The equation `sin^4x+cos^4x+sin2x+alpha=0` is solvable for `-5/2lt=alphalt=1/2` (b) `-3lt=alpha<1` `-3/2lt=alphalt=1/2` (d) `-1lt=alphalt=1`

Options

A. `-5//2 le alpha le 1//2`
B. `-3 le alpha le 1`
C. `-3//2 le alpha le 1//2`
D. `-1 le alpha le 1`

Answer

C

Explanation

`sin^(4) x+cos^(4) x+sin 2x+alpha =0`
or `(sin^(2)x+cos^(2)x)^(2)-2 sin^(2) x cos^(2) x+sin 2x+alpha=0`
`rArr sin^(2) 2x-2 sin 2x-2-2alpha =0`
Let `sin 2x =y`. Then the given equation becomes
`y^(2)-2y-2(1+alpha)=0`, where `-1 le y le 1," "( :' -1 le sin 2x le 1)`
For real, y, discriminant `ge 0`
`rArr 3+2 alpha ge 0" "or alpha ge - 3/2`
Also `-1 le y le 1" "or -1 le 1-sqrt(3+2alpha) le 1`
`rArr 3+2 alpha le 4" "or alpha le 1/2`. Thus `-3/2 le alpha le 1/2`