Question

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

• A. $\left|x\right|\le 1$
• B. $xϵR$
• C. $0\le x\le 1$
• D. $-1\le x\le 0$

Answer 1

Answer: (C)

$0\le x\le 1$

For the domain of the following function,
$1-x^2\ge 0$ and $1-\frac{x^2}{4}\ge 0$
Therefore
$1\ge x^2$ and $4\ge x^2$
$|x|\le 1$ and $|x|\le 2$
Hence $xϵ[-1,1]$
However
$co{s}^{-1}(-x)=\pi -co{s}^{-1}\left(x\right)$
Therefore
The above equation is valid for
$0\le x\le 1$