Difference between revisions of "1995 AHSME Problems/Problem 14"
Talkinaway (talk | contribs) (→Solution) |
Talkinaway (talk | contribs) |
||
| Line 4: | Line 4: | ||
<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</math> | <math>f(-x) = a(-x)^4 - b(-x)^2 - x + 5</math> | ||
| Line 14: | Line 14: | ||
Thus <math>f(3) = f(-3)-2(-3) = 8 \Rightarrow \mathrm{(E)}</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> | ||
== See also == | == See also == | ||
Revision as of 16:41, 18 August 2011
Contents
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 
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 | ||
