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2004 AMC 12B Problems/Problem 13 - Revision history
2024-03-29T13:46:22Z
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Chikafujiwara: /* Solution */
2021-03-02T17:56:09Z
<p><span dir="auto"><span class="autocomment">Solution</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 17:56, 2 March 2021</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l7" >Line 7:</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\qquad\mathrm{(D)}\ 1</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\qquad\mathrm{(D)}\ 1</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\qquad\mathrm{(E)}\ 2</math></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\qquad\mathrm{(E)}\ 2</math></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div>== Solution ==</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div>== Solution <ins class="diffchange diffchange-inline">(Alcumus)</ins>==</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Since <math>f(f^{-1}(x))=x</math>, it follows that <math>a(bx+a)+b=x</math>, which implies <math>abx + a^2 +b = x</math>. This equation holds for all values of <math>x</math> only if <math>ab=1</math> and <math>a^2+b=0</math>.</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>Since <math>f(f^{-1}(x))=x</math>, it follows that <math>a(bx+a)+b=x</math>, which implies <math>abx + a^2 +b = x</math>. This equation holds for all values of <math>x</math> only if <math>ab=1</math> and <math>a^2+b=0</math>.</div></td></tr>
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Chikafujiwara
https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_13&diff=129699&oldid=prev
Jalenluorion: /* Solution */
2020-07-29T02:34:08Z
<p><span dir="auto"><span class="autocomment">Solution</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 02:34, 29 July 2020</td>
</tr><tr><td colspan="2" class="diff-lineno" id="mw-diff-left-l8" >Line 8:</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\qquad\mathrm{(E)}\ 2</math></div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>\qquad\mathrm{(E)}\ 2</math></div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Solution ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== Solution ==</div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">By the definition of an [[inverse function]], </del><math><del class="diffchange diffchange-inline">x = </del>f(f^{-1}(x)) = a(bx+a)+b = abx + a^2 + b</math>. <del class="diffchange diffchange-inline">By comparing coefficients, we have </del><math>ab = 1 <del class="diffchange diffchange-inline">\Longrightarrow b = \frac 1a</del></math> and <math>a^2 + b = a^2 <del class="diffchange diffchange-inline">+ \frac{</del>1<del class="diffchange diffchange-inline">}{</del>a<del class="diffchange diffchange-inline">} </del>= <del class="diffchange diffchange-inline"> 0</del></math>. <del class="diffchange diffchange-inline">Simplifying</del>,</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">Since </ins><math>f(f^{-1}(x))=<ins class="diffchange diffchange-inline">x</math>, it follows that <math></ins>a(bx+a)+b=<ins class="diffchange diffchange-inline">x</math>, which implies <math></ins>abx + a^2 +b <ins class="diffchange diffchange-inline">= x</ins></math>. <ins class="diffchange diffchange-inline">This equation holds for all values of <math>x</math> only if </ins><math>ab=1</math> and <math>a^2+b=<ins class="diffchange diffchange-inline">0</math>.</ins></div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><cmath><del class="diffchange diffchange-inline">a</del>^<del class="diffchange diffchange-inline">3 </del>+ 1 = <del class="diffchange diffchange-inline">0</del></cmath></div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div> </div></td></tr>
<tr><td class='diff-marker'>−</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #ffe49c; vertical-align: top; white-space: pre-wrap;"><div><del class="diffchange diffchange-inline">and </del><<del class="diffchange diffchange-inline">math</del>><del class="diffchange diffchange-inline">a </del>= <del class="diffchange diffchange-inline">b </del>= -1</<del class="diffchange diffchange-inline">math</del>><del class="diffchange diffchange-inline">. Thus </del><math>a+b = -2 <del class="diffchange diffchange-inline">\Rightarrow \mathrm{(A)</del>}</math>.</div></td><td class='diff-marker'>+</td><td style="color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #a3d3ff; vertical-align: top; white-space: pre-wrap;"><div><ins class="diffchange diffchange-inline">Then <math>b = -</ins>a^2<ins class="diffchange diffchange-inline"></math>. Substituting into the equation <math>ab = </ins>1<ins class="diffchange diffchange-inline"></math>, we get <math>-</ins>a<ins class="diffchange diffchange-inline">^3 </ins>= <ins class="diffchange diffchange-inline">1</ins></math>. <ins class="diffchange diffchange-inline">Then <math>a = -1</math></ins>, <ins class="diffchange diffchange-inline">so <math>b = -1</math>, and<cmath>f(x)=-x-1.</cmath>Likewise<cmath>f^{-1}(x)=-x-1.</cmath>These are inverses to one another since</ins><cmath><ins class="diffchange diffchange-inline">f(f</ins>^<ins class="diffchange diffchange-inline">{-1}(x))=-(-x-1)-1=x</ins>+<ins class="diffchange diffchange-inline">1-</ins>1=<ins class="diffchange diffchange-inline">x.</ins></cmath><<ins class="diffchange diffchange-inline">cmath</ins>><ins class="diffchange diffchange-inline">f^{-1}(f(x))</ins>=<ins class="diffchange diffchange-inline">-(-x-1)-1</ins>=<ins class="diffchange diffchange-inline">x+1</ins>-1<ins class="diffchange diffchange-inline">=x.</ins></<ins class="diffchange diffchange-inline">cmath</ins>><ins class="diffchange diffchange-inline">Therefore </ins><math>a+b=<ins class="diffchange diffchange-inline">\boxed{</ins>-2}</math>.</div></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"></td></tr>
<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>== See also ==</div></td></tr>
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Jalenluorion
https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_13&diff=53508&oldid=prev
Nathan wailes: /* See also */
2013-07-03T23:57:38Z
<p><span dir="auto"><span class="autocomment">See also</span></span></p>
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<td colspan="2" style="background-color: #fff; color: #222; text-align: center;">Revision as of 23:57, 3 July 2013</td>
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<tr><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>[[Category:Introductory Algebra Problems]]</div></td><td class='diff-marker'> </td><td style="background-color: #f8f9fa; color: #222; font-size: 88%; border-style: solid; border-width: 1px 1px 1px 4px; border-radius: 0.33em; border-color: #eaecf0; vertical-align: top; white-space: pre-wrap;"><div>[[Category:Introductory Algebra Problems]]</div></td></tr>
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Nathan wailes
https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_12B_Problems/Problem_13&diff=23141&oldid=prev
Azjps: solution
2008-02-10T22:57:59Z
<p>solution</p>
<p><b>New page</b></p><div>== Problem ==<br />
If <math>f(x) = ax+b</math> and <math>f^{-1}(x) = bx+a</math> with <math>a</math> and <math>b</math> real, what is the value of <math>a+b</math>? <br />
<br />
<math>\mathrm{(A)}\ -2<br />
\qquad\mathrm{(B)}\ -1<br />
\qquad\mathrm{(C)}\ 0<br />
\qquad\mathrm{(D)}\ 1<br />
\qquad\mathrm{(E)}\ 2</math><br />
== Solution ==<br />
By the definition of an [[inverse function]], <math>x = f(f^{-1}(x)) = a(bx+a)+b = abx + a^2 + b</math>. By comparing coefficients, we have <math>ab = 1 \Longrightarrow b = \frac 1a</math> and <math>a^2 + b = a^2 + \frac{1}{a} = 0</math>. Simplifying,<br />
<cmath>a^3 + 1 = 0</cmath><br />
and <math>a = b = -1</math>. Thus <math>a+b = -2 \Rightarrow \mathrm{(A)}</math>.<br />
<br />
== See also ==<br />
{{AMC12 box|year=2004|ab=B|num-b=12|num-a=14}}<br />
<br />
[[Category:Introductory Algebra Problems]]</div>
Azjps