Question

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

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

Answer 1

Answer: (B)

${\cos }^{-1}(2x)-2{\cos }^{-1}\sqrt{1-x^2}=\pi$
${\cos }^{-1}(2x)-{\cos }^{-1}\left(2\left(1-x^2\right)-1\right)=\pi$
${\cos }^{-1}(2x)-{\cos }^{-1}\left(1-2x^2\right)=\pi$
$-{\cos }^{-1}\left(1-2x^2\right)=\pi -{\cos }^{-1}(2x)$
Taking $\cos$ both sides we get
$\mathrm{Cos}\left(-{\cos }^{-1}\left(1-2x^2\right)\right)=\cos \left(\pi -{\cos }^{-1}(2x)\right)$
$1-2x^2=-2x$
$2x^2-2x-1=0$
On solving, $x=\frac{1-\sqrt{3}}{2},\frac{1+\sqrt{3}}{2}$
As $x=[-1/2,1/2],x=\frac{1+\sqrt{3}}{2}=$ rejected
So $x=\frac{1-\sqrt{3}}{2}\Rightarrow x^2-1=-\sqrt{3}/2$
$=2{\sin }^{-1}\left(x^2-1\right)=2{\sin }^{-1}\left(\frac{-\sqrt{3}}{2}\right)=\frac{-2\pi }{3}$