Answer:
Option A,B,C,D
Explanation:
$f(x)=\begin{cases}-x-\frac{\pi}{2}, & x \leq -\frac{\pi}{2}\\-\cos x,&-\frac{\pi}{2}< x \leq0 \\ x-1, & 0 <x \leq1\\\log x , & x>1\end{cases}$,
continuity at x=-$\frac{\pi}{2}$
$f\left(-\frac{\pi}{2}\right)=-\left(-\frac{\pi}{2}\right)-\frac{\pi}{2}=0$
RHL= $\Rightarrow$ $\lim_{h \rightarrow 0}-\cos \left(-\frac{\pi}{2}+h\right)=0$
$\therefore$ continuous at x=0
Continuity at x=0 $\Rightarrow$ f(0)=-1$
$RHL \Rightarrow \lim_{h \rightarrow 0}(0+h)-1=-1$
$\therefore$ continuous at x=0
Continuity at x=1;f(1)=0
RHL$\Rightarrow$ $\lim_{h \rightarrow 0}\log (1+h)=0$
$\therefore$ continuous at x=1
Here, $f(x)=\begin{cases}-1, & x \leq -\frac{\pi}{2}\\\sin x,&-\frac{\pi}{2}< x \leq0 \\ 1, & 0 <x \leq1\\\frac{1}{x} , & x>1\end{cases}$,
Differentiable at x=0
LHD=0, RHD=1
$\therefore$ not differentiable at x=0
Differentiable at x=1
LHD=1,RHD=1
$\therefore$ Differentiable at x=1
also, for x$=-\frac{3}{2} \Rightarrow f(x)=-x-\frac{3}{2}$
$\therefore$ differentiable at x= $-\frac{3}{2}$
