The set of values of `lambda` for which the equation `"sin"^(4) x + "cos"^(4) x =lambda` has a solution, is

Options

A. `(0, 1)`
B. `(1, 3//2)`
C. `[-1, 1]`
D. `[1//2, 1]`

Answer

D

Explanation

We have,
`"sin"^(4) x + "cos"^(4) x = lambda`
`rArr 1-2" sin"^(2) x "cos"^(2) x = lambda`
`rArr "sin"^(2) 2x = 2 (1- lambda)`
This is equation will have a solution if
`0 le 2 (1-lambda) le 1 " " [therefore 0 le "sin"^(2) 2x le 1]`
`rArr 0 le 1 - lambda le (1)/(2) rArr - 1 le - lambda- (1)/(2) rArr lambda in [1//2, 1]`