The equation is sin4⁡x−2cos2⁡x+a2=0 is solvable if

A
$- \sqrt{3} \leq a \leq \sqrt{3}$
B
$- \sqrt{2} \leq a \leq \sqrt{2}$
C
$- 1 \leq a \leq 1$
D
none of these

Solution:

We have,

$\sin^{4} x - 2 \cos^{2} x + a^{2} = 0$

$\Rightarrow y^{2} - 2 (1 - y) + a^{2} = 0,$ where $\sin^{2} x = y$

$\begin{aligned} \Rightarrow y^{2} + 2 y + a^{2} - 2 = 0 \\ \Rightarrow y = - 1 \pm \sqrt{3 - a^{2}} \end{aligned}$

For $y$ to be real, we must have

Disc. $> 0 \Rightarrow 4 - 4 (a^{2} - 2) > 0 \Rightarrow a^{2} \leq 3$ ………………..….(i)

But, $\sin^{2} x = y$ Therefore,

$\begin{array}{l} 0 \leq y \leq 1 \\ \Rightarrow 0 \leq - 1 + \sqrt{3 - a^{2}} \leq 1 \\ \Rightarrow 1 \leq \sqrt{3 - a^{2}} \leq 2 \\ \Rightarrow 1 \leq 3 - a^{2} \leq 4 \\ \Rightarrow 2 - a^{2} > 0 \Rightarrow a^{2} \leq 2 . . . . . . . . . . . . . . . . . . . . . . . . . (i i) \end{array}$

From (i) and (ii), we have

$a^{2} \leq 2 \Rightarrow - \sqrt{2} \leq a \leq \sqrt{2}$

The equation is $\sin^{4} x - 2 \cos^{2} x + a^{2} = 0$ is solvable if

Solution:

Related content