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Let , and A = vmatrix -2 & 4+d & ( ) - 2 1 & ( + 2) & d 5 & (2 ) - d & (- ) + 2 + 2d…
MATHEMATICS · Single correct · CBSE · Mathematics · Determinants · Determinants
Problem Let d ∈ R d \in R d ∈ R , and
A = ∣ − 2 4 + d ( sin θ ) − 2 1 ( sin θ + 2 ) d 5 ( 2 sin θ ) − d ( − sin θ ) + 2 + 2 d ∣ , θ ∈ [ 0 , 2 π ] A = \begin{vmatrix} -2 & 4+d & (\sin \theta) - 2 \\ 1 & (\sin \theta + 2) & d \\ 5 & (2 \sin \theta) - d & (-\sin \theta) + 2 + 2d \end{vmatrix}, \theta \in [0, 2\pi] A = − 2 1 5 4 + d ( sin θ + 2 ) ( 2 sin θ ) − d ( sin θ ) − 2 d ( − sin θ ) + 2 + 2 d , θ ∈ [ 0 , 2 π ]
If the minimum value of det ( A ) \det(A) det ( A ) is 8, then a value of d d d is:
(a) -5
(b) -7
(c) 2 ( 2 + 1 ) 2(\sqrt{2} + 1) 2 ( 2 + 1 )
(d) 2 ( 2 − 2 ) 2(\sqrt{2} - 2) 2 ( 2 − 2 )
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