(5) Given $f(x)=x^{2}+\left(a^{2}+b^{2}-1\right) x+a^{2}+2 a b-b^{2}$ is an even function, then the maximum value of the y-coordinate of the intersection point of the function graph with the $y$-axis is ( ).
(A) $\sqrt{2}$
(B) 2
(C) $2 \sqrt{2}$
(D) 4
5 From the given conditions, we know that $a^{2}+b^{2}-1=0$, and the y-intercept of the function graph is $a^{2}+2 a b-b^{2}$. Let $a=\cos \theta, b=\sin \theta$, then
$$
\begin{aligned}
a^{2}+2 a b-b^{2} & =\cos ^{2} \theta+2 \sin \theta \cos \theta-\sin ^{2} \theta \\
& =\cos 2 \theta+\sin 2 \theta \\
& \leqslant \sqrt{2} .
\end{aligned}
$$

Therefore, the answer is A.