JEE Advanced 2023 Paper 1Question 1
Let $\tan^{-1}(x)\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, for $x\in\mathbb{R}$. Then the number of real solutions of the equation $\sqrt{1+\cos(2x)}=\sqrt{2}\tan^{-1}(\tan x)$ in the set $\left(-\frac{3\pi}{2},-\frac{\pi}{2}\right)\cup\left(-\frac{\pi}{2},\frac{\pi}{2}\right)\cup\left(\frac{\pi}{2},\frac{3\pi}{2}\right)$ is equal to ______.
Inverse Trigonometric Functions
Answer$3$
Solution
$\sqrt{1+\cos2x}=\sqrt2|\cos x|$, so $|\cos x|=\tan^{-1}(\tan x)$. On each of the three intervals this becomes $\cos u=u$ with $u=x,\,x-\pi,\,x+\pi\in[0,\frac\pi2)$, one root each. Total $3$.