Q: 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 $(0, \pi)$

It is given that $f(x)=\sin x$ Hence, $f^{\prime}(x)=\cos x$ (a) Here, $x \in\left(0, \frac{\pi}{2}\right)$ $$ \begin{aligned} & \Rightarrow \cos x>0 \\ & \Rightarrow f^{\prime}(x)>0 \end{aligned} $$ Thus, $f$ is strictly increasing in $\left(0, \frac{\pi}{2}\right)$. (b) Here, $x \in\left(\frac{\pi}{2}, \pi\right)$ $$ \begin{aligned} & \Rightarrow \cos x<0 \\ & \Rightarrow f^{\prime}(x)<0 \end{aligned} $$ Thus, $f$ is strictly decreasing in $\left(\frac{\pi}{2}, \pi\right)$. (c) Here, $x \in(0, \pi)$ The results obtained in (a) and (b) are sufficient to state that $f$ is neither increasing nor decreasing in $(0, \pi)$.

Q: Find the intervals in which the function $f$ given by $f(x)=2 x^{2}-3 x$ is (a) Strictly increasing (b) Strictly decreasing

The given function is $f(x)=2 x^{2}-3 x$ Hence, $$ f^{\prime}(x)=4 x-3 $$ Therefore, $$ \begin{aligned} & f^{\prime}(x)=0 \\ & \Rightarrow x=\frac{3}{4} \end{aligned} $$ ![](https://cdn.mathpix.com/cropped/2025_10_31_68cca1da1b09b0eabfa4g-016.jpg?height=163&width=822&top_left_y=2135&top_left_x=621) In $\left(-\infty, \frac{3}{4}\right), f^{\prime}(x)=4 x-3 0$ Hence, $f$ is strictly increasing in $\left(\frac{3}{4}, \infty\right)$.

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

We have, $$ \begin{aligned} y & =[x(x-2)]^{2} \\ & =\left[x^{2}-2 x\right]^{2} \end{aligned} $$ Therefore, $$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}\left[x^{2}-2 x\right]^{2} \\ & =2\left(x^{2}-2 x\right)(2 x-2) \\ & =4 x(x-2)(x-1) \end{aligned} $$ Now, $\frac{d y}{d x}=0$ Hence, $$ \begin{aligned} & \Rightarrow 4 x(x-2)(x-1) \\ & \Rightarrow x=0, x=2, x=1 \end{aligned} $$ $x=0, x=1$ and $x=2$ divide the number line intervals $(-\infty, 0),(0,1),(1,2)$ and $(2, \infty)$ In $(-\infty, 0)$ and $(1,2), \frac{d y}{d x} 0$ Hence, $y$ is strictly increasing in intervals $(0,1)$ and $(2, \infty)$

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

We have, $y=\frac{4 \sin \theta}{(2+\cos \theta)}-\theta$ Therefore, $$ \begin{aligned} \frac{d y}{d \theta} & =\frac{(2+\cos \theta)(4 \cos \theta)-4 \sin \theta(-\sin \theta)}{(2+\cos \theta)^{2}}-1 \\ & =\frac{8 \cos \theta+4 \cos ^{2} \theta+4 \sin ^{2} \theta}{(2+\cos \theta)^{2}}-1 \\ & =\frac{8 \cos \theta+4}{(2+\cos \theta)^{2}}-1 \end{aligned} $$ Now, $\frac{d y}{d \theta}=0$ Hence, $$ \begin{aligned} & \Rightarrow \frac{8 \cos \theta}{(2+\cos \theta)^{2}}=1 \\ & \Rightarrow 8 \cos \theta+4=4+\cos ^{2} \theta+4 \cos \theta \\ & \Rightarrow \cos ^{2} \theta-4 \cos \theta=0 \\ & \Rightarrow \cos \theta(\cos \theta-4)=0 \\ & \Rightarrow \cos \theta=0 \text { or } \cos \theta=4 \end{aligned} $$ Since, $\cos \theta \neq 4$ Therefore, $$ \begin{aligned} & \cos \theta=0 \\ & \Rightarrow \theta=\frac{\pi}{2} \end{aligned} $$ Now, $$ \begin{aligned} \frac{d y}{d \theta} & =\frac{8 \cos \theta+4-\left(4+\cos ^{2} \theta+4 \cos \theta\right)}{(2+\cos \theta)^{2}} \\ & =\frac{4 \cos \theta-\cos ^{2} \theta}{(2+\cos \theta)^{2}} \\ & =\frac{\cos (4-\cos \theta)}{(2+\cos \theta)^{2}} \end{aligned} $$ In interval $\left[0, \frac{\pi}{2}\right]$, we have $\cos \theta>0$ Also, $$ \begin{aligned} & 4>\cos \theta \\ & \Rightarrow 4-\cos \theta>0 \end{aligned} $$ Hence, $\cos \theta(4-\cos \theta)>0$ and also $(2+\cos \theta)^{2}>0$ Therefore, $$ \frac{\cos \theta(4-\cos \theta)}{(2+\cos \theta)^{2}}>0 $$ Hence, $\frac{d y}{d \theta}>0$ So, $y$ is strictly increasing in $\left(0, \frac{\pi}{2}\right)$ and the given function is continuous at $x=0$ and $x=\frac{\pi}{2}$ Thus, $y$ is increasing in interval $\left[0, \frac{\pi}{2}\right]$.

Question 1

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

Accepted Answer

Let $x_{1}$ and $x_{2}$ be any two numbers in $\mathbf{R}$.

Then,

$$

x_{1}<x_{2} \Rightarrow 3 x_{1}+17<3 x_{2}+17=f\left(x_{1}\right)<f\left(x_{2}\right)

$$

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

Question 2

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

Accepted Answer

Let $x_{1}$ and $x_{2}$ be any two numbers in $\mathbf{R}$.

Then,

$$

x_{1}<x_{2} \Rightarrow 2 x_{1}<2 x_{2} \Rightarrow e^{2 x_{1}}<e^{2 x_{2}}=f\left(x_{1}\right)<f\left(x_{2}\right)

$$

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

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 $(0, \pi)$

Accepted Answer

It is given that $f(x)=\sin x$ Hence, $f^{\prime}(x)=\cos x$ (a) Here, $x \in\left(0, \frac{\pi}{2} ight)$ $$ \begin{aligned} & \Rightarrow \cos x>0 \ & \Rightarrow f^{\prime}(x)>0 \end{aligned} $$ Thus, $f$ is strictly increasing in $\left(0, \frac{\pi}{2} ight)$. (b) Here, $x \in\left(\frac{\pi}{2}, \pi ight)$ $$ \begin{aligned} & \Rightarrow \cos x<0 \ & \Rightarrow f^{\prime}(x)<0 \end{aligned} $$ Thus, $f$ is strictly decreasing in $\left(\frac{\pi}{2}, \pi ight)$. (c) Here, $x \in(0, \pi)$ The results obtained in (a) and (b) are sufficient to state that $f$ is neither increasing nor decreasing in $(0, \pi)$.

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

Accepted Answer

The given function is $f(x)=2 x^{2}-3 x$

Hence,

$$

f^{\prime}(x)=4 x-3

$$

Therefore,

$$

\begin{aligned}

& f^{\prime}(x)=0 \

& \Rightarrow x=\frac{3}{4}

\end{aligned}

$$

![](https://cdn.mathpix.com/cropped/2025_10_31_68cca1da1b09b0eabfa4g-016.jpg?height=163&width=822&top_left_y=2135&top_left_x=621)

In $\left(-\infty, \frac{3}{4}ight), f^{\prime}(x)=4 x-3<0$

Hence, $f$ is strictly decreasing in $\left(-\infty, \frac{3}{4}ight)$.

In $\left(\frac{3}{4}, \inftyight), f^{\prime}(x)=4 x-3>0$

Hence, $f$ is strictly increasing in $\left(\frac{3}{4}, \inftyight)$.

Question 5

Find the intervals in which the function $f$ given $f(x)=2 x^{3}-3 x^{2}-36 x+7$ is

(a) Strictly increasing

(b) Strictly decreasing

Accepted Answer

The given function is $f(x)=2 x^{3}-3 x^{2}-36 x+7$

Hence,

$$

\begin{aligned}

f^{\prime}(x) & =6 x^{2}-6 x-36 \

& =6\left(x^{2}-x-6ight) \

& =6(x+2)(x-3)

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

& f^{\prime}(x)=0 \

& \Rightarrow x=-2,3

\end{aligned}

$$

![](https://cdn.mathpix.com/cropped/2025_10_31_68cca1da1b09b0eabfa4g-017.jpg?height=113&width=860&top_left_y=1544&top_left_x=608)

In $(-\infty,-2)$ and $(3, \infty), f^{\prime}(x)>0$

In $(-2,3), f^{\prime}(x)<0$

Hence, $f$ is strictly increasing in $(-\infty,-2),(3, \infty)$ and strictly decreasing in $(-2,3)$.

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) $6-9 x-9 x^{2}$

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

Accepted Answer

(a) $f(x)=x^{2}+2 x-5$ Hence, $$ f^{\prime}(x)=2 x+2 $$ Therefore, $$ \begin{aligned} & \Rightarrow f^{\prime}(x)=0 \ & \Rightarrow x=-1 \end{aligned} $$ $x=-1$ divides the number line into intervals $(-\infty, 1)$ and $(-1, \infty)$ In $(-\infty, 1), f^{\prime}(x)=2 x+2<0$ Thus, $f$ is strictly decreasing in $(-\infty, 1)$ In $(-1, \infty), f^{\prime}(x)=2 x+2>0$ Thus, $f$ is strictly increasing in ( $-1, \infty$ ) (b) $\quad f(x)=10-6 x-2 x^{2}$ Hence, $$ f^{\prime}(x)=-6-4 x $$ Therefore, $$ \begin{aligned} & \Rightarrow f^{\prime}(x)=0 \ & \Rightarrow x=-\frac{3}{2} \end{aligned} $$ $x=-\frac{3}{2}$, divides the number line into two intervals $\left(-\infty,-\frac{3}{2} ight)$ and $\left(-\frac{3}{2}, \infty ight)$ In $\left(-\infty,-\frac{3}{2} ight), f^{\prime}(x)=-6-4 x<0$ Hence, $f$ is strictly increasing for $x<-\frac{3}{2}$ In $\left(-\frac{3}{2}, \infty ight), f^{\prime}(x)=-6-4 x>0$ Hence, $f$ is strictly increasing for $x>-\frac{3}{2}$ (c) $\quad f(x)=-2 x^{3}-9 x^{2}-12 x+1$ Hence, $$ \begin{aligned} f^{\prime}(x) & =-6 x^{2}-18 x-12 \ & =-6\left(x^{2}+3 x+2 ight) \ & =-6(x+1)(x+2) \end{aligned} $$ Therefore, $$ \begin{aligned} & \Rightarrow f^{\prime}(x)=0 \ & \Rightarrow x=-1,2 \end{aligned} $$ $x=-1$ and $x=-2$ divide the number line into intervals $(-\infty,-2),(-2,-1)$ and $(-1, \infty)$. In $(-\infty,-2)$ and $(-1, \infty), f^{\prime}(x)=-6(x+1)(x+2)<0$ Hence, $f$ is strictly decreasing for $x<-2$ and $x>-1$ In $(-2,-1), f^{\prime}(x)=-6(x+1)(x+2)>0$ Hence, $f$ is strictly increasing in $-2-\frac{9}{2}$ In $\left(-\infty,-\frac{9}{2} ight), f^{\prime}(x)>0$ Hence, $f$ is strictly decreasing in $x>-\frac{9}{2}$ (e) $\quad f(x)=(x+1)^{3}(x-3)^{3}$ Hence, $$ \begin{aligned} f^{\prime}(x) & =3(x+1)^{2}(x-3)^{3}+3(x-3)^{2}(x+1)^{3} \ & =3(x+1)^{2}(x-3)^{2}[x-3+x+1] \ & =3(x+1)^{2}(x-3)^{2}(2 x-2) \ & =6(x+1)^{2}(x-3)^{2}(x-1) \end{aligned} $$ Therefore, $$ \begin{aligned} & f^{\prime}(x)=0 \ & \Rightarrow x=-1,3,1 \end{aligned} $$ $x=-1,3,1$ divides the number line into four intervals $(-\infty,-1),(-1,1),(1,3)$ and $(3, \infty)$ In $(-\infty,-1)$ and $(-1,1), f^{\prime}(x)=6(x+1)^{2}(x-3)^{2}(x-1)<0$ Hence, $f$ is strictly decreasing in $(-\infty,-1)$ and $(-1,1)$ In $(1,3)$ and $(3, \infty), f^{\prime}(x)=6(x+1)^{2}(x-3)^{2}(x-1)>0$ Hence, $f$ is strictly increasing in $(1,3)$ and $(3, \infty)$

Question 7

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

Accepted Answer

It is given that $y=\log (1+x)-\frac{2 x}{2+x}$ Therefore, $$ \begin{aligned} \frac{d y}{d x} & =\frac{1}{1+x}-\frac{(2+x)(2)-2 x(1)}{(2+x)^{2}} \ & =\frac{1}{1+x}-\frac{4}{(2+x)^{2}} \ & =\frac{x^{2}}{(1+x)(2+x)^{2}} \end{aligned} $$ Now, $\frac{d y}{d x}=0$ Hence, $$ \begin{aligned} & \Rightarrow \frac{x^{2}}{(2+x)^{2}}=0 \ & \Rightarrow x^{2}=0 \ & \Rightarrow x=0 \end{aligned} $$ Since, $x>-1, x=0$ divides domain $(-1, \infty)$ in two intervals $-10$ When, $-10 \ & x>-1 \Rightarrow(2+x)>0 \ & \Rightarrow(2+x)^{2}>0 \end{aligned} $$ Hence, $$ y=\frac{x^{2}}{(2+x)^{2}}>0 $$ When, $x>0$ Then, $$ \begin{aligned} & x>0 \Rightarrow x^{2}>0 \ & \Rightarrow(2+x)^{2}>0 \end{aligned} $$ Hence, $$ y=\frac{x^{2}}{(2+x)^{2}}>0 $$ Thus, $f$ is increasing throughout the domain.

Question 8

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

Accepted Answer

We have,

$$

\begin{aligned}

y & =[x(x-2)]^{2} \

& =\left[x^{2}-2 xight]^{2}

\end{aligned}

$$

Therefore,

$$

\begin{aligned}

\frac{d y}{d x} & =\frac{d}{d x}\left[x^{2}-2 xight]^{2} \

& =2\left(x^{2}-2 xight)(2 x-2) \

& =4 x(x-2)(x-1)

\end{aligned}

$$

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

Hence,

$$

\begin{aligned}

& \Rightarrow 4 x(x-2)(x-1) \

& \Rightarrow x=0, x=2, x=1

\end{aligned}

$$

$x=0, x=1$ and $x=2$ divide the number line intervals $(-\infty, 0),(0,1),(1,2)$ and $(2, \infty)$

In $(-\infty, 0)$ and $(1,2), \frac{d y}{d x}<0$

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

In $(0,1)$ and $(2, \infty), \frac{d y}{d x}>0$

Hence, $y$ is strictly increasing in intervals $(0,1)$ and $(2, \infty)$

Question 9

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

Accepted Answer

We have, $y=\frac{4 \sin 	heta}{(2+\cos 	heta)}-	heta$

Therefore,

$$

\begin{aligned}

\frac{d y}{d 	heta} & =\frac{(2+\cos 	heta)(4 \cos 	heta)-4 \sin 	heta(-\sin 	heta)}{(2+\cos 	heta)^{2}}-1 \

& =\frac{8 \cos 	heta+4 \cos ^{2} 	heta+4 \sin ^{2} 	heta}{(2+\cos 	heta)^{2}}-1 \

& =\frac{8 \cos 	heta+4}{(2+\cos 	heta)^{2}}-1

\end{aligned}

$$

Now, $\frac{d y}{d 	heta}=0$

Hence,

$$

\begin{aligned}

& \Rightarrow \frac{8 \cos 	heta}{(2+\cos 	heta)^{2}}=1 \

& \Rightarrow 8 \cos 	heta+4=4+\cos ^{2} 	heta+4 \cos 	heta \

& \Rightarrow \cos ^{2} 	heta-4 \cos 	heta=0 \

& \Rightarrow \cos 	heta(\cos 	heta-4)=0 \

& \Rightarrow \cos 	heta=0 	ext { or } \cos 	heta=4

\end{aligned}

$$

Since, $\cos 	heta 
eq 4$

Therefore,

$$

\begin{aligned}

& \cos 	heta=0 \

& \Rightarrow 	heta=\frac{\pi}{2}

\end{aligned}

$$

Now,

$$

\begin{aligned}

\frac{d y}{d 	heta} & =\frac{8 \cos 	heta+4-\left(4+\cos ^{2} 	heta+4 \cos 	hetaight)}{(2+\cos 	heta)^{2}} \

& =\frac{4 \cos 	heta-\cos ^{2} 	heta}{(2+\cos 	heta)^{2}} \

& =\frac{\cos (4-\cos 	heta)}{(2+\cos 	heta)^{2}}

\end{aligned}

$$

In interval $\left[0, \frac{\pi}{2}ight]$, we have $\cos 	heta>0$

Also,

$$

\begin{aligned}

& 4>\cos 	heta \

& \Rightarrow 4-\cos 	heta>0

\end{aligned}

$$

Hence, $\cos 	heta(4-\cos 	heta)>0$ and also $(2+\cos 	heta)^{2}>0$

Therefore,

$$

\frac{\cos 	heta(4-\cos 	heta)}{(2+\cos 	heta)^{2}}>0

$$

Hence, $\frac{d y}{d 	heta}>0$

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

Question 10

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

Accepted Answer

The given function is $f(x)=\log x$

Therefore, $f^{\prime}(x)=\frac{1}{x}$

For, $x>0, f^{\prime}(x)=\frac{1}{x}>0$

Thus, the logarithmic function is strictly increasing in interval $(0, \infty)$.

Question 11

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

Accepted Answer

The given function is $f(x)=x^{2}-x+1$

Therefore,

$$

f^{\prime}(x)=2 x-1

$$

Now,

$$

\begin{aligned}

& f^{\prime}(x)=0 \

& \Rightarrow x=\frac{1}{2}

\end{aligned}

$$

$x=\frac{1}{2}$ divides the interval $v$ into $\left(-1, \frac{1}{2}ight)$ and $\left(\frac{1}{2}, 1ight)$

In interval $\left(\frac{1}{2}, 1ight), f^{\prime}(x)=2 x-1>0$

Hence, $f$ is strictly decreasing in $\left(-1, \frac{1}{2}ight)$

In interval $\left(\frac{1}{2}, 1ight), f^{\prime}(x)=2 x-1>0$

Hence, $f$ is strictly increasing in $\left(\frac{1}{2}, 1ight)$

Thus, $f$ is strictly increasing nor strictly decreasing in interval $(-1,1)$

Question 12

Which of the following principles are strictly decreasing on $\left(0, \frac{\pi}{2}ight)$ ?

(A) $\cos x$

(B) $\cos 2 x$

(C) $\cos 3 x$

(D) $	an x$

Accepted Answer

(A) Let $f_{1}(x)=\cos x$ Therefore, $f_{1}^{\prime}(x)=-\sin x$ In interval $\left(0, \frac{\pi}{2} ight), f_{1}^{\prime}(x)=-\sin x<0$ Thus, $\cos x$ is strictly decreasing in $\left(0, \frac{\pi}{2} ight)$. (B) Let $f_{2}(x)=\cos 2 x$ Therefore, $f_{2}^{\prime}(x)=-2 \sin 2 x$ Now, $$ \begin{aligned} & \Rightarrow 00 \ & \Rightarrow-2 \sin 2 x<0 \end{aligned} $$ Hence, $f_{2}^{\prime}(x)=-2 \sin 2 x<0$ in $\left(0, \frac{\pi}{2} ight)$ Thus, $\cos 2 x$ is strictly decreasing in $\left(0, \frac{\pi}{2} ight)$ (C) Let $f_{3}(x)=\cos 3 x$ Therefore, $f_{3}^{\prime}(x)=-3 \sin 3 x$ Now, $$ \begin{aligned} & f_{3}^{\prime}(x)=0 \ & \Rightarrow \sin 3 x=0 \ & \Rightarrow 3 x=\pi \quad\left[\because x \in\left(0, \frac{\pi}{2} ight) ight] \ & \Rightarrow x=\frac{\pi}{3} \end{aligned} $$ The point $x=\frac{\pi}{3}$, divides $\left(0, \frac{\pi}{2} ight)$ into $\left(0, \frac{\pi}{3} ight)$ and $\left(\frac{\pi}{3}, \frac{\pi}{2} ight)$ In interval $\left(0, \frac{\pi}{3} ight), f_{3}(x)=-3 \sin 3 x<0 \quad\left[\begin{array}{l}00 \quad\left[\frac{\pi}{3}0$ Thus, $ an x$ is strictly increasing in $\left(0, \frac{\pi}{2} ight)$ Thus, the correct options are $\mathbf{A}$ and $\mathbf{B}$.

Question 13

On which of the following intervals is the function $f$ is given by $f(x)=x^{100}+\sin x-1$ is strictly decreasing?

(A) $(0,1)$

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

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

(D) None of these

Accepted Answer

We have,

$$

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

$$

Therefore,

$$

f^{\prime}(x)=100 x^{99}+\cos x

$$

In interval $(0,1), \cos x>0$ and $100 x^{99}>0$

Hence, $f^{\prime}(x)>0$

Thus, $f$ is strictly increasing in $(0,1)$

In interval $\left(\frac{\pi}{2}, \pi\right), \cos x<0$ and $100 x^{99}>0$

Hence, $f^{\prime}(x)>0$

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

Now, in interval $\left(0, \frac{\pi}{2}\right), \cos x>0$ and $100 x^{99}>0$

Hence, $f^{\prime}(x)>0$

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

Hence, $f$ is strictly decreasing in none of the intervals.

Thus, the correct option is $\mathbf{D}$.

Question 14

For what values of $a$ the function $f$ given $f(x)=x^{2}+a x+1$ is increasing on $[1,2]$ ?

Accepted Answer

We have

$$

f(x)=x^{2}+a x+1

$$

Therefore,

$$

f^{\prime}(x)=2 x+a

$$

Now, the function $f$ is strictly increasing on $[1,2]$

Therefore,

$$

\begin{aligned}

& \Rightarrow f^{\prime}(x)>0 \

& \Rightarrow 2 x+a>0 \

& \Rightarrow 2 x>-a \

& \Rightarrow x>\frac{-a}{2}

\end{aligned}

$$

Here, we have $1 \leq x \leq 2$

Thus,

$$

\begin{aligned}

& \frac{-a}{2}>1 \

& a>-2

\end{aligned}

$$

Question 15

Let $\mathbf{I}$ be any interval disjoint from $[-1,1]$. Prove that the function $f$ given by $f(x)=x+\frac{1}{x}$ is increasing on $\mathbf{I}$.

Accepted Answer

We have $$ f(x)=x+\frac{1}{x} $$ Therefore, $$ f^{\prime}(x)=1-\frac{1}{x^{2}} $$ Now, $$ \begin{aligned} & f^{\prime}(x)=0 \ & \Rightarrow 1-\frac{1}{x^{2}}=0 \ & \Rightarrow x^{2}=1 \ & \Rightarrow x= \pm 1 \end{aligned} $$ The points $x=1$ and $x=-1$ divide the real line intervals $(-\infty, 1),(-1,1)$ and $(1, \infty)$ In interval $(-1,1),-11 \ & \Rightarrow 1>\frac{1}{x^{2}} \ & \Rightarrow 1-\frac{1}{x^{2}}>0 \end{aligned} $$ Therefore, $f^{\prime}(x)=1-\frac{1}{x^{2}}>0$ on $(-\infty,-1)$ and $(1, \infty)$ Hence, $f$ is strictly increasing on $(-\infty,-1)$ and $(1, \infty)$ Thus, $f$ is strictly increasing in $\mathbf{I}$ in $[-1,1]$

Question 16

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

Accepted Answer

We have

$$

f(x)=\log \sin x

$$

Therefore,

$$

\begin{aligned}

f^{\prime}(x) & =\frac{1}{\sin x} \cos x \

& =\cot x

\end{aligned}

$$

In interval $\left(0, \frac{\pi}{2}ight), f^{\prime}(x)=\cot x>0$

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

In interval $\left(\frac{\pi}{2}, \piight), f^{\prime}(x)=\cot x<0$

Hence, $f$ is strictly decreasing in $\left(\frac{\pi}{2}, \piight)$.

Question 17

Prove that the function $f$ given by $f(x)=\log |\cos x|$ is decreasing on $\left(0, \frac{\pi}{2}\right)$ and increasing on $\left(\frac{3 \pi}{2}, 2 \pi\right)$.

Accepted Answer

We have $f(x)=\log |\cos x|$ Therefore, $$ \begin{aligned} f^{\prime}(x) & =\frac{1}{\cos x}(-\sin x) \ & =- an x \end{aligned} $$ In interval $\left(0, \frac{\pi}{2} ight), an x>0 \Rightarrow- an x<0$ Hence, $f^{\prime}(x)<0$ Thus, $f$ is strictly decreasing on $\left(0, \frac{\pi}{2} ight)$. In interval $\left(\frac{3 \pi}{2}, 2 \pi ight), an x<0 \Rightarrow- an x>0$ Hence, $f^{\prime}(x)>0$ Thus, $f$ is strictly increasing on $\left(\frac{3 \pi}{2}, 2 \pi ight)$.

Question 18

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

Accepted Answer

We have

$$

f(x)=x^{3}-3 x^{2}+3 x-100

$$

Therefore,

$$

\begin{aligned}

f^{\prime}(x) & =3 x^{2}-6 x+3 \

& =3\left(x^{2}-2 x+1ight) \

& =3(x-1)^{2}

\end{aligned}

$$

For $x \in \mathbf{R},(x-1)^{2} \geq 0$

So, $f^{\prime}(x)$ is always positive in $\mathbf{R}$.

Thus, the function is increasing in $\mathbf{R}$.

Question 19

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

(A) $(-\infty, \infty)$

(B) $(-2,0)$

(C) $(2, \infty)$

(D) $(0,2)$

Accepted Answer

We have $y=x^{2} e^{-x}$

Therefore,

$$

\begin{aligned}

\frac{d y}{d x} & =2 x e^{-x}-x^{2} e^{-x} \

& =x e^{-x}(2-x)

\end{aligned}

$$

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

Hence, $x=0$ and $x=2$

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.

Hence, $f$ is decreasing on $(-\infty, 0)$ and $(2, \infty)$

In interval $(0,2), f^{\prime}(x)>0$

Hence, $f$ is strictly increasing in ( 0,2 )

Thus, the correct option is $\mathbf{D}$.

\section*{EXERCISE 6.3}