Difference between revisions of "1983 AHSME Problems/Problem 18"
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| − | Problem | + | ==Problem== |
| + | |||
Let <math>f</math> be a polynomial function such that, for all real <math>x</math>, | Let <math>f</math> be a polynomial function such that, for all real <math>x</math>, | ||
| − | < | + | <math>f(x^2 + 1) = x^4 + 5x^2 + 3</math>. |
| − | For all real <math>x</math> | + | For all real <math>x, f(x^2-1)</math> is |
| + | |||
| + | <math>\textbf{(A)}\ x^4+5x^2+1\qquad | ||
| + | \textbf{(B)}\ x^4+x^2-3\qquad | ||
| + | \textbf{(C)}\ x^4-5x^2+1\qquad | ||
| + | \textbf{(D)}\ x^4+x^2+3\qquad | ||
| + | \textbf{(E)}\ \text{none of these} </math> | ||
| + | |||
| + | ==Solution== | ||
| − | + | 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 \\ | |
| − | Let <math>y = x^2 + 1</math>. Then <math>x^2 = y - 1</math>, so we can write the given equation as | ||
| − | \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> |
| − | \end{align*} | + | Then substituting <math>x^2 - 1</math> for <math>y</math>, we get |
| − | Then substituting <math>x^2 - 1</math>, we get | + | <cmath>\begin{align*}f(x^2 - 1) &= (x^2 - 1)^2 + 3(x^2 - 1) - 1 \\ |
| − | \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> |
| − | \end{align*} | + | The answer is therefore <math>\boxed{\textbf{(B)}}</math>. |
| − | The answer is (B). | + | |
| + | ==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== | ||
| + | {{AHSME box|year=1983|num-b=17|num-a=19}} | ||
| + | |||
| + | {{MAA Notice}} | ||
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. 
