To the equation `2^2pi//cos^((-1)x)-(a+1/2)2^pi//cos^((-1)x)-a^2=0` has only one real root, then `1lt=alt=3` (b) `ageq1` `alt=-3` (d) `ageq3`

Options

A. `1 le a le 3`
B. `a ge 1`
C. `a le -3`
D. `a ge 3`

Answer

B, C

Explanation

`1 le (pi)/(cos^(-1) x) lt oo`
`rArr 2 le 2^((pi)/(cos^(-1)x)) lt oo`
Hence, 2 should lie between or on the roots of
`t^(2) - (a + (1)/(2)) t - a^(2) = 0, " where " t = 2^(pi//cos^(-1)x)`
`rArr f(2) le 0 rArr a^(2) + 2a -3 ge 0`
`rArr a in (-oo, -3] uu [1, oo)`