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

Get Free NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 PDF in Hindi and English Medium. Sets Class 12 Maths NCERT Solutions are extremely helpful while doing your homework. Application of Derivatives Exercise 6.2 Class 12 Maths NCERT Solutions were prepared by Experienced LearnCBSE.in Teachers. Detailed answers of all the questions in Chapter 5 Class 12 Application of Derivatives Ex 6.2 provided in NCERT Textbook.

Free download NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives Ex 6.2 PDF in Hindi Medium as well as in English Medium for CBSE, Uttarakhand, Bihar, MP Board, Gujarat Board, BIE, Intermediate and UP Board students, who are using NCERT Books based on updated CBSE Syllabus for the session 2019-20.

The topics and sub-topics included in the Applications of Derivatives chapter are the following:

Section Name	Topic Name
6	Applications of Derivatives
6.1	Introduction
6.2	Rate of Change of Quantities
6.3	Increasing and Decreasing Functions
6.4	Tangents and Normals
6.5	Approximations
6.6	Maxima and Minima
6.7	Maximum and Minimum Values of a Function in a Closed Interval
6.8	Summary

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

Ex 6.2 Class 12 Maths 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.

Ex 6.2 Class 12 Maths 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
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 2

Ex 6.2 Class 12 Maths 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)

Ex 6.2 Class 12 Maths 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
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 4

Ex 6.2 Class 12 Maths 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).

Ex 6.2 Class 12 Maths 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.

Ex 6.2 Class 12 Maths Question 7.
Show that $y=log(1+x)-\frac { 2x }{ 2+x } x>-1$ , is an increasing function of x throughout its domain.
Solution:
let $f(x)=log(1+x)-\frac { 2x }{ 2+x } x>-1$
f’ (x) = $\frac { { x }^{ 2 } }{ { (x+1)(x+2) }^{ 2 } }$
For f (x) to be increasing f’ (x) > 0
$\Rightarrow \frac { 1 }{ x+1 } >0\Rightarrow x>-1$
Hence, $y=log(1+x)-\frac { 2x }{ 2+x }$ is an increasing function of x for all values of x > – 1.

Ex 6.2 Class 12 Maths Question 8.
Find the values of x for which y = [x (x – 2)]² is an increasing function.
Solution:
y = x⁴ – 4x³ + 4x²
∴ $\frac { dy }{ dx }$ = 4x³ – 12x² + 8x
For the function to be increasing $\frac { dy }{ dx }$ >0
4x³ – 12x² + 8x>0
⇒ 4x(x – 1)(x – 2)>0
For 0 < x < 1, $\frac { dy }{ dx }$ = (+)(-)(-) = +ve and for x > 2, $\frac { dy }{ dx }$ = (+) (+) (+) = +ve
Thus, the function is increasing for 0 < x < 1 and x > 2.

Ex 6.2 Class 12 Maths 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:
$\frac { dy }{ dx } =\frac { 8cos\theta +4 }{ { (2+cos\theta ) }^{ 2 } } -1=\frac { cos\theta (4-cos\theta ) }{ { (2+cos\theta ) }^{ 2 } }$
For the function to be increasing $\frac { dy }{ dx }$ > 0
⇒ cosθ(4-cos²θ)>0
⇒ cosθ>0
⇒ θ∈ $\left[ 0,\frac { \pi }{ 2 } \right]1$

Ex 6.2 Class 12 Maths Question 10.
Prove that the logarithmic function is strictly increasing on (0, ∞).
Solution:
Let f (x) = log x
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

Ex 6.2 Class 12 Maths 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
vedantu class 12 maths Chapter 6 Application of Derivatives 11
∴ f (x) is neither increasing nor decreasing on (-1,1).

Ex 6.2 Class 12 Maths Question 12.
Which of the following functions are strictly decreasing on $\left[ 0,\frac { \pi }{ 2 } \right]$
(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 $\left[ 0,\frac { \pi }{ 2 } \right]$
∴ f’ (x) is a decreasing function.
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 12

Ex 6.2 Class 12 Maths 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) $\left[ \frac { \pi }{ 2 } ,\pi \right]$
(c) $\left[ 0,\frac { \pi }{ 2 } \right]$
(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)
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 13

Ex 6.2 Class 12 Maths 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.

Ex 6.2 Class 12 Maths 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:
Given
$f(x)=x+\frac { 1 }{ x }$
NCERT Solutions for Class 12 Maths Chapter 6 Application of Derivatives 15
Hence, f’ (x) is strictly increasing on I.

Ex 6.2 Class 12 Maths Question 16.
Prove that the function f given by f (x) = log sin x is strictly increasing on $\left( 0,\frac { \pi }{ 2 } \right)$ and strictly decreasing on
$\left( \frac { \pi }{ 2 } ,\pi \right)$
Solution:
f’ (x) = $\frac { 1 }{ sin\quad x } .cos\quad x\quad cot\quad x\quad$
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).

Ex 6.2 Class 12 Maths Question 17.
Prove that the function f given by f(x) = log cos x is strictly decreasing on $\left( 0,\frac { \pi }{ 2 } \right)$ and strictly increasing on $\left( \frac { \pi }{ 2 } ,\pi \right)$
Solution:
$f(x)=log\quad cosx$
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.

Ex 6.2 Class 12 Maths 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.

Ex 6.2 Class 12 Maths 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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