If $$f(x) = \begin{cases} x^3 \sin\left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x = 0 \end{cases}$$ then

We must find the second derivative of the function $$f(x)=\begin{cases}x^{3}\sin\!\left(\dfrac{1}{x}\right), & x\neq 0\\[4pt]0,&x=0\end{cases}$$ at two points: $$x=\dfrac{2}{\pi}$$ and $$x=0$$, and then match with the given options.

Case 1: Evaluation of $$f''\!\left(\dfrac{2}{\pi}\right)$$ (point away from zero, so usual differentiation rules apply).

Write $$f(x)=x^{3}\sin\!\left(\dfrac{1}{x}\right)$$ for $$x\neq 0$$.
First derivative (product rule):
$$f'(x)=\frac{d}{dx}\!\bigl[x^{3}\bigr]\sin\!\left(\frac{1}{x}\right)+x^{3}\,\frac{d}{dx}\!\left[\sin\!\left(\frac{1}{x}\right)\right]$$

Using $$\dfrac{d}{dx}\!\bigl[x^{3}\bigr]=3x^{2}$$ and $$\dfrac{d}{dx}\!\left[\sin\!\left(\frac{1}{x}\right)\right]=\cos\!\left(\frac{1}{x}\right)\!\left(-\frac{1}{x^{2}}\right)$$, we get
$$f'(x)=3x^{2}\sin\!\left(\frac{1}{x}\right)-x\cos\!\left(\frac{1}{x}\right).$$

Second derivative:
Differentiate each term separately.

• For $$3x^{2}\sin\!\left(\frac{1}{x}\right)$$:
$$\frac{d}{dx}\Bigl[3x^{2}\sin\!\left(\frac{1}{x}\right)\Bigr]=6x\sin\!\left(\frac{1}{x}\right)+3x^{2}\!\left[\cos\!\left(\frac{1}{x}\right)\!\left(-\frac{1}{x^{2}}\right)\right]$$ $$=6x\sin\!\left(\frac{1}{x}\right)-3\cos\!\left(\frac{1}{x}\right).$$

• For $$-x\cos\!\left(\frac{1}{x}\right)$$:
$$\frac{d}{dx}\Bigl[-x\cos\!\left(\frac{1}{x}\right)\Bigr]=-\,\Bigl[\cos\!\left(\frac{1}{x}\right)+x\!\left(\sin\!\left(\frac{1}{x}\right)\frac{1}{x^{2}}\right)\Bigr]$$ $$= -\cos\!\left(\frac{1}{x}\right)-\frac{\sin\!\left(\frac{1}{x}\right)}{x}.$$

Combine the two results to get
$$f''(x)=\bigl[6x\sin\!\left(\frac{1}{x}\right)-3\cos\!\left(\frac{1}{x}\right)\bigr]-\Bigl[\cos\!\left(\frac{1}{x}\right)+\frac{\sin\!\left(\frac{1}{x}\right)}{x}\Bigr]$$ $$=6x\sin\!\left(\frac{1}{x}\right)-\frac{\sin\!\left(\frac{1}{x}\right)}{x}-4\cos\!\left(\frac{1}{x}\right).$$

Factor the sine term:
$$f''(x)=\left(\frac{6x^{2}-1}{x}\right)\sin\!\left(\frac{1}{x}\right)-4\cos\!\left(\frac{1}{x}\right).$$

Now put $$x=\dfrac{2}{\pi}$$:

• $$\dfrac{1}{x}=\dfrac{\pi}{2}\quad\Longrightarrow\quad \sin\!\left(\frac{1}{x}\right)=\sin\!\left(\frac{\pi}{2}\right)=1,$$ $$\cos\!\left(\frac{1}{x}\right)=\cos\!\left(\frac{\pi}{2}\right)=0.$$ • $$x^{2}=\dfrac{4}{\pi^{2}}\quad\Longrightarrow\quad 6x^{2}-1=\frac{24}{\pi^{2}}-1.$$ • Evaluate the prefactor:
$$\frac{6x^{2}-1}{x}=\frac{\dfrac{24}{\pi^{2}}-1}{\dfrac{2}{\pi}}=\left(\frac{24}{\pi^{2}}-1\right)\!\left(\frac{\pi}{2}\right)=\frac{24}{2\pi}-\frac{\pi}{2}=\frac{12}{\pi}-\frac{\pi}{2}=\frac{24-\pi^{2}}{2\pi}.$$

Hence
$$f''\!\left(\frac{2}{\pi}\right)=\left(\frac{24-\pi^{2}}{2\pi}\right)\cdot 1-4\cdot 0=\frac{24-\pi^{2}}{2\pi}.$$
This matches Option A.

Case 2: Does $$f''(0)$$ exist?

First, find $$f'(0)$$ by definition:
$$f'(0)=\lim_{h\to 0}\frac{f(h)-f(0)}{h}=\lim_{h\to 0}\frac{h^{3}\sin\!\left(\frac{1}{h}\right)}{h}= \lim_{h\to 0}h^{2}\sin\!\left(\frac{1}{h}\right)=0,$$ because $$|h^{2}\sin(1/h)|\le h^{2}\to 0.$$

For the second derivative, use
$$f''(0)=\lim_{h\to 0}\frac{f'(h)-f'(0)}{h}=\lim_{h\to 0}\frac{f'(h)}{h}.$$

Recall $$f'(h)=3h^{2}\sin\!\left(\frac{1}{h}\right)-h\cos\!\left(\frac{1}{h}\right).$$ Hence
$$\frac{f'(h)}{h}=3h\sin\!\left(\frac{1}{h}\right)-\cos\!\left(\frac{1}{h}\right).$$

As $$h\to 0$$, the term $$3h\sin\!\left(\frac{1}{h}\right)\to 0,$$ but $$\cos\!\left(\frac{1}{h}\right)$$ oscillates between $$-1$$ and $$1$$ without settling at a single value. Therefore the limit does not exist, so $$f''(0)$$ is undefined.

Thus both statements “$$f''(0)=1$$” and “$$f''(0)=0$$” are false.

Only Option A is correct.