JEE Main 2024 (Online) 6th April Morning ShiftQuestion 1
If $f(x)=\begin{cases}x^{3}\sin\left(\frac{1}{x}\right), & x\neq0\\ 0, & x=0\end{cases}$, then
Differentiation and its Application
Solution
Concept. Compute $f'$ for $x\ne0$ and at $0$ separately; check whether $f''(0)$ exists; then evaluate $f''\left(\frac2\pi\right)$. Formulas used. (1) $\dfrac{d}{dx}\sin\frac1x=-\dfrac{1}{x^2}\cos\frac1x$ and $\dfrac{d}{dx}\cos\frac1x=\dfrac{1}{x^2}\sin\frac1x$Step 1 - $f'$. For $x\ne0$, by (1), $f'(x)=3x^2\sin\frac1x-x\cos\frac1x$. At $0$, $f'(0)=\lim x^2\sin\frac1x=0$. Step 2 - $f''(0)$. $$\frac{f'(x)-f'(0)}{x}=3x\sin\frac1x-\cos\frac1x,$$ which has no limit as $x\to0$. So $f''(0)$ does not exist: (A), (C) are false. Step 3 - $f''$ for $x\ne0$. Differentiating again with (1): $$f''(x)=6x\sin\frac1x-4\cos\frac1x-\frac1x\sin\frac1x.$$ Step 4 - At $x=\frac2\pi$. $\frac1x=\frac\pi2$, $\sin=1$, $\cos=0$: $$f''\left(\tfrac2\pi\right)=\frac{12}{\pi}-\frac{\pi}{2}=\frac{24-\pi^2}{2\pi}.$$ Answer: (B)