Solution

Let ${ }^{x_1}$ and $x_2$ be any two numbers in $\mathbf{R}$.

Then, we have:

$x_1<x_2 \Rightarrow 3 x_1<3 x_2 \Rightarrow 3 x_1+17<3 x_2+17 \Rightarrow f(x_1)<f(x_2)$

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

Alternate method:

$f(x)=3>0$, in every interval of $\mathbf{R}$.

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

Solution

Let ${ }^{x_1}$ and $x_2$ be any two numbers in $\mathbf{R}$.

Then, we have:

$x_1<x_2 \Rightarrow 2 x_1<2 x_2 \Rightarrow e^{2 x_1}<e^{2 x_2} \Rightarrow f(x_1)<f(x_2)$

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

Solution

The given function is $f(x)=\sin x$.

$\therefore f^{\prime}(x)=\cos x$ (a) Since for each $x \in(0, \frac{\pi}{2}), \cos x>0$, we have $f^{\prime}(x)>0$.

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

(b) Since for each $x \in(\frac{\pi}{2}, \pi), \cos x<0$, we have $f^{\prime}(x)<0$.

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

(c) From the results obtained in (a) and (b), it is clear that $f$ is neither increasing nor decreasing in $(0, \pi)$.

Solution

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

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

$\therefore f^{\prime}(x)=0 \Rightarrow x=\frac{3}{4}$

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

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

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

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

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

Solution

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

$f^{\prime}(x)=6 x^{2}-6 x-36=6(x^{2}-x-6)=6(x+2)(x-3)$

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

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

$(-2,3)$.

Solution

(a) We have,

$f(x)=x^{2}+2 x-5$

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

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,

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

$\therefore f^{\prime}(x)=-6-4 x$

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., $(-\infty,-\frac{3}{2})$ and $(-\frac{3}{2}, \infty)$.

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

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

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

$\therefore f$ is strictly decreasing for $x>-\frac{3}{2}$.

(c) We have,

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

$\therefore f^{\prime}(x)=-6 x^{2}-18 x-12=-6(x^{2}+3 x+2)=-6(x+1)(x+2)$

Now,

$f^{\prime}(x)=0 \Rightarrow x=-1$ and $x=-2$

Points $x=-1$ and $x=-2$ divide the real line into three disjoint intervals

i.e., $(-\infty,-2),(-2,-1)$, and $(-1, \infty)$.

In intervals $(-\infty,-2)$ and $(-1, \infty)$ i.e., when $x<-2$ and $x>-1$,

$f^{\prime}(x)=-6(x+1)(x+2)<0$. $\therefore f$ is strictly decreasing for $x<-2$ and $x>-1$.

Now, in interval $(-2,-1)$ i.e., when $-2<x<-1, f^{\prime}(x)=-6(x+1)(x+2)>0$.

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

(d) We have,

$ \begin{aligned} & f(x)=6-9 x-x^{2} \\ & \therefore f^{\prime}(x)=-9-2 x \end{aligned} $

Now, $f^{\prime}$

$(x)=0$ gives $x=-\frac{9}{2}$

The point $x=-\frac{9}{2}$ divides the real line into two disjoint intervals i.e.,

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

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

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

In interval $(-\frac{9}{2}, \infty)$ i.e., for $x>-\frac{9}{2}, f^{\prime}(x)=-9-2 x<0$. $\therefore f$ is strictly decreasing for $x>-\frac{9}{2}$.

(e) We have,

$ \begin{aligned} f(x)= & (x+1)^{3}(x-3)^{3} \\ 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} $

Now,

$f^{\prime}(x)=0 \Rightarrow x=-1,3,1$

The points $x=-1, x=1$, and $x=3$ divide the real line into four disjoint intervals

i.e., $(-\infty,-1),(-1,1),(1,3)$, and $(3, \infty)$.

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

$\therefore f$ is strictly decreasing in intervals $(-\infty,-1)$ and $(-1,1)$.

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

$\therefore f$ is strictly increasing in intervals $(1,3)$ and $(3, \infty)$.

Solution

We have,

$ \begin{aligned} & y=\log (1+x)-\frac{2 x}{2+x} \\ & \therefore \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}}{(2+x)^{2}} \end{aligned} $

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

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

$\Rightarrow x^{2}=0 \quad[(2+x) \neq 0$ as $x>-1]$

$\Rightarrow x=0$

Since $x>-1$, point $x=0$ divides the domain $(-1, \infty)$ in two disjoint intervals i.e., $-1<$ $x<0$ and $x>0$.

When $-1<x<0$, we have:

$x<0 \Rightarrow x^{2}>0$

$x>-1 \Rightarrow(2+x)>0 \Rightarrow(2+x)^{2}>0$

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

Also, when $x>0$ :

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

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

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

Solution

We have,

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

$\therefore \frac{d y}{d x}=y^{\prime}=2(x^{2}-2 x)(2 x-2)=4 x(x-2)(x-1)$

$\therefore \frac{d y}{d x}=0 \Rightarrow x=0, x=2, x=1$.

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 $0<x<1$ and $x>2$.

Solution

We have,

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

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

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

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

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

$\Rightarrow \cos \theta(\cos \theta-4)=0$

$\Rightarrow \cos \theta=0$ or $\cos \theta=4$

Since $\cos \theta \neq 4, \cos \theta=0$.

$\cos \theta=0 \Rightarrow \theta=\frac{\pi}{2}$

Now,

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

In interval $(0, \frac{\pi}{2})$, we have $\cos \theta>0$. Also, $4>\cos \theta \Rightarrow 4-\cos \theta>0$.

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

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

$\Rightarrow \frac{d y}{d x}>0$

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

Also, the given function is continuous at $x=0$ and $x=\frac{\pi}{2}$.

Hence, $y$ is increasing in interval $[0, \frac{\pi}{2}]$.

Solution

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 $(0, \infty)$.

Solution

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

Now, in interval $(-1, \frac{1}{2}), f^{\prime}(x)=2 x-1<0$.

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

However, in interval $(\frac{1}{2}, 1), f^{\prime}(x)=2 x-1>0$.

Therefore, $f$ is strictly increasing in interval $(\frac{1}{2}, 1)$.

Hence, $f$ is neither strictly increasing nor decreasing in interval $(-1,1)$.

Solution

(A) Let $f_1(x)=\cos x$.

$\therefore f_1^{\prime}(x)=-\sin x$

In interval $(0, \frac{\pi}{2}), f_1^{\prime}(x)=-\sin x<0$.

$\therefore f_1(x)=\cos x _{\text{is strictly decreasing in interval }}(0, \frac{\pi}{2})$.

(B) Let $f_2(x)=\cos 2 x$.

$\therefore f_2^{\prime}(x)=-2 \sin 2 x$

Now, $0<x<\frac{\pi}{2} \Rightarrow 0<2 x<\pi \Rightarrow \sin 2 x>0 \Rightarrow-2 \sin 2 x<0$

$\therefore f_2^{\prime}(x)=-2 \sin 2 x<0$ on $(0, \frac{\pi}{2})$

$\therefore f_2(x)=\cos 2 x$ is strictly decreasing in interval $(0, \frac{\pi}{2})$.

(C) Let $f_3(x)=\cos 3 x$.

$\therefore f_3^{\prime}(x)=-3 \sin 3 x$

Now, $f_3^{\prime}(x)=0$.

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

$\Rightarrow x=\frac{\pi}{3}$

The point $x=\frac{\pi}{3}$ divides the interval $(0, \frac{\pi}{2})$ into two disjoint intervals

i.e., $0(0, \frac{\pi}{3})$ and $(\frac{\pi}{3}, \frac{\pi}{2})$.

Now, in interval $(0, \frac{\pi}{3}), f_3(x)=-3 \sin 3 x<0[.$ as $.0<x<\frac{\pi}{3} \Rightarrow 0<3 x<\pi]$.

$\therefore f_3$ is strictly decreasing in interval $(0, \frac{\pi}{3})$.

However, in interval $(\frac{\pi}{3}, \frac{\pi}{2}), f_3(x)=-3 \sin 3 x>0[.$ as $.\frac{\pi}{3}<x<\frac{\pi}{2} \Rightarrow \pi<3 x<\frac{3 \pi}{2}]$. $\therefore f_3$ is strictly increasing in interval $(\frac{\pi}{3}, \frac{\pi}{2})$.

Hence, $f_3$ is neither increasing nor decreasing in interval $(0, \frac{\pi}{2})$.

(D) Let $f_4(x)=\tan x$.

$\therefore f_4^{\prime}(x)=\sec ^{2} x$

In interval $(0, \frac{\pi}{2}), f_4^{\prime}(x)=\sec ^{2} x>0$.

$\therefore f_4$ is strictly increasing in interval $(0, \frac{\pi}{2})$.

Therefore, functions $\cos x$ and $\cos 2 x$ are strictly decreasing in $(0, \frac{\pi}{2})$. Hence, the correct answers are A and B.

Solution

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 $100 x^{99}>0$.

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

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

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

$\therefore f^{\prime}(x)>0$ in $(\frac{\pi}{2}, \pi)$.

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

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

$\therefore 100 x^{99}+\cos x>0$

$\Rightarrow f^{\prime}(x)>0$ on $(0, \frac{\pi}{2})$

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

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

The correct answer is D.

Solution

We have, $f(x)=x^{2}+a x+1$

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

Now, function $f$ will be increasing in $(1,2)$, if $f^{\prime}(x)>0$ in $(1,2)$.

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

$\Rightarrow 2 x+a>0$

$\Rightarrow 2 x>-a$

$\Rightarrow \quad x>\frac{-a}{2}$

Therefore, we have to find the least value of $a$ such that

$x>\frac{-a}{2}$, when $x \in(1,2)$.

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

Thus, the least value of $a$ for $f$ to be increasing on $(1,2)$ is given by,

$\frac{-a}{2}=1$

$\frac{-a}{2}=1 \Rightarrow a=-2$

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

Solution

We have,

$f(x)=x+\frac{1}{x}$

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

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:

$-1<x<1$

$\Rightarrow x^{2}<1$

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

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

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

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

In intervals $(-\infty,-1)$ and $(1, \infty)$, it is observed that: $x<-1$ or $1<x$

$\Rightarrow x^{2}>1$

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

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

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

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

Hence, function $f$ is strictly increasing in interval $\mathbf{I}$ disjoint from $(-1,1)$. Hence, the given result is proved.

Solution

We have,

$f(x)=\log \sin x$

$\therefore f^{\prime}(x)=\frac{1}{\sin x} \cos x=\cot x$

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

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

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

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

Solution

We have,

$f(x)=\log \cos x$

$\therefore f^{\prime}(x)=\frac{1}{\cos x}(-\sin x)=-\tan x$

In interval $(0, \frac{\pi}{2}), \tan x>0 \Rightarrow-\tan x<0$.

$\therefore f^{\prime}(x)<0$ on $(0, \frac{\pi}{2})$

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

In interval $(\frac{\pi}{2}, \pi), \tan x<0 \Rightarrow-\tan x>0$.

$\therefore f^{\prime}(x)>0$ on $(\frac{\pi}{2}, \pi)$ $\therefore f$ is strictly increasing on $(\frac{\pi}{2}, \pi)$.

Solution

We have,

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

$ \begin{aligned} f^{\prime}(x) & =3 x^{2}-6 x+3 \\ & =3(x^{2}-2 x+1) \\ & =3(x-1)^{2} \end{aligned} $

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

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

Hence, the given function $(f)$ is increasing in $\mathbf{R}$.

Solution

We have,

$y=x^{2} e^{-x}$ $\therefore \frac{d y}{d x}=2 x e^{-x}-x^{2} e^{-x}=x e^{-x}(2-x)$

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

$\Rightarrow x=0$ 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.

$\therefore f$ 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.