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Question: The equation \(2\cos^{- 1}x = \sin^{- 1}\left( 2x\sqrt{1 - x^{2}} \right)\) is valid for all values ...

The equation 2cos⁡−1x=sin⁡−1(2x1−x2)2\cos^{- 1}x = \sin^{- 1}\left( 2x\sqrt{1 - x^{2}} \right) is valid for all values of xx satisfying

A

−1≤x≤1- 1 \leq x \leq 1

B

0≤x≤10 \leq x \leq 1

C

0≤x≤120 \leq x \leq \frac{1}{\sqrt{2}}

D

12≤x≤1\frac{1}{\sqrt{2}} \leq x \leq 1

Answer

12≤x≤1\frac{1}{\sqrt{2}} \leq x \leq 1

Explanation

Solution

If we denote cos⁡−1x\cos^{- 1}x by y, then

Since 0≤cos⁡−1x≤π0 \leq \cos^{- 1}x \leq \pi ⇒ 0≤2y≤2π0 \leq 2y \leq 2\pi ………(1)

Also since −π2≤sin⁡−1(2x1−x2)≤π2- \frac{\pi}{2} \leq \sin^{- 1}\left( 2x\sqrt{1 - x^{2}} \right) \leq \frac{\pi}{2}

⇒ −π2≤sin⁡−1sin⁡(2y)≤π2- \frac{\pi}{2} \leq \sin^{- 1}{\sin(2y)} \leq \frac{\pi}{2}

⇒ −π2≤2y≤π2- \frac{\pi}{2} \leq 2y \leq \frac{\pi}{2} ………..(2)

From (1) and (2) we find 0≤2y≤π20 \leq 2y \leq \frac{\pi}{2}

⇒ 0≤y≤π40 \leq y \leq \frac{\pi}{4}

⇒ 0≤cos⁡−1x≤π40 \leq \cos^{- 1}x \leq \frac{\pi}{4}

which holds if 12≤x≤1\frac{1}{\sqrt{2}} \leq x \leq 1