Correct Answer - C
`(c )` For the `1^(st)` equation `(cos^(4)alpha+k)-(sin^(4)alpha+k)=sqrt(4lambda^(2)-4lambda)`
i.e., `cos^(4)alpha-sin^(4)alpha=sqrt(4lambda^(2)-4lambda)`
For the `2^(nd)` equation `cos^(2)alpha-sin^(2)alpha=sqrt(64-16)=sqrt(48)`
But `cos^(4)alpha-sin^(4)alpha=(cos^(2)alpha-sin^(2)alpha)(cos^(2)alpha+sin^(2)alpha)`
`=cos^(2)alpha-sin^(2)alpha`
`implies 4lambda^(2)-4lambda=48`
`implies lambda=4` or `-3`
Sum `=1`