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

Sevenoptimus (talk | contribs) (Fixed LaTeX and formatting) |
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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 |

− | < | + | <cmath>\begin{align*}f(y) &= x^4 + 5x^2 + 3 \\ |

&= (x^2)^2 + 5x^2 + 3 \\ | &= (x^2)^2 + 5x^2 + 3 \\ | ||

&= (y - 1)^2 + 5(y - 1) + 3 \\ | &= (y - 1)^2 + 5(y - 1) + 3 \\ | ||

&= y^2 - 2y + 1 + 5y - 5 + 3 \\ | &= y^2 - 2y + 1 + 5y - 5 + 3 \\ | ||

− | &= y^2 + 3y - 1.</ | + | &= y^2 + 3y - 1.\end{align*}</cmath> |

− | Then substituting <math>x^2 - 1</math>, we get | + | Then substituting <math>x^2 - 1</math> for <math>y</math>, we get |

− | < | + | <cmath>\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 - 2x^2 + 1 + 3x^2 - 3 - 1 \\ | ||

− | &= | + | &= x^4 + x^2 - 3.\end{align*}</cmath> |

− | The answer is (B). | + | The answer is therefore <math>\boxed{\textbf{(B)}}</math>. |

## Revision as of 22:12, 26 January 2019

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 Then substituting for , we get The answer is therefore .