# NCERT Solutions for Exercise 6.2 Class 12 Maths Chapter 10 - Application of Derivatives

NCERT solutions for exercise 6.2 Class 12 Maths chapter 6 gives an insight into topic 6.3 increasing and decreasing functions. Before exercise 6.2 Class 12 Maths, NCERT has explained the questions and examples related to the rate of change of quantities. After the NCERT solutions for Class 12 Maths chapter 6 exercise 6.2 the concepts of decreasing and increasing functions is introduced in the NCERT book and then certain theorems are discussed followed by example questions and Class 12th Maths chapter 6 exercise 6.2.

The NCERT solutions for Class 12 Maths chapter 6 exercise 6.2 gives practice on topic 6.3 of Class 12 Maths NCERT syllabus. Solving the Class 12 Maths chapter 6 exercise 6.2 gives more knowledge of the concepts of increasing and decreasing functions. The following exercises are also discussed in the chapter application of derivatives.

• Application of Derivatives Exercise 6.1

• Application of Derivatives Exercise 6.3

• Application of Derivatives Exercise 6.4

• Application of Derivatives Exercise 6.5

• Application of Derivatives Miscellaneous Exercise

## Application of Derivatives Class 12 Exercise 6.2

Question:1 . Show that the function given by f (x) = 3x + 17 is increasing on R.

Let $x_1 and x_2$ are two numbers in R
$x_1 < x_2 \Rightarrow 3x_1 < 3 x_2 \Rightarrow 3x_1 + 17 < 3x_2+17 \Rightarrow f(x_1)< f(x_2)$
Hence, f is strictly increasing on R

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

Let $x_1 \ and \ x_2$ are two numbers in R
$x_1 \ < \ x_2 \Rightarrow 2x_1 < 2x_2 \Rightarrow e^{2x_1} < e^{2x_2} \Rightarrow f(x_1) < f(x_2)$
Hence, the function $f(x) = e^{2x}$ is strictly increasing in R

Question:3 a) Show that the function given by f (x) = $\sin x$ is increasing in $\left ( 0 , \pi /2 \right )$

Given f(x) = sinx
$f^{'}(x) = \cos x$
Since, $\cos x > 0 \ for \ each \ x\ \epsilon \left ( 0,\frac{\pi}{2} \right )$
$f^{'}(x) > 0$
Hence, f(x) = sinx is strictly increasing in $\left ( 0,\frac{\pi}{2} \right )$

Question:3 b) Show that the function given by f (x) = $\sin x$ is

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

f(x) = sin x
$f^{'}(x) = \cos x$
Since, $\cos x < 0$ for each $x \ \epsilon \left ( \frac{\pi}{2},\pi \right )$
So, we have $f^{'}(x) < 0$
Hence, f(x) = sin x is strictly decreasing in $\left ( \frac{\pi}{2},\pi \right )$

Question:3 c) Show that the function given by f (x) = $\sin x$ is neither increasing nor decreasing in $( 0 , \pi )$

We know that sin x is strictly increasing in $\left ( 0,\frac{\pi}{2} \right )$ and strictly decreasing in $\left ( \frac{\pi}{2},\pi \right )$
So, by this, we can say that f(x) = sinx is neither increasing or decreasing in range $\left ( 0,\pi \right )$

Question:4(a). Find the intervals in which the function f given by $f ( x) = 2x ^2 - 3 x$ is increasing

$f ( x) = 2x ^2 - 3 x$
$f^{'}(x) = 4x - 3$
Now,
$f^{'}(x) = 0$
4x - 3 = 0
$x = \frac{3}{4}$
So, the range is $\left ( -\infty, \frac{3}{4} \right ) \ and \ \left ( \frac{3}{4}, \infty \right )$
So,
$f(x)< 0$ when $x \ \epsilon \left ( -\infty,\frac{3}{4} \right )$ Hence, f(x) is strictly decreasing in this range
and
$f(x) > 0$ when $x \epsilon \left ( \frac{3}{4},\infty \right )$ Hence, f(x) is strictly increasing in this range
Hence, $f ( x) = 2x ^2 - 3 x$ is strictly increasing in $x \epsilon \left ( \frac{3}{4},\infty \right )$

Question:4(b) Find the intervals in which the function f given by $f ( x) = 2 x ^2 - 3 x$ is
decreasing

$f ( x) = 2x ^2 - 3 x$
$f^{'}(x) = 4x - 3$
Now,
$f^{'}(x) = 0$
4x - 3 = 0
$x = \frac{3}{4}$
So, the range is $\left ( -\infty, \frac{3}{4} \right ) \ and \ \left ( \frac{3}{4}, \infty \right )$
So,
$f(x)< 0$ when $x \ \epsilon \left ( -\infty,\frac{3}{4} \right )$ Hence, f(x) is strictly decreasing in this range
and
$f(x) > 0$ when $x \epsilon \left ( \frac{3}{4},\infty \right )$ Hence, f(x) is strictly increasing in this range
Hence, $f ( x) = 2x ^2 - 3 x$ is strictly decreasing in $x \epsilon \left ( -\infty ,\frac{3}{4}\right )$

Question:5(a) Find the intervals in which the function f given by $f (x) = 2x^3 - 3x ^2 - 36 x + 7$ is
increasing

It is given that
$f (x) = 2x^3 - 3x ^2 - 36 x + 7$
So,
$f^{'}(x)= 6x^{2} - 6x - 36$
$f^{'}(x)= 0$
$6x^{2} - 6x -36 =0 \Rightarrow 6 (x^{2} - x-6)$
$x^{2} - x-6 = 0$
$x^{2} - 3x+2x-6 = 0$
$x(x-3) + 2(x-3) = 0\\$
$(x+2)(x-3) = 0$
x = -2 , x = 3

So, three ranges are there $(-\infty,-2) , (-2,3) \ and \ (3,\infty)$
Function $f^{'}(x)= 6x^{2} - 6x - 36$ is positive in interval $(-\infty,-2) , (3,\infty)$ and negative in the interval (-2,3)
Hence, $f (x) = 2x^3 - 3x ^2 - 36 x + 7$ is strictly increasing in $(-\infty,-2) \cup (3,\infty)$
and strictly decreasing in the interval (-2,3)

Question:5(b) Find the intervals in which the function f given by $f ( x) = 2x ^3 - 3x ^2 - 36x + 7$ is
decreasing

We have $f ( x) = 2x ^3 - 3x ^2 - 36x + 7$

Differentiating the function with respect to x, we get :

$f' ( x) = 6x ^2 - 6x - 36$

or $= 6\left ( x-3 \right )\left ( x+2 \right )$

When $f'(x)\ =\ 0$ , we have :

$0\ = 6\left ( x-3 \right )\left ( x+2 \right )$

or $\left ( x-3 \right )\left ( x+2 \right )\ =\ 0$

So, three ranges are there $(-\infty,-2) , (-2,3) \ and \ (3,\infty)$
Function $f^{'}(x)= 6x^{2} - 6x - 36$ is positive in the interval $(-\infty,-2) , (3,\infty)$ and negative in the interval (-2,3)

So, f(x) is decreasing in (-2, 3)

Question:6(a) Find the intervals in which the following functions are strictly increasing or
decreasing: $x ^2 + 2x -5$

f(x) = $x ^2 + 2x -5$
$f^{'}(x) = 2x + 2 = 2(x+1)$
Now,
$f^{'}(x) = 0 \\ 2(x+1) = 0\\ x = -1$

The range is from $(-\infty,-1) \ and \ (-1,\infty)$
In interval $(-\infty,-1)$ $f^{'}(x)= 2(x+1)$ is -ve
Hence, function f(x) = $x ^2 + 2x -5$ is strictly decreasing in interval $(-\infty,-1)$
In interval $(-1,\infty)$ $f^{'}(x)= 2(x+1)$ is +ve
Hence, function f(x) = $x ^2 + 2x -5$ is strictly increasing in interval $(-1,\infty)$

Question:6(b) Find the intervals in which the following functions are strictly increasing or
decreasing

$10 - 6x - 2x^2$

Given function is,
$f(x) = 10 - 6x - 2x^2$
$f^{'}(x) = -6 - 4x$
Now,
$f^{'}(x) = 0$
$6+4x= 0$
$x= -\frac{3}{2}$
So, the range is $(-\infty , -\frac{3}{2}) \ and \ (-\frac{3}{2},\infty)$
In interval $(-\infty , -\frac{3}{2})$ , $f^{'}(x) = -6 - 4x$ is +ve
Hence, $f(x) = 10 - 6x - 2x^2$ is strictly increasing in the interval $(-\infty , -\frac{3}{2})$
In interval $( -\frac{3}{2},\infty)$ , $f^{'}(x) = -6 - 4x$ is -ve
Hence, $f(x) = 10 - 6x - 2x^2$ is strictly decreasing in interval $( -\frac{3}{2},\infty)$

Question:6(c) Find the intervals in which the following functions are strictly increasing or
decreasing:

$- 2 x^3 - 9x ^2 - 12 x + 1$

Given function is,
$f(x) = - 2 x^3 - 9x ^2 - 12 x + 1^{}$
$f^{'}(x) = - 6 x^2 - 18x - 12$
Now,
$f^{'}(x) = 0\\ - 6 x^2 - 18x - 12 = 0\\ -6(x^{2}+3x+2) = 0 \\ x^{2}+3x+2 = 0 \\x^{2} + x + 2x + 2 = 0\\ x(x+1) + 2(x+1) = 0\\ (x+2)(x+1) = 0\\ x = -2 \ and \ x = -1$

So, the range is $(-\infty , -2) \ , (-2,-1) \ and \ (-1,\infty)$
In interval $(-\infty , -2) \cup \ (-1,\infty)$ , $f^{'}(x) = - 6 x^2 - 18x - 12$ is -ve
Hence, $f(x) = - 2 x^3 - 9x ^2 - 12 x + 1^{}$ is strictly decreasing in interval $(-\infty , -2) \cup \ (-1,\infty)$
In interval (-2,-1) , $f^{'}(x) = - 6 x^2 - 18x - 12$ is +ve
Hence, $f(x) = - 2 x^3 - 9x ^2 - 12 x + 1^{}$ is strictly increasing in the interval (-2,-1)

Question:6(d) Find the intervals in which the following functions are strictly increasing or
decreasing:

$6- 9x - x ^2$

Given function is,
$f(x) = 6- 9x - x ^2$
$f^{'}(x) = - 9 - 2x$
Now,
$f^{'}(x) = 0\\ - 9 - 2x = 0 \\ 2x = -9\\ x = -\frac{9}{2}$

So, the range is $(-\infty, - \frac{9}{2} ) \ and \ ( - \frac{9}{2}, \infty )$
In interval $(-\infty, - \frac{9}{2} )$ , $f^{'}(x) = - 9 - 2x$ is +ve
Hence, $f(x) = 6- 9x - x ^2$ is strictly increasing in interval $(-\infty, - \frac{9}{2} )$
In interval $( - \frac{9}{2},\infty )$ , $f^{'}(x) = - 9 - 2x$ is -ve
Hence, $f(x) = 6- 9x - x ^2$ is strictly decreasing in interval $( - \frac{9}{2},\infty )$

Question:6(e) Find the intervals in which the following functions are strictly increasing or
decreasing:

$( x+1) ^3 ( x-3) ^3$

Given function is,
$f(x) = ( x+1) ^3 ( x-3) ^3$
$f^{'}(x) = 3( x+1) ^2(x-3)^{3} + 3( x-3) ^2(x+1)^3$
Now,
$f^{'}(x) = 0 \\ 3( x+1) ^2(x-3)^{3} + 3( x-3) ^2(x+1)^{3} \\ 3(x+1)^{2}(x-3)^{2}((x-3) + (x+1) ) = 0 \\ (x+1)(x-3) = 0 \ \ \ \ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \ (2x-2) = 0\\ x=-1 \ and \ x = 3 \ \ \ \ \ \ \ \ \ \ \ \ \ \ or \ \ \ \ \ \ \ \ \ \ \ \ x = 1$
So, the intervals are $(-\infty,-1), (-1,1), (1,3) \ and \ (3,\infty)$

Our function $f^{'}(x) = 3( x+1) ^2(x-3)^{3} + 3( x-3) ^2(x+1)^3$ is +ve in the interval $(1,3) \ and \ (3,\infty)$
Hence, $f(x) = ( x+1) ^3 ( x-3) ^3$ is strictly increasing in the interval $(1,3) \ and \ (3,\infty)$
Our function $f^{'}(x) = 3( x+1) ^2(x-3)^{3} + 3( x-3) ^2(x+1)^3$ is -ve in the interval $(-\infty,-1) \ and \ (-1,1)$
Hence, $f(x) = ( x+1) ^3 ( x-3) ^3$ is strictly decreasing in interval $(-\infty,-1) \ and \ (-1,1)$

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

Given function is,
$f(x)\Rightarrow y = \log( 1+ x ) - \frac{2 x }{2+x }$
$f^{'}(x)\Rightarrow \frac{dy}{dx} = \frac{1}{1+x} - \frac{2 (2+x) - (1)(2x)} {(2+x)^{2} } = \frac{1}{1+x} - \frac{4+2x-2x}{(2+x)^{2}}$
$= \frac{1}{1+x} - \frac{4}{(2+x)^2} = \frac{(2+x)^2 - 4(x+1)}{(x+1)(2+x)^{2}}$
$= \frac{4+x^{2} +4x -4x - 4}{(x+1)(2+x)^{2}} = \frac{x^{2} }{(x+1)(2+x)^{2}}$
$f^{'}(x) = \frac{x^{2}}{(x+1)(x+2)^2}$
Now, for $x > -1$ , is is clear that $f^{'}(x) = \frac{x^{2}}{(x+1)(x+2)^2} > 0$
Hence, $f(x)\Rightarrow y = \log( 1+ x ) - \frac{2 x }{2+x }$ strictly increasing when $x > -1$

Question:8 Find the values of x for which $y = [x(x-2)]^{2}$ is an increasing function.

Given function is,
$f(x)\Rightarrow y = [x(x-2)]^{2}$
$f^{'}(x)\Rightarrow \frac{dy}{dx} = 2[x(x-2)][(x-2)+x]$
$= 2(x^2-2x)(2x-2)$
$= 4x(x-2)(x-1)$
Now,
$f^{'}(x) = 0\\ 4x(x-2)(x-1) = 0\\ x=0 , x= 2 \ and \ x = 1$
So, the intervals are $(-\infty,0),(0,1),(1,2) \ and \ (2,\infty)$
In interval $(0,1)and \ (2,\infty)$ , $f^{'}(x)> 0$
Hence, $f(x)\Rightarrow y = [x(x-2)]^{2}$ is an increasing function in the interval $(0,1)\cup (2,\infty)$

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

Given function is,
$f(x) = y = \frac{4 \sin \theta }{(2+ \cos \theta )} - \theta$

$f^{'}(x) = \frac{dy}{d\theta} = \frac{4 \cos \theta(2+\cos \theta) - (-\sin \theta)4\sin \theta) }{(2+ \cos \theta )^2} - 1$
$= \frac{8 \cos \theta+4\cos^2 \theta + 4\sin^2 \theta - (2+ \cos \theta )^2 }{(2+ \cos \theta )^2}$
$= \frac{8 \cos \theta+4(\cos^2 \theta + \sin^2 \theta) - 4- \cos^2 \theta - 4\cos \theta }{(2+ \cos \theta )^2}$
$= \frac{8 \cos \theta+4 - 4- \cos^2 \theta - 4\cos \theta }{(2+ \cos \theta )^2}$
$= \frac{4 \cos \theta-\cos^2 \theta }{(2+ \cos \theta )^2}$
Now, for $\theta \ \epsilon \ [0,\frac{\pi}{2}]$
$\\ 4 \cos \theta \geq \cos^2 \theta\\ 4 \cos \theta - \cos^2 \geq 0\\ and \ (2+\cos \theta)^2 > 0$
So, $f^{'}(x) > 0 \ for \ \theta \ in \ [0,\frac{\pi}{2}]$
Hence, $f(x) = y = \frac{4 \sin \theta }{(2+ \cos \theta )} - \theta$ is increasing function in $\theta \ \epsilon \ [0,\frac{\pi}{2}]$

Question:10 Prove that the logarithmic function is increasing on $( 0 , \infty )$

Let logarithmic function is log x
$f(x) = log x$
$f^{'}(x) = \frac{1}{x}$
Now, for all values of x in $( 0 , \infty )$ , $f^{'}(x) > 0$
Hence, the logarithmic function $f(x) = log x$ is increasing in the interval $( 0 , \infty )$

Question:11 Prove that the function f given by $f ( x) = x ^2 - x + 1$ is neither strictly increasing nor decreasing on (– 1, 1).

Given function is,
$f ( x) = x ^2 - x + 1$
$f^{'}(x) = 2x - 1$
Now, for interval $(-1,\frac{1}{2})$ , $f^{'}(x) < 0$ and for interval $(\frac{1}{2},1),f^{'}(x) > 0$
Hence, by this, we can say that $f ( x) = x ^2 - x + 1$ is neither strictly increasing nor decreasing in the interval (-1,1)

Question:12 Which of the following functions are decreasing on $0 , \pi /2$ $(A) \cos x \\(B) \cos 2x \\ (C) \cos 3x \\ (D) \tan x$

(A)
$f(x) = \cos x \\ f^{'}(x) = -\sin x$
$f^{'}(x) < 0$ for x in $(0,\frac{\pi}{2})$
Hence, $f(x) = \cos x$ is decreasing function in $(0,\frac{\pi}{2})$

(B)
$f(x) = \cos 2x \\ f^{'}(x) = -2\sin2 x$
Now, as
$0 < x < \frac{\pi}{2}\\ 0 < 2x < \pi$
$f^{'}(x) < 0$ for 2x in $(0,\pi)$
Hence, $f(x) = \cos 2x$ is decreasing function in $(0,\frac{\pi}{2})$

(C)
$f(x) = \cos 3x \\ f^{'}(x) = -3\sin3 x$
Now, as
$0 < x < \frac{\pi}{2}\\ 0 < 3x < \frac{3\pi}{2}$
$f^{'}(x) < 0$ for $x \ \epsilon \ \left ( 0,\frac{\pi}{3} \right )$ and $f^{'}(x) > 0 \ x \ \epsilon \ \left ( \frac{\pi}{3} , \frac{\pi}{2}\right )$
Hence, it is clear that $f(x) = \cos 3x$ is neither increasing nor decreasing in $(0,\frac{\pi}{2})$

(D)
$f(x) = \tan x\\ f^{'}(x) = \sec^{2}x$
$f^{'}(x) > 0$ for x in $(0,\frac{\pi}{2})$
Hence, $f(x) = \tan x$ is strictly increasing function in the interval $(0,\frac{\pi}{2})$

So, only (A) and (B) are decreasing functions in $(0,\frac{\pi}{2})$

Question:13 On which of the following intervals is the function f given by $f ( x) = x ^{100} + \sin x - 1$ decreasing ?
(A) (0,1) (B) $\frac{\pi}{2},\pi$ (C) $0,\frac{\pi}{2}$ (D) None of these

(A) Given function is,
$f ( x) = x ^{100} + \sin x - 1$
$f^{'}(x) = 100x^{99} + \cos x$
Now, in interval (0,1)
$f^{'}(x) > 0$
Hence, $f ( x) = x ^{100} + \sin x - 1$ is increasing function in interval (0,1)

(B) Now, in interval $\left ( \frac{\pi}{2},\pi \right )$
$100x^{99} > 0 \ but \ \cos x < 0$
$100x^{99} > \cos x \\ 100x^{99} - \cos x > 0$ , $f^{'}(x) > 0$
Hence, $f ( x) = x ^{100} + \sin x - 1$ is increasing function in interval $\left ( \frac{\pi}{2},\pi \right )$

(C) Now, in interval $\left ( 0,\frac{\pi}{2} \right )$
$100x^{99} > 0 \ and \ \cos x > 0$
$100x^{99} > \cos x \\ 100x^{99} - \cos x > 0$ , $f^{'}(x) > 0$
Hence, $f ( x) = x ^{100} + \sin x - 1$ is increasing function in interval $\left ( 0,\frac{\pi}{2} \right )$

So, $f ( x) = x ^{100} + \sin x - 1$ is increasing for all cases
Hence, correct answer is (D) None of these

Question:14 For what values of a the function f given by $f (x) = x^2 + ax + 1$ is increasing on
[1, 2]?

Given function is,
$f (x) = x^2 + ax + 1$
$f^{'}(x) = 2x + a$
Now, we can clearly see that for every value of $a > -2$
$f^{'}(x) = 2x + a$ $> 0$
Hence, $f (x) = x^2 + ax + 1$ is increasing for every value of $a > -2$ in the interval [1,2]

Question:15 Let I be any interval disjoint from [–1, 1]. Prove that the function f given by $f ( x) = x + 1/x$ is increasing on I.

Given function is,
$f ( x) = x + 1/x$
$f^{'}(x) = 1 - \frac{1}{x^2}$
Now,
$f^{'}(x) = 0\\ 1 - \frac{1}{x^2} = 0\\ x^{2} = 1\\ x = \pm1$

So, intervals are from $(-\infty,-1), (-1,1) \ and \ (1,\infty)$
In interval $(-\infty,-1), (1,\infty)$ , $\frac{1}{x^2} < 1 \Rightarrow 1 - \frac{1}{x^2} > 0$
$f^{'}(x) > 0$
Hence, $f ( x) = x + 1/x$ is increasing in interval $(-\infty,-1)\cup (1,\infty)$
In interval (-1,1) , $\frac{1}{x^2} > 1 \Rightarrow 1 - \frac{1}{x^2} < 0$
$f^{'}(x) < 0$
Hence, $f ( x) = x + 1/x$ is decreasing in interval (-1,1)
Hence, the function f given by $f ( x) = x + 1/x$ is increasing on I disjoint from [–1, 1]

Question:16 Prove that the function f given by $f (x) = \log \sin x$ is increasing on

$\left ( 0 , \pi /2 \right )\: \: and \: \: decreasing \: \: on \: \: \left ( \pi/2 , \pi \right )$

Given function is,
$f (x) = \log \sin x$
$f^{'}(x) = \frac{1}{\sin x}\cos x = \cot x$
Now, we know that cot x is+ve in the interval $\left ( 0 , \pi /2 \right )$ and -ve in the interval $\left ( \pi/2 , \pi \right )$
$f^{'}(x) > 0 \ in \ \left ( 0,\frac{\pi}{2} \right ) \ and \ f^{'}(x) < 0 \ in \ \left ( \frac{\pi}{2} , \pi \right )$
Hence, $f (x) = \log \sin x$ is increasing in the interval $\left ( 0 , \pi /2 \right )$ and decreasing in interval $\left ( \pi/2 , \pi \right )$

Question:17 Prove that the function f given by f (x) = log |cos x| is decreasing on $( 0 , \pi /2 )$
and increasing on $( 3 \pi/2 , 2\pi )$

Given function is,
f(x) = log|cos x|
value of cos x is always +ve in both these cases
So, we can write log|cos x| = log(cos x)
Now,
$f^{'}(x) = \frac{1}{\cos x}(-\sin x) = -\tan x$
We know that in interval $\left ( 0,\frac{\pi}{2} \right )$ , $\tan x > 0 \Rightarrow -\tan x< 0$
$f^{'}(x) < 0$
Hence, f(x) = log|cos x| is decreasing in interval $\left ( 0,\frac{\pi}{2} \right )$

We know that in interval $\left ( \frac{3\pi}{2},2\pi \right )$ , $\tan x < 0 \Rightarrow -\tan x> 0$
$f^{'}(x) > 0$
Hence, f(x) = log|cos x| is increasing in interval $\left ( \frac{3\pi}{2},2\pi \right )$

Question:18 Prove that the function given by $f (x) = x^3 - 3x^2 + 3x - 100$ is increasing in R.

Given function is,
$f (x) = x^3 - 3x^2 + 3x - 100$
$f^{'}(x) = 3x^2 - 6x + 3$
$= 3(x^2 - 2x + 1) = 3(x-1)^2$
$f^{'}(x) = 3(x-1)^2$
We can clearly see that for any value of x in R $f^{'}(x) > 0$
Hence, $f (x) = x^3 - 3x^2 + 3x - 100$ is an increasing function in R

Question:19 The interval in which $y = x ^2 e ^{-x}$ is increasing is

(A) $( - \infty , \infty )$ (B) $( - 2 , 0 )$ (C) $( - 2 , \infty )$ (D) $( 0, 2 )$

Given function is,
$f(x) \Rightarrow y = x ^2 e ^{-x}$
$f^{'}(x) \Rightarrow \frac{dy}{dx} = 2x e ^{-x} + -e^{-x}(x^{2})$
$xe ^{-x}(2 -x)$
$f^{'}(x) = xe ^{-x}(2 -x)$
Now, it is clear that $f^{'}(x) > 0$ only in the interval (0,2)
So, $f(x) \Rightarrow y = x ^2 e ^{-x}$ is an increasing function for the interval (0,2)

## More About NCERTSolutions for Class 12 Maths Chapter 6 Exercise 6.2

The questions discussed in the Class 12th Maths chapter 6 exercise 6.2 uses differentiation to find out the increasing and decreasing function. The NCERT Class 12 Maths Book explains the increasing and decreasing functions with suitable examples and graphical representations. All the examples in the NCERT Book and the NCERT solutions for Class 12 Maths chapter 6 exercise 6.2 are important from the exam point of view.

Also Read| Application of Derivatives Class 12 Notes

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

• Exercise 6.2 Class 12 Maths helps students to grasp the concepts in a better way.

• NCERT solutions for Class 12 Maths chapter 6 exercise 6.2 is useful for the preparation of board exams that follows the NCERT Syllabus

• Along with this students can also refer to the NCERT exemplar solutions of the same chapter for a good score.

Also see-

• NCERT Exemplar Solutions Class 12 Maths Chapter 6

• NCERT Solutions for Class 12 Maths Chapter 6

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