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. $|x|\le 1$
• B. $x\in R$
• C. $0\le x\le 1$
• D. $-1\le x\le 0$

Answer 1

Answer: (C)

$0\le x\le 1$

The range of inverse cosine function is $[0,\pi ]$.
Hence, for the equation to be satisfied,
$co{s}^{-1}\left(x\right)-co{s}^{-1}\left(\frac{x}{2}\right)\ge 0$
$\Rightarrow x\ge 0$ (Considering the graph of cos inverse function)
Hence, $x\ge 0$.
Considering the domain of inverse cosine function,
$1-x^2\ge 0$
$\Rightarrow |x|\le 1$
Hence, $0\le x\le 1$. (Considering the earlier set of values of $x$)
Let $cos\alpha =\frac{x}{2}$ and $cos\beta =x$ for $0\le x\le 1$
Let $0\le \alpha \le \frac{\pi }{2}$, $0\le \beta \le \frac{\pi }{2}$
Substituting the values of $\alpha$ and $\beta$ in the given equation we get,
$co{s}^{-1}(cos\alpha cos\beta +sin\alpha sin\beta )=\alpha -\beta$
$co{s}^{-1}\left(cos\right(\alpha -\beta \left)\right)=\alpha -\beta$
Since, $\alpha >\beta$
$\alpha -\beta =\alpha -\beta$
Hence, it is true for $0\le x\le 1$.
Hence, option C is correct.