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Mathematics · Methods of Differentiation

JEE Main 2024 — 6 April, Shift 1 — Question 1

If f(x)={x3sin(1x),x≠00,x=0f\left( x \right)=\left\{ \begin{matrix}{{x}^{3}}\text{sin}\left( \frac{1}{x} \right) & ,x\ne 0 \\0, & x=0 \\\end{matrix} \right., then

  1. Option A:

    f′′(0)=1\mathrm{f}^{\prime \prime}(0)=1

  2. Option B:

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

    Correct
  3. Option C:

    f" (2π)=12−π22π\left(\frac{2}{\pi}\right)=\frac{12-\pi^{2}}{2 \pi}

  4. Option D:

    f′′(0)=0\mathrm{f}^{\prime \prime}(0)=0

Answer: B

Step-by-step solution

f′(x)=3x2sin⁡(1x)−xcos⁡(1x)\quad f'(x)=3 x^{2} \sin \left(\frac{1}{x}\right)-x \cos \left(\frac{1}{x}\right)

f′′(x)=6xsin⁡(1x)−3cos⁡(1x)−cos⁡(1x)−sin⁡(1x)xf^{\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′′(2π)=12π−π2=24−π22π\mathrm{f}^{\prime \prime}\left(\frac{2}{\pi}\right)=\frac{12}{\pi}-\frac{\pi}{2}=\frac{24-\pi^{2}}{2 \pi}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
6 April, Shift 1
Subject
Mathematics
Chapter
Methods of Differentiation
Topic
Methods of Differentiation

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