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

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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
<math>\begin{align*}
+
\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*}</math>
+
\end{align*}
 
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 13:20, 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 f(y)=x4+5x2+3=(x2)2+5x2+3=(y1)2+5(y1)+3=y22y+1+5y5+3=y2+3y1. Then substituting $x^2 - 1$, we get f(x21)=(x21)2+3(x21)1=x42x2+1+3x231=x4+x23. The answer is (B).