The set of values of `alpha` for which the equation
`"sin"^(4) x + "cos"^(4) x + "sin" 2x + alpha =0` possesses a solution, is

Options

A. `[-3//2, 1]`
B. `[0, 1//2]`
C. `[-3//2, 1//2]`
D. none of these

Answer

C

Explanation

We have,
`"sin"^(4) x + "cos"^(4) + "sin" 2x + alpha = 0`
`rArr ("sin"^(2) x + "cos"^(2) x)^(2) - (1)/(2) "sin"^(2) 2x + "sin" 2x + alpha= 0`
`rArr "sin"^(2) 2x-2"sin"2x = 2+2alpha`
`rArr ("sin" 2x-1)^(2) = 3+2 alpha`
`rArr "sin" 2x- 1 = +- sqrt(3 + 2alpha)`
`rArr "sin" 2x = 1 +- sqrt(3 + 2alpha)`
This equation will have a solution if `3 + 2alpha ge 0` and in that case it reduces to
`"sin" 2x = 1 - sqrt(3 + 2alpha) " " [because 1+ sqrt(3 + 2 alpha) gt 1]`
This equation is valid, if
`-1 le 1 -sqrt(3 + 2 alpha) le 1`
`rArr -2 le - sqrt(3 + 2 alpha) le 0`
`rArr 0 le sqrt(3 + 2alpha) le 2`
`rArr 0 le 3 + 2 alpha le 4`
`rArr (-3)/(2) le alpha le (1)/(2)`
`"Also," 3 + 2 alpha ge 0 rArr alpha ge - (3)/(2)`
Hence, the given equation will possess a solution if `a in [-3//2, 1//2]`