2. As shown in the figure, in the regular tetrahedron $A B C D$, $E$ is on edge $A B$, and $F$ is on edge $C D$, such that $\frac{A E}{E B}=\frac{C F}{F D}=\lambda(0<\lambda<+\infty)$.
Let $f(\lambda)=\alpha_{\gamma}+\beta_{\gamma}$, where $\alpha_{\gamma}$ is the angle between $E F$ and $A C$, and $\beta_{\gamma}$ is the angle between $E F$ and $B D$, 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 $\frac{C G}{G B}=$ $\frac{A E}{E B}=\frac{C F}{F D}$, so $G F / / B D$. Therefore, $G F \perp E G$, and $\beta=\lambda=\angle E F G$. Thus,
$$
\begin{aligned}
f(\lambda) & =\alpha_{\lambda}+\beta_{\lambda}=\angle G E F+\angle E F G \\
& =180^{\circ}-\angle E G F=90^{\circ} .
\end{aligned}
$$