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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
MCQ
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
    $f^{\prime \prime}(0)=1$
  • ✓
    $\mathrm{f}^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{24-\pi^2}{2 \pi}$
  • C
    $f^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{12-\pi^2}{2 \pi}$
  • D
    $f^{\prime \prime}(0)=0$
✓

Answer

Correct option: B.
$\mathrm{f}^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{24-\pi^2}{2 \pi}$
b
$ 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}$

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