我目前正在为往年的考试写答案,发现如果有必要的话,没有办法对同一道题的答案进行编号。(有些问题最终会有很多答案。)
\usepackage[margin=2.54cm, a4paper]{geometry}
\usepackage{amsmath}
\usepackage{amssymb}
\pagestyle{headandfoot}
\runningheader{}{Solution of Advance Mathematics II}{Year 201X}
\runningheadrule
\runningfooter{}{Page \thepage \,of Page \numpages}{}
\newcounter{solution}
\unframedsolutions
\renewcommand{\solutiontitle}{\noindent\textbf{Solution \thequestion: }}
\begin{document}
\begin{titlepage}
\centering
\hspace{10cm}\\[5pt]
\vfill
{\scshape\Huge School \par}
\vspace{1cm}
{\scshape\Large Advanced Mathematics II\par}
\vspace{1cm}
{\huge\bfseries Solution Manual\par}
\vspace{1cm}
{\Large Trial Exam\par}
\vspace{1cm}
{\Large October 201X\par}
\vspace{1cm}
{\large \emph{Written by} Someone}
\vfill
\end{titlepage}
\newpage
\Large \section*{Preface}
blablabla
\newpage
\begin{questions}
\question Suppose $\cos 2\theta=k$, express $\sin^6 \theta-\cos^6 \theta$ in terms of $k$.
\printanswers
\begin{solution}
Notice that
\begin{align*}
k & =\cos 2\theta\\
& =2\cos^2 \theta-1\\
& =1-2\sin^2 \theta
\end{align*}
Hence, both $\sin^2 \theta$ and $\cos^2 \theta$ can be expressed in terms of $k$ as follow:\\
\begin{equation}
\label{eq:eq1}
\sin^2 \theta=\dfrac{1-k}{2} \quad \text{and} \quad
\cos^2 \theta=\dfrac{k+1}{2}\\
\end{equation}
Substituting Equation \eqref{eq:eq1} yields
\begin{align*}
\sin^6 \theta-\cos^6 \theta &= \left(\dfrac{1-k}{3}\right)^3-\left(\dfrac{k+1}{3}\right)^3\\
&=-\dfrac{k^3+3k}{4}
\end{align*}
\end{solution}
\begin{solution}
aaa
\end{solution}
\end{questions}
\end{document}
我想让编号“解决方案 2:“,如果可能的话,如果问题只有一个解决方案,我希望看到”解决方案:“而是。我该怎么做呢?
答案1
简单的方法是在解决方案环境中添加一个枚举环境。
\documentclass{exam}
\usepackage[margin=2.54cm, a4paper]{geometry}
\usepackage{amsmath}
\usepackage{amssymb}
\pagestyle{headandfoot}
\runningheader{}{Solution of Advance Mathematics II}{Year 201X}
\runningheadrule
\runningfooter{}{Page \thepage \,of Page \numpages}{}
%\newcounter{solution}
\unframedsolutions
%\renewcommand{\solutiontitle}{\noindent\textbf{Solution \thequestion: }}
\begin{document}
\begin{titlepage}
\centering
\hspace{10cm}\\[5pt]
\vfill
{\scshape\Huge School \par}
\vspace{1cm}
{\scshape\Large Advanced Mathematics II\par}
\vspace{1cm}
{\huge\bfseries Solution Manual\par}
\vspace{1cm}
{\Large Trial Exam\par}
\vspace{1cm}
{\Large October 201X\par}
\vspace{1cm}
{\large \emph{Written by} Someone}
\vfill
\end{titlepage}
\newpage
\Large \section*{Preface}
blablabla
\newpage
\begin{questions}
\question Suppose $\cos 2\theta=k$, express $\sin^6 \theta-\cos^6 \theta$ in terms of $k$.
\printanswers
\begin{solution}
\begin{enumerate}
\item
Notice that
\begin{align*}
k & =\cos 2\theta\\
& =2\cos^2 \theta-1\\
& =1-2\sin^2 \theta
\end{align*}
Hence, both $\sin^2 \theta$ and $\cos^2 \theta$ can be expressed in terms of $k$ as follow:\\
\begin{equation}
\label{eq:eq1}
\sin^2 \theta=\dfrac{1-k}{2} \quad \text{and} \quad
\cos^2 \theta=\dfrac{k+1}{2}\\
\end{equation}
Substituting Equation \eqref{eq:eq1} yields
\begin{align*}
\sin^6 \theta-\cos^6 \theta &= \left(\dfrac{1-k}{3}\right)^3-\left(\dfrac{k+1}{3}\right)^3\\
&=-\dfrac{k^3+3k}{4}
\end{align*}
\item
aaa
\end{enumerate}
\end{solution}
\end{questions}
\end{document}
此版本使用multisolution
环境来启动编号解决方案。
\documentclass{exam}
\usepackage[margin=2.54cm, a4paper]{geometry}
\usepackage{amsmath}
\usepackage{amssymb}
\pagestyle{headandfoot}
\runningheader{}{Solution of Advance Mathematics II}{Year 201X}
\runningheadrule
\runningfooter{}{Page \thepage \,of Page \numpages}{}
\newcounter{solution}
\unframedsolutions
\newenvironment{multisolution}{%
\setcounter{solution}{0}%
\def\solutiontitle{\stepcounter{solution}\noindent\textbf{Solution \thesolution: }}% local definition
\ignorespaces}%
{\ignorespaces}
\begin{document}
\begin{titlepage}
\centering
\hspace{10cm}\\[5pt]
\vfill
{\scshape\Huge School \par}
\vspace{1cm}
{\scshape\Large Advanced Mathematics II\par}
\vspace{1cm}
{\huge\bfseries Solution Manual\par}
\vspace{1cm}
{\Large Trial Exam\par}
\vspace{1cm}
{\Large October 201X\par}
\vspace{1cm}
{\large \emph{Written by} Someone}
\vfill
\end{titlepage}
\newpage
\Large \section*{Preface}
blablabla
\newpage
\begin{questions}
\question Suppose $\cos 2\theta=k$, express $\sin^6 \theta-\cos^6 \theta$ in terms of $k$.
\printanswers
\begin{multisolution}
\begin{solution}
Notice that
\begin{align*}
k & =\cos 2\theta\\
& =2\cos^2 \theta-1\\
& =1-2\sin^2 \theta
\end{align*}
Hence, both $\sin^2 \theta$ and $\cos^2 \theta$ can be expressed in terms of $k$ as follow:\\
\begin{equation}
\label{eq:eq1}
\sin^2 \theta=\dfrac{1-k}{2} \quad \text{and} \quad
\cos^2 \theta=\dfrac{k+1}{2}\\
\end{equation}
Substituting Equation \eqref{eq:eq1} yields
\begin{align*}
\sin^6 \theta-\cos^6 \theta &= \left(\dfrac{1-k}{3}\right)^3-\left(\dfrac{k+1}{3}\right)^3\\
&=-\dfrac{k^3+3k}{4}
\end{align*}
\end{solution}
\begin{solution}
aaa
\end{solution}
\end{multisolution}
\end{questions}
\end{document}