Question

Let $f\left(x\right)=3x^2.\sin \frac{1}{x}-x\cos \frac{1}{x},x\ne 0,f\left(0\right)=0f\left(\frac{1}{\pi }\right)=0$, then which of the following is/are not correct.

• A. $f\left(x\right)$ is continuous at $x=0$
• B. $f\left(x\right)$ is non-differentiable at $x=0$
• C. $f\left(x\right)$ is discontinuous at $x=0$
• D. $f\left(x\right)$ is differentiable at $x=0$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $f\left(x\right)$ is non-differentiable at $x=0$
C $f\left(x\right)$ is differentiable at $x=0$
D $f\left(x\right)$ is discontinuous at $x=0$

$f\left(x\right)=3x^2.\sin \frac{1}{x}-x.\cos \frac{1}{x}$
$\Rightarrow f\left(x\right)=\int (3x^2.\sin \frac{1}{x}-\cos \frac{1}{x})dx$
$=x^3\sin \frac{1}{x}.-\int \cos \frac{1}{x}(-\frac{1}{x^2})x^{3}dx-\int x\cos \frac{1}{x}dx$
$=x^3\sin \frac{1}{x}+C$
Since $f\left(\frac{1}{\pi }\right)=0+C\Rightarrow C=0$
$\Rightarrow f\left(x\right)\left\{\begin{array}{ccc}{x}^3\sin \frac{a}{x}& ,& x\ne 0\\ 0& ,& x=0\end{array}\right\}$

$F\left(x\right)$ is clearly continuous and differentiable at $x=0$ zero with $f‘\left(0\right)=0$.

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

This limit doesn't exist, hence $f\left(x\right)$ is non-differentiable at $x=0$.

Also $\underset{x\to 0}{lim}f‘\left(x\right)=0$.

Thus $f‘\left(x\right)$ is continuous at $x=0$.