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JEE Main 20246 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Differentiation

JEE Main 2024 · 6 April · Morning Shift · Mathematics · Differentiation

If f(x)=\ array l x^3 ((1)/(x)), x ≠ 0 0 , x=0 array . then

If f(x)=\left\{x3sin⁡(1x),x≠00,x=0\begin{array}{l}x^3 \sin \left(\frac{1}{x}\right), x \neq 0 \\ 0 \quad, x=0\end{array}\right. then

Options

  1. Af′′(2π)=24−π22πf^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{24-\pi^2}{2 \pi}
  2. Bf′′(2π)=12−π22πf^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{12-\pi^2}{2 \pi}
  3. Cf′′(0)=1f^{\prime \prime}(0)=1
  4. Df′′(0)=0f^{\prime \prime}(0)=0

Correct answer

A. f′′(2π)=24−π22πf^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{24-\pi^2}{2 \pi}

Step-by-step solution

f′(x)=3x2sin⁡(1x)−xcos⁡(1x)f′′(x)=6xsin⁡(1x)−3cos⁡(1x)−cos⁡(1x)−sin⁡(1x)xf′′(2π)=12π−π2=24−π22π\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}
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