| \[A=\left[ \begin{matrix} -2 & 4+d & (sin\,\theta )-2 \\ 1 & (sin\,\theta )+2 & d \\ 5 & (2\,sin\,\theta )-d & (-sin\,\theta )+2+2d \\ \end{matrix} \right]\] |
| \[\theta \in [0,\,\,2\,\pi ]\]. If the minimum value of det is 8, then a value of d is - |
A) -7
B) \[2(\sqrt{2}+2)\]
C) -5
D) \[2(\sqrt{2}+1)\]
Correct Answer: C
Solution :
\[{{R}_{3}}\text{ }\to \text{ }{{R}_{3}}+{{R}_{1}}-2{{R}_{2}}\] \[\det \,(A)\,\,=\,\,\left| \begin{matrix} -2 \\ 1 \\ 1 \\ \end{matrix}\,\,\,\,\begin{matrix} 4+d \\ 2+\sin \,\theta \\ 0 \\ \end{matrix}\,\,\,\,\begin{matrix} \sin \,\theta -2 \\ d \\ 0 \\ \end{matrix} \right|\] \[={{d}^{2}}+4d+4-si{{n}^{2}}\theta \] \[det(A)={{(d+2)}^{2}}\,-si{{n}^{\text{2}}}\theta \] min of det \[\left( A \right)=8={{(d+2)}^{2}}-1\] \[{{(d+2)}^{2}}=9\,\,\Rightarrow \,\,d+2=\pm 3\] \[\left( + \right),d=1,;\text{ }\left( - \right)d=-5\]
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