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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 39

  • question_answer
    Let \[f:R\to R\] be denned as \[f(x)\,=7{{e}^{{{\sin }^{2}}x}}-{{e}^{{{\cos }^{2}}x}}+2.\] Then the value of \[7{{f}_{\min .}}+{{f}_{\max .}}\]is

    A) \[36\]                    

    B)        \[49\]                    

    C) \[64\]            

    D)        \[81\]

    Correct Answer: C

    Solution :

    [c] \[f(x)=7{{e}^{{{\sin }^{2}}x}}-{{e}^{{{\cos }^{2}}x}}+2\] Clearly, period of \[f(x)\]is \[\pi .\] Now, \[f'(x)=(7{{e}^{{{\sin }^{2}}x}}+{{e}^{{{\cos }^{2}}x}})\sin 2x\] \[f'(x)=0\] for \[2x=n\pi \] or \[x=n\pi /2,\] \[n\in Z.\] Now,  \[f(0)=9-e\] \[f\left( \frac{\pi }{2} \right)=7e+11\] and  \[f(\pi )=9-e\] So,  \[7{{f}_{\min .}}+{{f}_{\max .}}=7(9-e)+7e+1=64\]      


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