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

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NCERT Solutions for Class 12 Maths Chapter 6 (Ex 6.2)

NCERT Solutions for Class 12 Math Chapter 6 Application of Derivatives Exercise 6.2 pdf is given on the page. The pdf  contains solutions for all questions given in the exercise  These NCERT solutions given in the pdf here are prepared by Vedantu’s subject experts based on the latest NCERT syllabus issued by the CBSE board . Students can download the Class 12 NCERT Math Chapter 6 Exercise 6.2 solution just by clicking once on the pdf link below to improve their skills. The questions in this exercise are based on the topic “ Distance Between Two Points”.


What Is Distance Between Two Points?

In three dimensional Geometry, the distance between two points is defined as the shortest line segment joining them.  The distance between two points ( x1, x1,x1) and (y2, y2, y2) can be calculated using the following formula:


$\sqrt{(x_{2}- x_{1})^{2} + (y_{2}- y_{1})^{2}}$


Free PDF download of NCERT Solutions for Class 12 Maths Chapter 6 Exercise 6.2 (Ex 6.2) and all chapter exercises at one place prepared by expert teacher as per NCERT (CBSE) books guidelines. Class 12 Maths Chapter 6 Application of Derivatives Exercise 6.2 Questions with Solutions to help you to revise complete Syllabus and Score More marks. Register and get all exercise solutions in your emails.

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Access NCERT Solutions for Class 12 Maths Chapter 6 – Application of Derivatives

Exercise 6.2

1. Show, that the function given by $f\left( x \right)=3x+17$ is strictly increasing on $R$.

Ans: Let the two numbers in $R$be ${{x}_{1}}$ and ${{x}_{2}}$.

$ {{x}_{1}}<{{x}_{_{2}}}\Rightarrow 3{{x}_{1}}<3{{x}_{2}} $ 

$ \Rightarrow 3{{x}_{1}}+17<3{{x}_{2}}+17 $ 

$ =f\left( {{x}_{1}} \right)<f\left( {{x}_{2}} \right)$ 

Thus, the function is strictly increasing on $R.$ 

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

Ans: Let the two numbers in $R$be ${{x}_{1}}$ and ${{x}_{2}}$.

Then, we have: 

$ {{x}_{1}}<{{x}_{2}}\Rightarrow 2{{x}_{1}}<2{{x}_{2}} $ 

$ \Rightarrow {{e}^{2{{x}_{1}}}}<{{e}^{2{{x}_{2}}}} $ 

$ \Rightarrow f\left( {{x}_{1}} \right)<f\left( {{x}_{2}} \right)$ 

Hence, $f$ is strictly increasing on $R.$ 

3. Show that the function given by $f\left( x \right)=\sin x$ 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 $\left( 0,n \right)$ 

Ans: The given function is $f\left( x \right)=\sin x$.

$\therefore f'\left( x \right)=\cos x$ 

A) Since for each $x\in \left( 0,\frac{\pi }{2} \right)$, $\cos x>0$, we have $f'\left( x \right)>0$.

Hence, $f$ is strictly increasing in $\left( 0,\frac{\pi }{2} \right)$.

B) Since for each $x\in \left( \frac{\pi }{2},\pi  \right),\cos x<0,$ we have $f'\left( x \right)<0$.

Hence, $f$ is strictly increasing in $\left( \frac{\pi }{2},\pi  \right)$ 

C) It is clear from the results obtained in (A) and (B) that $f$ is neither increasing nor decreasing in $\left( 0,n \right)$.

4. Find the intervals in which the function $f$ given by $f\left( x \right)=2{{x}^{2}}-3x$ is

(A) Strictly increasing

(B) Strictly decreasing

Ans: The function provided is $f\left( x \right)=2{{x}^{2}}-3x$.

  $ f'\left( x \right)=4x-3 $ 

  $ \therefore f'\left( x \right)=0 $ 

  $ \Rightarrow x=\frac{3}{4} $ 

Now, the point $\frac{3}{4}$ divides the real line into two disjoint intervals i.e.,  

$\left( -\infty ,\frac{3}{4} \right)$  and $\left( \frac{3}{4},\infty  \right)$ .

divides the real line into two disjoint intervals

In interval, $\left( -\infty ,\frac{3}{4} \right),f'\left( x \right)=4x-3<0$ 

Hence, the given function $\left( f \right)$ is strictly decreasing in interval $\left( -\infty ,\frac{3}{4} \right)$

In interval, $\left( \frac{3}{4},\infty  \right),f'\left( x \right)=4x-3>0$ 

Hence, the given function $\left( f \right)$ is strictly decreasing in interval $\left( \frac{3}{4},\infty  \right)$

5. Find the intervals in which the function $f$ is given as $f\left( x \right)=2{{x}^{2}}-3{{x}^{2}}-36x+7$ 

(A) strictly increasing

(B) strictly decreasing

 Ans: The given function is

 $ f\left( x \right)=2{{x}^{3}}-3{{x}^{2}}-36x+7 $ 

 $ f'\left( x \right)=6{{x}^{2}}-6x-36 $ 

 $ =6\left( {{x}^{2}}-x-6 \right) $ 

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

 $ \therefore f'\left( x \right)=0 $ 

 $ \Rightarrow x=-2,3 $ 

The points $x=-2,3$ divide the real line into three disjoints intervals 

i.e., $\left( -\infty ,-2 \right),\left( -2,3 \right),$and $\left( 3,\infty  \right)$.

divide the real line into three disjoints intervals

In intervals $\left( -\infty ,-2 \right)$ and $\left( 3,\infty  \right)$, ${f}'\left( x \right)$ is positive while in the interval $\left( -2,3 \right),f'\left( x \right)$ is negative.

Hence, the given function $\left( f \right)$is strictly increasing in intervals $\left( -\infty ,-2 \right)\cup \left( 3,\infty  \right)$ while the function $\left( f \right)$ is strictly decreasing in interval $\left( -2,3 \right)$.

6. Find the intervals in which the following functions are strictly 

increasing or decreasing:

 $ \left( a \right){{x}^{2}}+2x-5 $ 

 $ \left( b \right)10-6x-2{{x}^{2}} $ 

 $ \left( c \right)-2{{x}^{3}}-9{{x}^{2}}-12x+1 $ 

 $ \left( d \right)6-9x-{{x}^{2}} $ 

 $ \left( e \right){{\left( x+1 \right)}^{3}}{{\left( x-3 \right)}^{3}} $ 

Ans:  

a) Given that

$ f\left( x \right)={{x}^{2}}+2x-5 $ 

$ \therefore f'\left( x \right)=2x+2 $ 

Now, 

 $ f'\left( x \right)=0 $ 

 $ \Rightarrow x=-1 $

The real line is divided into two disjoint intervals $\left( -\infty ,-1 \right)and\left( -1,\infty  \right)$ by the point $x=-1$.

In interval $\left( -\infty ,-1 \right),$ 

$\therefore f'\left( x \right)=2x+2<0$ 

Therefore, $f$ is strictly decreasing in the interval $\left( -\infty ,-1 \right)$.

Thus $f$ is strictly decreasing for $x<-1$ 

In interval $\left( -1,\infty  \right),$ 

$\therefore f'\left( x \right)=2x+2>0$ 

Therefore, $f$ is strictly decreasing in interval $\left( -1,\infty  \right)$ 

Thus $f$ is strictly increasing for $x>-1$.

b) Given that,

$ f\left( x \right)=10-6x-2{{x}^{2}} $ 

$ \therefore f'\left( x \right)=-6-4x $ 

Now,

$ f'\left( x \right)=0 $ 

$ \Rightarrow x=\frac{-3}{2} $ 

The real line is divided into two disjoint intervals  $\left( -\infty ,-\frac{3}{2} \right)and\left( -\frac{3}{2},\infty  \right)$ by the point $x=-\frac{3}{2}$.

In interval $\left( -\infty ,-\frac{3}{2} \right)$ i.e., when $x<-\frac{3}{2},f'\left( x \right)=-6-4x<0.$ 

$\therefore f$ is strictly increasing for $x<-\frac{3}{2}$ 

In interval $\left( -\frac{3}{2},\infty  \right)$ i.e., when $x>\frac{3}{2},f'\left( x \right)=-6-4x<0$.

$\therefore f$ is strictly increasing for $x<-\frac{3}{2}$ 

c) Given that,

 $ f\left( x \right)=-2{{x}^{3}}-9{{x}^{2}}-12x+1 $ 

 $ \therefore f'\left( x \right)=-6{{x}^{2}}-18x-12 $ 

 $ =-6\left( {{x}^{2}}+3x+2 \right) $ 

 $ =-6\left( x+1 \right)\left( x+2 \right) $ 

Now,

 $ f'\left( x \right)=0 $ 

 $ \Rightarrow x=-1 $ 

and $x=-2$ 

The real line is divided into three disjoint interval $\left( -\infty ,-2 \right),\left( -2,-1 \right)and\left( -1,\infty  \right)$ by the points $x=-1,-2$  when $x<-2$ and $x>-1$,

$f'\left( x \right)=-6\left( x+1 \right)\left( x+2 \right)<0$ 

$\therefore f$ is strictly increasing for $x<-2<x>-1$ 

Now, in interval $\left( -2,-1 \right)$ i.e., when $-2<x<-1$,

$f'\left( x \right)=-6\left( x+1 \right)\left( x+2 \right)>0$ 

$\therefore f$ is strictly increasing for $-2<x<-1$

d) Given that,

$ f\left( x \right)=6-9x-{{x}^{2}} $ 

$ \therefore f'\left( x \right)=-9-2x $ 

Now, $f'\left( x \right)=0$ gives $x=-\frac{9}{2}$ 

The real line is divided into two disjoint intervals $\left( -\infty ,-\frac{9}{2} \right)and\left( -\frac{9}{2},\infty  \right)$ by the point $x=-\frac{9}{2}$.

In interval $\left( -\infty ,-\frac{9}{2} \right)$ i.e., for $x<\frac{9}{2},$ 

$\therefore f$  is strictly increasing for $x<\frac{9}{2}$

In interval $\left( \frac{9}{2},\infty  \right)$ i.e., for $x<-\frac{9}{2},f'\left( x \right)=-9-2x>0$ 

$\therefore f$  is strictly increasing for $x<-\frac{9}{2}$

In interval $\left( -\frac{9}{2},\infty  \right)$ i.e., for $x>-\frac{9}{2},f'\left( x \right)=-9-2x<0$ 

$\therefore f$  is strictly increasing for $x>-\frac{9}{2}$

e) Given that,

$ f\left( x \right)={{\left( x+1 \right)}^{3}}{{\left( x-3 \right)}^{3}} $ 

$ f'\left( x \right)=3x{{\left( x+1 \right)}^{2}}{{\left( x-3 \right)}^{3}}+3{{\left( x-3 \right)}^{2}}{{\left( x+1 \right)}^{3}} $ 

$ =3{{\left( x+1 \right)}^{2}}{{\left( x-3 \right)}^{2}}\left[ x-3+x+1 \right] $ 

$ =6{{\left( x+1 \right)}^{2}}{{\left( x-3 \right)}^{2}}\left( x-1 \right) $ 

Now, 

$ f'\left( x \right)=0 $ 

$ \Rightarrow x=-1,3,1 $  

The points $x=-1,1,3$ divided the real line into four disjoint intervals. i.e., $\left( -\infty ,-1 \right),\left( -1,1 \right),\left( 1,3 \right)and\left( 3,\infty  \right)$ 

In intervals $\left( -\infty ,-1 \right)and\left( -1,1 \right),f'\left( x \right)=6{{\left( x+1 \right)}^{2}}{{\left( x-3 \right)}^{2}}\left( x-1 \right)<0$ 

$\therefore f$ is strictly decreasing in interval $\left( -\infty ,-1 \right)and\left( -1,1 \right)$ 

In intervals $\left( 1,3 \right)and\left( 3,\infty  \right),f'\left( x \right)=6{{\left( x+1 \right)}^{2}}{{\left( x-3 \right)}^{2}}\left( x-1 \right)>0$ 

$\therefore f$ is strictly decreasing in the interval $\left( 1,3 \right)and\left( 3,\infty  \right)$.

7. Show that $y=\log \left( 1+x \right)-\frac{2x}{2+x},x>-1,$ is an increasing function of 

$x$ throughout its domain.

Ans: We are given that,

$ y=\log \left( 1+x \right)-\frac{2x}{2+x} $ 

$ \therefore \frac{dy}{dx}=\frac{1}{1+x}-\frac{\left( 2+x \right)\left( 2 \right)-2x\left( 1 \right)}{{{\left( 2+x \right)}^{2}}} $ 

 $ =\frac{{{x}^{2}}}{{{\left( 2+x \right)}^{2}}} $ 

Now,

$ \frac{dy}{dx}=0 $ 

 $ \Rightarrow \frac{{{x}^{2}}}{{{\left( 2+x \right)}^{2}}}=0 $ 

 $ \Rightarrow {{x}^{2}}=0 $ 

$ \Rightarrow x=0 $  

Since $x>-1$ point $x=0$ divides the domain $\left( -1,\infty  \right)$ in two disjoint 

intervals i.e., $-1<x<0$ and $x>0.$ 

 When $-1<x<0$we have:

  $ x<0\Rightarrow {{x}^{2}}>0 $ 

 $ x>-1\Rightarrow \left( 2+x \right)>0 $ 

 $ \Rightarrow {{\left( 2+x \right)}^{2}}>0 $ 

 \[\therefore y=\frac{{{x}^{2}}}{{{\left( 2+x \right)}^{2}}}>0\]

Also, when $x>0$ 

 $ x<0\Rightarrow {{x}^{2}}>0, $ 

 $ {{\left( 2+x \right)}^{2}}>0 $ 

 $ \therefore y=\frac{{{x}^{2}}}{{{\left( 2+x \right)}^{2}}}>0 $ 

 Hence, the function $f$ is increasing throughout this domain.

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

Ans: We have,         

  $ y={{\left[ x\left( x-2 \right) \right]}^{2}} $ 

 $ ={{\left[ {{x}^{2}}-2x \right]}^{2}} $ 

 $ \therefore \frac{dy}{dx}=2\left( {{x}^{2}}-2x \right)\left( 2x-2 \right) $ 

 $ =4x\left( x-2 \right)\left( x-1 \right) $ 

 $ \therefore \frac{dy}{dx}=0 $ 

 $ \Rightarrow x=0,1,2 $ 

The points $x=0,1,2$ divide the real line into four disjoint intervals i.e., 

$\left( -\infty ,0 \right),\left( 0,1 \right),\left( 1,2 \right)and\left( 2,\infty  \right).$ 

In intervals $\left( -\infty ,0 \right)and\left( 1,2 \right),\frac{dy}{dx}<0$ 

$\therefore y$ is strictly decreasing in intervals $\left( -\infty ,0 \right)and\left( 1,2 \right)$ 

However, in intervals $\left( 0,1 \right)and\left( 2,\infty  \right)$, $\frac{dy}{dx}>0$ 

$\therefore y$ is strictly decreasing in intervals $\left( 0,1 \right)and\left( 2,\infty  \right)$

$\therefore y$ is strictly decreasing in intervals $0<x<1$  and $x>2$.

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

Ans: We have,

 $ y=\frac{4\sin \theta }{\left( 2+\cos \theta  \right)}-\theta  $ 

 $ \therefore \frac{dy}{d\theta }=\frac{\left( 2+\cos \theta  \right)\left( 4\cos \theta  \right)-4\sin \theta \left( -\sin \theta  \right)}{{{\left( 2+\cos \theta  \right)}^{2}}}-1 $ 

 $ =\frac{8\cos \theta +4{{\cos }^{2}}\theta +4{{\sin }^{2}}\theta }{{{\left( 2+\cos \theta  \right)}^{2}}}-1 $ 

 $ =\frac{8\cos \theta +4}{{{\left( 2+\cos \theta  \right)}^{2}}}-1 $ 

Now,           

$ \frac{dy}{d\theta }=0 $ 

$ \Rightarrow \frac{8\cos \theta +4}{{{\left( 2+\cos \theta  \right)}^{2}}}=1 $ 

$ \Rightarrow 8\cos \theta +4=4+{{\cos }^{2}}\theta +4\cos \theta  $ 

$ \Rightarrow {{\cos }^{2}}\theta -4\cos \theta =0 $ 

$ \Rightarrow \cos \theta =0,4 $ 

Since, 

$ \cos \theta \ne 4, $ 

$ \cos \theta =0 $ 

$ \Rightarrow \theta =\frac{\pi }{2} $ 

Now, $\frac{dy}{d\theta }=\frac{\cos \left( 4-\cos \theta  \right)}{{{\left( 2+\cos  \right)}^{2}}}$ 

In the interval $\left[ 0,\frac{\pi }{2} \right]$, we have $\cos \theta >0$,

Also, $4>\cos \theta \Rightarrow 4-\cos \theta >0$ 

$\therefore \cos \theta \left( 4-\cos \theta  \right)>0$ and also ${{\left( 2+\cos \theta  \right)}^{2}}>0$ 

$ \Rightarrow \frac{\cos \theta \left( 4-\cos \theta  \right)}{{{\left( 2+\cos \theta  \right)}^{2}}}>0 $ 

$ \Rightarrow \frac{dy}{dx}>0 $  

Therefore $y$ is strictly increasing in the interval $\left( 0,\frac{\pi }{2} \right)$ 

Hence, $y$ is increasing in the interval $\left( 0,\frac{\pi }{2} \right)$.

10. Prove that the logarithmic function is strictly increasing on $\left( 0,\infty  \right)$.

Ans: The given function is $f\left( x \right)\log x$.

\[f'\left( x \right)=\frac{1}{x}\] 

It is clear that for $x>0,f'\left( x \right)=\frac{1}{x}>0$ 

Hence, $f\left( x \right)\log x$ is strictly increasing in the interval $\left( 0,\infty  \right)$.

11. Prove that the function $f$ is given by $f\left( x \right)={{x}^{2}}-x+1$ is neither strictly increasing nor strictly decreasing on $\left( -1,1 \right)$.

Ans: The given function is $f\left( x \right)={{x}^{2}}-x+1$ 

$\therefore f'\left( x \right)=2x-1$ 

Now,

$ f'\left( x \right)=0 $ 

 $ \Rightarrow x=\frac{1}{2}. $ 

The point $\frac{1}{2}$ divides the interval $\left( -1,1 \right)$ into two disjoint intervals i.e.,    

$\left( -1,\frac{1}{2} \right)and\left( \frac{1}{2},1 \right).$ 

Now, in the interval $\left( -1,\frac{1}{2} \right),f'\left( x \right)=2x-1<0$ 

Therefore, $f$ is strictly decreasing in the interval $\left( -1,\frac{1}{2} \right)$,

However, in the interval $\left( \frac{1}{2},1 \right),f'\left( x \right)=2x-1>0$     

Therefore, $f$ is strictly decreasing in the interval $\left( \frac{1}{2},1 \right)$,

Hence, $f$ is neither strictly increasing nor decreasing in interval 

$\left( -1,1 \right)$.

12. Which of the following functions are strictly decreasing on $\left( 0,\frac{\pi }{2} \right)$?

(A) \[\mathbf{cosx}\] (B) $\cos 2x$  (C) $\cos 3x$  (D) $\tan x$ 

 Ans. 

A) Let ${{f}_{1}}\left( x \right)=\cos x$ 

$\therefore {{f}_{1}}^{\prime }\left( x \right)=-\sin x$ 

In interval, $\left( 0,\frac{\pi }{2} \right),{{f}_{1}}^{\prime }\left( x \right)=-\sin x<0$ 

$\therefore {{f}_{1}}\left( x \right)=\cos x$ is strictly decreasing in the interval $\left( 0,\frac{\pi }{2} \right)$.

B) Let ${{f}_{2}}\left( x \right)=\cos 2x$ 

$\therefore {{f}_{2}}^{\prime }\left( x \right)=-2\sin 2x$ 

Now,

$ 0<x<\frac{\pi }{2}\Rightarrow 0<2x<\pi  $ 

$ \Rightarrow \sin 2x>0\Rightarrow -2\sin 2x<0 $ 

$\therefore {{f}_{2}}^{\prime }\left( x \right)=-2\sin 2x<0$  on $\left( 0,\frac{\pi }{2} \right)$ 

$\therefore {{f}_{2}}\left( x \right)=\cos 2x$ is strictly decreasing in interval $\left( 0,\frac{\pi }{2} \right)$$$ 

C) Let ${{f}_{3}}\left( x \right)=\cos 3x$ 

$\therefore {{f}_{3}}^{\prime }\left( x \right)=-3\sin 3x$ 

Now,

  $ {{f}_{3}}^{\prime }\left( x \right)=0 $ 

 $ \Rightarrow \sin 3x=0 $ 

 $ \Rightarrow 3x=\pi ,x\in \left( 0,\frac{\pi }{2} \right) $ 

 $ \Rightarrow x=\frac{\pi }{3} $ 

The point $x=\frac{\pi }{3}$  divides the interval $\left( 0,\frac{\pi }{2} \right)$  into two disjoint intervals i.e., $\left( 0,\frac{\pi }{3} \right)$  and $\left( \frac{\pi }{3},\frac{\pi }{2} \right)$.

Now, in the interval $\left( 0,\frac{\pi }{3} \right)$, ${{f}_{3}}\left( x \right)=-3\sin 3x<0\left[ 0<x<\frac{\pi }{3}\Rightarrow 0<3x<\pi  \right]$ 

$\therefore {{f}_{3}}$ is strictly decreasing in interval x$\left( 0,\frac{\pi }{3} \right)$ 

However, in the interval $\left( \frac{\pi }{3},\frac{\pi }{2} \right)$, ${{f}_{3}}\left( x \right)=-3\sin 3x>0\left[ \frac{\pi }{3}<x<\frac{\pi }{2}\Rightarrow \pi <3x<\frac{3\pi }{2} \right]$ 

$\therefore {{f}_{3}}$ is strictly increasing in the interval $\left( \frac{\pi }{3},\frac{\pi }{2} \right)$ 

Hence, ${{f}_{3}}$ is neither increasing nor decreasing in interval $\left( 0,\frac{\pi }{2} \right)$

D) Let ${{f}_{4}}\left( x \right)=\tan x$ 

$\therefore {{f}_{4}}^{\prime }\left( x \right)={{\sec }^{2}}x$ 

In interval $\left( 0,\frac{\pi }{2} \right)$,${{f}_{4}}^{\prime }\left( x \right)={{\sec }^{2}}x>0$ 

\[∴f_4\] is strictly increasing in the interval $\left( 0,\frac{\pi }{2} \right)$.

Therefore, functions $\cos x$and $\cos 2x$  are strictly decreasing in $\left( 0,\frac{\pi }{2} \right)$.

Hence, the correct options are A and B.

13. On which of the following intervals is the function $f$ is given by 

$f\left( x \right)={{x}^{100}}+\sin x-1$ strictly decreasing?

A. $\left( 0,1 \right)$ 

B. $\left( \frac{\pi }{2},\pi  \right)$ 

C. $\left( 0,\frac{\pi }{2} \right)$ 

D. None of these

Ans: We have,

$ f\left( x \right)={{x}^{100}}+\sin x-1 $ 

$ \therefore {f}'\left( x \right)=100{{x}^{90}}+\cos x $ 

In interval $\left( 0,1 \right),\cos x>0$ and $100{{x}^{90}}>0$ 

$\therefore f'\left( x \right)>0$ 

Thus, the function $f$ is strictly increasing in the interval $\left( 0,1 \right)$.

In interval $\left( \frac{\pi }{2},\pi  \right),\cos x<0$  and \[\text{100}{{\text{x}}^{90}}>0\].

Also, $100{{x}^{90}}>\cos x$ 

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

Thus, the function $f$ is strictly increasing in the interval $\left( \frac{\pi }{2},\pi  \right)$.

In the interval $\left( 0,\frac{\pi }{2} \right)$, $\cos x<0$ and $100{{x}^{90}}>0$ 

$ \therefore 100{{x}^{90}}+\cos x>0 $ 

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

$\therefore f$ is strictly increasing in the interval $\left( 0,\frac{\pi }{2} \right)$

Hence, the function $f$ is strictly decreasing in none of the intervals. The correct option is D.

14. Find the least value of $a$ such that the function $f$ given 

$f\left( x \right)={{x}^{2}}+ax+1$ is strictly increasing on $\left( 1,2 \right)$.

Ans: We have,

$ f\left( x \right)={{x}^{2}}+ax+1 $ 

$ \therefore {f}'\left( x \right)=2x+a $ 

Now, the function $f$ will be increasing in $\left( 1,2 \right)$ if ${f}'\left( x \right)>0$ in $\left( 1,2 \right)$.

  $ \Rightarrow 2x+a>0 $ 

 $ \Rightarrow 2x>-a $ 

 $ \Rightarrow x>\frac{-a}{2} $ 

 As a result, we must determine the smallest value of $a$ such that 

 $x>\frac{-a}{2},x\in \left( 1,2 \right)$ 

 $\Rightarrow x>\frac{-a}{2}\left( 1<x<2 \right)$ 

 Thus, the least value is given by

  $ \frac{-a}{2}=1 $ 

 $ \Rightarrow a=-2 $ 

Hence, the required value of $a$ is $-2$.

15. Let I be any interval disjoint from $\left( -1,1 \right)$, prove that the function $f$ given by $f\left( x \right)=x+\frac{1}{x}$ is strictly increasing on I.

Ans: We have,

$ f\left( x \right)=x+\frac{1}{x} $ 

 $ \therefore {f}'\left( x \right)=1-\frac{1}{{{x}^{2}}} $ 

Now, 

$ f\left( x \right)=0 $ 

$ \Rightarrow x=\pm 1 $ 

The real line is divided into three disjoint intervals i.e., 

$\left( -\infty ,-1 \right),\left( -1,1 \right),\left( 1,\infty  \right)$ by the points $x=1,-1$.

In the interval $\left( -1,1 \right)$ we observe,

$ -1<x<1 $ 

$ \Rightarrow {{x}^{2}}<1 $ 

$ \Rightarrow 1<\frac{1}{{{x}^{2}}},x\ne 0 $ 

$ \therefore f'\left( x \right)=1-\frac{1}{{{x}^{2}}}<0on\left( -1,1 \right)\sim \left\{ 0 \right\}. $ 

$\therefore f$ is strictly decreasing on $\left( -1,1 \right)\sim \left\{ 0 \right\}$.

In the interval, $\left( -\infty ,-1 \right)and\left( 1,\infty  \right),$ it is observed that:

$ x<-1 $ 

$ \Rightarrow {{x}^{2}}>1 $ 

$ \Rightarrow 1>\frac{1}{{{x}^{2}}} $ 

$ \Rightarrow 1-\frac{1}{{{x}^{2}}}>0 $ 

$\therefore f'\left( x \right)=1-\frac{1}{{{x}^{2}}}>0$  on $\left( -\infty ,-1 \right)and\left( 1,\infty  \right).$

$\therefore f$ is strictly increasing on $\left( -\infty ,-1 \right)and\left( 1,\infty  \right).$

Hence, the function $f$is strictly increasing in the interval I disjoint  

from $\left( -1,1 \right)$.

Hence, the given result is proved.

16. Prove that the function $f$ given by $f\left( x \right)=\log \sin x$ is strictly increasing on $\left( 0,\frac{\pi }{2} \right)$ and strictly decreasing on $\left( \frac{\pi }{2},\pi  \right)$.

Ans: We have,

$ f\left( x \right)=\log \sin x $ 

 $ \therefore f'\left( x \right)=\frac{1}{\sin x}\cos x=\cot x $ 

In interval, $\left( 0,\frac{\pi }{2} \right),f'\left( x \right)=\cot x>0$ 

$\therefore f$ is strictly increasing in $\left( 0,\frac{\pi }{2} \right)$ 

In interval, $\left( \frac{\pi }{2},\pi  \right),f'\left( x \right)=\cot x<0$ 

$\therefore f$ is strictly increasing in $\left( \frac{\pi }{2},\pi  \right)$.

17. Prove that the function $f$ is given by $f\left( x \right)=\log \cos x$ is strictly decreasing on $\left( 0,\frac{\pi }{2} \right)$ and strictly increasing on $\left( \frac{\pi }{2},\pi  \right)$.

Ans: We have,

$ f\left( x \right)=\log \cos x $ 

$ \therefore f'\left( x \right)=\frac{1}{\cos x}\left( -\sin x \right)=-\tan x $ 

In interval $\left( 0,\frac{\pi }{2} \right)$,

$ \tan x>0\Rightarrow -\tan x<0 $ 

$ \therefore f'\left( x \right)<0on\left( 0,\frac{\pi }{2} \right) $ 

$\therefore f$ is strictly decreasing on $\left( 0,\frac{\pi }{2} \right)$ 

In interval $\left( \frac{\pi }{2},\pi  \right)$,

$ \tan x<0\Rightarrow -\tan x>0 $ 

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

$\therefore f$ is strictly decreasing on $\left( \frac{\pi }{2},\pi  \right)$. 

18. Prove that the function given by $f\left( x \right)={{x}^{3}}-3{{x}^{2}}+3x=100$ is increasing in R.

Ans: We have,

$ f\left( x \right)={{x}^{3}}-3{{x}^{2}}+3x=100 $ 

$ f'\left( x \right)=3{{x}^{2}}-6x+3 $ 

$ =3\left( {{x}^{2}}-2x+1 \right) $ 

$ =3{{\left( x-1 \right)}^{2}} $ 

For any $x\in R$ 

${{\left( x-1 \right)}^{2}}>0$ 

Thus, $f'\left( x \right)$ is always positive in R.

Hence, the given function $f$ is increasing in R.

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

A) $\left( -\infty ,\infty  \right)$ 

B) $\left( -2,0 \right)$ 

C) $\left( 2,\infty  \right)$ 

D) $\left( 0,2 \right)$ 

Ans: We have,        

$ y={{x}^{2}}{{e}^{-x}} $ 

 $ \therefore \frac{dy}{dx}=2x{{e}^{-x}}-{{x}^{2}}{{e}^{-x}}=x{{e}^{-x}}\left( 2-x \right) $ 

Now,

$ \frac{dy}{dx}=0 $ 

 $ \Rightarrow x=0,2 $ 

 The points $x=0,2$ divided the real line into the three disjoint intervals 

 i.e., $\left( -\infty ,0 \right),\left( 0,2 \right),\left( 2,\infty  \right).$ 

 In intervals $\left( -\infty ,0 \right)and\left( 2,\infty  \right),f'\left( x \right)<0$ as ${{e}^{-x}}$ is always positive.

 $\therefore f$ is decreasing on $\left( -\infty ,0 \right)and\left( 2,\infty  \right)$.

 In interval $\left( 0,2 \right),f'\left( x \right)>0$ 

 $\therefore f$is strictly increasing on $\left( 0,2 \right)$.

Hence $f$ is strictly increasing in the interval $\left( 0,2 \right)$

Thus, D is the correct option.

NCERT Solutions for Class 12 Maths PDF Download

 

NCERT Solution Class 12 Maths of Chapter 6 All Exercises

Chapter 6 - Application of Derivatives Exercises in PDF Format

Exercise 6.1

18 Questions & Solutions (6 Short Answers, 10 Long Answers, 2 MCQs)

Exercise 6.2

19 Questions & Solutions (7 Short Answers, 10 Long Answers, 2 MCQs)

Exercise 6.3 

27 Questions & Solutions (25 Short Answers, 2 MCQs)

Exercise 6.4

9 Questions & Solutions (7 Short Answers, 2 MCQs)

Exercise 6.5 

29 Questions & Solutions (11 Short Answers, 15 Long Answers, 3 MCQs)


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

Opting for the NCERT solutions for Ex 6.2 Class 12 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 6.2 Class 12 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

Vedantu in-house subject matter experts have solved the problems/ questions from the exercise with the utmost care and by following all the guidelines by CBSE. Class 12 students who are thorough with all the concepts from the Maths textbook and quit well-versed with all the problems from the exercises given in it, then any student can easily score the highest possible marks in the final exam. With the help of this Class 12 Maths Chapter 6 Exercise 6.2 solutions, students can easily understand the pattern of questions that can be asked in the exam from this chapter and also learn the marks weightage of the chapter. So that they can prepare themselves accordingly for the final exam.

Besides these NCERT solutions for Class 12 Maths Chapter 6 Exercise 6.2, there are plenty of exercises in this chapter which contain innumerable questions as well. All these questions are solved/answered by our in-house subject experts as mentioned earlier. Hence all of these are bound to be of superior quality and anyone can refer to these during the time of exam preparation. In order to score the best possible marks in the class, it is really important to understand all the concepts of the textbooks and solve the problems from the exercises given next to it. 

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