1983 AHSME Problems/Problem 18

Problem: Let $f$ be a polynomial function such that, for all real $x$ , $f(x^2 + 1) = x^4 + 5x^2 + 3.$ For all real $x$ , $f(x^2 - 1)$ is

(A) $x^4 + 5x^2 + 1$ (B) $x^4 + x^2 - 3$ (C) $x^4 - 5x^2 + 1$ (D) $x^4 + x^2 + 3$ (E) none of these

Solution:

Let $y = x^2 + 1$ . Then $x^2 = y - 1$ , so we can write the given equation as $\begin{align*}f(y) &= x^4 + 5x^2 + 3 \\ &= (x^2)^2 + 5x^2 + 3 \\ &= (y - 1)^2 + 5(y - 1) + 3 \\ &= y^2 - 2y + 1 + 5y - 5 + 3 \\ &= y^2 + 3y - 1.\end{align*}$ Then substituting $x^2 - 1$ for $y$ , we get $\begin{align*}f(x^2 - 1) &= (x^2 - 1)^2 + 3(x^2 - 1) - 1 \\ &= x^4 - 2x^2 + 1 + 3x^2 - 3 - 1 \\ &= x^4 + x^2 - 3.\end{align*}$ The answer is therefore $\boxed{\textbf{(B)}}$ .

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1983 AHSME Problems/Problem 18