25) For every real number $x$, let $m(x)$ be the minimum of the numbers $2x, x+1, 10-2x$. The maximum value of $m(x)$ is:
(A)
(B) 2
(C) 4 (D) 6
(E) 8.
25) The answer is (C). Indeed, by drawing the three lines with equations $y=2 x, y=x+1$ and $y=10-2 x$, it can be seen that the graph corresponding to the equation $y=m(x)$ is a broken line that is consequently drawn. Its point of maximum ordinate is the intersection of the second and third lines, i.e., $(3,4)$, so 4 is the required maximum value.

SECOND SOLUTION

We have $2 x \leq x+13\end{cases}
$$

It can be observed that $m(x) \geq 0$ for $0 \leq x \leq 5$, the maximum value will occur for $x$ within this interval, more precisely the only possibilities are $x=1$ or $x=3$. A direct comparison shows that $m(1)=2, m(3)=4$, which is the sought maximum value.