Question

The equation ${\tan }^{4}x-2{\sec }^{2}x+a^2=0$ 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

Answer: (C)

${\tan }^{4}x-2{\sec }^{2}x+a^2=0$
$\Rightarrow$ ${\tan }^{4}x-2(1+{\tan }^{2}x)+a^2=0$
$\Rightarrow$ ${\tan }^{4}x-2{\tan }^{2}x+1=3-a^2$
$\Rightarrow$ ${({\tan }^{2}x-1)}^2=3-a^2$
$\Rightarrow$ $3-a^2\ge 0$
$\Rightarrow$ $a^2\le 3$
$\Rightarrow$ $|a|\le \sqrt{3}$