Chapter 6: Applications of Derivatives

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 

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

 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.

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.

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 $-2$.

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.

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

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

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.

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.