Question

$f\left(x\right)=min\{\cos x,1-\sin x\},-\pi \le x\le \pi .$ Then which among the following options is/are correct

• A. $f\left(x\right)$ is not differentiable at $x=0$
• B. $f\left(x\right)$ is differentiable at $x=\frac{\pi }{2}$
• C. $f\left(x\right)$ is discontinuous at $x=0$
• D. $f\left(x\right)$ is continuous at $x=\frac{\pi }{2}$

Answer 1

Answer: (A), (D)

$f\left(x\right)$ is continuous at $x=\frac{\pi }{2}$

$f\left(x\right)=min\{\cos x,1-\sin x\},-\pi \le x\le \pi$



$\Rightarrow f\left(x\right)=$$\{\begin{array}{c}\cos x,-\pi \le x<0\\ 1-\sin x,0\le x<\frac{\pi }{2}\\ \cos x,\frac{\pi }{2}\le x\le \pi \end{array}$

Clearly, from the graph, $f\left(x\right)$ is continuous everywhere, but has sharp corners at the points where the definition changes i.e. at $x=0$ and $x=\frac{\pi }{2}$
$\therefore f\left(x\right)$ is not differentiable at $0$ and $\frac{\pi }{2}.$