Let f(x)=min 1, cos x, 1-sin x,-π ≤ x ≤ π. Then, f(x), is

We have,

$f(x)=\min \{1, \cos x, 1-\sin x\}$

$\Rightarrow f(x)= \begin{cases}\cos x, & -\pi / 2 \leq x \leq 0 \\ 1-\sin x, & 0<x \leq \pi / 2 \\ \cos x, & \pi / 2<x \leq \pi\end{cases}$

We find that

$\lim\limits_{x \rightarrow 0^{-}} f(x)=\lim\limits_{x \rightarrow 0^{-}} \cos x=1$

$\lim\limits_{x \rightarrow 0^{+}} f(x)=\lim\limits_{x \rightarrow 0} 1-\sin x=1$

and, $f(0)=\cos 0=1$

∴  $\lim\limits_{x \rightarrow 0^{-}} f(x)=\lim\limits_{x \rightarrow 0^{+}} f(x)=f(0)$

So, f(x) is continuous at x = 0.

Now,

(LHD at x = 0) = $\left(\frac{d}{d x}(\cos x)\right)_{x=0}=0$

(RHD at x = 0) = $\left(\frac{d}{d x}(1-\sin x)\right)_{x=0}=-1$

∴  (LHD at x = 0) ≠ (RHD at x = 0)

Hence, f(x) is not differentiable at x = 0. However, it is continuous there at.

At $x=\pi / 2$ also, f(x) is continuous but not differentiable.