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KVPY Sample Paper KVPY Stream-SX Model Paper-7

  • question_answer
      Let de\[d\in R,\] and  \[A=\left[ \begin{matrix}    -2 & 4+d & (sin\theta )-2  \\    1 & (sin\theta )+2 & d  \\    5 & (2sin\theta )-d & (-sin\theta )+2+2d  \\ \end{matrix} \right],\] \[\theta \in [0,2\pi ].\] If the minimum value of det [A is 8. then   a value of d is:

    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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