JEE Main 2022 · 28 Jul · Shift 2 · Mathematics Q67

MathematicsSingle correct+4 / -1Question 67 of 90

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 :

\tan^{-1}\left(\dfrac1{\sqrt2}\right)-\dfrac\pi4

\cos^{-1}\left(\dfrac1{\sqrt3}\right)-\dfrac\pi4

\dfrac{-\pi}{12}

Answer and worked solution

Official answer: C

$g=\sin x-\cos x=\sqrt2\sin\left(x-\frac\pi4\right)$ runs from $-1$ to $\sqrt2$ on $[0,\pi]$ .

Sum $\tan^{-1}\sqrt2-\frac\pi4=\cos^{-1}\frac1{\sqrt3}-\frac\pi4$ .

(C)

Topics:Maxima and minima Inverse trigonometric functions

JEE Main 2022 · 28 Jul · Shift 2, Mathematics question 67