JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
Let \(S\) be the set of all solutions of the equation \(\cos ^{-1}(2 x)-2 \cos ^{-1}\left(\sqrt{1-x^2}\right)=\pi, \quad x \in\left[-\frac{1}{2}, \frac{1}{2}\right]\).Then \(\sum_{x \in S} 2 \sin ^{-1}\left(x^2-1\right)\) is equal to
- A \(0\)
- B \(\frac{-2 \pi}{3}\)
- C \(\pi-\sin ^{-1}\left(\frac{\sqrt{3}}{4}\right)\)
- D \(\pi-2 \sin ^{-1}\left(\frac{\sqrt{3}}{4}\right)\)
Answer & Solution
Correct Answer
(B) \(\frac{-2 \pi}{3}\)
Step-by-step Solution
Detailed explanation
\(\cos ^{-1}(2 x)-2 \cos ^{-1} \sqrt{1-x^2}=\pi\) \(\cos ^{-1}(2 x)-\cos ^{-1}\left(2\left(1-x^2\right)-1\right)=\pi\) \(\cos ^{-1}(2 x)-\cos ^{-1}\left(1-2 x^2\right)=\pi\) \(-\cos ^{-1}\left(1-2 x^2\right)=\pi-\cos ^{-1}(2 x)\) Taking \(\cos\) both sides we get…
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