\documentclass{article} \usepackage[fleqn]{amsmath} %\usepackage[pointsonleft,nototals,forpaper,useforms,vspacewithsolns]{eqexam} \usepackage[pointsonleft,nototals,forpaper,useforms,solutionsonly]{eqexam} \solAtEndFormatting{\eqequesitemsep{3pt}} \subject[MC]{My Course} \title[T1]{Test 1} \author{Dr.\ D. P. Story} \date{\thisterm, \the\year} \duedate{2012/04/28} \keywords{My Course, Exam \nExam, {\thisterm} semester, \theduedate, at AcroTeX.Net} % % Compile the file with the vspacewithsolns option to create the .sol auxiliary file % that contains a listing of all the solutions, like so % \usepackage[pointsonleft,nototals,forpaper,useforms,vspacewithsolns]{eqexam} % then compile with the solutionsonly option, like so % \usepackage[pointsonleft,nototals,forpaper,useforms,solutionsonly]{eqexam} % \encloseProblemsWith{theseproblems} \begin{document} \maketitle \begin{exam}{myProblems} \ifsolutionsonly \begin{instructions}[Solutions.] The solutions to Test~1. \end{instructions} \belowsqskip{\par} % removes the skip after the exam env \else \begin{instructions} Solve each problem and box in your final $\boxed{\text{answer}}$. \end{instructions} \fi \begin{theseproblems} \begin{problem*}[3ea] \leadinitem It is well known that \fillin{1in}{Newton} and \fillin{1in}{Leibniz} are jointly credited as the founders of modern calculus. \ifkeyalt \begin{solution} It is well known that \fillin{1in}{Newton} and \fillin{1in}{Leibniz} are jointly credited as the founders of modern calculus. \end{solution} \fi \begin{parts} \item \TF{T} (True `T' or False `F') The area of a circle is $\pi r^2$. \ifkeyalt \begin{solution} \item\TF{T} The area of a circle is $\pi r^2$. \end{solution} \fi \item Suppose the \emph{discriminant} of a quadratic equation is \emph{negative}, which statement describes the roots to the equation? \begin{answers}{2} \bChoices \Ans0 There is only one real root\eAns \Ans0 There are two distinct real roots\eAns \Ans1 There are two complex roots\eAns \Ans0 None of these\eAns \eChoices \end{answers} \ifkeyalt \begin{solution}[] \parbox[t]{\linewidth}{\sqTabPos{t}% \begin{answers}{2} \bChoices \Ans0 There is only one real root\eAns \Ans0 There are two distinct real roots\eAns \Ans1 There are two complex roots\eAns \Ans0 None of these\eAns \eChoices \end{answers}}% \adjDisplayBelow \end{solution} \fi \end{parts} \end{problem*} \begin{problem}[8] In the space below, solve the quadratic equation $ 2x^2 - 3x + 2 = 0 $ using any valid method. \begin{solution}[1.5in] We use the quadratic formula: \begin{equation*} x = \frac{3 \pm \sqrt{9-4\cdot2\cdot2}}{2\cdot2} = \frac{3 \pm \sqrt{-7}}{4} = \boxed{ \frac{3 \pm \sqrt{7}\,\imath}{4} } \end{equation*} \end{solution} \end{problem} \begin{problem}[8] Find the equation of the line perpendicular to $ 3x - 5y = 2 $ and passing through the point $P(1,7)$. Leave your answer in the general form. \[ \text{Ans:\quad} \fillin[boxed,boxsize=Large]{2in}{5x+3y=26} \] \begin{solution}[.5in]\ifkeyalt$ \text{Ans:\quad} \fillin[boxed,boxsize=Large]{2in}{5x+3y=26}$\\[3pt]\fi Apparently the given line has slope $m=3/5$, so $ m_{\perp}=-5/3$. The rest is left to the reader. \end{solution} \end{problem} \end{theseproblems} \end{exam} \end{document}