The number of critical points of $f\left(x\right)=\max \{\sin x,\cos x\},\forall x\in (-2\pi ,2\pi )$ is

Given : $\mathrm{f}\left(\mathrm{x}\right)=\max \{\mathrm{sinx},\mathrm{cosx}\},\forall \mathrm{x}\in (-2\mathrm{\pi },2\mathrm{\pi })$

Critical points are the points where either $\mathrm{f}\text{'}\left(\mathrm{x}\right)=0$ or $f\text{'}\left(x\right)$ is not defined or $f\left(x\right)$ is non-differentiable.

The graph of $\mathrm{f}\left(\mathrm{x}\right)=\max \{\mathrm{sinx},\mathrm{cosx}\}$ is,

As we know that a function $f\left(x\right)$ is non-differentiable at points that have sharp edge points in graph.

The points where the function $\mathrm{f}\left(\mathrm{x}\right)=\max \{\mathrm{sinx},\mathrm{cosx}\}$ has sharp points in the graph are $\mathrm{x}=\frac{\mathrm{\pi }}{4},\frac{5\mathrm{\pi }}{4},-\frac{3\mathrm{\pi }}{4},-\frac{7\mathrm{\pi }}{4}$.

Thus, $\mathrm{f}\left(\mathrm{x}\right)=\max \{\mathrm{sinx},\mathrm{cosx}\}$ is non-differentiable at points $\mathrm{x}=\frac{\mathrm{\pi }}{4},\frac{5\mathrm{\pi }}{4},-\frac{3\mathrm{\pi }}{4},-\frac{7\mathrm{\pi }}{4}$

Also, $f^{\text{'}}\left(x\right)=0$ at $x=-\frac{3\pi }{4},0,\frac{\pi }{2}$

Hence, there are $7$ critical points in $(-2\pi ,2\pi )$ that are $\mathrm{x}=\frac{\mathrm{\pi }}{4},\frac{5\mathrm{\pi }}{4},-\frac{3\mathrm{\pi }}{4},-\frac{7\mathrm{\pi }}{4},0,\frac{\mathrm{\pi }}{2},-\frac{3\mathrm{\pi }}{2}$

Hence option-$3$ is the correct answer.