Question

The value of $a$ for which the equation $2\sin 2\theta -\sqrt{a}\cos 2\theta =\sqrt{2}+\sqrt{2-a}$ has solution in $\theta$ is

• A. $(3,\infty )$
• B. $(0,\infty )$
• C. $[1,2]$
• D. $[\sqrt{5}-1,2]$

Answer 1

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

From the equation
$a\ge 0,a\le 2\Rightarrow a\in [0,2]$

We know that
$p\sin \theta +q\cos \theta \in [-\sqrt{p^2+q^2},\sqrt{p^2+q^2}]$

$\Rightarrow 2\sin 2\theta -\sqrt{a}\cos 2\theta \in [-\sqrt{4+a},\sqrt{4+a}]$

$\Rightarrow -\sqrt{4+a}\le \sqrt{2}+\sqrt{2-a}\le \sqrt{4+a}$

$\because \sqrt{2}+\sqrt{2-a}$ is positive,
$\therefore \sqrt{2}+\sqrt{2-a}\le \sqrt{4+a}$

On squaring both the sides, we have
$2+2-a+2\sqrt{2}\times \sqrt{2-a}\le 4+a$
$\Rightarrow a\ge \sqrt{2}\times \sqrt{2-a}$

On squaring both the sides, we have
$a^2\ge 2(2-a)$
$\Rightarrow a^2+2a-4\ge 0$
$\Rightarrow a\ge \sqrt{5}-1$,
or $a\le -\sqrt{5}-1\text{(Not possible)}$
$\therefore a\in [\sqrt{5}-1,2]$