`"If " 2"tan"^(2) theta - 5 "sec" theta = 1` has exactly 7 solution in the interval `[0, n pi//2], n in N`, then the least and greatest values of n are

Options

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

Answer

C

Explanation

We have,
`2 "tan"^(2) theta - 5 "sec" theta =1`
`rArr 2("sec"^(2) theta-1)-5"sec"theta = 1`
`rArr 2 "sec"^(2) theta -5 "sec" theta -3 =0`
`rArr (2 "sec" theta +1) ("sec"theta -3) = 0`
`rArr "sec" theta = 3 " " [because |"sec" theta| ge 1 therefore 2 "sec" theta + 1 ne 0]`
We observe that the curves `y= "sec"x " and " y = 3` intersects at two points in `[0, 2 pi]` one point lying in `(0, pi//2)` and the other in `(3pi//2, 2pi)`. Since sec x is periodic with period `2 pi`. So the curves will intersect at two points in each of the intervals `[2 pi, 4pi]` and `[4 pi, 6 pi]` Thus, y = sec x and y = 3 intersect in 6 points in `[0, 6 pi]` Clearly, 7th point point of intersection lies in the interval `(6 pi, 6pi + (pi)/(2)) " or " (6 pi, 6 pi + (3pi)/(2))`.
Thus, the two curves will intersects at 7 points in `[0, 13 pi//2] " and also in [0, 15 pi//2]`.
Hence, the least value of n is 13 and the greatest value of n is 15.