1995 AHSME Problems/Problem 14

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
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.

Page

Toolbox

Search

1995 AHSME Problems/Problem 14

Contents

Problem

Solution 1

Solution 2

Solution 3

See also