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

The correct answer is (B) : \(\frac{-2\pi}{3}\)
\(cos^{−1}(2x)−2cos^{−1} \sqrt{1−x^{2}} =π\)
\(cos^{−1}(2x)−cos^{−1} (2(1−x^{2})-1) =π\)
\(cos^{−1}(2x)−cos^{−1} (1−2x^{2}) =π\)
\(−cos^{−1} (1−2x^{2}) =π-cos^{−1}(2x)\)
Taking cos both sides we get
\(cos(−cos^{−1} (1−2x^{2})) = cos(π-cos^{−1}(2x))\)
\(1−2x^2=−2x\)
\(2x^{2}−2x−1=0\)
\(\text{On solving, } x= \frac{1− \sqrt3}{2}, \frac{1+ \sqrt3}{2}\)
\(\text{As }x=[\frac{−1}{2}, \frac{1}{2}],x= \frac{1+ \sqrt3}{2} = \text{rejected}\)
\(\text{So }x= \frac{1−\sqrt3}{2} ⇒x^2 −1=−  \frac{\sqrt3}{2}\)
\(=2sin^{−1}(x^{2}−1)=2sin^{−1}(\frac{−\sqrt{3}}{2})= \frac{−2π}{3}\)

Answer 2

Rewriting the equation, \[ \cos^{-1} (2x) = \pi + 2\cos^{-1} (\sqrt{1 - x^2}) \] Since the LHS belongs to the interval \([0, \pi]\), the equation must be meaningful. Solving, \[ \cos^{-1} (2x) = \pi, \quad \cos^{-1} (\sqrt{1 - x^2}) = 0 \] which gives \[ x = -\frac{1}{2}, \quad x = 0 \] Now sum over the set: \[ \sum_{x \in S} 2 \sin^{-1} (x^2 - 1) = 0 \]