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+\documentclass{article}
+\usepackage[fleqn]{amsmath}
+\usepackage[pdf,cfg=quiz,forpaper,pointsonleft,
+% compile with exactly one of the following three
+ nosolutions
+% answerkey
+% vspacewithsolns
+]{eqexam}
+
+\examNum{1}\numVersions{2}\forVersion{a}
+\longTitleText
+ {Quiz~\nExam--003}
+ {Quiz~\nExam--007}
+\endlongTitleText
+\shortTitleText
+ {Q{\nExam}s3}
+ {Q{\nExam}s7}
+\endshortTitleText
+
+\title[\sExam]{\bfseries\Exam}
+\author{D. P. Story}
+\subject[C1]{Calculus I}
+\date{Spring \the\year}
+\keywords{Test~\nExam, Section \ifAB{003}{007}}
+\email{dpstory@uakron.edu}
+
+\vspacewithkeyOn
+\solAtEndFormatting{\eqequesitemsep{3pt}}
+\everymath{\displaystyle}
+
+\begin{document}
+
+\maketitle
+
+\begin{exam}{Part1}
+
+\begin{instructions}[Instructions:]
+Solve each of the following problems without error. \textit{Show all
+details.} Box in your $\boxed{\text{answers}}$. Use good notation, you
+\emph{will} be marked off for bad notation. \textbf{Note:} The value of a
+limit can be a number, the symbol $+\infty$, the symbol $-\infty$, or may
+be labeled DNE (for ``does not exist'').
+\end{instructions}
+
+\begin{problem}[4]
+Compute $ \vA{\lim_{x\to-1}\frac{4x^2+x}{x}}\vB{\lim_{x\to2}\frac{1-3x}{x+1}}$
+\begin{solution}[2in]
+As discussed in class, this is a ``Skill Level 0'' limit problem:
+\[
+\begin{verA}
+\lim_{x\to-1}\frac{4x^2+x}{x}
+ = \frac{4(-1)^2+(-1)}{-1}
+ = \boxed{-3}
+\end{verA}
+\begin{verB}
+ \lim_{x\to2}\frac{1-3x}{x+1}
+ \lim_{x\to2}\frac{1-3(2)}{2+1}
+ = \boxed{-\frac{5}{3}}
+\end{verB}
+\]
+\ifkeyalt\adjDisplayBelow\fi
+\end{solution}
+\end{problem}
+
+\begin{problem}[3]
+Define the function $ f(x) = \begin{cases} 2x^3 - 1 & x < -2\\ 2- x^2 & x
+\ge -2\end{cases}$. Compute $\lim_{x\to\vA{-2^-}\vB{-2^+}} f(x) $, show the
+details of your reasoning.
+
+\begin{solution}[2in]
+We use standard techniques:
+\begin{verA}
+\begin{alignat*}{2}
+ \lim_{x\to-2^-} f(x) &
+ = \lim_{x\to-2^-} (2x^3-1) &&\qquad\text{since $ x < -2$}\\&
+ = 2(-2)^3 - 1&&\qquad\text{now a skill level 0 problem}\\&
+ = \boxed{-17}
+\end{alignat*}
+\end{verA}
+\begin{verB}
+\begin{alignat*}{2}
+ \lim_{x\to-2^+} f(x) &
+ = \lim_{x\to-2^+} (2- x^2) &&\qquad\text{since $ x < -2$}\\&
+ = 2 - (-2)^2&&\qquad\text{now a skill level 0 problem}\\&
+ = \boxed{-2}
+\end{alignat*}
+\end{verB}
+\ifkeyalt\adjDisplayBelow\fi
+\end{solution}
+\end{problem}
+
+\begin{problem}[3]
+Compute $\vA{\lim_{x\to2} \frac{1-x}{(x-2)^2}}
+ \vB{\lim_{x\to3} \frac{x-2}{(3-x)^2}}$
+
+\begin{solution}[1in]
+\begin{verA}
+Notice the denominator goes to zero, but the numerator does not;
+this indicates a vertical asymptote usually. Because the
+denominator is squared, it's always positive. When $x$ is
+``close'' to $2$, $1 - x < 0$, that is, when $x$ is ``close'' to
+$2$ the numerator is \emph{negative}. The ratio of the numerator and
+denominator is \emph{negative} when $x$ is ``close'' to $2$. Thus, we
+conclude,
+\[
+ \boxed{\lim_{x\to2} \frac{1-x}{(x-2)^2} = -\infty}
+\]
+\end{verA}
+\begin{verB}
+Notice the denominator goes to zero, but the numerator does not;
+this indicates a vertical asymptote usually. Because the
+denominator is squared, it's always positive. When $x$ is
+``close'' to $3$, $x - 2 > 0$, that is, when $x$ is ``close'' to
+$3$ the numerator is \emph{positive}. The ratio of the numerator and
+denominator is \emph{positive} when $x$ is ``close'' to $3$. Thus, we
+conclude,
+\[
+ \boxed{\lim_{x\to3} \frac{x-2}{(3-x)^2} = +\infty}
+\]
+\end{verB}
+\ifkeyalt\adjDisplayBelow\fi
+\end{solution}
+\end{problem}
+
+\end{exam}
+\end{document}