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

  • question_answer
    The maximum value of the expression \[3\cos \theta +5\sin \left( \theta -\frac{\pi }{6} \right)\]for any real value of \[\theta \]is [JEE Main Online Paper Held On 12-Jan-2019 Morning]

    A) \[\frac{\sqrt{79}}{2}\]                                       

    B) \[\sqrt{31}\]

    C) \[\sqrt{34}\]                              

    D)   \[\sqrt{19}\]

    Correct Answer: D

    Solution :

    Let \[x=3\cos \theta +5\sin \left( \theta -\frac{\pi }{6} \right)\] \[=3\cos \theta +5\left( \frac{\sqrt{3}}{2}\sin \theta -\frac{1}{2}\cos \theta  \right)\] \[=\frac{5\sqrt{3}}{2}\sin \theta +\frac{1}{2}\cos \theta \] So, maximum value of x for real \[\theta \] is\[\sqrt{\frac{75}{4}+\frac{1}{4}}=\sqrt{19}\]


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