jee-main 2023 Q72

jee-main · India · session1_01feb_shift1 Not Maths

Let $S$ be the set of all solutions of the equation $\cos^{-1}(2x) - 2\cos^{-1}\left(\sqrt{1 - x^2}\right) = \pi$, $x \in \left[-\frac{1}{2}, \frac{1}{2}\right]$. Then $\sum_{x \in S} 2\sin^{-1}(x^2 - 1)$ is equal to
(1) 0
(2) $\frac{-2\pi}{3}$
(3) $\pi - \sin^{-1}\frac{\sqrt{3}}{4}$
(4) $\pi - 2\sin^{-1}\frac{\sqrt{3}}{4}$

Let $S$ be the set of all solutions of the equation $\cos^{-1}(2x) - 2\cos^{-1}\left(\sqrt{1 - x^2}\right) = \pi$, $x \in \left[-\frac{1}{2}, \frac{1}{2}\right]$. Then $\sum_{x \in S} 2\sin^{-1}(x^2 - 1)$ is equal to\\
(1) 0\\
(2) $\frac{-2\pi}{3}$\\
(3) $\pi - \sin^{-1}\frac{\sqrt{3}}{4}$\\
(4) $\pi - 2\sin^{-1}\frac{\sqrt{3}}{4}$

Paper Questions

Q61 Q62 Q63 Q64 Q65 Q66 Q67 Q68 Q69 Q70 Q71 Q72 Q73 Q74 Q75 Q76 Q77 Q78 Q79 Q80 Q81 Q82 Q83 Q84 Q85 Q86 Q87 Q88 Q89 Q90