2. (2002 Beijing College Entrance Examination for Science) As shown in the figure, $f_i(x) (i=1,2,3,4)$ are four functions defined on $[0,1]$, among which only those satisfying the property: “for any $x_{1}$ and $x_{2}$ in $[0,1]$, and any $\lambda \in[0,1], f\left(\lambda x_{1}+(1-\lambda) x_{2}\right) \leqslant$ $\lambda f\left(x_{1}\right)+(1-\lambda) f\left(x_{2}\right)$ always holds” are
A. $f_{1}(x), f_{3}(x)$
B. $f_{2}(x)$
C. $f_{2}(x), f_{3}(x)$
D. $f_{4}(x)$
2. Utilize the method of special values. Since $\lambda \in[0,1]$, let's assume $\lambda=\frac{1}{2}$. Then the inequality becomes
$$
f\left(\frac{x_{1}+x_{2}}{2}\right) \leqslant \frac{f\left(x_{1}\right)+f\left(x_{2}\right)}{2},
$$

its geometric characteristic is: the midpoint $M$ of the line segment connecting two points $P\left(x_{1}, y_{1}\right), Q\left(x_{2}, y_{2}\right)$ is above or on the point $N$ on the curve $y=f(x)$ with the x-coordinate $\frac{x_{1}+x_{2}}{2}$, which means the function should be a concave function or a linear function. Therefore, the functions that meet the requirement are $f_{1}(x)$ and $f_{3}(x)$. The answer should be A.