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KVPY Sample Paper KVPY Stream-SX Model Paper-29

  • question_answer
    Let \[A=\left[ \begin{matrix}    2 & b & 1  \\    b & {{b}^{2}}+1 & b  \\    1 & b & 2  \\ \end{matrix} \right]\] where\[b>0.\] Then the minimum value of \[\frac{\det (A)}{b}\] is:

    A) \[2\sqrt{3}\]

    B) \[-2\sqrt{3}\]

    C) \[-\sqrt{3}\]

    D) \[\sqrt{3}\]

    Correct Answer: A

    Solution :

    Det \[A={{b}^{2}}+3\]
    \[\frac{\det A}{b}=b+\frac{3}{b}\]
    \[\therefore \]      Least value \[=2\sqrt{3}.\]


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