JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
If \(S=\left\{x \in R : \sin ^{-1}\left(\frac{x+1}{\sqrt{x^2+2 x+2}}\right)-\sin ^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right)=\frac{\pi}{4}\right\}\) then \(\sum_{x \in R }\left(\sin \left(\left(x^2+x+5\right) \frac{\pi}{2}\right)-\cos \left(\left(x^2+x+5\right) \pi\right)\right)\) is equal to \(........\).
- A \(3\)
- B \(2\)
- C \(4\)
- D \(1\)
Answer & Solution
Correct Answer
(C) \(4\)
Step-by-step Solution
Detailed explanation
\(\sin ^{-1}\left(\frac{(x+1)}{\sqrt{(x+1)^2+1}}\right)-\sin ^{-1}\left(\frac{x}{\sqrt{x^2+1}}\right)=\frac{\pi}{4}\) \(\because \frac{t}{\sqrt{t^2+1}} \in(-1,1)\)…
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