5 Let $f(x)=x^{3}+\log _{2}\left(x+\sqrt{x^{2}+1}\right)$, then for any real numbers $\dot{a}, b$, $a+b \geqslant 0$ is a ( ) for $f(a)+f(b) \geqslant 0$.
(A) Sufficient and necessary condition
(B) Sufficient but not necessary condition
(C) Necessary but not sufficient condition
(D) Neither sufficient nor necessary condition
Solution: It is evident that $f(x)=x^{3}+\log _{2}\left(x+\sqrt{x^{2}+1}\right)$ is an odd function and monotonically increasing. Therefore, if $a+b \geqslant 0$, then $a \geqslant -b$, which implies $f(a) \geqslant f(-b)$, i.e., $f(a) \geqslant -f(b)$, thus we have $f(a)+f(b) \geqslant 0$. Conversely, if $f(a)+f(b) \geqslant 0$, then $f(a) \geqslant -f(b)=f(-b)$, leading to $a \geqslant -b$, i.e., $a+b \geqslant 0$. Therefore, the correct choice is A.