Mathematics - Indefinite Integration Question with Solution | TestHub

$f^{\prime }(x)=3x^2\sin \frac{1}{x}-x\cos \frac{1}{x}$

$$\begin{array}{rl}& \Rightarrow f(x)=\int \left(3x^2\sin \frac{1}{x}-x\cos \frac{1}{x}\right)dx=x^3\sin \frac{1}{x}-\int \cos \frac{1}{x}\left(-\frac{1}{x^2}\right)x^{3}dx\\ & \\ & \end{array}$$

$$-\int x\cos \frac{1}{x}dx=x^3\sin \frac{1}{x}+C$$

$$sincef\left(\frac{1}{\pi }\right)=0+C\Rightarrow C=0\Rightarrow f(x)=\{\begin{array}{ll}{x}^3\sin \frac{1}{x},& x\ne 0\\ 0,& x=0\end{array}$$

$f(x)$ is clearly continuous and differentiable at $x=0$ with $f^{\prime }(0)=0$.

$$f^{\prime }(0)=\underset{h\to 0}{\lim }\frac{3h^2\sin \frac{1}{h}-h\cos \frac{1}{h}}{h}=3h\sin \frac{1}{h}-\cos \frac{1}{h}$$

This limit does not exist, hence $f^{\prime }(x)$ is non-differentiable at $x=0$.

Also ${\lim }_{x\to 0}{f}^{\prime }(x)=0$. Thus $f^{\prime }(x)$ is continuous at $x=0$.