Consider the system of equations
`sin x cos 2y=(a^(2)-1)^(2)+1, cos x sin 2y = a+1`
The number of values of `y in [0, 2pi]`, when the system has solution for permissible values of `a,` are

Options

A. 2
B. 3
C. 4
D. 5

Answer

D

Explanation

The given system is
`sin x cos 2y=(a^(@)-1)^(2)+1`,
and `cos x sin 2y=a+1` ...(i)
Since the L.H.S. of both the equations does not exceed 1, the given system may have solutions only for a's such that
`(a^(2)-1)^(2)+1 le 1 and -1 le a +1 le 1` ...(ii)
`(a^(2)-1)^(2)+1 le 1`
or `(a^(2)-1)^(2) le 0`
or `(a^(2)-1)^(2)=0`
or `a=1`
For `a=1`, equation `cos x sin 2y=a+1` does not hold.
Thus, `a=-1` only and we get
`sin x cos 2y=1`
`cos x sin 2y =0` ...(iii)
`sin x cos 2y =1`
`rArr sin x=1, cos 2y =1`
or `sin x=-1, cos 2y=-1`
for which `cos x sin 2y=0`.