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

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

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

We have, $$f(x)=|\sin x+\cos x| \text { at } x=\pi$$

Let $g(x)=\sin x+\cos x$

and $h(x)=|x|$

$$\begin{aligned} \therefore\quad \operatorname{hog}(x) & =h[g(x)] \\ & =h(\sin x+\cos x) \\ & =|\sin x+\cos x| \end{aligned}$$

Since, $g(x)=\sin x+\cos x$ is a continuous function as it is forming with addition of two continuous functions $\sin x$ and $\cos x$.

Also, $h(x)=|x|$ is also a continuous function. Since, we know that composite functions of two continuous functions is also a continuous function.

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

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

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

At $x=2$,

$$L f^{\prime}(2)=\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h}$$

$$=\lim _\limits{h \rightarrow 0} \frac{(2-h)[2-h]-(2-1) 2}{-h}$$

$\{\because[a-h]=[a-1]$, where $a$ is any positive number $\}$

$$=\lim _\limits{h \rightarrow 0} \frac{(2-h)(1)-2}{-h}$$

$$=\lim _\limits{h \rightarrow 0} \frac{2-h-2}{-h}=\lim _\limits{h \rightarrow 0} \frac{-h}{-h}=1$$

$$R f^{\prime}(2)=\lim _\limits{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}$$

$$=\lim _\limits{h \rightarrow 0} \frac{(2+h-1)(2+h)-(2-1) \cdot 2}{h}$$

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

$$=\lim _\limits{h \rightarrow 0} \frac{2+h+2 h+h^2-2}{h}$$

$$=\lim _\limits{h \rightarrow 0} \frac{h^2+3 h}{h}=\lim _\limits{h \rightarrow 0} \frac{h(h+3)}{h}=3$$

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

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

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

$$\begin{aligned} &\text { For differentiability at } x=0 \text {, }\\ &L f^{\prime}(0)=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0^{-}} \frac{x^2 \sin \frac{1}{x}-0}{x-0} \end{aligned}$$

$$\begin{aligned} & =\lim _{h \rightarrow 0} \frac{(0-h)^2 \sin \left(\frac{1}{0-h}\right)}{0-h}=\lim _{h \rightarrow 0} \frac{h^2 \sin \left(\frac{-1}{h}\right)}{-h} \\ & =\lim _{h \rightarrow 0}+h \sin \left(\frac{1}{h}\right) \quad[\because \sin (-\theta)=-\sin \theta] \\ & =0 \times[\text { an oscillating number between }-1 \text { and } 1]=0 \\ R f^{\prime}(0) & =\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}=\lim _{x \rightarrow 0^{+}} \frac{x^2 \sin \frac{1}{x}-0}{x-0} \\ & =\lim _{h \rightarrow 0} \frac{(0+h)^2 \sin \left(\frac{1}{0+h}\right)}{0+h}=\lim _{h \rightarrow 0} \frac{h^2 \sin (1 / h)}{h} \\ & =\lim _{h \rightarrow 0} h \sin (1 / h) \quad \\ & =0 \times[\text { an oscillating number between }-1 \text { and } 1]=0 \end{aligned}$$

$$\begin{aligned} &\because \quad L f^{\prime}(0)=R f^{\prime}(0)\\ &\text { So, } f(x) \text { is differentiable at } x=0 \text {. } \end{aligned}$$

We have, $f(x)=\left\{\begin{array}{l}1+x, \text { if } x \leq 2 \\ 5-x, \text { if } x>2\end{array}\right.$ at $x=2$.

For differentiability at $x=2$,

$$ \begin{aligned} L f^{\prime}(2) & =\lim _{x \rightarrow 2^{-}} \frac{f(x)-f(2)}{x-2}=\lim _{x \rightarrow 2^{-}} \frac{(1+x)-(1+2)}{x-2} \\ & =\lim _{h \rightarrow 0} \frac{(1+2-h)-3}{2-h-2}=\lim _{h \rightarrow 0} \frac{-h}{-h}=1 \\ R f^{\prime}(2) & =\lim _{x \rightarrow 2^{+}} \frac{f(x)-f(2)}{x-2}=\lim _{x \rightarrow 2^{+}} \frac{(5-x)-3}{x-2} \\ & =\lim _{h \rightarrow 0} \frac{5-(2+h)-3}{2+h-2} \\ & =\lim _{h \rightarrow 0} \frac{5-2-h-3}{h}=\lim _{h \rightarrow 0} \frac{-h}{+h} \\ & =-1 \\ \because\quad L f^{\prime}(2) & \neq R f^{\prime}(2) \end{aligned}$$

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