Question

The set of values of $a$ for which the equation $\sqrt{a}\cos x-2\sin x=\sqrt{2}+\sqrt{2-a}$ possesses a solution, is

• A. $[-1-\sqrt{5},-1+\sqrt{5}]$
• B. $[\sqrt{5}-1,2]$
• C. $R$
• D. $(-\infty ,-1-\sqrt{5})$

Answer 1

The correct option is B $[\sqrt{5}-1,2]$

Clearly, $a\ge 0$ and $2-a\ge 0$
$\Rightarrow 0\le a\le 2\cdots \left(1\right)$
Now, for solution to exist,
$\sqrt{2}+\sqrt{2-a}\le \sqrt{a+4}\Rightarrow 2+2-a+2\sqrt{2}\sqrt{2-a}\le a+4\Rightarrow \sqrt{2}\sqrt{2-a}\le a\Rightarrow 2(2-a)\le a^2\Rightarrow a^2+2a-4\ge 0$
$\Rightarrow a\le -1-\sqrt{5}$ or $a\ge -1+\sqrt{5}\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$
$\sqrt{5}-1\le a\le 2$