Question

If $2{\tan }^{2}x-5\sec x$ is equal to $1$ for exactly $7$ distinct values of $x\in [0,\frac{n\pi }{2}],n\in N$, then the greatest value of $n$ is

• A. $4$
• B. $10$
• C. $13$
• D. $15$

Answer 1

The correct option is A $4$

$2ta{n}^{2}x-5secx=1$

$2(se{c}^{2}x-1)-5secx=1$

$2se{c}^{2}x-5secx-3=0$

$(2secx+1)(secx-3)=0$

$secx=-1/2$ or 3, but $\mid secx\mid \ge 1$

hence $secx=3$ is only solution

or, $cosx=1/3$

therefore general solution is given as
$x=2n\pi \pm co{s}^{-1}\left(1/3\right)$, suppose $co{s}^{-1}\left(1/3\right)=\alpha$

or, $x=.....,\alpha ,2\pi -\alpha ,2\pi +\alpha ,4\pi -\alpha ,4\pi +\alpha ,6\pi -\alpha ,6\pi +\alpha$

so there are two value of n$ϵN$ for which there is 7 distinct roots in given interval

these are, $n=13$ and 15. (This can be easily seen by drawing $cosx=1/3$ graph)

Therefore, the value of $n$ is $15$