Question

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

• A. $1<a\le 4$
• B. $a\ge 2$
• C. $a\le 3$
• D. None of these

Answer 1

Answer: (C)

$a\le 3$

We have, ${\tan }^{4}x-2{\sec }^{2}x+a=0$

$\Rightarrow {\tan }^{4}x-2(1+{\tan }^{2}x)+a=0$

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

$\Rightarrow ({\tan }^{2}x-1{)}^2-3+a=0$

$\Rightarrow a=3-({\tan }^{2}x-1{)}^2$

so the range of a for at least one solution is,
$a\le 3$