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

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<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 14:55, 1 July 2017

Problem: Let be a polynomial function such that, for all real , 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).