JEE Main 2021 (Online) 17th March Evening ShiftQuestion 1
Consider the function $f:\mathbf{R}\to\mathbf{R}$ defined by $f(x)=\begin{cases}\left(2-\sin\left(\frac{1}{x}\right)\right)|x|, & x\neq0\\ 0, & x=0\end{cases}$. Then $f$ is :
Differentiation and its Application
Solution
Concept. On each side of $0$, $f'$ contains a term $\pm\frac1x\cos\frac1x$ that swings between large positive and large negative values near $0$. So $f'$ changes sign on both sides. Formulas used. (1) $\dfrac{d}{dx}\left(x\sin\dfrac1x\right)=\sin\dfrac1x-\dfrac1x\cos\dfrac1x$Step 1 - $x>0$. $f=2x-x\sin\frac1x$, so by (1) $$f'(x)=2-\sin\frac1x+\frac1x\cos\frac1x.$$ At $\frac1x=2k\pi$: $f'=2+2k\pi>0$. At $\frac1x=(2k+1)\pi$: $f'=2-(2k+1)\pi<0$. So $f$ is not monotonic on $(0,\infty)$. Step 2 - $x<0$. $f=-2x+x\sin\frac1x$, so by (1) $$f'(x)=-2+\sin\frac1x-\frac1x\cos\frac1x.$$ At $\frac1x=-2k\pi$: $f'=-2+2k\pi>0$ for $k\ge1$. At $\frac1x=-(2k+1)\pi$: $f'=-2-(2k+1)\pi<0$. So $f$ is not monotonic on $(-\infty,0)$ either. Answer: (D)