2. As shown in the figure, in the regular tetrahedron $ABCD$, $E$ is on edge $AB$, and $F$ is on edge $CD$, such that
$$
\begin{array}{l} 
\frac{AE}{EB}=\frac{CF}{FD}=\lambda \\
(0<\lambda<\infty)
\end{array}
$$

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
(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 a constant on $(0,+\infty)$.
2. (D)

Since $A B C D$ is a regular tetrahedron, $A C \perp B D$. Draw $E G / / A C$ intersecting $B C$ at $G$, and connect $G F$. Then $\alpha_{\lambda}=\angle G E F$, and $\frac{C G}{G B}=\frac{A E}{E B}=\frac{C F}{F D}$. Therefore, $G F / / B D$, so $G F \perp E G$, and $\beta_{\lambda}=\angle E F G$. Thus,
$$
\begin{aligned}
f(\lambda) & =\alpha_{\lambda}+\beta_{\lambda}=\angle G E F+\angle E F G \\
& =180^{\circ}-\angle E G F=90^{\circ}
\end{aligned}
$$