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Q.
If $f(x)=x+|x|+\cos \left(\left[\pi^{2}\right] x\right)$ and $g(x)=\sin x$ (where [ $\left.\cdot\right]$ denotes the greatest integer function), then
Continuity and Differentiability
Solution:
$f(x)=x+|x|+\cos 9 x, g(x)=\sin x$
Since $f(x)$ and $g(x)$ both are continuous everywhere, $f(x)+g(x)$ is also continuous everywhere $f(x)$ is non-differentiable and $x=0$.
Hence, $f(x)+g(x)$ is non-differentiable at $x=0$
Now $h(x)=f(x) \times g(x)$
$ =
\begin{cases}
(cos\,9x)(sin\,x), & \text{ $x<\,0$ } \\[2ex]
(2x+cos 9x). (sin\,x), & \text{ $x \geq 0$}
\end{cases}$
$\Rightarrow h'(x)$
$ =
\begin{cases}
cos\,x. cos\,9x-9 sin \,x\,sin\,9x, & \text{ $x<\,0$ } \\[2ex]
(2-9\,sin\,9x)sin\,x+cos\,x(2x+cos\,9x), & \text{$x \geq 0$}
\end{cases}$
$h'\left(0^{-}\right)=1, h'\left(0^{+}\right)=1 $
$\Rightarrow f(x) \times g(x)$ is differentiable everywhere