Question

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

• A. 0
• B. $\frac{-2 \pi}{3}$
• C. $\pi-2 \sin ^{-1}\left(\frac{\sqrt{3}}{4}\right)$
• D. $\pi-\sin ^{-1}\left(\frac{\sqrt{3}}{4}\right)$

Answer 1

The Correct Option is B

To solve this problem, we need to determine the set of all solutions of the given equation and find the sum for those solutions. Let us go step-by-step through the solution:

The given equation is:

\cos^{-1}(2x) - 2 \cos^{-1}\left(\sqrt{1-x^2}\right) = \pi.

This equation involves inverse trigonometric functions. Let z = \cos^{-1}(2x) and y = \cos^{-1}\left(\sqrt{1-x^2}\right) . Then we can rewrite the equation as:

z - 2y = \pi.

The range for y is [0, \frac{\pi}{2}] because \sqrt{1-x^2} is maximized at x = 0 . Let's analyze the transformation of characters in this equation.

The following identity is essential:

y = \cos^{-1}\left(\sqrt{1-x^2}\right) \Rightarrow \cos(y) = \sqrt{1-x^2} \Rightarrow 1-x^2 = \cos^2(y) .

For x in the domain [-\frac{1}{2}, \frac{1}{2}] , we have 2x \in [-1, 1] .

Now, rearrange the equation z = 2y + \pi . As inverse cosine is defined only within [0, \pi] , this means 2x = \cos(z) \leq 1 . Therefore, \cos(2y + \pi) = -\cos(2y) \Rightarrow 2x = -\cos(2y) \Rightarrow x = -\frac{1}{2} \cos(2y) .

Rewriting this, we need to solve \sin(2y) = 0 or \cos(2y) = -2x .

Substitute x = 0, y = \frac{\pi}{4} . Solutions in this domain confirm that x = 0 \Rightarrow z = \pi \Rightarrow 2y = \frac{\pi}{2} \Rightarrow y = \frac{\pi}{4} .

Finally, calculate the sum: 2 \sin^{-1}\left(x^2 - 1\right) = 2 \sin^{-1}(-1) \Rightarrow -\pi \right.. \text{Thus, the sum} \sum_{x \in S} 2 \sin^{-1}\left(x^2 - 1\right) = \frac{-2\pi}{3}.

Thus, the answer is: \frac{-2\pi}{3}.