Assertion: The function f (x) = |cos x| is continuous function.
Reason: The function f (x) = cos|x| is a continuous function.

Assertion We have, f(x)=|cos x|
$=\left\{\begin{array}{cl}\mathrm{cosx},& \mathrm{x}\ne 0\\ 1,& \mathrm{x}=0\end{array}\right.$
Continuity at x=0,
$\mathrm{LHL} =\underset{\mathrm{h}\to 0}{lim} \mathrm{f}(0-\mathrm{h})=\underset{\mathrm{h}\to 0}{lim} \cos (0-\mathrm{h})=\cos 0=1$
$\mathrm{RHL}=\underset{\mathrm{h}\to 0}{lim} \mathrm{f}(0+\mathrm{h})=\underset{\mathrm{h}\to 0}{lim} \cos (0+\mathrm{h})$
$=\underset{\mathrm{h}\to 0}{lim} \cosh =\cos 0=1$
and f(0)=1
$\therefore \mathrm{LHL}=\mathrm{RHL}=\mathrm{f}\left(0\right)$
So, f(x) is continuous at x=0.
Hence, f(x) is continuous everywhere.

Reason We have, $\mathrm{f}\left(\mathrm{x}\right)=\cos \left|\mathrm{x}\right|$
$\begin{array}{l}=\left\{\begin{array}{lc}\cos \mathrm{x},& \mathrm{x}\ge 0\\ \cos (-\mathrm{x}),& \mathrm{x}<0\end{array}\right.\\ =\left\{\begin{array}{ll}\cos \mathrm{x},& \mathrm{x}\ge 0\\ \cos \mathrm{x},& \mathrm{x}<0\end{array}\right.\\ =\cos \mathrm{x},\mathrm{x}\in \mathrm{R}\end{array}$
But cos x is always continuous in their domain. Hence, f ( x ) is continuous everywhere. Hence, both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.