$\left(\mathrm{a}\right)$ $y=\left|x\right|+\left|x-2\right|$ at $x=2,\frac{dy}{dx}$ is

at $x=2$

$y=\left|2\right|+0$

$\Rightarrow \frac{dy}{dx}=\left|2\right|$ [Does not exist. $\left(\mathrm{iv}\right)$]

$\left(\mathrm{b}\right)$ $f\left(x\right)=\left|\cos 2x\right|$ then  is,

using first principle,

$f^{\text{'}}\left({\frac{\pi }{4}}^{+}\right)=\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{f\left(x\right)-f\left(\frac{\mathrm{\pi }}{4}\right)}{x-\frac{\pi }{4}}$

$\Rightarrow f^{\text{'}}\left({\frac{\pi }{4}}^{+}\right)=\underset{x\to \frac{\mathrm{\pi }}{4}}{\lim }\frac{\cos 2x-0}{x-{\displaystyle \frac{\pi }{4}}}$

Applying L'Hospital Rule, we get

$f^{\text{'}}\left({\frac{\pi }{4}}^{+}\right)=\frac{2\sin 2x}{1}$

$\Rightarrow f^{\text{'}}\left({\frac{\pi }{4}}^{+}\right)=2$

[option $\left(\mathrm{i}\right)$]

$\left(\mathrm{c}\right)$ $f\left(x\right)=\sin \pi \left|x\right|$ then $f^1\left(1-\right)$ is,

using first principle,

$f^{\text{'}}\left(1^{-}\right)=\underset{x\to 1^{-}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1}$

$\Rightarrow f^{\text{'}}\left(1^{-}\right)=\underset{x\to 1^{-}}{\lim }\frac{\sin \left(0\right)-\sin \left(\pi \right)}{x-1}$

$\Rightarrow f^{\text{'}}\left(1^{-}\right)=0$    [option$\left(\mathrm{ii}\right)$]

$\left(\mathrm{iv}\right)$ $f\left(x\right)=\log \left|x-1\right|, x\ne 1,$

$f^{\text{'}}\left(x\right)=\frac{1}{x-1}$

$f^{\text{'}}\left(\frac{1}{2}\right)=\frac{1}{\frac{1}{2}-1}=\frac{1}{\frac{-1}{2}}=-2$
[Option $\left(\mathrm{iii}\right)$]