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> |
+ | |||
+ | <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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | If <math>f(-3) = 2</math>, then <math>81a - 9b + -3 + 5 = 2</math>. Simplifying, we get <math>81a - 9b = 0</math>. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | 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 But , so , which means . Then
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 |
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