Q. $\cos^{- 1}\left\{ \frac{1}{2}x^{2} + \sqrt{1 - x^{2}}\sqrt{1 - \frac{x^{2}}{4}} \right\} = \cos^{- 1}{\frac{x}{2} - \cos^{- 1}x}$ holds for
A) $|x| \leq 1$
B) $x \in R$
C) $0 \leq x \leq 1$
D) $- 1 \leq x \leq 0$
Explanation
(a)
Since, $0 \leq \cos^{- 1}{\left( \frac{x^{2}}{2} + \sqrt{1 - x^{2}}\sqrt{1 - \frac{x^{2}}{4}} \right) \leq \frac{\pi}{2}}$
Because $\cos^{- 1}x$ is in first quadrant when $x\$is positive
And ${\cos^{- 1}\ }{\frac{x}{2} - \cos^{- 1}{x \geq 0}}$
So, ${\cos^{- 1}\ }{\frac{x}{2} \geq \cos^{- 1}x}$
Also, $\left| \frac{x}{2} \right| \leq 1,|x| \leq 1 \Rightarrow |x| \leq 1$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.