Solution

We know that $y=f(x)$ will be continuous at $x=a$ if

$\lim _{x \to a^{-}} f(x)=\lim _{x \to a} f(x)=\lim _{x \to a^{+}} f(x)$

Given: $\quad f(x)=x^{3}+2 x^{2}-1$

$ \begin{aligned} & \lim _{x \to 1^{-}} f(x)=\lim _{h \to 0}(1+h)^{3}+2(1+h)^{2}-1=1+2-1=2 \\ & \lim _{x \to 1} f(x)=(1)^{3}+2(1)^{2}-1 \\ & =1+2-1=2 \\ & \lim _{x \to 1^{+}} f(x)=\lim _{arrow}(1+h)^{3}+2(1+h)^{2}-1 \\ & =1+2-1=2 \\ & \lim _{x \to 1^{-}} f(x)=\lim _{x \to 1} f(x)=\lim _{x \to 1^{+}} f(x)=2 . \end{aligned} $

Hence, $f(x)$ is continuous at $x=1$.

Solution

$ \begin{aligned} \lim _{x \to 2^{+}} f(x) & =3 x+5 \\ & =\lim _{h \to 0} 3(2+h)+5=11 \\ \lim _{x \to 2} f(x) & =3 x+5=3(2)+5=11 \\ \lim _{x \to 2^{-}} f(x) & =x^{2}=\lim _{h \to 0}(2-h)^{2} \\ & =\lim _{h \to 0}(2)^{2}+h^{2}-4 h=(2)^{2}=4 \end{aligned} $

Since

$ \lim _{x \to 2^{-}} f(x)=\lim _{x \to 2} f(x) \neq \lim _{x \to 2} f(x) $

Hence $f(x)$ is discontinuous at $x=2$.

Solution

$\quad \lim _{x \to 0^{-}} f(x)=\frac{1-\cos 2 x}{x^{2}}$

$ \begin{aligned} & =\lim _{h \to 0} \frac{1-\cos 2(0-h)}{(0-h)^{2}}=\lim _{h \to 0} \frac{1-\cos (-2 h)}{h^{2}} \\ & =\lim _{h \to 0} \frac{1-\cos 2 h}{h^{2}} \\ & =\lim _{h \to 0} \frac{2 \sin ^{2} h}{h^{2}} \quad[\because 1-\cos \theta=2 \sin ^{2} \frac{\theta}{2}] \end{aligned} $

$ \begin{aligned} & =\lim _{h \to 0} \frac{2 \sin h}{h} \cdot \frac{\sin h}{h}=2 \cdot 1 \cdot 1=2 \quad[\lim _{x \to 0} \frac{\sin x}{x}=1] \\ \lim _{x \to 0^{+}} f(x) & =\frac{1-\cos 2 x}{x^{2}} \\ & =\lim _{h \to 0} \frac{1-\cos 2(0+h)}{(0+h)^{2}}=\lim _{h \to 0} \frac{1-\cos 2 h}{h^{2}} \\ & =\lim _{h \to 0} \frac{2 \sin ^{2} h}{h^{2}}=\frac{2 \sin h}{h} \cdot \frac{\sin h}{h}=2 \cdot 1.1=2 \end{aligned} $

$\lim _{x \to 0} f(x)=5$

As $\lim _{x \to 0^{-}} f(x)=\lim _{x \to 0^{+}} f(x) \neq \lim _{x \to 0} f(x)$

$\therefore f(x)$ is discontinuous at $x=0$.

Solution

$\quad f(x)=\frac{2 x^{2}-3 x-2}{x-2}$

$ \begin{aligned} & =\frac{2 x^{2}-4 x+x-2}{x-2}=\frac{2 x(x-2)+1(x-2)}{x-2} \\ & =\frac{(2 x+1)(x-2)}{x-2}=2 x+1 \\ \lim _{x \to 2^{-}} f(x) & =2 x+1 \\ & =\lim _{h \to 0} 2(2-h)+1=4+1=5 \\ \lim _{x \to 2^{+}} f(x) & =2 x+1 \\ & =\lim _{h \to 0} 2(2+h)+1=4+1=5 \\ \lim _{x \to 2} f(x) & =5 \end{aligned} $

As $\lim _{x \to 2^{-}} f(x)=\lim _{x \to 2^{+}} f(x)=\lim _{x \to 2} f(x)=5$

Hence, $f(x)$ is continuous at $x=2$.

Solution

$\quad \lim _{x \to 4^{-}} f(x)=\frac{|x-4|}{2(x-4)} \quad \begin{cases} \text{ for } x<4,|x-4|=-(x-4) \\ \text{ for } x>4,|x-4|=(x-4) \end{cases} $

$ =\lim _{h \to 0} \frac{-[4-h-4]}{2[4-h-4]}=\lim _{h \to 0} \frac{h}{-2 h}=-\frac{1}{2} $

$\lim _{x \to 4^{+}} f(x)=\frac{|x-4|}{2(x-4)}=\lim _{h \to 0} \frac{[4+h-4]}{2[4+h-4]}=\frac{1}{2}$

$\lim _{x \to 4} f(x)=0$

$\therefore \lim _{x \to 4^{-}} f(x) \neq \lim _{x \to 4^{+}} f(x) \neq \lim _{x \to 4} f(x)$

Hence, $f(x)$ is discontinuous at $x=4$.

Solution

$\quad \lim _{x \to 0^{-}} f(x)=|x| \cos \frac{1}{x}$

$ \begin{aligned} & =\lim _{h \to 0}|0-h| \cos \frac{1}{(0-h)}=\lim _{h \to 0} h \cos \frac{1}{h} \\ & =0 \quad[\because \cos \frac{1}{x} \text{ oscillate between }-1 \text{ and } 1] \end{aligned} $

$\lim _{x \to 0^{+}} f(x)=|x| \cos \frac{1}{x}$

$=\lim _{h \to 0}|0+h| \cos \frac{1}{(0+h)}=\lim _{h \to 0} h \cdot \cos \frac{1}{h}=0$

$\lim _{x \to 0} f(x)=0$

$\lim _{x \to 0^{-}} f(x)=\lim _{x \to 0^{+}} f(x)=\lim _{x \to 0} f(x)=0$

Hence, $f(x)$ is continuous at $x=0$.

Solution

$\quad \lim _{x \to a^{-}} f(x)=|x-a| \sin \frac{1}{x-a}$

$=\lim _{h \to 0}|a-h-a| \cdot \sin \frac{1}{a-h-a}=\lim _{h \to 0} h \cdot \sin \frac{1}{-h}$

$=\lim _{h \to 0}-h \cdot \sin \frac{1}{h} \quad[\because \sin (-\theta)=-\sin \theta]$

$=0 \times[$ a number oscillating between -1 and 1$]$

$=0$

$\lim _{x \to a^{+}} f(x)=|x-a| \sin \frac{1}{x-a}$

$=\lim _{h \to 0}|a+h-a| \cdot \sin \frac{1}{a+h-a}=\lim _{h \to 0} h \cdot \sin \frac{1}{h}$

$=0 \times[$ a number oscillating between -1 and 1$]$

$\lim _{x \to a} f(x)=0$

As $\lim _{x \to a^{-}} f(x)=\lim _{x \to a^{+}} f(x)=\lim _{x \to a} f(x)=0$

Hence, $f(x)$ is continuous at $x=a$.

Solution

$\quad \lim _{x \to 0^{-}} f(x)=\frac{e^{1 / x}}{1+e^{1 / x}}$

$ \begin{aligned} & =\lim _{h \to 0} \frac{e^{\frac{1}{0-h}}}{1+e^{\frac{1}{0-h}}}=\lim _{h \to 0} \frac{e^{-1 / h}}{1+e^{-1 / h}} \\ & =\lim _{h \to 0} \frac{1}{e^{1 / h}(1-e^{-1 / h})}=\lim _{h \to 0} \frac{1}{e^{1 / h}-1}=\frac{1}{e^{1 / 0}-1} \\ & =\frac{1}{e^{\infty}-1}=\frac{1}{0-1}=-1 \quad \quad[\because e^{\infty}=0] \\ \lim _{x \to 0^{+}} f(x) & =\frac{e^{1 / x}}{1+e^{1 / x}} \\ & =\lim _{h \to 0} \frac{e^{\frac{1}{0+h}}}{1+e^{\frac{1}{0+h}}}=\lim _{h \to 0} \frac{e^{1 / h}}{1+e^{1 / h}} \end{aligned} $

$ \begin{aligned} & =\lim _{h \to 0} \frac{1}{e^{-1 / h}(1+e^{1 / h})}=\lim _{h \to 0} \frac{1}{e^{-1 / h}+1} \\ & =\frac{1}{e^{-\infty}+1}=\frac{1}{0+1}=1 \quad[e^{-\infty}=0] \end{aligned} $

$ \lim _{x \to 0} f(x)=0 $

As $\lim _{x \to 0^{-}} f(x) \neq \lim _{x \to 0^{+}} f(x) \neq \lim _{x \to 0} f(x)$

Hence, $f(x)$ is discontinuous at $x=0$.

Solution

$\lim _{x \to 1^{-}} f(x)=\frac{x^{2}}{2}=\lim _{h \to 0} \frac{(1-h)^{2}}{2}=\frac{1}{2}$

$ \begin{aligned} & \lim _{x \to 1} f(x)=\frac{x^{2}}{2}=\frac{(1)^{2}}{2}=\frac{1}{2} \\ & \lim _{x \to 1^{+}} f(x)=2 x^{2}-3 x+\frac{3}{2}=2(1)^{2}-3(1)+\frac{3}{2}=2-3+\frac{3}{2}=\frac{1}{2} \end{aligned} $

As $\lim _{x \to 1^{-}} f(x)=\lim _{x \to 1^{+}} f(x)=\lim _{x \to 1} f(x)=\frac{1}{2}$

Hence, $f(x)$ is continuous at $x=1$.

Solution

$\quad \lim _{x \to 1^{-}} f(x)=|x|+|x-1|=\lim _{h \to 0}|1-h|+|1-h-1|$

$ =|1-0|+|1-0-1|=1+0=1 $

$ \begin{aligned} \lim _{x \to 1^{+}} f(x) & =|x|+|x-1| \\ & =\lim _{h \to 0}|1+h|+|1+h-1|=1+0=1 \\ \lim _{x \to 1} f(x) & =|x|+|x-1|=|1|+|1-1|=1+0=1 \end{aligned} $

As $\lim _{x \to 1^{-}} f(x)=\lim _{x \to 1^{+}} f(x)=\lim _{x \to 1} f(x)$

Hence, $f(x)$ is continuous at $x=1$.

Solution

$\lim _{x \to 5} f(x)=3 x-8$

$ =\lim _{h \to 0} 3(5-h)-8=15-8=7 $

$ \lim _{x \to 5^{+}} f(x)=2 k $

As the function is continuous at $x=5$

$ \begin{aligned} & \therefore \lim _{x \to 5^{-}} f(x)=\lim _{x \to 5^{+}} f(x) \\ & \therefore \quad 7=2 k \Rightarrow k=\frac{7}{2} \end{aligned} $

Hence, the value of $k$ is $\frac{7}{2}$.

Solution

$\quad f(x)=\frac{2^{x+2}-16}{4^{x}-16}=\frac{2^{2} \cdot 2^{x}-16}{(2^{x})^{2}-(4)^{2}}=\frac{4(2^{x}-4)}{(2^{x}-4)(2^{x}+4)}$

$ f(x)=\frac{4}{2^{x}+4} $

$ \begin{aligned} \lim _{x \to 2^{-}} f(x) & =\lim _{h \to 0} \frac{4}{2^{2-h}+4}=\frac{4}{2^{2}+4}=\frac{4}{4+4}=\frac{4}{8}=\frac{1}{2} \\ \lim _{x \to 2} f(x) & =k \end{aligned} $

As the function is continuous at $x=2$.

$\therefore \lim _{x \to 2^{-}} f(x) =\lim _{x \to 2} f(x)$

$\therefore k =\frac{1}{2}$

Hence, value of $k$ is $\frac{1}{2}$.

Solution

$\lim _{x \to 0^{-}} f(x)=\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x}$

$ =\lim _{x \to 0^{-}} \frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} \times \frac{\sqrt{1+k x}+\sqrt{1-k x}}{\sqrt{1+k x}+\sqrt{1-k x}} $

$ \begin{aligned} = & \lim _{x \to 0^{-}} \frac{(1+k x)-(1-k x)}{x[\sqrt{1+k x}+\sqrt{1-k x}]} \\ = & \lim _{x \to 0^{-}} \frac{1+k x-1+k x}{x[\sqrt{1+k x}+\sqrt{1-k x}]} \\ = & \lim _{x \to 0^{-}} \frac{2 k x}{x[\sqrt{1+k x}+\sqrt{1-k x}]} \\ = & \lim _{x \to 0^{-}} \frac{2 k}{\sqrt{1+k x}+\sqrt{1-k x}} \\ = & \lim _{h \to 0} \frac{2 k}{\sqrt{1+k(0-h)}+\sqrt{1-k(0-h)}} \\ = & \frac{2 k}{\sqrt{1}+\sqrt{1}}=\frac{2 k}{2}=k \\ \lim _{x \to 0} f(x)= & \frac{2 x+1}{x-1}=\frac{2(0)+1}{0-1}=\frac{1}{-1}=-1 \end{aligned} $

As the function is continuous at $x=0$.

$ \begin{aligned} \therefore \lim _{x \to 0^{-}} f(x) & =\lim _{x \to 0} f(x) \\ k & =-1 \end{aligned} $

Hence, the value of $k$ is -1 .

Solution

$\lim _{x \to 0^{-}} f(x)=\frac{1-\cos k x}{x \sin x}$

$ \begin{aligned} & =\lim _{h \to 0} \frac{1-\cos k(0-h)}{(0-h) \sin (0-h)}=\lim _{h \to 0} \frac{1-\cos (-k h)}{-h \cdot \sin (-h)} \\ & =\lim _{h \to 0} \frac{1-\cos k h}{h \sin h} \quad \begin{bmatrix} \because \sin (-\theta)=-\sin \theta \\ \cos (-\theta)=\cos \theta \end{bmatrix} \end{aligned} $

$ \begin{aligned} & =\lim _{h \to 0} \frac{2 \sin ^{2} \frac{k h}{2}}{h \sin h} \\ & =\lim _{\substack{h \to 0 \\ k h \to 0}} \frac{2 \sin \frac{k h}{2}}{\frac{k h}{2}} \times \frac{k h}{2} \times \frac{\sin \frac{k h}{2}}{\frac{k h}{2}} \times \frac{k h}{2} \cdot \frac{1}{h \cdot \frac{\sin h}{h} \cdot h} \end{aligned} $

$ \begin{aligned} \begin{bmatrix} =2 \cdot 1 \cdot \frac{k h}{2} \cdot 1 \cdot \frac{k h}{2} \cdot \frac{1}{h^2} \cdot 1 \\ =k^{2} \\ \end{bmatrix} \begin{bmatrix} \lim _{h \to 0} \frac{\sin h}{h}=1 \text{ and } \\ \lim _{k h \to 0} \frac{\sin k h}{k h}=1 \end{bmatrix} \\ \lim _{x \to 0} f(x)=\frac{1}{2} \\ \text{ As } \lim _{x \to 0^{-}} f(x)=\lim _{x \to 0} f(x) \\ \therefore \quad \frac{k^{2}}{2}=\frac{1}{2} \\ \Rightarrow \quad k^{2}=1 \Rightarrow k= \pm 1 \end{aligned} $

Hence, the value of $k$ is $\pm 1$.

Solution

$\quad \lim _{x \to 0^{-}} f(x)=\frac{x}{|x|+2 x^{2}}=\lim _{h \to 0} \frac{0-h}{|0-h|+2(0-h)^{2}}$

$ =\lim _{h \to 0} \frac{-h}{h+2 h^{2}}=\lim _{h \to 0} \frac{-h}{h(1+2 h)} $

$ \begin{aligned} & =\lim _{h \to 0} \frac{-1}{1+2 h}=\frac{-1}{1+2(0)}=-1 \\ \lim _{x \to 0^{+}} f(x) & =\frac{x}{|x|+2 x^{2}}=\lim _{h \to 0} \frac{0+h}{|0+h|+2(0+h)^{2}} \\ & =\lim _{h \to 0} \frac{h}{h+2 h^{2}}=\lim _{h \to 0} \frac{h}{h(1+2 h)}=\frac{1}{1+0}=1 \\ \lim _{x \to 0^{-}} f(x) & \neq \lim _{x \to 0^{+}} f(x) \end{aligned} $

Hence, $f(x)$ is discontinuous at $x=0$ regardless the choice of $k$.

Solution

$\lim _{x \to 4^{-}} f(x)=\frac{x-4}{|x-4|}+a$

$ \begin{aligned} & =\lim _{h \to 0} \frac{4-h-4}{|4-h-4|}+a=\lim _{h \to 0} \frac{-h}{h}+a=-1+a \\ \lim _{x \to 4} f(x) & =a+b \\ \lim _{x \to 4^{+}} f(x) & =\frac{x-4}{|x-4|}+b \\ & =\lim _{h \to 0} \frac{4+h-4}{|4+h-4|}+b=\lim _{h \to 0} \frac{h}{h}+b=1+b \end{aligned} $

As the function is continuous at $x=4$.

$ \begin{matrix} \therefore \quad & \lim _{arrow} f(x)=\lim _{x \to 4} f(x)=\lim _{x \to 4^{+}} f(x) \\ & -1+a=a+b=1+b \\ \therefore \quad & -1+a=a+b \Rightarrow b=-1 \\ & 1+b=a+b \Rightarrow a=1 \end{matrix} $

Hence, the value of $a=1$ and $b=-1$.

Solution

$ \begin{aligned} f(x) & =\frac{1}{x+2} \\ f[f(x)] & =\frac{1}{f(x)+2}=\frac{1}{\frac{1}{x+2}+2}=\frac{1}{\frac{1+2 x+4}{x+2}}=\frac{x+2}{2 x+5} \\ \therefore \quad f[f(x)] & =\frac{x+2}{2 x+5} \end{aligned} $

This function will not be defined and continuous where $2 x+5=0 \Rightarrow x=\frac{-5}{2}$.

Hence, $x=\frac{-5}{2}$ is the point of discontinuity.

Solution

We have $f(t)=\frac{1}{t^{2}+t-2}$

$ \Rightarrow \quad f(t)=\frac{1}{\frac{1}{(x-1)^{2}}+\frac{1}{(x-1)}-2} \quad[\text{ putting } t=\frac{1}{x-1}] $

$ \begin{aligned} & =\frac{1}{\frac{1+x-1-2(x-1)^{2}}{(x-1)^{2}}}=\frac{(x-1)^{2}}{x-2 x^{2}-2+4 x} \\ & =\frac{(x-1)^{2}}{-2 x^{2}+5 x-2}=\frac{(x-1)^{2}}{-(2 x^{2}-5 x+2)} \\ & =\frac{(x-1)^{2}}{-[2 x^{2}-4 x-x+2]}=\frac{(x-1)^{2}}{-[2 x(x-2)-1(x-2)]} \\ & =\frac{(x-1)^{2}}{-(x-2)(2 x-1)}=\frac{(x-1)^{2}}{(2-x)(2 x-1)} \end{aligned} $

So, if $f(t)$ is discontinuous, then $2-x=0 \quad \therefore x=2$

and $2 x-1=0 \quad \therefore x=\frac{1}{2}$

Hence, the required points of discontinuity are 2 and $\frac{1}{2}$.

Solution

Given that $f(x)=|\sin x+\cos x| \quad$ at $x=\pi$

Put $g(x)=\sin x+\cos x$ and $h(x)=|x|$

$\therefore \quad h[g(x)]=h(\sin x+\cos x)=|\sin x+\cos x|$

Now, $g(x)=\sin x+\cos x$ is a continuous function since $\sin x$ and $\cos x$ are two continuous functions at $x=\pi$.

We know that every modulus function is a continuous function everywhere.

Hence, $f(x)=|\sin x+\cos x|$ is continuous function at $x=\pi$.

Solution

We know that a function $f$ is differentiable at a point ’ $a$ ’ in its domain if

$ L f^{\prime}(c)=R f^{\prime}(c) $

where $L f^{\prime}(c)=\lim _{h \to 0} \frac{f(a-h)-f(a)}{-h}$ and

$R f^{\prime}(c)=\lim _{h \to 0} \frac{f(a+h)-f(a)}{h}$

Here, $ f(x)=\begin{cases} x[x], & \text{ if } 0 \leq x<2 \\ (x-1) x, & \text{ if } 2 \leq x<3 \end{cases} \text{ at } x=2. $

$\text{ at } x=2$

$ \begin{aligned} L f^{\prime}(c) & =\lim _{h \to 0} \frac{f(2-h)-f(2)}{-h}=\lim _{h \to 0} \frac{(2-h)[2-h]-(2-1) 2}{-h} \quad[\because[2-h]=1] \\ & =\lim _{h \to 0} \frac{(2-h) \cdot 1-2}{-h} \\ = & \lim _{h \to 0} \frac{2-h-2}{-h}=1 \\ R f^{\prime}(c) & =\lim _{h \to 0} \frac{f(2+h)-f(2)}{h} \\ = & \lim _{h \to 0} \frac{(2+h-1)(2+h)-(2-1) \cdot 2}{h} \\ & =\lim _{h \to 0} \frac{(1+h)(2+h)-2}{h}=\lim _{h \to 0} \frac{2+h+2 h+h^{2}-2}{h} \\ & =\lim _{h \to 0} \frac{3 h+h^{2}}{h}=\lim _{h \to 0} \frac{h(3+h)}{h}=3 \end{aligned} $

$ L f^{\prime}(2) \neq R f^{\prime}(2) $

Hence, $f(x)$ is not differentiable at $x=2$.

Solution

Given that:

$ f(x)=\begin{cases} x^{2} \sin \frac{1}{x}, \text{ if } x \neq 0 \\ 0, \text{ if } x=0 \end{cases} \text{ at } x=0. $

For differentiability we know that:

$ \begin{aligned} & L f^{\prime}(c)=R f^{\prime}(c) \\ & \therefore L f^{\prime}(0)=\lim _{h \to 0} \frac{f(0-h)-f(0)}{-h} \\ & =\lim _{h \to 0} \frac{(0-h)^{2} \sin \frac{1}{(0-h)}-0}{-h}=\frac{h^{2} \cdot \sin (-\frac{1}{h})}{-h} \\ & \begin{matrix} =h \cdot \sin (\frac{1}{h})=0 \times[-1 \leq \sin (\frac{1}{h}) \leq 1] \\ =0 \end{matrix} \\ & R f^{\prime}(0)=\lim _{h \to 0} \frac{f(0+h)-f(0)}{h}=\lim _{h \to 0} \frac{(0+h)^{2} \sin (\frac{1}{0+h})-0}{h} \end{aligned} $

$ \begin{aligned} & =\lim _{h \to 0} \frac{h^{2} \sin (\frac{1}{h})}{h}=\lim _{h \to 0} h \cdot \sin (\frac{1}{h}) \\ & =0 \times[-1 \leq \sin (\frac{1}{h}) \leq 1]=0 \end{aligned} $

So, $L f^{\prime}(0)=R f^{\prime}(0)=0$

Hence, $f(x)$ is differentiable at $x=0$.

Solution

$f(x)$ is differentiable at $x=2$ if

$ \begin{aligned} L f^{\prime}(2) & =R f^{\prime}(2) \\ \therefore \quad L f^{\prime}(2) & =\lim _{h \to 0} \frac{f(2-h)-f(2)}{-h} \\ & =\lim _{h \to 0} \frac{(1+2-h)-(1+2)}{-h}=\lim _{h \to 0} \frac{3-h-3}{-h}=\frac{-h}{-h}=1 \\ R f^{\prime}(2) & =\lim _{h \to 0} \frac{f(2+h)-f(2)}{h} \\ & =\lim _{h \to 0} \frac{[5-(2+h)]-(1+2)}{h}=\lim _{h \to 0} \frac{3-h-3}{h} \\ & =\frac{-h}{h}=-1 \end{aligned} $

So, $\quad L f^{\prime}(2) \neq R f^{\prime}(2)$

Hence, $f(x)$ is not differentiable at $x=2$.

Solution

We have $f(x)=|x-5|$

$ \Rightarrow \quad f(x)=\begin{cases} -(x-5) \text{ if } x-5<0 \text{ or } x<5 \\ x-5 \text{ if } x-5>0 \text{ or } x>5 \end{cases} . $

For continuity at $x=5$

$ \begin{aligned} \text{ L.H.L. } \lim _{h \to 5^{-}} f(x) & =-(x-5) \\ & =\lim _{h \to 0}-(5-h-5)=\lim _{h \to 0} h=0 \\ \text{ R.H.L. } \lim _{x \to 5^{+}} f(x) & =x-5 \\ & =\lim _{h \to 0}(5+h-5)=\lim _{h \to 0} h=0 \end{aligned} $

$ \text{ L.H.L. = R.H.L. } $

So, $f(x)$ is continuous at $x=5$.

Now, for differentiability

$ \begin{aligned} L f^{\prime}(5) & =\lim _{h \to 0} \frac{f(5-h)-f(5)}{-h} \\ & =\lim _{h \to 0} \frac{-(5-h-5)-(5-5)}{-h}=\lim _{h \to 0} \frac{h}{-h}=-1 \\ R f^{\prime}(5) & =\lim _{h \to 0} \frac{f(5+h)-f(5)}{h} \\ & =\lim _{h \to 0} \frac{(5+h-5)-(5-5)}{h}=\lim _{h \to 0} \frac{h-0}{h}=1 \\ \because \quad L f^{\prime}(5) & \neq R f^{\prime}(5) \end{aligned} $

Hence, $f(x)$ is not differentiable at $x=5$.

Solution

Given that: $f: R \to R$ satisfies the equation $f(x+y)=f(x) \cdot f(y)$ $\forall x, y \in R, f(x) \neq 0$.

Let us take any point $x=0$ at which the function $f(x)$ is differentiable.

$\therefore \quad f^{\prime}(0)=\lim _{h \to 0} \frac{f(0+h)-f(0)}{h}$

$$ \begin{align*} 2 & =\lim _{h \to 0} \frac{f(0) \cdot f(h)-f(0)}{h}[\because f(0)=f(h)] \tag{i}\\ \Rightarrow \quad 2 & =\lim _{h \to 0} \frac{f(0)[f(h)-1]}{h} \end{align*} $$

Now $\quad f^{\prime}(x)=\lim _{h \to 0} \frac{f(x+h)-f(x)}{h}$

$ \begin{aligned} & =\lim _{h \to 0} \frac{f(x) \cdot f(h)-f(x)}{h} \quad[\because f(x+y)=f(x) \cdot f(y)] \\ & =\lim _{h \to 0} \frac{f(x)[f(h)-1]}{h}=2 f(x) \quad \text{ from eqn. (i) } \end{aligned} $

Hence, $f^{\prime}(x)=2 f(x)$.

Solution

Let $y=2^{\cos ^{2} x}$

Taking $\log$ on both sides, we get

$ \log y=\log 2^{\cos ^{2} x} \Rightarrow \log y=\cos ^{2} x \cdot \log 2 $

Differentiating both sides w.r.t. $x$

$\Rightarrow \quad \frac{1}{y} \cdot \frac{d y}{d x}=\log 2 \cdot \frac{d}{d x} \cos ^{2} x$

$\Rightarrow \quad \frac{1}{y} \cdot \frac{d y}{d x}=\log 2[2 \cos x \cdot \frac{d}{d x} \cos x]$

$\Rightarrow \quad \frac{1}{y} \cdot \frac{d y}{d x}=\log 2[2 \cos x(-\sin x)]$

$\Rightarrow \quad \frac{1}{y} \cdot \frac{d y}{d x}=\log 2(-\sin 2 x)$

$\frac{d y}{d x}=-y \cdot \log 2 \sin 2 x$

Hence, $\frac{d y}{d x}=-2^{\cos ^{2} x}(\log 2 \sin 2 x)$

Solution

Let $\quad y=\frac{8^{x}}{x^{8}}$

Taking $\log$ on both sides, we get, $\log y=\log \frac{8^{x}}{x^{8}}$

$\Rightarrow \quad \log y=\log 8^{x}-\log x^{8} \Rightarrow \log y=x \log 8-8 \log x$

Differentiating both sides w.r.t. $x$

$\Rightarrow \quad \frac{1}{y} \cdot \frac{d y}{d x}=\log 8.1-\frac{8}{x} \Rightarrow \frac{d y}{d x}=y[\log 8-\frac{8}{x}]$

Hence, $\frac{d y}{d x}=\frac{8^{x}}{x^{8}}[\log 8-\frac{8}{x}]$

Solution

Let $\quad y=\log (x+\sqrt{x^{2}+a})$

Differentiating both sides w.r.t. $x$

$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x} \log (x+\sqrt{x^{2}+a}) \\ & =\frac{1}{x+\sqrt{x^{2}+a}} \cdot \frac{d}{d x}(x+\sqrt{x^{2}+a}) \\ & =\frac{1}{x+\sqrt{x^{2}+a}} \cdot[1+\frac{1}{2 \sqrt{x^{2}+a}} \times \frac{d}{d x}(x^{2}+a)] \end{aligned} $

$ \begin{aligned} & =\frac{1}{x+\sqrt{x^{2}+a}} \cdot[1+\frac{1}{2 \sqrt{x^{2}+a}} \cdot 2 x] \\ & =\frac{1}{x+\sqrt{x^{2}+a}} \cdot[1+\frac{x}{\sqrt{x^{2}+a}}] \\ & =\frac{1}{x+\sqrt{x^{2}+a}} \cdot(\frac{\sqrt{x^{2}+a}+x}{\sqrt{x^{2}+a}})=\frac{1}{\sqrt{x^{2}+a}} \end{aligned} $

Hence, $\frac{d y}{d x}=\frac{1}{\sqrt{x^{2}+a}}$.

Solution

Let $\quad y=\log [\log (\log x^{5})]$

Differentiating both sides w.r.t. $x$

$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x} \log [\log (\log x^{5})] \\ & =\frac{1}{\log (\log x^{5})} \times \frac{d}{d x} \log (\log x^{5}) \\ & =\frac{1}{\log (\log x^{5})} \times \frac{1}{\log (x^{5})} \times \frac{d}{d x} \log x^{5} \\ & =\frac{1}{\log (\log x^{5})} \cdot \frac{1}{\log (x^{5})} \cdot \frac{1}{x^{5}} \cdot \frac{d}{d x} x^{5} \\ & =\frac{1}{\log (\log x^{5})} \cdot \frac{1}{\log (x^{5})} \cdot \frac{1}{x^{5}} \cdot 5 x^{4} \\ & =\frac{5}{x \log (x^{5}) \cdot \log (\log x^{5})} \end{aligned} $

Hence, $\frac{d y}{d x}=\frac{5}{x \log (x^{5}) \cdot \log (\log x^{5})}$.

Solution

Let $\quad y=\sin \sqrt{x}+\cos ^{2} \sqrt{x}$

Differentiating both sides w.r.t. $x$

$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}(\sin \sqrt{x})+\frac{d}{d x}(\cos ^{2} \sqrt{x}) \\ & =\cos \sqrt{x} \cdot \frac{d}{d x}(\sqrt{x})+2 \cos \sqrt{x} \cdot \frac{d}{d x}(\cos \sqrt{x}) \end{aligned} $

$ \begin{aligned} & =\cos \sqrt{x} \cdot \frac{1}{2 \sqrt{x}}+2 \cos \sqrt{x}(-\sin \sqrt{x}) \cdot \frac{d}{d x} \sqrt{x} \\ & =\frac{1}{2 \sqrt{x}} \cdot \cos \sqrt{x}-2 \cos \sqrt{x} \cdot \sin \sqrt{x} \cdot \frac{1}{2 \sqrt{x}} \\ & =\frac{\cos \sqrt{x}}{2 \sqrt{x}}-\frac{\sin 2 \sqrt{x}}{2 \sqrt{x}} \end{aligned} $

Hence, $\frac{d y}{d x}=\frac{\cos \sqrt{x}}{2 \sqrt{x}}-\frac{\sin 2 \sqrt{x}}{2 \sqrt{x}}$.

Solution

Let $y=\sin ^{n}(a x^{2}+b x+c)$

Differentiating both sides w.r.t. $x$

$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x} \sin ^{n}(a x^{2}+b x+c) \\ & =n \cdot \sin ^{n-1}(a x^{2}+b x+c) \cdot \frac{d}{d x} \sin (a x^{2}+b x+c) \\ & =n \cdot \sin ^{n-1}(a x^{2}+b x+c) \cdot \cos (a x^{2}+b x+c) \cdot \frac{d}{d x}(a x^{2}+b x+c) \\ & =n \cdot \sin ^{n-1}(a x^{2}+b x+c) \cdot \cos (a x^{2}+b x+c) \cdot(2 a x+b) \end{aligned} $

Hence, $\frac{d y}{d x}=n(2 a x+b) \cdot \sin ^{n-1}(a x^{2}+b x+c) \cdot \cos (a x^{2}+b x+c)$

Solution

Let $\quad y=\cos (\tan \sqrt{x+1})$

Differentiating both sides w.r.t. $x$

$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x} \cos (\tan \sqrt{x+1}) \\ & =-\sin (\tan \sqrt{x+1}) \cdot \frac{d}{d x}(\tan \sqrt{x+1}) \\ & =-\sin (\tan \sqrt{x+1}) \cdot \sec ^{2} \sqrt{x+1} \cdot \frac{d}{d x} \sqrt{x+1} \\ & =-\sin (\tan \sqrt{x+1}) \cdot \sec ^{2} \sqrt{x+1} \cdot \frac{1}{2 \sqrt{x+1}} \cdot 1 \end{aligned} $

Hence, $\frac{d y}{d x}=-\frac{1}{2 \sqrt{x+1}} \sin (\tan \sqrt{x+1}) \cdot \sec ^{2} \sqrt{x+1}$

Solution

Let $\quad y=\sin x^{2}+\sin ^{2} x+\sin ^{2}(x^{2})$

Differentiating both sides w.r.t. $x$,

$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x} \sin (x^{2})+\frac{d}{d x} \sin ^{2} x+\frac{d}{d x} \sin ^{2}(x^{2}) \\ & =\cos x^{2} \cdot \frac{d}{d x}(x^{2})+2 \sin x \cdot \frac{d}{d x}(\sin x)+2 \sin (x^{2}) \frac{d}{d x} \sin (x^{2}) \\ & =\cos x^{2} \cdot 2 x+2 \sin x \cdot \cos x+2 \sin x^{2} \cdot \cos x^{2} \cdot \frac{d}{d x}(x^{2}) \\ & =2 x \cdot \cos x^{2}+\sin 2 x+2 \sin x^{2} \cdot \cos x^{2} \cdot 2 x \end{aligned} $

Hence, $\frac{d y}{d x}=2 x \cdot \cos x^{2}+\sin 2 x+2 x \sin 2 x^{2}$

Solution

Let

$ y=\sin ^{-1}(\frac{1}{\sqrt{x+1}}) $

Differentiating both sides w.r.t. $x$

$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x} \sin ^{-1}(\frac{1}{\sqrt{x+1}})=\frac{1}{\sqrt{1-(\frac{1}{\sqrt{x+1}})^{2}}} \cdot \frac{d}{d x}(\frac{1}{\sqrt{x+1}}) \\ & =\frac{1}{\sqrt{1-\frac{1}{x+1}}} \cdot \frac{d}{d x}(x+1)^{-1 / 2} \\ & =\frac{1}{\sqrt{\frac{x+1-1}{x+1}}} \cdot \frac{-1}{2}(x+1)^{-3 / 2} \cdot \frac{d}{d x}(x+1) \\ & =\frac{\sqrt{x+1}}{\sqrt{x}} \cdot \frac{-1}{2}(x+1)^{-3 / 2} \cdot 1 \\ & =\frac{-1}{2} \cdot \frac{\sqrt{x+1}}{\sqrt{x}} \cdot \frac{1}{(x+1)^{3 / 2}}=-\frac{1}{2 \sqrt{x}(x+1)} \end{aligned} $

Hence, $\frac{d y}{d x}=-\frac{1}{2 \sqrt{x}(x+1)}$

Solution

Let $y=(\sin x)^{\cos x}$

Taking $\log$ on both sides, $\log y=\log (\sin x)^{\cos x}$

$\Rightarrow \quad \log y=\cos x \cdot \log (\sin x) \quad[\because \log x^{y}=y \log x]$

Differentiating both sides w.r.t. $x$,

$ \begin{aligned} \frac{1}{y} \cdot \frac{d y}{d x} & =\frac{d}{d x} \cos x \cdot \log (\sin x) \\ \Rightarrow \quad \frac{1}{y} \cdot \frac{d y}{d x} & =\cos x \cdot \frac{d}{d x} \log (\sin x)+\log (\sin x) \cdot \frac{d}{d x} \cos x \\ \Rightarrow \quad \frac{1}{y} \cdot \frac{d y}{d x} & =\cos x \cdot \frac{1}{\sin x} \cdot \frac{d}{d x}(\sin x)+\log (\sin x) \cdot(-\sin x) \\ \Rightarrow \quad \frac{1}{y} \cdot \frac{d y}{d x} & =\cot x \cdot \cos x-\sin x \cdot \log (\sin x) \\ \frac{d y}{d x} & =y[\cot x \cdot \cos x-\sin x \cdot \log (\sin x)] \\ \text{ Hence, } \frac{d y}{d x} & =(\sin x)^{\cos x}[\frac{\cos ^{2} x}{\sin x}-\sin x \cdot \log (\sin x)] \end{aligned} $

Solution

Let $y=\sin ^{m} x \cdot \cos ^{n} x$

Differentiating both sides w.r.t. $x$

$ \begin{aligned} \frac{d y}{d x}= & \frac{d}{d x}(\sin ^{m} x \cdot \cos ^{n} x) \\ = & \sin ^{m} x \cdot \frac{d}{d x}(\cos ^{n} x)+\cos ^{n} x \cdot \frac{d}{d x} \sin ^{m} x \\ = & \sin ^{m} x \cdot n \cdot \cos ^{n-1} x \frac{d}{d x}(\cos x)+\cos ^{n} x \cdot m \cdot \sin ^{m-1} x \\ & \frac{d}{d x}(\sin x) \\ = & n \cdot \sin ^{m} x \cdot \cos ^{n-1} x \cdot(-\sin x)+m \cdot \cos ^{n} x \cdot \sin ^{m-1} x \cdot \cos x \\ = & -n \cdot \sin ^{m+1} x \cdot \cos ^{n-1} x+m \cos ^{n+1} x \cdot \sin ^{m-1} x \\ = & \sin ^{m} x \cdot \cos ^{n} x[-n \frac{\sin x}{\cos x}+m \cdot \frac{\cos x}{\sin x}] \end{aligned} $

Hence, $\frac{d y}{d x}=\sin ^{m} x \cdot \cos ^{n} x[-n \tan x+m \cdot \cot x]$

Solution

Let $\quad y=(x+1)^{2}(x+2)^{3}(x+3)^{4}$

Taking $\log$ on both sides,

$ \begin{aligned} \log y & =\log [(x+1)^{2} \cdot(x+2)^{3} \cdot(x+3)^{4}] \\ \Rightarrow \quad \log y & =\log (x+1)^{2}+\log (x+2)^{3}+\log (x+3)^{4} \\ & {[\because \log x y=\log x+\log y] } \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \log y=2 \log (x+1)+3 \log (x+2)+4 \log (x+3) \\ & {[\because \log x^{y}=y \log x] } \end{aligned} $

Differentiating both sides w.r.t. $x$,

$ \begin{aligned} \frac{1}{y} \cdot \frac{d y}{d x}= & 2 \cdot \frac{d}{d x} \log (x+1)+3 \cdot \frac{d}{d x} \log (x+2)+4 \cdot \frac{d}{d x} \log (x+3) \\ \Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}= & 2 \cdot \frac{1}{x+1}+3 \cdot \frac{1}{x+2}+4 \cdot \frac{1}{x+3} \\ \Rightarrow \quad \frac{d y}{d x}= & y[\frac{2}{x+1}+\frac{3}{x+2}+\frac{4}{x+3}] \\ \Rightarrow \quad \frac{d y}{d x}= & (x+1)^{2}(x+2)^{3}(x+3)^{4}[\frac{2}{x+1}+\frac{3}{x+2}+\frac{4}{x+3}] \\ = & (x+1)^{2}(x+2)^{3}(x+3)^{4} \\ & {[\frac{2(x+2)(x+3)+3(x+1)(x+3)+4(x+1)(x+2)}{(x+1)(x+2)(x+3)}] } \\ = & (x+1)(x+2)^{2}(x+3)^{3}(2 x^{2}+10 x+12+3 x^{2}+12 x+9. \\ & .+4 x^{2}+12 x+8) \\ = & (x+1)(x+2)^{2}(x+3)^{3}(9 x^{2}+34 x+29) \end{aligned} $

Hence, $\frac{d y}{d x}=(x+1)(x+2)^{2}(x+3)^{3}(9 x^{2}+34 x+29)$

Solution

Let $y=\cos ^{-1}(\frac{\sin x+\cos x}{\sqrt{2}})$

$ \begin{aligned} & =\cos ^{-1}[\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x] \\ & =\cos ^{-1}[\sin \frac{\pi}{4} \sin x+\cos \frac{\pi}{4} \cdot \cos x]=\cos ^{-1}[\cos (\frac{\pi}{4}-x)] \\ y & =\frac{\pi}{4}-x \quad[\because-\frac{\pi}{4}<x<\frac{\pi}{4}] \end{aligned} $

Differentiating both sides w.r.t. $x$

$\frac{d y}{d x}=-1$

Solution

Let $y=\tan ^{-1}[\sqrt{\frac{1-\cos x}{1+\cos x}}]$

$ \begin{aligned} & =\tan ^{-1}[\sqrt{\frac{2 \sin ^{2} x / 2}{2 \cos ^{2} x / 2}}] \begin{bmatrix} \because 1-\cos x=2 \sin ^{2} x / 2 \\ 1+\cos x=2 \cos ^{2} x / 2 \end{bmatrix} \\ & =\tan ^{-1}[\frac{\sin x / 2}{\cos x / 2}]=\tan ^{-1}[\tan \frac{x}{2}] \\ \therefore \quad y & =\frac{x}{2} \end{aligned} $

Differentiating both sides w.r.t. $x$

$ \frac{d y}{d x}=\frac{1}{2} \frac{d}{d x}(x)=\frac{1}{2} \cdot 1=\frac{1}{2} $

Hence, $\frac{d y}{d x}=\frac{1}{2}$

Solution

Let

$ y=\tan ^{-1}(\sec x+\tan x) $

Differentiating both sides w.r.t. $x$

$ \begin{aligned} & \frac{d y}{d x}=\frac{d}{d x}[\tan ^{-1}(\sec x+\tan x)] \\ & =\frac{1}{1+(\sec x+\tan x)^{2}} \cdot \frac{d}{d x}(\sec x+\tan x) \\ & =\frac{1}{1+\sec ^{2} x+\tan ^{2} x+2 \sec x \tan x} \cdot(\sec x \tan x+\sec ^{2} x) \end{aligned} $

$\begin{aligned}=\frac{1}{1+\sec ^{2} x+\tan ^{2} x+2 \sec x \tan x} \cdot \sec x(\tan x+\sec x)\end{aligned}$

$ \begin{aligned} & =\frac{1}{\sec ^{2} x+\sec ^{2} x+2 \sec x \tan x} \cdot \sec x(\tan x+\sec x) \\ & =\frac{1}{2 \sec ^{2} x+2 \sec x \tan x} \cdot \sec x(\tan x+\sec x) \\ & =\frac{1}{2 \sec x(\sec x+\tan x)} \cdot \sec x(\tan x+\sec x)=\frac{1}{2} \end{aligned} $

Hence, $\frac{d y}{d x}=\frac{1}{2}$

Alternate solution

$ \text{ Let } \begin{aligned} & y=\tan ^{-1}(\sec x+\tan x), \frac{-\pi}{2}<x<\frac{\pi}{2} \\ &=\tan ^{-1}(\frac{1}{\cos x}+\frac{\sin x}{\cos x})=\tan ^{-1}(\frac{1+\sin x}{\cos x}) \\ &=\tan ^{-1}[\frac{\cos ^{2} x / 2+\sin ^{2} x / 2+2 \sin x / 2 \cos x / 2}{\cos ^{2} x / 2-\sin ^{2} x / 2}] \\ &=\tan ^{-1}[\frac{(\because \sin 2 x=2 \sin x \cos x}{\cos 2 x=\cos ^{2} x-\sin ^{2} x}] \\ &..=\tan ^{-1}[\frac{\cos x / 2+\sin x / 2}{\cos x / 2-\sin x / 2}] \sin x / 2)^{2}] \\ &.=\tan ^{-1}[\frac{1+\tan x / 2}{1-\tan x / 2}] \quad \text{ Den. by cos } x / 2] \\ &=\tan ^{-1}[\frac{\tan x / 4+\tan x / 2}{1-\tan \pi / 4 \cdot \tan x / 2}]=\tan ^{-1}[\tan (\frac{\pi}{4}+\frac{x}{2})] \\ & \therefore \quad y=\frac{\pi}{4}+\frac{x}{2} \\ & \therefore \quad \end{aligned} $

Differentiating both sides w.r.t. $x$

$ \frac{d y}{d x}=\frac{1}{2} \frac{d}{d x}(x)=\frac{1}{2} \cdot 1=\frac{1}{2} $

Hence, $\frac{d y}{d x}=\frac{1}{2}$.

Solution

Let $\quad y=\tan ^{-1}(\frac{a \cos x-b \sin x}{b \cos x+a \sin x})$

$ \Rightarrow \quad y=\tan ^{-1}[\frac{\frac{a \cos x}{b \cos x}-\frac{b \sin x}{b \cos x}}{\frac{b \cos x}{b \cos x}+\frac{a \sin x}{b \cos x}}] $

$ \begin{matrix} \Rightarrow & y=\tan ^{-1}[\frac{\frac{a}{b}-\tan x}{1+\frac{a}{b} \tan x}] \\ \Rightarrow \quad y & =\tan ^{-1} \frac{a}{b}-\tan ^{-1}(\tan x) \\ & {[\because \tan ^{-1}(\frac{x-y}{1+x y})=\tan ^{-1} x-\tan ^{-1} y]} \\ \Rightarrow \quad y & =\tan ^{-1} \frac{a}{b}-x \end{matrix} $

Differentiating both sides with respect to $x$

$ \frac{d y}{d x}=\frac{d}{d x}(\tan ^{-1} \frac{a}{b})-\frac{d}{d x}(x)=0-1=-1 $

Hence, $\frac{d y}{d x}=-1$.

Solution

Let

$ y=\sec ^{-1}(\frac{1}{4 x^{3}-3 x}) $

Put $x=\cos \theta \quad \therefore \theta=\cos ^{-1} x$

$ \begin{aligned} y & =\sec ^{-1}(\frac{1}{4 \cos ^{3} \theta-3 \cos \theta}) \\ \Rightarrow \quad y & =\sec ^{-1}(\frac{1}{\cos 3 \theta}) \quad[\because \cos 3 \theta=4 \cos ^{3} \theta-3 \cos \theta] \\ \Rightarrow \quad y & =\sec ^{-1}(\sec 3 \theta) \Rightarrow y=3 \theta \\ y & =3 \cos ^{-1} x \end{aligned} $

Differentiating both sides w.r.t. $x$

Hence, $\frac{d y}{d x}=\frac{-3}{\sqrt{1-x^{2}}}$.

$ \frac{d y}{d x}=3 \cdot \frac{d}{d x} \cos ^{-1} x=3(\frac{-1}{\sqrt{1-x^{2}}})=\frac{-3}{\sqrt{1-x^{2}}} $

Solution

Let

$ y=\tan ^{-1}[\frac{3 a^{2} x-x^{3}}{a^{3}-3 a x^{2}}] $

$ \begin{matrix} \text{ Put } x=a \tan \theta \quad \therefore \theta=\tan ^{-1} \frac{x}{a} \\ y=\tan ^{-1}[\frac{3 a^{2} \cdot a \tan \theta-a^{3} \tan ^{3} \theta}{a^{3}-3 a \cdot a^{2} \tan ^{2} \theta}] \\ \Rightarrow \quad y=\tan ^{-1}[\frac{3 a^{3} \tan \theta-a^{3} \tan ^{3} \theta}{a^{3}-3 a^{3} \tan ^{2} \theta}] \\ \Rightarrow \quad y=\tan ^{-1}[\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}] \\ \Rightarrow \quad y=\tan ^{-1}[\tan 3 \theta] \quad[\because \tan 3 \theta=\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}] \\ \Rightarrow \quad y=3 \theta \Rightarrow y=3 \tan ^{-1} \frac{x}{a} \end{matrix} $

Differentiating both sides w.r.t. $x$

$ \begin{aligned} \frac{d y}{d x} & =3 \cdot \frac{d}{d x}(\tan ^{-1} \frac{x}{a}) \\ & =3 \cdot \frac{1}{1+\frac{x^{2}}{a^{2}}} \cdot \frac{d}{d x} \cdot(\frac{x}{a})=3 \cdot \frac{a^{2}}{a^{2}+x^{2}} \cdot \frac{1}{a}=\frac{3 a}{a^{2}+x^{2}} \end{aligned} $

Hence, $\frac{d y}{d x}=\frac{3 a}{a^{2}+x^{2}}$.

Solution

Let $\quad y=\tan ^{-1}(\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}})$

Putting $x^{2}=\cos 2 \theta \quad \therefore \theta=\frac{1}{2} \cos ^{-1} x^{2}$

$ \begin{aligned} & y=\tan ^{-1}(\frac{\sqrt{1+\cos 2 \theta}+\sqrt{1-\cos 2 \theta}}{\sqrt{1+\cos 2 \theta}-\sqrt{1-\cos 2 \theta}}) \\ \Rightarrow \quad & y=\tan ^{-1}(\frac{\sqrt{2 \cos ^{2} \theta}+\sqrt{2 \sin ^{2} \theta}}{\sqrt{2 \cos ^{2} \theta}-\sqrt{2 \sin ^{2} \theta}}) \\ \Rightarrow \quad & y=\tan (\frac{\sqrt{2} \cos \theta+\sqrt{2} \sin \theta}{\sqrt{2} \cos \theta-\sqrt{2} \sin \theta}) \end{aligned} $

$ \begin{matrix} \Rightarrow & y=\tan ^{-1}(\frac{\cos \theta+\sin \theta}{\cos \theta-\sin \theta}) \\ \Rightarrow & y=\tan ^{-1}[\frac{\frac{\cos \theta}{\cos \theta}+\frac{\sin \theta}{\cos \theta}}{\frac{\cos \theta}{\cos \theta}-\frac{\sin \theta}{\cos \theta}}] \\ \Rightarrow \quad y & =\tan ^{-1}[\frac{1+\tan \theta}{1-\tan \theta}] \\ \Rightarrow \quad y & =\tan ^{-1}[\frac{\tan \frac{\pi}{4}+\tan \theta}{1-\tan \frac{\pi}{4} \cdot \tan \theta}] \\ \Rightarrow \quad y & =\tan ^{-1}[\tan (\frac{\pi}{4}+\theta)] \\ \Rightarrow \quad y & =\frac{\pi}{4}+\theta \quad \Rightarrow y=\frac{\pi}{4}+\frac{1}{2} \cos ^{-1} x^{2} \end{matrix} $

Differentiating both sides w.r.t. $x$

$ \begin{aligned} \frac{d y}{d x} & =\frac{d}{d x}(\frac{\pi}{4})+\frac{1}{2} \frac{d}{d x}(\cos ^{-1} x^{2}) \\ & =0+\frac{1}{2} \times \frac{-1}{\sqrt{1-x^{4}}} \cdot \frac{d}{d x}(x^{2})=\frac{-1.2 x}{2 \sqrt{1-x^{4}}}=-\frac{x}{\sqrt{1-x^{4}}} \end{aligned} $

Hence, $\frac{d y}{d x}=-\frac{x}{\sqrt{1-x^{4}}}$.

Solution

Given that:

$x=t+\frac{1}{t}, y=t-\frac{1}{t}$

Differentiating both the given parametric functions w.r.t. $t$

$ \frac{d x}{d t}=1-\frac{1}{t^{2}}, \frac{d y}{d t}=1+\frac{1}{t^{2}} $

$\therefore \quad \frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{1+\frac{1}{t^{2}}}{1-\frac{1}{t^{2}}}=\frac{t^{2}+1}{t^{2}-1}$

Hence, $\frac{d y}{d x}=\frac{t^{2}+1}{t^{2}-1}$.

Solution

Given that:

$ x=e^{\theta}(\theta+\frac{1}{\theta}), y=e^{-\theta}(\theta-\frac{1}{\theta}) $

Differentiating both the parametric functions w.r.t. $\theta$.

$ \begin{aligned} & \frac{d x}{d \theta}=e^{\theta}(1-\frac{1}{\theta^{2}})+(\theta+\frac{1}{\theta}) \cdot e^{\theta} \\ & \frac{d x}{d \theta}=e^{\theta}(1-\frac{1}{\theta^{2}}+\theta+\frac{1}{\theta}) \Rightarrow e^{\theta}(\frac{\theta^{2}-1+\theta^{3}+\theta}{\theta^{2}}) \\ & =\frac{e^{\theta}(\theta^{3}+\theta^{2}+\theta-1)}{\theta^{2}} \\ & y=e^{-\theta}(\theta-\frac{1}{\theta}) \\ & \frac{d y}{d \theta}=e^{-\theta}(1+\frac{1}{\theta^{2}})+(\theta-\frac{1}{\theta}) \cdot(-e^{-\theta}) \\ & \frac{d y}{d \theta}=e^{-\theta}(1+\frac{1}{\theta^{2}}-\theta+\frac{1}{\theta}) \Rightarrow e^{-\theta}(\frac{\theta^{2}+1-\theta^{3}+\theta}{\theta^{2}}) \\ & =e^{-\theta} \frac{(-\theta^{3}+\theta^{2}+\theta+1)}{\theta^{2}} \\ & \therefore \quad \frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{e^{-\theta}(\frac{-\theta^{3}+\theta^{2}+\theta+1}{\theta^{2}})}{e^{\theta}(\frac{\theta^{3}+\theta^{2}+\theta-1}{\theta^{2}})} \\ & =e^{-2 \theta}(\frac{-\theta^{3}+\theta^{2}+\theta+1}{\theta^{3}+\theta^{2}+\theta-1}) \end{aligned} $

Hence, $\frac{d y}{d x}=e^{-2 \theta}(\frac{-\theta^{3}+\theta^{2}+\theta+1}{\theta^{3}+\theta^{2}+\theta-1})$.

Solution

Given that: $x=3 \cos \theta-2 \cos ^{3} \theta$ and $y=3 \sin \theta-2 \sin ^{3} \theta$.

Differentiating both the parametric functions w.r.t. $\theta$

$ \begin{aligned} \frac{d x}{d \theta} & =-3 \sin \theta-6 \cos ^{2} \theta \cdot \frac{d}{d \theta}(\cos \theta) \\ & =-3 \sin \theta-6 \cos ^{2} \theta \cdot(-\sin \theta) \\ & =-3 \sin \theta+6 \cos ^{2} \theta \cdot \sin \theta \\ \frac{d y}{d \theta} & =3 \cos \theta-6 \sin ^{2} \theta \cdot \frac{d}{d \theta}(\sin \theta) \\ & =3 \cos \theta-6 \sin ^{2} \theta \cdot \cos \theta \\ \therefore \quad \frac{d y}{d x} & =\frac{d y / d \theta}{d x / d \theta}=\frac{3 \cos \theta-6 \sin ^{2} \theta \cos \theta}{-3 \sin \theta+6 \cos ^{2} \theta \cdot \sin \theta} \\ \Rightarrow \quad \frac{d y}{d x} & =\frac{\cos \theta(3-6 \sin ^{2} \theta)}{\sin \theta(-3+6 \cos ^{2} \theta)}=\frac{\cos \theta[3-6(1-\cos ^{2} \theta)]}{\sin \theta[-3+6 \cos ^{2} \theta]} \\ & =\cot \theta(\frac{3-6+6 \cos ^{2} \theta}{-3+6 \cos ^{2} \theta})=\cot \theta(\frac{-3+6 \cos ^{2} \theta}{-3+6 \cos ^{2} \theta}) \\ & =\cot \theta \end{aligned} $

Hence, $\frac{d y}{d x}=\cot \theta$.

Solution

Given that $\sin x=\frac{2 t}{1+t^{2}}$ and $\tan y=\frac{2 t}{1-t^{2}}$

$\therefore$ Taking $\sin x=\frac{2 t}{1+t^{2}}$

Differentiating both sides w.r.t $t$, we get

$ \begin{aligned} \cos x \cdot \frac{d x}{d t} & =\frac{(1+t^{2}) \cdot \frac{d}{d t}(2 t)-2 t \cdot \frac{d}{d t}(1+t^{2})}{(1+t^{2})^{2}} \\ \Rightarrow \quad \cos x \cdot \frac{d x}{d t} & =\frac{2(1+t^{2})-2 t \cdot 2 t}{(1+t^{2})^{2}} \\ \Rightarrow \quad \frac{d x}{d t} & =\frac{2+2 t^{2}-4 t^{2}}{(1+t^{2})^{2}} \times \frac{1}{\cos x} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \frac{d x}{d t}=\frac{2-2 t^{2}}{(1+t^{2})^{2}} \times \frac{1}{\sqrt{1-\sin ^{2} x}} \\ & \Rightarrow \quad \frac{d x}{d t}=\frac{2(1-t^{2})}{(1+t^{2})^{2}} \times \frac{1}{\sqrt{1-(\frac{2 t}{1+t^{2}})^{2}}} \\ & \Rightarrow \quad \frac{d x}{d t}=\frac{2(1-t^{2})}{(1+t^{2})^{2}} \times \frac{1}{\sqrt{\frac{(1+t^{2})^{2}-4 t^{2}}{(1+t^{2})^{2}}}} \\ & \Rightarrow \quad \frac{d x}{d t}=\frac{2(1-t^{2})}{(1+t^{2})^{2}} \times \frac{1+t^{2}}{\sqrt{1+t^{4}+2 t^{2}-4 t^{2}}} \\ & \Rightarrow \quad \frac{d x}{d t}=\frac{2(1-t^{2})}{(12^{2})^{2}} \times \frac{(1+t^{2})}{\sqrt{1+t^{4}-2 t^{2}}} \\ & \Rightarrow \quad \frac{d x}{d t}=\frac{2(1-t^{2})}{(1+t^{2})} \times \frac{1}{\sqrt{(1-t^{2})^{2}}} \\ & \Rightarrow \quad \frac{d x}{d t}=\frac{2(1-t^{2})}{(1+t^{2})} \times \frac{1}{(1-t^{2})} \Rightarrow \frac{d x}{d t}=\frac{2}{1+t^{2}} \end{aligned} $

Now taking, $\tan y=\frac{2}{1-t^{2}}$

Differentiating both sides w.r.t, $t$, we get

$ \begin{aligned} \frac{d}{d t}(\tan y) & =\frac{d}{d t}(\frac{2 t}{1-t^{2}}) \\ \Rightarrow \sec ^{2} y \frac{d y}{d t} & =\frac{(1-t^{2}) \cdot \frac{d}{d t}(2 t)-2 t \cdot \frac{d}{d t}(1-t^{2})}{(1-t^{2})^{2}} \\ \Rightarrow \sec ^{2} y \frac{d y}{d t} & =\frac{(1-t^{2}) \cdot 2-2 t \cdot(-2 t)}{(1-t^{2})^{2}} \\ \Rightarrow \sec ^{2} y \frac{d y}{d t} & =\frac{2-2 t^{2}+4 t^{2}}{(1-t^{2})^{2}} \\ \Rightarrow \quad \frac{d y}{d t} & =\frac{2+2 t^{2}}{(1-t^{2})^{2}} \times \frac{1}{\sec ^{2} y} \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \frac{d y}{d t}=\frac{2(1+t^{2})}{(1-t^{2})^{2}} \times \frac{1}{1+\tan ^{2} y} \\ & \Rightarrow \quad \frac{d y}{d t}=\frac{2(1+t^{2})}{(1-t^{2})^{2}} \times \frac{1}{1+(\frac{2 t}{1-t^{2}})^{2}} \\ & \Rightarrow \quad \frac{d y}{d t}=\frac{2(1+t^{2})}{(1-t^{2})^{2}} \times \frac{1}{\frac{(1-t^{2})^{2}+4 t^{2}}{(1-t^{2})^{2}}} \\ & \Rightarrow \quad \frac{d y}{d t}=\frac{2(1+t^{2})}{(1-t^{2})^{2}} \times \frac{(1-t^{2})^{2}}{1+t^{2}+2 t^{2}+4 t^{2}} \\ & \Rightarrow \quad \frac{d y}{d t}=\frac{2(1+t^{2})}{(1-t^{2})^{2}} \times \frac{(1-t^{2})^{2}}{1+t^{4}+2 t^{2}} \\ & \Rightarrow \quad \frac{d y}{d t}=\frac{2(1+t^{2})}{(1-t^{2})^{2}} \times \frac{(1-t^{2})^{2}}{(1+t^{2})^{2}} \Rightarrow \frac{d y}{d t}=\frac{2}{1+t^{2}} \\ & \therefore \quad \frac{d y}{d t}=\frac{d y / d t}{d x / d t}=\frac{\frac{2}{1+t^{2}}}{\frac{2}{1+t^{2}}}=1 \end{aligned} $

Solution

Given that: $x=\frac{1+\log t}{t^{2}}, y=\frac{3+2 \log t}{t}$.

Differentiating both the parametric functions w.r.t. $t$

$ \begin{aligned} \frac{d x}{d t} & =\frac{t^{2} \cdot \frac{d}{d t}(1+\log t)-(1+\log t) \cdot \frac{d}{d t}(t^{2})}{t^{4}} \\ & =\frac{t^{2} \cdot(\frac{1}{t})-(1+\log t) \cdot 2 t}{t^{4}}=\frac{t-(1+\log t) \cdot 2 t}{t^{4}} \\ & =\frac{t[1-2-2 \log t]}{t^{4}}=\frac{-(1+2 \log t)}{t^{3}} \\ y & =\frac{3+2 \log t}{t} \end{aligned} $

$ \begin{aligned} & \frac{d y}{d t}=\frac{t \cdot \frac{d}{d t}(3+2 \log t)-(3+2 \log t) \cdot \frac{d}{d t}(t)}{t^{2}} \\ & =\frac{t(2 / t)-(3+2 \log t) \cdot 1}{t^{2}} \\ & =\frac{2-3-2 \log t}{t^{2}}=\frac{-(1+2 \log t)}{t^{2}} \\ & \therefore \quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{\frac{-(1+2 \log t)}{t^{2}}}{-(1+2 \log t)}=\frac{t^{3}}{t^{2}}=t \\ & \text{ Hence, } \frac{d y}{d x}=t \text{. } \end{aligned} $

Solution

Given that: $x=e^{\cos 2 t}$ and $y=e^{\sin 2 t}$

$\Rightarrow \cos 2 t=\log x$ and $\sin 2 t=\log y$.

Differentiating both the parametric functions w.r.t. $t$

$ \begin{aligned} \frac{d x}{d t} & =e^{\cos 2 t} \cdot \frac{d}{d t}(\cos 2 t)=e^{\cos 2 t}(-\sin 2 t) \cdot \frac{d}{d t}(2 t) \\ & =-e^{\cos 2 t} \cdot \sin 2 t \cdot 2=-2 e^{\cos 2 t} \cdot \sin 2 t \end{aligned} $

Now $\quad y=e^{\sin 2 t}$

$ \begin{aligned} \frac{d y}{d t} & =e^{\sin 2 t} \cdot \frac{d}{d t}(\sin 2 t)=e^{\sin 2 t} \cdot \cos 2 t \cdot \frac{d}{d t}(2 t) \\ & =e^{\sin 2 t} \cdot \cos 2 t \cdot 2=2 e^{\sin 2 t} \cdot \cos 2 t \end{aligned} $

$\therefore \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{2 e^{\sin 2 t} \cdot \cos 2 t}{-2 e^{\cos 2 t} \cdot \sin 2 t}=\frac{e^{\sin 2 t} \cdot \cos 2 t}{-e^{\cos 2 t} \cdot \sin 2 t}=\frac{y \cos 2 t}{-x \sin 2 t}$

$ =\frac{y \log x}{-x \log y} \quad \begin{bmatrix} \because \cos 2 t=\log x \\ \sin 2 t=\log y \end{bmatrix} $

Hence, $\frac{d y}{d x}=-\frac{y \log x}{x \log y}$.

Solution

Given that: $x=a \sin 2 t(1+\cos 2 t)$ and $y=b \cos 2 t(1-\cos 2 t)$.

Differentiating both the parametric functions w.r.t. $t$

$ \begin{aligned} & \frac{d x}{d t}=a[\sin 2 t \cdot \frac{d}{d t}(1+\cos 2 t)+(1+\cos 2 t) \cdot \frac{d}{d t} \sin 2 t] \\ & =a[\sin 2 t \cdot(-\sin 2 t) \cdot 2+(1+\cos 2 t)(\cos 2 t) \cdot 2] \\ & =a[-2 \sin ^{2} 2 t+2 \cos 2 t+2 \cos ^{2} 2 t] \\ & =a[2(\cos ^{2} 2 t-\sin ^{2} 2 t)+2 \cos 2 t] \\ & =a[2 \cos 4 t+2 \cos 2 t] \quad[\because \cos 2 x=\cos ^{2} x-\sin ^{2} x] \\ & =2 a[\cos 4 t+\cos 2 t] \\ & y=b \cos 2 t(1-\cos 2 t) \\ & \frac{d y}{d t}=b[\cos 2 t \cdot \frac{d}{d t}(1-\cos 2 t)+(1-\cos 2 t) \cdot \frac{d}{d t}(\cos 2 t)] \\ & =b[\cos 2 t \cdot \sin 2 t \cdot 2+(1-\cos 2 t) \cdot(-\sin 2 t) \cdot 2] \\ & =b[2 \sin 2 t \cdot \cos 2 t-2 \sin 2 t+2 \sin 2 t \cos 2 t] \\ & =b[\sin 4 t-2 \sin 2 t+\sin 4 t][\because \sin 2 x=2 \sin x \cos x] \\ & =b[2 \sin 4 t-2 \sin 2 t]=2 b(\sin 4 t-\sin 2 t) \\ & \therefore \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{2 b[\sin 4 t-\sin 2 t]}{2 a[\cos 4 t+\cos 2 t]}=\frac{b}{a}[\frac{\sin 4 t-\sin 2 t}{\cos 4 t+\cos 2 t}] \\ & \text{ Put } \quad t=\frac{\pi}{4} \\ & \therefore(\frac{d y}{d x}) _{a t t=\frac{\pi}{4}}=\frac{b}{a}[\frac{\sin 4(\frac{\pi}{4})-\sin 2 \cdot(\frac{\pi}{4})}{\cos 4(\frac{\pi}{4})+\cos 2 \cdot(\frac{\pi}{4})}]=\frac{b}{a}[\frac{\sin \pi-\sin \frac{\pi}{2}}{\cos \pi+\cos \frac{\pi}{2}}] \\ & =\frac{b}{a}[\frac{0-1}{-1+0}]=\frac{b}{a}(\frac{-1}{-1})=\frac{b}{a} \end{aligned} $

Solution

Given that: $x=3 \sin t-\sin 3 t, y=3 \cos t-\cos 3 t$.

Differentiating both parametric functions w.r.t. $t$

$ \begin{aligned} & \frac{d x}{d t}=3 \cos t-\cos 3 t .3=3(\cos t-\cos 3 t) \\ & \underline{d y}=-3 \sin t+\sin 3 t .3=3(-\sin t+\sin 3 t) \end{aligned} $

$ \therefore \quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{3(-\sin t+\sin 3 t)}{3(\cos t-\cos 3 t)}=\frac{-\sin t+\sin 3 t}{\cos t-\cos 3 t} $

Put $\quad t=\frac{\pi}{3}$

$ \begin{aligned} \frac{d y}{d x} & =\frac{-\sin \frac{\pi}{3}+\sin 3(\frac{\pi}{3})}{\cos \frac{\pi}{3}-\cos 3(\frac{\pi}{3})} \\ & =\frac{-\frac{\sqrt{3}}{2}+\sin \pi}{\frac{1}{2}-\cos \pi}=\frac{-\frac{\sqrt{3}}{2}+0}{\frac{1}{2}-(-1)}=\frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}+1}=\frac{-\frac{\sqrt{3}}{2}}{\frac{3}{2}}=\frac{-1}{\sqrt{3}} \end{aligned} $

Hence, $\frac{d y}{d x}=\frac{-1}{\sqrt{3}}$.

Solution

Let

$ y=\frac{x}{\sin x} \quad \text{ and } \quad z=\sin x $

Differentiating both the parametric functions w.r.t. $x$,

$ \begin{aligned} \frac{d y}{d x} & =\frac{\sin x \cdot \frac{d}{d x}(x)-x \cdot \frac{d}{d x}(\sin x)}{(\sin x)^{2}} \\ & =\frac{\sin x \cdot 1-x \cdot \cos x}{\sin ^{2} x}=\frac{\sin x-x \cos x}{\sin ^{2} x} \\ \frac{d z}{d x} & =\cos x \\ \therefore \quad \frac{d y}{d z} & =\frac{d y / d x}{d z / d x}=\frac{\frac{\sin x-x \cos x}{\sin ^{2} x}}{\cos x}=\frac{\sin x-x \cos x}{\sin ^{2} x \cos x} \\ & =\frac{\sin x}{\sin ^{2} x \cos x}-\frac{x \cos x}{\sin ^{2} x \cos x} \\ & =\frac{\tan ^{2} x}{\sin ^{2} x}-\frac{x}{\sin ^{2} x}=\frac{\tan x-x}{\sin ^{2} x} \end{aligned} $

Hence, $\frac{d y}{d z}=\frac{\tan x-x}{\sin ^{2} x}$.

Solution

Let $\quad y=\tan ^{-1}(\frac{\sqrt{1+x^{2}}-1}{x})$ and $z=\tan ^{-1} x$.

Put $\quad x=\tan \theta$.

$\therefore \quad y=\tan ^{-1}(\frac{\sqrt{1+\tan ^{2} \theta}-1}{\tan \theta})$ and $z=\tan ^{-1}(\tan \theta)=\theta$.

$\Rightarrow \quad \tan (\frac{\sqrt{\sec \theta}-1}{\tan })=\tan ^{-1}(\frac{\sec \theta-1}{\tan \theta})$

$\Rightarrow \quad \tan ^{-1}(\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}})=\tan ^{-1}(\frac{1-\cos \theta}{\sin \theta})$

$\Rightarrow \quad \tan ^{-1}(\frac{2 \sin ^{2} \theta / 2}{2 \sin \theta / 2 \cos \theta / 2})=\tan ^{-1}(\frac{\sin \theta / 2}{\cos \theta / 2})$

$\Rightarrow \quad y=\tan ^{-1}(\tan \frac{\theta}{2}) \Rightarrow y=\frac{\theta}{2}$

Differentiating both parametric functions w.r.t. $\theta$

$ \begin{aligned} \frac{d y}{d \theta} & =\frac{1}{2} \cdot \frac{d}{d \theta}(\theta) \quad \text{ and } \quad \frac{d z}{d \theta}=\frac{d}{d \theta}(\theta) \\ & =\frac{1}{2} \cdot 1=\frac{1}{2} \quad \text{ and } \quad \frac{d z}{d \theta}=1 \\ \therefore \quad \frac{d y}{d z} & =\frac{d y / d \theta}{d z / d \theta}=\frac{1 / 2}{1}=\frac{1}{2} . \end{aligned} $

Solution

Given that: $\sin x y+\frac{x}{y}=x^{2}-y$.

Differentiating both sides w.r.t. $x$

$ \begin{aligned} & \frac{d}{d x} \sin (x y)+\frac{d}{d x}(\frac{x}{y})=\frac{d}{d x}(x^{2})-\frac{d}{d x}(y) \\ & \Rightarrow \quad \cos x y \cdot \frac{d}{d x}(x y)+\frac{y \cdot \frac{d}{d x} \cdot x-x \cdot \frac{d y}{d x}}{y^{2}}=2 x-\frac{d y}{d x} \end{aligned} $

$ \begin{gathered} \Rightarrow \quad \cos x y[x \cdot \frac{d y}{d x}+y \cdot 1]+\frac{y \cdot 1}{y^{2}}-\frac{x}{y^{2}} \cdot \frac{d y}{d x}=2 x-\frac{d y}{d x} \\ \Rightarrow \quad x \cos x y \cdot \frac{d y}{d x}+y \cos x y+\frac{1}{y}-\frac{x}{y^{2}} \frac{d y}{d x}=2 x-\frac{d y}{d x} \\ \Rightarrow \quad x \cos x y \cdot \frac{d y}{d x}-\frac{x}{y^{2}} \cdot \frac{d y}{d x}+\frac{d y}{d x}=-y \cos x y-\frac{1}{y}+2 x \\ \Rightarrow \quad \quad[x \cos x y-\frac{x}{y^{2}}+1] \frac{d y}{d x}=2 x-y \cos x y-\frac{1}{y} \\ \Rightarrow \quad \frac{[x y^{2} \cos x y-x+y^{2}]}{y^{2}} \frac{d y}{d x}=\frac{2 x y-y^{2} \cos x y-1}{y} \\ \Rightarrow \quad \frac{d y}{d x}=\frac{2 x y-y^{2} \cos x y-1}{y} \times \frac{y^{2}}{x y^{2} \cos x y-x+y^{2}} \\ =\frac{2 x y^{2}-y^{3} \cos (x y)-y}{x y^{2} \cos (x y)-x+y^{2}} \end{gathered} $

Hence, $\frac{d y}{d x}=\frac{2 x y^{2}-y^{3} \cos (x y)-y}{x y^{2} \cos (x y)-x+y^{2}}$.

Solution

Given that: $\sec (x+y)=x y$

Differentiating both sides w.r.t. $x$

$ \begin{aligned} & \frac{d}{d x} \sec (x+y)=\frac{d}{d x}(x y) \\ & \Rightarrow \quad \sec (x+y) \tan (x+y) \cdot \frac{d}{d x}(x+y)=x \cdot \frac{d y}{d x}+y \cdot 1 \\ & \Rightarrow \quad \sec (x+y) \cdot \tan (x+y)(1+\frac{d y}{d x})=x \cdot \frac{d y}{d x}+y \\ & \Rightarrow \quad \sec (x+y) \cdot \tan (x+y)+\sec (x+y) \cdot \tan (x+y) \cdot \frac{d y}{d x}=x \cdot \frac{d y}{d x}+y \\ & \Rightarrow \quad \sec (x+y) \cdot \tan (x+y) \cdot \frac{d y}{d x}-x \frac{d y}{d x}=y-\sec (x+y) \cdot \tan (x+y) \\ & \Rightarrow \quad[\sec (x+y) \cdot \tan (x+y)-x] \frac{d y}{d x}=y-\sec (x+y) \cdot \tan (x+y) \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{y-\sec (x+y) \cdot \tan (x+y)}{\sec (x+y) \cdot \tan (x+y)-x} \\ & \text{ Hence, } \quad \frac{d y}{d x}=\frac{y-\sec (x+y) \cdot \tan (x+y)}{\sec (x+y) \cdot \tan (x+y)-x} . \end{aligned} $

Solution

Given that: $\tan ^{-1}(x^{2}+y^{2})=a$ $\Rightarrow \quad x^{2}+y^{2}=\tan a$.

Differentiating both sides w.r.t. $x$.

$\frac{d}{d x}(x^{2}+y^{2}) =\frac{d}{d x}(\tan a)$

$\Rightarrow \quad 2 x+2 y \cdot \frac{d y}{d x} =0 \Rightarrow 2 y \cdot \frac{d y}{d x}=-2 x$

$\Rightarrow \quad \frac{d y}{d x} =\frac{-2 x}{2 y}=\frac{-x}{y}$

$\text{ Hence, } \quad \frac{d y}{d x} =\frac{-x}{y} $

Solution

Given that: $(x^{2}+y^{2})^{2}=x y$

$ \Rightarrow \quad x^{4}+y^{4}+2 x^{2} y^{2}=x y $

Differentiating both sides w.r.t. $x$

$ \begin{aligned} & \quad \frac{d}{d x}(x^{4})+\frac{d}{d x}(y^{4})+2 \cdot \frac{d}{d x}(x^{2} y^{2})=\frac{d}{d x}(x y) \\ & \Rightarrow \quad 4 x^{3}+4 y^{3} \cdot \frac{d y}{d x}+2[x^{2} \cdot 2 y \cdot \frac{d y}{d x}+y^{2} \cdot 2 x]=x \frac{d y}{d x}+y \cdot 1 \\ & \Rightarrow \quad 4 x^{3}+4 y^{3} \cdot \frac{d y}{d x}+4 x^{2} y \cdot \frac{d y}{d x}+4 x y^{2}=x \frac{d y}{d x}+y \\ & \Rightarrow \quad 4 y^{3} \frac{d y}{d x}+4 x^{2} y \frac{d y}{d x}-x \frac{d y}{d x}=y-4 x^{3}-4 x y^{2} \\ & \Rightarrow \quad(4 y^{3}+4 x^{2} y-x) \frac{d y}{d x}=y-4 x^{3}-4 x y^{2} \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{y-4 x^{3}-4 x y^{2}}{4 y^{3}+4 x^{2} y-x} \\ & \text{ Hence, } \quad \frac{d y}{d x}=\frac{y-4 x^{3}-4 x y^{2}}{4 x^{2} y+4 y^{3}-x} . \end{aligned} $

Solution

Given that: $a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0$.

Differentiating both sides w.r.t. $x$

$ \frac{d}{d x}(a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c)=\frac{d}{d x}(0) $

$ \begin{aligned} & \Rightarrow \quad a \cdot 2 x+2 h(x \cdot \frac{d y}{d x}+y \cdot 1)+b \cdot 2 y \cdot \frac{d y}{d x}+2 g \cdot 1+2 f \cdot \frac{d y}{d x}+0=0 \\ & \Rightarrow \quad 2 a x+2 h x \cdot \frac{d y}{d x}+2 h y+2 b y \cdot \frac{d y}{d x}+2 g+2 f \cdot \frac{d y}{d x}=0 \\ & \Rightarrow \quad 2 h x \cdot \frac{d y}{d x}+2 b y \frac{d y}{d x}+2 f \frac{d y}{d x}=-2 a x-2 h y-2 g \\ & \Rightarrow \quad(2 h x+2 b y+2 f) \frac{d y}{d x}=-2(a x+h y+g) \\ & \Rightarrow \quad 2(h x+b y+f) \frac{d y}{d x}=-2(a x+h y+g) \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{-2(a x+h y+g)}{2(h x+b y+f)} \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{-(a x+h y+g)}{(h x+b y+f)} \end{aligned} $

Now, differentiating the given equation w.r.t. $y$.

$ \begin{aligned} & \frac{d}{d y}(a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c)=\frac{d}{d y}(0) \\ & \Rightarrow \quad 2 a x \cdot \frac{d x}{d y}+2 h(y \cdot \frac{d x}{d y}+x \cdot 1)+2 b y+2 g \cdot \frac{d x}{d y}+2 f \cdot 1+0=0 \\ & \Rightarrow \quad 2 a x \cdot \frac{d x}{d y}+2 h y \cdot \frac{d x}{d y}+2 h x+2 b y+2 g \cdot \frac{d x}{d y}+2 f=0 \\ & \Rightarrow \quad 2 a x \frac{d x}{d y}+2 h y \cdot \frac{d x}{d y}+2 g \cdot \frac{d x}{d y}=-2 h x-2 b y-2 f \\ & \Rightarrow \quad(2 a x+2 h y+2 g) \frac{d x}{d y}=-2 h x-2 b y-2 f \\ & \Rightarrow \quad \frac{d x}{d y}=\frac{-2 h x-2 b y-2 f}{2 a x+2 h y+2 g} \\ & \Rightarrow \quad \frac{d x}{d y}=\frac{-2(h x+b y+f)}{2(a x+h y+g)} \Rightarrow \frac{d x}{d y}=\frac{-(h x+b y+f)}{(a x+h y+g)} \\ & \therefore \quad \frac{d y}{d x} \cdot \frac{d x}{d y}=[\frac{-(a x+h y+g)}{(h x+b y+f)}][\frac{-(h x+b y+f)}{(a x+h y+g)}]=1 \end{aligned} $

Hence, $\frac{d y}{d x} \cdot \frac{d x}{d y}=1$. Hence, proved.

Solution

Given that: $\quad x=e^{x / y}$

Taking $\log$ on both the sides,

$$ \begin{align*} \log x & =\log e^{x / y} \\ \Rightarrow \quad \log x & =\frac{x}{y} \log e \quad \Rightarrow \log x=\frac{x}{y} \quad[\because \log e=1] \tag{i} \end{align*} $$

Differentiating both sides w.r.t. $x$

$ \begin{aligned} \frac{d}{d x} \log x & =\frac{d}{d x}(\frac{x}{y}) \\ \Rightarrow \quad \frac{1}{x} & =\frac{y \cdot 1-x \cdot \frac{d y}{d x}}{y^{2}} \\ \Rightarrow \quad y^{2} & =x y-x^{2} \cdot \frac{d y}{d x} \Rightarrow x^{2} \cdot \frac{d y}{d x}=x y-y^{2} \\ \Rightarrow \quad \frac{d y}{d x} & =\frac{y(x-y)}{x^{2}} \Rightarrow \frac{d y}{d x}=\frac{y}{x} \cdot(\frac{x-y}{x}) \\ \Rightarrow \quad \frac{d y}{d x} & =\frac{1}{\log x} \cdot(\frac{x-y}{x}) \quad(\because \log x=\frac{x}{y} \text{ from eqn. }(i)) \end{aligned} $

Hence, $\frac{d y}{d x}=\frac{x-y}{x \log x}$.

Solution

Given that: $y^{x}=e^{y-x}$

Taking $\log$ on both sides $\log y^{x}=\log e^{y-x}$

$ \begin{aligned} & \Rightarrow \quad x \log y=(y-x) \log e \\ & \Rightarrow \quad x \log y=y-x \quad[\because \log e=1] \\ & \Rightarrow \quad x \log y+x=y \\ & \Rightarrow \quad x(\log y+1)=y \\ & \Rightarrow \quad x=\frac{y}{\log y+1} . \end{aligned} $

Differentiating both sides w.r.t. $y$

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

$ =\frac{\log y+1-y \cdot \frac{1}{y}}{(\log y+1)^{2}}=\frac{\log y+1-1}{(\log y+1)^{2}}=\frac{\log y}{(\log y+1)^{2}} $

We know that

$ \frac{d y}{d x}=\frac{1}{d x / d y}=\frac{1}{\frac{\log y}{(\log y+1)^{2}}}=\frac{(\log y+1)^{2}}{\log y} $

Hence, $\frac{d y}{d x}=\frac{(\log y+1)^{2}}{\log y}$.

Solution

Given that $y=(\cos x)^{(\cos x)^{(\cos x) \ldots \infty}}$

$ \Rightarrow \quad y=(\cos x)^{y} \quad[\because y=(\cos x)^{(\cos x)^{(\cos x) \ldots \infty}}] $

Taking $\log$ on both sides $\log y=y \cdot \log (\cos x)$

Differentiating both sides w.r.t. $x$

$ \begin{aligned} & \frac{1}{y} \cdot \frac{d y}{d x}=y \cdot \frac{d}{d x} \log (\cos x)+\log (\cos x) \cdot \frac{d y}{d x} \\ & \Rightarrow \quad \frac{1}{y} \cdot \frac{d y}{d x}=y \cdot \frac{1}{\cos x} \cdot \frac{d}{d x}(\cos x)+\log (\cos x) \cdot \frac{d y}{d x} \\ & \Rightarrow \quad \frac{1}{y} \cdot \frac{d y}{d x}=y \cdot \frac{1}{\cos x} \cdot(-\sin x)+\log (\cos x) \cdot \frac{d y}{d x} \\ & \Rightarrow \frac{1}{y} \cdot \frac{d y}{d x}-\log (\cos x) \frac{d y}{d x}=-y \tan x \\ & \Rightarrow[\frac{1}{y}-\log (\cos x)] \frac{d y}{d x}=-y \tan x \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{-y \tan x}{\frac{1}{y}-\log (\cos x)}=\frac{y^{2} \tan x}{y \log \cos x-1} \\ & \frac{d y}{d x}=\frac{y^{2} \tan x}{y \log \cos x-1} \text{. Hence, proved. } \end{aligned} $

Solution

Given that: $x \sin (a+y)+\sin a \cos (a+y)=0$

$ \begin{aligned} & \Rightarrow \quad x \sin (a+y)=-\sin a \cos (a+y) \\ & \Rightarrow \quad x=\frac{-\sin a \cdot \cos (a+y)}{\sin (a+y)} \Rightarrow x=-\sin a \cdot \cot (a+y) \end{aligned} $

Differentiating both sides w.r.t. $y$

$ \begin{matrix} \Rightarrow & \frac{d x}{d y}=-\sin a \cdot \frac{d}{d y} \cot (a+y) \\ \Rightarrow & \frac{d x}{d y}=-\sin a[-cosec^{2}(a+y)] \\ \Rightarrow & \frac{d x}{d y}=\frac{\sin a}{\sin ^{2}(a+y)} \\ \therefore & \frac{d y}{d x}=\frac{1}{d x / d y}=\frac{1}{\frac{\sin a}{\sin ^{2}(a+y)}} \end{matrix} $

Hence, $\quad \frac{d y}{d x}=\frac{\sin ^{2}(a+y)}{\sin a}$. Hence proved.

Solution

Given that: $\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)$

$\text{ Put } \quad x=\sin \theta \text{ and } y=\sin \phi \text{. }$

$\therefore \quad \theta=\sin ^{-1} x \text{ and } \phi =\sin ^{-1} y$

$\sqrt{1-\sin ^{2} \theta}+\sqrt{1-\sin ^{2} \phi} =a(\sin \theta-\sin \phi)$

$\Rightarrow \quad \sqrt{\cos ^{2} \theta}+\sqrt{\cos ^{2} \phi} =a(\sin \theta-\sin \phi)$

$\Rightarrow \quad \cos \theta+\cos \phi =a(\sin \theta-\sin \phi)$

$\Rightarrow \quad \frac{\cos \theta+\cos \phi}{\sin \theta-\sin \phi} =a \Rightarrow \frac{2 \cos \frac{\theta+\phi}{2} \cdot \cos \frac{\theta-\phi}{2}}{2 \cos \frac{\theta+\phi}{2} \cdot \sin \frac{\theta-\phi}{2}}=a$

$ \begin{bmatrix} \because \cos A+\cos B=2 \cos \frac{A+B}{2} \cdot \cos \frac{A-B}{2} \\ \sin A-\sin B=2 \cos \frac{A+B}{2} \cdot \sin \frac{A-B}{2} \end{bmatrix} $

$ \begin{aligned} & \Rightarrow \quad \frac{\cos (\frac{\theta-\phi}{2})}{\sin (\frac{\theta-\phi}{2})}=a \Rightarrow \cot (\frac{\theta-\phi}{2})=a \\ & \Rightarrow \quad \frac{\theta-\phi}{2}=\cot ^{-1} a \Rightarrow \theta-\phi=2 \cot ^{-1} a \\ & \Rightarrow \quad \sin ^{-1} x-\sin ^{-1} y=2 \cot ^{-1} a \end{aligned} $

Differentiating both sides w.r.t. $x$

$ \frac{d}{d x}(\sin ^{-1} x)-\frac{d}{d x}(\sin ^{-1} y)=2 \cdot \frac{d}{d x} \cot ^{-1} a $

$ \begin{aligned} & \Rightarrow \frac{1}{\sqrt{1-x^{2}}}-\frac{1}{\sqrt{1-y^{2}}} \cdot \frac{d y}{d x}=0 \\ & \Rightarrow \quad \frac{1}{\sqrt{1-y^{2}}} \cdot \frac{d y}{d x}=\frac{1}{\sqrt{1-x^{2}}} \\ & \therefore \quad \frac{d y}{d x}=\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}} \\ & \text{ Hence, } \quad \frac{d y}{d x}=\sqrt{\frac{1-y^{2}}{1-x^{2}}} \text{. } \end{aligned} $

Solution

Given that: $y=\tan ^{-1} x \Rightarrow x=\tan y$

Differentiating both sides w.r.t. $y$

$ \frac{d x}{d y}=\sec ^{2} y \Rightarrow \frac{d y}{d x}=\frac{1}{\sec ^{2} y}=\cos ^{2} y $

Again differentiating both sides w.r.t. $x$

$ \begin{aligned} \frac{d}{d x}(\frac{d y}{d x}) & =\frac{d}{d x}(\cos ^{2} y) \\ \Rightarrow \quad \frac{d^{2} y}{d x^{2}} & =2 \cos y \cdot \frac{d}{d x}(\cos y) \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \frac{d^{2} y}{d x^{2}}=2 \cos y(-\sin y) \cdot \frac{d y}{d x} \\ & \Rightarrow \quad \frac{d^{2} y}{d x^{2}}=-2 \sin y \cos y \cdot \cos ^{2} y \\ & \therefore \quad \frac{d^{2} y}{d x^{2}}=-2 \sin y \cos ^{3} y \end{aligned} $

Verify the Rolle’s Theorem for each of the functions in Exercises 65 to 69 :

Solution

Given that: $f(x)=x(x-1)^{2}$ in $[0,1]$

(i) $f(x)=x(x-1)^{2}$, being an algebraic polynomial, is continuous in $[0,1]$.

(ii)

$ \begin{aligned} f^{\prime}(x) & =x .2(x-1)+(x-1)^{2} .1 \\ & =2 x^{2}-2 x+x^{2}+1-2 x \\ & =3 x^{2}-4 x+1 \text{ which exists in }(0,1) \\ f(x) & =x(x-1)^{2} \\ f(0) & =0(0-1)^{2}=0 ; f(1)=1(1-1)^{2}=0 \\ \Rightarrow \quad f(0) & =f(1)=0 \end{aligned} $

(iii)

As the above conditions are satisfied, then there must exist at least one point $c \in(0,1)$ such that $f^{\prime}(c)=0$

$ \begin{aligned} \therefore \quad f^{\prime}(c)=3 c^{2}-4 c+1=0 & \Rightarrow 3 c^{2}-3 c-c+1=0 \\ \Rightarrow 3 c(c-1)-1(c-1)=0 & \Rightarrow(c-1)(3 c-1)=0 \\ \Rightarrow \quad c-1=0 & \Rightarrow c=1 \\ 3 c-1=0 & \Rightarrow 3 c=1 \quad \therefore c=\frac{1}{3} \in(0,1) \end{aligned} $

Hence, Rolle’s Theorem is verified.

Solution

Given that: $f(x)=\sin ^{4} x+\cos ^{4} x$ in $[0, \frac{\pi}{2}]$

(i) $f(x)=\sin ^{4} x+\cos ^{4} x$, being sine and cosine functions, $f(x)$ is continuous function in $[0, \frac{\pi}{2}]$.

(ii) $\quad f^{\prime}(x)=4 \sin ^{3} x \cdot \cos x+4 \cos ^{3} x(-\sin x)$

$ =4 \sin ^{3} x \cdot \cos x-4 \cos ^{3} x \cdot \sin x $

$ \begin{aligned} & =4 \sin x \cos x(\sin ^{2} x-\cos ^{2} x) \\ & =-4 \sin x \cos x(\cos ^{2} x-\sin ^{2} x) \\ & =-2 \cdot 2 \sin x \cos x \cdot \cos 2 x \begin{bmatrix} \because \cos 2 x=\cos ^{2} x-\sin ^{2} x \\ \sin 2 x=2 \sin x \cos x \end{bmatrix} \\ & =-2 \sin 2 x \cdot \cos 2 x \\ & =-\sin 4 x \quad \text{ which exists in }(0, \frac{\pi}{2}) . \end{aligned} $

So, $f(x)$ is differentiable in $(0, \frac{\pi}{2})$.

(iii)

$ \begin{aligned} f(0) & =\sin ^{4}(0)+\cos ^{4}(0)=1 \\ f(\frac{\pi}{2}) & =\sin ^{4}(\frac{\pi}{2})+\cos ^{2}(\frac{\pi}{2})=1 \\ \therefore \quad f(0) & =f(\frac{\pi}{2})=1 \end{aligned} $

As the above conditions are satisfied, there must exist at least one point $c \in(0, \frac{\pi}{2})$ such that $f^{\prime}(c)=0$

$ \begin{aligned} & \Rightarrow \quad-\sin 4 c=0 \\ & \Rightarrow \quad \sin 4 c=0 \quad \Rightarrow \quad \sin 4 c=\sin 0 \\ & \Rightarrow \quad 4 c=n \pi \\ & \therefore \quad c=\frac{n \pi}{4}, n \in I \\ & \text{ For } n=1, \quad c=\frac{\pi}{4} \in(0, \frac{\pi}{2}) \end{aligned} $

Hence, the Rolle’s Theorem is verified.

Solution

Given that: $f(x)=\log (x^{2}+2)-\log 3$ in $[-1,1]$

(i) $f(x)=\log (x^{2}+2)-\log 3$, being a logarithm function, is continuous in $[-1,1]$.

(ii) $f^{\prime}(x)=\frac{1}{x^{2}+2} \cdot 2 x-0=\frac{2 x}{x^{2}+2}$ which exists in $(-1,1)$

So, $f(x)$ is differentiable in $(-1,1)$.

(iii) $f(-1)=\log (1+2)-\log 3 \Rightarrow \log 3-\log 3=0$

$ f(1)=\log (1+2)-\log 3 \Rightarrow \log 3-\log 3=0 $

$\therefore f(-1)=f(1)=0$

As the above conditions are satisfied, then there must exist atleast one point $c \in(-1,1)$ such that $f^{\prime}(c)=0$.

$ \therefore \frac{2 c}{c^{2}+2}=0 \quad \Rightarrow \quad 2 c=0 \quad \therefore c=0 \in(-1,1) $

Hence, Rolle’s Theorem is verified.

Solution

Given that: $f(x)=x(x+3) e^{-x / 2}$ in $[-3,0]$

(i) Algebraic functions and exponential functions are continuous in their domains.

$\therefore f(x)$ is continuous in $[-3,0]$

(ii) $f^{\prime}(x)=x(x+3) \cdot \frac{d}{d x} e^{-x / 2}+x \cdot e^{-x / 2} \cdot \frac{d}{d x}(x+3)+(x+3) \cdot e^{-x / 2} \frac{d}{d x} \cdot x$

$=x(x+3) \cdot e^{-x / 2} \cdot(-\frac{1}{2})+x \cdot e^{-x / 2} \cdot 1+(x+3) \cdot e^{-x / 2} \cdot 1$

$=e^{-x / 2}[\frac{-x(x+3)}{2}+x+x+3]$

$=e^{-x / 2}[\frac{-x(x+3)}{2}+2 x+3]=e^{-x / 2}[\frac{-x^{2}-3 x+4 x+6}{2}]$

$=e^{-x / 2}[\frac{-x^{2}+x+6}{2}]$ which exists in $(-3,0)$.

So, $f(x)$ is differentiable in $(-3,0)$.

(iii) $\quad f(-3)=(-3)(-3+3) e^{-3 / 2}=0$

$ f(0)=(0)(0+3) e^{-0 / 2}=0 $

$\therefore f(-3)=f(0)=0$

As the above conditions are satisfied, then there must exist atleast one point $c \in(-3,0)$ such that

$ \begin{aligned} f^{\prime}(c)=0 & \Rightarrow e^{-c / 2}[\frac{-c^{2}+c+6}{2}]=0 \\ & \Rightarrow-\frac{e^{-c / 2}}{2}[c^{2}-c-6]=0 \\ & \Rightarrow-\frac{e^{-c / 2}}{2}(c-3)(c+2)=0 \\ & \Rightarrow e^{-c / 2} \neq 0 \quad \therefore(c-3)(c+2)=0 \end{aligned} $

Which gives $c=3, c=-2 \in(-3,0)$.

Hence, Rolle’s Theorem is verified.

Solution

Given that: $f(x)=\sqrt{4-x^{2}}$ in $[-2,2]$

(i) Since algebraic polynomials are continuous,

$\therefore f(x)$ is continuous in $[-2,2]$

(ii) $f^{\prime}(x)=\frac{d}{d x} \sqrt{4-x^{2}}=\frac{1}{2 \sqrt{4-x^{2}}} \times-2 x=\frac{-x}{\sqrt{4-x^{2}}}$ which exists

in $(-2,2)$

So, $f^{\prime}(x)$ is differentiable in $(-2,2)$.

(iii)

$ \begin{gathered} f(-2)=\sqrt{4-(-2)^{2}}=\sqrt{4-4}=0 \\ f(2)=\sqrt{4-(2)^{2}}=\sqrt{4-4}=0 \end{gathered} $

So $f(-2)=f(2)=0$

As the above conditions are satisfied, then there must exist atleast one point $c \in(-2,2)$ such that

$ f^{\prime}(c)=0 \Rightarrow \frac{-c}{\sqrt{4-c^{2}}}=0 \quad \Rightarrow \quad c=0 \in(-2,2) $

Hence, Rolle’s Theorem is verified.

Solution

(i) $f(x)$ being an algebraic polynomial, is continuous everywhere.

(ii) $f(x)$ must be differentiable at $x=1$

$ \begin{aligned} & \text{ L.H.L. }=\lim _{x \to 1^{-}} \frac{f(x)-f(1)}{x-1} \\ & =\lim _{x \to 1} \frac{(x^{2}+1)-(1+1)}{x-1} \\ & =\lim _{x \to 1} \frac{x^{2}+1-2}{x-1}=\lim _{x \to 1} \frac{x^{2}-1}{x-1} \\ & =\lim _{x \to 1} \frac{(x-1)(x+1)}{x-1}=\lim _{x \to 1}(x+1)=(1+1)=2 \\ & \text{ and R.H.L. }=\lim _{x \to 1^{+}} \frac{f(x)-f(1)}{x-1} \end{aligned} $

$ \begin{aligned} & =\lim _{x \to 1} \frac{(3-x)-(1+1)}{x-1} \\ & =\lim _{x \to 1} \frac{(3-x)-2}{x-1}=\lim _{x \to 1} \frac{1-x}{x-1}=-1 \end{aligned} $

$\therefore \quad$ L.H.L. $\neq$ R.H.L.

So, $f(x)$ is not differentiable at $x=1$.

Hence, Rolle’s Theorem is not applicable in [0,2].

Solution

Given that: $y=\cos x-1$ on $[0,2 \pi]$

We have to find a point $c$ on the given curve $y=\cos x-1$ on $[0,2 \pi]$ such that the tangent at $c \in[0,2 \pi]$ is parallel to $x$-axis i.e., $f^{\prime}(c)=0$ where $f^{\prime}(c)$ is the slope of the tangent.

So, we have to verify the Rolle’s Theorem.

(i) $y=\cos x-1$ is the combination of cosine and constant functions. So, it is continuous on $[0,2 \pi]$.

(ii) $\frac{d y}{d x}=-\sin x$ which exists in $(0,2 \pi)$.

So, it is differentiable on $(0,2 \pi)$.

(iii) Let $f(x)=\cos x-1$

$ f(0)=\cos 0-1=1-1=0 ; f(2 \pi)=\cos 2 \pi-1=1-1=0 $

$\therefore \quad f(0)=f(2 \pi)=0$

As the above conditions are satisfied, then there lies a point $c \in(0,2 \pi)$ such that $f^{\prime}(c)=0$.

$\therefore-\sin c=0 \Rightarrow \sin c=0$

$\therefore c=n \pi, n \in I$

$\Rightarrow c=\pi \in(0,2 \pi)$

Hence, $c=\pi$ is the point on the curve in $(0,2 \pi)$ at which the tangent is parallel to $x$-axis.

Solution

Given that: $y=x(x-4), x \in[0,4]$

Let $\quad f(x)=x(x-4), x \in[0,4]$

(i) $f(x)$ being an algebraic polynomial, is continuous function everywhere.

So, $f(x)=x(x-4)$ is continuous in $[0,4]$.

(ii) $f^{\prime}(x)=2 x-4$ which exists in $(0,4)$.

So, $f(x)$ is differentiable. (iii) $\quad f(0)=0(0-4)=0$

$ f(4)=4(4-4)=0 $

So $f(0)=f(4)=0$

As the above conditions are satisfied, then there must exist at least one point $c \in(0,4)$ such that $f^{\prime}(c)=0$

$\therefore 2 c-4=0 \Rightarrow c=2 \in(0,4)$

Hence, $c=2$ is the point in $(0,4)$ on the given curve at which the tangent is parallel to the $x$-axis.

Solution

Given that: $f(x)=\frac{1}{4 x-1}$ in $[1,4]$.

(i) $f(x)$ is an algebraic function, so it is continuous in $[1,4]$.

(ii) $f^{\prime}(x)=\frac{-4}{(4 x-1)^{2}}$ which exists in $(1,4)$.

So, $f(x)$ is differentiable.

As the above conditions are satisfied then there must exist a point $c \in(1,4)$ such that

$ \begin{matrix} f^{\prime}(c) =\frac{f(b)-f(a)}{b-a} \\ \frac{-4}{(4 c-1)^{2}} =\frac{\frac{1}{4(4)-1}-\frac{1}{4(1)-1}}{4-1} \\ \Rightarrow \quad \frac{-4}{(4 c-1)^{2}} =\frac{\frac{1}{15}-\frac{1}{3}}{3}=\frac{1-5}{15 \times 3}=\frac{-4}{45}=\frac{1}{(4 c-1)^{2}}=\frac{1}{45} \\ \Rightarrow \quad(4 c-1)^{2} =45 \quad 4 c-1 = \pm 3 \sqrt{5} \Rightarrow 4 c=+1 \pm 3 \sqrt{5} \\ \Rightarrow \quad \quad c =\frac{+1 \pm 3 \sqrt{5}}{4} \end{matrix} $

$ \therefore \quad c=\frac{+1 \pm 3 \sqrt{5}}{4} \in(1,4) $

Hence, Mean Value Theorem is verified.

Solution

Given that: $f(x)=x^{3}-2 x^{2}-x+3$ in $[0,1]$

(i) Being an algebraic polynomial, $f(x)$ is continuous in $[0,1]$

(ii) $f^{\prime}(x)=3 x^{2}-4 x-1$ which exists in $(0,1)$.

So, $f(x)$ is differentiable.

As the above conditions are satisfied, then there must exist atleast one point $c \in(0,1)$ such that

$ \begin{aligned} & f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \\ & \Rightarrow \quad 3 c^{2}-4 c-1=\frac{[(1)^{3}-2(1)^{2}-(1)+3]-[0-0-0+3]}{1-0} \\ & \Rightarrow \quad 3 c^{2}-4 c-1=\frac{(1-2-1+3)-(3)}{1} \\ & \Rightarrow \quad 3 c^{2}-4 c-1=1-3 \Rightarrow 3 c^{2}-4 c-1=-2 \\ & \Rightarrow \quad 3 c^{2}-4 c+1=0 \Rightarrow 3 c^{2}-3 c-c+1=0 \\ & \Rightarrow \quad 3 c(c-1)-1(c-1)=0 \Rightarrow(c-1)(3 c-1)=0 \\ & \Rightarrow \quad c-1=0 \quad \therefore c=1 \\ & 3 c-1=0 \quad \therefore c=\frac{1}{3} \in(0,1) \end{aligned} $

Hence, Mean Value Theorem is verified.

Solution

Given that: $f(x)=\sin x-\sin 2 x$ in $[0, \pi]$

(i) Since trigonometric functions are always continuous on their domain.

So, $f(x)$ is continuous on $[0, \pi]$.

(ii) $f^{\prime}(x)=\cos x-2 \cos 2 x$ which exists in $(0, \pi)$

So, $f(x)$ is differentiable on $(0, \pi)$.

Since the above conditions are satisfied, then there must exist atleast one point $c \in(0, \pi)$ such that

$ \begin{aligned} f^{\prime}(c) & =\frac{f(b)-f(a)}{b-a} \\ \cos c-2 \cos 2 c & =\frac{(\sin \pi-\sin 2 \pi)-(\sin 0-\sin 0)}{\pi-0} \\ \Rightarrow \quad \cos c-2(2 \cos ^{2} c-1) & =0 \Rightarrow \cos c-4 \cos ^{2} c+2=0 \\ \Rightarrow \quad 4 \cos ^{2} c-\cos c-2 & =0 \end{aligned} $

$ \begin{aligned} & \Rightarrow \quad \cos c=\frac{-(-1) \pm \sqrt{(-1)^{2}-4 \times 4 \times-2}}{2 \times 4} \\ & \Rightarrow \quad \cos c=\frac{1 \pm \sqrt{1+32}}{8}=\frac{1 \pm \sqrt{33}}{8} \\ & \Rightarrow \quad c=\cos ^{-1}(\frac{1 \pm \sqrt{33}}{8}) \in(0, \pi) . \end{aligned} $

Hence, Mean Value Theorem is verified.

Solution

Given that: $f(x)=\sqrt{25-x^{2}}$ in $[1,5]$

(i) $f(x)$ is continuous if $25-x^{2} \geq 0 \Rightarrow-x^{2} \geq-25$

$\Rightarrow x^{2} \leq 25 \Rightarrow x \leq \pm 5 \Rightarrow-5 \leq x \leq 5$

So, $f(x)$ is continuous on $[1,5]$.

(ii) $f^{\prime}(x)=\frac{1}{2 \sqrt{25-x^{2}}} \times(-2 x)=\frac{-x}{\sqrt{25-x^{2}}}$ which exists in $(1,5)$.

So, $f(x)$ is differentiable in $[1,5]$.

Since the above conditions are satisfied then there must exist atleast one point $c \in(1,5)$ such that

$ \begin{aligned} f^{\prime}(c) & =\frac{f(b)-f(a)}{b-a} \\ \frac{-c}{\sqrt{25-c^{2}}} & =\frac{\sqrt{25-25}-\sqrt{25-1}}{5-1} \\ \Rightarrow \quad \frac{-c}{\sqrt{25-c^{2}}} & =\frac{0-\sqrt{24}}{4} \\ \Rightarrow \quad \frac{c}{\sqrt{25-c^{2}}} & =\frac{2 \sqrt{6}}{4} \Rightarrow \frac{c}{\sqrt{25-c^{2}}}=\frac{\sqrt{6}}{2} \end{aligned} $

Squaring both sides

$\frac{c^{2}}{25-c^{2}} =\frac{6}{4}=\frac{3}{2}$

$\Rightarrow 2 c^{2} =75-3 c^{2} \Rightarrow 5 c^{2}=75 \Rightarrow c^{2}=15$

$\therefore c = \pm \sqrt{15} \in(1,5) $

Hence, Mean Value Theorem is verified.

Solution

Given that: $y=(x-3)^{2}$

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

(i) Being an algebraic polynomial, $f(x)$ is continuous at $x_1=3$ and $x_2=4$ i.e. in $[3,4]$.

(ii) $f^{\prime}(x)=2(x-3)$ which exists in $(3,4)$.

Hence, by mean value theorem, there must exist a point $c$ on the curve at which the tangent is parallel to the chord joining the points $(3,0)$ and $(4,1)$.

$\therefore \quad f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \quad$ where $b=4$ and $a=3$

$\Rightarrow \quad 2(c-3)=\frac{(4-3)^{2}-(3-3)^{2}}{4-3}$

$\Rightarrow \quad 2 c-6=\frac{1-0}{1}=1 \quad \Rightarrow 2 c=6+1=7$

$\therefore \quad c=\frac{7}{2}$

If $x=\frac{7}{2} \quad \therefore y=(\frac{7}{2}-3)^{2}=\frac{1}{4}$.

Hence, $(\frac{7}{2}, \frac{1}{4})$ is the point on the curve at which the tangent is parallel to the chord joining the points $(3,0)$ and $(4,1)$.

Solution

Given that: $y=2 x^{2}-5 x+3$

Let $\quad f(x)=2 x^{2}-5 x+3$

(i) Being an algebraic polynomial, $f(x)$ is continuous in [1,2].

(ii) $f^{\prime}(x)=4 x-5$ which exists in $(1,2)$.

As per the Mean Value Theorem, there must exist a point $c \in(1,2)$ on the curve at which the tangent is parallel to the chord joining the points $A(1,0)$ and $B(2,1)$.

So $\quad f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$

$ \begin{aligned} 4 c-5 & =\frac{(8-10+3)-(2-5+3)}{2-1} \\ \Rightarrow \quad 4 c-5 & =\frac{1-0}{1}=1 \Rightarrow 4 c=1+5 \Rightarrow 4 c=6 \end{aligned} $

$ \begin{aligned} & \therefore \quad c=\frac{6}{4}=\frac{3}{2} \\ & \therefore \quad y=2(\frac{3}{2})^{2}-5(\frac{3}{2})+3 \\ & =2 \times \frac{9}{4}-\frac{15}{2}+3=\frac{9}{2}-\frac{15}{2}+3=\frac{9-15+6}{2}=0 \end{aligned} $

Hence, $(\frac{3}{2}, 0)$ is the point on the curve at which the tangent is parallel to the chord joining the points $A(1,0)$ and $B(2,1)$.

Solution

Given that:

$ \begin{aligned} f(x) & =\begin{cases} x^{2}+3 x+p, \text{ if } x \leq 1 \\ q x+2, \text{ if } x>1 \end{cases} \text{ at } x=1 .. \\ \text{ L.H.L. } f^{\prime}(c) & =\lim _{x \to 1^{-}} \frac{f(x)-f(c)}{x-c} \\ \Rightarrow \quad f^{\prime}(1) & =\lim _{x \to 1^{-}} \frac{f(x)-f(1)}{x-1} \\ & =\lim _{x \to 1^{-}} \frac{(x^{2}+3 x+p)-(1+3+p)}{x-1} \\ & =\lim _{h \to 0} \frac{[(1-h)^{2}+3(1-h)+p]-[4+p]}{1-h-1} \\ & =\lim _{h \to 0} \frac{[1+h^{2}-2 h+3-3 h+p]-[4+p]}{-h} \\ & =\lim _{h \to 0} \frac{[h^{2}-5 h+4+p]-[4+p]}{-h} \\ & =\lim _{h \to 0} \frac{h^{2}-5 h+4+p-4-p}{-h} \\ & =\lim _{h \to 0} \frac{h^{2}-5 h}{-h}=\lim _{h \to 0} \frac{h[h-5]}{-h}=5 \\ \text{ R.H.L. } f^{\prime}(1) & =\lim _{x \to 1^{+}} \frac{f(x)-f(1)}{x-1} \end{aligned} $

$ \begin{aligned} & =\lim _{x \to 1^{+}} \frac{(q x+2)-(1+3+p)}{x-1} \\ & =\lim _{h \to 0} \frac{[q(1+h)+2]-[4+p]}{1+h-1} \\ & =\lim _{h \to 0} \frac{q+q h+2-4-p}{h}=\lim _{h \to 0} \frac{q h+q-2-p}{h} \end{aligned} $

For existing the limit

$$ \begin{align*} q-2-p=0 \Rightarrow q-p=2 \tag{i}\\ \Rightarrow \lim _{h \to 0} \frac{q h-0}{h}=q \end{align*} $$

If L.H.L. $f^{\prime}(1)=$ R.H.L. $f^{\prime}(1)$ then $q=5$.

Now putting the value of $q$ in eqn. (i)

$ 5-p=2 \Rightarrow p=3 . $

Hence, value of $p$ is 3 and that of $q$ is 5 .

Solution

(i) Given that: $x^{m} \cdot y^{n}=(x+y)^{m+n}$

Taking $\log$ on both sides

$ \begin{aligned} \log x^{m} \cdot y^{n} & =\log (x+y)^{m+n} \quad[\because \log x y=\log x+\log y] \\ \Rightarrow \quad \log x^{m}+\log y^{n} & =(m+n) \log (x+y) \\ \Rightarrow m \log x+n \log y & =(m+n) \log (x+y) \end{aligned} $

Differentiating both sides w.r.t. $x$

$ \begin{matrix} \Rightarrow m \cdot \frac{d}{d x} \log x+n \cdot \frac{d}{d x} \log y =(m+n) \frac{d}{d x} \log (x+y) \\ \Rightarrow m \cdot \frac{1}{x}+n \cdot \frac{1}{y} \cdot \frac{d y}{d x}= (m+n) \cdot \frac{1}{x+y}(1+\frac{d y}{d x}) \\ \Rightarrow \frac{m}{x}+\frac{n}{y} \cdot \frac{d y}{d x} =\frac{m+n}{x+y} \cdot(1+\frac{d y}{d x}) \\ \Rightarrow \frac{m}{x}+\frac{n}{y} \cdot \frac{d y}{d x} =\frac{m+n}{x+y}+\frac{m+n}{x+y} \cdot \frac{d y}{d x} \\ \Rightarrow \frac{n y}{y x}-\frac{m+n}{x+y} \cdot \frac{d y}{d x} =\frac{m+n}{x+y}-\frac{m}{x} \end{matrix} $

$ \begin{aligned} \Rightarrow (\frac{n}{y}-\frac{m+n}{x+y}) \frac{d y}{d x} =\frac{m+n}{x+y}-\frac{m}{x} \\ \Rightarrow (\frac{n x+n y-m y-n y}{y(x+y)}) \frac{d y}{d x} =(\frac{m x+n x-m x-m y}{x(x+y)}) \\ \Rightarrow (\frac{n x-m y}{y(x+y)}) \frac{d y}{d x} =(\frac{n x-m y}{x(x+y)}) \\ \Rightarrow \frac{d y}{d x} =\frac{n x-m y}{x(x+y)} \times \frac{y(x+y)}{n x-m y} \\ \Rightarrow \frac{d y}{d x} =\frac{y}{x} \text{ Hence proved. } \end{aligned} $

(ii) Given that: $\frac{d y}{d x}=\frac{y}{x}$

Differentiating both sides w.r.t. $x$

$ \begin{matrix} \frac{d}{d x}(\frac{d y}{d x}) =\frac{d}{d x}(\frac{y}{x}) \\ \Rightarrow \quad \frac{d^{2} y}{d x^{2}} =\frac{x \cdot \frac{d y}{d x} y \cdot 1}{x \cdot \frac{y}{x}-y} \\ =\frac{y-y}{x^{2}}=\frac{0}{x^{2}}=0 {[\because \frac{d y}{d x}=\frac{y}{x}]} \\ \text{ Hence, } \quad \frac{d^{2} y}{d x^{2}}=0 . \text{ Hence, proved. } \end{matrix} $

Solution

Given that: $x=\sin t$ and $y=\sin p t$

Differentiating both the parametric functions w.r.t. $t$

$ \begin{aligned} \frac{d x}{d t} & =\cos t \text{ and } \frac{d y}{d t}=\cos p t \cdot p=p \cos p t \\ \frac{d y}{d x} & =\frac{d y / d t}{d x / d t}=\frac{p \cos p t}{\cos t} \\ \therefore \quad \frac{d y}{d x} & =\frac{p \cos p t}{\cos t} \end{aligned} $

Again differentiating w.r.t. $x$,

$ \begin{aligned} \frac{d}{d x}(\frac{d y}{d x}) & =p \cdot \frac{d}{d x}(\frac{\cos p t}{\cos t}) \\ \Rightarrow \quad \frac{d^{2} y}{d x^{2}} & =p[\frac{\cos t \cdot \frac{d}{d x}(\cos p t)-\cos p t \cdot \frac{d}{d x}(\cos t)}{\cos ^{2} t}] \\ & =p[\frac{\cos t(-\sin p t) \cdot p \frac{d t}{d x}-\cos p t(-\sin t) \cdot \frac{d t}{d x}}{\cos ^{2} t}] \\ & =p[\frac{-p \cos t \sin p t+\cos p t \sin t}{\cos ^{2} t}] \frac{d t}{d x} \\ & =p[\frac{-p \cos t \sin p t+\cos p t \sin t}{\cos ^{2} t}] \cdot \frac{1}{\cos t} \\ & =p[\frac{-p \cos t \sin p t+\cos p t \sin t}{\cos ^{3} t}] \end{aligned} $

Now we have to prove that

$ (1-x^{2}) \cdot \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}+p^{2} y=0 $

L.H.S. $=(1-x^{2})[p(\frac{-p \cos t \sin p t+\cos p t \sin t}{\cos ^{3} t})]-x . p \frac{\cos p t}{\cos t}+p^{2} y$

$\Rightarrow(1-\sin ^{2} t)[p(\frac{-p \cos t \sin p t+\cos p t \sin t}{\cos ^{3} t})]-\frac{p \sin t \cdot \cos p t}{\cos t}$

$+p^{2} \cdot \sin p t$

$\Rightarrow \cos ^{2} t[\frac{-p^{2} \cos t \sin p t+p \cos p t \sin t}{\cos ^{3} t}]-\frac{p \sin t \cdot \cos p t}{\cos t}$

$+p^{2} \cdot \sin p t$

$\Rightarrow \frac{-p^{2} \cos t \sin p t+p \cos p t \sin t}{\cos t}-\frac{p \sin t \cos p t}{\cos t}+p^{2} \sin p t$

$\Rightarrow \frac{-p^{2} \cos t \sin p t+p \cos p t \sin t-p \sin t \cos p t+p^{2} \sin p t \cos t}{\cos t}$

$\Rightarrow \frac{0}{\cos t}=0=$ R.H.S.

Hence, proved.

Solution

Given that: $y=x^{\tan x}+\sqrt{\frac{x^{2}+1}{2}}$

Let $u=x^{\tan x} \quad$ and $v=\sqrt{\frac{x^{2}+1}{2}}$

$\therefore \quad y=u+v$

Differentiating both sides w.r.t. $x$

$$ \begin{equation*} \frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x} \tag{i} \end{equation*} $$

Now taking $u=x^{\tan x}$

Taking $\log$ on both sides $\log u=\log (x^{\tan x})$

$ \log u=\tan x \cdot \log x $

Differentiating both sides w.r.t. $x$

$ \begin{aligned} & \frac{1}{u} \cdot \frac{d u}{d x}=\frac{d}{d x}(\tan x \cdot \log x) \\ & \Rightarrow \quad \frac{1}{u} \cdot \frac{d u}{d x}=\tan x \cdot \frac{d}{d x}(\log x)+\log x \cdot \frac{d}{d x}(\tan x) \\ & \Rightarrow \quad \frac{1}{u} \cdot \frac{d u}{d x}=\tan x \cdot \frac{1}{x}+\log x \cdot \sec ^{2} x \\ & \Rightarrow \quad \frac{d u}{d x}=u[\frac{\tan x}{x}+\log x \cdot \sec ^{2} x] \\ & \therefore \quad \frac{d u}{d x}=x^{\tan x}[\frac{\tan x}{x}+\log x \sec ^{2} x] \\ & \text{ Taking } \quad v=\sqrt{\frac{x^{2}+1}{2}} \Rightarrow v=\frac{1}{\sqrt{2}} \sqrt{x^{2}+1} \end{aligned} $

Differentiating both sides w.r.t. $x$

$ \frac{d v}{d x}=\frac{1}{\sqrt{2}} \cdot \frac{1}{2 \sqrt{x^{2}+1}} \cdot 2 x=\frac{x}{\sqrt{2} \sqrt{x^{2}+1}} $

Putting the values of $\frac{d u}{d x}$ and $\frac{d v}{d x}$ in eqn. (i)

$ \frac{d y}{d x}=x^{\tan x}[\log x \sec ^{2} x+\frac{\tan x}{x}]+\frac{x}{\sqrt{2} \sqrt{x^{2}+1}} $

Solution

We know that the algebraic polynomials are continuous functions everywhere.

$\therefore f(x)+g(x)$ is continuous $\quad[\because$ Sum, difference and product of two continuous functions is continuous also continuous]

$f(x)-g(x)$ is continuous

$f(x) \cdot g(x)$ is continuous

$\frac{g(x)}{f(x)}$ is only continuous if $g(x) \neq 0$

$\therefore \frac{f(x)}{g(x)}=\frac{2 x}{\frac{x^{2}}{2}+1}=\frac{4 x}{x^{2}+2}$

Here, $\frac{g(x)}{f(x)}=\frac{\frac{x^{2}}{2}+1}{2 x}=\frac{x^{2}+2}{4 x}$ which is discontinuous at $x=0$.

Hence, the correct option is (d).

• Option (a): ( f(x) + g(x) ) is incorrect because the sum of two continuous functions is always continuous. Since both ( f(x) = 2x ) and ( g(x) = \frac{x^2}{2} + 1 ) are continuous everywhere, their sum ( f(x) + g(x) ) is also continuous.

• Option (b): ( f(x) - g(x) ) is incorrect because the difference of two continuous functions is always continuous. Since both ( f(x) = 2x ) and ( g(x) = \frac{x^2}{2} + 1 ) are continuous everywhere, their difference ( f(x) - g(x) ) is also continuous.

• Option (c): ( f(x) \cdot g(x) ) is incorrect because the product of two continuous functions is always continuous. Since both ( f(x) = 2x ) and ( g(x) = \frac{x^2}{2} + 1 ) are continuous everywhere, their product ( f(x) \cdot g(x) ) is also continuous.

Solution

Given that: $f(x)=\frac{4-x^{2}}{4 x-x^{3}}$

For discontinuous function

$ \begin{aligned} & 4 x-x^{3}=0 \\ \Rightarrow & x(4-x^{2})=0 \\ \Rightarrow & x(2-x)(2+x)=0 \\ \Rightarrow & x=0, x=-2, x=2 \end{aligned} $

Hence, the given function is discontinuous exactly at three points. Hence, the correct option is (c).

• Option (a) is incorrect because the function ( f(x) = \frac{4 - x^2}{4x - x^3} ) is discontinuous at three points, not just one. The points of discontinuity are ( x = 0 ), ( x = -2 ), and ( x = 2 ).

• Option (b) is incorrect because the function ( f(x) = \frac{4 - x^2}{4x - x^3} ) is discontinuous at three points, not exactly two. The points of discontinuity are ( x = 0 ), ( x = -2 ), and ( x = 2 ).

• Option (d) is incorrect because there is a correct option among the given choices. The function ( f(x) = \frac{4 - x^2}{4x - x^3} ) is discontinuous at exactly three points, which is correctly identified by option (c).

Solution

Given that: $f(x)=|2 x-1| \sin x$

Clearly, $f(x)$ is not differentiable at $x=\frac{1}{2}$

$ \begin{aligned} & \text{ R.H.L. }=f^{\prime}(\frac{1}{2})=\lim _{h \to 0} \frac{f(\frac{1}{2}+h)-f(\frac{1}{2})}{h} \\ & =\lim _{h \to 0} \frac{|2(\frac{1}{2}+h)-1| \sin (\frac{1}{2}+h)-0}{h} \\ & =\lim _{h \to 0} \frac{|2 h| \sin (\frac{1+2 h}{2})}{h}=2 \sin (\frac{1}{2}) \\ & \text{ Also L.H.L. }=f^{\prime}(\frac{1}{2})=\lim _{h \to 0} \frac{f(\frac{1}{2}-h)-f(\frac{1}{2})}{-h} \\ & =\lim _{h \to 0} \frac{|2(\frac{1}{2}-h)-1|[-\sin (\frac{1}{2}-h)]-0}{-h} \\ & =\frac{|-2 h|[-\sin (\frac{1}{2}-h)]}{-h}=-2 \sin (\frac{1}{2}) \\ & \therefore \text{ R.H.L. }=f^{\prime}(\frac{1}{2}) \neq \text{ L.H.L. } f^{\prime}(\frac{1}{2}) \end{aligned} $

So, the given function $f(x)$ is not differentiable at $x=\frac{1}{2}$.

$\therefore f(x)$ is differentiable in $R-{\frac{1}{2}}$.

Hence, the correct option is $(b)$.

• Option (a) $R$: This option is incorrect because the function $f(x) = |2x - 1| \sin x$ is not differentiable at $x = (\frac{1}{2})$. For a function to be differentiable on the entire set of real numbers (R), it must be differentiable at every point in (R). Since (f(x)) is not differentiable at (x = (\frac{1}{2}), it cannot be differentiable on (R).

• Option (c) $(0, \infty)$: This option is incorrect because it excludes negative values and zero, where the function (f(x)) is actually differentiable. The function (f(x)) is differentiable for all (x \in R) except at (x = (\frac{1}{2}). Therefore, the correct set should include all real numbers except (x = (\frac{1}{2}), not just the positive real numbers.

• Option (d) none of these: This option is incorrect because there is a correct option provided, which is (R - {\frac{1}{2}}). The function (f(x)) is differentiable everywhere except at (x = (\frac{1}{2}), making option (b) the correct choice.

Solution

Given that: $f(x)=\cot x$

$ \Rightarrow \quad f(x)=\frac{\cos x}{\sin x} $

We know that $\sin x=0$ if $f(x)$ is discontinuous.

$\therefore$ If $\quad \sin x=0$

$\therefore \quad x=n \pi, n \in n \pi$.

So, the given function $f(x)$ is discontinuous on the set $\{x=n \pi$; $n \in Z\}$.

Hence, the correct option is $(a)$.

• Option (b) $\{x=2 n \pi ; n \in Z\}$ is incorrect because $\sin x = 0$ not only at $2n\pi$ but also at $n\pi$ for all integers $n$. Therefore, the set of discontinuities is broader than just $2n\pi$.

• Option (c) $\{x=(2 n+1) \frac{\pi}{2} ; n \in Z\}$ is incorrect because $\sin x = 0$ at $n\pi$, not at $(2n+1)\frac{\pi}{2}$. The points $(2n+1)\frac{\pi}{2}$ are where $\cos x = 0$, which are not the points of discontinuity for $\cot x$.

• Option (d) $\{x=\frac{n \pi}{2} ; n \in Z\}$ is incorrect because $\sin x = 0$ at $n\pi$, not at $\frac{n\pi}{2}$. The points $\frac{n\pi}{2}$ include both points where $\sin x = 0$ and where $\cos x = 0$, but $\cot x$ is only discontinuous where $\sin x = 0$.

Solution

Given that: $f(x)=e^{|x|}$

We know that modulus function is continuous but not differentiable in its domain.

Let $\quad g(x)=|x|$ and $t(x)=e^{x}$

$\therefore \quad f(x)=got(x)=g[t(x)]=e^{|x|}$

Since $g(x)$ and $t(x)$ both are continuous at $x=0$ but $f(x)$ is not differentiable at $x=0$.

Hence, the correct option is $(a)$.

• Option (b) is incorrect because the function $f(x) = e^{|x|}$ is not differentiable at $x=0$. The modulus function $|x|$ is continuous everywhere but not differentiable at $x=0$, and since $f(x)$ involves $|x|$, it inherits this non-differentiability at $x=0$.

• Option (c) is incorrect because the function $f(x) = e^{|x|}$ is continuous everywhere, including at $x=0$. The exponential function and the modulus function are both continuous, and their composition is also continuous.

• Option (d) is incorrect because option (a) correctly describes the properties of the function $f(x) = e^{|x|}$, making it the correct choice.

Solution

Given that: $f(x)=x^{2} \sin \frac{1}{x}$ where $x \neq 0$.

So, the value of the function $f$ at $x=0$, so that $f(x)$ is continuous is 0 .

Hence, the correct option is $(a)$.

• Option (b) -1: If the value of the function $f$ at $x=0$ were -1, the function would not be continuous at $x=0$ because the limit of $f(x)$ as $x$ approaches 0 is 0, not -1. Therefore, setting $f(0) = -1$ would create a discontinuity at $x=0$.

• Option (c) 1: If the value of the function $f$ at $x=0$ were 1, the function would not be continuous at $x=0$ because the limit of $f(x)$ as $x$ approaches 0 is 0, not 1. Therefore, setting $f(0) = 1$ would create a discontinuity at $x=0$.

• Option (d) none of these: This option is incorrect because there is a correct value that makes the function continuous at $x=0$, which is 0. Therefore, “none of these” is not the correct answer.

Solution

Given that: $f(x)=\begin{cases} m x+1, \text{ if } x \leq \frac{\pi}{2} \\ \sin x+n, \text{ if } x>\frac{\pi}{2} \end{cases} .$ is continuous at $x=\frac{\pi}{2}$

L.H.L. $=\lim _{x \to \frac{\pi^{-}}{2}}(m x+1)=\lim _{h \to 0}[m(\frac{\pi}{2}-h)+1]=\frac{m \pi}{2}+1$

$ \begin{aligned} \text{ R.H.L. }= & \lim _{x \to \frac{\pi^{+}}{2}}(\sin x+n)=\quad \lim _{h \to 0}[\sin (\frac{\pi}{2}+h)+n] \\ & =\lim _{h \to 0} \cos h+n=1+n \end{aligned} $

When $f(x)$ is continuous at $x=\frac{\pi}{2}$

$\therefore \quad$ L.H.L. $=$ R.H.L.

$ \frac{m \pi}{2}+1=1+n \Rightarrow n=\frac{m \pi}{2} $

Hence, the correct option is (c).

• Option (a) $m=1, n=0$: This option is incorrect because substituting $m=1$ and $n=0$ into the equation $\frac{m \pi}{2} + 1 = 1 + n$ does not satisfy the equation. Specifically, $\frac{1 \cdot \pi}{2} + 1 = 1 + 0$ simplifies to $\frac{\pi}{2} + 1 = 1$, which is not true.

• Option (b) $m=\frac{n \pi}{2}+1$: This option is incorrect because it does not directly satisfy the condition derived from the continuity requirement. The correct relationship derived is $n = \frac{m \pi}{2}$, not $m = \frac{n \pi}{2} + 1$. Substituting $m = \frac{n \pi}{2} + 1$ into the continuity condition would not simplify correctly to the required form.

• Option (d) $m=n=\frac{\pi}{2}$: This option is incorrect because substituting $m = \frac{\pi}{2}$ and $n = \frac{\pi}{2}$ into the equation $\frac{m \pi}{2} + 1 = 1 + n$ does not satisfy the equation. Specifically, $\frac{\frac{\pi}{2} \cdot \pi}{2} + 1 = 1 + \frac{\pi}{2}$ simplifies to $\frac{\pi^2}{4} + 1 = 1 + \frac{\pi}{2}$, which is not true.

Solution

Given that: $f(x)=|\sin x|$

Let $g(x)=\sin x$ and $t(x)=|x|$

$\therefore \quad f(x)=tog(x)=t[g(x)]=t(\sin x)=|\sin x|$

where $g(x)$ and $t(x)$ both are continuous.

$\therefore f(x)=got(x)$ is continuous but $t(x)$ is not differentiable at $x=0$.

So, $f(x)$ is not continuous at $\sin x=0 \Rightarrow x=n \pi, n \in Z$.

Hence, the correct option is $(b)$.

• Option (a) is incorrect because ( f(x) = |\sin x| ) is not differentiable at points where ( \sin x = 0 ), which occurs at ( x = n\pi ) for ( n \in \mathbb{Z} ). At these points, the function has a cusp, and the derivative does not exist.

• Option (c) is incorrect because ( f(x) = |\sin x| ) is differentiable at ( x = (2n+1)\frac{\pi}{2} ) for ( n \in \mathbb{Z} ). At these points, ( \sin x ) reaches its maximum or minimum values of ( \pm 1 ), but the absolute value function does not introduce any non-differentiability at these points.

• Option (d) is incorrect because option (b) correctly describes the behavior of ( f(x) = |\sin x| ). The function is continuous everywhere but not differentiable at ( x = n\pi ) for ( n \in \mathbb{Z} ).

Solution

Given that: $y=\log (\frac{1-x^{2}}{1+x^{2}})$

$\Rightarrow y=\log (1-x^{2})-\log (1+x^{2})$

$[\because \log \frac{x}{y}=\log x-\log y]$

Differentiating both sides w.r.t. $x$

$ \begin{aligned} \frac{d y}{d x} & =\frac{1}{1-x^{2}} \cdot \frac{d}{d x}(1-x^{2})-\frac{1}{1+x^{2}} \cdot \frac{d}{d x}(1+x^{2}) \\ & =\frac{-2 x}{1-x^{2}}-\frac{2 x}{1+x^{2}}=\frac{-2 x-2 x^{3}-2 x+2 x^{3}}{(1-x^{2})(1+x^{2})}=\frac{-4 x}{1-x^{4}} \end{aligned} $

Hence, the correct option is $(b)$.

• Option (a) $\frac{4 x^{3}}{1-x^{4}}$: This option is incorrect because the numerator should be $-4x$ instead of $4x^3$. The correct differentiation results in a term involving $-2x$ and $-2x^3$, which simplifies to $-4x$ in the numerator.

• Option (c) $\frac{1}{4-x^{4}}$: This option is incorrect because the structure of the denominator is incorrect. The correct denominator should be $1 - x^4$, not $4 - x^4$. Additionally, the numerator should be $-4x$, not 1.

• Option (d) $\frac{-4 x^{3}}{1-x^{4}}$: This option is incorrect because the numerator should be $-4x$ instead of $-4x^3$. The differentiation process results in a term involving $-2x$ and $-2x^3$, which simplifies to $-4x$ in the numerator, not $-4x^3$.

Solution

Given that: $y=\sqrt{\sin x+y}$

Differentiating both sides w.r.t. $x$

$ \begin{aligned} & \frac{d y}{d x}=\frac{1}{2 \sqrt{\sin x+y}} \cdot \frac{d}{d x}(\sin x+y) \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{1}{2 \sqrt{\sin x+y}} \cdot(\cos x+\frac{d y}{d x}) \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{1}{2 y} \cdot[\cos x+\frac{d y}{d x}] \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{\cos x}{2 y}+\frac{1}{2 y} \cdot \frac{d y}{d x} \\ & \Rightarrow \quad \frac{d y}{d x}-\frac{1}{2 y} \cdot \frac{d y}{d x}=\frac{\cos x}{2 y} \\ & \Rightarrow \quad(1-\frac{1}{2 y}) \frac{d y}{d x}=\frac{\cos x}{2 y} \Rightarrow(\frac{2 y-1}{2 y}) \frac{d y}{d x}=\frac{\cos x}{2 y} \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{\cos x}{2 y} \times \frac{2 y}{2 y-1} \Rightarrow \frac{d y}{d x}=\frac{\cos x}{2 y-1} \end{aligned} $

Hence, the correct option is $(a)$.

• Option (b) $\frac{\cos x}{1-2 y}$: This option is incorrect because the denominator is incorrectly structured. The correct differentiation process shows that the denominator should be (2y - 1), not (1 - 2y). The sign and order of terms in the denominator are crucial for the correct result.

• Option (c) $\frac{\sin x}{1-2 y}$: This option is incorrect for two reasons. First, the numerator should be (\cos x) as derived from the differentiation of (\sin x). Second, the denominator is incorrectly structured as (1 - 2y) instead of (2y - 1).

• Option (d) $\frac{\sin x}{2 y-1}$: This option is incorrect because the numerator should be (\cos x) instead of (\sin x). The differentiation of (\sin x) with respect to (x) yields (\cos x), not (\sin x).

Solution

Let $y=\cos ^{-1}(2 x^{2}-1)$ and $t=\cos ^{-1} x$

Differentiating both the functions w.r.t. $x$

$ \frac{d y}{d x}=\frac{d}{d x} \cos ^{-1}(2 x^{2}-1) \text{ and } \frac{d t}{d x}=\frac{d}{d x} \cos ^{-1} x $

$ \begin{aligned} & \Rightarrow \quad \frac{d y}{d x}=\frac{-1}{\sqrt{1-(2 x^{2}-1)^{2}}} \cdot \frac{d}{d x}(2 x^{2}-1) \text{ and } \frac{d t}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \\ & =\frac{-1.4 x}{\sqrt{1-(4 x^{4}+1-4 x^{2})}} \text{ and } \frac{d t}{d x}=\frac{-1}{\sqrt{1-x^{2}}} \\ & =\frac{-4 x}{\sqrt{1-4 x^{4}-1+4 x^{2}}}=\frac{-4 x}{\sqrt{4 x^{2}-4 x^{4}}}=\frac{-4 x}{2 x \sqrt{1-x^{2}}} \\ & \Rightarrow \quad \frac{d y}{d x}=\frac{-2}{\sqrt{1-x^{2}}} \end{aligned} $

Now $\frac{d y}{d t}=\frac{d y / d x}{d t / d x}=\frac{\frac{-2}{\sqrt{1-x^{2}}}}{\frac{-1}{\sqrt{1-x^{2}}}}=2$

Hence, the correct option is (a).

• Option (b): The expression (\frac{-1}{2 \sqrt{1-x^{2}}}) is incorrect because it does not match the derived value of (\frac{d y}{d t}). The correct derivative is (\frac{d y}{d t} = 2), not a fraction involving (\sqrt{1-x^{2}}).

• Option (c): The expression (\frac{2}{x}) is incorrect because it does not align with the derived value of (\frac{d y}{d t}). The correct derivative is (\frac{d y}{d t} = 2), and there is no term involving (x) in the denominator.

• Option (d): The expression (1-x^{2}) is incorrect because it does not match the derived value of (\frac{d y}{d t}). The correct derivative is (\frac{d y}{d t} = 2), and there is no polynomial expression involving (x) in the result.

Solution

Given that $x=t^{2}$ and $y=t^{3}$

Differentiating both the parametric functions w.r.t. $t$

$\frac{d x}{d t}=2 t$ and $\frac{d y}{d t}=3 t^{2}$

$\therefore \quad \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{3 t^{2}}{2 t}=\frac{3}{2} t \Rightarrow \frac{d y}{d x}=\frac{3}{2} t$

Now differentiating again w.r.t. $x$

$ \frac{d}{d x}(\frac{d y}{d x})=\frac{3}{2} \cdot \frac{d t}{d x} \Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{3}{2} \cdot \frac{1}{2 t}=\frac{3}{4 t} $

Hence, the correct option is (b).

• Option (a) $\frac{3}{2}$ is incorrect because it does not account for the dependency on the parameter ( t ). The second derivative (\frac{d^2 y}{d x^2}) should involve ( t ) since both ( x ) and ( y ) are functions of ( t ).

• Option (c) $\frac{3}{2 t}$ is incorrect because it incorrectly simplifies the second derivative. The correct simplification involves dividing by ( 2t ) again, leading to (\frac{3}{4t}), not (\frac{3}{2t}).

• Option (d) $\frac{2}{3 t}$ is incorrect because it does not follow from the correct differentiation process. The correct second derivative involves the factor (\frac{3}{2}) and the correct simplification leads to (\frac{3}{4t}), not (\frac{2}{3t}).

Solution

Given that: $f(x)=x^{3}-3 x$ in $[0, \sqrt{3}]$

We know that if $f(x)=x^{3}-3 x$ satisfies the conditions of Rolle’s

Theorem in $[0, \sqrt{3}]$, then

$ \begin{matrix} f^{\prime}(c) =0 \\ \Rightarrow 3 c^{2}-3 =0 \Rightarrow 3 c^{2}=3 \Rightarrow c^{2}=1 \\ \therefore c = \pm 1 \Rightarrow 1 \in(0, \sqrt{3}) \end{matrix} $

Hence, the correct option is $(a)$.

• Option (b) -1: This value is incorrect because -1 is not within the interval [0, √3]. Rolle’s Theorem requires that the value of c must lie within the given interval.

• Option (c) $\frac{3}{2}$: This value is incorrect because when we solve for c in the equation $3c^2 = 3$, we get $c^2 = 1$, which gives $c = \pm 1$. $\frac{3}{2}$ does not satisfy this equation.

• Option (d) $\frac{1}{3}$: This value is incorrect because when we solve for c in the equation $3c^2 = 3$, we get $c^2 = 1$, which gives $c = \pm 1$. $\frac{1}{3}$ does not satisfy this equation.

Solution

Given that: $f(x)=x+\frac{1}{x}, x \in[1,3]$

We know that if $f(x)=x+\frac{1}{x}, x \in[1,3]$ satisfies all the conditions of mean value theorem then

$ \begin{aligned} & f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \text{ where } a=1 \text{ and } b=3 \\ \Rightarrow \quad & 1-\frac{1}{c^{2}}=\frac{(3+\frac{1}{3})-(1+\frac{1}{1})}{3-1} \\ \Rightarrow \quad & 1-\frac{1}{c^{2}}=\frac{\frac{10}{3}-2}{2} \Rightarrow 1-\frac{1}{c^{2}}=\frac{4}{6}=\frac{2}{3} \Rightarrow-\frac{1}{c^{2}}=\frac{2}{3}-1 \\ \Rightarrow \quad & -\frac{1}{c^{2}}=-\frac{1}{3} \Rightarrow \frac{1}{c^{2}}=\frac{1}{3} \Rightarrow c= \pm \sqrt{3} . \end{aligned} $

Here $c=\sqrt{3} \in(1,3)$.

Hence, the correct option is (b).

• Option (a) 1: This option is incorrect because when we solve for ( c ) using the mean value theorem, we find that ( c = \sqrt{3} ). The value ( c = 1 ) does not satisfy the equation derived from the mean value theorem for the given function and interval.

• Option (c) 2: This option is incorrect because when we solve for ( c ) using the mean value theorem, we find that ( c = \sqrt{3} ). The value ( c = 2 ) does not satisfy the equation derived from the mean value theorem for the given function and interval.

• Option (d) none of these: This option is incorrect because we have found that ( c = \sqrt{3} ) satisfies the mean value theorem for the given function and interval. Therefore, there is a correct option among the given choices, which is (b).

Solution

$|x|+|x-1|$ is the function which is continuous everywhere but fails to be differentiable at $x=0$ and $x=1$.

We can have more such examples.

Solution

Let $y=x^{2}$ and $t=x^{3}$

Differentiating both the parametric functions w.r.t. $x$

$ \frac{d y}{d x}=2 x \text{ and } \frac{d t}{d x}=3 x^{2} $

$ \therefore \quad \frac{d y}{d t}=\frac{d y / d x}{d t / d x}=\frac{2 x}{3 x^{2}}=\frac{2}{3 x} $

So, the derivative of $x^{2}$ w.r.t. $x^{3}$ is $\frac{2}{3 x}$

Solution

Given that: $f(x)=|\cos x|$

$ \Rightarrow \quad f(x)=\cos x \text{ if } x \in(0, \frac{\pi}{2}) $

Differentiating both sides w.r.t. $x$, we get $f^{\prime}(x)=-\sin x$

at $x=\frac{\pi}{4}, f^{\prime}(\frac{\pi}{4})=-\sin \frac{\pi}{4}=-\frac{1}{\sqrt{2}}$

Solution

Given that: $f(x)=|\cos x-\sin x|$

We know that $\sin x>\cos x$ if $x \in(\frac{\pi}{4}, \frac{\pi}{2})$

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

$ \begin{aligned} \therefore \quad f(x) & =-(\cos x-\sin x) \\ f^{\prime}(x) & =-(-\sin x-\cos x) \Rightarrow f^{\prime}(x)=(\sin x+\cos x) \\ \therefore \quad f^{\prime}(\frac{\pi}{3}) & =\sin \frac{\pi}{3}+\cos \frac{\pi}{3}=\frac{\sqrt{3}}{2}+\frac{1}{2}=\frac{\sqrt{3}+1}{2} \end{aligned} $

Solution

Given that: $\sqrt{x}+\sqrt{y}=1$

Differentiating both sides w.r.t. $x$

$\frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \cdot \frac{d y}{d x} =0$

$\Rightarrow \quad \frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}} \frac{d y}{d x} =0$

$\Rightarrow \quad \frac{1}{\sqrt{y}} \frac{d y}{d x} =\frac{-1}{\sqrt{x}} \Rightarrow \frac{d y}{d x}=\frac{-\sqrt{y}}{\sqrt{x}}$

$\therefore \quad \frac{d y}{d x} \text{ at }(\frac{1}{4}, \frac{1}{4}) =-\frac{\sqrt{\frac{1}{4}}}{\sqrt{\frac{1}{4}}}=-1$

Solution

False. Given that $f(x)=|x-1|$ in $[0,2]$

We know that modulus function is not differentiable. So, it is false.

Solution

True. We know that modulus function is continuous function on its domain. So, it is true.

Solution

True. We know that the sum and difference of two or more functions is always continuous. So, it is true.

Solution

True.

Solution

False. Let us take an example: $f(x)=\sin x$ and $g(x)=\cot x$ $\therefore f(x) \cdot g(x)=\sin x \cdot \cot x=\sin x \cdot \frac{\cos x}{\sin x}=\cos x$ which is continuous at $x=0$ but $\cot x$ is not continuous at $x=0$.