If m and M are the minimum and the maximum values of `4+1/2 sin^2 2x-2cos^4x, x in R` then

Options

A. `7/4`
B. `15/4`
C. `9/4`
D. `1/4`

Answer

C

Explanation

Let `f(x) =4 + 1/2 sin ^2 2x -2 cos^4 x`
Then
`f(x)=4+1/2(1-cos^(2)2x)-2((1+cos 2x)/(2))^2`
`rArr f(x)=4+1/2(1-cos^(2)2x)-1/2 (1+ cos 2x)^2`
Now `-1 le cos 2x le 1` for all x
`rArr -1/2 le 1/2 + cos 2x le 3/2` for all x
`rArr 0 le (1/2+cos^2x )-1/2(1+cos 2x)^2 le 9/4` for all x
`=- 9/4 le -(1/2 + cos 2x)^2 le 0 ` for all x
`rArr 17/4 9/4 lt 17/4 (1/2 + cos 2x )^2 le 17/4 ` for all x
`rArr 2 le f(x) le 17/4 -2 `
Hence `M-m = 17/4-2 = 9/4`