Difference between revisions of "1983 AHSME Problems/Problem 18"
Sevenoptimus (talk | contribs) m (Added box at the bottom) |
Mathboy282 (talk | contribs) (solution 2) |
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&= x^4 + x^2 - 3.\end{align*}</cmath> | &= x^4 + x^2 - 3.\end{align*}</cmath> | ||
The answer is therefore <math>\boxed{\textbf{(B)}}</math>. | The answer is therefore <math>\boxed{\textbf{(B)}}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Let <math>y=x^2.</math> We have that | ||
+ | |||
+ | <cmath>\begin{align*}f(x^2 + 1) &= x^4 + 5x^2 + 3 \\ | ||
+ | &= y^2 + 5y + 3 \\ | ||
+ | &= y^2 + 5y + 4 - 1 \\ | ||
+ | &= (y+1)(y+4) -1 \\ | ||
+ | &= (x^2 + 1)(x^2 + 4) - 1\end{align*}</cmath> | ||
+ | |||
+ | Thus, we have <math>f(n) = n(n+3)-1.</math> | ||
+ | |||
+ | If we plug in <math>n = x^2-1</math> we have | ||
+ | |||
+ | <cmath>\begin{align*}f(x^2 - 1) &= (x^2 -1)(x^2 + 2)-1 \\ | ||
+ | &= (x^4 + x^2 - 2) - 1 \\ | ||
+ | &= \boxed{x^4 + x^2 -3} \end{align*}</cmath> | ||
==See Also== | ==See Also== |
Latest revision as of 22:57, 22 September 2021
Contents
Problem
Let be a polynomial function such that, for all real , . For all real is
Solution
Let . Then , so we can write the given equation as Then substituting for , we get The answer is therefore .
Solution 2
Let We have that
Thus, we have
If we plug in we have
See Also
1983 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 17 |
Followed by Problem 19 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.