Question

If $f\left(x\right)=x+|x|+\cos \left(\right[{\pi }^2\left]x\right)$ and $g\left(x\right)=\sin x,$ then which of the following option is INCORRECT ?

(where [.] denotes the greatest integer function)

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

Answer 1

The correct option is D $f\left(x\right)\times g\left(x\right)$ is continous but not differentiable at $x=0$

$f\left(x\right)=x+|x|+\cos \left(\right[{\pi }^2\left]x\right),g\left(x\right)=\sin x$
$\Rightarrow f\left(x\right)=x+|x|+\cos \left(9x\right)(\because \pi =3.14)$
Since, both $f\left(x\right)$ and $g\left(x\right)$ are continuous everywhere, hence $f\left(x\right)+g\left(x\right)$ is also continuous everywhere.
Since, $f\left(x\right)$ is non-differentiable at $x=0.$
Hence, $f\left(x\right)+g\left(x\right)$ is also non-differentiable at $x=0.$

Now,
$h\left(x\right)=f\left(x\right)\times g\left(x\right)=\{\begin{array}{cc}\left(\cos 9x\right)\left(\sin x\right),& x<0\\ (2x+\cos 9x)\left(\sin x\right),& x\ge 0\end{array}$
Clearly, $h\left(x\right)$ is continuous at $x=0.$

Also,
$h^{\prime }\left(x\right)=\{\begin{array}{cc}\cos x\cos 9x-9\sin x\sin 9x,& x<0\\ (2-9\sin 9x)\left(\sin x\right)+\cos x(2x+\cos 9x),& x>0\end{array}$

$\Rightarrow h^{\prime }\left(0^{-}\right)=h^{\prime }\left(0^{+}\right)=1$
Hence, $f\left(x\right)\times g\left(x\right)$ is differentiable everywhere.