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1995 AHSME Problems/Problem 12

Revision as of 12:25, 4 May 2022 by Hastapasta (talk | contribs)

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 $a<b$, or for all $b<a$. 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 (

It should be very clear that $f(1)<f(2)$ 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

See also

1995 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

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