Discussion Forum

If $f(x)=\begin{cases}-x-\frac{\pi}{2}, & x \leq -\frac{\pi}{2}\\-\cos x,&-\frac{\pi}{2}< x \leq0 \\ x-1, & 0 <x \leq1\\\log x , & x>1\end{cases}$, then

Answer:

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= $-\frac{3}{2}$