Let $f'(x)=3x^2.\sin\frac{1}{x}-x\cos\frac{1}{x},\,x≠0,\,f(0)=0,\,f(\frac{1}{π})=0$, then which of the following is not correct.
Answer & explanation
Correct answer: option 3
$f'(x)=3x^2.\sin\frac{1}{x}-x\cos\frac{1}{x}$
$⇒f(x)=\int\left(3x^2.\sin\frac{1}{x}-x\cos\frac{1}{x}\right)dx=\sin\frac{1}{x}.x^3-\int\cos\frac{1}{x}\left(-\frac{1}{x^2}\right)x^3dx-\int x\cos\frac{1}{x}dx$
$=x^3.\sin\frac{1}{x}+c$, since $f(\frac{1}{π})=0+c⇒f(x)=\left\{\begin{matrix}x^3\sin\frac{1}{x},&x≠0\\0,&x=0\end{matrix}\right.$
f(x) is clearly continuous and differentiable at x = 0 zero with f'(0) = 0.
$f''(0)=\underset{h→0}{\lim}\frac{3h^2\sin\frac{1}{h}-h\cos\frac{1}{h}}{h}=3h\sin \frac{1}{h}-\cos\frac{1}{h}$
This limit doesn’t exist, hence f'(x) is non-differentiable at x = 0 also $\underset{x→0}{\lim}f'(x)=0$
Thus f'(x) is continuous at x = 0.