Difference between revisions of "1983 AHSME Problems/Problem 18"
| Line 3: | Line 3: | ||
<cmath>f(x^2 + 1) = x^4 + 5x^2 + 3.</cmath> | <cmath>f(x^2 + 1) = x^4 + 5x^2 + 3.</cmath> | ||
For all real <math>x</math>, <math>f(x^2 - 1)</math> is | For all real <math>x</math>, <math>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 | ||
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 | ||
\begin{align*} | \begin{align*} | ||
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*}
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*}
The answer is (B).
