Chapter
stringclasses 18
values | sentence_range
stringlengths 3
9
| Text
stringlengths 7
7.34k
|
|---|---|---|
1 | 2618-2621 | Sand is pouring from a pipe at the rate of 12 cm3/s The falling sand forms a cone
on the ground in such a way that the height of the cone is always one-sixth of the
radius of the base How fast is the height of the sand cone increasing when the
height is 4 cm 15 |
1 | 2619-2622 | The falling sand forms a cone
on the ground in such a way that the height of the cone is always one-sixth of the
radius of the base How fast is the height of the sand cone increasing when the
height is 4 cm 15 The total cost C (x) in Rupees associated with the production of x units of an
item is given by
C (x) = 0 |
1 | 2620-2623 | How fast is the height of the sand cone increasing when the
height is 4 cm 15 The total cost C (x) in Rupees associated with the production of x units of an
item is given by
C (x) = 0 007x3 – 0 |
1 | 2621-2624 | 15 The total cost C (x) in Rupees associated with the production of x units of an
item is given by
C (x) = 0 007x3 – 0 003x2 + 15x + 4000 |
1 | 2622-2625 | The total cost C (x) in Rupees associated with the production of x units of an
item is given by
C (x) = 0 007x3 – 0 003x2 + 15x + 4000 Find the marginal cost when 17 units are produced |
1 | 2623-2626 | 007x3 – 0 003x2 + 15x + 4000 Find the marginal cost when 17 units are produced 16 |
1 | 2624-2627 | 003x2 + 15x + 4000 Find the marginal cost when 17 units are produced 16 The total revenue in Rupees received from the sale of x units of a product is
given by
R (x) = 13x2 + 26x + 15 |
1 | 2625-2628 | Find the marginal cost when 17 units are produced 16 The total revenue in Rupees received from the sale of x units of a product is
given by
R (x) = 13x2 + 26x + 15 Find the marginal revenue when x = 7 |
1 | 2626-2629 | 16 The total revenue in Rupees received from the sale of x units of a product is
given by
R (x) = 13x2 + 26x + 15 Find the marginal revenue when x = 7 Choose the correct answer for questions 17 and 18 |
1 | 2627-2630 | The total revenue in Rupees received from the sale of x units of a product is
given by
R (x) = 13x2 + 26x + 15 Find the marginal revenue when x = 7 Choose the correct answer for questions 17 and 18 17 |
1 | 2628-2631 | Find the marginal revenue when x = 7 Choose the correct answer for questions 17 and 18 17 The rate of change of the area of a circle with respect to its radius r at r = 6 cm is
(A) 10π
(B) 12π
(C) 8π
(D) 11π
Rationalised 2023-24
MATHEMATICS
152
18 |
1 | 2629-2632 | Choose the correct answer for questions 17 and 18 17 The rate of change of the area of a circle with respect to its radius r at r = 6 cm is
(A) 10π
(B) 12π
(C) 8π
(D) 11π
Rationalised 2023-24
MATHEMATICS
152
18 The total revenue in Rupees received from the sale of x units of a product is
given by
R(x) = 3x2 + 36x + 5 |
1 | 2630-2633 | 17 The rate of change of the area of a circle with respect to its radius r at r = 6 cm is
(A) 10π
(B) 12π
(C) 8π
(D) 11π
Rationalised 2023-24
MATHEMATICS
152
18 The total revenue in Rupees received from the sale of x units of a product is
given by
R(x) = 3x2 + 36x + 5 The marginal revenue, when x = 15 is
(A) 116
(B) 96
(C) 90
(D) 126
6 |
1 | 2631-2634 | The rate of change of the area of a circle with respect to its radius r at r = 6 cm is
(A) 10π
(B) 12π
(C) 8π
(D) 11π
Rationalised 2023-24
MATHEMATICS
152
18 The total revenue in Rupees received from the sale of x units of a product is
given by
R(x) = 3x2 + 36x + 5 The marginal revenue, when x = 15 is
(A) 116
(B) 96
(C) 90
(D) 126
6 3 Increasing and Decreasing Functions
In this section, we will use differentiation to find out whether a function is increasing or
decreasing or none |
1 | 2632-2635 | The total revenue in Rupees received from the sale of x units of a product is
given by
R(x) = 3x2 + 36x + 5 The marginal revenue, when x = 15 is
(A) 116
(B) 96
(C) 90
(D) 126
6 3 Increasing and Decreasing Functions
In this section, we will use differentiation to find out whether a function is increasing or
decreasing or none Consider the function f given by f (x) = x2, x ∈ R |
1 | 2633-2636 | The marginal revenue, when x = 15 is
(A) 116
(B) 96
(C) 90
(D) 126
6 3 Increasing and Decreasing Functions
In this section, we will use differentiation to find out whether a function is increasing or
decreasing or none Consider the function f given by f (x) = x2, x ∈ R The graph of this function is a
parabola as given in Fig 6 |
1 | 2634-2637 | 3 Increasing and Decreasing Functions
In this section, we will use differentiation to find out whether a function is increasing or
decreasing or none Consider the function f given by f (x) = x2, x ∈ R The graph of this function is a
parabola as given in Fig 6 1 |
1 | 2635-2638 | Consider the function f given by f (x) = x2, x ∈ R The graph of this function is a
parabola as given in Fig 6 1 Fig 6 |
1 | 2636-2639 | The graph of this function is a
parabola as given in Fig 6 1 Fig 6 1
First consider the graph (Fig 6 |
1 | 2637-2640 | 1 Fig 6 1
First consider the graph (Fig 6 1) to the right of the origin |
1 | 2638-2641 | Fig 6 1
First consider the graph (Fig 6 1) to the right of the origin Observe that as we
move from left to right along the graph, the height of the graph continuously increases |
1 | 2639-2642 | 1
First consider the graph (Fig 6 1) to the right of the origin Observe that as we
move from left to right along the graph, the height of the graph continuously increases For this reason, the function is said to be increasing for the real numbers x > 0 |
1 | 2640-2643 | 1) to the right of the origin Observe that as we
move from left to right along the graph, the height of the graph continuously increases For this reason, the function is said to be increasing for the real numbers x > 0 Now consider the graph to the left of the origin and observe here that as we move
from left to right along the graph, the height of the graph continuously decreases |
1 | 2641-2644 | Observe that as we
move from left to right along the graph, the height of the graph continuously increases For this reason, the function is said to be increasing for the real numbers x > 0 Now consider the graph to the left of the origin and observe here that as we move
from left to right along the graph, the height of the graph continuously decreases Consequently, the function is said to be decreasing for the real numbers x < 0 |
1 | 2642-2645 | For this reason, the function is said to be increasing for the real numbers x > 0 Now consider the graph to the left of the origin and observe here that as we move
from left to right along the graph, the height of the graph continuously decreases Consequently, the function is said to be decreasing for the real numbers x < 0 We shall now give the following analytical definitions for a function which is
increasing or decreasing on an interval |
1 | 2643-2646 | Now consider the graph to the left of the origin and observe here that as we move
from left to right along the graph, the height of the graph continuously decreases Consequently, the function is said to be decreasing for the real numbers x < 0 We shall now give the following analytical definitions for a function which is
increasing or decreasing on an interval Definition 1 Let I be an interval contained in the domain of a real valued function f |
1 | 2644-2647 | Consequently, the function is said to be decreasing for the real numbers x < 0 We shall now give the following analytical definitions for a function which is
increasing or decreasing on an interval Definition 1 Let I be an interval contained in the domain of a real valued function f Then f is said to be
(i)
increasing on I if x1 < x2 in I ⇒ f (x1) < f (x2) for all x1, x2 ∈ I |
1 | 2645-2648 | We shall now give the following analytical definitions for a function which is
increasing or decreasing on an interval Definition 1 Let I be an interval contained in the domain of a real valued function f Then f is said to be
(i)
increasing on I if x1 < x2 in I ⇒ f (x1) < f (x2) for all x1, x2 ∈ I (ii)
decreasing on I, if x1, x2 in I ⇒ f (x1) < f (x2) for all x1, x2 ∈ I |
1 | 2646-2649 | Definition 1 Let I be an interval contained in the domain of a real valued function f Then f is said to be
(i)
increasing on I if x1 < x2 in I ⇒ f (x1) < f (x2) for all x1, x2 ∈ I (ii)
decreasing on I, if x1, x2 in I ⇒ f (x1) < f (x2) for all x1, x2 ∈ I (iii)
constant on I, if f(x) = c for all x ∈ I, where c is a constant |
1 | 2647-2650 | Then f is said to be
(i)
increasing on I if x1 < x2 in I ⇒ f (x1) < f (x2) for all x1, x2 ∈ I (ii)
decreasing on I, if x1, x2 in I ⇒ f (x1) < f (x2) for all x1, x2 ∈ I (iii)
constant on I, if f(x) = c for all x ∈ I, where c is a constant x
f (x) = x2
–2
4
−23
9
4
–1
1
−21
1
4
0
0
Values left to origin
as we move from left to right, the
height of the graph decreases
x
f (x) = x2
0
0
1
2
1
4
1
1
3
2
9
4
2
4
Values right to origin
as we move from left to right, the
height of the graph increases
Rationalised 2023-24
APPLICATION OF DERIVATIVES
153
(iv)
decreasing on I if x1 < x2 in I ⇒ f (x1) ≥ f (x2) for all x1, x2 ∈ I |
1 | 2648-2651 | (ii)
decreasing on I, if x1, x2 in I ⇒ f (x1) < f (x2) for all x1, x2 ∈ I (iii)
constant on I, if f(x) = c for all x ∈ I, where c is a constant x
f (x) = x2
–2
4
−23
9
4
–1
1
−21
1
4
0
0
Values left to origin
as we move from left to right, the
height of the graph decreases
x
f (x) = x2
0
0
1
2
1
4
1
1
3
2
9
4
2
4
Values right to origin
as we move from left to right, the
height of the graph increases
Rationalised 2023-24
APPLICATION OF DERIVATIVES
153
(iv)
decreasing on I if x1 < x2 in I ⇒ f (x1) ≥ f (x2) for all x1, x2 ∈ I (v)
strictly decreasing on I if x1 < x2 in I ⇒ f (x1) > f (x2) for all x1, x2 ∈ I |
1 | 2649-2652 | (iii)
constant on I, if f(x) = c for all x ∈ I, where c is a constant x
f (x) = x2
–2
4
−23
9
4
–1
1
−21
1
4
0
0
Values left to origin
as we move from left to right, the
height of the graph decreases
x
f (x) = x2
0
0
1
2
1
4
1
1
3
2
9
4
2
4
Values right to origin
as we move from left to right, the
height of the graph increases
Rationalised 2023-24
APPLICATION OF DERIVATIVES
153
(iv)
decreasing on I if x1 < x2 in I ⇒ f (x1) ≥ f (x2) for all x1, x2 ∈ I (v)
strictly decreasing on I if x1 < x2 in I ⇒ f (x1) > f (x2) for all x1, x2 ∈ I For graphical representation of such functions see Fig 6 |
1 | 2650-2653 | x
f (x) = x2
–2
4
−23
9
4
–1
1
−21
1
4
0
0
Values left to origin
as we move from left to right, the
height of the graph decreases
x
f (x) = x2
0
0
1
2
1
4
1
1
3
2
9
4
2
4
Values right to origin
as we move from left to right, the
height of the graph increases
Rationalised 2023-24
APPLICATION OF DERIVATIVES
153
(iv)
decreasing on I if x1 < x2 in I ⇒ f (x1) ≥ f (x2) for all x1, x2 ∈ I (v)
strictly decreasing on I if x1 < x2 in I ⇒ f (x1) > f (x2) for all x1, x2 ∈ I For graphical representation of such functions see Fig 6 2 |
1 | 2651-2654 | (v)
strictly decreasing on I if x1 < x2 in I ⇒ f (x1) > f (x2) for all x1, x2 ∈ I For graphical representation of such functions see Fig 6 2 Fig 6 |
1 | 2652-2655 | For graphical representation of such functions see Fig 6 2 Fig 6 2
We shall now define when a function is increasing or decreasing at a point |
1 | 2653-2656 | 2 Fig 6 2
We shall now define when a function is increasing or decreasing at a point Definition 2 Let x0 be a point in the domain of definition of a real valued function f |
1 | 2654-2657 | Fig 6 2
We shall now define when a function is increasing or decreasing at a point Definition 2 Let x0 be a point in the domain of definition of a real valued function f Then f is said to be increasing, decreasing at x0 if there exists an open interval I
containing x0 such that f is increasing, decreasing, respectively, in I |
1 | 2655-2658 | 2
We shall now define when a function is increasing or decreasing at a point Definition 2 Let x0 be a point in the domain of definition of a real valued function f Then f is said to be increasing, decreasing at x0 if there exists an open interval I
containing x0 such that f is increasing, decreasing, respectively, in I Let us clarify this definition for the case of increasing function |
1 | 2656-2659 | Definition 2 Let x0 be a point in the domain of definition of a real valued function f Then f is said to be increasing, decreasing at x0 if there exists an open interval I
containing x0 such that f is increasing, decreasing, respectively, in I Let us clarify this definition for the case of increasing function Example 7 Show that the function given by f(x) = 7x – 3 is increasing on R |
1 | 2657-2660 | Then f is said to be increasing, decreasing at x0 if there exists an open interval I
containing x0 such that f is increasing, decreasing, respectively, in I Let us clarify this definition for the case of increasing function Example 7 Show that the function given by f(x) = 7x – 3 is increasing on R Solution Let x1 and x2 be any two numbers in R |
1 | 2658-2661 | Let us clarify this definition for the case of increasing function Example 7 Show that the function given by f(x) = 7x – 3 is increasing on R Solution Let x1 and x2 be any two numbers in R Then
x1 < x2 ⇒ 7x1 < 7x2 ⇒ 7x1 – 3 < 7x2 – 3 ⇒ f (x1) < f (x2)
Thus, by Definition 1, it follows that f is strictly increasing on R |
1 | 2659-2662 | Example 7 Show that the function given by f(x) = 7x – 3 is increasing on R Solution Let x1 and x2 be any two numbers in R Then
x1 < x2 ⇒ 7x1 < 7x2 ⇒ 7x1 – 3 < 7x2 – 3 ⇒ f (x1) < f (x2)
Thus, by Definition 1, it follows that f is strictly increasing on R We shall now give the first derivative test for increasing and decreasing functions |
1 | 2660-2663 | Solution Let x1 and x2 be any two numbers in R Then
x1 < x2 ⇒ 7x1 < 7x2 ⇒ 7x1 – 3 < 7x2 – 3 ⇒ f (x1) < f (x2)
Thus, by Definition 1, it follows that f is strictly increasing on R We shall now give the first derivative test for increasing and decreasing functions The proof of this test requires the Mean Value Theorem studied in Chapter 5 |
1 | 2661-2664 | Then
x1 < x2 ⇒ 7x1 < 7x2 ⇒ 7x1 – 3 < 7x2 – 3 ⇒ f (x1) < f (x2)
Thus, by Definition 1, it follows that f is strictly increasing on R We shall now give the first derivative test for increasing and decreasing functions The proof of this test requires the Mean Value Theorem studied in Chapter 5 Theorem 1 Let f be continuous on [a, b] and differentiable on the open interval
(a,b) |
1 | 2662-2665 | We shall now give the first derivative test for increasing and decreasing functions The proof of this test requires the Mean Value Theorem studied in Chapter 5 Theorem 1 Let f be continuous on [a, b] and differentiable on the open interval
(a,b) Then
(a)
f is increasing in [a,b] if f ′(x) > 0 for each x ∈ (a, b)
(b)
f is decreasing in [a,b] if f ′(x) < 0 for each x ∈ (a, b)
(c)
f is a constant function in [a,b] if f ′(x) = 0 for each x ∈ (a, b)
Strictly Increasing function
(i)
Neither Increasing nor
Decreasing function
(iii)
Strictly Decreasing function
(ii)
Rationalised 2023-24
MATHEMATICS
154
Proof (a) Let x1, x2 ∈ [a, b] be such that x1 < x2 |
1 | 2663-2666 | The proof of this test requires the Mean Value Theorem studied in Chapter 5 Theorem 1 Let f be continuous on [a, b] and differentiable on the open interval
(a,b) Then
(a)
f is increasing in [a,b] if f ′(x) > 0 for each x ∈ (a, b)
(b)
f is decreasing in [a,b] if f ′(x) < 0 for each x ∈ (a, b)
(c)
f is a constant function in [a,b] if f ′(x) = 0 for each x ∈ (a, b)
Strictly Increasing function
(i)
Neither Increasing nor
Decreasing function
(iii)
Strictly Decreasing function
(ii)
Rationalised 2023-24
MATHEMATICS
154
Proof (a) Let x1, x2 ∈ [a, b] be such that x1 < x2 Then, by Mean Value Theorem (Theorem 8 in Chapter 5), there exists a point c
between x1 and x2 such that
f(x2) – f(x1) = f ′(c) (x2 – x1)
i |
1 | 2664-2667 | Theorem 1 Let f be continuous on [a, b] and differentiable on the open interval
(a,b) Then
(a)
f is increasing in [a,b] if f ′(x) > 0 for each x ∈ (a, b)
(b)
f is decreasing in [a,b] if f ′(x) < 0 for each x ∈ (a, b)
(c)
f is a constant function in [a,b] if f ′(x) = 0 for each x ∈ (a, b)
Strictly Increasing function
(i)
Neither Increasing nor
Decreasing function
(iii)
Strictly Decreasing function
(ii)
Rationalised 2023-24
MATHEMATICS
154
Proof (a) Let x1, x2 ∈ [a, b] be such that x1 < x2 Then, by Mean Value Theorem (Theorem 8 in Chapter 5), there exists a point c
between x1 and x2 such that
f(x2) – f(x1) = f ′(c) (x2 – x1)
i e |
1 | 2665-2668 | Then
(a)
f is increasing in [a,b] if f ′(x) > 0 for each x ∈ (a, b)
(b)
f is decreasing in [a,b] if f ′(x) < 0 for each x ∈ (a, b)
(c)
f is a constant function in [a,b] if f ′(x) = 0 for each x ∈ (a, b)
Strictly Increasing function
(i)
Neither Increasing nor
Decreasing function
(iii)
Strictly Decreasing function
(ii)
Rationalised 2023-24
MATHEMATICS
154
Proof (a) Let x1, x2 ∈ [a, b] be such that x1 < x2 Then, by Mean Value Theorem (Theorem 8 in Chapter 5), there exists a point c
between x1 and x2 such that
f(x2) – f(x1) = f ′(c) (x2 – x1)
i e f(x2) – f(x1) > 0
(as f ′(c) > 0 (given))
i |
1 | 2666-2669 | Then, by Mean Value Theorem (Theorem 8 in Chapter 5), there exists a point c
between x1 and x2 such that
f(x2) – f(x1) = f ′(c) (x2 – x1)
i e f(x2) – f(x1) > 0
(as f ′(c) > 0 (given))
i e |
1 | 2667-2670 | e f(x2) – f(x1) > 0
(as f ′(c) > 0 (given))
i e f(x2) > f (x1)
Thus, we have
1
2
1
2
1
2
(
)
(
), for all
,
[ , ]
x
x
f x
f x
x x
a b
<
Hence, f is an increasing function in [a,b] |
1 | 2668-2671 | f(x2) – f(x1) > 0
(as f ′(c) > 0 (given))
i e f(x2) > f (x1)
Thus, we have
1
2
1
2
1
2
(
)
(
), for all
,
[ , ]
x
x
f x
f x
x x
a b
<
Hence, f is an increasing function in [a,b] The proofs of part (b) and (c) are similar |
1 | 2669-2672 | e f(x2) > f (x1)
Thus, we have
1
2
1
2
1
2
(
)
(
), for all
,
[ , ]
x
x
f x
f x
x x
a b
<
Hence, f is an increasing function in [a,b] The proofs of part (b) and (c) are similar It is left as an exercise to the reader |
1 | 2670-2673 | f(x2) > f (x1)
Thus, we have
1
2
1
2
1
2
(
)
(
), for all
,
[ , ]
x
x
f x
f x
x x
a b
<
Hence, f is an increasing function in [a,b] The proofs of part (b) and (c) are similar It is left as an exercise to the reader Remarks
There is a more generalised theorem, which states that if f¢(x) > 0 for x in an interval
excluding the end points and f is continuous in the interval, then f is increasing |
1 | 2671-2674 | The proofs of part (b) and (c) are similar It is left as an exercise to the reader Remarks
There is a more generalised theorem, which states that if f¢(x) > 0 for x in an interval
excluding the end points and f is continuous in the interval, then f is increasing Similarly,
if f¢(x) < 0 for x in an interval excluding the end points and f is continuous in the
interval, then f is decreasing |
1 | 2672-2675 | It is left as an exercise to the reader Remarks
There is a more generalised theorem, which states that if f¢(x) > 0 for x in an interval
excluding the end points and f is continuous in the interval, then f is increasing Similarly,
if f¢(x) < 0 for x in an interval excluding the end points and f is continuous in the
interval, then f is decreasing Example 8 Show that the function f given by
f (x) = x3 – 3x2 + 4x, x ∈ R
is increasing on R |
1 | 2673-2676 | Remarks
There is a more generalised theorem, which states that if f¢(x) > 0 for x in an interval
excluding the end points and f is continuous in the interval, then f is increasing Similarly,
if f¢(x) < 0 for x in an interval excluding the end points and f is continuous in the
interval, then f is decreasing Example 8 Show that the function f given by
f (x) = x3 – 3x2 + 4x, x ∈ R
is increasing on R Solution Note that
f ′(x) = 3x2 – 6x + 4
= 3(x2 – 2x + 1) + 1
= 3(x – 1)2 + 1 > 0, in every interval of R
Therefore, the function f is increasing on R |
1 | 2674-2677 | Similarly,
if f¢(x) < 0 for x in an interval excluding the end points and f is continuous in the
interval, then f is decreasing Example 8 Show that the function f given by
f (x) = x3 – 3x2 + 4x, x ∈ R
is increasing on R Solution Note that
f ′(x) = 3x2 – 6x + 4
= 3(x2 – 2x + 1) + 1
= 3(x – 1)2 + 1 > 0, in every interval of R
Therefore, the function f is increasing on R Example 9 Prove that the function given by f (x) = cos x is
(a)
decreasing in (0, π)
(b)
increasing in (π, 2π), and
(c)
neither increasing nor decreasing in (0, 2π) |
1 | 2675-2678 | Example 8 Show that the function f given by
f (x) = x3 – 3x2 + 4x, x ∈ R
is increasing on R Solution Note that
f ′(x) = 3x2 – 6x + 4
= 3(x2 – 2x + 1) + 1
= 3(x – 1)2 + 1 > 0, in every interval of R
Therefore, the function f is increasing on R Example 9 Prove that the function given by f (x) = cos x is
(a)
decreasing in (0, π)
(b)
increasing in (π, 2π), and
(c)
neither increasing nor decreasing in (0, 2π) Rationalised 2023-24
APPLICATION OF DERIVATIVES
155
Fig 6 |
1 | 2676-2679 | Solution Note that
f ′(x) = 3x2 – 6x + 4
= 3(x2 – 2x + 1) + 1
= 3(x – 1)2 + 1 > 0, in every interval of R
Therefore, the function f is increasing on R Example 9 Prove that the function given by f (x) = cos x is
(a)
decreasing in (0, π)
(b)
increasing in (π, 2π), and
(c)
neither increasing nor decreasing in (0, 2π) Rationalised 2023-24
APPLICATION OF DERIVATIVES
155
Fig 6 4
Solution Note that f ′(x) = – sin x
(a)
Since for each x ∈ (0, π), sin x > 0, we have f ′(x) < 0 and so f is decreasing in
(0, π) |
1 | 2677-2680 | Example 9 Prove that the function given by f (x) = cos x is
(a)
decreasing in (0, π)
(b)
increasing in (π, 2π), and
(c)
neither increasing nor decreasing in (0, 2π) Rationalised 2023-24
APPLICATION OF DERIVATIVES
155
Fig 6 4
Solution Note that f ′(x) = – sin x
(a)
Since for each x ∈ (0, π), sin x > 0, we have f ′(x) < 0 and so f is decreasing in
(0, π) (b)
Since for each x ∈ (π, 2π), sin x < 0, we have f ′(x) > 0 and so f is increasing in
(π, 2π) |
1 | 2678-2681 | Rationalised 2023-24
APPLICATION OF DERIVATIVES
155
Fig 6 4
Solution Note that f ′(x) = – sin x
(a)
Since for each x ∈ (0, π), sin x > 0, we have f ′(x) < 0 and so f is decreasing in
(0, π) (b)
Since for each x ∈ (π, 2π), sin x < 0, we have f ′(x) > 0 and so f is increasing in
(π, 2π) (c)
Clearly by (a) and (b) above, f is neither increasing nor decreasing in (0, 2π) |
1 | 2679-2682 | 4
Solution Note that f ′(x) = – sin x
(a)
Since for each x ∈ (0, π), sin x > 0, we have f ′(x) < 0 and so f is decreasing in
(0, π) (b)
Since for each x ∈ (π, 2π), sin x < 0, we have f ′(x) > 0 and so f is increasing in
(π, 2π) (c)
Clearly by (a) and (b) above, f is neither increasing nor decreasing in (0, 2π) Example 10 Find the intervals in which the function f given by f(x) = x2 – 4x + 6 is
(a) increasing
(b) decreasing
Solution We have
f (x)
= x2 – 4x + 6
or
f ′(x) = 2x – 4
Therefore, f ′(x) = 0 gives x = 2 |
1 | 2680-2683 | (b)
Since for each x ∈ (π, 2π), sin x < 0, we have f ′(x) > 0 and so f is increasing in
(π, 2π) (c)
Clearly by (a) and (b) above, f is neither increasing nor decreasing in (0, 2π) Example 10 Find the intervals in which the function f given by f(x) = x2 – 4x + 6 is
(a) increasing
(b) decreasing
Solution We have
f (x)
= x2 – 4x + 6
or
f ′(x) = 2x – 4
Therefore, f ′(x) = 0 gives x = 2 Now the point x = 2 divides the real line into two
disjoint intervals namely, (– ∞, 2) and (2, ∞) (Fig 6 |
1 | 2681-2684 | (c)
Clearly by (a) and (b) above, f is neither increasing nor decreasing in (0, 2π) Example 10 Find the intervals in which the function f given by f(x) = x2 – 4x + 6 is
(a) increasing
(b) decreasing
Solution We have
f (x)
= x2 – 4x + 6
or
f ′(x) = 2x – 4
Therefore, f ′(x) = 0 gives x = 2 Now the point x = 2 divides the real line into two
disjoint intervals namely, (– ∞, 2) and (2, ∞) (Fig 6 3) |
1 | 2682-2685 | Example 10 Find the intervals in which the function f given by f(x) = x2 – 4x + 6 is
(a) increasing
(b) decreasing
Solution We have
f (x)
= x2 – 4x + 6
or
f ′(x) = 2x – 4
Therefore, f ′(x) = 0 gives x = 2 Now the point x = 2 divides the real line into two
disjoint intervals namely, (– ∞, 2) and (2, ∞) (Fig 6 3) In the interval (– ∞, 2), f ′(x) = 2x
– 4 < 0 |
1 | 2683-2686 | Now the point x = 2 divides the real line into two
disjoint intervals namely, (– ∞, 2) and (2, ∞) (Fig 6 3) In the interval (– ∞, 2), f ′(x) = 2x
– 4 < 0 Therefore, f is decreasing in this interval |
1 | 2684-2687 | 3) In the interval (– ∞, 2), f ′(x) = 2x
– 4 < 0 Therefore, f is decreasing in this interval Also, in the interval (2,
)
∞ ,
( )
0
f
x >
′
and so the function f is increasing in this interval |
1 | 2685-2688 | In the interval (– ∞, 2), f ′(x) = 2x
– 4 < 0 Therefore, f is decreasing in this interval Also, in the interval (2,
)
∞ ,
( )
0
f
x >
′
and so the function f is increasing in this interval Example 11 Find the intervals in which the function f given by f (x) = 4x3 – 6x2 – 72x
+ 30 is (a) increasing (b) decreasing |
1 | 2686-2689 | Therefore, f is decreasing in this interval Also, in the interval (2,
)
∞ ,
( )
0
f
x >
′
and so the function f is increasing in this interval Example 11 Find the intervals in which the function f given by f (x) = 4x3 – 6x2 – 72x
+ 30 is (a) increasing (b) decreasing Solution We have
f (x)
= 4x3 – 6x2 – 72x + 30
or
f ′(x) = 12x2 – 12x – 72
= 12(x2 – x – 6)
= 12(x – 3) (x + 2)
Therefore, f ′(x) = 0 gives x = – 2, 3 |
1 | 2687-2690 | Also, in the interval (2,
)
∞ ,
( )
0
f
x >
′
and so the function f is increasing in this interval Example 11 Find the intervals in which the function f given by f (x) = 4x3 – 6x2 – 72x
+ 30 is (a) increasing (b) decreasing Solution We have
f (x)
= 4x3 – 6x2 – 72x + 30
or
f ′(x) = 12x2 – 12x – 72
= 12(x2 – x – 6)
= 12(x – 3) (x + 2)
Therefore, f ′(x) = 0 gives x = – 2, 3 The
points x = – 2 and x = 3 divides the real line into
three disjoint intervals, namely, (– ∞, – 2), (– 2, 3)
and (3, ∞) |
1 | 2688-2691 | Example 11 Find the intervals in which the function f given by f (x) = 4x3 – 6x2 – 72x
+ 30 is (a) increasing (b) decreasing Solution We have
f (x)
= 4x3 – 6x2 – 72x + 30
or
f ′(x) = 12x2 – 12x – 72
= 12(x2 – x – 6)
= 12(x – 3) (x + 2)
Therefore, f ′(x) = 0 gives x = – 2, 3 The
points x = – 2 and x = 3 divides the real line into
three disjoint intervals, namely, (– ∞, – 2), (– 2, 3)
and (3, ∞) Fig 6 |
1 | 2689-2692 | Solution We have
f (x)
= 4x3 – 6x2 – 72x + 30
or
f ′(x) = 12x2 – 12x – 72
= 12(x2 – x – 6)
= 12(x – 3) (x + 2)
Therefore, f ′(x) = 0 gives x = – 2, 3 The
points x = – 2 and x = 3 divides the real line into
three disjoint intervals, namely, (– ∞, – 2), (– 2, 3)
and (3, ∞) Fig 6 3
Rationalised 2023-24
MATHEMATICS
156
In the intervals (– ∞, – 2) and (3, ∞), f ′(x) is positive while in the interval (– 2, 3),
f ′(x) is negative |
1 | 2690-2693 | The
points x = – 2 and x = 3 divides the real line into
three disjoint intervals, namely, (– ∞, – 2), (– 2, 3)
and (3, ∞) Fig 6 3
Rationalised 2023-24
MATHEMATICS
156
In the intervals (– ∞, – 2) and (3, ∞), f ′(x) is positive while in the interval (– 2, 3),
f ′(x) is negative Consequently, the function f is increasing in the intervals
(– ∞, – 2) and (3, ∞) while the function is decreasing in the interval (– 2, 3) |
1 | 2691-2694 | Fig 6 3
Rationalised 2023-24
MATHEMATICS
156
In the intervals (– ∞, – 2) and (3, ∞), f ′(x) is positive while in the interval (– 2, 3),
f ′(x) is negative Consequently, the function f is increasing in the intervals
(– ∞, – 2) and (3, ∞) while the function is decreasing in the interval (– 2, 3) However,
f is neither increasing nor decreasing in R |
1 | 2692-2695 | 3
Rationalised 2023-24
MATHEMATICS
156
In the intervals (– ∞, – 2) and (3, ∞), f ′(x) is positive while in the interval (– 2, 3),
f ′(x) is negative Consequently, the function f is increasing in the intervals
(– ∞, – 2) and (3, ∞) while the function is decreasing in the interval (– 2, 3) However,
f is neither increasing nor decreasing in R Interval
Sign of f ′(x)
Nature of function f
(– ∞, – 2)
(–) (–) > 0
f is increasing
(– 2, 3)
(–) (+) < 0
f is decreasing
(3, ∞)
(+) (+) > 0
f is increasing
Example 12 Find intervals in which the function given by f (x) = sin 3x, x ∈
0 2
, π is
(a) increasing (b) decreasing |
1 | 2693-2696 | Consequently, the function f is increasing in the intervals
(– ∞, – 2) and (3, ∞) while the function is decreasing in the interval (– 2, 3) However,
f is neither increasing nor decreasing in R Interval
Sign of f ′(x)
Nature of function f
(– ∞, – 2)
(–) (–) > 0
f is increasing
(– 2, 3)
(–) (+) < 0
f is decreasing
(3, ∞)
(+) (+) > 0
f is increasing
Example 12 Find intervals in which the function given by f (x) = sin 3x, x ∈
0 2
, π is
(a) increasing (b) decreasing Solution We have
f (x) = sin 3x
or
f ′(x) = 3cos 3x
Therefore, f′(x) = 0 gives cos 3x = 0 which in turn gives
3
3
2,
2
x
π
π
=
(as x ∈
0 2
, π
implies
3
3
0, 2
x
π
∈
) |
1 | 2694-2697 | However,
f is neither increasing nor decreasing in R Interval
Sign of f ′(x)
Nature of function f
(– ∞, – 2)
(–) (–) > 0
f is increasing
(– 2, 3)
(–) (+) < 0
f is decreasing
(3, ∞)
(+) (+) > 0
f is increasing
Example 12 Find intervals in which the function given by f (x) = sin 3x, x ∈
0 2
, π is
(a) increasing (b) decreasing Solution We have
f (x) = sin 3x
or
f ′(x) = 3cos 3x
Therefore, f′(x) = 0 gives cos 3x = 0 which in turn gives
3
3
2,
2
x
π
π
=
(as x ∈
0 2
, π
implies
3
3
0, 2
x
π
∈
) So
6
x
=π
and 2
π |
1 | 2695-2698 | Interval
Sign of f ′(x)
Nature of function f
(– ∞, – 2)
(–) (–) > 0
f is increasing
(– 2, 3)
(–) (+) < 0
f is decreasing
(3, ∞)
(+) (+) > 0
f is increasing
Example 12 Find intervals in which the function given by f (x) = sin 3x, x ∈
0 2
, π is
(a) increasing (b) decreasing Solution We have
f (x) = sin 3x
or
f ′(x) = 3cos 3x
Therefore, f′(x) = 0 gives cos 3x = 0 which in turn gives
3
3
2,
2
x
π
π
=
(as x ∈
0 2
, π
implies
3
3
0, 2
x
π
∈
) So
6
x
=π
and 2
π The point
6
x
=π
divides the interval 0 2
, π
into two disjoint intervals 0, 6
π
and π π
6 2
,
|
1 | 2696-2699 | Solution We have
f (x) = sin 3x
or
f ′(x) = 3cos 3x
Therefore, f′(x) = 0 gives cos 3x = 0 which in turn gives
3
3
2,
2
x
π
π
=
(as x ∈
0 2
, π
implies
3
3
0, 2
x
π
∈
) So
6
x
=π
and 2
π The point
6
x
=π
divides the interval 0 2
, π
into two disjoint intervals 0, 6
π
and π π
6 2
,
Now,
( )
0
f
′x >
for all
0, 6
x
π
∈
as 0
0
3
6
2
x
x
π
π
≤
<
⇒
≤
<
and
( )
0
f
′x <
for
all
,
6 2
x
π π
∈
as
3
3
6
2
2
2
x
x
π
π
π
π
<
<
⇒
<
< |
1 | 2697-2700 | So
6
x
=π
and 2
π The point
6
x
=π
divides the interval 0 2
, π
into two disjoint intervals 0, 6
π
and π π
6 2
,
Now,
( )
0
f
′x >
for all
0, 6
x
π
∈
as 0
0
3
6
2
x
x
π
π
≤
<
⇒
≤
<
and
( )
0
f
′x <
for
all
,
6 2
x
π π
∈
as
3
3
6
2
2
2
x
x
π
π
π
π
<
<
⇒
<
< Therefore, f is increasing in 0, 6
π
and decreasing in
,
6 2
π π
|
1 | 2698-2701 | The point
6
x
=π
divides the interval 0 2
, π
into two disjoint intervals 0, 6
π
and π π
6 2
,
Now,
( )
0
f
′x >
for all
0, 6
x
π
∈
as 0
0
3
6
2
x
x
π
π
≤
<
⇒
≤
<
and
( )
0
f
′x <
for
all
,
6 2
x
π π
∈
as
3
3
6
2
2
2
x
x
π
π
π
π
<
<
⇒
<
< Therefore, f is increasing in 0, 6
π
and decreasing in
,
6 2
π π
Fig 6 |
1 | 2699-2702 | Now,
( )
0
f
′x >
for all
0, 6
x
π
∈
as 0
0
3
6
2
x
x
π
π
≤
<
⇒
≤
<
and
( )
0
f
′x <
for
all
,
6 2
x
π π
∈
as
3
3
6
2
2
2
x
x
π
π
π
π
<
<
⇒
<
< Therefore, f is increasing in 0, 6
π
and decreasing in
,
6 2
π π
Fig 6 5
Rationalised 2023-24
APPLICATION OF DERIVATIVES
157
Also, the given function is continuous at x = 0 and
6
x
=π |
1 | 2700-2703 | Therefore, f is increasing in 0, 6
π
and decreasing in
,
6 2
π π
Fig 6 5
Rationalised 2023-24
APPLICATION OF DERIVATIVES
157
Also, the given function is continuous at x = 0 and
6
x
=π Therefore, by Theorem 1,
f is increasing on 0 6
, π
and decreasing on
π π
6 2
,
|
1 | 2701-2704 | Fig 6 5
Rationalised 2023-24
APPLICATION OF DERIVATIVES
157
Also, the given function is continuous at x = 0 and
6
x
=π Therefore, by Theorem 1,
f is increasing on 0 6
, π
and decreasing on
π π
6 2
,
Example 13 Find the intervals in which the function f given by
f (x) = sin x + cos x, 0 ≤ x ≤ 2π
is increasing or decreasing |
1 | 2702-2705 | 5
Rationalised 2023-24
APPLICATION OF DERIVATIVES
157
Also, the given function is continuous at x = 0 and
6
x
=π Therefore, by Theorem 1,
f is increasing on 0 6
, π
and decreasing on
π π
6 2
,
Example 13 Find the intervals in which the function f given by
f (x) = sin x + cos x, 0 ≤ x ≤ 2π
is increasing or decreasing Solution We have
f(x) = sin x + cos x,
or
f ′(x) = cos x – sin x
Now
( )
0
f
′x =
gives sin x = cos x which gives that
4
x
=π
, 5
4
π as 0
2
≤x
≤ π
The points
4
x
=π
and
5
4
x
=π
divide the interval [0, 2π] into three disjoint intervals,
namely, 0, 4
π
, π
π
4
5
,4
and 5 ,2
π4
π
|
1 | 2703-2706 | Therefore, by Theorem 1,
f is increasing on 0 6
, π
and decreasing on
π π
6 2
,
Example 13 Find the intervals in which the function f given by
f (x) = sin x + cos x, 0 ≤ x ≤ 2π
is increasing or decreasing Solution We have
f(x) = sin x + cos x,
or
f ′(x) = cos x – sin x
Now
( )
0
f
′x =
gives sin x = cos x which gives that
4
x
=π
, 5
4
π as 0
2
≤x
≤ π
The points
4
x
=π
and
5
4
x
=π
divide the interval [0, 2π] into three disjoint intervals,
namely, 0, 4
π
, π
π
4
5
,4
and 5 ,2
π4
π
Note that
5
( )
0 if
0,
,2
4
4
f
x
x
π
π
′
>
∈
∪
π
or
f is increasing in the intervals 0, 4
π
and
π45 ,2
π
Also
′
<
∈
f
x
x
( )
,
0
4
5
4
if
π
π
or
f is decreasing in
π
π
4
5
,4
Fig 6 |
1 | 2704-2707 | Example 13 Find the intervals in which the function f given by
f (x) = sin x + cos x, 0 ≤ x ≤ 2π
is increasing or decreasing Solution We have
f(x) = sin x + cos x,
or
f ′(x) = cos x – sin x
Now
( )
0
f
′x =
gives sin x = cos x which gives that
4
x
=π
, 5
4
π as 0
2
≤x
≤ π
The points
4
x
=π
and
5
4
x
=π
divide the interval [0, 2π] into three disjoint intervals,
namely, 0, 4
π
, π
π
4
5
,4
and 5 ,2
π4
π
Note that
5
( )
0 if
0,
,2
4
4
f
x
x
π
π
′
>
∈
∪
π
or
f is increasing in the intervals 0, 4
π
and
π45 ,2
π
Also
′
<
∈
f
x
x
( )
,
0
4
5
4
if
π
π
or
f is decreasing in
π
π
4
5
,4
Fig 6 6
Rationalised 2023-24
MATHEMATICS
158
Interval
Sign of
f( )
′x
Nature of function
0, 4
π
> 0
f is increasing
π
π
4
5
,4
< 0
f is decreasing
π45 ,2
π
> 0
f is increasing
EXERCISE 6 |
1 | 2705-2708 | Solution We have
f(x) = sin x + cos x,
or
f ′(x) = cos x – sin x
Now
( )
0
f
′x =
gives sin x = cos x which gives that
4
x
=π
, 5
4
π as 0
2
≤x
≤ π
The points
4
x
=π
and
5
4
x
=π
divide the interval [0, 2π] into three disjoint intervals,
namely, 0, 4
π
, π
π
4
5
,4
and 5 ,2
π4
π
Note that
5
( )
0 if
0,
,2
4
4
f
x
x
π
π
′
>
∈
∪
π
or
f is increasing in the intervals 0, 4
π
and
π45 ,2
π
Also
′
<
∈
f
x
x
( )
,
0
4
5
4
if
π
π
or
f is decreasing in
π
π
4
5
,4
Fig 6 6
Rationalised 2023-24
MATHEMATICS
158
Interval
Sign of
f( )
′x
Nature of function
0, 4
π
> 0
f is increasing
π
π
4
5
,4
< 0
f is decreasing
π45 ,2
π
> 0
f is increasing
EXERCISE 6 2
1 |
1 | 2706-2709 | Note that
5
( )
0 if
0,
,2
4
4
f
x
x
π
π
′
>
∈
∪
π
or
f is increasing in the intervals 0, 4
π
and
π45 ,2
π
Also
′
<
∈
f
x
x
( )
,
0
4
5
4
if
π
π
or
f is decreasing in
π
π
4
5
,4
Fig 6 6
Rationalised 2023-24
MATHEMATICS
158
Interval
Sign of
f( )
′x
Nature of function
0, 4
π
> 0
f is increasing
π
π
4
5
,4
< 0
f is decreasing
π45 ,2
π
> 0
f is increasing
EXERCISE 6 2
1 Show that the function given by f (x) = 3x + 17 is increasing on R |
1 | 2707-2710 | 6
Rationalised 2023-24
MATHEMATICS
158
Interval
Sign of
f( )
′x
Nature of function
0, 4
π
> 0
f is increasing
π
π
4
5
,4
< 0
f is decreasing
π45 ,2
π
> 0
f is increasing
EXERCISE 6 2
1 Show that the function given by f (x) = 3x + 17 is increasing on R 2 |
1 | 2708-2711 | 2
1 Show that the function given by f (x) = 3x + 17 is increasing on R 2 Show that the function given by f (x) = e2x is increasing on R |
1 | 2709-2712 | Show that the function given by f (x) = 3x + 17 is increasing on R 2 Show that the function given by f (x) = e2x is increasing on R 3 |
1 | 2710-2713 | 2 Show that the function given by f (x) = e2x is increasing on R 3 Show that the function given by f (x) = sin x is
(a)
increasing in 0, 2
π
(b)
decreasing in
π2,
π
(c) neither increasing nor decreasing in (0, π)
4 |
1 | 2711-2714 | Show that the function given by f (x) = e2x is increasing on R 3 Show that the function given by f (x) = sin x is
(a)
increasing in 0, 2
π
(b)
decreasing in
π2,
π
(c) neither increasing nor decreasing in (0, π)
4 Find the intervals in which the function f given by f (x) = 2x2 – 3x is
(a)
increasing
(b)
decreasing
5 |
1 | 2712-2715 | 3 Show that the function given by f (x) = sin x is
(a)
increasing in 0, 2
π
(b)
decreasing in
π2,
π
(c) neither increasing nor decreasing in (0, π)
4 Find the intervals in which the function f given by f (x) = 2x2 – 3x is
(a)
increasing
(b)
decreasing
5 Find the intervals in which the function f given by f(x) = 2x3 – 3x2 – 36x + 7 is
(a)
increasing
(b)
decreasing
6 |
1 | 2713-2716 | Show that the function given by f (x) = sin x is
(a)
increasing in 0, 2
π
(b)
decreasing in
π2,
π
(c) neither increasing nor decreasing in (0, π)
4 Find the intervals in which the function f given by f (x) = 2x2 – 3x is
(a)
increasing
(b)
decreasing
5 Find the intervals in which the function f given by f(x) = 2x3 – 3x2 – 36x + 7 is
(a)
increasing
(b)
decreasing
6 Find the intervals in which the following functions are strictly increasing or
decreasing:
(a)
x2 + 2x – 5
(b)
10 – 6x – 2x2
(c)
–2x3 – 9x2 – 12x + 1
(d)
6 – 9x – x2
(e)
(x + 1)3 (x – 3)3
7 |
1 | 2714-2717 | Find the intervals in which the function f given by f (x) = 2x2 – 3x is
(a)
increasing
(b)
decreasing
5 Find the intervals in which the function f given by f(x) = 2x3 – 3x2 – 36x + 7 is
(a)
increasing
(b)
decreasing
6 Find the intervals in which the following functions are strictly increasing or
decreasing:
(a)
x2 + 2x – 5
(b)
10 – 6x – 2x2
(c)
–2x3 – 9x2 – 12x + 1
(d)
6 – 9x – x2
(e)
(x + 1)3 (x – 3)3
7 Show that
2
log(1
)
2
x
y
x
x
=
+
−
+
, x > – 1, is an increasing function of x
throughout its domain |
1 | 2715-2718 | Find the intervals in which the function f given by f(x) = 2x3 – 3x2 – 36x + 7 is
(a)
increasing
(b)
decreasing
6 Find the intervals in which the following functions are strictly increasing or
decreasing:
(a)
x2 + 2x – 5
(b)
10 – 6x – 2x2
(c)
–2x3 – 9x2 – 12x + 1
(d)
6 – 9x – x2
(e)
(x + 1)3 (x – 3)3
7 Show that
2
log(1
)
2
x
y
x
x
=
+
−
+
, x > – 1, is an increasing function of x
throughout its domain 8 |
1 | 2716-2719 | Find the intervals in which the following functions are strictly increasing or
decreasing:
(a)
x2 + 2x – 5
(b)
10 – 6x – 2x2
(c)
–2x3 – 9x2 – 12x + 1
(d)
6 – 9x – x2
(e)
(x + 1)3 (x – 3)3
7 Show that
2
log(1
)
2
x
y
x
x
=
+
−
+
, x > – 1, is an increasing function of x
throughout its domain 8 Find the values of x for which y = [x(x – 2)]2 is an increasing function |
1 | 2717-2720 | Show that
2
log(1
)
2
x
y
x
x
=
+
−
+
, x > – 1, is an increasing function of x
throughout its domain 8 Find the values of x for which y = [x(x – 2)]2 is an increasing function 9 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.