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

Made in 2016 (talk | contribs) (Added the problem and solution) |
|||

Line 7: | Line 7: | ||

(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 | ||

− | \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 \\ | ||

Line 13: | Line 13: | ||

&= 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*} | \begin{align*} |

## Revision as of 14:20, 23 October 2016

Problem: Let be a polynomial function such that, for all real , For all real , is

Solution: (A) (B) (C) (D) (E) none of these 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. ! Package amsmath Error: \begin{align*} allowed only in paragraph mode.) 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).