# 1995 AHSME Problems/Problem 12

## Problem

Let $f$ be a linear function with the properties that $f(1) \leq f(2), f(3) \geq f(4),$ and $f(5) = 5$. Which of the following is true?

$\mathrm{(A) \ f(0) < 0 } \qquad \mathrm{(B) \ f(0) = 0 } \qquad \mathrm{(C) \ f(1) < f(0) < f( - 1) } \qquad \mathrm{(D) \ f(0) = 5 } \qquad \mathrm{(E) \ f(0) > 5 }$

## Solution 1

A linear function has the property that $f(a)\leq f(b)$ either for all $a, or for all $b. Since $f(3)\geq f(4)$, $f(1)\geq f(2)$. Since $f(1)\leq f(2)$, $f(1)=f(2)$. And if $f(a)=f(b)$ for $a\neq b$, then $f(x)$ is a constant function. Since $f(5)=5$, $f(0)=5\Rightarrow \mathrm{(D)}$

## Solution 2

If $f$ is a linear function, the statement $f(1) \leq f(2)$ states that the slope of the line $\frac{f(2) - f(1)}{2 - 1} = f(2) - f(1)$ is nonnegative: it is either positive or zero.

Similarly, the statement $f(3) \geq f(4)$ states that the slope of the line $\frac{f(4) - f(3)}{4 - 3} = f(4) - f(3)$ is nonpositive: it is either negative or zero.

Since the slope of a linear function can only have one value, it must be zero, and thus the function is a constant. The statement $f(5) = 5$ tells us that the value of the constant is $5$, and thus that $f(x) = 5$. This leads to $f(0)=5\Rightarrow \mathrm{(D)}$

## Solution 3 (Common Sense)

It should be very clear that $f(1) and $f(3)>f(4)$ is contradictory because of the fact that linear functions are monotonic. The only thing that makes sense is $f(1)=f(2)$, and $f(3)=f(4)$. This means that $f(x)$ has a slope of $0$. So $f(x)=5$. So $f(0)=5$. Select $\boxed{D}$.

~hastapasta