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Let d be real and A = [[-2, 4+d, (sinθ)-2], [1, (sinθ)+2, d], [5, (2 sinθ)-d, (-sinθ)+2+2d]], θ in [0, 2π]. If the minimum value of det(A) is 8, find a value of d.

1. -7
2. 2(sqrt(2) + 2)
3. -5
4. 2(sqrt(2) + 1)

Correct answer: -5

Solution

Simplifying the determinant yields a value linear in sinθ, whose minimum over [-1,1] set equal to 8 gives a quadratic in d satisfied by d = -5.

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