JEE Main 2023 (Online) 13th April Morning ShiftQuestion 1
If $S=\left\{x\in\mathbb{R}:\sin^{-1}\left(\frac{x+1}{\sqrt{x^{2}+2x+2}}\right)-\sin^{-1}\left(\frac{x}{\sqrt{x^{2}+1}}\right)=\frac{\pi}{4}\right\}$, then $\sum_{x\in S}\left(\sin\left(\left(x^{2}+x+5\right)\frac{\pi}{2}\right)-\cos\left(\left(x^{2}+x+5\right)\pi\right)\right)$ is equal to $\_\_\_\_\_\_$ .
Inverse Trigonometric Functions
Answer$4$
Solution
$\frac{x+1}{\sqrt{(x+1)^2+1}}=\sin(\tan^{-1}(x+1))$, $\frac x{\sqrt{x^2+1}}=\sin(\tan^{-1}x)$, and $\tan^{-1}\in(-\frac\pi2,\frac\pi2)$ so $\sin^{-1}\sin$ returns the angle. Equation: $\tan^{-1}(x+1)-\tan^{-1}x=\frac\pi4\Rightarrow\frac1{1+x(x+1)}=1\Rightarrow x^2+x=0$, $x=0,-1$. Both check (differences $\frac\pi4$). For each, $x^2+x+5=5$: $\sin\frac{5\pi}2-\cos5\pi=1+1=2$. Sum over two values $=4$.