Question

Let $f$ be defined on $R$ by $f\left(x\right)=\{\begin{array}{c}{x}^4\sin \left(1/x\right);x\ne 0\\ 0;x=0\end{array}$, then

• A. $f^{\prime }\left(0\right)$ doesn't exist
• B. $f^{\prime \prime }\left(0\right)$ doesn't exist
• C. $f^{\prime \prime }\left(x\right)$ is continuous at $x=0$
• D. $f^{\prime \prime }\left(0\right)$ does not exist but $f^{\prime \prime }\left(x\right)$ is not continuous at $x=0$

Answer 1

The correct option is D $f^{\prime \prime }\left(0\right)$ does not exist but $f^{\prime \prime }\left(x\right)$ is not continuous at $x=0$

$f\left(x\right)=\{\begin{array}{c}{x}^4\sin \frac{1}{x}x\ne 0\\ 0x=0\end{array}$
For $x\ne 0$
$f^{\prime }\left(x\right)={3x}^2\sin \frac{1}{x}+x^4\cos \left(\frac{1}{x}\right)(-\frac{1}{x^2})$
$={3x}^2\sin \frac{1}{x}-x^2\cos \frac{1}{x}$
Now
$\frac{f\left(x\right)-f\left(0\right)}{x-0}=\frac{x^4\sin \frac{1}{x}}{x}=x^3\sin \frac{1}{x}\Rightarrow f^{\prime }\left(0\right)=0$
$f^{\prime \prime }\left(x\right)={9x}^2\sin \frac{1}{x}+{3x}^2\cos \frac{1}{x}\left(\frac{-1}{x^2}\right)$
$+2x\cos \frac{1}{x}-3\cos \frac{1}{x}+2x\cos \frac{1}{x}+\sin \frac{1}{x}$
$\Rightarrow f^{\prime \prime }\left(0\right)$
does not exits
And hence not continuous at $x=0$