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).

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).

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 $(b)$.

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 $(a)$.

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 $(a)$.

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 $(a)$.

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).

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 $(b)$.

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 $(b)$.

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 $(a)$.

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).

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).

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 $(a)$.

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).

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 $x=0$.