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 $\sum_\limits{x \in S} 2 \sin ^{-1}\left(x^{2}-1\right)$ is equal to :
Solution
$$
\begin{aligned}
& \cos ^{-1}(2 \mathrm{x})=\pi+2 \cos ^{-1} \sqrt{1-\mathrm{x}^2} \\\\
& \text { Since } \cos ^{-1}(2 \mathrm{x}) \in[0, \pi] \\\\
& \text { R.H.S. } \geq \pi \\\\
& \pi+2 \cos ^{-1} \sqrt{1-\mathrm{x}^2}=\pi \\\\
& \Rightarrow \cos ^{-1} \sqrt{1-\mathrm{x}^2}=0 \\\\
& \Rightarrow \sqrt{1-\mathrm{x}^2}=1 \\\\
& \Rightarrow \mathrm{x}=0 \\\\
& \text { but at } \mathrm{x}=0 \\\\
& \cos ^{-1}(2 \mathrm{x})=\cos ^{-1}(0)=\frac{\pi}{2}
\end{aligned}
$$
<br/><br/>$\therefore$ No solution possible for given equation.
<br/><br/>$\mathrm{x} \in \phi$
About this question
Subject: Mathematics · Chapter: Inverse Trigonometric Functions · Topic: Domain and Range
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