Q39 / 90 · JEE Main 2024 (06 Apr Shift 1) · Mathematics

If $f(x)=\left\{\begin{array}{l}x^3 \sin \left(\frac{1}{x}\right), x \neq 0 \\ 0 \quad, x=0\end{array}\right.$ then

  • Correct answer
    A
    $f^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{24-\pi^2}{2 \pi}$
  • B
    $f^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{12-\pi^2}{2 \pi}$
  • C
    $f^{\prime \prime}(0)=1$
  • D
    $f^{\prime \prime}(0)=0$

Solution

$\begin{aligned} & f^{\prime}(x)=3 x^2 \sin \left(\frac{1}{x}\right)-x \cos \left(\frac{1}{x}\right) \\ & f^{\prime \prime}(x)=6 x \sin \left(\frac{1}{x}\right)-3 \cos \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right)-\frac{\sin \left(\frac{1}{x}\right)}{x} \\ & f^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{12}{\pi}-\frac{\pi}{2}=\frac{24-\pi^2}{2 \pi}\end{aligned}$

Hence, the correct answer is option: 'A'.

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