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diff --git a/texmf-dist/doc/latex/eqexam/examples/quiz01.tex b/texmf-dist/doc/latex/eqexam/examples/quiz01.tex new file mode 100644 index 00000000..8682623c --- /dev/null +++ b/texmf-dist/doc/latex/eqexam/examples/quiz01.tex @@ -0,0 +1,126 @@ +\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} |