2. As the pen, correctly illuminated

Body $A B C D$, $E$ is on
edge $A B$, $F$ is on the
edge $C D$, such that
$$
\begin{array}{l} 
\therefore \frac{A E}{E B}=\frac{C F}{F D}=\lambda \\
(0<\lambda<+\infty) .
\end{array}
$$

Let $f(\lambda)=\alpha_{\lambda}+\beta_{\lambda}$, where
$\alpha_{\lambda}$ represents the angle between $E F$ and $A C$,
$\beta_{2}$ represents 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 $(0,+\infty)$
(Proposed by the Question Committee)
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, $G F / / B D$. Hence $G F \perp E G$, and $\beta_{\lambda}=\angle E F G$. So,
$$
\begin{aligned}
f(\lambda) & =\alpha_{\lambda}+\beta_{\lambda}=\angle G E F+\angle E E_{\lambda} G \\
& =180^{\circ}-\angle E G Y=90^{\circ} .
\end{aligned}
$$