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+2 y+a^2-2=0$
$y^2+2 y+a^2-2=0$ ....(1)
for solution, $ D \geq 0$
$4-4\left(a^2-2\right) \geq 0 $
$1-a^2+2 \geq 0 \Rightarrow a^2 \leq 3$
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 \leq y \leq 1 \Rightarrow 0 \leq-1 \pm \sqrt{3- a ^2} \leq 1 $
$\Rightarrow 1 \leq \pm \sqrt{3- a ^2} \leq 2 \Rightarrow 1 \leq 3- a ^2 \leq 4 $
$\Rightarrow-2 \leq- a ^2 \leq+1 \Rightarrow-1 \leq a ^2 \leq 2$
$\Rightarrow a ^2 \leq 2 $
$a \in[-\sqrt{2}, \sqrt{2}]$
Number of integral values $=3$.