Question

If $2{\tan }^{2}x-5\sec x-1=0$ has 7 different roots in $\left[0,\frac{n\pi }{2}\right],n\in N$, then greatest value of $n$ is

• A. 8
• B. 10
• C. 13
• D. 15

Answer 1

Answer: (D)

$2{\tan }^{2}x-5\sec x-1=0$
$\Rightarrow 2\left({\sec }^{2}x-1\right)-5\sec x-1=0$
$\Rightarrow 2{\sec }^{2}x-5\sec x-3=0$
$\Rightarrow \sec x=\frac{6}{2},\frac{-1}{2}=3,\frac{-1}{2}$
$\Rightarrow \sec x=3\left(\sec x\ne \frac{-1}{2}\right)$

$\Rightarrow \cos x=\frac{1}{3}\Rightarrow 7$ solutions in $\left[0,\frac{15\pi }{2}\right]$
$\therefore n=15$