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

NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 are part of NCERT Solutions for Class 12 Maths. Here we have given NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2.

• Application of Derivatives Class 12 Ex 6.1
• Application of Derivatives Class 12 Ex 6.3
• Application of Derivatives Class 12 Ex 6.4
• Application of Derivatives Class 12 Ex 6.5

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Maths
Chapter	Chapter 6
Chapter Name	Application of Derivatives
Exercise	Ex 6.2
Number of Questions Solved	19
Category	NCERT Solutions

NCERT Solutions for 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>0 ∀ x∈R
⇒ 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
⇒ f’ (x) = 2e^2x
Case I When x > 0, then f’ (x) = 2e^2x

Question 3.
Show that the function given by f (x) = sin x is
(a) strictly increasing in
(b) strictly decreasing in
(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
⇒ f(x) is strictly increasing on
(b) f’ (x) = cos x is a -ve in the interval
⇒ f (x) is strictly decreasing in
(c) f’ (x) = cos x is +ve in the interval
while f’ (x) is -ve in the interval
∴ 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 =
The point divides the real

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) = 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) – 2x³ – 9x² – 12x + 1
(d) 6 – 9x – x²
(e) (x + 1)³(x – 3)³
Solution:
(c) Let f(x) = – 2x³ – 9x² – 12x + 1
∴ f’ (x) = – 6x² – 18x – 12
= – 6(x² + 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 ∞).
In the interval (-∞,-2) i.e.,-∞<x<-2 (x+ 1) (x+2) are -ve.
∴f’ (x) = (-) (-) (- ) = – ve.
⇒ f (x) is decreasing in (-∞,-2)
In the interval (-2, -1) i.e., – 2 < x < -1,
(x + 1) is -ve and (x + 2) is + ve.
∴ f'(x) = (-)(-) (+) = + ve.
⇒ 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.

Question 7.
Show that , is an increasing function of x throughout its domain.
Solution:
let
f’ (x) =
For f (x) to be increasing f’ (x) > 0

Hence, is an increasing function of x for all values of x > – 1.

Question 8.
Find the values of x for which y = [x (x – 2)]² is an increasing function.
Solution:
y = x⁴ – 4x³ + 4x²
∴ = 4x³ – 12x² + 8x
For the function to be increasing >0
4x³ – 12x² + 8x>0
⇒ 4x(x – 1)(x – 2)>0
For 0 < x < 1, = (+)(-)(-) = +ve and for x > 2, = (+) (+) (+) = +ve
Thus, the function is increasing for 0 < x < 1 and x > 2.

Question 9.
Prove that is an increasing function of θ in
Solution:

For the function to be increasing > 0
⇒ cosθ(4-cos²θ)>0
⇒ cosθ>0
⇒ θ∈

Question 10.
Prove that the logarithmic function is strictly increasing on (0, ∞).
Solution:
Let f (x) = log x
Now, f’ (x) = ; When takes the
values x > 0, > 0, when x > 0,
∵ f’ (x) > 0
Hence, f (x) is an increasing function for x > 0 i.e

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:
Given
f (x) = x² – x + 1

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

Question 12.
Which of the following functions are strictly decreasing on
(a) cos x
(b) cos 2x
(c) cos 3x
(d) tan x
Solution:
(a) We have f (x) = cos x
∴ f’ (x) = – sin x < 0 in
∴ f’ (x) is a decreasing function.

Question 13.
On which of the following intervals is the function f given by f (x )= x¹⁰⁰ + sin x – 1 strictly decreasing ?
(a) (0,1)
(b)
(c)
(d) none of these
Solution:
(d) f(x) = x¹⁰⁰ + sin x – 1
∴ f’ (x)= 100x⁹⁹+ cos x
(a) for(-1, 1)i.e.,- 1 <x< 1,-1 <x⁹⁹< 1
⇒ -100<100x⁹⁹<100;
Also 0 ⇒ f’ (x) can either be +ve or -ve 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 +ve ∴ f’ (x) > 0
⇒ f (x) is increasing on(0,1)

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 is strictly increasing on I.
Solution:
Given


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 and strictly decreasing on

Solution:
f’ (x) =
when 0 < x < , f’ (x) is +ve; i.e., increasing
When < 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 and strictly increasing on
Solution:

f’ (x) =
In the interval ,f’ (x) = -ve
∴ f is strictly decreasing.
In the interval , f’ (x) is + ve.
∴ 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) = 3x² – 6x + 3
= 3 (x² – 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 = x² e^-x is increasing is
(a) (-∞,∞)
(b) (-2,0)
(c) (2,∞)
(d) (0,2)
Solution:
(d) f’ (x) = 2xe^-x + x²( – e^-x) = xe^-x(2-x) = e^-xx(2-x)
Now e^-x is positive for all x ∈ R f’ (x) = 0 at x = 0,2
x = 0, x = 2 divide the number line into three disjoint intervals, viz. (-∞, 0), (0,2), (2, ∞)
(a) Interval (-∞,0) x is +ve and (2-x) is +ve
∴ f’ (x) = e^-xx (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)
(c) Interval (2, ∞) f’ (x) = e^-x x (2 – x) = (+) (+) (-)
= – ve
⇒ f is decreasing in the interval (2, ∞)

 

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