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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 15

  • question_answer
    The function \[f(x)={{e}^{ax}}+{{e}^{-ax}},a>0\] is monotonically increasing for

    A) \[-1<x<1\]      

    B) \[x<-1\]

    C) \[x>-1\]          

    D)  \[x>0\]

    Correct Answer: D

    Solution :

    [d]: \[f(x)={{e}^{ax}}+{{e}^{-ax}}\] \[f'(x)=a{{e}^{ax}}-a{{e}^{-ax}}=a({{e}^{ax}}-{{e}^{-ax}})\] If f(x) is monotonically increasing, then \[f'(x)>0\] \[\Rightarrow \] \[a({{e}^{ax}}-{{e}^{-ax}})>0=({{e}^{ax}}-{{e}^{-ax}})>0(\because a>0)\] \[\Rightarrow \] \[{{e}^{ax}}>{{e}^{-ax}}={{e}^{2ax}}>1={{e}^{0}}\Rightarrow 2ax>0\] \[\Rightarrow \] \[x>0\]                \[(\because a>0)\] \[\therefore \] Function f (x) is increasing when x > 0.


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