Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 Textbook Questions and Answers.

## 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.

âˆ´ f'(x) = 3 = + ve for all x âˆˆ R.

â‡’ f is strictly increasing on R.

Question 2.

Show that the function given by f(x) = e^{2x} is strictly increasing on R.

Solution:

We have: f(x) = e^{2x}

â‡’ f'(x) = 2^{2x}

Case I : When x > 0, then

f’ (x) = 2^{2x}

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

âˆ´ f'(x) > 0 for all x > 0.

Case II : When x = 0, then

f'(x) = 2e^{0} = 2

âˆ´ f'(x) > 0 for x = 0.

Case III : When x < 0.

Let x = – y, where y is a positive quantity.

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

= \(\frac{2}{\text { Some +ve quantity }}\) > 0

â‡’ f'(x) > 0 for x < 0.

Thus, f'(x) > 0 for all x âˆˆR.

Hence, e^{2x} 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, Ï€).

Solution:

We have: f(x) = sin x.

âˆ´ f'(x) = cos x

(a) f'(x) = cos x is positive in the interval (0, \(\frac{\pi}{2}\))

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

(b) f(x) cos x is negative in the interval (\(\frac{\pi}{2}\), Ï€)

â‡’ f(x) is strictly decreasing in (\(\frac{\pi}{2}\), Ï€)

(c) f’ (x) – cos x is +ve in the interval (0, \(\frac{\pi}{2}\)),

while f'(x) is -ve in the interval (\(\frac{\pi}{2}\), Ï€)

â‡’ f’ (x) does not have the same sign in the interval (0, Ï€)

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

Question 4.

Find the intervals in which the function f given by, f(x) = 2x^{2} – 3x is (a) strictly increasing, (b) strictly decreasing.

Solution:

f(x) = 2x^{2} – 3x

âˆ´ 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 (-âˆž, \(\frac{3}{4}\)), f’ (x) is negative. Therefore, f is strictly decreasing in (-âˆž, \(\frac{3}{4}\))

In the interval (-âˆž, \(\frac{3}{4}\)), f’ is positive. Hence, f is strictly increasing in the interval (\(\frac{3}{4}\), âˆž)

Question 5.

Find the intervals in which the function f given by f(x) = 2x^{3} – 3x^{2} – 36x + 7 is (a)strictly increasing, (b) strictly decreasing.

Solution:

f(x) = 2x^{3} – 3x^{2} – 36x + 7

âˆ´ f (x) = 6x^{2} – 6x – 36 = 6(x^{2} – x – 6)

= 6(x -3)(x + 2)

â‡’ f'(x) = 0 at x = 3 and x = – 2.

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

Now, f'(x) is +ve in the intervals (-âˆž, – 2) and (3, âˆž), since in the interval (-âˆž, – 2), each factor x – 3, x + 2 is negative.

âˆ´ f’ (x) = + ve

(a) f is strictly increasing in (-âˆž, – 2) u (3, âˆž).

(b) In the interval (- 2, 3). x + 2 is +ve and x – 3 is -ve.

âˆ´ f’ (x) = 6 (x – 3)(x + 2) = + x – = -ve.

âˆ´ 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) x^{2} + 2x – 5

(b) 10 – 6x – 2x^{2}

(c) -2x^{2} – 9x^{2} – 12x + 1

(d) 6 – 9x – x^{2}

(e) (x + 1)^{2}(x – 3)^{3}

Solution:

(a) We have : f(x) = x^{2} + 2x – 5.

âˆ´ f’ (x) = 2x + 2 = 2(x + 1)

The function/(x) will be increasing, if f ‘(x) > 0,

i.e., if 2(x + 1) > 0

â‡’ x + 1 > 0.

â‡’ x > – 1.

The function/(x) will be decreasing if f'(x) < 0,

i.e., 2(x + 1) < 0 â‡’ (x + 1) < 0.

â‡’ x < -1.

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

(b) We have: f(x) = 10 – 6x – 2x^{2}.

âˆ´ 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 â‡’ x < –\(\frac{3}{2}\)

and f'{x) is decreasing, if f'(x) < 0, i.e., if – 6 – 4x < 0,

i.e., – 4x < 6 â‡’ x > \(\frac{-3}{2}\)

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

(c) Let f(x) = – 2x^{3} – 9x^{2} – 12x + 1.

âˆ´ f'(x) = -6x^{2} – 18x – 12

= – 6(x^{2} + 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, (- âˆž, – 2), (- 2, -1) and (-1, âˆž).

In the interval (-âˆž, – 2), i.e., -âˆž < x < – 2,

(x + 1) and (x + 2) are negative.

âˆ´ f'(x) = (-)(-)(-) = -ve.

â‡’ f(x) is strictly decreasing in (-âˆž, – 2).

In the interval (- 2, – 1), i.e., – 2 < x < – 1,

(x + 1) is -ve and (x + 2) is positive.

âˆ´ f'(x) = (-)(-)(+) = +ve.

â‡’ f(x) is strictly increasing in (- 2, -1).

In the interval (- 1, âˆž), i.e., – 1 < x < âˆž,

(x + 1) and (x + 2) are both positive.

âˆ´ f(‘x) = (-)(+)(+) = -ve.

â‡’ f(x) is strictly decreasing in (- 1, âˆž).

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 – x^{2}

âˆ´ f'(x) = – 9 – 2x

Now, f(x) is increasing, if f'(x) > 0, i.e., if – 9 – 2x > 0,

i.e., if -2x > 9 â‡’ x < \(\frac{-9}{2}\)

and f(x) is decreasing, if f{x) < 0, i.e., if – 9 – 2x < 0,

i.e., -2x < 9 â‡’ 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}

âˆ´ 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.

â‡’ 6(x + 1)^{2} (x – 3)^{2} (x – 1) > 0

â‡’ (x – 1) > 0 [âˆµ 6(x + 1)^{2}(x – 3)^{2} > 0]

â‡’ x > 1

But f ‘(x) = 0 at x = 3 .

â‡’ f is strictly increasing in (1, 3) and (3, âˆž)

So, f is strictly increasing in (1, 3) âˆª (3, âˆž).

For f(x) to be decreasing :

f’ (x) < 0

â‡’ 6(x + 1)^{2} (x – 3)^{2} (x – 1) < 0

â‡’ (x – 1) < 0 [âˆµ 6(x + 1)^{2}(x – 3)^{2} > 0]

â‡’ x < 1. But f ‘(x) = 0 at x = -1.

â‡’ f is strictly decreasing in (-âˆž, – 1) and (-1, 1). So, f is strictly decreasing in (-âˆž, – 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} = x^{2}(x^{2} – 4x + 4)

= x^{4} – 4x^{3} + 4x^{2}

âˆ´ \(\frac{d y}{d x}\) = 4x^{3} – 12x^{2} + 8x.

For the function to be increasing :

\(\frac{d y}{d x}\) > 0

â‡’ 4x^{3} – 12x^{2} + 8x > 0

â‡’ 4x(x^{2} – 3x + 2) > 0

â‡’ 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.

âˆ´ 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 = \(\frac{4 \sin \theta}{(2+\cos \theta)}\) – Î¸ is an increasing function of Î¸ in [0, \(\frac{\pi}{2}\)]

Solution:

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

Question 10.

Prove that the logarthmic function is strictly increasing in (0, âˆž).

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

âˆ´ \(\frac{1}{2}\) > 0.

â‡’ fâ€™(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, âˆž).

Question 11.

Prove that the function f given by f(x) = x^{2} – x + 1 is neither strictly increasing nor strictly decreasing on (- 1, 1).

Solution:

f(x) = x^{2} – x + 1

âˆ´ f'(x) = 2x – 1 = 2(x – \(\frac{1}{2}\))

Now, – 1 < x < \(\frac{1}{2}\) â‡’ (x – \(\frac{1}{2}\))< 0.

â‡’ 2(x – \(\frac{1}{2}\)) < 0 â‡’ f'(x)<0.

and, \(\frac{1}{2}\) < x < 1 â‡’ x – \(\frac{1}{2}\) > 0.

2(x – \(\frac{1}{2}\)) > 0 â‡’ 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

âˆ´ f’ (x) = – sin x < 0 in (0, \(\frac{\pi}{2}\))

âˆ´ 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 < Ï€, sin 2x is positive.

âˆ´ f(x) is a decreasing function.

(C) We have : f(x) = cos 3x f'(x) = – 3 sin 3x

For 0 < x < \(\frac{\pi}{2}\) â‡’ 0 < 3x < \(\frac{3 \pi}{2}\).

Now, sin 3x is positive in 0 < 3x < Ï€.

âˆ´ f’ (x) < 0 â‡’ f(x) is decreasing.

And sin 3x is negative in Ï€ < 3x < \(\frac{3 \pi}{2}\).

âˆ´ f'(x) > 0 â‡’ f(x) is increasing.

âˆ´ 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 ‘

âˆ´ f’ (x) = sec^{2}x > 0 for all x âˆˆ (0, \(\frac{\pi}{2}\))

âˆ´ 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) = x^{100} + sin x – 1 strictly decreasing?

(A) (- 1, 1)

(B) (0, 1)

(C) (\(\frac{\pi}{2}\), Ï€]

(D) (o, \(\frac{\pi}{2}\)).

Solution:

Let f(x) = x^{100} + sinx – 1.

âˆ´ f’ (x) = 100 x^{99} + cos x

(A) For (- 1, 1), i.e., – 1 < x < 1, -1 < x^{99} < 1

â‡’ -100 < 100x^{99} <100

Also, 0 < cosx < 1

â‡’ f’ (x) can either be positive or negative on (- 1, 1).

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

(B) For (0, 1), i.e., 0 < x < 1,

x^{99} and cos x are both positive.

âˆ´ f'(x)> 0.

â‡’ f(x) is increasing on (0, 1).

(C) For (\(\frac{\pi}{2}\), Ï€), i.e., \(\frac{\pi}{2}\) < x < Ï€, x^{99} & positive and – 1 < cos x < 0.

âˆ´ f'(x) > 0.

â‡’ f(x) is increasing on (\(\frac{\pi}{2}\), Ï€)

(D) For (0, \(\frac{\pi}{2}\)), i.e., 0 < x < \(\frac{\pi}{2}\), x^{99} and cos x are both positive.

âˆ´ f’ (x) > 0.

â‡’ 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) = x^{2} + ax + 1 is strictly increasing on (1, 2).

Solution:

We have : f(x) = x^{2} + ax + 1

âˆ´ 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 âˆˆ (1, 2) â‡’ f “(x) > 0 for all x âˆˆ (1, 2)

â‡’ f’ (x) is an increasing function on (1, 2).

â‡’ f’ (1) is the least value of f’ (x) on (1, 2).

But f’ (x) > 0 for all x âˆˆ (1, 2).Therefore,

f’ (1) > 0 â‡’ 2 + a > 0

â‡’ 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 âˆˆ 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 (\(\frac{\pi}{2}\), Ï€).

Solution:

We have: f(x) = log sin x

âˆ´ f’ (x) = \(\frac{1}{\sin x}\) . cos x = cot x.

When 0 < x < \(\frac{\pi}{2}\), f'(x) is +ve.

âˆ´ f(x) is increasing.

When \(\frac{\pi}{2}\) < x < Ï€, f'(x) is -ve.

âˆ´ f(x) is decreasing.

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

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 (\(\frac{\pi}{2}\), Ï€).

Solution:

f(x) = log cos x

âˆ´ f’ (x) = \(\frac{1}{\cos x}\) (- sin x) = – tan x.

In the interval (0, \(\frac{\pi}{2}\)), f’ (x) is -ve.

âˆ´ f is strictly decreasing in the interval.

In the interval (\(\frac{\pi}{2}\), Ï€), f’ (x) is +ve.

[Since tan x is -ve in this interval]

âˆ´ f is strictly increasing in the interval.

Question 18.

Prove that the function given by f(x) = x^{3} – 3x^{2} + 3x -100 is increasing in R.

Solution:

f(x) = x^{3} – 3x^{2} + 3x – 100

âˆ´ f’ (x) = 3x^{2} – 6x + 3 = 3(x^{2} – 2x + 1)

= 3(x – 1)^{2}

Now for x âˆˆ R, f'(x) = (x – 1)^{2} â‰¥ 0

i. e., f’ (x) â‰¥ 0 for all x âˆˆ R.

Hence, f(a) is increasing in R.

Question 19.

The interval in which y = x^{2} e^{-x} is increasing with respect to x is

(A) (-âˆž, âˆž)

(B) (-2, 0)

(C) (2, âˆž)

(D) (0, 2)

Solution:

f(x) = x^{2}e^{-x}

âˆ´ f’ (x) = 2x e^{-x} + x^{2}(- e^{-x})

= xe^{-x}(2 – x) = e^{-x} x(2 – x).

Now e^{-x} is positive for all x âˆˆ R

f’ (x) = 0 at x = 0, 2.

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

(a) Interval (-âˆž, 0)

x is -ve and 2 – x is +ve

âˆ´ f'(x) = e^{-x} x(2 – x) = (+)(-)(+) = -ve.

â‡’ f is decreasing in (-âˆž, 0).

(b) Interval (0, 2)

f’ (x) = e^{-x} x(2 – x) = (+)(+)(+) = +ve.

â‡’ f is increasing in (0, 2).

(c) Interval (2, âˆž)

f'(x) = e^{-x} x(2 – x) = (+)(+)(-) = -ve.

â‡’ f is decreasing in the interval (2, âˆž).

âˆ´ Part (D) is the correct answer.