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

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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
<math>\begin{align*}
+
<math>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 \\
 
&= (y - 1)^2 + 5(y - 1) + 3 \\
 
&= (y - 1)^2 + 5(y - 1) + 3 \\
 
&= y^2 - 2y + 1 + 5y - 5 + 3 \\
 
&= y^2 - 2y + 1 + 5y - 5 + 3 \\
&= y^2 + 3y - 1.
+
&= y^2 + 3y - 1.</math>
\end{align*}</math>
 
 
Then substituting <math>x^2 - 1</math>, we get
 
Then substituting <math>x^2 - 1</math>, we get
<math>\begin{align*}
+
<math>f(x^2 - 1) &= (x^2 - 1)^2 + 3(x^2 - 1) - 1 \\
f(x^2 - 1) &= (x^2 - 1)^2 + 3(x^2 - 1) - 1 \\
 
 
&= x^4 - 2x^2 + 1 + 3x^2 - 3 - 1 \\
 
&= x^4 - 2x^2 + 1 + 3x^2 - 3 - 1 \\
&= \boxed{x^4 + x^2 - 3}.
+
&= \boxed{x^4 + x^2 - 3}.</math>
\end{align*}</math>
 
 
The answer is (B).
 
The answer is (B).

Revision as of 14:56, 1 July 2017

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 $f(y) &= x^4 + 5x^2 + 3 \\ &= (x^2)^2 + 5x^2 + 3 \\ &= (y - 1)^2 + 5(y - 1) + 3 \\ &= y^2 - 2y + 1 + 5y - 5 + 3 \\ &= y^2 + 3y - 1.$ (Error compiling LaTeX. ! Misplaced alignment tab character &.) Then substituting $x^2 - 1$, we get $f(x^2 - 1) &= (x^2 - 1)^2 + 3(x^2 - 1) - 1 \\ &= x^4 - 2x^2 + 1 + 3x^2 - 3 - 1 \\ &= \boxed{x^4 + x^2 - 3}.$ (Error compiling LaTeX. ! Misplaced alignment tab character &.) The answer is (B).

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