Question

If $2{\tan }^2\theta -5\sec \theta =1$ has exactly $7$ solutions in the interval $[0,\frac{n\pi }{2}],n\in N$ then the least and greatest values of $n$ are

• A. $6,8$
• B. $12,14$
• C. $13,15$
• D. $15,17$

Answer 1

The correct option is C $13,15$

R.E.F image

$2ta{n}^2\theta -5\sec \theta =1\to$ 7 solu.in $[0,\frac{n\pi }{2}]$

$\Rightarrow 2({\sec }^2\theta -1)5\sec \theta -1=0$

$\Rightarrow 2{\sec }^2\theta -5\sec \theta -3=0$

$sec\theta =\frac{5\pm \sqrt{25+24}}{4}$

$=\frac{5\pm 7}{4}=\frac{-1}{2},3$

$\Rightarrow \sec \theta =3$

$\Rightarrow \cos \theta =1/3\to 7solu.in[0,\frac{n\pi }{2}]$

$(\frac{n\pi }{2}{)}_{min}=6\pi +\pi /2=\frac{13\pi }{2}$

$(\frac{n\pi }{2}{)}_{max}=7\pi +\pi /2=\frac{15\pi }{2}$

$\Rightarrow n_{min}=13$

$n_{max}=15$