A2. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be any even function and $g: \mathbb{R} \rightarrow \mathbb{R}$ be any odd function. Define the function $\varphi: \mathbb{R} \rightarrow \mathbb{R}$ by the rule $\varphi(x)=f(g(x))+g(f(x))$ for all $x \in \mathbb{R}$. Which of the following equalities is definitely satisfied for all real numbers $x$?
(A) $\varphi(x)=f(f(x))$
(B) $\varphi(x)=g(g(x))$
(C) $\varphi(x)=\varphi(-x)$
(D) $\varphi(-x)=-\varphi(x)$
(E) $\varphi(x)=2 f(g(x))$
A2. Since $f(x)$ is an even function and $g(x)$ is an odd function, it holds that $f(-x)=f(x)$ and $g(-x)=-g(x)$ for all real numbers $x$. Therefore, $\varphi(-x)=f(g(-x))+g(f(-x))=f(-g(x))+g(f(x))=f(g(x))+$ $g(f(x))=\varphi(x)$. The correct answer is (C). The example $f(x)=x^{2}$ and $g(x)=2x$ shows that none of the other equalities are necessarily satisfied, since in this case $\varphi(x)=6x^{2}$, $f(f(x))=x^{4}$, $g(g(x))=4x$ and $2f(g(x))=8x^{2}$.