Question

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

• A. $-\sqrt{3}\ge a\ge \sqrt{3}$
• B. $-\sqrt{2}\ge a\ge \sqrt{2}$
• C. $-1\ge a\ge 1$
• D. $Noneofthese$

Answer 1

Answer: (B)

$-\sqrt{2}\ge a\ge \sqrt{2}$

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

If ${\sin }^{2}x=y,$ then the equation reduce to $y^2+2y+a^2-2=0$

Now, $\Delta =4-4(a^2-2)\ge 0\Rightarrow 3-a^2\ge 0\Rightarrow a^2\le 3\Rightarrow y={\sin }^{2}x=-1\pm \sqrt{3-a^2}$ ...(1)

Now, $y={\sin }^{2}x$ is positive

$\therefore 0\le y\le 1\Rightarrow 0\le -1\pm \sqrt{3-a^2}\le 1\Rightarrow 1\le \pm \sqrt{3-a^2}\le 2$

$\Rightarrow 1\le 3-a^2\le 4\le -2\le -a^2\le 1\Rightarrow a^2-2\le 0$ and $a^2+1\ge 0$ ...(2)

Using (1) and (2), we have

$a^2-2\le 0\Rightarrow (a+\sqrt{2})(a-\sqrt{2})\le 0\Rightarrow -\sqrt{2}\le a\le \sqrt{2}$