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 $x=5$.

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

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

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

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

We have, $f(x)=\left\{\begin{array}{cll}\frac{2^{x+2}-16}{4^x-16}, & \text { if } & x \neq 2 \\ k, & \text { if } & x=2\end{array}\right.$ at $x=2$.

$$ \begin{aligned} &\text { Since, } f(x) \text { is continuous at } x=2 \text {. }\\ &\begin{aligned} & \therefore \quad \mathrm{LHL=RHL=}f(2)\\ & \text { At } x=2, \quad \lim _{x \rightarrow 2} \frac{2^x \cdot 2^2-2^4}{4^x-4^2}=\lim _{x \rightarrow 2} \frac{4 \cdot\left(2^x-4\right)}{\left(2^x\right)^2-(4)^2} \end{aligned} \end{aligned}$$

$$\begin{aligned} & =\lim _{x \rightarrow 2} \frac{4 \cdot\left(2^x-4\right)}{\left(2^x-4\right)\left(2^x+4\right)} \quad\left[\because a^2-b^2=(a+b)(a-b)\right] \\ & =\lim _{x \rightarrow 2} \frac{4}{2^x+4}=\frac{4}{8}=\frac{1}{2} \end{aligned}$$

$$\begin{aligned} \text{But}\quad f(2) & =k \\ \therefore\quad k & =\frac{1}{2} \end{aligned}$$

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

$$\begin{aligned} \therefore\quad \mathrm{LHL} & =\lim _{x \rightarrow 0^{-}} \frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} \\ & =\lim _{x \rightarrow 0^{-}}\left(\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x}\right) \cdot\left(\frac{\sqrt{1+k x}+\sqrt{1-k x}}{\sqrt{1+k x}+\sqrt{1-k x}}\right) \\ & =\lim _{x \rightarrow 0^{-}} \frac{1+k x-1+k x}{x[\sqrt{1+k x}+\sqrt{1-k x}]} \\ & =\lim _{x \rightarrow 0^{-}} \frac{2 k x}{x \sqrt{1+k x}+\sqrt{1-k x}} \\ & =\lim _{h \rightarrow 0} \frac{2 k}{\sqrt{1+k(0-h)+\sqrt{1-k(0-h)}}} \\ & =\lim _{h \rightarrow 0} \frac{2 k}{\sqrt{1-k h}+\sqrt{1+k h}}=\frac{2 k}{2}=k \\ \text{and}\quad f(0) & =\frac{2 \times 0+1}{0-1}=-1 \\ \Rightarrow\quad k & =-1 \quad \quad[\because \mathrm{LHL}=\mathrm{RHL}=f(0)] \end{aligned}$$