JEE Main 2020 (Online) 8th January Morning ShiftQuestion 1
Let $f(x)=x\cos^{-1}(-\sin|x|)$, $x\in\left[-\frac\pi2,\frac\pi2\right]$, then which of the following is true ?
Differentiation and its Application
Solution
Concept. For $|x|\le\frac\pi2$, $\cos^{-1}(-\sin|x|)=\frac\pi2+|x|$; then $f=x\left(\frac\pi2+|x|\right)$. Formulas used. (1) $\cos^{-1}(-u)=\pi-\cos^{-1}u$ and $\cos^{-1}(\sin\theta)=\frac\pi2-\theta$ for $\theta\in\left[0,\frac\pi2\right]$Step 1 - Simplify. By (1), $\cos^{-1}(-\sin|x|)=\pi-\left(\frac\pi2-|x|\right)=\frac\pi2+|x|$. Step 2 - $f'$. For $x>0$: $f=\frac\pi2x+x^2$, $f'=\frac\pi2+2x$. For $x<0$: $f=\frac\pi2x-x^2$, $f'=\frac\pi2-2x$. Both give $\frac\pi2$ at $0$, so $f'(0)=\frac\pi2$. Step 3 - Conclude. $f'=\frac\pi2-2x$ decreases on $\left(-\frac\pi2,0\right)$ and $f'=\frac\pi2+2x$ increases on $\left(0,\frac\pi2\right)$. Answer: (C)