JEE Main & Advanced JEE Main Paper (Held On 10-Jan-2019 Evening)

  • 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- [JEE Main Online Paper (Held On 10-Jan-2019 Evening]

    A) \[\sqrt{3}\]                                

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

    C) \[-\sqrt{3}\]       

    D)                  \[2\sqrt{3}\]

    Correct Answer: D

    Solution :

    \[det\left( A \right)=2\left( 2{{b}^{2}}+2-{{b}^{2}} \right)-b\left( 2b-b \right)\]                                     \[+1\left( {{b}^{2}}-{{b}^{2}}-1 \right)\] \[=\,\,2\left( {{b}^{2}}+2 \right)-{{b}^{2}}-1\] \[=\text{ }{{b}^{2}}+3\] \[\frac{\det (A)}{b}\,\,=\,\,\frac{{{b}^{2}}+3}{b}\,\,=b+\frac{3}{b}\] Minimum value \[=\,\,2\sqrt{3}\]


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