The set of points where

We know

$|x|= \begin{cases}x, & x \geq 0 \\ -x, & x<0\end{cases}$

$\therefore f(x)=\cos |x|= \begin{cases}\cos x, & x \geq 0 \\ \cos (-x), & x<0\end{cases}$

$\Rightarrow f(x)=\cos |x|= \begin{cases}\cos x, & x \geq 0 \\ \cos x, & x<0\end{cases}$     $[\cos (-\theta)=\cos \theta]$

We know that, cosine function is differentiable in its domain. So, f(x) is differentiable for all x < 0 and x > 0.

Let us check the differentiability of $f(x)=\cos |x|$ at $x=0$.

Now,

$L f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}$

$\Rightarrow L f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{\cos (-h)-\cos 0}{-h}$

$\Rightarrow L f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{\cos h-1}{-h}$

$\Rightarrow L f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{-2 \sin ^{2} \frac{h}{2}}{-h}$

$\Rightarrow L f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \times \lim _{h \rightarrow 0} \sin \frac{h}{2}$

$\Rightarrow L f^{\prime}(0)=1 \times 0$

$\Rightarrow L f^{\prime}(0)=0$

And

$R f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$

$\Rightarrow R f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{\cos h-\cos 0}{h}$

$\Rightarrow R f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{\cos h-1}{h}$

$\Rightarrow R f^{\prime}(0)=\lim _{h \rightarrow 0} \frac{-2 \sin ^{2} \frac{h}{2}}{h}$

$\Rightarrow R f^{\prime}(0)=-\lim _{h \rightarrow 0} \frac{\sin \frac{h}{2}}{\frac{h}{2}} \times \lim _{h \rightarrow 0} \sin \frac{h}{2}$

$\Rightarrow R f^{\prime}(0)=-1 \times 0$

$\Rightarrow R f^{\prime}(0)=0$

$\therefore L f^{\prime}(0)=R f^{\prime}(0)$

So, f(x) is differentiable at x = 0. Thus, the function f(x) is differentiable everywhere.

Hence, the set of points where $f(x)=\cos |x|$ is differentiable is $\mathrm{R}$ (set of real real numbers).

The set of points where f(x) = cos |x| is differentiable, is _____R_____.