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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

f(-x) = a(-x)^4 - b(-x)^2 - x + 5 = ax^4 - bx^2 - x + 5 = f(x) - 2x. Thus f(3) = f(-3)-2(-3) = 8 \Rightarrow \mathrm{(E)}.

See also

1995 AHSME (Problems)
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
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