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1 | 3018-3021 | The value of f at this point is given by
f (1) = (1 – 3)2 + (1)4 = 5 Thus, the distance between the solider and the helicopter is
(1)
5
f
= Note that 5 is either a maximum value or a minimum value Since
f(0)
=
2
4
(0
3)
(0)
3
5
−
+
=
>
,
it follows that
5 is the minimum value of
( )
f x |
1 | 3019-3022 | Thus, the distance between the solider and the helicopter is
(1)
5
f
= Note that 5 is either a maximum value or a minimum value Since
f(0)
=
2
4
(0
3)
(0)
3
5
−
+
=
>
,
it follows that
5 is the minimum value of
( )
f x Hence,
5 is the minimum
distance between the soldier and the helicopter |
1 | 3020-3023 | Note that 5 is either a maximum value or a minimum value Since
f(0)
=
2
4
(0
3)
(0)
3
5
−
+
=
>
,
it follows that
5 is the minimum value of
( )
f x Hence,
5 is the minimum
distance between the soldier and the helicopter EXERCISE 6 |
1 | 3021-3024 | Since
f(0)
=
2
4
(0
3)
(0)
3
5
−
+
=
>
,
it follows that
5 is the minimum value of
( )
f x Hence,
5 is the minimum
distance between the soldier and the helicopter EXERCISE 6 3
1 |
1 | 3022-3025 | Hence,
5 is the minimum
distance between the soldier and the helicopter EXERCISE 6 3
1 Find the maximum and minimum values, if any, of the following functions
given by
(i) f (x) = (2x – 1)2 + 3
(ii) f (x) = 9x2 + 12x + 2
(iii) f (x) = – (x – 1)2 + 10
(iv) g(x) = x3 + 1
Rationalised 2023-24
APPLICATION OF DERIVATIVES
175
2 |
1 | 3023-3026 | EXERCISE 6 3
1 Find the maximum and minimum values, if any, of the following functions
given by
(i) f (x) = (2x – 1)2 + 3
(ii) f (x) = 9x2 + 12x + 2
(iii) f (x) = – (x – 1)2 + 10
(iv) g(x) = x3 + 1
Rationalised 2023-24
APPLICATION OF DERIVATIVES
175
2 Find the maximum and minimum values, if any, of the following functions
given by
(i) f (x) = |x + 2| – 1
(ii) g(x) = – |x + 1| + 3
(iii) h(x) = sin(2x) + 5
(iv) f (x) = |sin 4x + 3|
(v) h(x) = x + 1, x ∈ (– 1, 1)
3 |
1 | 3024-3027 | 3
1 Find the maximum and minimum values, if any, of the following functions
given by
(i) f (x) = (2x – 1)2 + 3
(ii) f (x) = 9x2 + 12x + 2
(iii) f (x) = – (x – 1)2 + 10
(iv) g(x) = x3 + 1
Rationalised 2023-24
APPLICATION OF DERIVATIVES
175
2 Find the maximum and minimum values, if any, of the following functions
given by
(i) f (x) = |x + 2| – 1
(ii) g(x) = – |x + 1| + 3
(iii) h(x) = sin(2x) + 5
(iv) f (x) = |sin 4x + 3|
(v) h(x) = x + 1, x ∈ (– 1, 1)
3 Find the local maxima and local minima, if any, of the following functions |
1 | 3025-3028 | Find the maximum and minimum values, if any, of the following functions
given by
(i) f (x) = (2x – 1)2 + 3
(ii) f (x) = 9x2 + 12x + 2
(iii) f (x) = – (x – 1)2 + 10
(iv) g(x) = x3 + 1
Rationalised 2023-24
APPLICATION OF DERIVATIVES
175
2 Find the maximum and minimum values, if any, of the following functions
given by
(i) f (x) = |x + 2| – 1
(ii) g(x) = – |x + 1| + 3
(iii) h(x) = sin(2x) + 5
(iv) f (x) = |sin 4x + 3|
(v) h(x) = x + 1, x ∈ (– 1, 1)
3 Find the local maxima and local minima, if any, of the following functions Find
also the local maximum and the local minimum values, as the case may be:
(i) f (x) = x2
(ii) g(x) = x3 – 3x
(iii) h(x) = sin x + cos x, 0
2
x
π
<
<
(iv) f (x) = sin x – cos x, 0
2
<x
< π
(v) f (x) = x3 – 6x2 + 9x + 15
(vi)
2
( )
,
0
x2
g x
x
x
=
+
>
(vii)
2
1
( )
2
g x
x
=
+
(viii)
( )
1
, 0
1
=
−
<
<
f x
x
x
x
4 |
1 | 3026-3029 | Find the maximum and minimum values, if any, of the following functions
given by
(i) f (x) = |x + 2| – 1
(ii) g(x) = – |x + 1| + 3
(iii) h(x) = sin(2x) + 5
(iv) f (x) = |sin 4x + 3|
(v) h(x) = x + 1, x ∈ (– 1, 1)
3 Find the local maxima and local minima, if any, of the following functions Find
also the local maximum and the local minimum values, as the case may be:
(i) f (x) = x2
(ii) g(x) = x3 – 3x
(iii) h(x) = sin x + cos x, 0
2
x
π
<
<
(iv) f (x) = sin x – cos x, 0
2
<x
< π
(v) f (x) = x3 – 6x2 + 9x + 15
(vi)
2
( )
,
0
x2
g x
x
x
=
+
>
(vii)
2
1
( )
2
g x
x
=
+
(viii)
( )
1
, 0
1
=
−
<
<
f x
x
x
x
4 Prove that the following functions do not have maxima or minima:
(i) f (x) = ex
(ii) g(x) = log x
(iii) h (x) = x3 + x2 + x +1
5 |
1 | 3027-3030 | Find the local maxima and local minima, if any, of the following functions Find
also the local maximum and the local minimum values, as the case may be:
(i) f (x) = x2
(ii) g(x) = x3 – 3x
(iii) h(x) = sin x + cos x, 0
2
x
π
<
<
(iv) f (x) = sin x – cos x, 0
2
<x
< π
(v) f (x) = x3 – 6x2 + 9x + 15
(vi)
2
( )
,
0
x2
g x
x
x
=
+
>
(vii)
2
1
( )
2
g x
x
=
+
(viii)
( )
1
, 0
1
=
−
<
<
f x
x
x
x
4 Prove that the following functions do not have maxima or minima:
(i) f (x) = ex
(ii) g(x) = log x
(iii) h (x) = x3 + x2 + x +1
5 Find the absolute maximum value and the absolute minimum value of the following
functions in the given intervals:
(i) f (x) = x3, x ∈ [– 2, 2]
(ii) f (x) = sin x + cos x , x ∈ [0, π]
(iii) f (x) =
12
9
4
,
2,
2
2
x
x
x
−
∈ −
(iv)
2
( )
(
1)
3,
[ 3,1]
f x
x
x
=
−
+
∈ −
6 |
1 | 3028-3031 | Find
also the local maximum and the local minimum values, as the case may be:
(i) f (x) = x2
(ii) g(x) = x3 – 3x
(iii) h(x) = sin x + cos x, 0
2
x
π
<
<
(iv) f (x) = sin x – cos x, 0
2
<x
< π
(v) f (x) = x3 – 6x2 + 9x + 15
(vi)
2
( )
,
0
x2
g x
x
x
=
+
>
(vii)
2
1
( )
2
g x
x
=
+
(viii)
( )
1
, 0
1
=
−
<
<
f x
x
x
x
4 Prove that the following functions do not have maxima or minima:
(i) f (x) = ex
(ii) g(x) = log x
(iii) h (x) = x3 + x2 + x +1
5 Find the absolute maximum value and the absolute minimum value of the following
functions in the given intervals:
(i) f (x) = x3, x ∈ [– 2, 2]
(ii) f (x) = sin x + cos x , x ∈ [0, π]
(iii) f (x) =
12
9
4
,
2,
2
2
x
x
x
−
∈ −
(iv)
2
( )
(
1)
3,
[ 3,1]
f x
x
x
=
−
+
∈ −
6 Find the maximum profit that a company can make, if the profit function is
given by
p(x) = 41 – 72x – 18x2
7 |
1 | 3029-3032 | Prove that the following functions do not have maxima or minima:
(i) f (x) = ex
(ii) g(x) = log x
(iii) h (x) = x3 + x2 + x +1
5 Find the absolute maximum value and the absolute minimum value of the following
functions in the given intervals:
(i) f (x) = x3, x ∈ [– 2, 2]
(ii) f (x) = sin x + cos x , x ∈ [0, π]
(iii) f (x) =
12
9
4
,
2,
2
2
x
x
x
−
∈ −
(iv)
2
( )
(
1)
3,
[ 3,1]
f x
x
x
=
−
+
∈ −
6 Find the maximum profit that a company can make, if the profit function is
given by
p(x) = 41 – 72x – 18x2
7 Find both the maximum value and the minimum value of
3x4 – 8x3 + 12x2 – 48x + 25 on the interval [0, 3] |
1 | 3030-3033 | Find the absolute maximum value and the absolute minimum value of the following
functions in the given intervals:
(i) f (x) = x3, x ∈ [– 2, 2]
(ii) f (x) = sin x + cos x , x ∈ [0, π]
(iii) f (x) =
12
9
4
,
2,
2
2
x
x
x
−
∈ −
(iv)
2
( )
(
1)
3,
[ 3,1]
f x
x
x
=
−
+
∈ −
6 Find the maximum profit that a company can make, if the profit function is
given by
p(x) = 41 – 72x – 18x2
7 Find both the maximum value and the minimum value of
3x4 – 8x3 + 12x2 – 48x + 25 on the interval [0, 3] 8 |
1 | 3031-3034 | Find the maximum profit that a company can make, if the profit function is
given by
p(x) = 41 – 72x – 18x2
7 Find both the maximum value and the minimum value of
3x4 – 8x3 + 12x2 – 48x + 25 on the interval [0, 3] 8 At what points in the interval [0, 2π], does the function sin 2x attain its maximum
value |
1 | 3032-3035 | Find both the maximum value and the minimum value of
3x4 – 8x3 + 12x2 – 48x + 25 on the interval [0, 3] 8 At what points in the interval [0, 2π], does the function sin 2x attain its maximum
value 9 |
1 | 3033-3036 | 8 At what points in the interval [0, 2π], does the function sin 2x attain its maximum
value 9 What is the maximum value of the function sin x + cos x |
1 | 3034-3037 | At what points in the interval [0, 2π], does the function sin 2x attain its maximum
value 9 What is the maximum value of the function sin x + cos x 10 |
1 | 3035-3038 | 9 What is the maximum value of the function sin x + cos x 10 Find the maximum value of 2x3 – 24x + 107 in the interval [1, 3] |
1 | 3036-3039 | What is the maximum value of the function sin x + cos x 10 Find the maximum value of 2x3 – 24x + 107 in the interval [1, 3] Find the
maximum value of the same function in [–3, –1] |
1 | 3037-3040 | 10 Find the maximum value of 2x3 – 24x + 107 in the interval [1, 3] Find the
maximum value of the same function in [–3, –1] Rationalised 2023-24
MATHEMATICS
176
11 |
1 | 3038-3041 | Find the maximum value of 2x3 – 24x + 107 in the interval [1, 3] Find the
maximum value of the same function in [–3, –1] Rationalised 2023-24
MATHEMATICS
176
11 It is given that at x = 1, the function x4 – 62x2 + ax + 9 attains its maximum value,
on the interval [0, 2] |
1 | 3039-3042 | Find the
maximum value of the same function in [–3, –1] Rationalised 2023-24
MATHEMATICS
176
11 It is given that at x = 1, the function x4 – 62x2 + ax + 9 attains its maximum value,
on the interval [0, 2] Find the value of a |
1 | 3040-3043 | Rationalised 2023-24
MATHEMATICS
176
11 It is given that at x = 1, the function x4 – 62x2 + ax + 9 attains its maximum value,
on the interval [0, 2] Find the value of a 12 |
1 | 3041-3044 | It is given that at x = 1, the function x4 – 62x2 + ax + 9 attains its maximum value,
on the interval [0, 2] Find the value of a 12 Find the maximum and minimum values of x + sin 2x on [0, 2π] |
1 | 3042-3045 | Find the value of a 12 Find the maximum and minimum values of x + sin 2x on [0, 2π] 13 |
1 | 3043-3046 | 12 Find the maximum and minimum values of x + sin 2x on [0, 2π] 13 Find two numbers whose sum is 24 and whose product is as large as possible |
1 | 3044-3047 | Find the maximum and minimum values of x + sin 2x on [0, 2π] 13 Find two numbers whose sum is 24 and whose product is as large as possible 14 |
1 | 3045-3048 | 13 Find two numbers whose sum is 24 and whose product is as large as possible 14 Find two positive numbers x and y such that x + y = 60 and xy3 is maximum |
1 | 3046-3049 | Find two numbers whose sum is 24 and whose product is as large as possible 14 Find two positive numbers x and y such that x + y = 60 and xy3 is maximum 15 |
1 | 3047-3050 | 14 Find two positive numbers x and y such that x + y = 60 and xy3 is maximum 15 Find two positive numbers x and y such that their sum is 35 and the product x2 y5
is a maximum |
1 | 3048-3051 | Find two positive numbers x and y such that x + y = 60 and xy3 is maximum 15 Find two positive numbers x and y such that their sum is 35 and the product x2 y5
is a maximum 16 |
1 | 3049-3052 | 15 Find two positive numbers x and y such that their sum is 35 and the product x2 y5
is a maximum 16 Find two positive numbers whose sum is 16 and the sum of whose cubes is
minimum |
1 | 3050-3053 | Find two positive numbers x and y such that their sum is 35 and the product x2 y5
is a maximum 16 Find two positive numbers whose sum is 16 and the sum of whose cubes is
minimum 17 |
1 | 3051-3054 | 16 Find two positive numbers whose sum is 16 and the sum of whose cubes is
minimum 17 A square piece of tin of side 18 cm is to be made into a box without top, by
cutting a square from each corner and folding up the flaps to form the box |
1 | 3052-3055 | Find two positive numbers whose sum is 16 and the sum of whose cubes is
minimum 17 A square piece of tin of side 18 cm is to be made into a box without top, by
cutting a square from each corner and folding up the flaps to form the box What
should be the side of the square to be cut off so that the volume of the box is the
maximum possible |
1 | 3053-3056 | 17 A square piece of tin of side 18 cm is to be made into a box without top, by
cutting a square from each corner and folding up the flaps to form the box What
should be the side of the square to be cut off so that the volume of the box is the
maximum possible 18 |
1 | 3054-3057 | A square piece of tin of side 18 cm is to be made into a box without top, by
cutting a square from each corner and folding up the flaps to form the box What
should be the side of the square to be cut off so that the volume of the box is the
maximum possible 18 A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top,
by cutting off square from each corner and folding up the flaps |
1 | 3055-3058 | What
should be the side of the square to be cut off so that the volume of the box is the
maximum possible 18 A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top,
by cutting off square from each corner and folding up the flaps What should be
the side of the square to be cut off so that the volume of the box is maximum |
1 | 3056-3059 | 18 A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top,
by cutting off square from each corner and folding up the flaps What should be
the side of the square to be cut off so that the volume of the box is maximum 19 |
1 | 3057-3060 | A rectangular sheet of tin 45 cm by 24 cm is to be made into a box without top,
by cutting off square from each corner and folding up the flaps What should be
the side of the square to be cut off so that the volume of the box is maximum 19 Show that of all the rectangles inscribed in a given fixed circle, the square has
the maximum area |
1 | 3058-3061 | What should be
the side of the square to be cut off so that the volume of the box is maximum 19 Show that of all the rectangles inscribed in a given fixed circle, the square has
the maximum area 20 |
1 | 3059-3062 | 19 Show that of all the rectangles inscribed in a given fixed circle, the square has
the maximum area 20 Show that the right circular cylinder of given surface and maximum volume is
such that its height is equal to the diameter of the base |
1 | 3060-3063 | Show that of all the rectangles inscribed in a given fixed circle, the square has
the maximum area 20 Show that the right circular cylinder of given surface and maximum volume is
such that its height is equal to the diameter of the base 21 |
1 | 3061-3064 | 20 Show that the right circular cylinder of given surface and maximum volume is
such that its height is equal to the diameter of the base 21 Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic
centimetres, find the dimensions of the can which has the minimum surface
area |
1 | 3062-3065 | Show that the right circular cylinder of given surface and maximum volume is
such that its height is equal to the diameter of the base 21 Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic
centimetres, find the dimensions of the can which has the minimum surface
area 22 |
1 | 3063-3066 | 21 Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic
centimetres, find the dimensions of the can which has the minimum surface
area 22 A wire of length 28 m is to be cut into two pieces |
1 | 3064-3067 | Of all the closed cylindrical cans (right circular), of a given volume of 100 cubic
centimetres, find the dimensions of the can which has the minimum surface
area 22 A wire of length 28 m is to be cut into two pieces One of the pieces is to be
made into a square and the other into a circle |
1 | 3065-3068 | 22 A wire of length 28 m is to be cut into two pieces One of the pieces is to be
made into a square and the other into a circle What should be the length of the
two pieces so that the combined area of the square and the circle is minimum |
1 | 3066-3069 | A wire of length 28 m is to be cut into two pieces One of the pieces is to be
made into a square and the other into a circle What should be the length of the
two pieces so that the combined area of the square and the circle is minimum 23 |
1 | 3067-3070 | One of the pieces is to be
made into a square and the other into a circle What should be the length of the
two pieces so that the combined area of the square and the circle is minimum 23 Prove that the volume of the largest cone that can be inscribed in a sphere of
radius R is 8
27 of the volume of the sphere |
1 | 3068-3071 | What should be the length of the
two pieces so that the combined area of the square and the circle is minimum 23 Prove that the volume of the largest cone that can be inscribed in a sphere of
radius R is 8
27 of the volume of the sphere 24 |
1 | 3069-3072 | 23 Prove that the volume of the largest cone that can be inscribed in a sphere of
radius R is 8
27 of the volume of the sphere 24 Show that the right circular cone of least curved surface and given volume has
an altitude equal to
2 time the radius of the base |
1 | 3070-3073 | Prove that the volume of the largest cone that can be inscribed in a sphere of
radius R is 8
27 of the volume of the sphere 24 Show that the right circular cone of least curved surface and given volume has
an altitude equal to
2 time the radius of the base 25 |
1 | 3071-3074 | 24 Show that the right circular cone of least curved surface and given volume has
an altitude equal to
2 time the radius of the base 25 Show that the semi-vertical angle of the cone of the maximum volume and of
given slant height is
tan1
2
− |
1 | 3072-3075 | Show that the right circular cone of least curved surface and given volume has
an altitude equal to
2 time the radius of the base 25 Show that the semi-vertical angle of the cone of the maximum volume and of
given slant height is
tan1
2
− 26 |
1 | 3073-3076 | 25 Show that the semi-vertical angle of the cone of the maximum volume and of
given slant height is
tan1
2
− 26 Show that semi-vertical angle of right circular cone of given surface area and
maximum volume is sin−
1 1
3 |
1 | 3074-3077 | Show that the semi-vertical angle of the cone of the maximum volume and of
given slant height is
tan1
2
− 26 Show that semi-vertical angle of right circular cone of given surface area and
maximum volume is sin−
1 1
3 Rationalised 2023-24
APPLICATION OF DERIVATIVES
177
Choose the correct answer in Questions 27 and 29 |
1 | 3075-3078 | 26 Show that semi-vertical angle of right circular cone of given surface area and
maximum volume is sin−
1 1
3 Rationalised 2023-24
APPLICATION OF DERIVATIVES
177
Choose the correct answer in Questions 27 and 29 27 |
1 | 3076-3079 | Show that semi-vertical angle of right circular cone of given surface area and
maximum volume is sin−
1 1
3 Rationalised 2023-24
APPLICATION OF DERIVATIVES
177
Choose the correct answer in Questions 27 and 29 27 The point on the curve x2 = 2y which is nearest to the point (0, 5) is
(A) (2 2,4)
(B) (2 2,0)
(C) (0, 0)
(D) (2, 2)
28 |
1 | 3077-3080 | Rationalised 2023-24
APPLICATION OF DERIVATIVES
177
Choose the correct answer in Questions 27 and 29 27 The point on the curve x2 = 2y which is nearest to the point (0, 5) is
(A) (2 2,4)
(B) (2 2,0)
(C) (0, 0)
(D) (2, 2)
28 For all real values of x, the minimum value of
2
2
1
1
x
x
x
x
−
+
+
+
is
(A) 0
(B) 1
(C) 3
(D) 1
3
29 |
1 | 3078-3081 | 27 The point on the curve x2 = 2y which is nearest to the point (0, 5) is
(A) (2 2,4)
(B) (2 2,0)
(C) (0, 0)
(D) (2, 2)
28 For all real values of x, the minimum value of
2
2
1
1
x
x
x
x
−
+
+
+
is
(A) 0
(B) 1
(C) 3
(D) 1
3
29 The maximum value of
31
[ (
1)
1]
x x −
+
, 0
1
≤x
≤ is
(A)
1
3
31
(B) 1
2
(C) 1
(D) 0
Miscellaneous Examples
Example 30 A car starts from a point P at time t = 0 seconds and stops at point Q |
1 | 3079-3082 | The point on the curve x2 = 2y which is nearest to the point (0, 5) is
(A) (2 2,4)
(B) (2 2,0)
(C) (0, 0)
(D) (2, 2)
28 For all real values of x, the minimum value of
2
2
1
1
x
x
x
x
−
+
+
+
is
(A) 0
(B) 1
(C) 3
(D) 1
3
29 The maximum value of
31
[ (
1)
1]
x x −
+
, 0
1
≤x
≤ is
(A)
1
3
31
(B) 1
2
(C) 1
(D) 0
Miscellaneous Examples
Example 30 A car starts from a point P at time t = 0 seconds and stops at point Q The
distance x, in metres, covered by it, in t seconds is given by
x
t
t
=
−
2 2
3
Find the time taken by it to reach Q and also find distance between P and Q |
1 | 3080-3083 | For all real values of x, the minimum value of
2
2
1
1
x
x
x
x
−
+
+
+
is
(A) 0
(B) 1
(C) 3
(D) 1
3
29 The maximum value of
31
[ (
1)
1]
x x −
+
, 0
1
≤x
≤ is
(A)
1
3
31
(B) 1
2
(C) 1
(D) 0
Miscellaneous Examples
Example 30 A car starts from a point P at time t = 0 seconds and stops at point Q The
distance x, in metres, covered by it, in t seconds is given by
x
t
t
=
−
2 2
3
Find the time taken by it to reach Q and also find distance between P and Q Solution Let v be the velocity of the car at t seconds |
1 | 3081-3084 | The maximum value of
31
[ (
1)
1]
x x −
+
, 0
1
≤x
≤ is
(A)
1
3
31
(B) 1
2
(C) 1
(D) 0
Miscellaneous Examples
Example 30 A car starts from a point P at time t = 0 seconds and stops at point Q The
distance x, in metres, covered by it, in t seconds is given by
x
t
t
=
−
2 2
3
Find the time taken by it to reach Q and also find distance between P and Q Solution Let v be the velocity of the car at t seconds Now
x =
2 2
t3
t
−
Therefore
v = dx
dt = 4t – t2 = t(4 – t)
Thus, v = 0 gives t = 0 and/or t = 4 |
1 | 3082-3085 | The
distance x, in metres, covered by it, in t seconds is given by
x
t
t
=
−
2 2
3
Find the time taken by it to reach Q and also find distance between P and Q Solution Let v be the velocity of the car at t seconds Now
x =
2 2
t3
t
−
Therefore
v = dx
dt = 4t – t2 = t(4 – t)
Thus, v = 0 gives t = 0 and/or t = 4 Now v = 0 at P as well as at Q and at P, t = 0 |
1 | 3083-3086 | Solution Let v be the velocity of the car at t seconds Now
x =
2 2
t3
t
−
Therefore
v = dx
dt = 4t – t2 = t(4 – t)
Thus, v = 0 gives t = 0 and/or t = 4 Now v = 0 at P as well as at Q and at P, t = 0 So, at Q, t = 4 |
1 | 3084-3087 | Now
x =
2 2
t3
t
−
Therefore
v = dx
dt = 4t – t2 = t(4 – t)
Thus, v = 0 gives t = 0 and/or t = 4 Now v = 0 at P as well as at Q and at P, t = 0 So, at Q, t = 4 Thus, the car will
reach the point Q after 4 seconds |
1 | 3085-3088 | Now v = 0 at P as well as at Q and at P, t = 0 So, at Q, t = 4 Thus, the car will
reach the point Q after 4 seconds Also the distance travelled in 4 seconds is given by
x]t = 4 =
2
4
2
32
4
2
16
m
3
3
3
−
=
=
Rationalised 2023-24
MATHEMATICS
178
Example 31 A water tank has the shape of an inverted right circular cone with its axis
vertical and vertex lowermost |
1 | 3086-3089 | So, at Q, t = 4 Thus, the car will
reach the point Q after 4 seconds Also the distance travelled in 4 seconds is given by
x]t = 4 =
2
4
2
32
4
2
16
m
3
3
3
−
=
=
Rationalised 2023-24
MATHEMATICS
178
Example 31 A water tank has the shape of an inverted right circular cone with its axis
vertical and vertex lowermost Its semi-vertical angle is tan–1(0 |
1 | 3087-3090 | Thus, the car will
reach the point Q after 4 seconds Also the distance travelled in 4 seconds is given by
x]t = 4 =
2
4
2
32
4
2
16
m
3
3
3
−
=
=
Rationalised 2023-24
MATHEMATICS
178
Example 31 A water tank has the shape of an inverted right circular cone with its axis
vertical and vertex lowermost Its semi-vertical angle is tan–1(0 5) |
1 | 3088-3091 | Also the distance travelled in 4 seconds is given by
x]t = 4 =
2
4
2
32
4
2
16
m
3
3
3
−
=
=
Rationalised 2023-24
MATHEMATICS
178
Example 31 A water tank has the shape of an inverted right circular cone with its axis
vertical and vertex lowermost Its semi-vertical angle is tan–1(0 5) Water is poured
into it at a constant rate of 5 cubic metre per hour |
1 | 3089-3092 | Its semi-vertical angle is tan–1(0 5) Water is poured
into it at a constant rate of 5 cubic metre per hour Find the rate at which the level of
the water is rising at the instant when the depth of water in the tank is 4 m |
1 | 3090-3093 | 5) Water is poured
into it at a constant rate of 5 cubic metre per hour Find the rate at which the level of
the water is rising at the instant when the depth of water in the tank is 4 m Solution Let r, h and α be as in Fig 6 |
1 | 3091-3094 | Water is poured
into it at a constant rate of 5 cubic metre per hour Find the rate at which the level of
the water is rising at the instant when the depth of water in the tank is 4 m Solution Let r, h and α be as in Fig 6 20 |
1 | 3092-3095 | Find the rate at which the level of
the water is rising at the instant when the depth of water in the tank is 4 m Solution Let r, h and α be as in Fig 6 20 Then |
1 | 3093-3096 | Solution Let r, h and α be as in Fig 6 20 Then tan
hr
α =
So
α =
tan1
r
h
−
|
1 | 3094-3097 | 20 Then tan
hr
α =
So
α =
tan1
r
h
−
But
α = tan–1(0 |
1 | 3095-3098 | Then tan
hr
α =
So
α =
tan1
r
h
−
But
α = tan–1(0 5) (given)
or
r
h = 0 |
1 | 3096-3099 | tan
hr
α =
So
α =
tan1
r
h
−
But
α = tan–1(0 5) (given)
or
r
h = 0 5
or
r = 2
h
Let V be the volume of the cone |
1 | 3097-3100 | But
α = tan–1(0 5) (given)
or
r
h = 0 5
or
r = 2
h
Let V be the volume of the cone Then
V =
2
3
2
1
1
3
3
2
12
h
h
r h
h
π
π
=
π
=
Therefore
dV
dt =
3
12
d
h
dh
dh
dt
π
⋅
(by Chain Rule)
=
2
4
dh
h dt
π
Now rate of change of volume, i |
1 | 3098-3101 | 5) (given)
or
r
h = 0 5
or
r = 2
h
Let V be the volume of the cone Then
V =
2
3
2
1
1
3
3
2
12
h
h
r h
h
π
π
=
π
=
Therefore
dV
dt =
3
12
d
h
dh
dh
dt
π
⋅
(by Chain Rule)
=
2
4
dh
h dt
π
Now rate of change of volume, i e |
1 | 3099-3102 | 5
or
r = 2
h
Let V be the volume of the cone Then
V =
2
3
2
1
1
3
3
2
12
h
h
r h
h
π
π
=
π
=
Therefore
dV
dt =
3
12
d
h
dh
dh
dt
π
⋅
(by Chain Rule)
=
2
4
dh
h dt
π
Now rate of change of volume, i e , V
5
d
dt =
m3/h and h = 4 m |
1 | 3100-3103 | Then
V =
2
3
2
1
1
3
3
2
12
h
h
r h
h
π
π
=
π
=
Therefore
dV
dt =
3
12
d
h
dh
dh
dt
π
⋅
(by Chain Rule)
=
2
4
dh
h dt
π
Now rate of change of volume, i e , V
5
d
dt =
m3/h and h = 4 m Therefore
5 =
(4)2
4
dh
dt
π
⋅
or
dh
dt = 5
35
22
m/h
4
88
7
=
π =
π
Thus, the rate of change of water level is 35 m/h
88 |
1 | 3101-3104 | e , V
5
d
dt =
m3/h and h = 4 m Therefore
5 =
(4)2
4
dh
dt
π
⋅
or
dh
dt = 5
35
22
m/h
4
88
7
=
π =
π
Thus, the rate of change of water level is 35 m/h
88 Example 32 A man of height 2 metres walks at a uniform speed of 5 km/h away from
a lamp post which is 6 metres high |
1 | 3102-3105 | , V
5
d
dt =
m3/h and h = 4 m Therefore
5 =
(4)2
4
dh
dt
π
⋅
or
dh
dt = 5
35
22
m/h
4
88
7
=
π =
π
Thus, the rate of change of water level is 35 m/h
88 Example 32 A man of height 2 metres walks at a uniform speed of 5 km/h away from
a lamp post which is 6 metres high Find the rate at which the length of his shadow
increases |
1 | 3103-3106 | Therefore
5 =
(4)2
4
dh
dt
π
⋅
or
dh
dt = 5
35
22
m/h
4
88
7
=
π =
π
Thus, the rate of change of water level is 35 m/h
88 Example 32 A man of height 2 metres walks at a uniform speed of 5 km/h away from
a lamp post which is 6 metres high Find the rate at which the length of his shadow
increases Fig 6 |
1 | 3104-3107 | Example 32 A man of height 2 metres walks at a uniform speed of 5 km/h away from
a lamp post which is 6 metres high Find the rate at which the length of his shadow
increases Fig 6 20
Rationalised 2023-24
APPLICATION OF DERIVATIVES
179
Solution In Fig 6 |
1 | 3105-3108 | Find the rate at which the length of his shadow
increases Fig 6 20
Rationalised 2023-24
APPLICATION OF DERIVATIVES
179
Solution In Fig 6 21, Let AB be the lamp-post, the
lamp being at the position B and let MN be the man at
a particular time t and let AM = l metres |
1 | 3106-3109 | Fig 6 20
Rationalised 2023-24
APPLICATION OF DERIVATIVES
179
Solution In Fig 6 21, Let AB be the lamp-post, the
lamp being at the position B and let MN be the man at
a particular time t and let AM = l metres Then, MS is
the shadow of the man |
1 | 3107-3110 | 20
Rationalised 2023-24
APPLICATION OF DERIVATIVES
179
Solution In Fig 6 21, Let AB be the lamp-post, the
lamp being at the position B and let MN be the man at
a particular time t and let AM = l metres Then, MS is
the shadow of the man Let MS = s metres |
1 | 3108-3111 | 21, Let AB be the lamp-post, the
lamp being at the position B and let MN be the man at
a particular time t and let AM = l metres Then, MS is
the shadow of the man Let MS = s metres Note that
∆MSN ~ ∆ASB
or
MS
AS = MN
AB
or
AS = 3s (as MN =
2 and AB = 6 (given))
Thus
AM = 3s – s = 2s |
1 | 3109-3112 | Then, MS is
the shadow of the man Let MS = s metres Note that
∆MSN ~ ∆ASB
or
MS
AS = MN
AB
or
AS = 3s (as MN =
2 and AB = 6 (given))
Thus
AM = 3s – s = 2s But AM = l
So
l = 2s
Therefore
dl
dt = 2 ds
dt
Since
5
dl
dt =
km/h |
1 | 3110-3113 | Let MS = s metres Note that
∆MSN ~ ∆ASB
or
MS
AS = MN
AB
or
AS = 3s (as MN =
2 and AB = 6 (given))
Thus
AM = 3s – s = 2s But AM = l
So
l = 2s
Therefore
dl
dt = 2 ds
dt
Since
5
dl
dt =
km/h Hence, the length of the shadow increases at the rate 5
2 km/h |
1 | 3111-3114 | Note that
∆MSN ~ ∆ASB
or
MS
AS = MN
AB
or
AS = 3s (as MN =
2 and AB = 6 (given))
Thus
AM = 3s – s = 2s But AM = l
So
l = 2s
Therefore
dl
dt = 2 ds
dt
Since
5
dl
dt =
km/h Hence, the length of the shadow increases at the rate 5
2 km/h Example 33 Find intervals in which the function given by
f (x) =
4
3
2
3
4
36
3
11
10
5
5
x
x
x
x
−
−
+
+
is (a) increasing (b) decreasing |
1 | 3112-3115 | But AM = l
So
l = 2s
Therefore
dl
dt = 2 ds
dt
Since
5
dl
dt =
km/h Hence, the length of the shadow increases at the rate 5
2 km/h Example 33 Find intervals in which the function given by
f (x) =
4
3
2
3
4
36
3
11
10
5
5
x
x
x
x
−
−
+
+
is (a) increasing (b) decreasing Solution We have
f (x) =
4
3
2
3
4
36
3
11
10
5
5
x
x
x
x
−
−
+
+
Therefore
f ′(x) =
3
2
3
4
36
(4
)
(3
)
3(2 )
10
5
5
x
x
x
−
−
+
= 6(
1)(
2)(
3)
5 x
x
x
−
+
−
(on simplification)
Fig 6 |
1 | 3113-3116 | Hence, the length of the shadow increases at the rate 5
2 km/h Example 33 Find intervals in which the function given by
f (x) =
4
3
2
3
4
36
3
11
10
5
5
x
x
x
x
−
−
+
+
is (a) increasing (b) decreasing Solution We have
f (x) =
4
3
2
3
4
36
3
11
10
5
5
x
x
x
x
−
−
+
+
Therefore
f ′(x) =
3
2
3
4
36
(4
)
(3
)
3(2 )
10
5
5
x
x
x
−
−
+
= 6(
1)(
2)(
3)
5 x
x
x
−
+
−
(on simplification)
Fig 6 21
Rationalised 2023-24
MATHEMATICS
180
Now f ′(x) = 0 gives x = 1, x = – 2, or x = 3 |
1 | 3114-3117 | Example 33 Find intervals in which the function given by
f (x) =
4
3
2
3
4
36
3
11
10
5
5
x
x
x
x
−
−
+
+
is (a) increasing (b) decreasing Solution We have
f (x) =
4
3
2
3
4
36
3
11
10
5
5
x
x
x
x
−
−
+
+
Therefore
f ′(x) =
3
2
3
4
36
(4
)
(3
)
3(2 )
10
5
5
x
x
x
−
−
+
= 6(
1)(
2)(
3)
5 x
x
x
−
+
−
(on simplification)
Fig 6 21
Rationalised 2023-24
MATHEMATICS
180
Now f ′(x) = 0 gives x = 1, x = – 2, or x = 3 The
points x = 1, – 2, and 3 divide the real line into four
disjoint intervals namely, (– ∞, – 2), (– 2, 1), (1, 3)
and (3, ∞) (Fig 6 |
1 | 3115-3118 | Solution We have
f (x) =
4
3
2
3
4
36
3
11
10
5
5
x
x
x
x
−
−
+
+
Therefore
f ′(x) =
3
2
3
4
36
(4
)
(3
)
3(2 )
10
5
5
x
x
x
−
−
+
= 6(
1)(
2)(
3)
5 x
x
x
−
+
−
(on simplification)
Fig 6 21
Rationalised 2023-24
MATHEMATICS
180
Now f ′(x) = 0 gives x = 1, x = – 2, or x = 3 The
points x = 1, – 2, and 3 divide the real line into four
disjoint intervals namely, (– ∞, – 2), (– 2, 1), (1, 3)
and (3, ∞) (Fig 6 22) |
1 | 3116-3119 | 21
Rationalised 2023-24
MATHEMATICS
180
Now f ′(x) = 0 gives x = 1, x = – 2, or x = 3 The
points x = 1, – 2, and 3 divide the real line into four
disjoint intervals namely, (– ∞, – 2), (– 2, 1), (1, 3)
and (3, ∞) (Fig 6 22) Consider the interval (– ∞, – 2), i |
1 | 3117-3120 | The
points x = 1, – 2, and 3 divide the real line into four
disjoint intervals namely, (– ∞, – 2), (– 2, 1), (1, 3)
and (3, ∞) (Fig 6 22) Consider the interval (– ∞, – 2), i e |
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