JEE Main 2024 — Differentiation Question with Solution

Given the function:

$f(x)=\{\begin{array}{ll}{x}^3\sin \left(\frac{1}{x}\right),& x\ne 0\\ 0,& x=0\end{array}$

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 $x=0$.