The equation `2("cos"^(8)theta - "sin"^(8)theta) "sec" 2 theta =a^(2)` has real solution if a lies in the interval

Options

A. `[-sqrt(2), sqrt(2)]`
B. `[-sqrt(2), -1) cup (1, sqrt(2)]`
C. `[-sqrt(2), -1] cup [1, sqrt(2)]`
D. none of these

Answer

B

Explanation

We have,
`2("cos"^(8)theta - "sin"^(8)theta) "sec" 2 theta =a^(2)`
`rArr 2("cos"^(2)theta + "sin"^(2)theta) ("cos" ^(4) + "sin"^(4)theta) ("cos"^(2)theta- "sin"^(2)theta )"sec"2theta = a^(2)`
`rArr 2{("cos"^(2)theta)^(2) + ("sin"^(2) theta)^(2) + 2 "cos"^(2)theta "sin"^(2)theta -2"sin" ^(2)theta "cos"^(2)theta} = a^(2)`
`rArr 2(1-2"sin"^(2) theta "cos"^(2) theta) =a^(2)`
`rArr 2 (1-(1)/(2)"sin"^(2) 2 theta) = a^(2)`
`rArr 2-"sin"^(2) 2 theta = a^(2) rArr 1 + "cos"^(2)2 theta =a^(2) rArr "cos"^(2) 2 theta =a^(2) -1`
Thus, the given equation will have a real solution iff
`0 lt a^(2) -1 le 1 [{:(because "For" a^(2) -1 = 0 i.e. "cos" 2 theta = 0.), ("For which sec " 2 theta " is not defined"):}]`
`hArr 1 lt a^(2) le 2 hArr a in [-sqrt(2) , -1)cup (1, sqrt(2)]`