Question

$f\left(x\right)=\left\{\begin{array}{cc}-2\sin x,& -\pi \le x\le \frac{-\pi }{2}\\ a\cos \left(x\right)+b,& \frac{-\pi }{2}<x<0\\ -2+\cos x,& 0\le x\le \frac{\pi }{2}\\ c\sin x,& \frac{\pi }{2}<x\le \pi \end{array}\right\}$ is continuous in $[-\pi ,\pi ],$ then number of points at which $f\left(x\right)$ is not differentiable is/are

• A. $x=\pi /2$
• B. $x=-\pi /2$
• C. $x=\pi$
• D. none of the above
  

Answer 1

The correct option is A $x=\pi /2$

$f^1\left(x\right)$ $=\{-2cosx,-\pi \le x\le -\pi /2-asinx,-\pi /2<x<0-sinx,0\le x\le \pi /2Ccosx,\pi /2<x\le \pi \}$

LHD$\to$ $f^1\left({-\pi /2}^{-}\right)=-2\cos (-\pi /2)=0$ , RHD, $f^1\left({-\pi /2}^{+}\right)=-asin(-\pi /2)=a$

at $x=-\pi /2$ at $x=\pi /2$

$\Rightarrow a=0$

RHD $f^1\left({\pi /2}^{+}\right)=C\cos \left(\pi /2\right)=0$ , LHD, $f^1\left({\pi /2}^{-}\right)=-\sin \pi /2=-1$

at $x=\pi /2$ at $x=\pi /2$

LHD $\ne$ RHD (Not diff.)