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

(Fixed LaTeX and formatting)
m (Fixed formatting)
Line 1: Line 1:
Problem:
+
==Problem==
 +
 
 
Let <math>f</math> be a polynomial function such that, for all real <math>x</math>,
 
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>
+
<math>f(x^2 + 1) = x^4 + 5x^2 + 3</math>.
For all real <math>x</math>, <math>f(x^2 - 1)</math> is
+
For all real <math>x, f(x^2-1)</math> is
  
(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
+
<math>\textbf{(A)}\ x^4+5x^2+1\qquad
 +
\textbf{(B)}\ x^4+x^2-3\qquad
 +
\textbf{(C)}\ x^4-5x^2+1\qquad
 +
\textbf{(D)}\ x^4+x^2+3\qquad
 +
\textbf{(E)}\ \text{none of these}  </math>
  
Solution:
+
==Solution==
  
 
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  

Revision as of 15:53, 27 January 2019

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

$\textbf{(A)}\ x^4+5x^2+1\qquad \textbf{(B)}\ x^4+x^2-3\qquad \textbf{(C)}\ x^4-5x^2+1\qquad \textbf{(D)}\ x^4+x^2+3\qquad \textbf{(E)}\ \text{none of these}$

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 $\boxed{\textbf{(B)}}$.