exam_template.tex

%
%  untitled
%
%  Created by Hans Sigrist on 2010-04-18. mail: 
%  Copyright (c) 2010 . All rights reserved.
%

\documentclass[%
				usepdftitle=false,%
				11pt,%
				a4paper,%
				addpoints,%
				%answers%
				]{exam}
\usepackage{hyperref}
\usepackage{listings}
\usepackage[spanish]{babel}
\usepackage[x11names]{xcolor}
\usepackage{graphicx}
\usepackage{fourier}
\usepackage[latin1]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{cclicenses}
\usepackage{pgf,tikz}
\usepackage{pstricks-add}
\usetikzlibrary{%
     decorations.fractals%
	,decorations.pathmorphing%
	,shadows%
	,mindmap%
	,calc%
	,through%
	,backgrounds%
	,arrows%
	,snakes%
	,fit%
	}

\def\NN{\mathbb{N}}
\def\RR{\mathbb{R}}
\def\ZZ{\mathbb{Z}}
\def{\displaystyle}
\def\bex{\begin{ex}}
\def\eex{\end{ex}}
\def\bsol{\begin{sol}}
\def\esol{\end{sol}}
\def\lp{\left(}
\def\rp{\right)}
\def\lpc{\left[}
\def\rpc{\right]}

\hypersetup{colorlinks=false,
pdftitle={Unidad 1: Ores},
pdfauthor={Sigrist, Hans},
pdfsubject={Lipsum Lorun},
pdfkeywords={lorem, ipsum}
}

\everymath{\displaystyle}


\usepackage{ifpdf}
\ifpdf
\usepackage{subfigure}
\usepackage[pdftex]{graphicx}
\else
\usepackage{graphicx}
\fi



\pagestyle{headandfoot} 
\firstpageheader{\large\bfseries Cálculo I\\
					Ingeniería Comercial(PCE)\\
					Prueba 1, Abril 4, 2010} 
					{}
					{\includegraphics[width=1in]{logouacb.pdf}\\ \\
					\large\bfseries Nombre:\enspace\makebox[2in]{\hrulefill}} \runningheader{\large\bfseries Cálculo I\\
				Primera Prueba, July 4, 1776}{}{} 
\firstpagefooter{}{}{}
\runningfooter{}{\thepage}{}



\begin{tikzpicture}[overlay] 
	\node [xshift=-2cm,yshift=-24cm] at (current page.south west)[text width=0.5cm,fill=white,rounded corners,above right]{\footnotesize{\rotatebox{90}{\cc\ccby\ccsa 2010 \scalebox{1.5}{ $\infty$ } hsigrist@gmail.com}}};
\end{tikzpicture}

\begin{document}

\begin{center} 
  \fbox{\fbox{\parbox{5.5in}{\centering Responda las preguntas en el espacio correspondiente. Justifique todas sus respuestas. No se corregirán respuestas sin justificación. Si le falta espacio continúe en el reverso.  Hay \numquestions\  preguntas con un total de \numpoints\ puntos.}}}
\end{center} 



\addpoints
\renewcommand{\solutiontitle}{\noindent\textbf{Solución:}\enspace}
\pointname{ puntos}
\pointpoints{ punto}{ puntos}
%\bracketedpoints
\boxedpoints


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%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\begin{questions}

\question Calcula el valor de las siguiente expresiones:
\begin{parts}
	\part[4] $\frac{ab}{a^{-1}+b^{-1}}$, si $a=2^{-1}$ y $b=0.5^{-1}$.
	\answerline[$2/5$]
	\begin{solution}[0.5in] $a=\frac{1}{2}, b=2\Rightarrow\frac{ab}{a^{-1}+b^{-1}}=\frac{\frac{1}{2}\cdot2}{2+\frac{1}{2}}=\frac{1}{\frac{5}{2}}=\frac{2}{5}.$
	\end{solution}
	
	\part[6] $\frac{a^{2}}{b}-\frac{a}{b^{2}}$, si $a=3^{-1}$ y $b=-1^{2}$.
	\answerline[$-2/3$]
	\begin{solution}[0.7in] $a=\frac{1}{3},b=-1\Rightarrow\frac{a^{2}}{b}-\frac{a}{b^{2}}= \frac{\lp\frac{1}{3}\rp^{2}}{-1}-\frac{\frac{1}{3}}{(-1)^{2}}=\frac{\frac{1}{9}}{-1}-\frac{\frac{1}{3}}{1}=-\frac{1}{9}-\frac{1}{3}=-\frac{1-3}{9}=-\frac{-2}{3}.$
	\end{solution}
\end{parts}





\question[15] Si $(0,01)^{x-5}=100$, entonces el valor de $x$ es
\answerline[$x=4$]
\begin{solution}[1in] $\lp\frac{1}{100}\rp^{x-5}=100\Rightarrow\lp10^{-2}\rp^{x-5}=100\Rightarrow10^{-2\cdot(x-5)}=100\Rightarrow10^{-2x+10}=10^{2}\\ \Rightarrow10^{-2x}\cdot10^{10}=10^{2}\Rightarrow10^{-2x}=\frac{10^{2}}{10^{10}}\Rightarrow10^{-2x}=10^{-8}\Rightarrow-2x=-8\Rightarrow x=4.$
\end{solution}





\question[10]  Simplifique $a^{m}\left(a^{2m}+a^{3-m}\right)$
\answerline[$a^{3m}+a^{3}$]
\begin{solution}[0.3in] $a^{m}\left(a^{2m}+a^{3-m}\right)=a^{m}\cdot a^{2m}+a^{m}\cdot a^{3-m}=a^{3m}+a^{m+3-m}=a^{3m}+a^{3}.$
\end{solution}






\question Calcule:
\begin{parts}
	\part[15] $\sqrt{3\sqrt{3}+\sqrt{2}}\cdot\sqrt{3\sqrt{3}-\sqrt{2}}=$
\answerline[5]
\begin{solution}[1in] $\sqrt{3\sqrt{3}+\sqrt{2}}\cdot\sqrt{3\sqrt{3}-\sqrt{2}}=\sqrt{\lp3\sqrt{3}+\sqrt{2}\rp\cdot\lp3\sqrt{3}-\sqrt{2}\rp}=\sqrt{3^{2}\cdot (\sqrt{3})^{2}-(\sqrt{2})^{2}}=\sqrt{9\cdot3-2}=\sqrt{27-2}=\sqrt{25}=5$.
\end{solution}
	
	\part[10] $\frac{\sqrt[6]{16}}{\sqrt{2\sqrt[3]{2}}}=$
	\answerline[1]
	\begin{solution}[0.3in] $\frac{\sqrt[6]{16}}{\sqrt{2\sqrt[3]{2}}}=\frac{\sqrt[6]{16}}{\sqrt{\sqrt[3]{2^{3}\cdot 2}}}=\frac{\sqrt[6]{16}}{\sqrt[6]{16}}=1$
	\end{solution}
\end{parts}


\end{questions}
\end{document}