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 $\sum_{x \in S}\left(\sin\left((x^{2}+x+5)\frac{\pi}{2}\right)-\cos\left((x^{2}+x+5)\pi\right)\right)$ is equal to ______.

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Inverse Trigonometric Functions

S = {x \in R : sin^{- 1} (\frac{x + 1}{x ^{2} + 2 x + 2}) - sin^{- 1} (\frac{x}{x ^{2} + 1}) = \frac{π}{4}}

, then

\sum_{x \in S} (sin ((x^{2} + x + 5) \frac{π}{2}) - cos ((x^{2} + x + 5) π))

is equal to ______.

Answer: 4

Since

\frac{x + 1}{x ^{2} + 2 x + 2} = \frac{x + 1}{( x + 1 ) ^{2} + 1}

, the identity

sin^{- 1} \frac{t}{t ^{2} + 1} = tan^{- 1} t

reduces the equation to

tan^{- 1} (x + 1) - tan^{- 1} x = \frac{π}{4}

. Hence

\frac{( x + 1 ) - x}{1 + x ( x + 1 )} = 1

, so

x^{2} + x + 1 = 1

, giving

x^{2} + x = 0

and

x = 0

x = - 1

. For each root

x^{2} + x + 5 = 5

, and

sin (5 \cdot \frac{π}{2}) - cos (5 π) = 1 - (- 1) = 2

. Adding the equal contributions of the two elements of

S

gives

2 + 2 = 4

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