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BCECE Engineering BCECE Engineering Solved Paper-2005

  • question_answer
    If\[f\]is any function, then \[\frac{1}{2}[f(x)+f(-x)]\]is  always:

    A)  odd

    B)  even

    C)  neither even nor odd

    D)  one-one

    Correct Answer: B

    Solution :

    Let          \[g(x)=\frac{1}{2}[f(x)+f(-x)]\] \[\Rightarrow \]    \[g(-x)=\frac{1}{2}[f(-x)+f(x)]\] \[\Rightarrow \]    \[g(x)=g(-x)\] \[\therefore \] It is an even function. Note: In any function, if we replace \[x\]by the function does not change the sign, that is an \[-x,\]even function.


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