Difference between revisions of "1995 AHSME Problems/Problem 12"
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<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 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== | ||
− | {{ | + | {{AHSME box|year=1995|num-b=11|num-a=13}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 12:26, 4 May 2022
Problem
Let be a linear function with the properties that and . Which of the following is true?
Solution 1
A linear function has the property that either for all , or for all . Since , . Since , . And if for , then is a constant function. Since ,
Solution 2
If is a linear function, the statement states that the slope of the line is nonnegative: it is either positive or zero.
Similarly, the statement states that the slope of the line 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 tells us that the value of the constant is , and thus that . This leads to
Solution 3 (Common Sense)
It should be very clear that and is contradictory because of the fact that linear functions are monotonic. The only thing that makes sense is , and . This means that has a slope of . So . So . Select .
~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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