Difference between revisions of "1995 AHSME Problems/Problem 14"
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<math> \mathrm{(A) \ -5 } \qquad \mathrm{(B) \ -2 } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 8 } </math> | <math> \mathrm{(A) \ -5 } \qquad \mathrm{(B) \ -2 } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 8 } </math> | ||
| − | == Solution == | + | == Solution 1== |
| − | <math>f(-x) = a(-x)^4 - b(-x)^2 - x + 5 = ax^4 - bx^2 - x + 5 = f(x) - 2x</math>. Thus <math>f(3) = f(-3)-2(-3) = 8 \Rightarrow \mathrm{(E)}</math>. | + | <math>f(-x) = a(-x)^4 - b(-x)^2 - x + 5</math> |
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| + | <math>f(-x) = ax^4 - bx^2 - x + 5</math> | ||
| + | |||
| + | <math>f(-x) = (ax^4 - bx^2 + x + 5) - 2x</math> | ||
| + | |||
| + | <math>f(-x) = f(x) - 2x</math>. | ||
| + | |||
| + | Thus <math>f(3) = f(-3)-2(-3) = 8 \Rightarrow \mathrm{(E)}</math>. | ||
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| + | ==Solution 2== | ||
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| + | If <math>f(-3) = 2</math>, then <math>81a - 9b + -3 + 5 = 2</math>. Simplifying, we get <math>81a - 9b = 0</math>. | ||
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| + | Getting an expression for <math>f(3)</math>, we find <math>f(3) = 81a - 9b + 3 + 5</math>. Since the first two terms sum up to zero, we get <math>f(3) = 8</math>, which is answer <math>\mathrm{(E)}</math> | ||
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| + | ==Solution 3== | ||
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| + | Substituting <math>x = -3</math>, we get | ||
| + | <cmath>f(-3) = 81a - 9b - 3 + 5 = 81a - 9b + 2.</cmath> | ||
| + | But <math>f(-3) = 2</math>, so <math>81a - 9b + 2 = 2</math>, which means <math>81a - 9b = 0</math>. Then | ||
| + | <cmath>f(3) = 81a - 9b + 3 + 5 = 0 + 3 + 5 = \boxed{8}.</cmath> | ||
== See also == | == See also == | ||
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[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 17:05, 6 April 2016
Problem
If 
and 
, then 
Solution 1
.
Thus 
.
Solution 2
If 
, then 
. Simplifying, we get 
.
Getting an expression for 
, we find 
. Since the first two terms sum up to zero, we get 
, which is answer 
Solution 3
Substituting 
, we get
![\[f(-3) = 81a - 9b - 3 + 5 = 81a - 9b + 2.\]](http://latex.artofproblemsolving.com/9/4/6/94649732be3c30ed54cd54f237243708f0d46560.png)
But , so 
, which means . Then
![\[f(3) = 81a - 9b + 3 + 5 = 0 + 3 + 5 = \boxed{8}.\]](http://latex.artofproblemsolving.com/d/5/f/d5fb2743b8131ca65b52f588d6580eb2969e44e3.png)
See also
| 1995 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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. 
