5. In a regular tetrahedron $ABCD$, $E$ is on edge $AB$, and $F$ is on edge $CD$, such that $\frac{AE}{EB}=\frac{CF}{FD}=\lambda(0<\lambda<+\infty)$. Let $f(\lambda)=\alpha_{\lambda}+\beta_{\lambda}$, where $\alpha_{\lambda}$ is the angle between $EF$ and $AC$, and $\beta_{\lambda}$ is the angle between $EF$ and $BD$. Then ( ).
(1997 National High School League Problem)
A. $f(\lambda)$ is monotonically increasing on $(0,+\infty)$
B. $f(\lambda)$ is monotonically decreasing on $(0,+\infty)$
C. $f(\lambda)$ is monotonically increasing on $(0,1)$ and monotonically decreasing on $(1,+\infty)$
D. $f(\lambda)$ is constant on $(1,+\infty)$
5. (D). Given $A C \perp B D$, draw $E G / / A C$ intersecting $B C$ at $G$, connect $G F$, then $\alpha_{\lambda}=\angle G E F$, and $\frac{C G}{G B}=\frac{A E}{E B}=\frac{C F}{F D}$, so $G F / / B D$, hence $G F \perp E G$ and $\beta_{\lambda}=\angle E F G$, thus $f(\lambda)=\alpha_{\lambda}+\beta_{\lambda}=\angle G E F+\angle E F G=180^{\circ}-\angle E G F=90^{\circ}$.