Question

If $f(x)=x+|x|+\cos \left(\left[{\pi }^2\right]x\right)$ and $g(x)=\sin x$ (where [ $\cdot ]$ denotes the greatest integer function), then

• A. $f(x)+g(x)$ is discontinuous
• B. $f(x)+g(x)$ is differentiable everywhere
• C. $f(x)\times g(x)$ is differentiable everywhere
• D. $f(x)\times g(x)$ is continuous but not differentiable at $x=0$

Answer 1

Answer: (C)

$f(x)=x+|x|+\cos 9x,g(x)=\sin x$
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 $h(x)=f(x)\times g(x)$
$=\{\begin{array}{ll}(cos9x)(sinx),& \text{ x<0 }\\ (2x+cos9x).(sinx),& \text{ x≥0}\end{array}$
$\Rightarrow h^{\prime }(x)$
$=\{\begin{array}{ll}cosx.cos9x-9sinxsin9x,& \text{ x<0 }\\ (2-9sin9x)sinx+cosx(2x+cos9x),& \text{x≥0}\end{array}$
$h^{\prime }\left(0^{-}\right)=1,h^{\prime }\left(0^{+}\right)=1$
$\Rightarrow f(x)\times g(x)$ is differentiable everywhere