These NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.2

Question 1.

Show that the function given by f(x) = 3x + 17 is strictly increasing on R.

Solution:

f(x) = 3x + 17

Since f(x) is a polynomial function, it is continuous and differentiable in R

∴ f'(x) = 3 > 0 for all x ∈ R

Hence f'(x) is strictly increasing on R

Question 2.

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

Solution:

f (x) = e^{2x}

Since f(x) is an exponential function, it is continuous and differentiable in R

∴ f (x) = e^{2x} > 0 for all x ∈ R

Hence f(x) is strictly increasing on R.

Question 3.

Show that the function given by f (x) = sin x is

(a) strictly increasing in \(\left( 0,\frac { \pi }{ 2 } \right) \)

(b) strictly decreasing in \(\left( \frac { \pi }{ 2 } ,\pi \right) \)

(c) neither increasing nor decreasing in (0, π)

Solution:

We have f(x) = sinx

∴ f’ (x) = cosx

(a) f’ (x) = cos x is + ve in the interval \(\left( 0,\frac { \pi }{ 2 } \right) \)

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

(b) f’ (x) = cos x is a -ve in the interval \(\left( \frac { \pi }{ 2 } ,\pi \right) \)

⇒ f (x) is strictly decreasing in \(\left( \frac { \pi }{ 2 } ,\pi \right) \)

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

while f’ (x) is -ve in the interval \(\left( \frac { \pi }{ 2 } ,\pi \right) \)

∴ 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

(i) f(x) is strictly increasing in (\(\frac { 3 }{ 4 }\), ∞)

(ii) f(x) is strictly increasing in (- ∞, \(\frac { 3 }{ 4 }\))

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

Differentiating w.r.t. x,

f (x) = 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), (3, ∞)

Now f’ (x) is +ve in the intervals (-∞, -2) and (3, ∞). Since in the interval (-∞, -2) each factor x – 3, x + 2 is -ve.

⇒ f’ (x) = + ve.

(a) f is strictly increasing in (-∞, -2)∪(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) – 2x^{3} – 9x² – 12x + 1

(d) 6 – 9x – x²

(e) (x + 1)^{3}(x – 3)^{3}

Solution:

(a) Let f(x) = x² + 2x – 5

Differentiating w.r.t. x,

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

f'(x) = 0 ⇒ 2(x + 1) = 0 = – 1

x = – 1 divides R into disjoint open intervals as (-∞, -1) and (-1, ∞)

∴ f is strictly decreasing in (-∞, -1)

f is strictly increasing in (-1, ∞)

(b) Let f(x) = 10 – 6x – 2x²

Differentiating w.r.t. x,

f'(x) = – 6 – 4x = – 2(2x + 3)

f'(x) = 0 ⇒ – 2(2x + 3) = 0

∴ f is strictly decreasing in (-∞, \(\frac { -3 }{ 2 }\)) and

f is strictly increasing in (\(\frac { 3 }{ 2 }\), ∞)

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

∴ f’ (x) = – 6x^{2} – 18x – 12

= – 6(x^{2} + 3x + 2)

f'(x) = – 6(x + 1)(x + 2), f’ (x) = 0 gives x = -1 or x = -2

The points x = – 2 and x = – 1 divide the real line into three disjoint intervals namely ( – ∞, – 2) ( – 2, – 1) and( – 1, ∞).

f (x) is 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 decreasing in (-1, ∞)

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

(d) Let f(x) = 6 – 9x – x²

Differentiating w.r.t. x,

(e) Let f(x) = (x + 1)^{3}(x – 3)^{3}

Differentiating w.r.t. x,

f'(x) = 3 (x² – 2x – 3)² (2x – 2)

= 6 [(x + 1) (x – 3)]² (x – 1)

= 6 (x + 1)² (x – 1 ) (x – 3)²

f'(x) = 0 ⇒ x = – 1, 1 or 3

x = – 1, x = 1 or x = 3 divide R into disjoint open intervals (-∞, -1), (-1, 1), (1, 3) and (-3, – ∞)

Thus f is strictly decreasing in (- ∞, – 1) and (- 1, 1) and f is strictly increasing in (1, 3) and (3, ∞)

Question 7.

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

Solution:

When x > – 1, we get y’ = (+)(+) = (+) i.e.,

y’ is positive for x > – 1

∴ When x > – 1, y = log(1 + x) – \(\frac { 2x }{ 2 + x }\) an increasing function.

Question 8.

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

Solution:

y = x^{4} – 4x^{3} + 4x^{2}

∴ \(\frac { dy }{ dx }\) = 4x^{3} – 12x^{2} + 8x

For the function to be increasing \(\frac { dy }{ dx }\) >0

4x^{3} – 12x^{2} + 8x>0

⇒ 4x(x – 1)(x – 2)>0

Thus, the function is increasing for 0 < x < 1 and x > 2.

Question 9.

Prove that y = \(\frac { 4sin\theta }{ (2+cos\theta ) } -\theta \) is an increasing function of θ in \(\left[ 0,\frac { \pi }{ 2 } \right]\)

Solution:

∴ \(\frac { dy }{ dθ }\) > 0 since (4 – cos θ) is positive.

Hence y is strictly increasing on (0, \(\frac { π }{ 2 }\)).

Also y is continuous at θ = 0 and θ = \(\frac { π }{ 2 }\).

Hence y is increasing on [0, \(\frac { π }{ 2 }\)]

Question 10.

Prove that the logarithmic function is strictly increasing on (0, ∞).

Solution:

Let f (x) = log x.

The domain of f is (0, ∞).

Now, f’ (x) = \(\frac { 1 }{ x }\)

When takes the values x > 0

\(\frac { 1 }{ x }\) > 0, when x > 0,

∵ f’ (x) > 0

Hence, f (x) is an increasing function for x > 0 i.e

Another method:

From the graph, we can observe that as x increases f(x) also increases. Hence logx is an increasing function 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:

let f (x) = x² – x + 1

Differentiating both sides w.r.t. x,

f'(x) = 2x – 1

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

f'(x) = 0 ⇒ 2(x – \(\frac { 1 }{ 2 }\)) = 0 ⇒ x = \(\frac { 1 }{ 2 }\)

∴ The domain of f(x) is (-1, 1)

∴ f(x) is strictly increasing in (\(\frac { 1 }{ 2 }\), 1) and strictly decreasing in (- 1, \(\frac { 1 }{ 2 }\))

In the interval (- 1, 1) the given function is neither strictly increasing nor strictly decreasing.

Question 12.

Which of the following functions are strictly decreasing on \(\left[ 0,\frac { \pi }{ 2 } \right] \)

i. cos x

ii. cos 2x

iii. cos 3x

iv. tan x

Solution:

sin 3x can be positive or negative.

∴ f'(x) can be positive or negative.

Hence cos 3x is neither increasing nor decreasing on (0, (\(\frac { π }{ 2 }\)).

iv. Let f(x) = tan x, x ∈ (0, \(\frac { π }{ 2 }\))

Differentiating w.r.t. x,

f'(x) = sec²x is positive

tan x is strictly increasing on (0, \(\frac { π }{ 2 }\))

Thus cos x and cos 2x are strictly decreasing on (0, \(\frac { π }{ 2 }\))

Question 13.

On which of the following intervals is the function f given by f (x) = x^{100} + sin x – 1 strictly decreasing ?

i. (0, 1)

ii. \(\left[ \frac { \pi }{ 2 } ,\pi \right] \)

iii. \(\left[ 0,\frac { \pi }{ 2 } \right] \)

iv. none of these

Solution:

f(x) = x^{100} + sin x – 1

Differentiating w.r.t x,

∴ f’ (x)= 100x^{99 }+ cos x

i. When x ∈ (0, 1), x^{99} > 0 and cosx > 0

∴ f'(x) is positive

i.e.,/(x) is strictly increasing on (0, 1)

ii. When x ∈ (\(\frac { π }{ 2 }\), π), 100x^{99} > 100 and -1 < cosx < 0

f'(x)= 100x^{99} + cosx > 100 + a negative number greater than – 1.

f'(x) is positive

∴ Hence f(x) is strictly increasing on (\(\frac { π }{ 2 }\), π)

iii. When x ∈ (0, \(\frac { π }{ 2 }\)), 100x^{99} > 0 and cosx > 0

Hence f'(x) is positive

f(x) is strictly increasing on (0, \(\frac { π }{ 2 }\))

∴ f(x) is not strictly decreasing in any of the intervals (0, 1), (\(\frac { π }{ 2 }\), π) and (0, \(\frac { π }{ 2 }\)).

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)

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

But f’ (x)>0 ∀ x∈ (1, 2)

∴ 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:

f(x) = x + \(\frac { 1 }{ x } \) and I = R – (-1, 1)

Differentiating w.r.t. x,

f'(x) = 1 – \(\frac{1}{x^{2}}\) = \(\frac{x^{2}-1}{x^{2}}\)

When x² > 1, f'(x) > 0

When x ∈ R – (- 1, 1), x² > 1

Hence f'(x) is positive.

i.e., when x ∈ 1, f(x) is strictly increasing

Question 16.

Prove that the function f given by f (x) = log sin x is strictly increasing on (0, \(\frac { π }{ 2 }\)) and strictly decreasing on (\(\frac { π }{ 2 }\), π)

Solution:

f'(x) = \(\frac { 1 }{ sin\quad x }\)cos x cot x

when 0 < x < \(\frac { \pi }{ 2 }\), f’ (x) is +ve; i.e., increasing

When \(\frac { \pi }{ 2 }\) < x < π, f’ (x) is – ve; i.e., decreasing,

∴ f (x) is decreasing. Hence, f is increasing on (0, π/2) and strictly decreasing on (π/2, π).

Question 17.

Prove that the function f given by f(x) = log cos x is strictly decreasing on (0, \(\frac { π }{ 2 }\))and strictly increasing on (\(\frac { π }{ 2 }\), π)

Solution:

Given f(x) = log cos x

f'(x) = \(\frac { 1 }{ cosx }\) (-sinx) = – tanx

In the interval \(\left( 0,\frac { \pi }{ 2 } \right)\), f’ (x) = -ve

∴ f is strictly decreasing.

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

∴ 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) = 3x^{2} – 6x + 3

= 3 (x^{2} – 2x + 1)

= 3 (x -1 )^{2}

Now x ∈ R, f'(x) = (x – 1)^{2} ≥ 0

i.e. f'(x) ≥ 0 ∀ x ∈ R; hence, f(x) is increasing on R.

Question 19.

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

(a) (-∞, ∞)

(b) (-2 0)

(c) (2, ∞)

(d) (0, 2)

Solution:

y = x^{2} e^{-x}

Differentiating w.r.t. x, we get

x = 0 and x = 2 divide the domain R into disjoint open intervals as (-∞, 0), (0, 2) and (2, ∞).

∴ f is strictly increasing on (0, 2)