cos−1⁡12x2+1−x21−x24=cos−1⁡x2−cos−1⁡x holds for

A
$| x | \leq 1$
B
$x \in R$
C
$0 \leq x \leq 1$
D
$- 1 \leq x \leq 0$

Solution:

Since, $0 \leq \cos^{- 1} (\frac{x^{2}}{2} + \sqrt{1 - x^{2}} \sqrt{1 - \frac{x^{2}}{4}}) \leq \frac{π}{2}$

Because cos^-1 x is in first quadrant when r is positive

$\begin{array}{l} and \cos^{- 1} \frac{x}{2} - \cos^{- 1} x \geq 0 \\ So, \cos^{- 1} \frac{x}{2} \geq \cos^{- 1} x \\ Also |\frac{x}{2}| \leq 1, | x | \leq 1 \Rightarrow | x | \leq 1 \end{array}$

$\cos^{- 1} \{\frac{1}{2} x^{2} + \sqrt{1 - x^{2}} \sqrt{1 - \frac{x^{2}}{4}}\} = \cos^{- 1} \frac{x}{2} - \cos^{- 1} x$ holds for

Solution:

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