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JEE Main & Advanced Sample Paper JEE Main Sample Paper-29

  • question_answer
    If \[{{\tan }^{-1}}({{x}^{2}}+3\left| x \right|-4)+{{\cot }^{-1}}(4\pi +{{\sin }^{-1}}(\sin \,14))\]\[=\frac{\pi }{2}\], then the value of \[{{\sin }^{-1}}(\sin 2\left| x \right|)\] is equal to

    A) \[6-2\pi \]                           

    B) \[2\pi -6\]

    C) \[\pi -3\]                                             

    D) \[3-\pi \]

    Correct Answer: A

    Solution :

    \[{{\sin }^{-1}}(\sin 14)=(14-4\pi )\] So,\[{{x}^{2}}+3|x|-4=4\pi +(14-4\pi )\] \[\Rightarrow \]\[{{x}^{2}}+3|x|-18=0\] \[\Rightarrow \]\[(|x|+6)(|x|-3)=0\] \[\therefore \,\,|x|\,\,=3\] \[{{\sin }^{-1}}(\sin 2|x|)={{\sin }^{-1}}(\sin 6)=(6-2\pi )\]


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