JEE Mains · Maths · STD 12 - 5. continuity and differentiation
If \(f(x)=\left\{\begin{array}{cl}x^3 \sin \left(\frac{1}{x}\right), & x \neq 0 \\ 0, & x=0\end{array}\right.\), then
- A \(f^{\prime \prime}(0)=1\)
- B \(\mathrm{f}^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{24-\pi^2}{2 \pi}\)
- C \(f^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{12-\pi^2}{2 \pi}\)
- D \(f^{\prime \prime}(0)=0\)
Answer & Solution
Correct Answer
(B) \(\mathrm{f}^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{24-\pi^2}{2 \pi}\)
Step-by-step Solution
Detailed explanation
\( f^{\prime}(x)=3 x^2 \sin \left(\frac{1}{x}\right)-x \cos \left(\frac{1}{x}\right) \) \( f^{\prime \prime}(x)=6 x \sin \left(\frac{1}{x}\right)-3 \cos \left(\frac{1}{x}\right)-\cos \left(\frac{1}{x}\right)-\frac{\sin \left(\frac{1}{x}\right)}{x} \)…
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