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

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

Revision as of 00:55, 20 February 2019

Problem

Let $f$ be a polynomial function such that, for all real $x$, $f(x^2 + 1) = x^4 + 5x^2 + 3$. For all real $x, f(x^2-1)$ is

$\textbf{(A)}\ x^4+5x^2+1\qquad \textbf{(B)}\ x^4+x^2-3\qquad \textbf{(C)}\ x^4-5x^2+1\qquad \textbf{(D)}\ x^4+x^2+3\qquad \textbf{(E)}\ \text{none of these}$

Solution

Let $y = x^2 + 1$. Then $x^2 = y - 1$, 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 $x^2 - 1$ for $y$, 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 \\ &= x^4 + x^2 - 3.\end{align*} The answer is therefore $\boxed{\textbf{(B)}}$.

See Also

1983 AHSME (ProblemsAnswer KeyResources)
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


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