(b) find the interval in which 2x ^ 3 + 9x ^ 2 + 12x - 1 is strictly increasing. (kerala b. 2022)

The interval on which the function f(x) = 2x3 + 9x2 + 12x – 1 is decreasing is [– 2, – 1].

Explanation:

The given function is f(x) = 2x3 + 9x2 + 12x – 1

f'(x) = 6x2 + 18x + 12

For increasing and decreasing f'(x) = 0

∴ 6x2 + 18x + 12 = 0

⇒ x2 + 3x + 2 = 0

⇒ x2 + 2x + x + 2 = 0

⇒ x(x + 2) + 1(x + 2) = 0

⇒ (x + 2)(x + 1) = 0

⇒ x = – 2, x = – 1

The possible intervals are `(–oo, – 2), (– 2, – 1), (– 1, oo)`

Now f'(x) = (x + 2) (x + 1)

⇒ `"f'"(x)_((-oo"," -2))` = (–) (–) = (+) increasing

⇒ `"f'"(x)_((-2"," -1))` = (+) (–) = (–) decreasing

⇒ `"f'"(x)_((-1"," oo))` = (+) (+) = (+) increasing.

Gujarat Board Textbook Solutions Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2

Question 1. Show that the function given by f(x) = 3x + 17 is strictly increasing on R. Solution: f(x) = 3x + 17. ∴ f'(x) = 3 = + ve for all x ∈ R.

⇒ f is strictly increasing on R.

Question 2.
Show that the function given by f(x) = e2x is strictly increasing on R. Solution:

We have: f(x) = e2x

⇒ f'(x) = 22x Case I : When x > 0, then

f’ (x) = 22x

= 2[ 1 + (2x) + \(\frac{(2 x)^{2}}{2 !}+\frac{(2 x)^{3}}{3 !}+\ldots\) ]

∴ f'(x) > 0 for all x > 0.

Case II : When x = 0, then
f'(x) = 2e0 = 2
∴ f'(x) > 0 for x = 0.

Case III : When x < 0. Let x = – y, where y is a positive quantity.

∴ f’ (x) = 2e-2y = \(\frac{2}{e^{2 y}}\)

= \(\frac{2}{\text { Some +ve quantity }}\) > 0 ⇒ f'(x) > 0 for x < 0. Thus, f'(x) > 0 for all x ∈R.

Hence, e2x is strictly increasing on R.

Question 3. Show that the function given by f(x) = sin x is (a) strictly increasing in (0, \(\frac{\pi}{2}\)) (b) strictly decreasing in (\(\frac{\pi}{2}\), 0) (c) neither increasing nor decreasing in (0, π). Solution: We have: f(x) = sin x. ∴ f'(x) = cos x (a) f'(x) = cos x is positive in the interval (0, \(\frac{\pi}{2}\))

⇒ f(x) is strictly increasing in (0, \(\frac{\pi}{2}\))

(b) f(x) cos x is negative in the interval (\(\frac{\pi}{2}\), π)
⇒ f(x) is strictly decreasing in (\(\frac{\pi}{2}\), π)

(c) f’ (x) – cos x is +ve in the interval (0, \(\frac{\pi}{2}\)), while f'(x) is -ve in the interval (\(\frac{\pi}{2}\), π) ⇒ f’ (x) does not have the same sign in the interval (0, π)

Hence, f(x) is neither increasing nor decreasing in (0, π).

Question 4.
Find the intervals in which the function f given by, f(x) = 2x2 – 3x is (a) strictly increasing, (b) strictly decreasing. Solution:

f(x) = 2x2 – 3x

∴ f'(x) = 4x – 3. f'(x) = 0 at x = \(\frac{3}{4}\)

The point x = \(\frac{3}{4}\) divides the real line into two disjoint intervals

In the interval (-∞, \(\frac{3}{4}\)), f’ (x) is negative. Therefore, f is strictly decreasing in (-∞, \(\frac{3}{4}\))
In the interval (-∞, \(\frac{3}{4}\)), f’ is positive. Hence, f is strictly increasing in the interval (\(\frac{3}{4}\), ∞)

Question 5.
Find the intervals in which the function f given by f(x) = 2x3 – 3x2 – 36x + 7 is (a)strictly increasing, (b) strictly decreasing. Solution:

f(x) = 2x3 – 3x2 – 36x + 7

∴ f (x) = 6x2 – 6x – 36 = 6(x2 – x – 6) = 6(x -3)(x + 2) ⇒ f'(x) = 0 at x = 3 and x = – 2.

The points x = 3, x = -2 divide the real line into three disjoint intervals viz (-∞, – 2), (- 2, 3) and (3, ∞).

Now, f'(x) is +ve in the intervals (-∞, – 2) and (3, ∞), since in the interval (-∞, – 2), each factor x – 3, x + 2 is negative. ∴ f’ (x) = + ve (a) f is strictly increasing in (-∞, – 2) u (3, ∞). (b) In the interval (- 2, 3). x + 2 is +ve and x – 3 is -ve. ∴ f’ (x) = 6 (x – 3)(x + 2) = + x – = -ve.

∴ f is strictly decreasing in the interval (- 2, 3).

Question 6. Find the intervals in which the following functions are strictly increasing or decreasing :

(a) x2 + 2x – 5

(b) 10 – 6x – 2x2
(c) -2x2 – 9x2 – 12x + 1
(d) 6 – 9x – x2
(e) (x + 1)2(x – 3)3 Solution:

(a) We have : f(x) = x2 + 2x – 5.

∴ f’ (x) = 2x + 2 = 2(x + 1) The function/(x) will be increasing, if f ‘(x) > 0, i.e., if 2(x + 1) > 0 ⇒ x + 1 > 0. ⇒ x > – 1. The function/(x) will be decreasing if f'(x) < 0, i.e., 2(x + 1) < 0 ⇒ (x + 1) < 0. ⇒ x < -1.

Hence, f(x) is strictly increasing in (- 1, ∞) and strictly decreasing in (-∞, – 1).

(b) We have: f(x) = 10 – 6x – 2x2. ∴ f'(x) = -6 – 4x = -2(3 + 2x) Now, f(x) is increasing, if f'(x) > 0, i. e., – 6 – 4x > 0. i. e., – 4x > 6 ⇒ x < –\(\frac{3}{2}\) and f'{x) is decreasing, if f'(x) < 0, i.e., if – 6 – 4x < 0, i.e., – 4x < 6 ⇒ x > \(\frac{-3}{2}\)

Hence, f(x) is strictly increasing for x < \(\frac{-3}{2}\), i.e., in the interval (-∞, \(\frac{-3}{2}\)) and strictly decreasing for x > \(\frac{-3}{2}\) , i.e., in the interval (\(\frac{-3}{2}\), ∞)

(c) Let f(x) = – 2x3 – 9x2 – 12x + 1.
∴ f'(x) = -6x2 – 18x – 12
= – 6(x2 + 3x + 2) = – 6((x + 1)(x + 2). f'(x) = 0 gives x = -1 or x = -2. The points x = -2 and x = -1 (arranged in ascending order) divide the real line into three disjoint intervals namely, (- ∞, – 2), (- 2, -1) and (-1, ∞). In the interval (-∞, – 2), i.e., -∞ < x < – 2, (x + 1) and (x + 2) are negative. ∴ f'(x) = (-)(-)(-) = -ve. ⇒ f(x) is strictly decreasing in (-∞, – 2). In the interval (- 2, – 1), i.e., – 2 < x < – 1, (x + 1) is -ve and (x + 2) is positive. ∴ f'(x) = (-)(-)(+) = +ve. ⇒ f(x) is strictly increasing in (- 2, -1). In the interval (- 1, ∞), i.e., – 1 < x < ∞, (x + 1) and (x + 2) are both positive. ∴ f(‘x) = (-)(+)(+) = -ve. ⇒ f(x) is strictly decreasing in (- 1, ∞).

Hence, f(x) is strictly increasing for – 2 < x < -1 and strictly decreasing for x < -2 and x > – 1.

(d) We have: f(x) – 6 – 9x – x2 ∴ f'(x) = – 9 – 2x Now, f(x) is increasing, if f'(x) > 0, i.e., if – 9 – 2x > 0, i.e., if -2x > 9 ⇒ x < \(\frac{-9}{2}\) and f(x) is decreasing, if f{x) < 0, i.e., if – 9 – 2x < 0, i.e., -2x < 9 ⇒ x > \(\frac{-9}{2}\).

Hence, f(x) is strictly increasing for x < \(\frac{-9}{2}\) and strictly decreasing for x > \(\frac{-9}{2}\).

(e) We have: f(x) = (x + 1)3 (x – 3)3
∴ f'(x) = 3(x + 1)2 [\(\frac{d}{d x}\) (x + 1)] . (x – 3)3 + (x + 1)3 . 3(x – 3)2 . \(\frac{d}{d x}\)(x – 3)
= 3(x + 1)2 (x – 3)3 + 3(x + 1)3(x – 3)2
= 3(x + 1)2(x – 3)2(x – 3 + x + 1)
= 6(x + 1)2(x – 3)2(x – 1). For f{x) to be increasing : . f’ (x) > o.

⇒ 6(x + 1)2 (x – 3)2 (x – 1) > 0

⇒ (x – 1) > 0 [∵ 6(x + 1)2(x – 3)2 > 0] ⇒ x > 1 But f ‘(x) = 0 at x = 3 . ⇒ f is strictly increasing in (1, 3) and (3, ∞) So, f is strictly increasing in (1, 3) ∪ (3, ∞). For f(x) to be decreasing : f’ (x) < 0

⇒ 6(x + 1)2 (x – 3)2 (x – 1) < 0

⇒ (x – 1) < 0 [∵ 6(x + 1)2(x – 3)2 > 0] ⇒ x < 1. But f ‘(x) = 0 at x = -1.

⇒ f is strictly decreasing in (-∞, – 1) and (-1, 1). So, f is strictly decreasing in (-∞, – 1) u (-1, 1).

Question 7. Show that y = log (1 + x) – \(\frac{2 x}{2+x}\), x > -1 is an increasing function of x throughout its domain.

Solution:

Hence, y = log x – \(\frac{2 x}{2+x}\) is an increasing function of x for all values of x > – 1.

Note : From now onwards, we shall be using the terms increasing and decreasing for strictly increasing and strictly decreasing respectively, unless stated otherwise.

Question 8.
Find the values of x for which y = [x(x – 2)]2 is an increasing function. Solution:

Let y = [x(x – 2)]2 = x2(x2 – 4x + 4)

= x4 – 4x3 + 4x2
∴ \(\frac{d y}{d x}\) = 4x3 – 12x2 + 8x. For the function to be increasing : \(\frac{d y}{d x}\) > 0

⇒ 4x3 – 12x2 + 8x > 0

⇒ 4x(x2 – 3x + 2) > 0 ⇒ 4x(x – 1)(x – 2) > 0 Now, \(\frac{d y}{d x}\) = 0 gives x = 0 or x = 1 or x = 2. These values divide the real line into three disjoint intervals, namely x < 0, 0 < x < 1 and x > 2. ∴ For 0 < x < 1, \(\frac{d y}{d x}\) = (+)(-)(-) = +ve and for x > 2, \(\frac{d y}{d x}\) = (+)(+)(+) = +ve Thus, the function is increasing for 0 < x < 1 and x > 2.

For x < 0, \(\frac{d y}{d x}\) = (-)(-)(-) = -ve so, the function is decreasing in x < 0.

Question 9. Prove that y = \(\frac{4 \sin \theta}{(2+\cos \theta)}\) – θ is an increasing function of θ in [0, \(\frac{\pi}{2}\)] Solution:

We have, y = \(\frac{4 \sin \theta}{(2+\cos \theta)}\) – θ

Question 10. Prove that the logarthmic function is strictly increasing in (0, ∞). Solution: Let f(x) = log x. So, we shall prove that f(x) is an increasing function for x > 0. Now, f'(x) = \(\frac{1}{x}\) Clearly, when x takes the values > 0, we observe that ∴ \(\frac{1}{2}\) > 0. ⇒ f’(x) > 0. Hence, f(Y) is an increasing function for x > 0. .

i.e., f(x) is an increasing function whenever it is defined, i.e., on (0, ∞).

Question 11.
Prove that the function f given by f(x) = x2 – x + 1 is neither strictly increasing nor strictly decreasing on (- 1, 1). Solution:

f(x) = x2 – x + 1

∴ f'(x) = 2x – 1 = 2(x – \(\frac{1}{2}\)) Now, – 1 < x < \(\frac{1}{2}\) ⇒ (x – \(\frac{1}{2}\))< 0. ⇒ 2(x – \(\frac{1}{2}\)) < 0 ⇒ f'(x)<0. and, \(\frac{1}{2}\) < x < 1 ⇒ x – \(\frac{1}{2}\) > 0. 2(x – \(\frac{1}{2}\)) > 0 ⇒ f'(x) > 0. Thus, f'(x) does not have the same sign throughout the interval (-1, 1).

Hence, f(x) is neither increasing nor decreasing in (-1, 1).

Question 12. Which of the following functions are strictly decreasing on (0, \(\frac{\pi}{2}\)) (A) cos x (B) cos 2x (C) cos 3x (D) tan x. Solution: (A) We have: f(x) = cos x f'(x) = – sin x For 0 < x < \(\frac{\pi}{2}\), sin x > 0 ∴ f’ (x) = – sin x < 0 in (0, \(\frac{\pi}{2}\))

∴ f'(x) is a decreasing function.

(B) We have : f(x) = cos 2x f'(x) = – 2 sin 2x For 0 < x < \(\frac{\pi}{2}\) or 0 < 2x < π, sin 2x is positive.

∴ f(x) is a decreasing function.

(C) We have : f(x) = cos 3x f'(x) = – 3 sin 3x For 0 < x < \(\frac{\pi}{2}\) ⇒ 0 < 3x < \(\frac{3 \pi}{2}\). Now, sin 3x is positive in 0 < 3x < π. ∴ f’ (x) < 0 ⇒ f(x) is decreasing. And sin 3x is negative in π < 3x < \(\frac{3 \pi}{2}\). ∴ f'(x) > 0 ⇒ f(x) is increasing. ∴ f(x) is neither increasing nor decreasing in (0, \(\frac{\pi}{2}\)).

Hence, f(x) is not a decreasing function in (0, \(\frac{\pi}{2}\))

(D) We have: f(x) = tan x ‘
∴ f’ (x) = sec2x > 0 for all x ∈ (0, \(\frac{\pi}{2}\)) ∴ f (x) is an increasing function.

Thus, (A) cos x and (B) cos 2x are strictly decreasing functions on (0, \(\frac{\pi}{2}\))

Question 13.
On which of the following intervals is the function f given by f(x) = x100 + sin x – 1 strictly decreasing? (A) (- 1, 1) (B) (0, 1) (C) (\(\frac{\pi}{2}\), π] (D) (o, \(\frac{\pi}{2}\)). Solution:

Let f(x) = x100 + sinx – 1.

∴ f’ (x) = 100 x99 + cos x
(A) For (- 1, 1), i.e., – 1 < x < 1, -1 < x99 < 1
⇒ -100 < 100x99 <100 Also, 0 < cosx < 1 ⇒ f’ (x) can either be positive or negative on (- 1, 1).

∴ f(x) is neither increasing nor decreasing on (- 1, 1).

(B) For (0, 1), i.e., 0 < x < 1,
x99 and cos x are both positive. ∴ f'(x)> 0.

⇒ f(x) is increasing on (0, 1).

⇒ f(x) is increasing on (\(\frac{\pi}{2}\), π)

(D) For (0, \(\frac{\pi}{2}\)), i.e., 0 < x < \(\frac{\pi}{2}\), x99 and cos x are both positive. ∴ f’ (x) > 0. ⇒ f(x) is increasing on (0, \(\frac{\pi}{2}\)).

So, none of the above options gives the correct answer.

Question 14.
Find the least value of a such that the function f given by f(x) = x2 + ax + 1 is strictly increasing on (1, 2). Solution:

We have : f(x) = x2 + ax + 1

∴ f’ (x) = 2x + a Since f(x) is an increasing function on (1, 2), therefore f’ (x) > 0 for all 1 < x < 2. Now, f “(x) = 2 for all x ∈ (1, 2) ⇒ f “(x) > 0 for all x ∈ (1, 2) ⇒ f’ (x) is an increasing function on (1, 2). ⇒ f’ (1) is the least value of f’ (x) on (1, 2). But f’ (x) > 0 for all x ∈ (1, 2).Therefore, f’ (1) > 0 ⇒ 2 + a > 0 ⇒ a > – 2

Thus, the least value of a is – 2.

Question 15. Let I be any interval disjoint from (-1, 1). Prove that the function f given by f(x) = x + \(\frac{1}{x}\) is strictly increasing on I.

Solution:

Thus, f'(x) > 0 for all x ∈ I.
Hence, f’ (x) is strictly increasing on I.

Question 16. Prove that the function f given by f(x) = log sin x is strictly increasing on (o, \(\frac{\pi}{2}\)) and strictly decreasing on (\(\frac{\pi}{2}\), π). Solution: We have: f(x) = log sin x ∴ f’ (x) = \(\frac{1}{\sin x}\) . cos x = cot x. When 0 < x < \(\frac{\pi}{2}\), f'(x) is +ve. ∴ f(x) is increasing. When \(\frac{\pi}{2}\) < x < π, f'(x) is -ve. ∴ f(x) is decreasing.

Hence, f is strictly increasing on (0, \(\frac{\pi}{2}\)) and strictly decreasing on (\(\frac{\pi}{2}\), π)

Question 17. Prove that the function f given by f(x) = log cos x is strictly decreasing on (0, \(\frac{\pi}{2}\)) and strictly increasing on (\(\frac{\pi}{2}\), π). Solution: f(x) = log cos x ∴ f’ (x) = \(\frac{1}{\cos x}\) (- sin x) = – tan x. In the interval (0, \(\frac{\pi}{2}\)), f’ (x) is -ve. ∴ f is strictly decreasing in the interval. In the interval (\(\frac{\pi}{2}\), π), f’ (x) is +ve. [Since tan x is -ve in this interval]

∴ f is strictly increasing in the interval.

Question 18.
Prove that the function given by f(x) = x3 – 3x2 + 3x -100 is increasing in R. Solution:

f(x) = x3 – 3x2 + 3x – 100

∴ f’ (x) = 3x2 – 6x + 3 = 3(x2 – 2x + 1)
= 3(x – 1)2
Now for x ∈ R, f'(x) = (x – 1)2 ≥ 0 i. e., f’ (x) ≥ 0 for all x ∈ R.

Hence, f(a) is increasing in R.

Question 19.
The interval in which y = x2 e-x is increasing with respect to x is (A) (-∞, ∞) (B) (-2, 0) (C) (2, ∞) (D) (0, 2) Solution:

f(x) = x2e-x

∴ f’ (x) = 2x e-x + x2(- e-x)
= xe-x(2 – x) = e-x x(2 – x).
Now e-x is positive for all x ∈ R f’ (x) = 0 at x = 0, 2.

x = 0 and x = 2 divide the number line into three disjoint intervals viz (-∞, 0), (0, 2) and (2, ∞).

(a) Interval (-∞, 0) x is -ve and 2 – x is +ve

∴ f'(x) = e-x x(2 – x) = (+)(-)(+) = -ve.

⇒ f is decreasing in (-∞, 0).

(b) Interval (0, 2)
f’ (x) = e-x x(2 – x) = (+)(+)(+) = +ve.
⇒ f is increasing in (0, 2).

∴ Part (D) is the correct answer.

(b) find the interval in which 2x ^ 3 + 9x ^ 2 + 12x - 1 is strictly increasing. (kerala b. 2022)

Gujarat Board Textbook Solutions Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2

zusammenhängende Posts

Werbung

NEUESTEN NACHRICHTEN

Toplisten

Werbung

Populer

Token Data

Werbung

Um

Legal

Hilfe

Sozial