Question

Set of values of ${}^{\prime }{a}^{\prime }$ for which the equation $\sqrt{a}\cos x-2\sin x=\sqrt{2}+\sqrt{2-a}$ possesses a solution is

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

Answer 1

The correct option is B $\sqrt{5}-1\le a\le 2$

$2-a\ge 0\&a\ge 0\Rightarrow 0\le a\le 2...........\left(i\right)$
Now for solution to exist
$\sqrt{2}+\sqrt{2-a}\le \sqrt{a+4}2+2-a+2\sqrt{2}\sqrt{2-a}\le a+4\sqrt{2}\sqrt{2-a}\le a2(2-a)\le a^2{a}^2+2a-4\ge 0$
$\Rightarrow a\le -1-\sqrt{5}$ or $a\ge -1+\sqrt{5}..........\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
$\sqrt{5}-1\le a\le 2$