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JEE Main & Advanced Mathematics Functions Question Bank Differentiability

  • question_answer
    Let \[f(x+y)=f(x)f(y)\] and \[f(x)=1+\sin (3x)g(x)\] where \[g(x)\] is continuous then \[f'(x)\] is [Kerala (Engg.) 2005]

    A)            \[f(x)g(0)\]

    B)            \[3g(0)\]

    C)            \[f(x)\cos 3x\]

    D)            \[3f(x)g(0)\]

    E)            \[3f(x)g(x)\]

    Correct Answer: C

    Solution :

                \[f(x)=1+\sin (3x)g(x)\]                    \[f'(x)=3\cos 3x\,g(x)+\sin 3x\,g'(x)\]\[=f(x)\cos 3x\].


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