A) \[-5\]
B) \[-7\]
C) \[2\left( \sqrt{2}+1 \right)\]
D) \[2\left( \sqrt{2}+2 \right)\]
Correct Answer: A
Solution :
\[\left| A \right|=\left| \begin{matrix} -2 & 4+d & (sin\theta -2) \\ 1 & (sin\theta )+2 & d \\ 5 & (2sin\theta )-d & (-sin\theta )+2+2d \\ \end{matrix} \right|\] |
\[=\left| \begin{matrix} -2 & 4+d & (sin\theta -2) \\ 1 & (sin\theta ) & d \\ 1 & 0 & 0 \\ \end{matrix} \right|\] |
(New \[(New\,\,{{R}_{3}}={{R}_{3}}-2{{R}_{2}}+{{R}_{1}})\] |
\[=(4+d)d-si{{n}^{2}}\theta +4\] |
\[={{(d+2)}^{2}}-{{\sin }^{2}}\theta \] |
Because minimum value of |A| = 8 |
\[\Rightarrow \] \[{{(d+2)}^{2}}=9\] |
\[\Rightarrow \] d = 1 or \[-5.\] |
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