JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer If\[{{\Delta }_{1}}=\left| \begin{matrix}    x & \sin \theta  & \cos \theta   \\    -\sin \theta  & -x & 1  \\    \cos \theta  & l & x  \\ \end{matrix} \right|\]and \[{{\Delta }_{2}}=\left| \begin{matrix}    x & \sin 2\theta  & \cos 2\theta   \\    -\sin 2\theta  & -x & 1  \\    \cos 2\theta  & \text{l} & x  \\ \end{matrix} \right|,x\ne 0;\]then for all\[\theta \in \left( 0,\frac{\pi }{2} \right):\] [JEE Main 10-4-2019 Morning]

    A) \[{{\Delta }_{1}}-{{\Delta }_{2}}=x(cos2\theta -cos4\theta )\]

    B) \[{{\Delta }_{1}}+{{\Delta }_{2}}=-2{{x}^{3}}\]

    C) \[{{\Delta }_{1}}-{{\Delta }_{2}}=-2{{x}^{3}}\]

    D) \[{{\Delta }_{1}}+{{\Delta }_{2}}=-2({{x}^{3}}+x-1)\]

    Correct Answer: B

    Solution :

    \[{{\Delta }_{1}}=f(\theta )=\left| \begin{matrix}    x & \sin \theta  & \cos \theta   \\    -\sin \theta  & -x & \text{l}  \\    \cos \theta  & \text{l} & x  \\ \end{matrix} \right|=-{{x}^{3}}\]             and\[{{\Delta }_{2}}=f(2\theta )=\left| \begin{matrix}    x & \sin 2\theta  & \cos 2\theta   \\    -\sin 2\theta  & -x & \text{l}  \\    \cos 2\theta  & \text{l} & x  \\ \end{matrix} \right|=-{{x}^{3}}\]             So\[{{\Delta }_{1}}+{{\Delta }_{2}}=-2{{x}^{3}}\]


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