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

At $x=a$,

$$\begin{aligned} \mathrm{LHL} & =\lim _{x \rightarrow a^{-}}|x-a| \sin \frac{1}{x-a} \\ & =\lim _{h \rightarrow 0}|a-h-a| \sin \left(\frac{1}{a-h-a}\right) \\ & =\lim _{h \rightarrow 0}-h \sin \left(\frac{1}{h}\right) \quad[\because \sin (-\theta)=-\sin \theta] \\ & =0 \times[\text { an oscillating number between }-1 \text { and } 1]=0 \\ \mathrm{RHL} & =\lim _{x \rightarrow a^{+}}|x-a| \sin \left(\frac{1}{x-a}\right) \\ & =\lim _{h \rightarrow 0}|a+h-a| \sin \left(\frac{1}{a+h-a}\right)=\lim _{h \rightarrow 0} h \sin \frac{1}{h} \\ & =0 \times[\text { an oscillating number between }-1 \text { and } 1]=0 \\ \end{aligned}$$

and $f(a)=0$

$$\therefore$$ $\mathrm{LHL = RHL}=f(a)$

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

We have, $f(x)=\left\{\begin{array}{ll}\frac{e^{1 / x}}{1+e^{1 / x}}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0\end{array}\right.$ at $x=0$.

$$\begin{aligned} &\text { At } x=0 \text {, }\\ &\begin{aligned} \mathrm{LHL} & =\lim _{x \rightarrow 0^{-}} \frac{e^{1 / x}}{1+e^{1 / x}}=\lim _{h \rightarrow 0} \frac{e^{1 / 0-h}}{1+e^{1 / 0-h}} \\ & =\lim _{h \rightarrow 0} \frac{e^{-1 / h}}{1+e^{-1 / h}}=\lim _{h \rightarrow 0} \frac{1}{e^{1 / h}\left(1+e^{-1 / h}\right)} \\ & =\lim _{h \rightarrow 0} \frac{1}{e^{1 / h}+1}=\frac{1}{e^{\infty}+1}=\frac{1}{\infty+1} \quad \quad\left[\because e^{\infty}=\infty\right] \\ & =\frac{1}{\frac{1}{0}}=0 \\ \mathrm{RHL} & =\lim _{x \rightarrow 0^{+}} \frac{e^{1 / x}}{1+e^{1 / x}} \\ & =\lim _{h \rightarrow 0} \frac{e^{1 / 0+h}}{1+e^{1 / 0+h}}=\lim _{h \rightarrow 0} \frac{e^{1 / h}}{1+e^{1 / h}} \end{aligned} \end{aligned}$$

$$\begin{aligned} &\begin{aligned} & =\lim _{h \rightarrow 0} \frac{1}{e^{-1 / h}+1}=\frac{1}{e^{-\infty}+1} \\ & =\frac{1}{0+1}=1 \end{aligned}\\ &\left[\because e^{-\infty}=0\right] \end{aligned}$$

Hence, $\quad \mathrm{LHL} \neq \mathrm{RHL}$ at $x=0$.

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

We have, $f(x)=\left\{\begin{array}{ll}\frac{x^2}{2}, & \text { if } 0 \leq x \leq 1 \\ 2 x^2-3 x+\frac{3}{2}, & \text { if } 1< x \leq 2\end{array}\right.$ at $x=1$.

At $x=1$,

$$\begin{aligned} \mathrm{HL} & =\lim _{x \rightarrow 1^{-}} \frac{x^2}{2}=\lim _{h \rightarrow 0} \frac{(1-h)^2}{2} \\ & =\lim _{h \rightarrow 0} \frac{1+h^2-2 h}{2}=\frac{1}{2} \\ \mathrm{RHL} & =\lim _{x \rightarrow 1^{+}}\left(2 x^2-3 x+\frac{3}{2}\right) \\ & =\lim _{h \rightarrow 0}\left[2(1+h)^2-3(1+h)+\frac{3}{2}\right] \\ & =\lim _{h \rightarrow 0}\left(2+2 h^2+4 h-3-3 h+\frac{3}{2}\right)=-1+\frac{3}{2}=\frac{1}{2} \end{aligned}$$

and $$f(1)=\frac{t^2}{2}=\frac{1}{2}$$

$$\therefore \quad \mathrm{LHL}=\mathrm{RHL}=f(1)$$

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

We have,

$$\begin{aligned} f(x) & =|x|+|x-1| \text { at } x=1 \\ \text{At }x=1,\quad \mathrm{LHL} & =\lim _{x \rightarrow 1^{-}}[|x|+|x-1|] \\ & =\lim _{h \rightarrow 0}[|1-h|+|1-h-1|]=1+0=1 \end{aligned}$$

$$\begin{aligned} \text{and}\quad \mathrm{RHL} & =\lim _{x \rightarrow 1^{-}}[|x|+|x-1|] \\ & =\lim _{h \rightarrow 0}[|1+h|+|1+h-1|]=1+0=1 \end{aligned}$$

and $$f(1)=|1|+|0|=1$$

$$\therefore \quad \mathrm{LHL}=\mathrm{RHL}=f(1)$$

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

We have, $f(x)=\left\{\begin{aligned} 3 x-8, & \text { if } x \leq 5 \\ 2 k, & \text { if } x>5\end{aligned}\right.$ at $x=5$.

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

$$\therefore \quad \mathrm{LHL}=\mathrm{RHL}=f(5)$$

$$\begin{aligned} &\text { Now, }\\ &\begin{aligned} \mathrm{LHL} & =\lim _{x \rightarrow 5^{-}}(3 x-8)=\lim _{h \rightarrow 0}[3(5-h)-8] \\ & =\lim _{h \rightarrow 0}[15-3 h-8]=7 \end{aligned} \end{aligned}$$

$\mathrm{RHL}=\lim _\limits{x \rightarrow 5^{+}} 2 k=\lim _\limits{h \rightarrow 0} 2 k=2 k=7 \quad[\because \mathrm{LHL}=\mathrm{RHL}]$

$$\begin{aligned} &\begin{aligned} \text { and }\quad f(5) & =3 \times 5-8=7 \\ \therefore\quad 2 k & =7 \quad \Rightarrow \quad k=\frac{7}{2} \end{aligned} \end{aligned}$$