QJEEM24Maebf3: JEE Main 2024 Mathematics - Class 12 JEEM24Maebf3 Solution:  If  f ( x ) = { x 3 sin ⁡ ( 1 x ) , x ≠ 0 0 , x = 0…

Given the function:

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we need to find its second derivative at specific points.

First, let’s compute the first derivative $f^{\prime}(x)$ :

$f^{\prime}(x) = 3x^2 \sin \left( \frac{1}{x} \right) - x \cos \left( \frac{1}{x} \right)$

Next, the second derivative $f^{\prime \prime}(x)$ is:

$f^{\prime \prime}(x) = 6x \sin \left(\frac{1}{x}\right) - 3x \cos \left(\frac{1}{x}\right) - \cos \left(\frac{1}{x}\right) - \frac{1}{x} \sin \left(\frac{1}{x}\right)$

Therefore, evaluating the second derivative at $x = \frac{2}{\pi}$ :

$f^{\prime \prime} \left(\frac{2}{\pi}\right) = 6 \left( \frac{2}{\pi} \right) \sin \left(\frac{\pi}{2}\right) - 3 \left( \frac{2}{\pi} \right) \cos \left(\frac{\pi}{2}\right) - \cos \left( \frac{\pi}{2} \right) - \frac{\pi}{2} \sin \left( \frac{\pi}{2} \right)$

Since $\sin(\frac{\pi}{2}) = 1$ and $\cos(\frac{\pi}{2}) = 0$ , this simplifies to:

$f^{\prime \prime} \left( \frac{2}{\pi} \right) = \frac{12}{\pi} - \frac{\pi}{2} = \frac{24 - \pi^2}{2\pi}$

Finally, note that $f^{\prime}(0)$ is not defined, as it involves terms like $\frac{1}{x}$ when $x = 0$ .