f^{\prime}(x)<br/>$$<br/>$\Rightarrow f^{\prime}(x)$ changes its sign from positive to negative in the neighbourhood of $x=2009$<br/>$\Rightarrow f(x)$ has local maxima at $x=2009$<br/>So, the number of local maximum is one.</div>

MarksBatch2_P1.db

let-f-be-a-nonnegative-function-defined-on-the-interval-0-1-if-0-x-1-f-t-2-d-t-0-x-f-t-d-t-0-x-1-and-f-0-0-then

let-f-be-a-nonnegative-function-defined-on-the-interval-0-1-if-0-x-1-f-t-2-d-t-0-x-f-t-d-t-0-x-1-and-f-0-0-then-72949

<div class="question">Let $f$ be a non-negative function defined on the interval $[0,1]$. If $\int_0^x \sqrt{1-\left\{f^{\prime}(t)\right\}^2} d t=\int_0^x f(t) d t, \quad 0 \leq x \leq 1$<br/>$$<br/>\text { and } f(0)=0 \text {, then }<br/>$$</div>

['Mathematics', 'Definite Integration', 'JEE Advanced', 'JEE Advanced 2009 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$f\left(\frac{1}{2}\right) &lt; \frac{1}{2}$ and $f\left(\frac{1}{3}\right)&gt;\frac{1}{3}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$f\left(\frac{1}{2}\right)&gt;\frac{1}{2}$ and $f\left(\frac{1}{3}\right)&gt;\frac{1}{3}$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$f\left(\frac{1}{2}\right) &lt; \frac{1}{2}$ and $f\left(\frac{1}{3}\right) &lt; \frac{1}{3}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$f\left(\frac{1}{2}\right)&gt;\frac{1}{2}$ and $f\left(\frac{1}{3}\right) &lt; \frac{1}{3}$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$f\left(\frac{1}{2}\right) &lt; \frac{1}{2}$ and $f\left(\frac{1}{3}\right) &lt; \frac{1}{3}$</span> </div>

<div class="solution">Given $\int_0^x \sqrt{1-\left(f^{\prime}(t)\right)^2} d t=\int_0^x f(t) d t$,<br/>$$<br/>0 \leq x \leq 1<br/>$$<br/>Applying Leibnitz theorem, we get<br/>$$<br/>\begin{array}{rlrl} <br/>&amp; &amp; \sqrt{1-\left(f^{\prime}(x)\right)^2} &amp; =f(x) \\<br/>\Rightarrow &amp; &amp; 1-\left(f^{\prime}(x)\right)^2 &amp; =f^2(x) \\<br/>\Rightarrow &amp; &amp; \left(f^{\prime}(x)\right)^2 &amp; =1-f^2(x) \\<br/>\Rightarrow &amp; &amp; f^{\prime}(x) &amp; =\pm \sqrt{1-f^2(x)} \\<br/>\Rightarrow &amp; &amp; \quad \frac{d y}{d x} &amp; =\pm \sqrt{1-y^2} \\<br/>&amp; \text { where } y=f(x) \Rightarrow &amp; \frac{d y}{\sqrt{1-y^2}}=\pm d x<br/>\end{array}<br/>$$<br/>On integrating both sides, we get<br/>$$<br/>\begin{aligned}<br/>&amp; \sin ^{-1}(y)=\pm x+C \\<br/>&amp; \because \quad f(0)=0 \Rightarrow C=0 \Rightarrow y=\pm \sin x \\<br/>&amp; y=\sin x=f(x) \text { given } f(x) \geq 0 \text { for } \\<br/>&amp; x \in[0,1]<br/>\end{aligned}<br/>$$<br/>It is known that $\sin x &lt; x, \forall x \in R^{+}$<br/>$\therefore \quad \sin \left(\frac{1}{2}\right) &lt; \frac{1}{2} \Rightarrow f\left(\frac{1}{2}\right) &lt; \frac{1}{2}$<br/>and $\sin \left(\frac{1}{3}\right) &lt; \frac{1}{3} \Rightarrow f\left(\frac{1}{3}\right) &lt; \frac{1}{3}$</div>

MarksBatch2_P1.db

let-f-be-a-realvalued-differentiable-function-on-r-the-set-of-all-real-numbers-such-that-f-1-1-if-the-y-intercept-of-the-tangent-at-any-point-p-x-y-on

let-f-be-a-realvalued-differentiable-function-on-r-the-set-of-all-real-numbers-such-that-f-1-1-if-the-y-intercept-of-the-tangent-at-any-point-p-x-y-on-54781

<div class="question">Let $f$ be a real-valued differentiable function on $R$ (the set of all real numbers) such that $f(1)=1$. If the $y$-intercept of the tangent at any point $P(x, y)$ on the curve $y=f(x)$ is equal to the cube of the abscissa of $P$, then the value of $f(-3)$ is equal to</div>

['Mathematics', 'Differential Equations', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)']

None

<div class="correct-answer"> The correct answer is: <span class="option-value">9</span> </div>

<div class="solution">The equation of the tangent at $(x, y)$ to the given curve $y=f(x)$ is<br/>$$<br/>\begin{gathered}<br/>Y-y=\frac{d y}{d x}(X-x) \\<br/>Y \text {-intercept }=y-x \frac{d y}{d x}<br/>\end{gathered}<br/>$$<br/>According to the question<br/>$$<br/>x^3=y-x \frac{d y}{d x}<br/>$$<br/><br/>$$<br/>\Rightarrow \quad \frac{d y}{d x}-\frac{y}{x}=-x^2<br/>$$<br/>which is linear in $x$.<br/>$$<br/>\text { IF }=e^{\int \frac{-1}{x} d x}=\frac{1}{x}<br/>$$<br/>$\therefore$ Required solution is<br/>$$<br/>\begin{aligned}<br/>&amp; y \cdot \frac{1}{x}=\int-x^2 \cdot \frac{1}{x} d x \\<br/>&amp; \Rightarrow \quad \frac{y}{x}=\frac{-x^2}{2}+c \\<br/>&amp; \Rightarrow \quad y=\frac{-x^3}{2}+c x \\<br/>&amp; \text { At } x=1, y=1 \text {, } \\<br/>&amp; 1=\frac{-1}{2}+c \\<br/>&amp; \Rightarrow \quad c=\frac{3}{2} \\<br/>&amp;<br/>\end{aligned}<br/>$$<br/>Now, $\begin{aligned} f(-3) &amp; =\frac{27}{2}+\frac{3}{2}(-3) \\ &amp; =\frac{27-9}{2}=9\end{aligned}$</div>

MarksBatch2_P1.db

let-f-be-a-realvalued-function-defined-on-the-interval-0-by-f-x-ln-x-0-x-1-sin-t-d-t-then-which-of-the-following-statements-is-are-true

let-f-be-a-realvalued-function-defined-on-the-interval-0-by-f-x-ln-x-0-x-1-sin-t-d-t-then-which-of-the-following-statements-is-are-true-89713

<div class="question">Let $f$ be a real-valued function defined on the interval $(0, \infty)$, by $f(x)=\ln x+\int_0^x \sqrt{1+\sin t} d t$. Then which of the following statement(s) is (are) true ?</div>

['Mathematics', 'Continuity and Differentiability', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$f^{\prime \prime}(x)$ exists for all $x \in(0, \infty)$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$f^{\prime}(x)$ exists for all $x \in(0, \infty)$ and $f^{\prime}$ is continuous on $(0, \infty)$, but not differentiable on $(0, \infty)$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>there exists $\alpha&gt;1$ such that $\left|f^{\prime}(x)\right| &lt; |f(x)|$ for all $x \in(\alpha, \infty)$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>there exists $\beta&gt;0$ such that $|f(x)|+\left|f^{\prime}(x)\right| \leq \beta \quad$ from all $x \in(0, \infty)$</span> </li> </ul>

<div class="correct-answer"> The correct answers are: <span class="option-value"><br/>$f^{\prime}(x)$ exists for all $x \in(0, \infty)$ and $f^{\prime}$ is continuous on $(0, \infty)$, but not differentiable on $(0, \infty)$<br/>, <br/>there exists $\alpha&gt;1$ such that $\left|f^{\prime}(x)\right| &lt; |f(x)|$ for all $x \in(\alpha, \infty)$</span> </div>

<div class="solution">Here, $f^{\prime}(x)=\frac{1}{x}+\sqrt{1+\sin x}, x&gt;0$ but $f(x)$ is not differentiable in $(0, \infty)$ as $\sin x$ may be $-1$ and then $f^{\prime \prime}(x)=-\frac{1}{x^2}+\frac{\cos x}{2 \sqrt{1+\sin x}}$ will not exists.<br/>$\Rightarrow f^{\prime}(x)$ is continuous for all $x \in(0, \infty)$ but $f^{\prime}(x)$ is not differentiable on $(0, \infty)$.<br/>$\therefore$ Option (b) is true.<br/>Also, $\quad f^{\prime}(x) \leq 3$, if $x&gt;1$ and $\quad f(x)&gt;3$, if $x&gt;e^3$<br/>$\therefore$ Let $\alpha=e^3$<br/>$\Rightarrow$ Option (c) is true.<br/>(d) is not possible as $f(x) \rightarrow \infty$ when $x \rightarrow \infty$.<br/>Hence, (b, c) is the correct option.</div>

MarksBatch2_P1.db

let-f-be-a-realvalued-function-defined-on-the-interval-1-1-such-that-e-x-f-x-2-0-x-t-4-1-d-t-for-all-x-1-1-and-let-f-1-be-the-inverse-function-of-f-th

let-f-be-a-realvalued-function-defined-on-the-interval-1-1-such-that-e-x-f-x-2-0-x-t-4-1-d-t-for-all-x-1-1-and-let-f-1-be-the-inverse-function-of-f-th-66400

<div class="question">Let $f$ be a real-valued function defined on the interval $(-1,1)$ such that $e^{-x} f(x)=2+\int_0^x \sqrt{t^4+1} d t, \quad$ for all $x \in(-1,1)$ and let $f^{-1}$ be the inverse function of $f$. Then $\left(f^{-1}\right)^{\prime}(2)$ is equal to</div>

['Mathematics', 'Differentiation', 'JEE Advanced', 'JEE Advanced 2010 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>1<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{1}{3}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{1}{2}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{1}{e}$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$\frac{1}{3}$<br/></span> </div>

<div class="solution">We have, $e^{-x} f(x)=2+\int_0^x \sqrt{t^4+1} d t x \in(-1,1)$<br/>On differentiating w.r.t. $x$, we get<br/>$$<br/>\begin{array}{ll} <br/>&amp; e^{-x}\left(f^{\prime}(x)-f(x)\right)=\sqrt{x^4+1} \\<br/>\Rightarrow \quad &amp; f^{\prime}(x)=f(x)+\sqrt{x^4+1} e^x \\<br/>\because &amp; f^{-1} \text { is the inverse of } f \\<br/>\therefore &amp; f^{-1}(f(x))=x \\<br/>\Rightarrow &amp; f^{-1^{\prime}}(f(x)) f^{\prime}(x)=1 \\<br/>\Rightarrow &amp; f^{-1^{\prime}}(f(x))=\frac{1}{f^{\prime}(x)} \\<br/>\Rightarrow \quad &amp; f^{-1^{\prime}}(f(x))=\frac{1}{f(x)+\sqrt{x^4+1} e^x} \\<br/>\text { At } \quad &amp; x=0, f(x)=2 \\<br/>&amp; f^{-1^{\prime}}(2)=\frac{1}{2+1}=\frac{1}{3}<br/>\end{array}<br/>$$</div>

MarksBatch2_P1.db

let-f-g-and-h-be-realvalued-functions-defined-on-the-interval-0-1-by-f-x-e-x-2-e-x-2-g-x-x-e-x-2-e-x-2-and-h-x-x-2-e-x-2-e-x-2-if-a-b-and-c-denote-res

let-f-g-and-h-be-realvalued-functions-defined-on-the-interval-0-1-by-f-x-e-x-2-e-x-2-g-x-x-e-x-2-e-x-2-and-h-x-x-2-e-x-2-e-x-2-if-a-b-and-c-denote-res-21179

<div class="question">Let $f, g$ and $h$ be real-valued functions defined on the interval $[0,1]$ by $f(x)=e^{x^2}+e^{-x^2}, \quad g(x)=x e^{x^2}+e^{-x^2}$ and $h(x)=x^2 e^{x^2}+e^{-x^2}$. If $a, b$ and $c$ denote respectively, the absolute maximum of $f, g$ and $h$ on $[0,1]$, then</div>

['Mathematics', 'Application of Derivatives', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$a=b$ and $c \neq b$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$a=c$ and $a \neq b$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$a \neq b$ and $c \neq b$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$a=b=c$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$a=b=c$</span> </div>

<div class="solution">Given function,<br/>$$<br/>\begin{aligned}<br/>&amp; f(x)=e^{x^2}+e^{-x^2}, \\<br/>&amp; g(x)=x e^{x^2}+e^{-x^2} \text { and } \\<br/>&amp; h(x)=x^2 e^{x^2}+e^{-x^2} \text { are } \quad \text { strictly }<br/>\end{aligned}<br/>$$<br/>increasing on $[0,1]$. Hence, at $x=1$, the given function attains absolute maximum all equal to $e+\frac{1}{e}$.<br/>$$<br/>\Rightarrow \quad a=b=c<br/>$$</div>

MarksBatch2_P1.db

let-f-i-r-i-r-be-defined-as-f-x-x-x-2-1-the-total-number-of-points-at-which-f-attains-either-a-local-maximum-or-a-local-minimum-is

let-f-i-r-i-r-be-defined-as-f-x-x-x-2-1-the-total-number-of-points-at-which-f-attains-either-a-local-maximum-or-a-local-minimum-is-74307

<div class="question">Let $f: I R \rightarrow I R$ be defined as $f(x)=|x|+\left|x^{2}-1\right|$. The total number of points at which $f$ attains either a local maximum or a local minimum is</div>

['Mathematics', 'Application of Derivatives', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)']

None

<div class="correct-answer"> The correct answer is: <span class="option-value">5</span> </div>

<div class="solution">$f(x)=|x|+\left|x^{2}-1\right|=\left\{\begin{array}{c}-x+x^{2}-1, x &lt; -1 \\ -x-x^{2}+1,-1 \leq x \leq 0 \\ x-x^{2}+1,0 &lt; x &lt; 1 \\ x^{2}+x-1, \quad x \geq 1\end{array}\right.$ <br/> <br/>$\therefore \quad f^{\prime}(x)=\left[\begin{array}{ccc} <br/> <br/>2 x-1 &amp; , &amp; x &lt; -1 \\ <br/> <br/>-2 x-1 &amp; , &amp; -1 \leq x \leq 0 \\ <br/> <br/>-2 x+1 &amp; , &amp; 0 &lt; x &lt; 1 \\ <br/> <br/>2 x+1 &amp; , &amp; x&gt;1 <br/> <br/>\end{array}\right.$ <br/> <br/>Critical points are $\frac{1}{2}, \frac{-1}{2},-1,0$ and 1 . <br/> <br/>We observe at five points $f^{\prime}(x)$ changes its sign $\therefore$ There are 5 points at which either local maximum or local minimum.</div>

MarksBatch2_P1.db

let-f-r-r-be-a-continuous-function-which-satisfies-f-x-0-x-f-t-d-t-then-the-value-of-f-ln-5-is

let-f-r-r-be-a-continuous-function-which-satisfies-f-x-0-x-f-t-d-t-then-the-value-of-f-ln-5-is-54869

<div class="question">Let $f: R \rightarrow R$ be a continuous function, which satisfies $f(x)=\int_0^x f(t) d t$. Then, the value of $f(\ln 5)$ is</div>

['Mathematics', 'Definite Integration', 'JEE Advanced', 'JEE Advanced 2009 (Paper 2)']

None

<div class="correct-answer"> The correct answer is: <span class="option-value">0</span> </div>

<div class="solution">From given integral equation, $f(0)=0$. Also, differentiating the given integral equation w.r.t. $x$, we get<br/>$$<br/>\begin{gathered}<br/>f^{\prime}(x)=f(x) \\<br/>\text { If } \quad f(x) \neq 0 \Rightarrow \frac{f^{\prime}(x)}{f(x)}=1 \\<br/>\Rightarrow \quad \log f(x)=x+C \Rightarrow f(x)=e^C e^x \\<br/>\because f(0)=0 \Rightarrow e^C=0, \text { a contradiction } \\<br/>\therefore f(x)=0, \forall x \in R \Rightarrow f(\ln 5)=0<br/>\end{gathered}<br/>$$</div>

MarksBatch2_P1.db

let-f-r-r-be-a-function-such-that-f-x-y-f-x-f-y-x-y-r-if-f-x-is-differentiable-at-x-0-then

let-f-r-r-be-a-function-such-that-f-x-y-f-x-f-y-x-y-r-if-f-x-is-differentiable-at-x-0-then-45424

<div class="question">Let $f: R \rightarrow R$ be a function such that $f(x+y)=f(x)+f(y), \forall x, y \in R$. If $f(x)$ is differentiable at $x=0$, then</div>

['Mathematics', 'Continuity and Differentiability', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$f(x)$ is differentiable only in a finite interval containing zero<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$f(x)$ is continuous, $\forall x \in R$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$f^{\prime}(x)$ is constant, $\forall x \in R$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$f(x)$ is differentiable except at finitely many points</span> </li> </ul>

<div class="correct-answer"> The correct answers are: <span class="option-value"><br/>$f(x)$ is continuous, $\forall x \in R$<br/>, <br/>$f^{\prime}(x)$ is constant, $\forall x \in R$<br/></span> </div>

<div class="solution">$f(x+y)=f(x)+f(y)$, as $f(x)$ is differentiable at $x=0$.<br/>$$<br/>\begin{aligned}<br/>\Rightarrow &amp; f^{\prime}(0)=k \\<br/>\text { Now, } f^{\prime}(x) &amp; =\lim _{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} \\<br/>&amp; =\lim _{h \rightarrow 0} \frac{f(x)+f(h)-f(x)}{h} \\<br/>&amp; =\lim _{h \rightarrow 0} \frac{f(h)}{h} \quad\left[\frac{0}{0} \text { from }\right]<br/>\end{aligned}<br/>$$<br/>Given, $f(x+y)=f(x)+f(y), \forall x, y$<br/>$$<br/>\begin{aligned}<br/>&amp; \therefore &amp; f(0) &amp; =f(0)+f(0), \\<br/>&amp; \text { when } &amp; x &amp; =y=0 \Rightarrow f(0)=0<br/>\end{aligned}<br/>$$<br/>Using L'Hospital's rule,<br/>$$<br/>\lim _{h \rightarrow 0} \frac{f^{\prime}(h)}{1}=f^{\prime}(0)=k<br/>$$<br/>$\Rightarrow f^{\prime}(x)=k$, on integrating both sides, $f(x)=k x+C$, as $f(0)=0 \Rightarrow C=0$<br/>So, $f(x)=k x$<br/>$\therefore f(x)$ is continuous for all $x \in R$ and $f^{\prime}(x)=k$, i.e. constant for all $x \in R$.<br/>Hence, both (b) and (c) are correct.<br/>Solutions (Q. Nos. 12-13)<br/>$$<br/>\left.\left(\begin{array}{l}<br/>3 \mathrm{~W} \\<br/>2 \mathrm{R}<br/>\end{array}\right)(\underbrace{1 \mathrm{~W}}_{u_1})\right\} \text { Initial }<br/>$$<br/>Head appears.<br/>$\left(\begin{array}{l}2 \mathrm{~W} \\ 2 \mathrm{R}\end{array}\right) \underset{u_1}{\mathrm{wW}}(\underbrace{2 \mathrm{~W}}_{u_2})$<br/>or $\left.\left(\begin{array}{l}3 \mathrm{~W} \\ 1 \mathrm{R}\end{array}\right) \underset{u_1}{\stackrel{1 \mathrm{R}}{u_1}}\left(\begin{array}{c}1 \mathrm{~W} \\ 1 \mathrm{R}\end{array}\right)\right)_{u_2}^{u_2}$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/GBdGyDD_0UwU3UqT52VW0IalBzIUtwdpq7GUfZVxGHE.original.fullsize.png"/><br/></div>

MarksBatch2_P1.db

let-f-sin-tan-1-c-o-s-2-s-i-n-where-4-4-then-the-value-of-d-t-a-n-d-f-is

let-f-sin-tan-1-c-o-s-2-s-i-n-where-4-4-then-the-value-of-d-t-a-n-d-f-is-70148

<div class="question">Let $f(\theta)=\sin \left[\tan ^{-1}\left(\frac{\sin \theta}{\sqrt{\cos 2 \theta}}\right)\right]$, where $-\frac{\pi}{4} &lt; \theta &lt; \frac{\pi}{4}$. Then, the value of $\frac{d}{d(\tan \theta)}(f(\theta))$ is</div>

['Mathematics', 'Inverse Trigonometric Functions', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)']

None

<div class="correct-answer"> The correct answer is: <span class="option-value">1</span> </div>

<div class="solution">$f(\theta)=\sin \left(\tan ^{-1} \frac{\sin \theta}{\sqrt{\cos 2 \theta}}\right),-\frac{\pi}{4} &lt; \theta &lt; \frac{\pi}{4}$<br/>Let $\tan ^{-1} \frac{\sin \theta}{\sqrt{\cos 2 \theta}}=\phi$<br/>$$<br/>\Rightarrow \quad \tan \phi=\frac{\sin \theta}{\sqrt{\cos 2 \theta}}<br/>$$<img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/bbRBtOL8gHVb5FgMH9HblcPKawLYnY879QtTMljq-9M.original.fullsize.png"/><br/><br/>$$<br/>\begin{aligned}<br/>\therefore \quad \sin \phi &amp; =\frac{\sin \theta}{\sqrt{\sin ^2 \theta+\cos 2 \theta}} \\<br/>&amp; =\frac{\sin \theta}{\sqrt{1-\sin ^2 \theta}}=\frac{\sin \theta}{\cos \theta}=\tan \theta<br/>\end{aligned}<br/>$$<br/><br/>$$<br/>\begin{array}{ll}<br/>\therefore &amp; f(\theta)=\sin \phi=\tan \theta \\<br/>\Rightarrow &amp; \frac{d f(\theta)}{d(\tan \theta)}=1<br/>\end{array}<br/>$$</div>

MarksBatch2_P1.db

let-f-x-1-x-2-sin-2-x-x-2-for-all-x-i-r-and-let-g-x-1-x-t-1-2-t-1-ln-t-f-t-d-t-for-all-x-1-question-consider-the-statements-p-there-exists-some-x-r-su

let-f-x-1-x-2-sin-2-x-x-2-for-all-x-i-r-and-let-g-x-1-x-t-1-2-t-1-ln-t-f-t-d-t-for-all-x-1-question-consider-the-statements-p-there-exists-some-x-r-su-55128

<div class="question">Let $f(x)=(1-x)^{2} \sin ^{2} x+x^{2}$ for all $x \in I R$ and let $g(x)=\int_{1}^{x}\left(\frac{2(t-1)}{t+1}-\ln t\right) f(t) d t$ for all $x \in(1, \infty)$. <br/> <br/><strong> Question:</strong> Consider the statements: <br/> <br/>$P$ : There exists some $x \in \mathrm{R}$ such that $f(x)+2 x$ <br/> <br/>$=2\left(1+x^{2}\right)$ <br/> <br/>$Q$ : There exists some $x \in \mathrm{R}$ such that $2 f(x)+1$ $=2 x(1+x)$ <br/> <br/>Then</div>

['Mathematics', 'Functions', 'JEE Advanced', 'JEE Advanced 2012 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">both $P$ and $Q$ are true</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$P$ is true and $Q$ is false</span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data">$P$ is false and $Q$ is true</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data">both $P$ and $Q$ are false</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value">$P$ is false and $Q$ is true</span> </div>

<div class="solution">For the statement $P, f(x)+2 x=2\left(1+x^{2}\right)$ <br/> <br/>$\Rightarrow(1-x)^{2} \sin ^{2} x+x^{2}+2 x=2\left(1+x^{2}\right)$ <br/> <br/>$\Rightarrow(1-x)^{2} \sin ^{2} x=x^{2}-2 x+1+1$ <br/> <br/>$\Rightarrow(1-x)^{2} \sin ^{2} x=(1-x)^{2}+1$ <br/> <br/>$\Rightarrow(1-x)^{2} \cos ^{2} x=-1$, which is not possible for any real value of $x$. <br/> <br/>Hence $P$ is not true. <br/> <br/>Let $H(x)=2 f(x)+1-2 x(1+x)$ <br/> <br/>$H(0)=2 f(0)+1-0=1$ <br/> <br/>and $H(1)=2 f(1)+1-4=-3$ <br/> <br/>Hence, $H(x)$ has a solution in $(0,1)$ <br/> <br/>Therefore, $Q$ is true.</div>

MarksBatch2_P1.db

let-f-x-1-x-2-sin-2-x-x-2-for-all-x-i-r-and-let-g-x-1-x-t-1-2-t-1-ln-t-f-t-d-t-for-all-x-1-question-which-of-the-following-is-true

let-f-x-1-x-2-sin-2-x-x-2-for-all-x-i-r-and-let-g-x-1-x-t-1-2-t-1-ln-t-f-t-d-t-for-all-x-1-question-which-of-the-following-is-true-48953

<div class="question">Let $f(x)=(1-x)^{2} \sin ^{2} x+x^{2}$ for all $x \in I R$ and let $g(x)=\int_{1}^{x}\left(\frac{2(t-1)}{t+1}-\ln t\right) f(t) d t$ for all $x \in(1, \infty)$. <br/> <br/><strong> Question:</strong> Which of the following is true?</div>

['Mathematics', 'Application of Derivatives', 'JEE Advanced', 'JEE Advanced 2012 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$g$ is increasing on $(1, \infty)$</span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data">$g$ is decreasing on $(1, \infty)$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$g$ is increasing on $(1,2)$ and decreasing on $(2, \infty)$</span> </li><li class=""> <span class="option-label">D</span> <span class="option-data">$g$ is decreasing on $(1,2)$ and increasing on $(2, \infty)$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value">$g$ is decreasing on $(1, \infty)$</span> </div>

<div class="solution">$g(x)=\int_{1}^{x}\left(\frac{2(t-1)}{t+1}-\ln t\right) f(t) d t$, $\therefore \quad g^{\prime}(x)=\left[\frac{2(x-1)}{x+1}-\ln x\right] f(x)$ <br/> <br/>Here $f(x)&gt;0, \forall x \in(1, \infty)$ <br/> <br/>Let $h(x)=\frac{2(x-1)}{x+1}-\ln x$ <br/> <br/>$\therefore \quad h^{\prime}(x)=\frac{4}{(x+1)^{2}}-\frac{1}{x}=\frac{-(x-1)^{2}}{(x+1)^{2} x} &lt; 0, x \in(1, \infty)$ <br/> <br/>$\Rightarrow h(x)$ is decreasing function. <br/> <br/>Hence, for $x&gt;1, h(x) &lt; h(1) \Rightarrow h(x) &lt; 0 \forall x&gt;1$ <br/> <br/>$\Rightarrow g^{\prime}(x) &lt; 0 \forall x \in(1, \infty)$ <br/> <br/>Therefore, $g(x)$ is decreasing on $(1, \infty)$.</div>

MarksBatch2_P1.db

let-f-x-1-x-n-1-n-x-for-n-2-and-g-x-f-occurs-n-times-f-f-o-o-f-x-then-x-n-2-g-x-d-x-equals

let-f-x-1-x-n-1-n-x-for-n-2-and-g-x-f-occurs-n-times-f-f-o-o-f-x-then-x-n-2-g-x-d-x-equals-83727

<div class="question">Let $f(x)=\frac{x}{\left(1+x^n\right)^{1 / n}}$ for $n \geq 2$ and $g(x)=\underbrace{(f \circ f o \ldots o f)}_{f \text { occurs } n \text { times }}(x)$. Then, $\int x^{n-2} g(x) d x$ equals</div>

['Mathematics', 'Indefinite Integration', 'JEE Advanced', 'JEE Advanced 2007 (Paper 2)']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{1}{n(n-1)}\left(1+n x^n\right)^{1-\frac{1}{n}}+k$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{1}{n-1}\left(1+n x^n\right)^{1-\frac{1}{n}}+k$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{1}{n(n+1)}\left(1+n x^n\right)^{1+\frac{1}{n}}+k$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{1}{n+1}\left(1+n x^n\right)^{1+\frac{1}{n}}+k$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$\frac{1}{n(n-1)}\left(1+n x^n\right)^{1-\frac{1}{n}}+k$<br/></span> </div>

<div class="solution">Here, $\begin{array}{rlrl} &amp; \text { and } &amp; f f(x) &amp; =\frac{f(x)}{\left[1+f(x)^n\right]^{1 / n}}=\frac{x}{\left(1+2 x^n\right)^{1 / n}} \\ &amp; \therefore \quad f f f(x) &amp; =\frac{x}{\left(1+3 x^n\right)^{1 / n}} \\ &amp; \text { Let } &amp; g(x) &amp; =\underbrace{(f \circ f o \ldots o f)}_{n \text { times }}(x)=\frac{x}{\left(1+n x^n\right)^{1 / n}} \\ I &amp; =\int x^{n-2} g(x) d x=\int \frac{x^{n-1} d x}{\left(1+n x^n\right)^{1 / n}} \\ &amp; &amp; =\frac{1}{n^2} \int \frac{n^2 x^{n-1} d x}{\left(1+n x^n\right)^{1 / n}}=\frac{1}{n^2} \int \frac{\frac{d}{d x}\left(1+n x^n\right)}{\left(1+n x^n\right)^{1 / n}} d x\end{array}$<br/>$\therefore \quad I=\frac{1}{n(n-1)}\left(1+n x^n\right)^{1-\frac{1}{n}}+k$.</div>

MarksBatch2_P1.db

let-f-x-2-cos-x-for-all-real-x-statement-i-for-each-real-t-there-exists-a-point-c-in-t-t-such-that-f-c-0-statement-ii-f-t-f-t-2-for-each-real-t

let-f-x-2-cos-x-for-all-real-x-statement-i-for-each-real-t-there-exists-a-point-c-in-t-t-such-that-f-c-0-statement-ii-f-t-f-t-2-for-each-real-t-67226

<div class="question">Let $f(x)=2+\cos x$ for all real $x$.<br/>Statement I For each real $t$, there exists a point $c$ in $[t, t+\pi]$ such that $f^{\prime}(c)=0$. <br/>Statement II $f(t)=f(t+2 \pi)$ for each real $t$.</div>

['Mathematics', 'Application of Derivatives', 'JEE Advanced', 'JEE Advanced 2007 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>Statement I is True, Statement II is true; Statement II is not a correct explanation for Statement I<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>Statement I is true, Statement II is false<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>Statement I is false, Statement II is true</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>Statement I is True, Statement II is true; Statement II is not a correct explanation for Statement I<br/></span> </div>

<div class="solution">$f(x)=2+\cos x \forall x \in R$<br/>Statement I There exists a point $c \in[t, t+\pi]$, where $f^{\prime}(c)=0$ Hence, Statement 1 is true.<br/>Statement II $f(t)=f(t+2 \pi)$ is true.<br/>But Statement II is not a correct explanation for Statement I.</div>

MarksBatch2_P1.db

let-f-x-be-a-nonconstant-twice-differentiable-function-defined-on-such-that-f-x-f-1-x-and-f-4-1-0-then

let-f-x-be-a-nonconstant-twice-differentiable-function-defined-on-such-that-f-x-f-1-x-and-f-4-1-0-then-78446

<div class="question">Let $f(x)$ be a non-constant twice differentiable function defined on $(-\infty, \infty)$, such that $f(x)=f(1-x)$ and $f^{\prime}\left(\frac{1}{4}\right)=0$. Then,</div>

['Mathematics', 'Application of Derivatives', 'JEE Advanced', 'JEE Advanced 2008 (Paper 1)']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$f^{\prime}(x)$ vanishes atleast twice on $[0,1]$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$f^{\prime}\left(\frac{1}{2}\right)=0$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$\int_{\frac{1}{2}}^{\frac{1}{2}} f\left(x+\frac{1}{2}\right) \sin x d x=0$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$\int_0^{\frac{1}{2}} f(t) e^{\sin \pi t} d t=\int_{\frac{1}{2}}^1 f(1-t) e^{\sin \pi t} d t$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

<div class="correct-answer"> The correct answers are: <span class="option-value"><br/>$f^{\prime}(x)$ vanishes atleast twice on $[0,1]$<br/>, <br/>$f^{\prime}\left(\frac{1}{2}\right)=0$<br/>, <br/>$\int_{\frac{1}{2}}^{\frac{1}{2}} f\left(x+\frac{1}{2}\right) \sin x d x=0$<br/>, <br/>$\int_0^{\frac{1}{2}} f(t) e^{\sin \pi t} d t=\int_{\frac{1}{2}}^1 f(1-t) e^{\sin \pi t} d t$</span> </div>

<div class="solution">Given that, $f(x)=f(1-x)$<br/>On differentiating w.r.t. $x$, we get<br/>$$<br/>f^{\prime}(x)=-f^{\prime}(1-x)<br/>$$<br/>Let us put $\quad x=\frac{1}{2}$<br/>$$<br/>\Rightarrow \quad 2 f^{\prime}\left(\frac{1}{2}\right)=0 \Rightarrow f^{\prime}\left(\frac{1}{2}\right)=0<br/>$$<br/>Since, $f^{\prime}\left(\frac{1}{2}\right)=0$ and $f^{\prime}\left(\frac{1}{4}\right)=0$<br/>$\Rightarrow f^{\prime \prime}(x)=0$ at two points in $[0,1]$.<br/>Now, $\int_{-1 / 2}^{1 / 2} f\left(x+\frac{1}{2}\right) \sin x d x=0$<br/>As, $f\left(x+\frac{1}{2}\right) \sin x$ is an odd function which is clear from the following explanation.<br/>Let $g(x)=f\left(x+\frac{1}{2}\right) \sin x$,<br/>$$<br/>g(-x)=f\left(\frac{1}{2}-x\right) \sin (-x)=-\sin x f\left(1-\left(\frac{1}{2}-x\right)\right)=-\sin x f\left(\frac{1}{2}+x\right)=-g(x)<br/>$$<br/>Moreover, $\int_{1 / 2}^1 f(1-t) e^{\sin (\pi t)} d t=\int_0^{1 / 2} f(u) \cdot e^{\sin \pi u} d u$ where, $1-t=u$.</div>

MarksBatch2_P1.db

let-f-x-be-an-indefinite-integral-of-sin-2-x-statement-i-the-function-f-x-satisfies-f-x-f-x-for-all-real-x-statement-ii-sin-2-x-sin-2-x-for-all-real-x

let-f-x-be-an-indefinite-integral-of-sin-2-x-statement-i-the-function-f-x-satisfies-f-x-f-x-for-all-real-x-statement-ii-sin-2-x-sin-2-x-for-all-real-x-64347

<div class="question">Let $F(x)$ be an indefinite integral of $\sin ^2 x$.<br/>Statement I The function $F(x)$ satisfies $F(x+\pi)=F(x)$ for all real $x$.<br/>Statement II $\sin ^2(x+\pi)=\sin ^2 x$ for all real $x$.</div>

['Mathematics', 'Indefinite Integration', 'JEE Advanced', 'JEE Advanced 2007 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I.<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>Statement I is true, Statement II is false<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>Statement I is false, Statement II is true</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>Statement I is false, Statement II is true</span> </div>

<div class="solution">$$<br/>\begin{aligned}<br/>&amp; F(x)=\int \sin ^2 x d x=\int \frac{1-\cos 2 x}{2} d x \\<br/>&amp; \Rightarrow F(x)=\frac{1}{4}(2 x-\sin 2 x)+c<br/>\end{aligned}<br/>$$<br/>Since, $F(x+\pi) \neq F(x)$.<br/>Hence, Statement I is false.<br/>But Statement II is true as $\sin ^2 x$ is periodic with period $\pi$.</div>

MarksBatch2_P1.db

let-f-x-be-differentiable-on-the-interval-0-such-that-f-1-1-and-lim-t-x-t-x-t-2-f-x-x-2-f-t-1-for-each-x-0-then-f-x-is

let-f-x-be-differentiable-on-the-interval-0-such-that-f-1-1-and-lim-t-x-t-x-t-2-f-x-x-2-f-t-1-for-each-x-0-then-f-x-is-83213

<div class="question">Let $f(x)$ be differentiable on the interval $(0, \infty)$ such that $f(1)=1$, and $\lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$ for each $x&gt;0$. Then, $f(x)$ is</div>

['Mathematics', 'Limits', 'JEE Advanced', 'JEE Advanced 2007 (Paper 1)']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{1}{3 x}+\frac{2 x^2}{3}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$-\frac{1}{3 x}+\frac{4 x^2}{3}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$-\frac{1}{x}+\frac{2}{x^2}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{1}{x}$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$\frac{1}{3 x}+\frac{2 x^2}{3}$<br/></span> </div>

<div class="solution">$$<br/>\text { } \lim _{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1<br/>$$<br/><br/>$$<br/>\begin{aligned}<br/>&amp; \Rightarrow &amp; x^2 f^{\prime}(x)-2 x f(x)+1 &amp; =0 \\<br/>&amp; \Rightarrow &amp; \frac{x^2 f^{\prime}(x)-2 x f(x)}{\left(x^2\right)^2}+\frac{1}{x^4} &amp; =0 \\<br/>&amp; \Rightarrow &amp; \frac{d}{d x}\left(\frac{f(x)}{x^2}\right) &amp; =-\frac{1}{x^4} \\<br/>\Rightarrow &amp; &amp; f(x) &amp; =c x^2+\frac{1}{3 x} \text { also } f(1)=1 \Rightarrow c=\frac{2}{3} . \\<br/>&amp; \text { Hence, } &amp; f(x) &amp; =\frac{2}{3} x^2+\frac{1}{3 x}<br/>\end{aligned}<br/>$$</div>

MarksBatch2_P1.db

let-f-x-x-2-5-x-6-x-2-6-x-5-1

let-f-x-x-2-5-x-6-x-2-6-x-5-1-60880

<div class="question">Let $f(x)=\frac{x^2-6 x+5}{x^2-5 x+6}$.<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/5LpElX78gcLAqJVL43i4iELoQ4DU2RBjX9Jh8inUrTA.original.fullsize.png"/><br/></div>

['Mathematics', 'Application of Derivatives', 'JEE Main']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>A-p; B-q; C-q; D-p<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>A-r; B-s; C-q; D-r<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>A-s; B-p; C-q; D-p<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>A-r; B-q; C-p; D-r</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>A-p; B-q; C-q; D-p<br/></span> </div>

<div class="solution">$$<br/>f(x)=\frac{(x-1)(x-5)}{(x-2)(x-3)}<br/>$$<br/>The graph of $f(x)$ is shown<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/wvp0nANowQRUUqEKwDE04wK9eFTqRImRXTt4B2hjHOM.original.fullsize.png"/><br/><br/>(A) If $-1 &lt; x &lt; 1 \Rightarrow 0 &lt; f(x) &lt; 1$<br/>(B) If $1 &lt; x &lt; 2 \Rightarrow f(x) &lt; 0$<br/>(C) If $3 &lt; x &lt; 5 \Rightarrow f(x) &lt; 0$<br/>(D) If $x&gt;5 \Rightarrow 0 &lt; f(x) &lt; 1$</div>

MarksBatch2_P1.db

let-f-x-x-2-5-x-6-x-2-6-x-5

let-f-x-x-2-5-x-6-x-2-6-x-5-70664

<div class="question">Let $f(x)=\frac{x^2-6 x+5}{x^2-5 x+6}$.<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/5LpElX78gcLAqJVL43i4iELoQ4DU2RBjX9Jh8inUrTA.original.fullsize.png"/><br/></div>

['Mathematics', 'Application of Derivatives', 'JEE Advanced', 'JEE Advanced 2007 (Paper 2)']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>A-p; B-q; C-q; D-p<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>A-r; B-s; C-q; D-r<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>A-s; B-p; C-q; D-p<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>A-r; B-q; C-p; D-r</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>A-p; B-q; C-q; D-p<br/></span> </div>

<div class="solution">$$<br/>f(x)=\frac{(x-1)(x-5)}{(x-2)(x-3)}<br/>$$<br/>The graph of $f(x)$ is shown<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/wvp0nANowQRUUqEKwDE04wK9eFTqRImRXTt4B2hjHOM.original.fullsize.png"/><br/><br/>(A) If $-1 &lt; x &lt; 1 \Rightarrow 0 &lt; f(x) &lt; 1$<br/>(B) If $1 &lt; x &lt; 2 \Rightarrow f(x) &lt; 0$<br/>(C) If $3 &lt; x &lt; 5 \Rightarrow f(x) &lt; 0$<br/>(D) If $x&gt;5 \Rightarrow 0 &lt; f(x) &lt; 1$</div>

MarksBatch2_P1.db

let-f-x-x-2-and-g-x-sin-x-for-all-x-r-then-the-set-of-all-x-satisfying-fogogof-x-gogof-x-where-f-g-x-f-g-x-is

let-f-x-x-2-and-g-x-sin-x-for-all-x-r-then-the-set-of-all-x-satisfying-fogogof-x-gogof-x-where-f-g-x-f-g-x-is-40956

<div class="question">Let $f(x)=x^2$ and $g(x)=\sin x$ for all $x \in R$. Then, the set of all $x$ satisfying $($ fogogof $)(x)=(\operatorname{gogof})(x)$, where $(f \circ g)(x)=f(g(x))$ is</div>

['Mathematics', 'Trigonometric Equations', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\pm \sqrt{n \pi}, n \in\{0,1,2, \ldots\}$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$\pm \sqrt{n \pi}, n \in\{1,2, \ldots\}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{\pi}{2}+2 n \pi, n \in\{\ldots,-2,-1,0,1,2, \ldots\}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$2 n \pi, n \in\{\ldots,-2,-1,0,1,2, \ldots\}$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$\pm \sqrt{n \pi}, n \in\{1,2, \ldots\}$<br/></span> </div>

<div class="solution">$f(x)=x^2, g(x)=\sin x$<br/>$(g \circ f)(x)=\sin x^2$<br/>go(gof) $(x)=\sin \left(\sin x^2\right)$<br/>$(f \circ g \circ g \circ f)(x)=\left(\sin \left(\sin x^2\right)\right)^2$<br/>Again, $(g \circ f)(x)=\sin x^2$<br/>(gogof) $(x)=\sin \left(\sin x^2\right)$<br/>Given, $($ fogogof $)(x)=($ gogof $)(x)$<br/>$\Rightarrow \quad\left(\sin \left(\sin x^2\right)\right)^2=\sin \left(\sin x^2\right)$<br/>$\Rightarrow \sin \left(\sin x^2\right)\left\{\sin \left(\sin x^2\right)-1\right\}=0$<br/>$\Rightarrow \sin \left(\sin x^2\right)=0$ or $\sin \left(\sin x^2\right)=1$<br/>$\Rightarrow \sin x^2=0$ or $\sin x^2=\frac{\pi}{2}$<br/>$\therefore \quad x^2=n \pi$<br/>(i.e. not possible as $-1 \leq \sin \theta \leq 1$ )</div>

MarksBatch2_P1.db

let-g-x-l-o-g-c-o-s-m-x-1-x-1-n-0-x-2-m-and-n-are-integers-m-0-n-0-and-let-p-be-the-left-hand-derivative-of-x-1-at-x-1-if-lim-x-1-g-x-p-then

let-g-x-l-o-g-c-o-s-m-x-1-x-1-n-0-x-2-m-and-n-are-integers-m-0-n-0-and-let-p-be-the-left-hand-derivative-of-x-1-at-x-1-if-lim-x-1-g-x-p-then-93537

<div class="question">Let $g(x)=\frac{(x-1)^n}{\log \cos ^m(x-1)} ; 0 &lt; x &lt; 2, m$ and $n$ are integers, $m \neq 0, n&gt;0$ and let $p$ be the left hand derivative of $|x-1|$ at $x=1$. If $\lim _{x \rightarrow 1^{+}} g(x)=p$, then</div>

['Mathematics', 'Limits', 'JEE Advanced', 'JEE Advanced 2008 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$n=1, m=1$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$n=1, m=-1$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$n=2, m=2$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$n&gt;2, m=n$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$n=2, m=2$<br/></span> </div>

<div class="solution">Given, $g(x)=\frac{(x-1)^n}{\log \cos ^m(x-1)} ; 0 &lt; x &lt; 2, m \neq 0, n$ are integers and $|x-1|=\left\{\begin{array}{l}x-1 ; x \geq 1 \\ 1-x ; x &lt; 1\end{array}\right.$<br/>The left hand derivative of $|x-1|$ at $x=1$ is $p=-1$.<br/>Also,<br/>$$<br/>\begin{aligned}<br/>&amp; \Rightarrow \quad \lim _{h \rightarrow 0} \frac{(1+h-1)^n}{\log \cos ^m(1+h-1)}=-1 \Rightarrow \lim _{h \rightarrow 0} \frac{h^n}{\log \cos ^m h}=-1 \\<br/>&amp; \Rightarrow \quad \lim _{h \rightarrow 0} \frac{h^n}{m \log \cos h}=-1<br/>\end{aligned}<br/>$$<br/>[Using L' Hospital rule]<br/>$$<br/>\begin{array}{ll}<br/>\Rightarrow &amp; \quad \lim _{h \rightarrow 0} \frac{n \cdot h^{n-1}}{m \frac{1}{\cos h}(-\sin h)}=-1 \\<br/>\Rightarrow \quad &amp; \quad \lim _{h \rightarrow 0}\left(-\frac{n}{m}\right) \cdot \frac{h^{n-2}}{\left(\frac{\tan h}{h}\right)}=-1 \\<br/>\Rightarrow \quad &amp; \quad\left(\frac{n}{m}\right) \lim _{h \rightarrow 0} \frac{h^{n-2}}{\left(\frac{\tan h}{h}\right)}=1 \\<br/>\Rightarrow &amp; n=2 \text { and } \frac{n}{m}=1 \\<br/>\therefore &amp; \quad m=n=2<br/>\end{array}<br/>$$</div>

MarksBatch2_P1.db

let-g-x-lo-g-f-x-where-f-x-is-a-twice-differentiable-positive-function-on-0-such-that-f-x-1-x-f-x-then-for-n-1-2-3-g-n-2-1-g-2-1-is-equal-to

let-g-x-lo-g-f-x-where-f-x-is-a-twice-differentiable-positive-function-on-0-such-that-f-x-1-x-f-x-then-for-n-1-2-3-g-n-2-1-g-2-1-is-equal-to-69403

<div class="question">Let $g(x)=\log f(x)$, where $f(x)$ is a twice differentiable positive function on $(0, \infty)$ such that $f(x+1)=x f(x)$. Then, for $N=1,2,3, \ldots . g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)$ is equal to</div>

['Mathematics', 'Functions', 'JEE Advanced', 'JEE Advanced 2008 (Paper 2)']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\}$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N+1)^2}\right\}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N+1)^2}\right\}$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\}$<br/></span> </div>

<div class="solution">Since, $\quad f(x)=e^{g(x)}$<br/>$$<br/>\begin{aligned}<br/>\Rightarrow \quad e^{g(x+1)} &amp; =f(x+1) \\<br/>&amp; =x f(x) \\<br/>&amp; =x e^{g(x)}<br/>\end{aligned}<br/>$$<br/>and<br/>$$<br/>g(x+1)=\log x+g(x)<br/>$$<br/>$$<br/>\Rightarrow \quad g(x+1)-g(x)=\log x<br/>$$<br/>Replacing $x$ by $x-\frac{1}{2}$, we get<br/>$$<br/>\begin{aligned}<br/>g\left(x+\frac{1}{2}\right)-g\left(x-\frac{1}{2}\right) &amp; =\log \left(x-\frac{1}{2}\right)=\log (2 x-1)-\log 2 \\<br/>\therefore g^{\prime \prime}\left(x+\frac{1}{2}\right)-g^{\prime \prime}\left(x-\frac{1}{2}\right) &amp; =-\frac{4}{(2 x-1)^2}<br/>\end{aligned}<br/>$$<br/>Substituting, $x=1,2,3, \ldots, N$ in Eq. (ii) and adding, we get<br/>$$<br/>g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)=-4\left\{1+\frac{1}{9}+\frac{1}{25}+\ldots+\frac{1}{(2 N-1)^2}\right\} .<br/>$$</div>

MarksBatch2_P1.db

let-h-1-h-2-h-n-be-mutually-exclusive-events-with-p-h-i-0-i-1-2-n-let-e-be-any-other-event-with-0-p-e-1-statement-i-p-h-i-e-p-e-h-i-p-h-i-for-i-1-2-n-

let-h-1-h-2-h-n-be-mutually-exclusive-events-with-p-h-i-0-i-1-2-n-let-e-be-any-other-event-with-0-p-e-1-statement-i-p-h-i-e-p-e-h-i-p-h-i-for-i-1-2-n-81492

<div class="question">Let $H_1, H_2, \ldots, H_n$ be mutually exclusive events with $P\left(H_i\right)&gt;0, i=1,2, \ldots, n$. Let $E$ be any other event with $0 &lt; P(E) &lt; 1$.<br/>Statement I $P\left(H_i / E\right)&gt;P\left(E / H_i\right) P\left(H_i\right)$ for $i=1,2, \ldots, n$.<br/>Statement II $\sum_{i=1}^n P\left(H_i\right)=1$</div>

['Mathematics', 'Probability', 'JEE Advanced', 'JEE Advanced 2007 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement II<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>Statement I is true, Statement II is false<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>Statement I is false, Statement II is true</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>Statement I is false, Statement II is true</span> </div>

<div class="solution">Statement : I If $P\left(H_i \cap E\right)=0$ for some $i$, then<br/>$$<br/>P\left(\frac{H_i}{E}\right)=P\left(\frac{E}{H_i}\right)=0<br/>$$<br/>If $P\left(H_i \cap E\right) \neq 0$ for $\forall i=1,2, \ldots . n$, then<br/>$$<br/>\begin{aligned}<br/>P\left(\frac{H_i}{E}\right) &amp; =\frac{P\left(H_i \cap E\right)}{P\left(H_i\right)} \times \frac{P\left(H_i\right)}{P(E)} \\<br/>&amp; \left.=\frac{P\left(\frac{E}{H_i}\right) \times P\left(H_i\right)}{P(E)}&gt;P\left(\frac{E}{H_i}\right) \cdot P\left(H_i\right) \quad \text { [as } 0 &lt; P(E) &lt; 1\right]<br/>\end{aligned}<br/>$$<br/>Hence, Statement I may not always be true.<br/>Statement II<br/>Clearly, $H_1 \cup H_2 \cup \ldots . \cup H_n=S$ (sample space)<br/>$$<br/>\Rightarrow \quad P\left(H_1\right)+P\left(H_2\right)+\ldots+P\left(H_n\right)=1 .<br/>$$<br/>Hence, Statement II is true.</div>

MarksBatch2_P1.db

let-i-e-4-x-e-2-x-1-e-x-d-x-j-e-4-x-e-2-x-1-e-x-d-x-then-for-an-arbitrary-constant-c-the-value-of-j-i-equals

let-i-e-4-x-e-2-x-1-e-x-d-x-j-e-4-x-e-2-x-1-e-x-d-x-then-for-an-arbitrary-constant-c-the-value-of-j-i-equals-56578

<div class="question">Let $I=\int \frac{e^x}{e^{4 x}+e^{2 x}+1} d x, J=\int \frac{e^{-x}}{e^{-4 x}+e^{-2 x}+1} d x$.<br/>Then, for an arbitrary constant $C$, the value of $J-I$ equals</div>

['Mathematics', 'Indefinite Integration', 'JEE Advanced', 'JEE Advanced 2008 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\frac{1}{2} \log \left|\frac{e^{4 x}-e^{2 x}+1}{e^{4 x}+e^{2 x}+1}\right|+C$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\frac{1}{2} \log \left|\frac{e^{2 x}+e^x+1}{e^{2 x}-e^x+1}\right|+C$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$\frac{1}{2} \log \left|\frac{e^{2 x}-e^x+1}{e^{2 x}+e^x+1}\right|+C$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\frac{1}{2} \log \left|\frac{e^{4 x}+e^{2 x}+1}{e^{4 x}-e^{2 x}+1}\right|+C$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$\frac{1}{2} \log \left|\frac{e^{2 x}-e^x+1}{e^{2 x}+e^x+1}\right|+C$<br/></span> </div>

<div class="solution">Since, $J=\int \frac{e^{3 x}}{1+e^{2 x}+e^{4 x}} d x$<br/>$$<br/>\begin{aligned}<br/>\therefore J-I &amp; =\int \frac{\left(e^{3 x}-e^x\right)}{1+e^{2 x}+e^{4 x}} d x=\int \frac{\left(u^2-1\right)}{1+u^2+u^4} d u \\<br/>&amp; =\int \frac{\left(1-\frac{1}{u^2}\right)}{1+\frac{1}{u^2}+u^2} d u=\int \frac{\left(1-\frac{1}{u^2}\right)}{\left(u+\frac{1}{u}\right)^2-1} d u \\<br/>&amp; =\int \frac{d t}{t^2-1} \\<br/>&amp; =\frac{1}{2} \log \left|\frac{t-1}{t+1}\right|+C \\<br/>&amp; =\frac{1}{2} \log \left|\frac{u^2-u+1}{u^2+u+1}\right|+C=\frac{1}{2} \log \left|\frac{e^{2 x}-e^x+1}{e^{2 x}+e^x+1}\right|+C<br/>\end{aligned}<br/>$$</div>

MarksBatch2_P1.db

let-k-be-a-positive-real-number-and-let-a-b-2-k-1-2-k-2-k-2-k-1-2-k-2-k-2-k-1-and-0-1-2-k-k-2-k-1-0-2-k-k-2-k-0-if-det-adj-a-det-adj-b-1-0-6-then-k-is

let-k-be-a-positive-real-number-and-let-a-b-2-k-1-2-k-2-k-2-k-1-2-k-2-k-2-k-1-and-0-1-2-k-k-2-k-1-0-2-k-k-2-k-0-if-det-adj-a-det-adj-b-1-0-6-then-k-is-39129

<div class="question">Let $k$ be a positive real number and let $\begin{aligned} A &amp; =\left[\begin{array}{ccc}2 k-1 &amp; 2 \sqrt{k} &amp; 2 \sqrt{k} \\ 2 \sqrt{k} &amp; 1 &amp; -2 k \\ -2 \sqrt{k} &amp; 2 k &amp; -1\end{array}\right] \text { and } \\ B &amp; =\left[\begin{array}{ccc}0 &amp; 2 k-1 &amp; \sqrt{k} \\ 1-2 k &amp; 0 &amp; 2 \sqrt{k} \\ -\sqrt{k} &amp; -2 \sqrt{k} &amp; 0\end{array}\right]\end{aligned}$<br/>If $\operatorname{det}(\operatorname{adj} A)+\operatorname{det}(\operatorname{adj} B)=10^6$, then $[k]$ is equal to<br/>[Note : adj $M$ denotes the adjoint of a square matrix $M$ and $[k]$ denotes the largest integer less than or equal to $k$ ].</div>

['Mathematics', 'Determinants', 'JEE Advanced', 'JEE Advanced 2010 (Paper 2)']

None

<div class="correct-answer"> The correct answer is: <span class="option-value">4</span> </div>

<div class="solution">$|A|=(2 k+1)^3,|B|=0$<br/>But $\operatorname{det}(\operatorname{adj} A)=\operatorname{det}(\operatorname{adj} B)=10^6$<br/>$$<br/>\begin{aligned}<br/>&amp; \Rightarrow(2 k+1)^6=10^6 \\<br/>&amp; \Rightarrow k=\frac{9}{2} \Rightarrow[k]=4<br/>\end{aligned}<br/>$$</div>

MarksBatch2_P1.db

let-l-be-a-normal-to-the-parabola-y-2-4-x-if-l-passes-through-the-point-9-6-then-l-is-given-by

let-l-be-a-normal-to-the-parabola-y-2-4-x-if-l-passes-through-the-point-9-6-then-l-is-given-by-41823

<div class="question">Let $L$ be a normal to the parabola $y^2=4 x$. If $L$ passes through the point $(9,6)$, then $L$ is given by</div>

['Mathematics', 'Parabola', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$y-x+3=0$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$y+3 x-33=0$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$y+x-15=0$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$y-2 x+12=0$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

<div class="correct-answer"> The correct answers are: <span class="option-value"><br/>$y-x+3=0$<br/>, <br/>$y+3 x-33=0$<br/>, <br/>$y-2 x+12=0$</span> </div>

<div class="solution">Normal to $y^2=4 x$, is $y-m x-2 m-m^3$ which passes through $(9,6)$.<br/>Now, $\quad 6=9 m-2 m-m^3$ $\Rightarrow m^3-7 m+6=0 \Rightarrow m=1,2,-3$<br/>$\therefore$ Equation of normals are<br/>$y-x+3=0$<br/>$y+3 x-33=0$ and $y-2 x+12=0$</div>

MarksBatch2_P1.db

let-l-lim-x-0-x-4-a-a-2-x-2-4-x-2-a-0-if-l-is-finite-then

let-l-lim-x-0-x-4-a-a-2-x-2-4-x-2-a-0-if-l-is-finite-then-27156

<div class="question">Let $L=\lim _{x \rightarrow 0} \frac{a-\sqrt{a^2-x^2}-\frac{x^2}{4}}{x^4}$, $a&gt;0$. If $L$ is finite, then</div>

['Mathematics', 'Limits', 'JEE Advanced', 'JEE Advanced 2009 (Paper 1)']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$a=2$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$a=1$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$L=\frac{1}{64}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$L=\frac{1}{32}$</span> </li> </ul>

<div class="correct-answer"> The correct answers are: <span class="option-value"><br/>$a=2$<br/>, <br/>$L=\frac{1}{64}$<br/></span> </div>

<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/OppXIo5ahAMcXNcH9uV52GAIpIQkQcAKb8c0qNyv8nQ.original.fullsize.png"/><br/></div>

MarksBatch2_P1.db

let-m-and-n-be-two-3-3-nonsingular-skewsymmetric-matrices-such-that-mn-nm-if-p-t-denotes-the-transpose-of-p-then-m-2-n-2-m-t-n-1-m-n-1-t-is-equal-to

let-m-and-n-be-two-3-3-nonsingular-skewsymmetric-matrices-such-that-mn-nm-if-p-t-denotes-the-transpose-of-p-then-m-2-n-2-m-t-n-1-m-n-1-t-is-equal-to-93774

<div class="question">Let $M$ and $N$ be two $3 \times 3$ non-singular skew-symmetric matrices such that $M N=N M$. If $P^T$ denotes the transpose of $P$, then $M^2 N^2\left(M^T N\right)^{-1}\left(M N^{-1}\right)^T$ is equal to</div>

['Mathematics', 'Matrices', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$M^2$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$-N^2$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$-M^2$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$M N$</span> </li> </ul>

<div class="correct-answer"> The correct answers are: <span class="option-value"><br/>$-M^2$<br/></span> </div>

<div class="solution">Given, $M^T=-M, N^T=-N$<br/>and $\quad M N=N M$<br/>$$<br/>\begin{aligned}<br/>\therefore &amp; M^2 N^2\left(M^T N\right)^{-1}\left(M N^{-1}\right)^T \\<br/>&amp; =M^2 N^2 N^{-1}\left(M^T\right)^{-1}\left(N^{-1}\right)^T \cdot M^T \\<br/>&amp; =M^2 N\left(N N^{-1}\right)(-M)^{-1}\left(N^T\right)^{-1}(-M) \\<br/>&amp; =M^2 N\left(-M^{-1}\right)(-N)^{-1}(-M) \\<br/>&amp; =-M^2 N M^{-1} N^{-1} M \\<br/>&amp; =-M \cdot(M N) M^{-1} N^{-1} M \\<br/>&amp; =-M(N M) M^{-1} N^{-1} M \\<br/>&amp; =-M N\left(N M^{-1}\right) N^{-1} M \\<br/>&amp; =-M\left(N N^{-1}\right) M=-M^2<br/>\end{aligned}<br/>$$<br/>Note Here, non-singular word should not be used, since there is no non-singular $3 \times 3$ skew-symmetric matrix.</div>

MarksBatch2_P1.db

let-m-be-a-3-3-matrix-satisfying-m-0-1-0-1-2-3-m-1-1-0-1-1-1-and-m-1-1-1-0-0-12-t-h-e-n-t-h-es-u-m-o-f-t-h-e-d-ia-g-o-na-l-e-n-t-r-i-eso-f-m-is

let-m-be-a-3-3-matrix-satisfying-m-0-1-0-1-2-3-m-1-1-0-1-1-1-and-m-1-1-1-0-0-12-t-h-e-n-t-h-es-u-m-o-f-t-h-e-d-ia-g-o-na-l-e-n-t-r-i-eso-f-m-is-39211

<div class="question">$$<br/>\text { Let } M \text { be a } 3 \times 3 \text { matrix satisfying }<br/>$$<br/><br/>$$<br/>M\left[\begin{array}{l}<br/>0 \\<br/>1 \\<br/>0<br/>\end{array}\right]=\left[\begin{array}{c}<br/>-1 \\<br/>2 \\<br/>3<br/>\end{array}\right], M\left[\begin{array}{c}<br/>1 \\<br/>-1 \\<br/>0<br/>\end{array}\right]=\left[\begin{array}{c}<br/>1 \\<br/>1 \\<br/>-1<br/>\end{array}\right] \text { and } M\left[\begin{array}{l}<br/>1 \\<br/>1 \\<br/>1<br/>\end{array}\right]=\left[\begin{array}{c}<br/>0 \\<br/>0 \\<br/>12<br/>\end{array}\right]<br/>$$<br/>Then, the sum of the diagonal entries of $M$ is</div>

['Mathematics', 'Matrices', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)']

None

<div class="correct-answer"> The correct answer is: <span class="option-value">9</span> </div>

<div class="solution">Let $M=\left[\begin{array}{lll}a_1 &amp; a_2 &amp; a_3 \\ b_1 &amp; b_2 &amp; b_3 \\ c_1 &amp; c_2 &amp; c_3\end{array}\right]$<br/>$\therefore M\left[\begin{array}{l}0 \\ 1 \\ 0\end{array}\right]=\left[\begin{array}{r}-1 \\ 2 \\ 3\end{array}\right], M\left[\begin{array}{r}1 \\ -1 \\ 0\end{array}\right]=\left[\begin{array}{r}1 \\ 1 \\ -1\end{array}\right]$,<br/>$M\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\left[\begin{array}{r}0 \\ 0 \\ 12\end{array}\right]$,<br/>$$<br/>\begin{aligned}<br/>\Rightarrow &amp; {\left[\begin{array}{l}<br/>a_2 \\<br/>b_2 \\<br/>c_2<br/>\end{array}\right]=\left[\begin{array}{r}<br/>-1 \\<br/>2 \\<br/>3<br/>\end{array}\right],\left[\begin{array}{l}<br/>a_1-a_2 \\<br/>b_1-b_2 \\<br/>c_1-c_2<br/>\end{array}\right]=\left[\begin{array}{r}<br/>1 \\<br/>1 \\<br/>-1<br/>\end{array}\right], } \\<br/>&amp; {\left[\begin{array}{l}<br/>a_1+a_2+a_3 \\<br/>b_1+b_2+b_3 \\<br/>c_1+c_2+c_3<br/>\end{array}\right]=\left[\begin{array}{r}<br/>0 \\<br/>0 \\<br/>12<br/>\end{array}\right] } \\<br/>\Rightarrow &amp; a_2=-1, b_2=2, c_2=3, a_1-a_2=1, \\<br/>\Rightarrow &amp; b_1-b_2=1, c_1-c_2=-1 \\<br/>\Rightarrow \quad &amp; a_1+a_2+a_3=0, b_1+b_2+b_3=0, \\<br/>\therefore &amp; c_1+c_2+c_3=12 \\<br/>\therefore &amp; a_1=0, b_2=2 \text { and } c_3=7<br/>\end{aligned}<br/>$$<br/>Hence, sum of diagonal elements<br/>$$<br/>=0+2+7=9<br/>$$</div>

MarksBatch2_P1.db

let-o-0-0-p-3-4-and-q-6-0-be-the-vertices-of-the-opq-the-point-r-inside-the-opq-is-such-that-the-opr-pqr-oqr-are-of-equal-area-the-coordinates-of-r-ar

let-o-0-0-p-3-4-and-q-6-0-be-the-vertices-of-the-opq-the-point-r-inside-the-opq-is-such-that-the-opr-pqr-oqr-are-of-equal-area-the-coordinates-of-r-ar-17740

<div class="question">Let $O(0,0), P(3,4)$ and $Q(6,0)$ be the vertices of the $\triangle O P Q$. The point $R$ inside the $\triangle O P Q$ is such that the $\triangle O P R, \triangle P Q R, \triangle O Q R$ are of equal area. The coordinates of $R$ are</div>

['Mathematics', 'Properties of Triangles', 'JEE Advanced', 'JEE Advanced 2007 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\left(\frac{4}{3}, 3\right)$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$\left(3, \frac{2}{3}\right)$<br/></span> </li><li class="correct"> <span class="option-label">C</span> <span class="option-data"><br/>$\left(3, \frac{4}{3}\right)$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\left(\frac{4}{3}, \frac{2}{3}\right)$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$\left(3, \frac{4}{3}\right)$<br/></span> </div>

<div class="solution">$$<br/>\text { 7. Since, triangle is on isosceles, hence centroid is the desired point. }<br/>$$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/mPt6BiU7se3e8XRpvnhJ0CQd1IFrMT3sobkeFYbTv-Y.original.fullsize.png"/><br/><br/>$\therefore$ Coordinates of $R\left(3, \frac{4}{3}\right)$</div>

MarksBatch2_P1.db

let-p-3-2-6-be-a-point-in-space-and-q-be-a-point-on-the-line-r-i-j-2-k-3-i-j-5-k-then-the-value-of-for-which-the-vector-pq-is-parallel-to-the-plane-x-

let-p-3-2-6-be-a-point-in-space-and-q-be-a-point-on-the-line-r-i-j-2-k-3-i-j-5-k-then-the-value-of-for-which-the-vector-pq-is-parallel-to-the-plane-x-38163

<div class="question">Let $P(3,2,6)$ be a point in space and $Q$ be a point on the line $\mathbf{r}=(\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})+\mu(-3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+5 \hat{\mathbf{k}})$.<br/>Then, the value of $\mu$ for which the vector $\mathbf{P Q}$ is parallel to the plane $x-4 y+3 z=1$ is</div>

['Mathematics', 'Three Dimensional Geometry', 'JEE Advanced', 'JEE Advanced 2009 (Paper 1)']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>$1 / 4$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$-1 / 4$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$1 / 8$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$-1 / 8$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$1 / 4$<br/></span> </div>

<div class="solution"><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/xhSyjGZAfvOYhd3cVOPzs1UpROnoGZGmyZU8hofeEj4.original.fullsize.png"/><br/><br/>$\mathbf{O Q}=(1-3 \mu) \mathbf{i}+(\mu-1) \mathbf{j}+(5 \mu+2) \mathbf{k}$ and $\mathbf{O P}=3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$, where $O$ is origin.<br/>$$<br/>\begin{aligned}<br/>&amp; \text { Now, } \mathbf{P Q}=(1-3 \mu-3) \hat{\mathbf{i}}+(\mu-1-2) \hat{\mathbf{j}} \\<br/>&amp; +(5 \mu+2-6) \hat{\mathbf{k}} \\<br/>&amp; =(2-3 \mu) \hat{\mathbf{i}}+(\mu-3) \hat{\mathbf{j}}+(5 \mu-4) \hat{\mathbf{k}} \\<br/>&amp;<br/>\end{aligned}<br/>$$<br/>$\because \mathbf{P Q}$ is parallel to the plane<br/>$$<br/>\begin{array}{rr} <br/>&amp; x-4 y+3 z=1 . \\<br/>\therefore &amp; -2-3 \mu-4 \mu+12+15 \mu-12=0 \\<br/>\Rightarrow &amp; 8 \mu=2 \Rightarrow \mu=\frac{1}{4}<br/>\end{array}<br/>$$</div>

MarksBatch2_P2.db

let-p-6-3-be-a-point-on-the-hyperbola-a-2-x-2-b-2-y-2-1-if-the-normal-at-the-point-p-intersects-the-x-axis-at-9-0-then-the-eccentricity-of-the-hyperbo

let-p-6-3-be-a-point-on-the-hyperbola-a-2-x-2-b-2-y-2-1-if-the-normal-at-the-point-p-intersects-the-x-axis-at-9-0-then-the-eccentricity-of-the-hyperbo-46862

<div class="question">Let $P(6,3)$ be a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If the normal at the point $P$ intersects the $X$-axis at $(9,0)$, then the eccentricity of the hyperbola is</div>

['Mathematics', 'Hyperbola', 'JEE Advanced', 'JEE Advanced 2011 (Paper 2)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\sqrt{\frac{5}{2}}$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$\sqrt{\frac{3}{2}}$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\sqrt{2}$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\sqrt{3}$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$\sqrt{\frac{3}{2}}$<br/></span> </div>

<div class="solution">Equation of normal to hyperbola at $\left(x_1, y_1\right)$ is<br/>$$<br/>\begin{gathered}<br/>\frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2 \\<br/>\therefore \text { At }(6,3), \frac{a^2 x}{6}+\frac{b^2 y}{3}=a^2+b^2<br/>\end{gathered}<br/>$$<br/>It passes throught $(9,0)$.<br/><br/>$$<br/>\begin{aligned}<br/>&amp; \text { Now, } \quad \frac{a^2 \cdot 9}{6}=a^2+b^2 \\<br/>&amp; \Rightarrow \quad \frac{3 a^2}{2}-a^2=b^2 \Rightarrow \frac{a^2}{b^2}=2 \\<br/>&amp; \therefore \quad e^2=1+\frac{b^2}{a^2}=1+\frac{1}{2} \Rightarrow e=\sqrt{\frac{3}{2}}<br/>\end{aligned}<br/>$$</div>

MarksBatch2_P2.db

let-p-a-ij-be-a-3-3-matrix-and-let-q-b-ij-where-b-ij-2-i-j-a-ij-for-1-i-j-3-if-the-determinant-of-p-is-2-then-the-determinant-of-the-matrix-q-is

let-p-a-ij-be-a-3-3-matrix-and-let-q-b-ij-where-b-ij-2-i-j-a-ij-for-1-i-j-3-if-the-determinant-of-p-is-2-then-the-determinant-of-the-matrix-q-is-86042

<div class="question">Let $P=\left[a_{i j}\right]$ be a $3 \times 3$ matrix and let $Q=\left[b_{i j}\right]$, where $b_{i j}=2^{i+j} a_{i j}$ for $1 \leq i, j \leq 3$. If the determinant of $P$ is 2 , then the determinant of the matrix $Q$ is</div>

['Mathematics', 'Determinants', 'JEE Advanced', 'JEE Advanced 2012 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data">$2^{10}$</span> </li><li class=""> <span class="option-label">B</span> <span class="option-data">$2^{11}$</span> </li><li class=""> <span class="option-label">C</span> <span class="option-data">$2^{12}$</span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data">$2^{13}$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value">$2^{13}$</span> </div>

<div class="solution">$|Q|=\left|\begin{array}{lll}2^{2} a_{11} &amp; 2^{3} a_{12} &amp; 2^{4} a_{13} \\ 2^{3} a_{21} &amp; 2^{4} a_{22} &amp; 2^{5} a_{23} \\ 2^{4} a_{31} &amp; 2^{5} a_{32} &amp; 2^{6} a_{33}\end{array}\right|$ <br/> <br/>$=2^{2} \cdot 2^{3} \cdot 2^{4}\left|\begin{array}{ccc}a_{11} &amp; a_{12} &amp; a_{13} \\ 2 a_{21} &amp; 2 a_{22} &amp; 2 a_{23} \\ 2^{2} a_{31} &amp; 2^{2} a_{32} &amp; 2^{2} a_{33}\end{array}\right|$ <br/> <br/>$=2^{9} \cdot 2 \cdot 2^{2}\left|\begin{array}{lll}a_{11} &amp; a_{12} &amp; a_{13} \\ a_{21} &amp; a_{22} &amp; a_{23} \\ a_{31} &amp; a_{32} &amp; a_{33}\end{array}\right|$ <br/> <br/>$=2^{12} \times|P|=2^{12} \times 2=2^{13}$</div>

MarksBatch2_P2.db

let-p-and-q-be-real-numbers-such-that-p-0-p-3-q-and-p-3-q-if-and-are-nonzero-complex-numbers-satisfying-p-and-3-3-q-then-a-quadratic-equation-having-a

let-p-and-q-be-real-numbers-such-that-p-0-p-3-q-and-p-3-q-if-and-are-nonzero-complex-numbers-satisfying-p-and-3-3-q-then-a-quadratic-equation-having-a-65635

<div class="question">Let $p$ and $q$ be real numbers such that $p \neq 0, p^3 \neq q$ and $p^3 \neq-q$. If $\alpha$ and $\beta$ are non-zero complex numbers satisfying $\alpha+\beta=-p$ and $\alpha^3+\beta^3=q$, then $a$ quadratic equation having $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ as its roots is</div>

['Mathematics', 'Quadratic Equation', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$\left(p^3+q\right) x^2-\left(p^3+2 q\right) x$<br/>$$<br/>+\left(p^3+q\right)=0<br/>$$<br/></span> </li><li class="correct"> <span class="option-label">B</span> <span class="option-data"><br/>$\left(p^3+q\right) x^2-\left(p^3-2 q\right) x$<br/>$$<br/>+\left(p^3+q\right)=0<br/>$$<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$\left(p^3-q\right) x^2-\left(5 p^3-2 q\right) x$<br/>$$<br/>+\left(p^3-q\right)=0<br/>$$<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>$\left(p^3-q\right) x^2-\left(5 p^3+2 q\right) x$<br/>$$<br/>+\left(p^3-q\right)=0<br/>$$</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$\left(p^3+q\right) x^2-\left(p^3-2 q\right) x$<br/>$$<br/>+\left(p^3+q\right)=0<br/>$$<br/></span> </div>

<div class="solution">Sum of roots $=\frac{\alpha^2+\beta^2}{\alpha \beta}$ and product $=1$<br/>Given, $\alpha+\beta=-p$ and $\alpha^3+\beta^3=q$<br/>$$<br/>\begin{aligned}<br/>&amp; \Rightarrow(\alpha+\beta)\left(\alpha^2-\alpha \beta+\beta^2\right)=q \\<br/>&amp; \therefore \quad \alpha^2+\beta^2-\alpha \beta=\frac{-q}{p}<br/>\end{aligned}<br/>$$<br/>and $\quad(\alpha+\beta)^2=p^2$<br/>$$<br/>\Rightarrow \quad \alpha^2+\beta^2+2 \alpha \beta=p^2<br/>$$<br/>From Eqs. (i) and (ii), we get<br/>$$<br/>\alpha^2+\beta^2=\frac{p^3-2 q}{3 p}<br/>$$<br/>and $\alpha \beta=\frac{p^3+q}{3 p}$<br/>$\therefore$ Required equation is<br/>$$<br/>\begin{gathered}<br/>x^2-\frac{\left(p^3-2 q\right) x}{\left(p^3+q\right)}+1=0 \\<br/>\Rightarrow\left(p^3+q\right) x^2-\left(p^3-2 q\right) x+\left(p^3+q\right)=0<br/>\end{gathered}<br/>$$</div>

MarksBatch2_P2.db

let-p-q-r-and-s-be-the-points-on-the-plane-with-position-vectors-2-i-j-4-i-3-i-3-j-an-d-3-i-2-j-respectively-the-quadrilateral-pqrs-must-be-a

let-p-q-r-and-s-be-the-points-on-the-plane-with-position-vectors-2-i-j-4-i-3-i-3-j-an-d-3-i-2-j-respectively-the-quadrilateral-pqrs-must-be-a-58541

<div class="question">Let $P, Q, R$ and $S$ be the points on the plane with position vectors $-2 \hat{\mathbf{i}}-\hat{\mathbf{j}}, 4 \hat{\mathbf{i}}, 3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}$ an $\mathrm{d}-3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}$ respectively. The quadrilateral $P Q R S$ must be a</div>

['Mathematics', 'Vector Algebra', 'JEE Advanced', 'JEE Advanced 2010 (Paper 1)']

<ul class="options"> <li class="correct"> <span class="option-label">A</span> <span class="option-data"><br/>parallelogram, which is neither a rhombus nor a rectangle<br/></span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>square<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>rectangle, but not a square<br/></span> </li><li class=""> <span class="option-label">D</span> <span class="option-data"><br/>rhombus, but not a square</span> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>parallelogram, which is neither a rhombus nor a rectangle<br/></span> </div>

<div class="solution">$m_{P Q}=\frac{1}{6}, m_{S R}=\frac{1}{6}, \quad m_{R Q}=-3$, $m_{S P}=-3$<br/><img src="https://cdn-question-pool.getmarks.app/pyq/jee_advanced/ZpLK0MicUbuHy0zRsNvkTlJbp96CIu7p_UQH6TC2DEQ.original.fullsize.png"/><br/><br/>$\Rightarrow$ Parallelogram<br/>But neither $P R=S Q$ nor $P R \perp S Q$.<br/>$\therefore$ Parallelogram, which is neither a rhombus nor a rectangle.</div>

MarksBatch2_P2.db

let-p-sin-cos-2-cos-and-q-sin-cos-2-sin-be-two-sets-then-1

let-p-sin-cos-2-cos-and-q-sin-cos-2-sin-be-two-sets-then-1-90689

<div class="question">Let $P=\{\theta: \sin \theta-\cos \theta=\sqrt{2} \cos \theta\}$ and $Q=\{\theta: \sin \theta+\cos \theta=\sqrt{2} \sin \theta\}$ be two sets. Then,</div>

['Mathematics', 'Trigonometric Equations', 'JEE Advanced', 'JEE Advanced 2011 (Paper 1)']

<ul class="options"> <li class=""> <span class="option-label">A</span> <span class="option-data"><br/>$P \subset Q$ and $Q-P \neq \Phi$<br/></span> </li><li class=""> <span class="option-label">B</span> <span class="option-data"><br/>$Q \not \subset P$<br/></span> </li><li class=""> <span class="option-label">C</span> <span class="option-data"><br/>$P \not \subset Q$<br/></span> </li><li class="correct"> <span class="option-label">D</span> <span class="option-data"><br/>$P=Q$</span> <svg fill="none" height="24" viewbox="0 0 24 24" width="24" xmlns="http://www.w3.org/2000/svg"> <path d="M12 2.25C10.0716 2.25 8.18657 2.82183 6.58319 3.89317C4.97982 4.96452 3.73013 6.48726 2.99218 8.26884C2.25422 10.0504 2.06114 12.0108 2.43735 13.9021C2.81355 15.7934 3.74215 17.5307 5.10571 18.8943C6.46928 20.2579 8.20656 21.1865 10.0979 21.5627C11.9892 21.9389 13.9496 21.7458 15.7312 21.0078C17.5127 20.2699 19.0355 19.0202 20.1068 17.4168C21.1782 15.8134 21.75 13.9284 21.75 12C21.7473 9.41498 20.7192 6.93661 18.8913 5.10872C17.0634 3.28084 14.585 2.25273 12 2.25ZM16.2806 10.2806L11.0306 15.5306C10.961 15.6004 10.8783 15.6557 10.7872 15.6934C10.6962 15.7312 10.5986 15.7506 10.5 15.7506C10.4014 15.7506 10.3038 15.7312 10.2128 15.6934C10.1218 15.6557 10.039 15.6004 9.96938 15.5306L7.71938 13.2806C7.57865 13.1399 7.49959 12.949 7.49959 12.75C7.49959 12.551 7.57865 12.3601 7.71938 12.2194C7.86011 12.0786 8.05098 11.9996 8.25 11.9996C8.44903 11.9996 8.6399 12.0786 8.78063 12.2194L10.5 13.9397L15.2194 9.21937C15.2891 9.14969 15.3718 9.09442 15.4628 9.0567C15.5539 9.01899 15.6515 8.99958 15.75 8.99958C15.8486 8.99958 15.9461 9.01899 16.0372 9.0567C16.1282 9.09442 16.2109 9.14969 16.2806 9.21937C16.3503 9.28906 16.4056 9.37178 16.4433 9.46283C16.481 9.55387 16.5004 9.65145 16.5004 9.75C16.5004 9.84855 16.481 9.94613 16.4433 10.0372C16.4056 10.1282 16.3503 10.2109 16.2806 10.2806Z" fill="#24A865"></path> </svg> </li> </ul>

<div class="correct-answer"> The correct answer is: <span class="option-value"><br/>$P=Q$</span> </div>

<div class="solution">$$<br/>\begin{aligned}<br/>&amp; P=\{\theta: \sin \theta-\cos \theta=\sqrt{2} \cos \theta\} \\<br/>&amp; \Rightarrow \quad \cos \theta(\sqrt{2}+1)=\sin \theta \\<br/>&amp; \Rightarrow \quad \tan \theta=\sqrt{2}+1 \\<br/>&amp; Q=\{\theta: \sin \theta+\cos \theta=\sqrt{2} \sin \theta\} \\<br/>&amp; \Rightarrow \quad \sin \theta(\sqrt{2}-1)=\cos \theta \\<br/>&amp; \Rightarrow \quad \tan \theta=\frac{1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} \\<br/>&amp; =(\sqrt{2}+1) \\<br/>&amp; \therefore \quad P=Q \\<br/>&amp;<br/>\end{aligned}<br/>$$</div>

MarksBatch2_P2.db