If f is any function, then 12[f(2x)+f(–2x)] is always

g(x)=12[f(2x)+f(−2x)] g(−x)=12[f(−2x)+f(2x)]=g(x) Even function

If f is any function, then 12[f(2x)+f(–2x)] is always

g(x)=12[f(2x)+f(−2x)] g(−x)=12[f(−2x)+f(2x)]=g(x) Even function

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

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Neither even nor odd

Odd

Even

Symmetric about $x = 2$

answer is A.

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Detailed Solution

$g (x) = \frac{1}{2} [f (2 x) + f (- 2 x)]$

$g (- x) = \frac{1}{2} [f (- 2 x) + f (2 x)] = g (x)$

Even function

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