Question

Number of integral values of ' $a$ ' for which the equation ${\sin }^{4}x-2{\cos }^{2}x+a^2=0$ has a solution, is

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

${\sin }^{4}x-2{\cos }^{2}x+a^2=0$
${\sin }^{4}x+2{\sin }^{2}x+a^2-2=0$
$\text{ Put }{\sin }^{2}x=y,y\in [0,1]$
$y^2+2y+a^2-2=0$
$y^2+2y+a^2-2=0$ ....(1)
for solution, $D\ge 0$
$4-4\left(a^2-2\right)\ge 0$
$1-a^2+2\ge 0\Rightarrow a^2\le 3$
Solution of equation (1),
$y=\frac{-2\pm \sqrt{4-4\left(a^2-2\right)}}{2}=-1\pm \sqrt{3-a^2}$
But,
$0\le y\le 1\Rightarrow 0\le -1\pm \sqrt{3-a^2}\le 1$
$\Rightarrow 1\le \pm \sqrt{3-a^2}\le 2\Rightarrow 1\le 3-a^2\le 4$
$\Rightarrow -2\le -a^2\le +1\Rightarrow -1\le a^2\le 2$
$\Rightarrow a^2\le 2$
$a\in [-\sqrt{2},\sqrt{2}]$
Number of integral values $=3$.