JEE Main & Advanced JEE Main Paper (Held On 12-Jan-2019 Evening)

  • question_answer
    If \[a=\left[ \begin{matrix}    1 & \sin \theta  & 1  \\    -\sin \theta  & 1 & \sin \theta   \\    -1 & -\sin \theta  & 1  \\ \end{matrix} \right];\]then for all \[\theta \in \left( \frac{3\pi }{4},\frac{5\pi }{4} \right),\]det lies in the interval [JEE Main Online Paper Held On 12-Jan-2019 Evening]

    A) \[\left( \frac{3}{2},3 \right]\]                               

    B) \[\left[ \frac{5}{2},4 \right)\]

    C) \[\left( 1,\frac{5}{2} \right]\]                   

    D)   \[\left( 0,\frac{3}{2} \right]\]

    Correct Answer: A

    Solution :

    Here, det \[(A)=\left| \begin{matrix}    1 & \sin \theta  & 1  \\    -\sin \theta  & 1 & \sin \theta   \\    -1 & -\sin \theta  & 1  \\ \end{matrix} \right|\] \[=(1+si{{n}^{2}}\theta )-sin\theta (0)+1(si{{n}^{2}}\theta +1)\] \[=2(1+si{{n}^{2}}\theta )\] \[\because \]\[\theta \in \left( \frac{3\pi }{4},\frac{5\pi }{4} \right)\Rightarrow \sin \theta \in \left( -\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right)\] \[\Rightarrow \]\[{{\sin }^{2}}\theta \in \left[ 0,\frac{1}{2} \right)\] \[\therefore \]\[\det (A)\in [2,3)\]


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