Question

The equation ${\tan }^{4}x-2{\sec }^{2}x+a=0$ will have at least one solution if :

• A. $a\ge -3$
• B. $a\ge 2$
• C. $a\le 3$
• D. $a\ge 3$

Answer 1

The correct option is C $a\le 3$

${\tan }^{4}x-2{\sec }^{2}x+a=0\Rightarrow {\tan }^{4}x-2(1+{\tan }^{2}x)+a=0\Rightarrow {\tan }^{4}x-2{\tan }^{2}x+1=3-a\Rightarrow ({\tan }^{2}x-1{)}^2=3-a$
Since, $({\tan }^{2}x-1{)}^2\ge 0$
So, equation to have solution
$3-a\ge 0\Rightarrow a\le 3$