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diff --git a/texmf-dist/doc/latex/eqexam/examples/quiz02.tex b/texmf-dist/doc/latex/eqexam/examples/quiz02.tex new file mode 100644 index 00000000..091d2030 --- /dev/null +++ b/texmf-dist/doc/latex/eqexam/examples/quiz02.tex @@ -0,0 +1,146 @@ +\documentclass{article} +\usepackage[fleqn]{amsmath} +\usepackage[pdf,cfg=quiz,forpaper,pointsonleft, +% compile with exactly one of the following three + nosolutions +% answerkey +% vspacewithsolns +]{eqexam} + +\examNum{2}\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 \vA{003}\vB{007}} +\email{dpstory@uakron.edu} + +\vspacewithkeyOn +\solAtEndFormatting{\eqequesitemsep{3pt}} +\everymath{\displaystyle} + +\begin{document} + +\maketitle + +\begin{exam}{qz02} + +\begin{instructions}[Global 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. +\end{instructions} + +\begin{problem}[3] +Identify all numbers $x$ at which the function $ f(x) = \frac{x+2}{\sqrt{\vA{x-1}\vB{2-x}}} $ is continuous. + +\begin{solution}[.75in] +We require $ \vA{x - 1}\vB{2-x} >0 $ or $ \vA{x > 1}\vB{x<2} $. In +interval notation, the set of all numbers at which $f$ is continuous is +$\boxed{\vA{( 1, \infty )}\vB{(-\infty, 2)} }$. +\end{solution} +\end{problem} + +\begin{problem}[3] +Given $ f(x) = \begin{cases} + 3x^2 - 2x & x < -1 \\ + 6x^2 + x \vB{+1} & x \ge -1 +\end{cases}$. Is this function (a) continuous at $ x = -1 $;, (b)~discontinuous with a removable discontinuity +at $ x = -1 $; or (c)~discontinuous with a jump discontinuity at $ x = -1 $? Justify your response. + +\begin{solution}[2in] +Look at the left and right limits: +\begin{align*} + \lim_{x\to-1^-}f(x) &= \lim_{x\to-1^-} 3x^2 - 2x = 5\\ + \lim_{x\to-1^+}f(x) &= \lim_{x\to-1^+} 6x^2 + x \vB{+1} = \vA{5}\vB{6} \vA{=}\vB{\neq} f(-1) +\end{align*} +Thus, $\lim_{x\to-1^-}f(x) \vA{=}\vB{\neq} +\lim_{x\to-1^+}f(x)\vA{=f(-1)}$. The two sided limit \vA{exists}\vB{does +not exist}\vA{ and $\lim_{x\to-1}f(x)=f(-1)$}. This function \vA{is}\vB{is +not} continuous at $x=-1$, \vB{it has a jump discontinuity, since +$\lim_{x\to-1^-}f(x) \neq \lim_{x\to-1^+}f(x)$}; as a result, the answer +is \vA{(a)}\vB{(c)}. +\end{solution} +\end{problem} + +\begin{problem}[4] +Define the function $ f(x) = 3x^2 - 2x $. Use one of the formulas: +\[ + m = \lim_{x\to a} \frac{f(x) - f(a)}{x-a}\quad\text{or}\quad + m = \lim_{h\to 0} \frac{f(a+h) - f(a)}{h} +\] +Then the slope of the line tangent to the graph of $f$ at the point $ +\vA{( 1, 1 )}\vB{(-1,5)} $. + +\renameSolnAfterTo{} +\begin{solution}[2in]\ifkeyalt We make the following calculations:\fi +\begin{multicols}{2} +\noindent\textbf{Calculations} +\begin{verA} +\begin{alignat*}{2} + m &= \lim_{x\to1} \frac{f(x)-f(1)}{x-1}\\& + = \lim_{x\to1} 3x+1&&\quad\text{from side calc}\\& + = \boxed4 +\end{alignat*} +\end{verA} +\begin{verB} +\begin{alignat*}{2} + m &= \lim_{x\to-1} \frac{f(x)-f(-1)}{x+1}\\& + = \lim_{x\to-1} 3x-5&&\quad\text{from side calc}\\& + = \boxed{-8} +\end{alignat*} +\end{verB} + +\columnbreak +\noindent\textbf{Side Calculations} +\begin{verA} +\begin{align*} + f(x)-f(1) &= 3x^2 - 2x - 1\\& + = (x-1)(3x+1) +\intertext{thus, the difference quotient is} + \frac{f(x)-f(1)}{x-1} &= 3x+1 +\end{align*} +\end{verA} +\begin{verB} +\begin{align*} + f(x)-f(-1) &= 3x^2 - 2x - 5\\& + = (x+1)(3x-5) +\intertext{thus, the difference quotient is} + \frac{f(x)-f(-1)}{x+1} &= 3x-5 +\end{align*} +\end{verB} + + +\vfill +\vspace*{\sameVspace} +\vfill +\end{multicols} +\end{solution} +\begin{workarea}{\sameVspace}\parindent0pt\bfseries +\begin{multicols}{2} +\textbf{Calculations} + +\vfil\vspace*{\sameVspace}\vfil + + +\columnbreak +\textbf{Side Calculations} + +\vfil\vspace*{1.9in}\vfil + +\end{multicols} +\end{workarea} + +\end{problem} + +\end{exam} +\end{document} |