1995 AHSME Problems/Problem 14

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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 \[f(3) = 81a - 9b + 3 + 5 = 0 + 3 + 5 = \boxed{8}.\]

See also

1995 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
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