Difference between revisions of "1995 AHSME Problems/Problem 14"

Latest revision as of 17:05, 6 April 2016

Problem

If $f(x) = ax^4 - bx^2 + x + 5$ and $f( - 3) = 2$ , then $f(3) =$

$\mathrm{(A) \ -5 } \qquad \mathrm{(B) \ -2 } \qquad \mathrm{(C) \ 1 } \qquad \mathrm{(D) \ 3 } \qquad \mathrm{(E) \ 8 }$

Solution 1

$f(-x) = a(-x)^4 - b(-x)^2 - x + 5$

$f(-x) = ax^4 - bx^2 - x + 5$

$f(-x) = (ax^4 - bx^2 + x + 5) - 2x$

$f(-x) = f(x) - 2x$ .

Thus $f(3) = f(-3)-2(-3) = 8 \Rightarrow \mathrm{(E)}$ .

Solution 2

If $f(-3) = 2$ , then $81a - 9b + -3 + 5 = 2$ . Simplifying, we get $81a - 9b = 0$ .

Getting an expression for $f(3)$ , we find $f(3) = 81a - 9b + 3 + 5$ . Since the first two terms sum up to zero, we get $f(3) = 8$ , which is answer $\mathrm{(E)}$

Solution 3

Substituting $x = -3$ , we get $f(-3) = 81a - 9b - 3 + 5 = 81a - 9b + 2.$ But $f(-3) = 2$ , so $81a - 9b + 2 = 2$ , which means $81a - 9b = 0$ . Then $f(3) = 81a - 9b + 3 + 5 = 0 + 3 + 5 = \boxed{8}.$

1995 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 20: / Line 20: @@
 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>
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