Question

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

• A. $x\in [-1,1]$
• B. $x\in R$
• C. $x\in [0,1]$
• D. $x\in [-1,0]$

Answer 1

The correct option is C $x\in [0,1]$

Given , ${\cos }^{-1}\{\frac{1}{2}{x}^2+\sqrt{1-x^2}\sqrt{1-\frac{x^2}{4}}\}={\cos }^{-1}\frac{x}{2}-{\cos }^{-1}x$
Since, $\frac{1}{2}{x}^2+\sqrt{1-x^2}\sqrt{1-\frac{x^2}{4}}$ is positive and defined for all $x\in [-1,1]$

Now, ${\cos }^{-1}\frac{x}{2}$ is defined for $-1\le \frac{x}{2}\le 1$
$\Rightarrow -2\le x\le 2$
Also, ${\cos }^{-1}x$ is defined for $-1\le x\le 1$
Also, LHS$\ge 0$
$\Rightarrow {\cos }^{-1}\frac{x}{2}-{\cos }^{-1}x\ge 0\Rightarrow {\cos }^{-1}\frac{x}{2}\ge {\cos }^{-1}x$
$\Rightarrow x\in [0,1]$
So, LHS is defined for $x\in [0,1]$