Difference between revisions of "1995 AHSME Problems/Problem 12"

Latest revision as of 13:26, 4 May 2022

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 (Common Sense)

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

1995 AHSME (Problems • Answer Key • Resources)
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 16: / Line 16: @@
 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 <math>f(5) = 5</math> tells us that the value of the constant is <math>5</math>, and thus that <math>f(x) = 5</math>.  This leads to <math>f(0)=5\Rightarrow \mathrm{(D)}</math>
-==Solution 3 (==
+==Solution 3 (Common Sense)==
-It should be very clear that <math>f(1)<f(2)</math> and <math>f(3)>f(4)</math> is contradictory because of the fact that linear functions are monotonic. The only thing that makes sense is <math>f(1)=f(2)</math>, and <math>f(3)=f(4)</math>. This means that <math>f(x)</math> has a slope of 0<math>. So </math>f(x)=5<math>. So </math>f(0)=5<math>. Select </math>\boxed{D}$.
+It should be very clear that <math>f(1)<f(2)</math> and <math>f(3)>f(4)</math> is contradictory because of the fact that linear functions are monotonic. The only thing that makes sense is <math>f(1)=f(2)</math>, and <math>f(3)=f(4)</math>. This means that <math>f(x)</math> has a slope of <math>0</math>. So <math>f(x)=5</math>. So <math>f(0)=5</math>. Select <math>\boxed{D}</math>.
 ~hastapasta

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