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

(Added the problem and solution)
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(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
 
(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
 
Let <math>y = x^2 + 1</math>. Then <math>x^2 = y - 1</math>, so we can write the given equation as
\begin{align*}
+
<math>\begin{align*}
 
f(y) &= x^4 + 5x^2 + 3 \\
 
f(y) &= x^4 + 5x^2 + 3 \\
 
&= (x^2)^2 + 5x^2 + 3 \\
 
&= (x^2)^2 + 5x^2 + 3 \\
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&= y^2 - 2y + 1 + 5y - 5 + 3 \\
 
&= y^2 - 2y + 1 + 5y - 5 + 3 \\
 
&= y^2 + 3y - 1.
 
&= y^2 + 3y - 1.
\end{align*}
+
\end{align*}</math>
 
Then substituting <math>x^2 - 1</math>, we get
 
Then substituting <math>x^2 - 1</math>, we get
 
\begin{align*}
 
\begin{align*}

Revision as of 14:20, 23 October 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 $\begin{align*} f(y) &= x^4 + 5x^2 + 3 \\ &= (x^2)^2 + 5x^2 + 3 \\ &= (y - 1)^2 + 5(y - 1) + 3 \\ &= y^2 - 2y + 1 + 5y - 5 + 3 \\ &= y^2 + 3y - 1. \end{align*}$ (Error compiling LaTeX. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.) 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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