If $f\left(x\right)=\left{\begin{matrix}{x}^{3}\sin \left(\frac{1}{x}\right), & x\neq 0 \ 0 & ,x=0\end{matrix}\right.$, then [JEE Main 2024, 6 Apr (Shift 1)]

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YESPYQ · Accepted Answer

Correct answer: d) $f^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{24-\pi^2}{2 \pi}$. Given $f'(x)=3x^2\sin\left(\dfrac{1}{x}\right)-x\cos\left(\dfrac{1}{x}\right)$. Differentiating again, $f''(x)=6x\sin\left(\dfrac{1}{x}\right)-3\cos\left(\dfrac{1}{x}\right)-\cos\left(\dfrac{1}{x}\right)-\dfrac{\sin\left(\dfrac{1}{x}\right)}{x}$ $\Rightarrow f''(x)=6x\sin\left(\dfrac{1}{x}\right)-4\cos\left(\dfrac{1}{x}\right)-\dfrac{\sin\left(\dfrac{1}{x}\right)}{x}$ At $x=\dfrac{2}{\pi}$, $f''\left(\dfrac{2}{\pi}\right)=6\cdot\dfrac{2}{\pi}\sin\dfrac{\pi}{2}-4\cos\dfrac{\pi}{2}-\dfrac{\sin\dfrac{\pi}{2}}{\dfrac{2}{\pi}}$ $\Rightarrow f''\left(\dfrac{2}{\pi}\right)=\dfrac{12}{\pi}-0-\dfrac{\pi}{2}$ $\Rightarrow f''\left(\dfrac{2}{\pi}\right)=\dfrac{24-\pi^2}{2\pi}$

If \(f\left(x\right)=\left{\begin{matrix}{x}^{3}\sin \left(\frac{1}{x}\right), & x\neq 0 \\ 0 & ,x=0\end{matrix}\right.\…

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