texlive[65192] Master: hfutexam (4dec22)
commits+karl at tug.org
commits+karl at tug.org
Sun Dec 4 22:07:44 CET 2022
Revision: 65192
http://tug.org/svn/texlive?view=revision&revision=65192
Author: karl
Date: 2022-12-04 22:07:43 +0100 (Sun, 04 Dec 2022)
Log Message:
-----------
hfutexam (4dec22)
Modified Paths:
--------------
trunk/Master/tlpkg/bin/tlpkg-ctan-check
trunk/Master/tlpkg/tlpsrc/collection-publishers.tlpsrc
Added Paths:
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trunk/Master/texmf-dist/doc/latex/hfutexam/
trunk/Master/texmf-dist/doc/latex/hfutexam/CHANGELOG.md
trunk/Master/texmf-dist/doc/latex/hfutexam/LICENSE
trunk/Master/texmf-dist/doc/latex/hfutexam/README.md
trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam.pdf
trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam.tex
trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_cankaodaan.pdf
trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_cankaodaan.tex
trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_datizhi.pdf
trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_datizhi.tex
trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_shijuan.pdf
trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_shijuan.tex
trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_simple.pdf
trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_simple.tex
trunk/Master/texmf-dist/tex/latex/hfutexam/
trunk/Master/texmf-dist/tex/latex/hfutexam/hfutexam.cls
trunk/Master/tlpkg/tlpsrc/hfutexam.tlpsrc
Added: trunk/Master/texmf-dist/doc/latex/hfutexam/CHANGELOG.md
===================================================================
--- trunk/Master/texmf-dist/doc/latex/hfutexam/CHANGELOG.md (rev 0)
+++ trunk/Master/texmf-dist/doc/latex/hfutexam/CHANGELOG.md 2022-12-04 21:07:43 UTC (rev 65192)
@@ -0,0 +1,18 @@
+# 更新日志
+
+## [1.5] - 2022/12/04
++ 现在`\XiZhuRenQianMing`不设置或设置为空时即可不显示签名.
+
+## [1.4] - 2022/12/04
++ 修复了有时选择题选项未能正确计算行数的问题.
++ 修复了有时填空题答案会被横线截断的问题.
+
+## [1.3] - 2022/11/27
++ 移除了简易模式下学期的下划线.
++ 打分框的宽度减少了`10%`.
+
+## [1.2] - 2022/10/31
++ 选择题选项后增加了`5pt`的垂直间距.
+
+## [1.1] - 2022/10/30
++ 首次提交.
Property changes on: trunk/Master/texmf-dist/doc/latex/hfutexam/CHANGELOG.md
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Added: trunk/Master/texmf-dist/doc/latex/hfutexam/LICENSE
===================================================================
--- trunk/Master/texmf-dist/doc/latex/hfutexam/LICENSE (rev 0)
+++ trunk/Master/texmf-dist/doc/latex/hfutexam/LICENSE 2022-12-04 21:07:43 UTC (rev 65192)
@@ -0,0 +1,416 @@
+The LaTeX Project Public License
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Added: trunk/Master/texmf-dist/doc/latex/hfutexam/README.md
===================================================================
--- trunk/Master/texmf-dist/doc/latex/hfutexam/README.md (rev 0)
+++ trunk/Master/texmf-dist/doc/latex/hfutexam/README.md 2022-12-04 21:07:43 UTC (rev 65192)
@@ -0,0 +1,7 @@
+# hfutexam: exam class for Hefei University of Technology (China)
+
+The package provides an exam class for Hefei University of Technology (China).
+
++ Package: An exam class for Jinan University
++ Author: Shenxing Zhang <zhangshenxing at hfut.edu.cn>
++ License: The LaTeX Project Public License 1.3c
Property changes on: trunk/Master/texmf-dist/doc/latex/hfutexam/README.md
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===================================================================
--- trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam.pdf 2022-12-04 21:06:57 UTC (rev 65191)
+++ trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam.pdf 2022-12-04 21:07:43 UTC (rev 65192)
Property changes on: trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam.pdf
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===================================================================
--- trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam.tex (rev 0)
+++ trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam.tex 2022-12-04 21:07:43 UTC (rev 65192)
@@ -0,0 +1,270 @@
+\documentclass{hfutexam}
+
+%% 示例所需的自定义命令
+\newcommand{\diff}{\,\mathrm{d}}
+\usetikzlibrary{arrows.meta, overlay-beamer-styles}
+\newfontfamily\couriernew{Courier New}
+\usepackage{enumitem}
+\usepackage{tcolorbox}
+\usepackage{listings}
+\makeatletter
+\definecolor{winered}{rgb}{0.5,0,0}
+\definecolor{lightgrey}{rgb}{0.9,0.9,0.9}
+\definecolor{frenchplum}{RGB}{190,20,83}
+\lstset{language=[LaTeX]TeX,
+ basicstyle=\couriernew,
+ texcsstyle=*\color{winered},
+ mathescape,
+ breaklines=true,
+ keywordstyle=\color{winered},
+ commentstyle=\color{green!70!black},
+ stringstyle=\color{green!50!blue},
+ frame=single,
+ tabsize=3,
+ framerule=0.5pt,
+ columns=flexible,
+ backgroundcolor=\color{black!5},
+ morekeywords={\diff, \maketitle, \titlesep, \BiaoTi, \XueNian, \XueQi, \KeChengDaiMa, \KeChengMingCheng, \XueFen, \KeChengXingZhi, \KaoShiXingShi, \ZhuanYeBanJi, \KaoShiRiQi, \MingTiJiaoShi, \XiZhuRenQianMing, \tigan, \scorebox, \score, \Score, \fillblank, \xx, \notice, \xuanzeti, \yihang, \erhang, \sihang, XeLaTeX},
+ keywordstyle=\color{winered},
+ morekeywords=[2]{hfutexam, shijuan, datizhi, cankaodaan, simple, nofangzheng, flalign, 5cm, enumerate, align},
+ keywordstyle=[2]\color{blue},
+}
+\tcbset{
+ colback=white,
+ colframe=blue,
+ boxrule=0.5pt,
+ arc=0pt,
+}
+\makeatother
+
+\begin{document}
+\BiaoTi{合肥工业大学试卷（A）}
+\XueNian{2021}{2022}
+\XueQi{二}
+\KeChengDaiMa{034Y01}
+\KeChengMingCheng{数学（下）}
+\XueFen{5}
+\KeChengXingZhi{必修}
+\KaoShiXingShi{闭卷}
+\ZhuanYeBanJi{少数民族预科班}
+\KaoShiRiQi{2022年6月18日8:00-10:00}
+\MingTiJiaoShi{集体}
+\XiZhuRenQianMing{}
+
+
+\tigan{一、模板选项}
+
+\indent
+本模板 (2022/12/04 v1.5) 旨在为将合肥工业大学试卷的 word 格式转为\LaTeX{}格式.
+使用时, 只需在文档开头写上
+\begin{lstlisting}
+\documentclass[shijuan]{hfutexam}
+\end{lstlisting}
+即可使用.
+需要使用~{\color{blue}{\lstinline|UTF-8|}} 编码, 并使用 \lstinline|XeLaTeX| 至少编译两次, 以正确生成页码.
+
+\indent
+可使用的选项为: \lstinline|shijuan| (试卷), \lstinline|datizhi| (答题纸), \lstinline|cankaodaan| (参考答案) 和 \lstinline|simple| (简易模式). 如果留空则为默认值 \lstinline|shijuan| (试卷).
+\begin{enumerate}
+\item 试卷/答题纸/参考答案三个选项下页面会设置为 A3 大小, 三种情形的页眉页脚显示的内容以及标题的文字间隔有所不同.
+\item 简易模式选项下页面会设置为 A4 大小, 页眉页脚也较为简单. 此时需要使用命令 \lstinline|\maketitle| 来生成标题.
+一般用于保存(多张)试卷的内容，或者便于打印使用.
+\item 标题默认使用方正字体, 因此请在使用前先安装字体: {\bfseries\titlesongti 方正小标宋}和{\bfseries\titlefangsong 方正仿宋}(右键选择为所有用户安装), 否则请使用选项 \lstinline|nofangzheng| (采用新宋体和仿宋代替).
+\end{enumerate}
+
+\tigan{二、试卷信息}
+
+\indent
+通过下述命令来设置试卷信息.
+
+\textit{\color{blue}{试卷信息示例:}}
+\begin{lstlisting}
+\BiaoTi{合肥工业大学试卷（A）} % 试卷标题, 一般为: 合肥工业大学试卷（A）或（B)
+\XueNian{2021}{2022} % 学年起始和结束, 一般为相差 1 的 4 位数字
+\XueQi{二} % 学期, 一般为: 一, 二
+\KeChengDaiMa{034Y01} % 课程代码
+\KeChengMingCheng{数学（下）} % 课程名称
+\XueFen{5} % 学分
+\KeChengXingZhi{必修} % 课程性质, 只能为: 必修, 选修, 限修
+\KaoShiXingShi{闭卷} % 考试形式, 只能为: 开卷, 闭卷
+\ZhuanYeBanJi{少数民族预科班} % 专业班级, 一般不需要填写
+\KaoShiRiQi{2022年6月18日8:00-10:00} % 考试日期
+\MingTiJiaoShi{集体} % 命题教师
+\XiZhuRenQianMing{dengbing.png} % 系主任签名
+\end{lstlisting}
+其中系主任签名处需要填写相应的图片名, 若不设置或设置为空则不显示.
+
+其它选项默认均为空, 可根据需要只填部分内容.
+
+\newpage
+\tigan{三、命令}
+\begin{enumerate}
+\item \lstinline|\tigan{三、命令}| 用于生成题干, 字体相对较大, 且为黑体. 小题建议使用~{\color{blue}\lstinline|enumerate|} 环境来生成.
+\item \hspace{-8mm}\scorebox\hspace{8mm}\lstinline|\scorebox| 用于生成打分框, 请放置在答题纸一行的开头使用.
+\vspace{-2mm}
+\item \lstinline|\notice| 用于生成答题纸提示信息, 请放置在答题纸的正文开始处.
+\item 答题纸中可能需要设置一定高度的空白, 使用命令 \lstinline|\hspace{5cm}| 之类的命令即可. 也可以使用 \lstinline|\newpage| 换到新的一页(或分栏).
+\end{enumerate}
+
+\tigan{填空题相关}
+\begin{enumerate}[resume]
+\item \lstinline|\fillblank[长度][最低高度]{内容}| 用于生成填空题的空白, 内容可以为空. 其中长度默认值是~{\color{blue}{\lstinline|3.5cm|}}, 最低高度默认值是~{\color{blue}{\lstinline|1cm|}} (答题纸和参考答案)或~{\color{blue}{\lstinline|0.5cm|}} (其它).
+\end{enumerate}
+
+\textit{\color{blue}{填空题示例:}}
+\begin{lstlisting}
+\textbf{请将你的答案对应填在横线上：}
+
+\textbf{1.} \fillblank{},
+\textbf{2.} \fillblank[5cm]{},
+\textbf{3.} \fillblank{}.
+\end{lstlisting}
+
+\begin{tcolorbox}
+\textbf{请将你的答案对应填在横线上：}
+
+\textbf{1.} \fillblank[3.5cm][1cm]{},
+\textbf{2.} \fillblank[5cm][1cm]{},
+\textbf{3.} \fillblank[3.5cm][1cm]{}.
+\end{tcolorbox}
+
+\tigan{选择题相关}
+\begin{enumerate}[resume]
+\item \lstinline|\xx{选项}{选项}{选项}{选项}| 用于生成选择题的选项, 直接在选择题题干后使用即可. 该命令会自动根据选项长度设置行数. 只支持四个选项, 选项会自动带上 ABCD.
+\item 如果想要手动改变每行显示的选项数, 可使用命令 \lstinline|\xx[每行显示的选项数]{选项}{选项}{选项}{选项}|, 每行只能显示 1, 2 或 4 个选项.
+\item \lstinline|\xuanzeti{题号}{答案}| 用于生成答题纸选择题的答题区域, 或参考答案选择题的答案区域.
+\end{enumerate}
+
+\textit{\color{blue}{选择题示例:}}
+\begin{lstlisting}
+\begin{enumerate}
+\item 柳宗元的《江雪》包含下面哪一句? (~~~~)
+\xx[2]{一山鸟飞绝}{百山鸟飞绝}{千山鸟飞绝}{亿山鸟飞绝}
+\item 张志和的《渔歌子》是(~~~~).
+\xx{东塞山前白鹭飞，桃花流水鳜鱼肥。青箬笠，绿蓑衣，斜风细雨不须归。}
+{南塞山前白鹭飞，桃花流水鳜鱼肥。青箬笠，绿蓑衣，斜风细雨不须归。}
+{西塞山前白鹭飞，桃花流水鳜鱼肥。青箬笠，绿蓑衣，斜风细雨不须归。}
+{北塞山前白鹭飞，桃花流水鳜鱼肥。青箬笠，绿蓑衣，斜风细雨不须归。}
+\end{enumerate}
+\end{lstlisting}
+
+\begin{tcolorbox}
+\begin{enumerate}
+\item 柳宗元的《江雪》包含下面哪一句? (~~~~).
+\xx[2]{一山鸟飞绝}{百山鸟飞绝}{千山鸟飞绝}{亿山鸟飞绝}
+\item 张志和的《渔歌子》是(~~~~).
+\xx{东塞山前白鹭飞，桃花流水鳜鱼肥。青箬笠，绿蓑衣，斜风细雨不须归。}
+{南塞山前白鹭飞，桃花流水鳜鱼肥。青箬笠，绿蓑衣，斜风细雨不须归。}
+{西塞山前白鹭飞，桃花流水鳜鱼肥。青箬笠，绿蓑衣，斜风细雨不须归。}
+{北塞山前白鹭飞，桃花流水鳜鱼肥。青箬笠，绿蓑衣，斜风细雨不须归。}
+\end{enumerate}
+\end{tcolorbox}
+
+%\newpage
+\textit{\color{blue}{选择题示例:}}
+\begin{lstlisting}
+\textbf{请将你所选择的字母 A, B, C, D 之一对应填在下列表格里：}
+
+\xuanzeti{\textbf{题号}}{\textbf{答案}}%
+\xuanzeti{1}{}\xuanzeti{2}{}\xuanzeti{3}{}\xuanzeti{4}{}
+\end{lstlisting}
+
+\begin{tcolorbox}
+\textbf{请将你所选择的字母 A, B, C, D 之一对应填在下列表格里：}
+
+\xuanzeti{\textbf{题号}}{\textbf{答案}}%
+\xuanzeti{1}{}\xuanzeti{2}{}\xuanzeti{3}{}\xuanzeti{4}{}
+\end{tcolorbox}
+
+\newpage
+\tigan{得分点相关}
+\begin{enumerate}[resume]
+\item \lstinline|\score{数值}| 用于在参考答案一行结尾处生成得分点的虚线.\score2
+\item \lstinline|\Score{(2分, 缺少常数得1分)}| 用于自定义得分说明.\Score{(2分, 缺少常数得1分)}
+\item 在公式中也可使用, 但是需要编译两次才会正常计算出虚线长度.
+\end{enumerate}
+\textit{\color{blue}{得分点示例:}}
+\begin{lstlisting}
+\[\int e^x\diff x=e^x+C. \Score{(4分, 缺少常数得2分)}\]
+\begin{align*}
+\int\sin x\diff x&=-\cos x+C, \Score{(4分, 缺少常数得2分)}\\
+\int_0^\pi(1+\sin x)\diff x&=\pi+2. \score5
+\end{align*}
+\end{lstlisting}
+
+\begin{tcolorbox}
+\[\int e^x\diff x=e^x+C. \Score{(4分, 缺少常数得2分)}\]
+\begin{align*}
+\int\sin x\diff x&=-\cos x+C, \Score{(4分, 缺少常数得2分)}\\
+\int_0^\pi(1+\sin x)\diff x&=\pi+2. \score5
+\end{align*}
+\end{tcolorbox}
+
+\textbf{如有疑问或建议, 欢迎联系我: {\color{red}{zhangshenxing at hfut.edu.cn}} 或 {\color{blue}{QQ362037052}}.}
+
+\textbf{CTAN: \color{blue}https://www.ctan.org/pkg/hfutexam}
+
+\newpage
+\tigan{一、填空题（每题3分，共18分）}
+\begin{enumerate}
+\item 如果 $f(x)>0$ 且 $\displaystyle\lim_{x\to\infty}f(x)=0$, 则 $\displaystyle\lim_{x\to\infty}\bigl[1+f(x)\bigr]^{1/f(x)}=$\fillblank{}.
+\item 设 $y=\sin(x^2+1)$, 则 $\diff y=$\fillblank{}.
+\item 极限 $\displaystyle\lim_{n\to\infty}\left(\frac1{n^2-1}+\frac2{n^2-2}+\cdots+\frac n{n^2-n}\right)=$\fillblank{}.
+\item 曲线 $y=2\ln(x+1)$ 在点 $(1,2\ln2)$ 处的切线方程为\fillblank{}.
+\item 若 $e^{y-1}=1+xy$, 则 $\dfrac{\diff y}{\diff x}\bigg|_{x=0}=$\fillblank{}.
+\item 如果函数 $f(x)$ 的定义域是 $(0,+\infty)$, 且 $x=0$ 是曲线 $y=f(x)$ 的垂直渐近线, 那么 $\displaystyle\lim_{x\to0^+}\frac1{f(x)}=$\fillblank{}.
+\end{enumerate}
+
+\tigan{二、选择题（每题3分，共18分）}
+\begin{enumerate}
+\item 当 $x\to+\infty$ 时, $\dfrac1x$ 和(~~~~)是等价无穷小.
+% 自动根据选项长度设置行数
+\xx{$\sin\dfrac1x$}{$\sin x$}{$e^{-x}$}{$e^{1/x}$}
+\item 若当 $x\to0$ 时, $\arctan(e^x-1)\cdot(\cos x-1)$ 和 $x^n$ 是同阶无穷小, 则 $n=$(~~~~).
+\xx{$0$}{$1$}{$2$}{$3$}
+\item 设 $f(x)=\arctan\dfrac1{x(x-1)^2}$, 则 $x=0$ 是 $f(x)$ 的(~~~~).
+\xx{可去间断点}{跳跃间断点}{第二类间断点}{连续点}
+\item
+\begin{tikzpicture}[overlay,xshift=12.5cm,yshift=-3cm]
+\draw[-Stealth,thick](-3,0)--(3,0);
+\draw[-Stealth,thick](0,-0.8)--(0,2.5);
+\draw[very thick,smooth,domain=-55:55] plot ({\x/50-1.3}, {tan(\x)*tan(\x)});
+\draw[very thick,smooth,domain=0.15:2] plot ({\x}, {-ln(\x)});
+\draw
+ (-0.3,-0.3) node {$O$}
+ (2.8,-0.3) node {$x$}
+ (-0.3,2.3) node {$y$};
+\end{tikzpicture}
+设 $f(x)$ 是定义在 $(-\infty,+\infty)$ 上的连续函数, 且 $f'(x)$ 的图像如下图所示, 则 $f(x)$ 有(~~~~).
+% 手动设置为每行1个
+\xx[1]{一个极大值点，没有极小值点}{没有极大值点，一个极小值点}{一个极大值点和一个极小值点}{一个极大值点和两个极小值点}
+\newpage
+\item 设函数 $f(x)$ 在点 $x=0$ 处可导, 且 $f(0)=0$, 则 $\displaystyle\lim_{x\to0}\frac{f(x^{2022})+x^{2021}f(x)}{x^{2022}}=$(~~~~).
+\xx{$0$}{$f'(0)$}{$2f'(0)$}{$2022f'(0)$}
+\item 如果点 $(x_0,y_0)$ 是曲线 $y=f(x)$ 的拐点, 则 $f''(x_0)=$(~~~~).
+\xx{$0$}{$\infty$}{不存在}{$0$ 或不存在}
+\end{enumerate}
+
+\tigan{三、解答题（每题8分，共64分）}
+\begin{enumerate}
+\item 求极限 $\displaystyle\lim_{x\to-1}\frac{x^2-1}{x^2+3x+2}$.
+\item 求极限 $\displaystyle\lim_{x\to0}\frac{e^x-1-x}{\arcsin x^2}$.
+\item 设 $\begin{cases}x=t^2+t&\\y=t^3+t&\end{cases}$, 求 $\dfrac{\diff y}{\diff x}$ 和 $\dfrac{\diff^2 y}{\diff x^2}$.
+\item 设 $f(x)=\begin{cases}x\arctan\dfrac1x,&x<0,\\x^2+ax+b,&x\ge0.\end{cases}$
+求常数 $a,b$ 使得函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导, 并求出此时曲线 $y=f(x)$ 的渐近线.
+\item 求函数 $f(x)=x^3-x^2-x$ 在区间 $[-2,2]$ 上的最大值和最小值.
+\item 证明: 当 $-\dfrac\pi2<x_1<x_2<\dfrac\pi2$ 时, $\tan x_2-\tan x_1\ge x_2-x_1$.
+\item 设函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导, 且 $f(1)=0$.
+证明: 存在 $\xi\in(0,1)$ 使得 $\xi f'(\xi)+2022f(\xi)=0$.
+\item 设函数 $f(x)=\ln x+\dfrac2{x^2}, x\in(0,+\infty)$. 求
+\begin{enumerate}
+\item[(1)] 函数 $f(x)$ 的增减区间及极值;
+\item[(2)] 曲线 $y=f(x)$ 的凹凸区间及拐点.
+\end{enumerate}
+\end{enumerate}
+
+\end{document}
+
+
+
+
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@@ -0,0 +1,158 @@
+\documentclass[cankaodaan]{hfutexam}
+\usepackage{extarrows}
+\newcommand{\diff}{\,\mathrm{d}}
+
+\begin{document}
+\BiaoTi{合肥工业大学试卷参考答案（A）}
+\XueNian{2021}{2022}
+\XueQi{二}
+\KeChengDaiMa{034Y01}
+\KeChengMingCheng{数学（下）}
+\XueFen{5}
+\KeChengXingZhi{必修}
+\KaoShiXingShi{闭卷}
+\ZhuanYeBanJi{少数民族预科班}
+\KaoShiRiQi{2022年6月18日8:00-10:00}
+\MingTiJiaoShi{集体}
+%\XiZhuRenQianMing{dengbing.png}
+
+\tigan{一、填空题（每小题3分，共18分）}
+
+\textbf{请将你的答案对应填在横线上：}
+
+\textbf{1.} \fillblank{$e$},
+\textbf{2.} \fillblank{$2x\cos(x^2+1)\diff x$},
+\textbf{3.} \fillblank{$\dfrac12$},
+
+\textbf{4.} \fillblank{$y=x-1+2\ln 2$},
+\textbf{5.} \fillblank{$1$},
+\textbf{6.} \fillblank{$0$}.
+
+\tigan{二、选择题（每小题3分，共18分）}
+
+\textbf{请将你所选择的字母 A, B, C, D 之一对应填在下列表格里：}
+
+\xuanzeti{\textbf{题号}}{\textbf{答案}}%
+\xuanzeti{1}{A}%
+\xuanzeti{2}{D}%
+\xuanzeti{3}{B}%
+\xuanzeti{4}{A}%
+\xuanzeti{5}{C}%
+\xuanzeti{6}{D}
+
+\tigan{三、解答题（每小题8分，共64分）}
+% 得分点命令 \score1, 得分点长度是自动的
+% \Score{(2分, 缺少常数得1分)} 用于自定义得分说明
+
+\textbf{1. （8分）【解】}
+\begin{align*}
+\lim_{x\to-1}\frac{x^2-1}{x^2+3x+2}&=\lim_{x\to-1}\frac{(x-1)(x+1)}{(x+2)(x+1)} \score3\\
+&=\lim_{x\to-1}\frac{x-1}{x+2} \score3\\
+&=\frac{-2}1=-2. \score2
+\end{align*}
+
+\textbf{2. （8分）【解】}
+\begin{align*}
+\lim_{x\to0}\frac{e^x-1-x}{\arcsin x^2}&=\lim_{x\to0}\frac{e^x-1-x}{x^2} \score3\\
+&\xlongequal{\text{洛必达}}\lim_{x\to0}\frac{e^x-1}{2x} \score3\\
+&=\lim_{x\to0}\frac{x}{2x}=\frac12. \score2
+\end{align*}
+
+\newpage
+
+\textbf{3. （8分）【解】}
+\begin{align*}
+\frac{\diff y}{\diff x}&=\frac{\diff y/\diff t}{\diff x/\diff t} \score2\\
+&=\frac{3t^2+1}{2t+1}, \score2\\
+\frac{\diff^2 y}{\diff x^2}&=\frac{\diff y'/\diff t}{\diff x/\diff t} \score2\\
+&=\frac{6t(2t+1)-(3t^2+1)2}{(2t+1)^3}=\frac{6t^2+6t-2}{(2t+1)^3}. \score2
+\end{align*}
+
+\textbf{4. （8分）【解】}
+
+\indent 由于 $f(x)$ 在 $x=0$ 处连续, 因此
+\begin{align*}
+f(0)&=f(0^+) \score1\\
+ &=b=\lim_{x\to0^-}x\arctan\frac1x=0\times\left(-\frac\pi2\right)=0. \score1
+\end{align*}
+
+\indent 由于 $f(x)$ 在 $x=0$ 处可导, 因此
+\begin{align*}
+f'_-(0)&=f'_+(0), \score1\\
+f'_-(0)&=\lim_{x\to0^-}\frac{x\arctan\frac1x}x=\lim_{x\to0^-}\arctan\frac1x=-\frac\pi2 \score1\\
+f'_+(0)&=(2x+a)|_{x=0}=a, \score1
+\end{align*}
+因此 $a=-\dfrac\pi2$. \score1
+
+\indent 由于
+\begin{align*}
+\lim_{x\to+\infty}\frac yx&=\lim_{x\to+\infty}\left(x-\frac\pi2\right)=+\infty, \score1\\
+\lim_{x\to-\infty}\frac yx&=\lim_{x\to-\infty}\arctan\frac1x=0,\\
+\lim_{x\to-\infty}y&=\lim_{x\to-\infty}x\arctan\frac1x=\lim_{t\to0^-}\frac{\arctan t}t=1,
+\end{align*}
+因此曲线 $y=f(x)$ 的渐近线只有 $y=1$. \score1
+
+\newpage
+
+\textbf{5. （8分）【解】}
+
+\indent 由
+\[f'(x)=3x^2-2x-1=(3x+1)(x-1)=0 \score2\]
+可得驻点 $x=-\dfrac13,1$. \score2
+
+\indent 由于
+\[f(-2)=-10,\quad f(2)=2,\quad f\left(-\frac13\right)=\frac5{27},\quad f(1)=-1, \score2\]
+因此最大值为 $2$, 最小值为 $-10$. \score2
+
+\textbf{6. （8分）【证明】}
+
+\textbf{证法一}: 设 $f(x)=\tan x-x$, 则 \score2
+\begin{align*}
+f'(x)=\frac1{\cos^2x}-1=\tan^2x\ge0. \score2
+\end{align*}
+因此 $f(x)$ 在 $\left(-\dfrac\pi2,\dfrac\pi2\right)$ 上单调递增, 从而 \score2
+\begin{align*}
+f(x_2)\ge f(x_1),\quad\tan x_2-\tan x_1\ge x_2-x_1. \score2
+\end{align*}
+
+\textbf{证法二}: 设 $f(x)=\tan x$, 则 $f(x)$ 在 $[x_1,x_2]$ 上连续, $(x_1,x_2)$ 内可导. \score2
+
+\indent 由拉格朗日中值定理, 存在 $\xi\in(x_1,x_2)$ 使得
+\begin{align*}
+\frac{f(x_2)-f(x_1)}{x_2-x_1}=f'(\xi), \score2
+\end{align*}
+即
+\begin{align*}
+\frac{\tan x_2-\tan x_1}{x_2-x_1}=\frac1{\cos^2\xi}\ge1. \score2
+\end{align*}
+所以 $\tan x_2-\tan x_1\ge x_2-x_1$. \score2
+\newpage
+
+\textbf{7. （8分）【证明】}
+
+\indent 设 $F(x)=x^{2022}f(x)$, \score2\\
+则 $F(x)$ 在 $[0,1]$ 上连续, $(0,1)$ 内可导, \score1\\
+且 $F(0)=0,F(1)=f(1)=0$. \score1
+
+\indent 由罗尔中值定理, 存在 $\xi\in(0,1)$ 使得 $F'(\xi)=0$. \score2\\
+由于 $F'(x)=x^{2022}f'(x)+2022x^{2021}f(x)$ 且 $\xi\neq0$, \score1\\
+所以 $\xi f'(\xi)+2022f(\xi)=1$. \score1
+
+
+\textbf{8. （8分）【解】}
+
+(1)
+\[f'(x)=\frac1x-\frac4{x^3}=\frac{x^2-4}{x^3}=\frac{(x+2)(x-2)}{x^3}. \score1\]
+当 $0<x<2$ 时, $f'(x)<0$. 当 $x>2$ 时, $f'(x)>0$. \score1\\
+因此 $(0,2]$ 是 $f(x)$ 的单减区间, $[2,+\infty)$ 是 $f(x)$ 的单增区间. \Score{(1分, 写成开区间不扣分)}\\
+所以 $f(x)$ 只有唯一的极小值 $f(2)=\ln2+\dfrac12$. \score1
+
+(2)
+\[f''(x)=-\frac1{x^2}+\frac{12}{x^4}=-\frac{x^2-12}{x^4}=-\frac{(x-2\sqrt3)(x+2\sqrt3)}{x^4}. \score1\]
+当 $0<x<2\sqrt3$ 时, $f''(x)>0$. 当 $x>2\sqrt3$ 时, $f''(x)<0$. \score1\\
+因此 $(0,2\sqrt3]$ 是曲线 $y=f(x)$ 的凹区间, \\
+$[2\sqrt3,+\infty)$ 是曲线 $y=f(x)$ 的凸区间, \Score{(1分, 写成开区间不扣分)}\\
+拐点为 $\left(2\sqrt3,\ln(2\sqrt3)+\dfrac16\right)$. \score1
+
+\end{document}
+
Property changes on: trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_cankaodaan.tex
___________________________________________________________________
Added: svn:eol-style
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+native
\ No newline at end of property
Added: trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_datizhi.pdf
===================================================================
(Binary files differ)
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===================================================================
--- trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_datizhi.pdf 2022-12-04 21:06:57 UTC (rev 65191)
+++ trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_datizhi.pdf 2022-12-04 21:07:43 UTC (rev 65192)
Property changes on: trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_datizhi.pdf
___________________________________________________________________
Added: svn:mime-type
## -0,0 +1 ##
+application/pdf
\ No newline at end of property
Added: trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_datizhi.tex
===================================================================
--- trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_datizhi.tex (rev 0)
+++ trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_datizhi.tex 2022-12-04 21:07:43 UTC (rev 65192)
@@ -0,0 +1,73 @@
+\documentclass[datizhi]{hfutexam}
+\newcommand{\diff}{\,\mathrm{d}}
+
+\begin{document}
+\BiaoTi{合肥工业大学考试专用答卷纸（A）}
+\XueNian{2021}{2022}
+\XueQi{二}
+\KeChengDaiMa{034Y01}
+\KeChengMingCheng{数学（下）}
+\XueFen{5}
+\KeChengXingZhi{必修}
+\KaoShiXingShi{闭卷}
+\ZhuanYeBanJi{少数民族预科班}
+\KaoShiRiQi{2022年6月18日8:00-10:00}
+\MingTiJiaoShi{集体}
+%\XiZhuRenQianMing{dengbing.png}
+
+
+\notice
+
+%\scorebox 为打分框, 必须放在行首
+\scorebox\tigan{一、填空题（每小题3分，共18分）}
+
+\textbf{请将你的答案对应填在横线上：}
+
+\textbf{1.} \fillblank{},
+\textbf{2.} \fillblank{},
+\textbf{3.} \fillblank{},
+
+\textbf{4.} \fillblank{},
+\textbf{5.} \fillblank{},
+\textbf{6.} \fillblank{}.
+
+\scorebox\tigan{二、选择题（每小题3分，共18分）}
+
+\textbf{请将你所选择的字母 A, B, C, D 之一对应填在下列表格里：}
+
+\xuanzeti{\textbf{题号}}{\textbf{答案}}%
+\xuanzeti{1}{}%
+\xuanzeti{2}{}%
+\xuanzeti{3}{}%
+\xuanzeti{4}{}%
+\xuanzeti{5}{}%
+\xuanzeti{6}{}
+
+\tigan{三、解答题（每小题8分，共64分）}
+
+\scorebox\textbf{1. （8分）【解】}
+% \vspace 用于生成一定高度的空白, \newpage 直接换页
+\vspace{3cm}
+
+\scorebox\textbf{2. （8分）【解】}
+\newpage
+
+\scorebox\textbf{3. （8分）【解】}
+\vspace{7cm}
+
+\scorebox\textbf{4. （8分）【解】}
+\newpage
+
+\scorebox\textbf{5. （8分）【解】}
+\vspace{7cm}
+
+\scorebox\textbf{6. （8分）【证明】}
+\newpage
+
+\scorebox\textbf{7. （8分）【证明】}
+\vspace{7cm}
+
+\scorebox\textbf{8. （8分）【解】}
+
+\end{document}
+
Property changes on: trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_datizhi.tex
___________________________________________________________________
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+native
\ No newline at end of property
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===================================================================
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===================================================================
--- trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_shijuan.pdf 2022-12-04 21:06:57 UTC (rev 65191)
+++ trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_shijuan.pdf 2022-12-04 21:07:43 UTC (rev 65192)
Property changes on: trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_shijuan.pdf
___________________________________________________________________
Added: svn:mime-type
## -0,0 +1 ##
+application/pdf
\ No newline at end of property
Added: trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_shijuan.tex
===================================================================
--- trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_shijuan.tex (rev 0)
+++ trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_shijuan.tex 2022-12-04 21:07:43 UTC (rev 65192)
@@ -0,0 +1,82 @@
+\documentclass{hfutexam}
+\newcommand{\diff}{\,\mathrm{d}}
+\usetikzlibrary{arrows.meta, overlay-beamer-styles}
+
+\begin{document}
+\BiaoTi{合肥工业大学试卷（A）}
+\XueNian{2021}{2022}
+\XueQi{二}
+\KeChengDaiMa{034Y01}
+\KeChengMingCheng{数学（下）}
+\XueFen{5}
+\KeChengXingZhi{必修}
+\KaoShiXingShi{闭卷}
+\ZhuanYeBanJi{少数民族预科班}
+\KaoShiRiQi{2022年6月18日8:00-10:00}
+\MingTiJiaoShi{集体}
+\XiZhuRenQianMing{}
+% \XiZhuRenQianMing{dengbing.png}
+
+\tigan{一、填空题（每题3分，共18分）}
+\begin{enumerate}
+\item 如果 $f(x)>0$ 且 $\displaystyle\lim_{x\to\infty}f(x)=0$, 则 $\displaystyle\lim_{x\to\infty}\bigl[1+f(x)\bigr]^{1/f(x)}=$\fillblank{}.
+\item 设 $y=\sin(x^2+1)$, 则 $\diff y=$\fillblank{}.
+\item 极限 $\displaystyle\lim_{n\to\infty}\left(\frac1{n^2-1}+\frac2{n^2-2}+\cdots+\frac n{n^2-n}\right)=$\fillblank{}.
+\item 曲线 $y=2\ln(x+1)$ 在点 $(1,2\ln2)$ 处的切线方程为\fillblank{}.
+\item 若 $e^{y-1}=1+xy$, 则 $\dfrac{\diff y}{\diff x}\bigg|_{x=0}=$\fillblank{}.
+\item 如果函数 $f(x)$ 的定义域是 $(0,+\infty)$, 且 $x=0$ 是曲线 $y=f(x)$ 的垂直渐近线, 那么 $\displaystyle\lim_{x\to0^+}\frac1{f(x)}=$\fillblank{}.
+\end{enumerate}
+
+\tigan{二、选择题（每题3分，共18分）}
+\begin{enumerate}
+\item 当 $x\to+\infty$ 时, $\dfrac1x$ 和(~~~~)是等价无穷小.
+% 自动根据选项长度设置行数
+\xx{$\sin\dfrac1x$}{$\sin x$}{$e^{-x}$}{$e^{1/x}$}
+\item 若当 $x\to0$ 时, $\arctan(e^x-1)\cdot(\cos x-1)$ 和 $x^n$ 是同阶无穷小, 则 $n=$(~~~~).
+\xx{$0$}{$1$}{$2$}{$3$}
+\item 设 $f(x)=\arctan\dfrac1{x(x-1)^2}$, 则 $x=0$ 是 $f(x)$ 的(~~~~).
+\xx{可去间断点}{跳跃间断点}{第二类间断点}{连续点}
+\item
+\begin{tikzpicture}[overlay,xshift=13cm,yshift=-3.5cm]
+\draw[-Stealth,thick](-3,0)--(3,0);
+\draw[-Stealth,thick](0,-1)--(0,3);
+\draw[very thick,smooth,domain=-55:55] plot ({\x/50-1.3}, {tan(\x)*tan(\x)});
+\draw[very thick,smooth,domain=0.15:2] plot ({\x}, {-ln(\x)});
+\draw
+ (-0.3,-0.3) node {$O$}
+ (2.8,-0.3) node {$x$}
+ (-0.3,2.8) node {$y$};
+\end{tikzpicture}
+设 $f(x)$ 是定义在 $(-\infty,+\infty)$ 上的连续函数, 且 $f'(x)$ 的图像如下图所示, 则 $f(x)$ 有(~~~~).
+% 手动设置为每行1个
+\xx[1]{一个极大值点，没有极小值点}{没有极大值点，一个极小值点}{一个极大值点和一个极小值点}{一个极大值点和两个极小值点}
+\newpage
+\item 设函数 $f(x)$ 在点 $x=0$ 处可导, 且 $f(0)=0$, 则 $\displaystyle\lim_{x\to0}\frac{f(x^{2022})+x^{2021}f(x)}{x^{2022}}=$(~~~~).
+\xx{$0$}{$f'(0)$}{$2f'(0)$}{$2022f'(0)$}
+\item 如果点 $(x_0,y_0)$ 是曲线 $y=f(x)$ 的拐点, 则 $f''(x_0)=$(~~~~).
+\xx{$0$}{$\infty$}{不存在}{$0$ 或不存在}
+\end{enumerate}
+
+\tigan{三、解答题（每题8分，共64分）}
+\begin{enumerate}
+\item 求极限 $\displaystyle\lim_{x\to-1}\frac{x^2-1}{x^2+3x+2}$.
+\item 求极限 $\displaystyle\lim_{x\to0}\frac{e^x-1-x}{\arcsin x^2}$.
+\item 设 $\begin{cases}x=t^2+t&\\y=t^3+t&\end{cases}$, 求 $\dfrac{\diff y}{\diff x}$ 和 $\dfrac{\diff^2 y}{\diff x^2}$.
+\item 设 $f(x)=\begin{cases}x\arctan\dfrac1x,&x<0,\\x^2+ax+b,&x\ge0.\end{cases}$
+求常数 $a,b$ 使得函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导, 并求出此时曲线 $y=f(x)$ 的渐近线.
+\item 求函数 $f(x)=x^3-x^2-x$ 在区间 $[-2,2]$ 上的最大值和最小值.
+\item 证明: 当 $-\dfrac\pi2<x_1<x_2<\dfrac\pi2$ 时, $\tan x_2-\tan x_1\ge x_2-x_1$.
+\item 设函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导, 且 $f(1)=0$.
+证明: 存在 $\xi\in(0,1)$ 使得 $\xi f'(\xi)+2022f(\xi)=0$.
+\item 设函数 $f(x)=\ln x+\dfrac2{x^2}, x\in(0,+\infty)$.
+\begin{enumerate}
+\item[(1)] 函数 $f(x)$ 的增减区间及极值;
+\item[(2)] 曲线 $y=f(x)$ 的凹凸区间及拐点.
+\end{enumerate}
+\end{enumerate}
+
+\end{document}
+
+
+
+
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___________________________________________________________________
Added: svn:eol-style
## -0,0 +1 ##
+native
\ No newline at end of property
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===================================================================
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===================================================================
--- trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_simple.pdf 2022-12-04 21:06:57 UTC (rev 65191)
+++ trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_simple.pdf 2022-12-04 21:07:43 UTC (rev 65192)
Property changes on: trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_simple.pdf
___________________________________________________________________
Added: svn:mime-type
## -0,0 +1 ##
+application/pdf
\ No newline at end of property
Added: trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_simple.tex
===================================================================
--- trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_simple.tex (rev 0)
+++ trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_simple.tex 2022-12-04 21:07:43 UTC (rev 65192)
@@ -0,0 +1,260 @@
+\documentclass[simple]{hfutexam}
+\newcommand{\diff}{\,\mathrm{d}}
+\usetikzlibrary{arrows.meta, overlay-beamer-styles}
+\RequirePackage{extarrows}
+
+\begin{document}
+\BiaoTi{合肥工业大学期中试卷}
+\XueNian{2021}{2022}
+\XueQi{二}
+\KeChengDaiMa{034Y01}
+\KeChengMingCheng{数学（下）}
+\XueFen{5}
+\KeChengXingZhi{必修}
+\KaoShiXingShi{闭卷}
+\ZhuanYeBanJi{少数民族预科班}
+\KaoShiRiQi{2022年5月13日8:00-10:00}
+\MingTiJiaoShi{集体}
+\maketitle
+
+\begin{enumerate}
+\item \textbf{(10分)} 求函数 $\displaystyle f(x)=\ln\frac1{\sqrt{x^2-1}}+\arctan\frac1x$ 的定义域.
+\item \textbf{(5分)} 求函数 $\displaystyle y=\begin{cases}
+1/x,&x<0,\\1,&x=0,\\1+e^{-x},&x>0\end{cases}$ 的反函数.
+\item \textbf{(10分)} 求极限 $\displaystyle\lim_{x\to0^-}(1-x)^{1/x}$.
+\item \textbf{(5分)} 求极限 $\displaystyle\lim_{x\to-2}\frac{x^2-4}{x^3+8}$.
+\item \textbf{(5分)} 求极限 $\displaystyle\lim_{x\to0}\frac{\sin(e^{-x}-1)}{\arctan(1-\cos x)}$.
+\item \textbf{(5分)} 求极限 $\displaystyle\lim_{x\to0}\frac{\sqrt{1+2x-x^2}-\sqrt{1-2x+x^2}}x$.
+\item \textbf{(5分)} 求极限 $\displaystyle\lim_{x\to\infty}\left(\cos\frac1x\right)^{\frac1{\ln(1+x^2)-2\ln x}}$.
+\item \textbf{(5分)} 求极限 $\displaystyle\lim_{x\to\infty}\left(\frac\pi{e^x-1}-\arctan\frac x2\right)$.
+\item \textbf{(5分)} 求极限 $\displaystyle\lim_{n\to\infty}\left(\frac1{n^2+2}+\frac2{n^2+4}+\cdots+\frac n{n^2+2n}\right)$.
+\item \textbf{(5分)} 设 $a_1=4,a_{n+1}=\sqrt{a_n+6}$, 证明 $\displaystyle\lim_{n\to\infty}a_n$ 存在并求之.
+\item \textbf{(10分)} 证明 $e^x+x=4$ 在 $(0,+\infty)$ 内有零点.
+\item \textbf{(5分)} 设函数 $f(x)$ 在 $[-1,1]$ 上连续, 且 $f(-1)\le1\le f(1)$. 证明存在 $\xi\in[-1,1]$, 使得 $f(\xi)=\xi^2$.
+\item \textbf{(10分)} 求 $y=e^{x+1}\sin x-e^2\sin1$ 的导数.
+\item \textbf{(5分)} 求 $y=\arctan e^x$ 的导数.
+\item \textbf{(5分)} 求曲线 $y=\tan x$ 在点 $\left(-\dfrac\pi4,-1\right)$ 处的切线方程和法线方程.
+\item \textbf{(5分)} 设
+$\displaystyle f(x)=\begin{cases}\dfrac{e^{3x}-1}{\arctan x},&x<0,\\2x+a,&x\ge0\end{cases}$
+在 $x=0$ 处连续, 求常数 $a$.
+\end{enumerate}
+\newpage
+
+\BiaoTi{合肥工业大学试卷（A）}
+\XueNian{2021}{2022}
+\XueQi{二}
+\KeChengDaiMa{034Y01}
+\KeChengMingCheng{数学（下）}
+\XueFen{5}
+\KeChengXingZhi{必修}
+\KaoShiXingShi{闭卷}
+\ZhuanYeBanJi{少数民族预科班}
+\KaoShiRiQi{2022年6月18日8:00-10:00}
+\MingTiJiaoShi{集体}
+\maketitle
+
+\tigan{一、填空题（每题3分，共18分）}
+\begin{enumerate}
+\item 如果 $f(x)>0$ 且 $\displaystyle\lim_{x\to\infty}f(x)=0$, 则 $\displaystyle\lim_{x\to\infty}\bigl[1+f(x)\bigr]^{1/f(x)}=$\fillblank{}.
+\item 设 $y=\sin(x^2+1)$, 则 $\diff y=$\fillblank{}.
+\item 极限 $\displaystyle\lim_{n\to\infty}\left(\frac1{n^2-1}+\frac2{n^2-2}+\cdots+\frac n{n^2-n}\right)=$\fillblank{}.
+\item 曲线 $y=2\ln(x+1)$ 在点 $(1,2\ln2)$ 处的切线方程为\fillblank{}.
+\item 若 $e^{y-1}=1+xy$, 则 $\dfrac{\diff y}{\diff x}\bigg|_{x=0}=$\fillblank{}.
+\item 如果函数 $f(x)$ 的定义域是 $(0,+\infty)$, 且 $x=0$ 是曲线 $y=f(x)$ 的垂直渐近线, 那么 $\displaystyle\lim_{x\to0^+}\frac1{f(x)}=$\fillblank{}.
+\end{enumerate}
+
+\tigan{二、选择题（每题3分，共18分）}
+\begin{enumerate}
+\item 当 $x\to+\infty$ 时, $\dfrac1x$ 和(~~~~)是等价无穷小.
+\xx{$\sin\dfrac1x$}{$\sin x$}{$e^{-x}$}{$e^{1/x}$}
+\item 若当 $x\to0$ 时, $\arctan(e^x-1)\cdot(\cos x-1)$ 和 $x^n$ 是同阶无穷小, 则 $n=$(~~~~).
+\xx{$0$}{$1$}{$2$}{$3$}
+\item 设 $f(x)=\arctan\dfrac1{x(x-1)^2}$, 则 $x=0$ 是 $f(x)$ 的(~~~~).
+\xx{可去间断点}{跳跃间断点}{第二类间断点}{连续点}
+\item
+\begin{tikzpicture}[overlay,xshift=12cm,yshift=-3cm]
+\draw[-Stealth,thick](-3,0)--(3,0);
+\draw[-Stealth,thick](0,-0.8)--(0,2.5);
+\draw[very thick,smooth,domain=-55:55] plot ({\x/50-1.3}, {tan(\x)*tan(\x)});
+\draw[very thick,smooth,domain=0.15:2] plot ({\x}, {-ln(\x)});
+\draw
+ (-0.3,-0.3) node {$O$}
+ (2.8,-0.3) node {$x$}
+ (-0.3,2.3) node {$y$};
+\end{tikzpicture}
+设 $f(x)$ 是定义在 $(-\infty,+\infty)$ 上的连续函数, 且 $f'(x)$ 的图像如下图所示, 则 $f(x)$ 有(~~~~).
+\xx[1]{一个极大值点，没有极小值点}{没有极大值点，一个极小值点}{一个极大值点和一个极小值点}{一个极大值点和两个极小值点}
+\item 设 $f(x)$ 在点 $x=0$ 处可导, 且 $f(0)=0$, 则 $\displaystyle\lim_{x\to0}\frac{f(x^{2022})+x^{2021}f(x)}{x^{2022}}=$(~~~~).
+\xx{$0$}{$f'(0)$}{$2f'(0)$}{$2022f'(0)$}
+\item 如果点 $(x_0,y_0)$ 是曲线 $y=f(x)$ 的拐点, 则 $f''(x_0)=$(~~~~).
+\xx{$0$}{$\infty$}{不存在}{$0$ 或不存在}
+\end{enumerate}
+
+\tigan{三、解答题（每题8分，共64分）}
+\begin{enumerate}
+\item 求极限 $\displaystyle\lim_{x\to-1}\frac{x^2-1}{x^2+3x+2}$.
+\item 求极限 $\displaystyle\lim_{x\to0}\frac{e^x-1-x}{\arcsin x^2}$.
+\item 设 $\begin{cases}x=t^2+t&\\y=t^3+t&\end{cases}$, 求 $\dfrac{\diff y}{\diff x}$ 和 $\dfrac{\diff^2 y}{\diff x^2}$.
+\item 设 $f(x)=\begin{cases}x\arctan\dfrac1x,&x<0,\\x^2+ax+b,&x\ge0.\end{cases}$
+求常数 $a,b$ 使得函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导, 并求出此时曲线 $y=f(x)$ 的渐近线.
+\item 求函数 $f(x)=x^3-x^2-x$ 在区间 $[-2,2]$ 上的最大值和最小值.
+\item 证明: 当 $-\dfrac\pi2<x_1<x_2<\dfrac\pi2$ 时, $\tan x_2-\tan x_1\ge x_2-x_1$.
+\item 设函数 $f(x)$ 在 $(-\infty,+\infty)$ 内可导, 且 $f(1)=0$.
+证明: 存在 $\xi\in(0,1)$ 使得 $\xi f'(\xi)+2022f(\xi)=0$.
+\item 设函数 $f(x)=\ln x+\dfrac2{x^2}, x\in(0,+\infty)$. 求
+\begin{enumerate}
+\item[(1)] 函数 $f(x)$ 的增减区间及极值；
+\item[(2)] 曲线 $y=f(x)$ 的凹凸区间及拐点.
+\end{enumerate}
+\end{enumerate}
+
+\newpage
+\BiaoTi{合肥工业大学试卷参考答案（A）}
+\XueNian{2021}{2022}
+\XueQi{二}
+\KeChengDaiMa{034Y01}
+\KeChengMingCheng{数学（下）}
+\XueFen{5}
+\KeChengXingZhi{必修}
+\KaoShiXingShi{闭卷}
+\ZhuanYeBanJi{少数民族预科班}
+\KaoShiRiQi{2022年6月18日8:00-10:00}
+\MingTiJiaoShi{集体}
+\maketitle
+
+\tigan{一、填空题（每小题3分，共18分）}
+
+\textbf{请将你的答案对应填在横线上：}
+
+\textbf{1.} \fillblank{$e$},
+\textbf{2.} \fillblank{$2x\cos(x^2+1)\diff x$},
+\textbf{3.} \fillblank{$\dfrac12$}, \\
+\textbf{4.} \fillblank{$y=x-1+2\ln 2$},
+\textbf{5.} \fillblank{$1$},
+\textbf{6.} \fillblank{$0$}.
+
+\tigan{二、选择题（每小题3分，共18分）}
+
+\textbf{请将你所选择的字母 A, B, C, D 之一对应填在下列表格里：}
+
+\xuanzeti{\textbf{题号}}{\textbf{答案}}%
+\xuanzeti{1}{A}%
+\xuanzeti{2}{D}%
+\xuanzeti{3}{B}%
+\xuanzeti{4}{A}%
+\xuanzeti{5}{C}%
+\xuanzeti{6}{D}
+
+\tigan{三、解答题（每小题8分，共64分）}
+
+\textbf{1. （8分）【解】}
+\vspace{-\baselineskip}
+
+\begin{align*}
+\lim_{x\to-1}\frac{x^2-1}{x^2+3x+2}&=\lim_{x\to-1}\frac{(x-1)(x+1)}{(x+2)(x+1)} \score3\\
+&=\lim_{x\to-1}\frac{x-1}{x+2} \score3\\
+&=\frac{-2}1=-2. \score2
+\end{align*}
+
+\textbf{2. （8分）【解】}
+\vspace{-\baselineskip}
+
+\begin{align*}
+\lim_{x\to0}\frac{e^x-1-x}{\arcsin x^2}&=\lim_{x\to0}\frac{e^x-1-x}{x^2} \score3\\
+&\xlongequal[]{\text{洛必达}}\lim_{x\to0}\frac{e^x-1}{2x} \score3\\
+&=\lim_{x\to0}\frac{x}{2x}=\frac12. \score2
+\end{align*}
+
+\textbf{3. （8分）【解】}
+\vspace{-\baselineskip}
+
+\begin{align*}
+\frac{\diff y}{\diff x}&=\frac{\diff y/\diff t}{\diff x/\diff t} \score2\\
+&=\frac{3t^2+1}{2t+1}, \score2\\
+\frac{\diff^2 y}{\diff x^2}&=\frac{\diff y'/\diff t}{\diff x/\diff t} \score2\\
+&=\frac{6t(2t+1)-(3t^2+1)2}{(2t+1)^3}=\frac{6t^2+6t-2}{(2t+1)^3}. \score2
+\end{align*}
+
+\newpage
+\textbf{4. （8分）【解】}
+
+\indent 由于 $f(x)$ 在 $x=0$ 处连续, 因此
+\begin{align*}
+f(0)&=f(0^+) \score1\\
+&=b=\lim_{x\to0^-}x\arctan\frac1x=0\times\left(-\frac\pi2\right)=0. \score1
+\end{align*}
+
+\indent 由于 $f(x)$ 在 $x=0$ 处可导, 因此
+\begin{align*}
+f'_-(0)&=f'_+(0), \score1\\
+f'_-(0)&=\lim_{x\to0^-}\frac{x\arctan\frac1x}x=\lim_{x\to0^-}\arctan\frac1x=-\frac\pi2 \score1\\
+f'_+(0)&=(2x+a)|_{x=0}=a, \score1
+\end{align*}
+因此 $a=-\dfrac\pi2$. \score1
+
+\indent 由于
+\begin{align*}
+\lim_{x\to+\infty}\frac yx&=\lim_{x\to+\infty}\left(x-\frac\pi2\right)=+\infty, \score1\\
+\lim_{x\to-\infty}\frac yx&=\lim_{x\to-\infty}\arctan\frac1x=0,\\
+\lim_{x\to-\infty}y&=\lim_{x\to-\infty}x\arctan\frac1x=\lim_{t\to0^-}\frac{\arctan t}t=1,
+\end{align*}
+因此曲线 $y=f(x)$ 的渐近线只有 $y=1$. \score1
+
+\textbf{5. （8分）【解】}
+
+\indent 由
+\[f'(x)=3x^2-2x-1=(3x+1)(x-1)=0 \score2\]
+可得驻点 $x=-\dfrac13,1$. \score2
+
+\indent 由于
+\[f(-2)=-10,\quad f(2)=2,\quad f\left(-\frac13\right)=\frac5{27},\quad f(1)=-1, \score2\]
+因此最大值为 $2$, 最小值为 $-10$. \score2
+
+\textbf{6. （8分）【证明】}
+
+\textbf{证法一}: 设 $f(x)=\tan x-x$, 则 \score2
+\[f'(x)=\frac1{\cos^2x}-1=\tan^2x\ge0. \score2\]
+因此 $f(x)$ 在 $\left(-\dfrac\pi2,\dfrac\pi2\right)$ 上单调递增, 从而 \score2
+\[f(x_2)\ge f(x_1),\quad\tan x_2-\tan x_1\ge x_2-x_1. \score2\]
+
+\newpage
+\textbf{证法二}: 设 $f(x)=\tan x$, 则 $f(x)$ 在 $[x_1,x_2]$ 上连续, $(x_1,x_2)$ 内可导. \score2
+
+\indent 由拉格朗日中值定理, 存在 $\xi\in(x_1,x_2)$ 使得
+\[\frac{f(x_2)-f(x_1)}{x_2-x_1}=f'(\xi), \score2\]
+即
+\[\frac{\tan x_2-\tan x_1}{x_2-x_1}=\frac1{\cos^2\xi}\ge1. \score2\]
+所以 $\tan x_2-\tan x_1\ge x_2-x_1$. \score2
+
+\textbf{7. （8分）【证明】}
+
+\indent 设 $F(x)=x^{2022}f(x)$, \score2\\
+则 $F(x)$ 在 $[0,1]$ 上连续, $(0,1)$ 内可导, \score1\\
+且 $F(0)=0,F(1)=f(1)=0$. \score1
+
+\indent 由罗尔中值定理, 存在 $\xi\in(0,1)$ 使得 $F'(\xi)=0$. \score2\\
+由于 $F'(x)=x^{2022}f'(x)+2022x^{2021}f(x)$ 且 $\xi\neq0$, \score1\\
+所以 $\xi f'(\xi)+2022f(\xi)=1$. \score1
+
+
+\textbf{8. （8分）【解】}
+
+(1)
+\[f'(x)=\frac1x-\frac4{x^3}=\frac{x^2-4}{x^3}=\frac{(x+2)(x-2)}{x^3}. \score1\]
+当 $0<x<2$ 时, $f'(x)<0$. 当 $x>2$ 时, $f'(x)>0$. \score1\\
+因此 $(0,2]$ 是 $f(x)$ 的单减区间,\\
+$[2,+\infty)$ 是 $f(x)$ 的单增区间. \Score{(1分, 写成开区间不扣分)}\\
+所以 $f(x)$ 只有唯一的极小值 $f(2)=\ln2+\dfrac12$. \score1
+
+(2)
+\[f''(x)=-\frac1{x^2}+\frac{12}{x^4}=-\frac{x^2-12}{x^4}=-\frac{(x-2\sqrt3)(x+2\sqrt3)}{x^4}. \score1\]
+当 $0<x<2\sqrt3$ 时, $f''(x)>0$. 当 $x>2\sqrt3$ 时, $f''(x)<0$. \score1\\
+因此 $(0,2\sqrt3]$ 是曲线 $y=f(x)$ 的凹区间,\\
+$[2\sqrt3,+\infty)$ 是曲线 $y=f(x)$ 的凸区间,\Score{(1分, 写成开区间不扣分)}\\
+拐点为 $\left(2\sqrt3,\ln(2\sqrt3)+\dfrac16\right)$. \score1
+
+\end{document}
+
+
+
+
Property changes on: trunk/Master/texmf-dist/doc/latex/hfutexam/hfutexam_simple.tex
___________________________________________________________________
Added: svn:eol-style
## -0,0 +1 ##
+native
\ No newline at end of property
Added: trunk/Master/texmf-dist/tex/latex/hfutexam/hfutexam.cls
===================================================================
--- trunk/Master/texmf-dist/tex/latex/hfutexam/hfutexam.cls (rev 0)
+++ trunk/Master/texmf-dist/tex/latex/hfutexam/hfutexam.cls 2022-12-04 21:07:43 UTC (rev 65192)
@@ -0,0 +1,346 @@
+% 文件 `hfutexam.cls'
+% !TEX TS-program = xelatex
+% !TEX encoding = UTF-8 Unicode
+% 合肥工业大学试卷模板
+% 作者: 张神星
+% 使用前请先安装字体: 方正小标宋、方正仿宋, 否则请使用选项 nofangzheng (采用新宋体和仿宋代替)
+% 编译模式： XeLaTeX
+% 你可以任意修改或再次分发该文件
+\NeedsTeXFormat{LaTeX2e}
+\ProvidesClass{hfutexam}[2022/12/04 v1.5 HFUTExam document class by Zhang Shenxing]
+% 文档选项 shijuan, datizhi, cankaodaan, simple
+\newif\ifHFUT at ShiJuan\HFUT at ShiJuantrue
+\newif\ifHFUT at DaTiZhi\HFUT at DaTiZhifalse
+\newif\ifHFUT at CanKaoDaAn\HFUT at CanKaoDaAnfalse
+\newif\ifHFUT at Simple\HFUT at Simplefalse
+\DeclareOption{datizhi}{\HFUT at ShiJuanfalse\HFUT at DaTiZhitrue}
+\DeclareOption{cankaodaan}{\HFUT at ShiJuanfalse\HFUT at CanKaoDaAntrue}
+\DeclareOption{simple}{\HFUT at ShiJuanfalse\HFUT at Simpletrue}
+% 文档选项 nofangzheng
+\newif\ifHFUT at Fandol\HFUT at Fandoltrue
+\DeclareOption{nofangzheng}{\HFUT at Fandolfalse}
+% 基于 ctexart 文档类
+\DeclareOption*{\PassOptionsToClass{\CurrentOption}{ctexart}}
+\ProcessOptions\relax
+\ifHFUT at Simple\else\PassOptionsToClass{twocolumn}{ctexart}\fi
+\ProcessOptions\relax
+\LoadClass[12pt,oneside]{ctexart}
+% 页面设置
+\RequirePackage{amsmath,amssymb,amsthm}
+\RequirePackage{graphicx}
+\RequirePackage{enumitem}
+\RequirePackage{geometry}
+\ifHFUT at Simple
+ \geometry{a4paper,scale=0.8}
+\else
+ \geometry{
+ paperheight=29.7cm,
+ paperwidth=42cm,
+ left=3.17cm,
+ right=3.17cm,
+ voffset=2.2cm,
+ headheight=77pt,
+ headsep=12pt,
+ footskip=1cm
+ }
+\fi
+% 试卷信息
+\newcommand{\BiaoTi}[1]{\gdef\HFUT at BiaoTi{#1}}
+\newcommand{\XueNian}[2]{\gdef\HFUT at XueNians{#1}\gdef\HFUT at XueNiane{#2}}
+\newcommand{\XueQi}[1]{\gdef\HFUT at XueQi{#1}}
+\newcommand{\KeChengDaiMa}[1]{\gdef\HFUT at KeChengDaiMa{#1}}
+\newcommand{\KeChengMingCheng}[1]{\gdef\HFUT at KeChengMingCheng{#1}}
+\newcommand{\XueFen}[1]{\gdef\HFUT at XueFen{#1}}
+\newcommand{\KeChengXingZhi}[1]{\gdef\HFUT at KeChengXingZhi{#1}}
+\newcommand{\KaoShiXingShi}[1]{\gdef\HFUT at KaoShiXingShi{#1}}
+\newcommand{\ZhuanYeBanJi}[1]{\gdef\HFUT at ZhuanYeBanJi{#1}}
+\newcommand{\KaoShiRiQi}[1]{\gdef\HFUT at KaoShiRiQi{#1}}
+\newcommand{\MingTiJiaoShi}[1]{\gdef\HFUT at MingTiJiaoShi{#1}}
+\newcommand{\XiZhuRenQianMing}[1]{\gdef\HFUT at XiZhuRenQianMing{#1}}
+\BiaoTi{}
+\XueNian{}{}
+\XueQi{}
+\KeChengDaiMa{}
+\KeChengMingCheng{}
+\XueFen{}
+\KeChengXingZhi{}
+\KaoShiXingShi{}
+\ZhuanYeBanJi{}
+\KaoShiRiQi{}
+\MingTiJiaoShi{}
+\XiZhuRenQianMing{}
+% 粗字体设定
+\newfontfamily\timesnewroman[AutoFakeBold={1.5}]{Times New Roman} % 粗罗马
+\ifHFUT at Fandol
+ \setCJKfamilyfont{titlesongti}[AutoFakeBold={1.5}]{FZXiaoBiaoSong-B05S} % 粗方正小标宋
+ \newfontfamily\entitlesongti[AutoFakeBold={1.5}]{FZXiaoBiaoSong-B05S}
+ \setCJKfamilyfont{titlefangsong}[AutoFakeBold={1.5}]{FZFangSong-Z02S} % 粗方正仿宋
+ \newfontfamily\entitlefangsong[AutoFakeBold={1.5}]{FZFangSong-Z02S}
+\else
+ \setCJKfamilyfont{titlesongti}[AutoFakeBold={4}]{NSimSun} % 粗新宋体
+ \newfontfamily\entitlesongti[AutoFakeBold={4}]{NSimSun}
+ \setCJKfamilyfont{titlefangsong}[AutoFakeBold={1.5}]{FangSong} % 粗仿宋
+ \newfontfamily\entitlefangsong[AutoFakeBold={1.5}]{FangSong}
+\fi
+\newcommand{\titlesongti}{\entitlesongti\CJKfamily{titlesongti}}
+\newcommand{\titlefangsong}{\entitlefangsong\CJKfamily{titlefangsong}}
+\setCJKfamilyfont{xinsongti}[AutoFakeBold={1.5}]{NSimSun} % 粗新宋体
+\newfontfamily\enxinsongti[AutoFakeBold={1.5}]{NSimSun}
+\newcommand{\xinsongti}{\enxinsongti\CJKfamily{xinsongti}}
+\setCJKfamilyfont{cusongti}[AutoFakeBold={1.5}]{SimSun} % 粗宋体
+\newcommand{\cusongti}{\CJKfamily{cusongti}}
+% 重新设定字体大小
+\renewcommand{\LARGE}{\fontsize{21}{21}}
+\renewcommand{\large}{\fontsize{14}{14}}
+\renewcommand{\normalsize}{\fontsize{12}{12}}
+\renewcommand{\small}{\fontsize{10.5}{10.5}}
+% 缩进
+\setlength{\parindent}{0em}
+\renewcommand{\indent}{\hspace*{2em}}
+% 分栏
+\setlength\columnsep{0.8cm} % 分栏间距
+\columnseprule=0.5pt % 分栏线宽度
+% 填空题
+\newlength{\ltemp}
+\RequirePackage[normalem]{ulem}
+\ifHFUT at ShiJuan%
+ \NewDocumentCommand\fillblank{O{3.5cm} O{0.5cm} m}{\uline{\makebox[#1]{\rule{0pt}{#2}#3}}}%
+ \else\ifHFUT at DaTiZhi%
+ \NewDocumentCommand\fillblank{O{3.5cm} O{1cm} m}{\uline{\makebox[#1]{\rule{0pt}{#2}#3}}}%
+ \else%
+ \NewDocumentCommand\fillblank{O{3.5cm} O{1cm} m}{%
+ \settodepth{\ltemp}{#3}%
+ \uline{\makebox[#1]{\rule{0pt}{#2}\raisebox{\ltemp}{#3}}}%
+ }%
+\fi\fi
+% 页眉页脚
+\RequirePackage{fancyhdr}
+\RequirePackage{lastpage}
+\RequirePackage{etoolbox}
+\renewcommand\headrulewidth{0.5pt} % 页眉线宽度
+\renewcommand\footrulewidth{0.5pt} % 页脚线宽度
+\pagestyle{fancy}
+\fancyhf{}
+\newcommand{\boxyes}{{\raisebox{-0.5mm}{\LARGE$\checkmark\hspace{-1.2em}\square$}}}
+\newcommand{\boxno}{{\raisebox{-0.5mm}{\LARGE$\square$}}}
+\newcommand{\filltitle}[2]{\uline{\makebox[#1]{#2}}}
+\newlength{\ltitle}
+\settowidth{\ltitle}{\HFUT at BiaoTi}
+\ifHFUT at ShiJuan%试卷页眉页脚
+ \fancyhead[C]{%
+ \hfill\bfseries\LARGE\titlesongti\ifdimcomp\ltitle>{0.5\linewidth}{\HFUT at BiaoTi}{\makebox[0.5\linewidth][s]{\HFUT at BiaoTi}}\hfill%
+ \large\cusongti 共\filltitle{1.5em}{\timesnewroman\pageref{LastPage}}页第\filltitle{1.5em}{\timesnewroman\thepage}页\\\vspace{5pt}%
+ \timesnewroman\HFUT at XueNians\titlefangsong～\timesnewroman\HFUT at XueNiane\titlefangsong 学年第\filltitle{2em}{\HFUT at XueQi}学期\hfill%
+ 课程代码\filltitle{5em}{\HFUT at KeChengDaiMa}\hfill%
+ 课程名称\filltitle{12em}{\HFUT at KeChengMingCheng}\hfill%
+ 学分\filltitle{2.5em}{\HFUT at XueFen}\hfill%
+ 课程性质: 必修\ifdefstring{\HFUT at KeChengXingZhi}{必修}{\boxyes}{\boxno}%
+ 选修\ifdefstring{\HFUT at KeChengXingZhi}{选修}{\boxyes}{\boxno}%
+ 限修\ifdefstring{\HFUT at KeChengXingZhi}{限修}{\boxyes}{\boxno}\hfill%
+ 考试形式: 开卷\ifdefstring{\HFUT at KaoShiXingShi}{开卷}{\boxyes}{\boxno}%
+ 闭卷\ifdefstring{\HFUT at KaoShiXingShi}{闭卷}{\boxyes}{\boxno}\\\vspace{2pt}%
+ 专业班级（教学班）\filltitle{11em}{\HFUT at ZhuanYeBanJi}\hfill%
+ 考试日期\filltitle{16em}{\HFUT at KaoShiRiQi}\hfill%
+ 命题教师\fillblank[5.5em][0.6cm]{\HFUT at MingTiJiaoShi}\hfill%
+ 系（所或教研室）主任审批签名%
+ \begin{tikzpicture}[overlay,xshift=3.25em,yshift=0.15cm]%
+ \node at (0,0) {\ifx\HFUT at XiZhuRenQianMing\@empty\else\includegraphics[height=0.9cm]{\HFUT at XiZhuRenQianMing}\fi};%
+ \end{tikzpicture}%
+ \filltitle{6.5em}{}\vspace{4pt}%
+ }
+ \fancyfoot[C]{\small\vspace{0.5\baselineskip}命题教师注意事项:
+ 1. 主考教师必须于考试一周前将“试卷A”、“试卷B”经教研室主任审批签字后送教务科印刷。\hspace{1em}%
+ 2. 请命题教师用黑色水笔工整地书写题目或用A4纸横式打印贴在试卷版芯中。
+ }
+\fi
+\ifHFUT at DaTiZhi%答题纸页眉页脚
+ \fancyhead[C]{%
+ \bfseries\LARGE\xinsongti\scalebox{2.0}[1.0]{%
+ \ifdimcomp\ltitle>{0.35\linewidth}{\HFUT at BiaoTi}{\makebox[0.35\linewidth][s]{\HFUT at BiaoTi}}%
+ }\\\vspace{8pt}%
+ \large\timesnewroman\HFUT at XueNians\titlefangsong～\timesnewroman\HFUT at XueNiane\titlefangsong 学年第\filltitle{2em}{\HFUT at XueQi}学期\hfill%
+ 课程代码\filltitle{6.5em}{\HFUT at KeChengDaiMa}\hfill%
+ 课程名称\filltitle{13em}{\HFUT at KeChengMingCheng}\hfill%
+ 命题教师\filltitle{7em}{\HFUT at MingTiJiaoShi}\hfill%
+ 系主任审批%
+ \begin{tikzpicture}[overlay,xshift=3.5em,yshift=0.2cm]%
+ \node at (0,0) {\ifx\HFUT at XiZhuRenQianMing\@empty\else\includegraphics[height=0.9cm]{\HFUT at XiZhuRenQianMing}\fi};%
+ \end{tikzpicture}%
+ \filltitle{7em}{}\\\vspace{2pt}%
+ 教学班级\filltitle{10.5em}{}\hfill%
+ 学生姓名\fillblank[8em][0.6cm]{}\hfill%
+ 学号\filltitle{8em}{}\hfill%
+ 考试日期\filltitle{16em}{\HFUT at KaoShiRiQi}\hfill%
+ 成绩\filltitle{6.3em}{}\vspace{4pt}%
+ }
+ \fancyfoot[C]{\small\vspace{0.5\baselineskip}
+ 第 \timesnewroman\thepage 页~共 \timesnewroman\pageref{LastPage} 页
+ }
+\fi
+\ifHFUT at CanKaoDaAn%参考答案页眉页脚
+ \fancyhead[C]{%
+ \bfseries\LARGE\titlesongti\ifdimcomp\ltitle>{0.6\linewidth}{\HFUT at BiaoTi}{\makebox[0.6\linewidth][s]{\HFUT at BiaoTi}}\\\vspace{5pt}%
+ \large\timesnewroman\HFUT at XueNians\titlefangsong～\timesnewroman\HFUT at XueNiane\titlefangsong 学年第\filltitle{2em}{\HFUT at XueQi}学期\hfill%
+ 课程代码\filltitle{6.5em}{\HFUT at KeChengDaiMa}\hfill%
+ 课程名称\filltitle{13em}{\HFUT at KeChengMingCheng}\hfill%
+ 命题教师\filltitle{7em}{\HFUT at MingTiJiaoShi}\hfill%
+ 系主任审批%
+ \begin{tikzpicture}[overlay,xshift=3.5em,yshift=0.2cm]%
+ \node at (0,0) {\ifx\HFUT at XiZhuRenQianMing\@empty\else\includegraphics[height=0.9cm]{\HFUT at XiZhuRenQianMing}\fi};%
+ \end{tikzpicture}%
+ \filltitle{7em}{}\\\vspace{2pt}%
+ 教学班级\filltitle{10.5em}{}\hfill%
+ 学生姓名\fillblank[8em][0.6cm]{}\hfill%
+ 学号\filltitle{8em}{}\hfill%
+ 考试日期\filltitle{16em}{\HFUT at KaoShiRiQi}\hfill%
+ 成绩\filltitle{6.3em}{}\vspace{4pt}%
+ }
+ \fancyfoot[C]{\small\vspace{0.5\baselineskip}
+ 第 \timesnewroman\thepage 页~共 \timesnewroman\pageref{LastPage} 页
+ }
+\fi
+\ifHFUT at Simple%简单模式页眉页脚
+ \renewcommand\headrulewidth{0pt}
+ \renewcommand\footrulewidth{0pt}
+ \renewcommand{\maketitle}{%
+ \begin{center}
+ \bfseries\LARGE\titlesongti%
+ \ifdimcomp\ltitle>{0.9\linewidth}{\HFUT at BiaoTi}{\makebox[0.9\linewidth][s]{\HFUT at BiaoTi}}\\%
+ \makebox[0.75\linewidth]{\large\timesnewroman\HFUT at XueNians\titlefangsong～\timesnewroman\HFUT at XueNiane\titlefangsong 学年%
+ 第{\HFUT at XueQi}学期\hfill\HFUT at KeChengMingCheng(\HFUT at KeChengDaiMa)}
+ \end{center}
+ }
+ \fancyfoot[C]{\small\vspace{0.5\baselineskip}
+ 第 \timesnewroman\thepage 页~共 \timesnewroman\pageref{LastPage} 页
+ }
+\fi
+% 题号加粗
+\renewcommand{\labelenumi}{{\bfseries \theenumi.}}
+% 题干
+\newcommand\tigan[1]{\noindent{\large\textbf{#1}}}
+% 打分栏
+\RequirePackage{tikz}
+\usetikzlibrary{overlay-beamer-styles}
+\newcommand\scorebox{%
+ \vspace{0.5\baselineskip}\noindent%
+ \begin{tikzpicture}[overlay,xshift=13.8cm,yshift=-1.6cm]%
+ \draw (0,0) rectangle (3.6,2);%
+ \draw (1.8,0)--(1.8,2);%
+ \draw (0,1)--(3.6,1);%
+ \draw (0.9,1.5) node {\textbf{\normalsize 得分}};%
+ \draw (2.7,1.5) node {\textbf{\normalsize 阅卷人}};%
+ \end{tikzpicture}%
+}
+% 答题纸提示信息
+\newcommand\notice{%
+ \noindent\textbf{\small 考生注意事项：\\%
+ \indent 1. 本试卷分试题与答卷两部分；\\%
+ \indent 2. 所有试题的解答（包括选择、填空）必须写在专用答卷纸上，在试题上直接作答一律无效；\\%
+ \indent 3. 考试结束后，必须将试题、答卷整理上交，不得将试题带离考场；\\%
+ \indent 4. 考生务必认真填写班级、姓名、学号等信息。}\par%
+ {\leavevmode\xleaders\hbox{\rule[4pt]{8pt}{0.5pt}\,}\hfill\null}%
+}
+% 选择题, 根据选项内容长度自动排版
+\newlength{\lxxmax}
+\newlength{\lquar}
+\newlength{\lhalf}
+\newlength{\lfull}
+\newcounter{lxxtype}
+\NewDocumentCommand\xx{O{0} m m m m}{%
+ \setlength{\lfull}{\columnwidth}%
+ \addtolength{\lfull}{-\leftmargin}%
+ \setlength{\lhalf}{0.5\lfull}%
+ \setlength{\lquar}{0.25\lfull}%
+ \setcounter{lxxtype}{0}%
+ \ifnum#1=1\setcounter{lxxtype}{1}\fi%
+ \ifnum#1=2\setcounter{lxxtype}{2}\fi%
+ \ifnum#1=4\setcounter{lxxtype}{4}\fi%
+ \settowidth{\lxxmax}{A.~#2~}% 获取最长选项长度
+ \settowidth{\ltemp}{B.~#3~}%
+ \ifdimcomp\ltemp>\lxxmax{\setlength{\lxxmax}{\ltemp}}{}%
+ \settowidth{\ltemp}{C.~#4~}%
+ \ifdimcomp\ltemp>\lxxmax{\setlength{\lxxmax}{\ltemp}}{}%
+ \settowidth{\ltemp}{D.~#5~}%
+ \ifdimcomp\ltemp>\lxxmax{\setlength{\lxxmax}{\ltemp}}{}%
+ \ifnum\value{lxxtype}=0%
+ \setcounter{lxxtype}{4}%
+ \ifdimcomp\lxxmax>\lquar{\setcounter{lxxtype}{2}}{}%
+ \ifnum\value{lxxtype}=2%
+ \ifdimcomp\lxxmax>\lhalf{\setcounter{lxxtype}{1}}{}%
+ \fi%
+ \fi%
+ \vspace{5pt}%
+ \ifnum\value{lxxtype}=1%
+ \\\makebox[\lfull][l]{A.~#2}%
+ \\\makebox[\lfull][l]{B.~#3}%
+ \\\makebox[\lfull][l]{C.~#4}%
+ \\\makebox[\lfull][l]{D.~#5}%
+ \fi%
+ \ifnum\value{lxxtype}=2%
+ \\\makebox[\lhalf][l]{A.~#2}%
+ \makebox[\lhalf][l]{B.~#3}%
+ \\\makebox[\lhalf][l]{C.~#4}%
+ \makebox[\lhalf][l]{D.~#5}%
+ \fi%
+ \ifnum\value{lxxtype}=4%
+ \\\makebox[\lquar][l]{A.~#2}%
+ \makebox[\lquar][l]{B.~#3}%
+ \makebox[\lquar][l]{C.~#4}%
+ \makebox[\lquar][l]{D.~#5}%
+ \fi%
+}
+% 选择题答题框
+\newcommand\xuanzeti[2]{%
+ \begin{tikzpicture}%
+ \draw (0,0) rectangle (1.8,2);%
+ \draw (0,1)--(1.8,1);%
+ \draw (0.9,0.5) node {#2} (0.9,1.5) node {\textbf{#1}};%
+ \end{tikzpicture}%
+}
+% 得分点命令
+% https://ask.latexstudio.net/ask/question/7557.html
+\RequirePackage{zref-savepos}
+\makeatletter
+\ExplSyntaxOn
+\zref at require@unique
+\NewDocumentCommand{\examscore}{O{} m}{
+ \mode_if_math:TF
+ { \@@_math_cdotfill:n {\text{#2}} }
+ { \__examzh_cdotfill: #2 }
+ \mode_if_math:F{
+ \ignorespaces
+ }
+}
+\cs_new:Npn \__examzh_cdotfill:
+{
+ \mode_leave_vertical:
+ \cleaders \hb at xt@ .44em {\hss $\cdot$ \hss} \hfill
+ \kern\z@
+}
+\cs_new_protected:Npn \@@_math_cdotfill:n #1
+{
+ \stepcounter { zref at unique }
+ \hbox_overlap_right:n
+ {
+ \zsaveposx { \thezref at unique L }
+ \zref at ifrefundefined { \thezref at unique R }
+ { }
+ {
+ \cleaders
+ \hbox_to_wd:nn { .44em } { \hss $\cdot$ \hss }
+ \skip_horizontal:n
+ {
+ \zposx { \thezref at unique R } sp
+ - \zposx { \thezref at unique L } sp
+ }
+ }
+ }
+ \tag * { \zsaveposx { \thezref at unique R } #1 }
+}
+\ExplSyntaxOff
+\makeatother
+\newcommand\Score[1]{~~~\examscore{#1}}
+\newcommand\score[1]{~~~\examscore{(#1分)}}
+\renewcommand\le{\leqslant}
+\renewcommand\ge{\geqslant}
+
Property changes on: trunk/Master/texmf-dist/tex/latex/hfutexam/hfutexam.cls
___________________________________________________________________
Added: svn:eol-style
## -0,0 +1 ##
+native
\ No newline at end of property
Modified: trunk/Master/tlpkg/bin/tlpkg-ctan-check
===================================================================
--- trunk/Master/tlpkg/bin/tlpkg-ctan-check 2022-12-04 21:06:57 UTC (rev 65191)
+++ trunk/Master/tlpkg/bin/tlpkg-ctan-check 2022-12-04 21:07:43 UTC (rev 65192)
@@ -396,7 +396,7 @@
hep-math hep-math-font hep-paper hep-reference hep-text hep-title hepnames
hepparticles hepthesis hepunits here hereapplies heros-otf hershey-mp
heuristica hexboard hexgame
- hf-tikz hfbright hfoldsty hfutthesis
+ hf-tikz hfbright hfoldsty hfutexam hfutthesis
hhtensor hideanswer highlightlatex hindawi-latex-template hindmadurai
histogr historische-zeitschrift hitec hitreport
hitszthesis hitszbeamer hithesis
Modified: trunk/Master/tlpkg/tlpsrc/collection-publishers.tlpsrc
===================================================================
--- trunk/Master/tlpkg/tlpsrc/collection-publishers.tlpsrc 2022-12-04 21:06:57 UTC (rev 65191)
+++ trunk/Master/tlpkg/tlpsrc/collection-publishers.tlpsrc 2022-12-04 21:07:43 UTC (rev 65192)
@@ -89,6 +89,7 @@
depend har2nat
depend hecthese
depend hep-paper
+depend hfutexam
depend hfutthesis
depend hithesis
depend hitszthesis
Added: trunk/Master/tlpkg/tlpsrc/hfutexam.tlpsrc
===================================================================
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