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

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

Answer:

$\text{Let } x_1 \text{ and } x_2 \text{ be two numbers in } \text{R}.$
$x_1 \ < \ x_2 \Rightarrow 2x_1 < 2x_2 \Rightarrow e^{2x_1} < e^{2x_2} \Rightarrow f(x_1) < f(x_2)$
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 $\left ( 0 , \pi /2 \right )$

Answer:

Given f(x) = sinx
$f^{'}(x) = \cos x$

$\text{Since, } \cos x > 0 \ \text{for each } x \in \left( 0, \frac{\pi}{2} \right)$

$f^{'}(x) > 0$
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 $\left ( \frac{\pi}{2},\pi \right )$

Answer:

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 $( 0 , \pi )$

Answer:

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

Answer:

$f ( x) = 2x ^2 - 3 x$
$f^{'}(x) = 4x - 3$
Now,
$f^{'}(x) = 0$
4x - 3 = 0
$x = \frac{3}{4}$

$\text{So, the range is } \left( -\infty, \frac{3}{4} \right) \ \text{and} \ \left( \frac{3}{4}, \infty \right)$

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

Answer:

$f ( x) = 2x ^2 - 3 x$
$f^{'}(x) = 4x - 3$
Now,
$f^{'}(x) = 0$
4x - 3 = 0
$x = \frac{3}{4}$

$\text{So, the range is } \left( -\infty, \frac{3}{4} \right) \ \text{and} \ \left( \frac{3}{4}, \infty \right)$

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

Answer:

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

$\text{So, three ranges are there } (-\infty, -2),\ (-2, 3)\ \text{and}\ (3, \infty)$

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

Answer:

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

$\text{So, three ranges are there } (-\infty, -2),\ (-2, 3),\ \text{and } (3, \infty)$
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: $x ^2 + 2x -5$

Answer:

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

$\text{The range is from } (-\infty, -1)\ \text{ and }\ (-1, \infty)$

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

$10 - 6x - 2x^2$

Answer:

Given function is,
$f(x) = 10 - 6x - 2x^2$
$f^{'}(x) = -6 - 4x$
Now,
$f^{'}(x) = 0$
$6+4x= 0$
$x= -\frac{3}{2}$

$\text{So, the range is } (-\infty , -\frac{3}{2})\ \text{ and }\ (-\frac{3}{2}, \infty)$
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:

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

Answer:

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

$f^{'}(x) = 0$

$-6x^2 - 18x - 12 = 0$

$-6(x^2 + 3x + 2) = 0$

$x^2 + 3x + 2 = 0$

$x^2 + x + 2x + 2 = 0$

$x(x + 1) + 2(x + 1) = 0$

$(x + 2)(x + 1) = 0$

$x = -2\ \text{and}\ x = -1$

So, the range is $\left( -\infty , -2 \right)$, $\left( -2 , -1 \right)$, and $\left( -1 , \infty \right)$.
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:

$6- 9x - x ^2$

Answer:

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 $\left( -\infty, -\frac{9}{2} \right)$ and $\left( -\frac{9}{2}, \infty \right)$.
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:

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

Answer:

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,

$f'(x) = 0$

$3(x + 1)^2(x - 3)^3 + 3(x - 3)^2(x + 1)^3$

$3(x + 1)^2(x - 3)^2 \left( (x - 3) + (x + 1) \right) = 0$

$(x + 1)(x - 3) = 0 \ \text{or} \ (2x - 2) = 0$

$x = -1 \ \text{and} \ x = 3 \ \text{or} \ x = 1$

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 negative in the interval $(-\infty, -1)$ and $(-1, 1)$.

Hence, $f(x) = (x + 1)^3(x - 3)^3$ is strictly decreasing in the 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.

Answer:

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.

Answer:

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,

$f'(x) = 0$

$4x(x - 2)(x - 1) = 0$

$x = 0,\ x = 2\ \text{and}\ x = 1$

So, the intervals are $(-\infty, 0),\ (0, 1),\ (1, 2)\ \text{and}\ (2, \infty)$

In interval $(0, 1)\ \text{and}\ (2, \infty)$, $f'(x) > 0$
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 $\theta\: \: in\: \: [ 0 , \pi /2 ]$

Answer:

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 $\theta \ \epsilon \ [0,\frac{\pi}{2}]$

$4 \cos \theta \geq \cos^2 \theta$

$4 \cos \theta - \cos^2 \theta \geq 0$ 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 $( 0 , \infty )$

Answer:

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).

Answer:

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 $0 , \pi /2$

$(A) \cos x \\(B) \cos 2x \\ (C) \cos 3x \\ (D) \tan x$

Answer:

(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

Answer:

(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 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]?

Answer:

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.

Answer:

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

Answer:

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 positive in the interval $\left( 0, \frac{\pi}{2} \right)$ and negative in the interval $\left( \frac{\pi}{2}, \pi \right)$.

$f'(x) > 0$ 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 $( 3 \pi/2 , 2\pi )$

Answer:

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.

Answer:

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) $( 0, 2 )$

Answer:

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)
Hence, $( 0, 2 )$ is the answer