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diff --git a/macros/latex/contrib/eqexam/examples/test03.tex b/macros/latex/contrib/eqexam/examples/test03.tex new file mode 100644 index 0000000000..717c41a4f8 --- /dev/null +++ b/macros/latex/contrib/eqexam/examples/test03.tex @@ -0,0 +1,581 @@ +\documentclass{article} +\usepackage[fleqn]{amsmath} +\usepackage[pointsonleft,nototals, + forcolorpaper,useforms, +% choose to compile with exactly one of the next 4 options +%------------------- + nosolutions, % compile with no solutions to get the exam document +% answerkey, % get answer key +% vspacewithsolns,% put solutions at end of document +% solutionsonly, % compile with vspacewithsolns several times, then compile with solutionsonly +%------------------- +% coverpage,coverpagesumry=bypages + showgrayletters]{eqexam} +\usepackage{graphicx} + +\forceNoColor +\vspacewithkeyOn + +\university +{% + NORTHWEST FLORIDA STATE COLLEGE\\ + Department of Mathematics +} +\email{storyd@nwfsc.edu} + +\examSIDLabel{Class: MAC 1105, \vA{12:30 pm, L-134}\vB{12:30 am, L-105}} +\coverpageSubjectFmt{\bfseries\LARGE} +\coverpageTitleFmt{\bfseries\LARGE} +\examNum{3}\numVersions{2}\forVersion{a} +\subject[MAC1105]{College Algebra} +\longTitleText + {Test~\nExam} + {Test~\nExam} +\endlongTitleText +\shortTitleText + {T\nExam} + {T\nExam} +\endshortTitleText +\altTitle{\vA{12:30 pm, L-134}\vB{12:30 pm, L-105}} +\title[\sExam]{\Exam} +\author{Dr.\ D. P. Story} +\date{\thisterm, \the\year} +\duedate{04/05/11} +\keywords{MAC 1105, Exam \nExam, {\thisterm} semester, \theduedate, at NWFSC} +\renewcommand{\fillInFormatDefault}{} +\DoNotFitItIn +\eqpartsitemsep{3pt} +\solAtEndFormatting{\eqequesitemsep{3pt}} + + +\everymath{\displaystyle} +%\renameSolnAfterTo{} +%\resetSolnAfterToDefault + + +\eqCommentsColor{gray} +\eqCommentsColorBody{gray} +\newcommand{\cs}[1]{\texttt{\char`\\#1}} +\def\qt#1{&&\qquad\text{#1}} + + +\encloseProblemsWith{theseproblems} + +\begin{document} + +\maketitle + + +\begin{exam}{Test\nExam} + +\ifsolutionsonly\NoPoints +\begin{instructions}[Solutions:] +The solutions to the test. +\end{instructions} +\else +\begin{instructions}[Instructions:] +This exam has {\nQuesInExam} questions distributed over {\nPagesOnExam} pages. +Solve each of the problem and box in your final $\boxed{\text{answer}}$, where applicable. +\end{instructions} +\fi + +\begin{theseproblems} + +\renameSolnAfterTo{} + +\begin{problem*}[2ea]\label{shortAns} +Answer each of the following, none of the problems shown below requires any +calculations. Respond to True/False questions with \texttt{T} (for True) or \texttt{F} (for +False). +\begin{parts} + \item When viewing the graph of a function, we may use the + \fillin[u]{1.5in}{Horizontal Line} Test to determine if it is a + one-to-one function. +\begin{solution}[]\ifvspacewithsolns +When viewing the graph of a function, we may use the +\fillin[u]{1.5in}{Horizontal Line} Test to determine if it is a +one-to-one function.\fi +\end{solution} + + \item \TF{F} (\texttt{T} or \texttt{F}) The graph of the function $ f(x) = + 2-4x-3x^2$ is a parabola that opens up. +\begin{solution}[]\ifvspacewithsolns +\TF{F} (\texttt{T} or \texttt{F}) The graph of the function $ f(x) = +2-4x-3x^2$ is a parabola that opens up.\fi +\end{solution} + + \item \TF{F} (\texttt{T} or \texttt{F}) For a quadratic function of the form + $f(x)=ax^2+bx+c$, if $a>0$, then the function has a \emph{maximum + value}. +\begin{solution}[]\ifvspacewithsolns +\TF{F} (\texttt{T} or \texttt{F}) For a quadratic function of the form + $f(x)=ax^2+bx+c$, if $a>0$, then the function has a \emph{maximum + value}.\fi +\end{solution} + +\pushProblem + +\begin{eqComments}[Comments:] +Questions like the three above (fill-in and True/False) often have no +solution; hence, normally, the \texttt{solution} environment is not used. When +using the \texttt{vspacewithsolns} or the \texttt{solutionsonly} options +you would like the ``answers'' to appear on the solutions pages. To +rectify this, we simply copy and past the item into a solutions +environment, like so, in the case of the last question above. +\begin{verbatim} +\begin{solution}[]\ifvspacewithsolns +\TF{F} (\texttt{T} or \texttt{F}) For a quadratic function of the form + $f(x)=ax^2+bx+c$, if $a>0$, then the function has a \emph{maximum + value}.\fi +\end{solution} +\end{verbatim} +The optional argument is empty (important). We don't want the student or instructor to +see this solution when the document is compiled using the \texttt{answerkey} +option, so we wrap this solution in a conditional +\verb~\ifvspacewithsolns...\fi~ This switch will be true if either the +options \texttt{vspacewithsolns} or \texttt{solutionsonly} options are +taken +\end{eqComments} + +\popProblem + + \item\label{whichRatFunc} Which rational function below has a horizontal asymptote of + $y=-2$, and has vertical asymptotes of $x=1$ (odd) and $ x=2 $ (even)? + \begin{answers}{3}\rowsep{6pt} + \bChoices[label=whichRat] + \Ans0 $ y = \frac{(x+2) (1-2x)}{(1-x)(x-2)^2} $\eAns + \Ans0 $ y = \frac{(x+2)^2 (2x-1)}{(x-1)(x-2)^2} $\eAns + \Ans1 $ y = \frac{(x+2)^2 (1-2x)}{(x-1)(x-2)^2} $\eAns + \Ans0 $ y = \frac{(x+2)^2 (2x-1)}{(x-1)^2(x-2)} $\eAns + \Ans0 $ y = \frac{(x+2) (1-2x)^2}{(x-1)^2(x-2)} $\eAns + \Ans0 none of these\eAns + \eChoices + \end{answers} +\begin{solution}[]\ifvspacewithsolns +We can access the ``answers'' to a multiple choice question in several +ways: +\begin{itemize} +\item The correct alternative is part~\useSavedAlts{whichRat}, +\useSavedAns{whichRat}. +\begin{verbatim} +The correct alternative is part~\useSavedAlts{whichRat}, +\useSavedAns{whichRat}. +\end{verbatim} +The command \verb!\useSavedAlts{whichRat}! expands to the letter alternative of the +correct response, \useSavedAlts{whichRat}, in this case. Similarly, +\verb!\useSavedAns{whichRat}! expands to the correct answer, here, the +correct answer is \useSavedAns{whichRat}. + +\item The correct answer is \useSavedAltsAns{whichRat}. +\begin{verbatim} +The correct answer is \useSavedAltsAns{whichRat}. +\end{verbatim} +The command \verb!\useSavedAltsAns{whichRat}! expands to the correct +letter followed by the correct answer. +\item You can now copy and paste the \texttt{answers} (or \texttt{manswers}) + environment into the \texttt{solutions} environment, like so. + +\item[] Which rational function below has a horizontal asymptote of + $y=-2$, and has vertical asymptotes of $x=1$ (odd) and $ x=2 $ (even)? + \begin{answers}{3}\rowsep{6pt} + \bChoices[label=whichRat] + \Ans0 $ y = \frac{(x+2) (1-2x)}{(1-x)(x-2)^2} $\eAns + \Ans0 $ y = \frac{(x+2)^2 (2x-1)}{(x-1)(x-2)^2} $\eAns + \Ans1 $ y = \frac{(x+2)^2 (1-2x)}{(x-1)(x-2)^2} $\eAns + \Ans0 $ y = \frac{(x+2)^2 (2x-1)}{(x-1)^2(x-2)} $\eAns + \Ans0 $ y = \frac{(x+2) (1-2x)^2}{(x-1)^2(x-2)} $\eAns + \Ans0 none of these\eAns + \eChoices + \end{answers} +\end{itemize}\fi +\end{solution} + +\pushProblem +\begin{eqComments}[Comments:] +Multiple choice and multiple selection questions were an especially +difficult problem to solve; the \texttt{answers} and \texttt{manswers} +environments are undefined outside of an \texttt{exam} environment so one +cannot simply copy and paste the choices into the \texttt{solution} environment. + +To resolve this issue, I added a key-value pair to the \cs{bChoices} command, +the key is \texttt{label}. The source code for the above question reads +\verb!\bChoices[label=whichRat]! The value of the label key is used to +build a series of macros that record the labels and text for the choices +that are marked correct by \cs{Ans1}. The information gathered by these +macros are accessible through \cs{useSavedAlts}, \cs{useSavedAns}, +\cs{useSavedAltsAns}, and \cs{useSavedNumAns}, as described in the \textsf{eqexam} +manual. See the solutions pages to see the answers to these multiple +choice questions and details on the use of these commands. +\end{eqComments} +\popProblem + + \item How many times can a quadratic equation cross the $x$-axis? + Check as many of the alternatives that are possibly correct for a + quadratic function. + \begin{manswers}{4} + \bChoices[label=nCrossings] + \Ans1 0\eAns + \Ans1 2\eAns + \Ans1 3\eAns + \Ans0 4\eAns + \Ans0 5\eAns + \Ans0 6\eAns + \Ans0 infinitely many\eAns + \Ans0 none of these\eAns + \eChoices + \end{manswers} +\begin{solution}[]\ifvspacewithsolns +Here is how these same macros expand for multiple selection problems. +\begin{itemize} +\item The correct alternatives are parts~\useSavedAlts{nCrossings}. +\begin{verbatim} +The correct alternatives are parts~\useSavedAlts{nCrossings}. +\end{verbatim} +\item The correct answers are \useSavedAns{nCrossings}. +\begin{verbatim} +The correct answers are \useSavedAns{nCrossings}. +\end{verbatim} +\item The correct responses are \useSavedAltsAns{nCrossings}. +\begin{verbatim} +The correct responses are \useSavedAltsAns{nCrossings}. +\end{verbatim} +\item[] End each case, the command expands to a comma-delimited list of correct +answers. +\end{itemize} +You can also access the answers individually, for example the second +correct response is part~\useSavedAlts[2]{nCrossings}, the answer for +part~\useSavedAlts[2]{nCrossings} is \useSavedAns[2]{nCrossings}. Or, +we can say, \useSavedAltsAns[2]{nCrossings} to get a combined listing of +the second correct response. +\begin{verbatim} +You can also access the answers individually, for example the second +correct response is part~\useSavedAlts[2]{nCrossings}, the answer for +part~\useSavedAlts[2]{nCrossings} is \useSavedAns[2]{nCrossings}. Or, +we can say, \useSavedAltsAns[2]{nCrossings} to get a combined listing of +the second correct response. +\end{verbatim} +\fi +\end{solution} +\end{parts} +\end{problem*} + +\begin{eqComments}[Comments:] +The above question is a multiple selection question. The student must +select all the correct choices. See the solutions pages to see the answers +to these multiple choice questions and details on the use of these +commands. +\end{eqComments} + +\begin{problem}[5] +Which rational function below has a horizontal asymptote of + $y=-2$, and has vertical asymptotes of $x=1$ (odd) and $ x=2 $ (even)? + \begin{answers}{3}\rowsep{6pt} + \bChoices[label=whichRat1] + \Ans0 $ y = \frac{(x+2) (1-2x)}{(1-x)(x-2)^2} $\eAns + \Ans0 $ y = \frac{(x+2)^2 (2x-1)}{(x-1)(x-2)^2} $\eAns + \Ans1 $ y = \frac{(x+2)^2 (1-2x)}{(x-1)(x-2)^2} $\eAns + \Ans0 $ y = \frac{(x+2)^2 (2x-1)}{(x-1)^2(x-2)} $\eAns + \Ans0 $ y = \frac{(x+2) (1-2x)^2}{(x-1)^2(x-2)} $\eAns + \Ans0 none of these\eAns + \eChoices + \end{answers} +\begin{solution}[]\ifvspacewithsolns +We can access the ``answers'' to a multiple choice question in several +ways: +\begin{itemize} +\item The correct alternative is part~\useSavedAlts{whichRat1}, +\useSavedAns{whichRat1}. +\begin{verbatim} +The correct alternative is part~\useSavedAlts{whichRat1}, +\useSavedAns{whichRat1}. +\end{verbatim} +The command \verb!\useSavedAlts{whichRat1}! expands to the letter alternative of the +correct response, \useSavedAlts{whichRat1}, in this case. Similarly, +\verb!\useSavedAns{whichRat1}! expands to the correct answer, here, the +correct answer is \useSavedAns{whichRat1}. + +\item The correct answer is \useSavedAltsAns{whichRat1}. +\begin{verbatim} +The correct answer is \useSavedAltsAns{whichRat1}. +\end{verbatim} +The command \verb!\useSavedAltsAns{whichRat1}! expands to the correct +letter followed by the correct answer. +\end{itemize}\fi +\end{solution} +\end{problem} + +\begin{eqComments}[Comments:] +This is the same question as Problem~\ref{shortAns} (\ref{whichRatFunc}), +but this one is a stand alone question. The lettering of the label can +change depending on the options you take, so, if you compile this document +without the \texttt{useforms} options, the choices listed in~\ref{shortAns} (\ref{whichRatFunc}) +will be numbers, (A), (B),\dots, and the choices of this question will be +letters, (a), (b),\dots. Check the solutions page, the references should +change to reflect the change in options, let's hope. +\end{eqComments} + +\resetSolnAfterToDefault + +\begin{problem*}[\auto] +Let $f(x) = 4x+3$ and $ g(x) = 2x^2 - 5 $. Compute each of the following, +simplify were appropriate. +\begin{multicols}{2} +\begin{parts} +\item \PTs{2} $ (fg)(-2) = \fillin[boxed,boxsize=LARGE,align=l]{1in}{-15} $ + +\begin{solution}[.65in] +We have \[ (fg)(-2)=f(-2)g(-2)=(-5)(3)=\boxed{-15}\] +\end{solution} + +\item \PTs{2} $\left(\frac{g}{f}\right)(x)= \fillin[boxed,boxsize=LARGE]{\ifNoSolutions{1in}{}}{\frac{2x^2-5}{4x+3}} $ + +\begin{solution}[\sameVspace] +$ \left(\frac{g}{f}\right)(x)=\frac{g(x)}{f(x)}=\boxed{\frac{2x^2 - 5}{4x+3}}$ +\end{solution} + +\item \PTs{2} $ (f\circ f )(x) = \fillin[boxed,boxsize=LARGE]{\ifNoSolutions{1in}{}}{16x+15} $ + +\begin{solution}[\sameVspace] +Composing, $(f\circ f )(x)=f(f(x))=f(4x+3)=4(4x+3)+3=\boxed{16x+15}$ +\end{solution} + +\item \PTs{4} $ (f\circ g )(x) = \fillin[boxed,boxsize=LARGE]{1in}{8x^2-17} $ + +\begin{solution}[\sameVspace] +Composing, $(f\circ g )(x)=f(g(x))=f(2x^2 - 5)=4(2x^2 -5)+3=\boxed{8x^2-17}$ +\end{solution} +\end{parts} +\end{multicols} +\end{problem*} + +\begin{eqComments}[Comments:] +Nothing new about the above problem, each has a solution, no special +attention is needed. In some of the answer boxes, \cs{ifNoSolutions} is +used to set the width then \texttt{nosolutions} is in effect, and to et the box to +its natural width otherwise. +\end{eqComments} + + +\begingroup + +\setlength{\columnsep}{30pt} + +\begin{multicols}{2} +\begin{problem}[5] +Use the \textbf{vertex formula} to find the $x$-coordinate, $h$, and the +$y$-coordinate, $k$, of the quadratic function $ f(x) = 2x^2 - 8x + 5 $. +\begin{equation*} + \fillin[boxed,boxsize=LARGE,align=l,boxpretext={h=}]{1in}{2}\quad + \fillin[boxed,boxsize=LARGE,align=l,boxpretext={k=}]{1in}{-3} +\end{equation*} +\begin{solution}[.5in] +We use the vertex formula, $ h = -b/(2a) = - (-8)/4 = 2 $, and +so $h=f(2) = 8 - 16 + 5 = -3$. +\end{solution} +\end{problem} + +\columnbreak + +\begin{problem}[] % +\PTs{3}\addtocounter{eqpointvalue}{3} The function $ f(x) = x^2 - x + 1 $ has a +\fillin[u]{.75in}{minimum} (max/min) at $x = \fillin[u]{.5in}{1/2}$. +\begin{solution}[\sameVspace] +We use the vertex formula, $ h = -b/(2a) = - (-1)/2 = +1/2 $. A \textbf{minimum} occurs since the leading coefficient is +positive, which means the parabola opens up, the vertex is a minimum. +\end{solution} +\end{problem} +\end{multicols} + +\endgroup + +\begin{eqComments}[Comments:] +I include this problem in this file, because it is a construct that +appeared in a test of mine. I wanted to conserve vertical space so I put +to problems into two column format. The problem is the points appear to +the left. So, for the problem on the left, the points appear as usual, for +the problem on the right, the points appear in-line, I had to explicitly +increment the points counter, like so +\verb~\addtocounter{eqpointvalue}{3}~. Some adjustment of the space +between the columns was necessary \verb~\setlength{\columnsep}{30pt}~. +\end{eqComments} + +\renameSolnAfterTo{} + +\begin{problem}[5] +For a polynomial of degree $12$, according to theory, the maximum number +of zeros is \fillin[u]{.5in}{12}, and the maximum number of turning points +is \fillin[u]{.5in}{11}. +\begin{solution}[]\ifvspacewithsolns +For a polynomial of degree $12$, according to theory, the maximum number +of zeros is \fillin[u]{.5in}{12}, and the maximum number of turning points +is \fillin[u]{.5in}{11}.\fi +\end{solution} +\end{problem} + +\begin{eqComments}[Comments:] +A fill-in the blank problem, just copy and paste it into the solution +environment, protected by \verb~\ifvspacewithsolns...\fi~. +\end{eqComments} + + +\begin{problem}[5] +In the boxes provided, list the laws of the exponents and the laws of +logarithms. + \begin{equation*}\def\bwidth{2.75in}\def\bheight{1.5in} + \begin{tabular}{cc} + \textbf{Laws of the Exponents} & \textbf{Laws of Logarithms}\\ + \multicolumn{1}{p{\bwidth}}{% + \fillin[boxed,enclosesoln,parbox={[c][\bheight][t]}]{\linewidth}{% +% + \begin{enumerate} + \item $a^x a^y = a^{x+y}$ + \item $a^x/a^y = a^{x-y}$ + \item $ (a^x)^y = a^{xy}$ + \end{enumerate} +% + }}& + \multicolumn{1}{p{\bwidth}}{% + \fillin[boxed,enclosesoln,parbox={[c][\bheight][t]}]{\linewidth}{% +% + \begin{enumerate} + \item $\log_a(xy) = \log_a(x)+\log_a(y)$ + \item $\log_a(x/y) = \log_a(x)-\log_a(y)$ + \item $\log_a(x^r) = r\log_a(x)$ + \end{enumerate} +% + }} + \end{tabular} + \end{equation*} +\begin{solution}[]\ifvspacewithsolns +Write sentences, in the provided boxes, describing, in laymen's terms, Type I + and Type II errors for this test of hypothesis. + \begin{equation*}\def\bwidth{2.75in}\def\bheight{1.5in} + \begin{tabular}{cc} + \textbf{Laws of the Exponents} & \textbf{Laws of Logarithms}\\ + \multicolumn{1}{p{\bwidth}}{% + \fillin[boxed,enclosesoln,parbox={[c][\bheight][t]}]{\linewidth}{% +% + \begin{enumerate} + \item $a^x a^y = a^{x+y}$ + \item $a^x/a^y = a^{x-y}$ + \item $ (a^x)^y = a^{xy}$ + \end{enumerate} +% + }}& + \multicolumn{1}{p{\bwidth}}{% + \fillin[boxed,enclosesoln,parbox={[c][\bheight][t]}]{\linewidth}{% +% + \begin{enumerate} + \item $\log_a(xy) = \log_a(x)+\log_a(y)$ + \item $\log_a(x/y) = \log_a(x)-\log_a(y)$ + \item $\log_a(x^r) = r\log_a(x)$ + \end{enumerate} +% + }} + \end{tabular} + \end{equation*}\fi +\end{solution} +\end{problem} + +\begin{eqComments}[Comments:] +The above pair of boxes use the \texttt{enclosesoln} key. When this key is +used, the vertical size of the box is adjusted to the vertical size the +solution uses when either \texttt{nosolutions} or \texttt{vspacewithsolns} +option are used. Note the dimensions of the \cs{parbox} are adjusted so +that the width and height are correct. The \cs{boxed} command adds +\texttt{2\cs{fboxesp}+2\cs{fboxrule}}, so we reduce the \cs{parbox} by +that amount so the boxes are the correct size. +\end{eqComments} + +\begin{problem}[12] +Define $ f(x) = -2x^2(x+1) $. Make a good sketch of the graph in the +coordinate plane below, taking into consideration the end-behavior of the +polynomial, and its intercepts. +\begin{solution}[3in] +The graph of $ f(x) = -2x^2(x+1) $ is seen below. +\par\nobreak\medskip\vskip-1.5\baselineskip +\begin{minipage}[t]{3.5in}\vskip\baselineskip\kern0pt +\includegraphics[width=3.5in]{graph} +\end{minipage}\hfill +\begin{minipage}[t]{\linewidth-3.5in-30pt}\vskip\baselineskip\kern0pt +\noindent\makebox[\linewidth][c]{\textbf{Work Area}} +\begin{itemize} +\item The end-behavior is like $y=-x^3$ +\item $x$-int: $x=0$ (even); $ x=-1 $ (odd) +\item $y$-int: $y=0$ (passes through origin) +\end{itemize} +\end{minipage} +\end{solution} +\begin{workarea}{\sameVspace}% +\begin{minipage}[t]{3.5in}\kern0pt +\includegraphics[width=3.5in]{coorplane} +\end{minipage}\hfill +\begin{minipage}[t]{\linewidth-3.5in-30pt}\kern0pt +\makebox[\linewidth][c]{\textbf{Work Area}} +\end{minipage} +\end{workarea} +\end{problem} + +\begin{eqComments}[Comments:] +Finally, we have the above problem. It uses the \texttt{workarea} +environment. Previously, \texttt{workarea} appeared with the \texttt{nosolutions} +option. Now it appears with the \texttt{vspacewithsolns} option as well. +On the actual test, I used \textsf{PSTricks} for the graphics, for this +demo file, I replace the \textsf{pstricks} code this a figure depicting what the +\textsf{pstricks} produced, that way users of pdflatex can compile this +file! \texttt{:-)} +\end{eqComments} + +\newpage + +\begin{eqComments}% +On this page, we more clearly demonstrate the new feature of preserving +the vertical space even when the \texttt{answerkey} option is used. In the +preamble, we have \cs{vspacewithkeyOn}. +\end{eqComments} + + +\resetSolnAfterToDefault +% try changing \vspacewithkeyOn to \vspacewithkeyOff and recompile, +% the 4 inches of vertical space are not preserved when you compile +% with the answerkey option. +\vspacewithkeyOn + +\begin{problem}[10] +Solve the equation $2x^2 - 5x + 10 = 0 $ using the quadratic formula. +\begin{solution}[4in] +Applying the quadratic formula with $a=2$, $ b = -5 $, and $ c = 10 $, +\begin{alignat*}{2} + x & = \frac{-b \pm \sqrt{b^2 -4ac}}{2a} \qt{The Quad.\ Formula}\\& + = \frac{5 \pm \sqrt{25 -4(2)(10)}}{2(2)} \qt{substitute}\\& + = \frac{5 \pm \sqrt{25 -80}}{4} \qt{arithmetic}\\& + = \frac{5 \pm \sqrt{-45}}{4} \qt{ditto}\\& + = \frac{5 \pm 3\sqrt{5}\,\imath}{4} \qt{simplify} +\end{alignat*} +The solution is $\boxed{x=\frac{5 \pm 3\sqrt{5}\,\imath}{4}}$ +\end{solution} +\end{problem} + +\begin{problem}[5] +Write the equation, in standard form, for the circle with center at +$C(1,-3)$ and radius of $2$ +\begin{solution}[1in] +We have $(x-1)^2 + (y+2)^2 = 4 $. Expanding and combining the equation, we +have\dots \[\boxed{x^2+y^2-2x+4y+1=0}\] +\end{solution} +\end{problem} + +\begin{eqComments}[Comments:] +The \texttt{solution} environments in the above problems declared 4 inches +and 1 inch of vertical space, respectively. With \cs{vspacewithkeyOn} we +should have about 4 inches (resp., 1 inch) of vertical space even with the +\texttt{answerkey} option. Try compiling the file with +\cs{vspacewithkeyOff}. +\end{eqComments} + +\end{theseproblems} + +\end{exam} + +\end{document} |