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Mathematics
Mathematics
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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