Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎See also)
(→‎Solution 3 ()
 
(3 intermediate revisions by 3 users not shown)
Line 5: Line 5:
 
<math> \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 }  </math>
 
<math> \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 }  </math>
  
−
==Solution==
+
==Solution 1==
−
A linear function has the property that <math>f(a)\leq f(b)</math> either for all a<b, or for all b<a. Since <math>f(3)\geq f(4)</math>, <math>f(1)\geq f(2)</math>. Since <math>f(1)\leq f(2)</math>, <math>f(1)=f(2)</math>. And if <math>f(a)=f(b)</math> for a≠b, then f(x) is a constant function. Since <math>f(5)=5</math>, <math>f(0)=5\Rightarrow \mathrm{(D)}</math>
+
A linear function has the property that <math>f(a)\leq f(b)</math> either for all <math>a<b</math>, or for all <math> b<a</math>. Since <math>f(3)\geq f(4)</math>, <math>f(1)\geq f(2)</math>. Since <math>f(1)\leq f(2)</math>, <math>f(1)=f(2)</math>. And if <math>f(a)=f(b)</math> for <math>a\neq b</math>, then <math>f(x)</math> is a constant function. Since <math>f(5)=5</math>, <math>f(0)=5\Rightarrow \mathrm{(D)}</math>
 +
 
 +
==Solution 2==
 +
 
 +
If <math>f</math> is a linear function, the statement <math>f(1) \leq f(2)</math> states that the slope of the line <math>\frac{f(2) - f(1)}{2 - 1} = f(2) - f(1)</math> is nonnegative:  it is either positive or zero.
 +
 
 +
Similarly, the statement <math>f(3) \geq f(4)</math> states that the slope of the line <math>\frac{f(4) - f(3)}{4 - 3} = f(4) - f(3)</math> 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 <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 (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 <math>0</math>. So <math>f(x)=5</math>. So <math>f(0)=5</math>. Select <math>\boxed{D}</math>.
 +
 
 +
~hastapasta
  
 
==See also==
 
==See also==
Line 12: Line 26:
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 12:26, 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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1995_AHSME_Problems/Problem_12&oldid=173775"