JEE Main 2022 (Online) 28th July Evening ShiftQuestion 1
The sum of the absolute maximum and absolute minimum values of the function $f(x)=\tan^{-1}(\sin x-\cos x)$ in the interval $[0,\pi]$ is :
Differentiation and its Application
Solution
Concept. $\tan^{-1}$ is increasing, so the extremes of $f$ come from the extremes of $\sin x-\cos x$ on $[0,\pi]$. Formulas used. (1) $\sin x-\cos x=\sqrt2\sin\left(x-\frac\pi4\right)$(2) $\tan\theta=\sqrt2$, $\theta$ acute $\Rightarrow$ $\cos\theta=\frac{1}{\sqrt3}$Step 1 - Range. By (1), $x-\frac\pi4\in\left[-\frac\pi4,\frac{3\pi}4\right]$, so $\sin x-\cos x\in\left[-1,\sqrt2\right]$. Step 2 - Extremes. Max $=\tan^{-1}\sqrt2$, min $=\tan^{-1}(-1)=-\frac\pi4$. Step 3 - Sum. By (2), $\tan^{-1}\sqrt2=\cos^{-1}\frac{1}{\sqrt3}$, so the sum is $\cos^{-1}\frac{1}{\sqrt3}-\frac\pi4$. Answer: (C)