Mathematics - Continuity - Differentiability Question with Solution | TestHub

$f(x)=\{\begin{array}{cc}-x-\frac{\pi }{2},& x\le -\frac{\pi }{2}\\ -\cos x,& -\frac{\pi }{2}<x\le 0\\ x-1,& 0<x\le 1\\ \log x,& x>1\end{array}$ Continuity at $x=-\frac{\pi }{2}$, $$\begin{array}{rl}& f\left(-\frac{\pi }{2}\right)=-\left(-\frac{\pi }{2}\right)-\frac{\pi }{2}=0\\ & \mathrm{RHL}\Rightarrow \underset{h\to 0}{\lim }-\cos \left(-\frac{\pi }{2}+h\right)=0\end{array}$$ $\therefore$ Continuous at $x=0$ Continuity at $x=0\Rightarrow f(0)=-1$ $\mathrm{RHL}\Rightarrow {\lim }_{h\to 0}(0+h)-1=-1$ $\therefore$ Continuous at $x=0$ Continuity at $x=1;f(1)=0$ $\mathrm{RHL}\Rightarrow {\lim }_{h\to 0}\log (1+h)=0$ $\therefore$ Continuous at $x=1$ Here, $f^{\prime }(x)=\{\begin{array}{cc}-1,& x\le -\frac{\pi }{2}\\ \sin x,& -\frac{\pi }{2}<x\le 0\\ 1,& 0<x\le 1\\ \frac{1}{x},& x>1\end{array}$ Differentiable at $x=0$, $\mathrm{LHD}=0,\mathrm{RHD}=1$ $\therefore$ Not differentiable at $x=0$ Differentiable at $x=1$, $\mathrm{LHD}=1,\mathrm{RHD}=1$ $\therefore$ Differentiable at $x=1$ Also, for $x=-\frac{3}{2}\Rightarrow f(x)=-x-\frac{3}{2}$ $\therefore$ Differentiable at $x=-\frac{3}{2}$