If `cos^4theta+alpha` are the roots of the equation `x^2+2b x+b=0a n dcos^2theta+beta,sin^2theta+betaa r e` the roots of the equation `x^2+4x+2=0,` then values of `b` are `2` b. `-1` c. `-2` d. `2`

Options

A. 2
B. -1
C. -2
D. 1

Answer

1,2

Explanation

We have ` cos ^(2) theta - sin^(2() theta = cos 2 theta`
`rArr cos ^(4) theta theta - sin ^(4) theta = cos 2 theta `
`rArr (-2b)^(2) - 4b = - (-4)^(2) - 4xx2`
(since L.H.S. is differnece of roots of first equation and
R.H.S. is diffenece of roots of second equation).
or ` 4b^(2) - 4b = 16 - 8= 8 `
or ` 4b^(2) - 4b - 8 = 0`
or ` b^(2) - b - 2 = 0`
or ` (b + 1)(b - 2) = 0`
or ` b = 2, - 1`