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

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Problem:
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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>,
 
Let <math>f</math> be a polynomial function such that, for all real <math>x</math>,
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<cmath>f(x^2 + 1) = x^4 + 5x^2 + 3.</cmath>
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<math>f(x^2 + 1) = x^4 + 5x^2 + 3</math>.
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For all real <math>x</math>, <math>f(x^2 - 1)</math> is
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For all real <math>x, f(x^2-1)</math> is
  
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(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
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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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==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  
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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>.
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==Solution 2==
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Let <math>y=x^2.</math> We have that
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<cmath>\begin{align*}f(x^2 + 1) &= x^4 + 5x^2 + 3 \\
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&= y^2 + 5y + 3 \\
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&= y^2 + 5y + 4 - 1 \\
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&= (y+1)(y+4) -1 \\
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&= (x^2 + 1)(x^2 + 4) - 1\end{align*}</cmath>
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Thus, we have <math>f(n) = n(n+3)-1.</math>
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If we plug in <math>n = x^2-1</math> we have
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<cmath>\begin{align*}f(x^2 - 1) &= (x^2 -1)(x^2 + 2)-1 \\
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&= (x^4 + x^2 - 2) - 1 \\
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&= \boxed{x^4 + x^2 -3} \end{align*}</cmath>
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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}}

Latest revision as of 22:57, 22 September 2021

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 $\boxed{\textbf{(B)}}$.

Solution 2

Let $y=x^2.$ We have that

\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*}

Thus, we have $f(n) = n(n+3)-1.$

If we plug in $n = x^2-1$ we have

\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*}

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


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