1983 AHSME Problems/Problem 18: Difference between revisions
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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 | ||
\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*} | <math>\begin{align*} | ||
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}. | ||
\end{align*} | \end{align*}</math> | ||
The answer is (B). | The answer is (B). | ||
Revision as of 13:55, 1 July 2017
Problem:
Let 
be a polynomial function such that, for all real 
,
![\[f(x^2 + 1) = x^4 + 5x^2 + 3.\]](http://latex.artofproblemsolving.com/c/8/f/c8f29164293f89007f92f723a7e33eec4bb44d3f.png)
For all real , 
is
(A) 
(B) 
(C) 
(D) 
(E) none of these
Solution:
Let 
. Then 
, 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. Unknown error_msg)
Then substituting 
, we get
$\begin{align*}
f(x^2 - 1) &= (x^2 - 1)^2 + 3(x^2 - 1) - 1 \\
&= x^4 - 2x^2 + 1 + 3x^2 - 3 - 1 \\
&= \boxed{x^4 + x^2 - 3}.
\end{align*}$ (Error compiling LaTeX. Unknown error_msg)
The answer is (B).
