1. (15th "Hope Cup" Grade 1 High School Competition Question) Given that the set $M$ consists of all functions $f(x)$ that satisfy the following conditions:
(1) When $x \in[0,+\infty)$, the function value is a non-negative real number;
(2) For any $s, t \in[0,+\infty)$, $f(s)+f(t) \leqslant f(s+t)$. Among the functions $f_{1}(x)=x, f_{2}(x)=$ $2^{x}-1, f_{3}(x)=\ln (x+1)$, those that belong to $M$ are
A. $f_{1}(x)$ and $f_{2}(x)$
B. $f_{1}(x)$ and $f_{3}(x)$
C. $f_{2}(x)$ and $f_{3}(x)$
D. $f_{1}(x), f_{2}(x)$ and $f_{3}(x)$
1. A Prompt: All three functions satisfy condition (1); $f_{1}(x)=x$ clearly satisfies condition (2).
$f_{2}(x)=2^{x}-1$ satisfies condition (2) $\Leftrightarrow 2^{x}-1+2^{y}-1 \leqslant 2^{x+y}-1 \Leftrightarrow 2^x-1+2^y-1 \leqslant 2^{x+y}-1 \Leftrightarrow 2^x-1 \leqslant 2^x(2^y-1)$.
Since $2^t \geqslant 1$, the above inequality obviously holds, i.e., $f_{2}(x)=2^x-1$ satisfies condition (2).
$f_{3}(x)=\ln (x+1)$ satisfies condition (2) $\Leftrightarrow \ln (s+1)+\ln (t+1) \leqslant \ln (s+t+1) \Leftrightarrow \ln (s+s+t+1) \leqslant \ln (s+t+1)$.
This conclusion is incorrect, so $f_{3}(x)$ does not satisfy condition (2).
Therefore, the answer is A.