We have, $\quad f(x)=\left\{\begin{array}{ll}\frac{1-\cos k x}{x \sin x}, & \text { if } x \neq 0 \\ \frac{1}{2}, & \text { if } x=0\end{array}\right.$ at $x=0$

At $x=0, \quad \mathrm{LHL}=\lim _\limits{x \rightarrow 0^{-}} \frac{1-\cos k x}{x \sin x}=\lim _\limits{h \rightarrow 0} \frac{1-\cos k(0-h)}{(0-h) \sin (0-h)}$

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

$$ \begin{array}{rlrl} & =\lim _\limits{h \rightarrow 0} \frac{1-1+2 \sin ^2 \frac{k h}{2}}{h \sin h} & {\left[\because \cos \theta=1-2 \sin ^2 \frac{\theta}{2}\right]} \\ & =\lim _\limits{h \rightarrow 0} \frac{2 \sin ^2 \frac{k h}{2}}{h \sin h} \\ & =\lim _\limits{h \rightarrow 0} \frac{2 \sin \frac{k h}{2}}{\frac{k h}{2}} \cdot \frac{\sin \frac{k h}{2}}{\frac{k h}{2}} \cdot \frac{1}{\frac{\sin h}{h}} \cdot \frac{k^2 h / 4}{h} \\ & =\frac{2 k^2}{4}=\frac{k^2}{2} \quad {\left[\because \lim _\limits{h \rightarrow 0} \frac{\sin h}{h}=1\right]} \\ \text{Also,}\quad f(0) & =\frac{1}{2} \Rightarrow \frac{k^2}{2}=\frac{1}{2} \Rightarrow k= \pm 1 \quad p \end{array}$$

We have, $\quad f(x)= \begin{cases}\frac{x}{|x|+2 x^2}, & \text { if } x \neq 0 \\ k, & \text { if } x=0\end{cases}$

At $x=0$, $$\mathrm{LHL}=\lim _\limits{x \rightarrow 0^{-}} \frac{x}{|x|+2 x^2}=\lim _\limits{h \rightarrow 0} \frac{(0-h)}{|0-h|+2(0-h)^2}$$

$$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{-h}{h+2 h^2}=\lim _{h \rightarrow 0} \frac{-h}{h(1+2 h)}=-1 \\ \mathrm{RHL} & =\lim _{x \rightarrow 0^{+}} \frac{x}{|x|+2 x^2}=\lim _{h \rightarrow 0} \frac{0+h}{|0+h|+2(0+h)^2} \\ & =\lim _{h \rightarrow 0} \frac{h}{h+2 h^2}=\lim _{h \rightarrow 0} \frac{h}{h(1+2 h)}=1 \end{aligned}$$

and $\quad f(0)=k$

Since, $\quad \mathrm{LHL} \neq \mathrm{RHL}$ for any value of $k$.

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

We have, $f(x)=\left\{\begin{array}{l}\frac{x-4}{|x-4|}+a, \text { if } x<4 \\ a+b, \quad \text { if } x=4 \\ \frac{x-4}{|x-4|}+b, \text { if } x>4\end{array}\right.$

$$\begin{aligned} &\text { At } x=4 \text {, }\\ &\begin{aligned} \mathrm{LHL} & =\lim _{x \rightarrow 4^{-}} \frac{x-4}{|x-4|}+a \\ & =\lim _{h \rightarrow 0} \frac{4-h-4}{|4-h-4|}+a=\lim _{h \rightarrow 0} \frac{-h}{h}+a \\ & =-1+a \end{aligned} \end{aligned}$$

$$\begin{aligned} \mathrm{RHL} & =\lim _{x \rightarrow 4^{+}} \frac{x-4}{|x-4|}+b \\ & =\lim _{h \rightarrow 0} \frac{4+h-4}{|4+h-4|}+b=\lim _{h \rightarrow 0} \frac{h}{h}+b=1+b \\ f(4) & =a+b \Rightarrow-1+a=1+b=a+b \\ \Rightarrow \quad-1+a & =a+b \text { and } 1+b=a+b \\ \therefore \quad b & =-1 \text { and } a=1 \end{aligned}$$

We have,

$$\begin{aligned} f(x) & =\frac{1}{x+2} \\ \therefore\quad y & =f\{f(x)\} \\ & =f\left(\frac{1}{x+2}\right)=\frac{1}{\frac{1}{x+2}+2} \\ & =\frac{1}{1+2 x+4} \cdot(x+2)=\frac{(x+2)}{(2 x+5)} \end{aligned}$$

So, the function $y$ will not be continuous at those points, where it is not defined as it is a rational function.

Therefore, $y=\frac{x+2}{(2 x+5)}$ is not defined, when $2 x+5=0$

$$\therefore \quad x=\frac{-5}{2}$$

Hence, $y$ is discontinuous at $x=\frac{-5}{2}$.

$$\begin{aligned} \text{We have,}\quad f(t) & =\frac{1}{t^2+t-2} \text { and } t=\frac{1}{x-1} \\ \therefore\quad f(t) & =\frac{1}{\left(\frac{1}{x^2+1-2 x}\right)+\left(\frac{1}{x-1}\right)-\frac{2}{1}} \\ & =\frac{1}{\left(\frac{1+x-1+\left[-2(x-1)^2\right]}{\left(x^2+1-2 x\right)}\right)} \\ & =\frac{x^2+1-2 x}{x-2 x^2-2+4 x} \\ & =\frac{x^2+1-2 x}{-2 x^2+5 x-2} \\ & =\frac{(x-1)^2}{-\left(2 x^2-5 x+2\right)} \\ & =\frac{(x-1)^2}{(2 x-1)(2-x)} \end{aligned}$$

So, $f(t)$ is discontinuous at $2 x-1=0 \Rightarrow x=1 / 2$

and $$2-x=0 \Rightarrow x=2$$