Difference between revisions of "1983 AHSME Problems/Problem 18"

Revision as of 20:17, 26 September 2016

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

Solution: (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 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 $x^2 - 1$ , 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 \\ &= \boxed{x^4 + x^2 - 3}. \end{align*} The answer is (B).

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Difference between revisions of "1983 AHSME Problems/Problem 18"

Revision as of 20:17, 26 September 2016

@@ Line 1: / Line 1: @@
-...
+Problem:
+Let <math>f</math> be a polynomial function such that, for all real <math>x</math>,
+<cmath>f(x^2 + 1) = x^4 + 5x^2 + 3.</cmath>
+For all real <math>x</math>, <math>f(x^2 - 1)</math> is
+Solution:
+(A) <math>x^4 + 5x^2 + 1</math> (B) <math>x^4 + x^2 - 3</math> (C) <math>x^4 - 5x^2 + 1</math> (D) <math>x^4 + x^2 + 3</math> (E) none of these
+Let <math>y = x^2 + 1</math>. Then <math>x^2 = y - 1</math>, 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 <math>x^2 - 1</math>, 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 \\
+&= \boxed{x^4 + x^2 - 3}.
+\end{align*}
+The answer is (B).