Exercise 6.2 (Revised) - Chapter 6 - Applications Of Derivatives - Ncert Solutions class 12 - Maths

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NCERT Class 12 Maths Solutions: Chapter 6 - Applications Of Derivatives

Ex 6.2 Question 1:

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

Answer:
Let $x_1$ and $x_2$ be any two numbers in $\mathbf{R}$.
Then, we have:
$
x_1 $

Hence, $f$ is strictly increasing on $\mathbf{R}$.

Ex 6.2 Question 2:

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

Answer :
Let $x_1$ and $x_2$ be any two numbers in $\mathrm{R}$.
Then, we have:
$
x_1 $

Hence, $f$ is strictly increasing on $\mathbf{R}$.

Ex 6.2 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 $(0, \pi)$

Answer:
The given function is $f(x)=\sin x$.
$
\therefore f^{\prime}(x)=\cos x
$
(a) Since for each $x \in\left(0, \frac{\pi}{2}\right), \cos x>0$, we have $f^{\prime}(x)>0$.

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 $f^{\prime}(x)<0$.

Hence, $f$ is strictly decreasing in $\left(\frac{\pi}{2}, \pi\right)$.
(c) From the results obtained in (a) and (b), it is clear that $f$ is neither increasing nor decreasing in ( 0 , $\pi)$.

Ex 6.2 Question 4:

Find the intervals in which the function $f$ given by $f(x)=2 x^2-3 x$ is
(a) strictly increasing (b) strictly decreasing

Answer :
The given function is $f(x)=2 x^2-3 x$.
$
\begin{aligned}
& f^{\prime}(x)=4 x-3 \\
& \therefore f^{\prime}(x)=0 \Rightarrow x=\frac{3}{4}
\end{aligned}
$

Now, the point $\frac{3}{4}$ divides the real line into two disjoint intervals i.e., $\left(-\infty, \frac{3}{4}\right)$ and $\left(\frac{3}{4}, \infty\right)$.

In interval $\left(-\infty, \frac{3}{4}\right), f^{\prime}(x)=4 x-3<0$.
Hence, the given function $(f)$ is strictly decreasing in interval $\left(-\infty, \frac{3}{4}\right)$.
In interval $\left(\frac{3}{4}, \infty\right), f^{\prime}(x)=4 x-3>0$.
Hence, the given function $(f)$ is strictly increasing in interval $\left(\frac{3}{4}, \infty\right)$.

Ex 6.2 Question 5 :

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

Answer :
The given function is $f(x)=2 x^3-3 x^2-36 x+7$.
$
\begin{aligned}
& f^{\prime}(x)=6 x^2-6 x-36=6\left(x^2-x-6\right)=6(x+2)(x-3) \\
& \therefore f^{\prime}(x)=0 \Rightarrow x=-2,3
\end{aligned}
$

The points $x=-2$ and $x=3$ divide the real line into three disjoint intervals i.e., $(-\infty,-2),(-2,3)$, and $(3, \infty)$.

In intervals $(-\infty,-2)$ and $(3, \infty), f^{\prime}(x)$ is positive while in interval $(-2,3), f^{\prime}(x)$ is negative.

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

Ex 6.2 Question 6:

Find the intervals in which the following functions are strictly increasing or decreasing:
(a) $x^2+2 x-5$
(b) $10-6 x-2 x^2$
(c) $-2 x^3-9 x^2-12 x+1$
(d) $6-9 x-x^2$
(e) $(x+1)^3(x-3)^3$

Answer :
(a) We have,
$
\begin{aligned}
& f(x)=x^2+2 x-5 \\
& \therefore f^{\prime}(x)=2 x+2
\end{aligned}
$

Now,
$
f^{\prime}(x)=0 \Rightarrow x=-1
$

Point $x=-1$ divides the real line into two disjoint intervals i.e., $(-\infty,-1)$ and $(-1, \infty)$.
In interval $(-\infty,-1), f^{\prime}(x)=2 x+2<0$.
$\therefore f$ is strictly decreasing in interval $(-\infty,-1)$.
Thus, $f$ is strictly decreasing for $x<-1$.
In interval $(-1, \infty), f^{\prime}(x)=2 x+2>0$.
$\therefore f$ is strictly increasing in interval $(-1, \infty)$.

Thus, $f$ is strictly increasing for $x>-1$.
(b) We have,
$
\begin{aligned}
& f(x)=10-6 x-2 x^2 \\
& \therefore f^{\prime}(x)=-6-4 x
\end{aligned}
$

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

The point $x=-\frac{3}{2}$ divides the real line into two disjoint intervals i.e., $\left(-\infty,-\frac{3}{2}\right)$ and $\left(-\frac{3}{2}, \infty\right)$.
In interval $\left(-\infty,-\frac{3}{2}\right)$ i.e., when $x<-\frac{3}{2}$,

Ex 6.2 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
$
y=\log (1+x)-\frac{2 x}{2+x}, x>-1
$

We need to show that $\mathrm{y}$ is strictly increasing function for $x>-1$ i.e. we need to show that $\frac{d y}{d x}>0$ for $x>-1$

Finding $\frac{d y}{d x}$
$
\begin{aligned}
& y=\log (1+x)-\frac{2 x}{2+x} \\
& \frac{d y}{d x}=\frac{d\left(\log (1+x)-\frac{2 x}{2+x}\right)}{d x}
\end{aligned}
$

$\begin{aligned}
& \frac{d y}{d x}=\frac{d(\log (1+x))}{d x}-\frac{d}{d x}\left(\frac{2 x}{2+x}\right) \\
& \frac{d y}{d x}=\frac{1}{1+x} \cdot(1+x)^{\prime}-\left[\frac{(2 x)^{\prime}(2+x)-(2+x)^{\prime} 2 x}{(2+x)^2}\right] \\
& \frac{d y}{d x}=\frac{1}{1+x} \cdot(0+1)-\left[\frac{2(2+x)-(0+1) 2 x}{(2+x)^2}\right] \\
& \frac{d y}{d x}=\frac{1}{1+x}-\left[\frac{4+2 x-2 x}{(2+x)^2}\right] \\
& \frac{d y}{d x}=\frac{1}{1+x}-\left[\frac{4}{(2+x)^2}\right] \\
& \frac{d y}{d x}=\frac{(2+x)^2-4(1+x)}{(1+x)(2+x)^2} \\
& \frac{d y}{d x}=\frac{(2)^2+(x)^2+2(2)(x)-4-4 x}{(1+x)(2+x)^2}
\end{aligned}$

$
\begin{aligned}
& \frac{d y}{d x}=\frac{4+x^2+4 x-4-4 x}{(1+x)(2+x)^2} \\
& \frac{d y}{d x}=\frac{x^2}{(1+x)(2+x)^2} \\
& \frac{d y}{d x}=\left(\frac{x}{2+x}\right)^2 \frac{1}{1+x}
\end{aligned}
$

Now,
$
\frac{d y}{d x}=\left(\frac{x}{2+x}\right)^2 \frac{1}{1+x}
$

Always positive (as square is always positive)

Now, finding value where $\frac{d y}{d x}>0$
$
\begin{aligned}
& \frac{d y}{d x}>0 \\
& \left(\frac{x}{2+x}\right)^2 \cdot\left(\frac{1}{1+x}\right)>0 \\
& \left(\frac{1}{1+x}\right)>0
\end{aligned}
$

This is possible only when
$
1+x>0
$
i.e. $x>-1$

So, $\frac{d y}{d x}>0$ for $x>-1$
Hence proved

Ex 6.2 Question 8:

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

Answer:
We have,
$
\begin{aligned}
& y=[x(x-2)]^2=\left[x^2-2 x\right]^2 \\
& \therefore \frac{d y}{d x}=y^{\prime}=2\left(x^2-2 x\right)(2 x-2)=4 x(x-2)(x-1) \\
& \therefore \frac{d y}{d x}=0 \Rightarrow x=0, x=2, x=1 .
\end{aligned}
$

The points $x=0, x=1$, and $x=2$ divide the real line into four disjoint intervals i.e., $(-\infty, 0),(0,1)(1,2)$, and $(2, \infty)$.

In intervals $(-\infty, 0)$ and $(1,2), \frac{d y}{d x}<0$.
$\therefore y$ is strictly decreasing in intervals $(-\infty, 0)$ and $(1,2)$.
However, in intervals $(0,1)$ and $(2, \infty), \frac{d y}{d x}>0$.
$\therefore y$ is strictly increasing in intervals $(0,1)$ and $(2, \infty)$.
$\therefore y$ is strictly increasing for $02$.

Ex 6.2 Question 9 :

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

Answer:
We have,
$
\begin{aligned}
& y=\frac{4 \sin \theta}{(2+\cos \theta)}-\theta \\
& \therefore \frac{d y}{d x}=\frac{(2+\cos \theta)(4 \cos \theta)-4 \sin \theta(-\sin \theta)}{(2+\cos \theta)^2}-1 \\
& =\frac{8 \cos \theta+4 \cos ^2 \theta+4 \sin ^2 \theta}{(2+\cos \theta)^2}-1 \\
& =\frac{8 \cos \theta+4}{(2+\cos \theta)^2}-1 \\
&
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow 8 \cos \theta+4=4+\cos ^2 \theta+4 \cos \theta \\
& \Rightarrow \cos ^2 \theta-4 \cos \theta=0 \\
& \Rightarrow \cos \theta(\cos \theta-4)=0 \\
& \Rightarrow \cos \theta=0 \text { or } \cos \theta=4 \\
& \text { Since } \cos \theta \neq 4, \cos \theta=0 . \\
& \cos \theta=0 \Rightarrow \theta=\frac{\pi}{2}
\end{aligned}
$

Now,
$
\frac{d y}{d x}=\frac{8 \cos \theta+4-\left(4+\cos ^2 \theta+4 \cos \theta\right)}{(2+\cos \theta)^2}=\frac{4 \cos \theta-\cos ^2 \theta}{(2+\cos \theta)^2}=\frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^2}
$

In interval $\left(0, \frac{\pi}{2}\right)$, we have $\cos \theta>0$. Also, $4>\cos \theta \Rightarrow 4-\cos \theta>0$.
$
\begin{aligned}
& \therefore \cos \theta(4-\cos \theta)>0 \text { and also }(2+\cos \theta)^2>0 \\
& \Rightarrow \frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^2}>0 \\
& \Rightarrow \frac{d y}{d x}>0
\end{aligned}
$

Therefore, $y$ is strictly increasing in interval $\left(0, \frac{\pi}{2}\right)$.
Also, the given function is continuous at $x=0$ and $x=\frac{\pi}{2}$.
Hence, $y$ is increasing in interval $\left[0, \frac{\pi}{2}\right]$.

Ex 6.2 Question 10:

Prove that the logarithmic function is strictly increasing on $\left(0, \angle Å^3 / 4\right)$.

Answer:
The given function is $f(x)=\log x$.
$
\therefore f^{\prime}(x)=\frac{1}{x}
$

It is clear that for $x>0, f^{\prime}(x)=\frac{1}{x}>0$.
Hence, $f(x)=\log x$ is strictly increasing in interval $(0, \infty)$.

Ex 6.2 Question 11:

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

Answer:
The given function is $f(x)=x^2-x+1$.
$
\therefore f^{\prime}(x)=2 x-1
$

Now, $f^{\prime}(x)=0 \Rightarrow x=\frac{1}{2}$.
The point $\frac{1}{2}$ divides the interval $(-1,1)$ into two disjoint intervals i.e., $\left(-1, \frac{1}{2}\right)$ and $\left(\frac{1}{2}, 1\right)$.
Now, in interval $\left(-1, \frac{1}{2}\right), f^{\prime}(x)=2 x-1<0$.
Therefore, $f$ is strictly decreasing in interval $\left(-1, \frac{1}{2}\right)$.
However, in interval $\left(\frac{1}{2}, 1\right), f^{\prime}(x)=2 x-1>0$.
Therefore, $f$ is strictly increasing in interval $\left(\frac{1}{2}, 1\right)$.
Hence, $f$ is neither strictly increasing nor decreasing in interval $(-1,1)$.

Ex 6.2 Question 12:

Which of the following functions are strictly decreasing on $\left(0, \frac{\pi}{2}\right)$ ?
(A) $\cos x(\mathrm{~B}) \cos 2 x(\mathrm{C}) \cos 3 x$ (D) $\tan x$

Answer :

(A) $f(x)=\cos x$
$
\Rightarrow f^{\prime}(x)=-\sin x
$

Since $0<x<\frac{\pi}{2}$
thus for $\mathrm{x}$ in $\left(0, \frac{\pi}{2}\right), \quad \sin \mathrm{x}$ is positive in first quadrent
$
\begin{aligned}
& \Rightarrow \sin x>0 \\
& \Rightarrow-\sin x<0
\end{aligned}
$

So, $\quad f^{\prime}(x)<0$
Therefore, $f(x)$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$.
(B) $f(x)=\cos 2 x$
$
\Rightarrow f^{\prime}(x)=-2 \sin 2 x
$

Since $0<x<\frac{\pi}{2}$
$\therefore 0<2 x<\pi$ therefore $\sin 2 x>0$
$
\Rightarrow-2 \sin 2 x<0
$

So, $f^{\prime}(x)<0$
Therefore, $f(x)$ is strictly decreasing on $\left(0, \frac{\pi}{2}\right)$.
(C) $f(x)=\cos 3 x$
$
\Rightarrow f^{\prime}(x)=-3 \sin 3 x
$

Since $0<x<\frac{\pi}{2}$
$
\therefore 0<3 x<\frac{3 \pi}{2}
$
thus two cases $0<3 x<\pi$ and $\pi<3 x<\frac{3 \pi}{2}$

For $0<3 x<\pi$
$
\sin 3 x>0
$

$
\Rightarrow-3 \sin 3 x<0
$

So, $f^{\prime}(x)<0$
Therefore, $f(x)$ is strictly decreasing on $\left(0, \frac{\pi}{3}\right)$.
For $\pi<3 x<\frac{3 \pi}{2}$
$
\begin{gathered}
\sin 3 x<0 \\
\Rightarrow-3 \sin 3 x>0
\end{gathered}
$

So, $\quad f^{\prime}(x)>0$
Therefore, $f(x)$ is strictly increasing on $\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$.
Hence, $f(x)$ is neither strictly increasing not strictly decreasing on $\left(0, \frac{\pi}{2}\right)$.
$
\begin{aligned}
& \text { (D) } f(x)=\tan x \\
& \Rightarrow f^{\prime}(x)=\sec ^2 x>0
\end{aligned}
$

Therefore, $f(x)$ is strictly increasing on $\left(0, \frac{\pi}{2}\right)$.

Ex 6.2 Question 13:

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

Answer:
We have,
$
\begin{aligned}
& f(x)=x^{100}+\sin x-1 \\
& \therefore f^{\prime}(x)=100 x^{99}+\cos x
\end{aligned}
$

In interval $(0,1), \cos x>0$ and $100 x^{99}>0$.
$
\therefore f^{\prime}(x)>0 \text {. }
$

Thus, function $f$ is strictly increasing in interval $(0,1)$.
In interval $\left(\frac{\pi}{2}, \pi\right), \cos x<0$ and $100 x^{99}>0$. Also, $100 x^{99}>\cos x$
$
\therefore f^{\prime}(x)>0 \text { in }\left(\frac{\pi}{2}, \pi\right) \text {. }
$

Thus, function $f$ is strictly increasing in interval $\left(\frac{\pi}{2}, \pi\right)$.
In interval $\left(0, \frac{\pi}{2}\right), \cos x>0$ and $100 x^{99}>0$.
$
\therefore 100 x^{99}+\cos x>0
$

$
\Rightarrow f^{\prime}(x)>0 \text { on }\left(0, \frac{\pi}{2}\right)
$
$\therefore f$ is strictly increasing in interval $\left(0, \frac{\pi}{2}\right)$.
Hence, function $f$ is strictly decreasing in none of the intervals.
The correct answer is D.

Ex 6.2 Question 14 :

Find the least value of a such that the function $f$ given $f(x)=x^2+a x+1$ is strictly increasing on $(1,2)$.

Answer :
We have,
$
\begin{aligned}
& f(x)=x^2+a x+1 \\
& \therefore f^{\prime}(x)=2 x+a
\end{aligned}
$

Now, function $f$ will be increasing in $(1,2)$, if $f^{\prime}(x)>0$ in $(1,2)$.
$
\begin{aligned}
& f^{\prime}(x)>0 \\
& \Rightarrow 2 x+a>0 \\
& \Rightarrow 2 x>-a \\
& \Rightarrow x>\frac{-a}{2}
\end{aligned}
$

Therefore, we have to find the least value of $a$ such that
$
\begin{aligned}
& x>\frac{-a}{2} \text {, when } x \in(1,2) . \\
& \Rightarrow x>\frac{-a}{2} \text { (when } 1 \end{aligned}
$

Thus, the least value of $a$ for $f$ to be increasing on $(1,2)$ is given by,
$
\begin{aligned}
& \frac{-a}{2}=1 \\
& \frac{-a}{2}=1 \Rightarrow a=-2
\end{aligned}
$

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

Ex 6.2 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.

Answer :
We have,
$
\begin{aligned}
& f(x)=x+\frac{1}{x} \\
& \therefore f^{\prime}(x)=1-\frac{1}{x^2}
\end{aligned}
$

Now,
$
f^{\prime}(x)=0 \Rightarrow \frac{1}{x^2}=1 \Rightarrow x= \pm 1
$

The points $x=1$ and $x=-1$ divide the real line in three disjoint intervals i.e., $(-\infty,-1),(-1,1)$, and $(1, \infty)$.

In interval $(-1,1)$, it is observed that:
$
\begin{aligned}
& -1 & \Rightarrow x^2<1 \\
& \Rightarrow 1<\frac{1}{x^2}, x \neq 0 \\
& \Rightarrow 1-\frac{1}{x^2}<0, x \neq 0 \\
& \therefore f^{\prime}(x)=1-\frac{1}{x^2}<0 \text { on }(-1,1) \sim\{0\} .
\end{aligned}
$
$\therefore f$ is strictly decreasing on $(-1,1) \sim\{0\}$.

In intervals $(-\infty,-1)$ and $(1, \infty)$, it is observed that:
$
\begin{aligned}
& x<-1 \text { or } 1 & \Rightarrow x^2>1 \\
& \Rightarrow 1>\frac{1}{x^2} \\
& \Rightarrow 1-\frac{1}{x^2}>0 \\
& \therefore f^{\prime}(x)=1-\frac{1}{x^2}>0 \text { on }(-\infty,-1) \text { and }(1, \infty) .
\end{aligned}
$
$\therefore f$ is strictly increasing on $(-\infty, 1)$ and $(1, \infty)$.
Hence, function $f$ is strictly increasing in interval I disjoint from ( - 1, 1).
Hence, the given result is proved.

Ex 6.2 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 $\left(\frac{\pi}{2}, \pi\right)$.

Answer :
We have,
$
\begin{aligned}
& f(x)=\log \sin x \\
& \therefore f^{\prime}(x)=\frac{1}{\sin x} \cos x=\cot x
\end{aligned}
$

In interval $\left(0, \frac{\pi}{2}\right), f^{\prime}(x)=\cot x>0$.
$\therefore f$ is strictly increasing in $\left(0, \frac{\pi}{2}\right)$.
In interval $\left(\frac{\pi}{2}, \pi\right), f^{\prime}(x)=\cot x<0$.
$\therefore f$ is strictly decreasing in $\left(\frac{\pi}{2}, \pi\right)$.

Ex 6.2 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 $\left(\frac{\pi}{2}, \pi\right)$.

Answer :
We have,
$
\begin{aligned}
& f(x)=\log \cos x \\
& \therefore f^{\prime}(x)=\frac{1}{\cos x}(-\sin x)=-\tan x
\end{aligned}
$

In interval $\left(0, \frac{\pi}{2}\right), \tan x>0 \Rightarrow-\tan x<0$.
$
\therefore f^{\prime}(x)<0 \text { on }\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^{\prime}(x)>0 \text { on }\left(\frac{\pi}{2}, \pi\right)
$
$\therefore f$ is strictly increasing on $\left(\frac{\pi}{2}, \pi\right)$.

Ex 6.2 Question 18:

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

Answer :
We have,
$
\begin{aligned}
f(x) & =x^3-3 x^2+3 x-100 \\
f^{\prime}(x) & =3 x^2-6 x+3 \\
& =3\left(x^2-2 x+1\right) \\
& =3(x-1)^2
\end{aligned}
$

For any $x \in \mathrm{R},(x-1)^2>0$.
Thus, $f^{\prime}(x)$ is always positive in $\mathrm{R}$.
Hence, the given function $(f)$ is increasing in $\mathbf{R}$.

Ex 6.2 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 :
We have,
$
\begin{aligned}
& y=x^2 e^{-x} \\
& \therefore \frac{d y}{d x}=2 x e^{-x}-x^2 e^{-x}=x e^{-x}(2-x) \\
& \text { Now, } \frac{d y}{d x}=0 . \\
& \Rightarrow x=0 \text { and } x=2
\end{aligned}
$

Now, $\frac{d y}{d x}=0$.

The points $x=0$ and $x=2$ divide the real line into three disjoint intervals i.e., $(-\infty, 0),(0,2)$, and $(2, \infty)$.

In intervals $(-\infty, 0)$ and $(2, \infty), f^{\prime}(x)<0$ as $e^{-x}$ is always positive.
$\therefore f$ is decreasing on $(-\infty, 0)$ and $(2, \infty)$.
In interval $(0,2), f^{\prime}(x)>0$.
$\therefore f$ is strictly increasing on $(0,2)$.
Hence, $f$ is strictly increasing in interval $(0,2)$.
The correct answer is D.