JEE Advanced 2023 Paper 1 · Q08 · Inverse Trigonometric Functions

Question

Let $	an^{-1}(x) \in \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}ight)$ for $x \in \mathbb{R}$. Then the number of real solutions of the equation $\sqrt{1 + \cos(2x)} = \sqrt{2}\, 	an^{-1}(	an x)$ in the set $\left(-\dfrac{3\pi}{2}, -\dfrac{\pi}{2}ight) \cup \left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}ight) \cup \left(\dfrac{\pi}{2}, \dfrac{3\pi}{2}ight)$ is equal to

SolveKar AI · Accepted Answer

$\sqrt{1+\cos 2x} = \sqrt{2}|\cos x|$, so the equation becomes $|\cos x| = 	an^{-1}(	an x)$. On each open interval the right side is a piecewise-linear branch ($x - k\pi$ for appropriate $k$), while the left side is the standard $|\cos x|$ wave. Plotting both: on $(-3\pi/2, -\pi/2)$ the line $y = x + \pi$ crosses $|\cos x|$ once; on $(-\pi/2, \pi/2)$ the line $y = x$ crosses $|\cos x|$ once at $x = 0$ (actually a tangent touch giving one solution); on $(\pi/2, 3\pi/2)$ the line $y = x - \pi$ crosses once. Total: $3$ real solutions.

JEE Advanced 2023 Paper 1 · Q08 · Inverse Trigonometric Functions

Solution