Question

The equation $ta{n}^{4}x-2se{c}^{2}x+a^2=0$ will have atleast one solution, if

• A. $|a|\le 4$
• B. $|a|\le 2$
• C. $|a|\le \sqrt{3}$
• D. $|a|\le \sqrt{2}$

Answer 1

The correct option is C $|a|\le \sqrt{3}$

$ta{n}^{4}x-2se{c}^{2}x+a^2=0$
$\Rightarrow ta{n}^{4}x-2(1+ta{n}^{2}x)+a^2=0$
$\Rightarrow ta{n}^{4}x-2ta{n}^{2}x+1=3-a^2$
$\Rightarrow (ta{n}^{2}x-1{)}^2=3-a^2$
$\Rightarrow 3-a^2\ge 0$
$\Rightarrow a^2\le 3$
$\Rightarrow |a|\le \sqrt{3}$