Question

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

Answer 1

Answer: (B)

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.