texlive[64093] Master/texmf-dist: bhcexam (10aug22)
commits+karl at tug.org
commits+karl at tug.org
Wed Aug 10 22:38:07 CEST 2022
Revision: 64093
http://tug.org/svn/texlive?view=revision&revision=64093
Author: karl
Date: 2022-08-10 22:38:07 +0200 (Wed, 10 Aug 2022)
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bhcexam (10aug22)
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+# BHCexam
+
+[English](./README.md)
+
+BHCexam 是一款为中国数学老师设计的试卷排版文档类，目前已被 [橘子数学开源题库社区](https://www.mathcrowd.cn) 选用为默认文档类导出试卷pdf文件.
+
+你可以使用该文档类实现：
+
+* 将内容与样式分离
+* 使用一个参数生成教师版和学生版试卷;
+* 排版3-6个选项的选择题，可根据选项的长度自适应对齐;
+* 排版填空题，可根据答案的长度自适应设置横线长度;
+* 排版简答题，并以列表形式展示小问，并控制是否展示小问、缩进;
+* 对试题分组，对组内试题以列表呈现，并控制是否展示分值、留空、是否重新开始编号;
+* 更多 (见 [BHCexam 文档](http://docs.mathcrowd.cn/advances/bhcexam.html) )
+
+## 版本历史
+
+* **version 1.7** (2022.8)
+ * 针对TeXLive 2022, 修正`ctex`的字号、字体设置.
+* **version 1.6** (2021.8)
+ * 不再使用`stix`数学字体
+ * `ctex` 设置 `punct = kaiming`
+* **version 1.5** (2020.6)
+ * `questions` 环境新增 `r` 选项，重置题组的编号;
+ * 在 `master` 分支中清理历史版本;
+ * 新增 `fandol` 宏包选项以支持 `fandol` 字体;
+* **version 1.4** (2020.5)
+ * 支持`subquestion`环境的嵌套;
+ * `\parallel` 命令重定义;
+* **version 1.3** (2020.3)
+ * 新增 `\sixchoices` , `\threechoices` 命令，以支持对3个和6个选项，并保持智能断行同及选项对齐.
+* **version 1.2** (2020.3)
+ * 支持苹果字体
+ * 使用 `stix` 数学字体
+ * 支持在选择题最后显示右对齐括号
+* **version 1.1** (2020.1)
+ * 新增对A3双栏版式的支持
+ * 新增列表样式的试题
+* **version 1.0** (2019.5)
+ * 弃用 `exam` 而改用 `article` 为基宏包类
+
+## 贡献者
+
+* Bao Hongchang - @mathedu4all, charles at mathcrowd.cn
+* CamuseCao - @ CamuseCao, camusecao at gmail.com
+
+------
+
+This work may be distributed and/or modified under the conditions of
+the LaTeX Project Public License, either version 1.3 of this license
+or (at your option) any later version.
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+# BHCexam
+
+[中文版](./README-zh.md)
+
+BHCexam is an exam class designed for mathematics teacher in China.
+
+Now it is used by [mathcrowd.cn](www.mathcrowd.cn) ( an opensource math exam database) as the default class to export exam papers in pdf.
+
+Using bhcexam you can
+
+* separate the format and the content very well;
+* export both teacher paper and student paper;
+* typeset multiple choice questions with 3-6 options keeping adaptively neat alignment;
+* typeset cloze questions with an adaptively underline;
+* typeset questions with subquestions in list;
+* group questions in list to control whether to show score, leave spacing, initialize question number;
+* and more (see [BHCexam Documentation](http://docs.mathcrowd.cn/advances/bhcexam.html)).
+
+## Authors and Contributors
+
+* Bao Hongchang <charles at mathcrowd.cn>
+* CamuseCao <camusecao at gmail.com>
+
+------
+
+This work may be distributed and/or modified under the conditions of
+the LaTeX Project Public License, either version 1.3 of this license
+or (at your option) any later version.
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+\documentclass[list, windows]{BHCexam}
+\pagestyle{fancy}
+\fancyfoot[C]{\kaishu \small 第 \thepage 页 共 \pageref{lastpage} 页}
+\fancyhead[L]{\includegraphics[width=2cm]{qrcode.png}}
+\fancyhead[R]{\raisebox{0.5\height}{\includegraphics[height=1cm]{logo.png}}}
+\begin{document}
+\title{上海某高中 2017-2018 学年度第一学期}
+\subtitle{高一数学期中试卷}
+\notice{满分150分，120分钟完成，允许使用计算器，答案一律写在答题纸上}
+\author{微信关注公众号：橘子数学}
+\date{2017.11}
+\maketitle
+\begin{groups}
+\group{填空}{第1-6题每题4分，第7-12题每题5分.}
+
+\begin{questions}[p]
+\begin{minipage}{\linewidth}
+\question [4] 已知集合$U=\{1,2,3,4\}$,集合$A=\{1,2\}$,$B=\{2,3\}$,则$(A\cap\complement_UB) \cup (\complement_UA\cap B)=$\key{$\{1,3\}$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$A\cap\complement_UB=\{1\}$, $B\cap\complement_UA=\{3\}$.
+
+$(A\cap\complement_UB) \cup (\complement_UA\cap B)=\{1,3\}$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [4] 设集合$M=\{x|0\lt x\leqslant{}3\}$,$N=\{x|0\lt x\leqslant{}2\}$,那么 “ $a\in{M}$ ” 是 “ $a\in{N}$ ” 的\key{必要非充分}条件.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+由$N \subsetneqq M$，可知 “ $a\in{M}$ ” 是 “ $a\in{N}$ ” 的必要非充分条件.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [4] 函数$f(x)=\sqrt{x+1}+\dfrac{1}{2-x}$的定义域为\key{$[-1,2)\cup{}(2,+\infty)$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+根据题意:$\left\{\begin{array}{l}{x+1\geqslant{}0}\\{2-x\neq{}0}\end{array}\right.$
+
+解得:$x\geqslant{}-1$且$x\neq{}2$
+
+定义域是:$[-1,2)\cup{}(2,+\infty)$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [4] 已知集合$A=\{x \big| |x-a| \lt 1,x\in\symbf{R}\}$,
+$B=\{x\Big|\dfrac{2x-a}{x+1} \lt 1,x\in\symbf{R}\}$,
+且$A\cap B=\varnothing$,
+ 则实数$a$的取值范围是\key{$a \in (-\infty,-2]$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$A=(a-1,a+1)$，
+
+由$\dfrac{2x-a}{x+1} \lt 1$化简得$\dfrac{x-a-1}{x+1} \lt 0$.
+
+若$a+1 \gt -1$，不满足$A \cap B=\varnothing$.
+
+若$a+1 \le -1$，满足$A \cap B=\varnothing$.
+
+故$a\in(-\infty,-2]$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [4] 已知$y=f(x), y=g(x)$是两个定义在$\symbf{R}$上的二次函数，其$x, y$的取值如下表所示:
+\[\begin{array}{|c|c|c|c|c|}
+\hline
+x & 1 & 2 & 3 & 4 \\
+\hline
+f(x) & -3 & -4 & -3 & 0\\
+\hline
+g(x) & 0 & 1 & 0 & -3\\
+\hline
+\end{array}\]
+则不等式$f(g(x))\ge0$的解集为\key{$(-\infty,1]\cup[3,+\infty)$.}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+由表格可得$y=f(x)$图像开口向上且关于$x=2$对称，其零点为$x=4$和$x=0$.
+
+故不等式$f(g(x))\ge0$, 即$g(x)\in(-\infty,0]\cup[4,+\infty)$.
+
+又$y=g(x)$的图像开口向下且关于$x=2$对称，其最大值为$1$.
+
+故$g(x)\in(-\infty,0]$，解得$x\in(-\infty,1]\cup[3,+\infty)$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [4] 关于$x$的不等式$2kx^2+kx+\frac{3}{8} \lt 0$的解集不为空集，则$k$的取值范围为\key{$k \in (-\infty,0)\cup(3,+\infty)$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+若$k \lt 0 $, 二次函数$y=2kx^2+kx+\frac{3}{8}$图像开口向下，解集恒不为空集. 满足.
+
+若$k=0$, 不等式解集为$\varnothing$，不满足.
+
+若$k \gt 0$，二次函数$y=2kx^2+kx+\frac{3}{8}$图像开口向上，要求$\Delta=k^2-4 \cdot (2k) \cdot \frac{3}{8}=k^2-3k \gt 0$，解得$k\in(3,+\infty)$.
+
+综上所属，$k \in (-\infty,0)\cup(3,+\infty)$
+
+\end{solution}
+\vfill
+\newpage
+\begin{minipage}{\linewidth}
+\question [5] 已知本张试卷的出卷人在公元$x^2$年时年龄为$x-8$岁，则出卷人的出生年份是\key{$1989$}.(假设出生当年的年龄为$1$岁)
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+设出卷人的出生年份是$y$.
+
+则有$x^2-y+1=x-8$.
+
+化简得$y=x^2-x+9$.
+
+由生活常识，
+
+$1952 \lt x^2-x+9 \lt 2000$, $x\in\symbf{N}$.
+
+解得$x=45$, 故$y=1989$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 若对任意$x\in\symbf{R}$,不等式$|x|\geqslant{}ax$恒成立,则实数$a$的取值范围是\key{$a \in [-1,1]$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+若$x \gt 0$，得$x \ge ax$， 解得$a \le 1$.
+
+若$x = 0$，得$0 \ge 0$成立.
+
+若$x \lt 0$，得$-x \ge ax$，解得$a \ge -1$.
+
+综上所述，$a \in [-1,1]$.
+\method
+画出$y=|x|$与$y=ax$的图像,
+
+要使$y=ax$的图像恒在$y=|x|$下方，
+
+则$a\in[-1,1]$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 设常数$a\gt \ 0$,若$9x+\dfrac{a^2}{x}\geqslant{}a+1$对一切正实数$x$成立,则$a$的取值范围为\key{$\left[\dfrac{1}{5}, + \infty \right)$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$\forall x\in{R}^{+},9x+\dfrac{a^2}{x}\geqslant{}2\sqrt{9x\cdot{}\dfrac{a^2}{x}}=6a$,
+
+$\therefore{}6a\geqslant{}a+1$,
+
+$\therefore{}a\geqslant{}\dfrac{1}{5}$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 设函数$f(x)=\begin{cases}x^2+2x+2,x\leqslant{}0\\-x^2,x\gt \ 0\end{cases}$,若$f(f(a))=2$,则$a=$\key{}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+若$a\leqslant{}0$,
+则$f(a)=a^2+2a+2=(a+1)^2+1 \gt 0$,
+而$f(f(a))=-(a^2+2a+2)^2=2$无解，舍去;
+
+若$a\gt \ 0$,
+则$f(a)=-a^2\lt 0$,
+故$f(f(a))=(-a^2)^2+2(-a^2)+2=2$,
+解得$a=\sqrt{2}$.
+
+综上所述, $a=\sqrt{2}$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 若二次函数$y=f(x)$对一切$x\in \text{R}$恒有${{x}^{2}}-2x+4\le f(x)\le 2{{x}^{2}}-4x+5$成立，且$f(5)=27$，则$f(11)=$\key{$153$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+画出不等式两边的两个二次函数图像，如图，可得未知的二次函数$y=f(x)$开口向上，以$(1,3)$为顶点，故可设函数解析式为$f(x)=a(x-1)^2+3$，将$f(5)=27$代入可得$a=\frac{3}{2}$，则$f(x)=\frac{3}{2}(x-1)^2+3$，故$f(11)=153$.
+\begin{center}
+\includegraphics[height=8cm]{./To7VxKXpZaOWO5Qq9qzeZGtlwYTwJ5KF.jpg}\\
+第11题
+\vspace{0.5cm}
+\end{center}
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 已知$f(x)=(a^2-5)x^2+2x+2$. 若不等式$f(x) \gt x$的解集为$A$. 已知$(0,1)\subseteq A$，则$a$的取值范围为\key{$a \in (-\infty,-\sqrt{2}]\cup[\sqrt{2},+\infty)$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+由题意得，$(a^2-5)x^2+x+2 \gt 0$在$(0,1)$上恒成立.
+
+即$a^2-5 \gt -\frac{2}{x^2}-\frac{1}{x}$在$(0,1)$上恒成立.
+
+设$f(x)=-\frac{2}{x^2}-\frac{1}{x}=-2(\frac{1}{x}+\frac{1}{4})^2+\frac{1}{8}$,
+
+又$\frac{1}{x}\in(1,+\infty)$, 故$f(x) \lt -3$.
+
+即$a^2-5 \ge -3$, 解得$a \in (-\infty,-\sqrt{2}]\cup[\sqrt{2},+\infty)$.
+
+\end{solution}
+\end{questions}
+
+\group{选择}{第13-16题每题5分.}
+
+\begin{questions}[p]
+\begin{minipage}{\linewidth}
+\question [5] 设$P,Q$为两个非空实数集,定义集合$P+Q=\{a+b|a\in{P},b\in{Q}\}$.若$P=\{0,2,5\}$,$Q=\{1,2,6\}$,则$P+Q$中元素的个数是\key{B}.
+\fourchoices{$9$}{$8$}{$7$}{$6$}
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$\because P=\{0,2,5\}$,$Q=\{1,2,6\}$,
+
+$\therefore P+Q=\{1,2,3,4,6,7,8,11\}$.
+
+故选B
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 不等式$(1+x)(1-|x|)\gt \ 0$的解集是\key{D}.
+\fourchoices{$\{x|0\leqslant{}x\lt 1\}$}{$\{x|x\lt 0$且$x\neq{}-1\}$}{$\{x|-1\lt x\lt 1\}$}{$\{x|x\lt 1$且$x\neq{}-1\}$}
+\end{minipage}
+\begin{solution}{4cm}
+\method
+求不等式$(1+x)(1-|x|)\gt \ 0$的解集,则分两种情况讨论:
+
+情况$1:\left\{\begin{array}{l}{1+x\gt \ 0\;\;}\\{1-|x|\gt \ 0}\end{array}\right.$
+
+即$\left\{\begin{array}{l}{x\gt \ -1}\\{-1\lt x\lt 1}\end{array}\right.$
+则$-1\lt x\lt 1$.
+
+情况$2:\left\{\begin{array}{l}{1+x\lt 0}\\{1-|x|\lt 0}\end{array}\right.$
+
+即$\left\{\begin{array}{l}{x\lt -1}\\{x\gt 1\text{或}x\lt -1}\end{array}\right.$
+则$x\lt -1$.
+
+两种情况取并集得$\{x|x\lt 1$且$x\neq{}-1\}$.
+
+故选D.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 已知三个不等式:$ab\gt \ 0$,$bc-ad\gt \ 0$,$\dfrac{c}{a}-\dfrac{d}{b}\gt \ 0$(其中$a,b,c,d$均为实数),用其中两个不等式作为条件,余下的一个不等式作为结论组成一个命题,可组成的正确命题的个数是\key{D}.
+\fourchoices{$0$}{$1$}{$2$}{$3$}
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$\begin{cases}ab\gt \ 0, &\cdots(1)\\
+bc-ad\gt \ 0,&\cdots(2)
+\end{cases}$
+
+$(2)\div(1)$得$\dfrac{bc-ad}{ab}=\dfrac{c}{a}-\dfrac{d}{b}\gt \ 0$.
+
+$\begin{cases}
+ab\gt \ 0,&\cdots(1)\\
+\dfrac{c}{a}-\dfrac{d}{b}\gt \ 0,&\cdots(3)
+\end{cases}$
+
+$(1)\times{}(3)$得$ab(\dfrac{c}{a}-\dfrac{d}{b})=bc-ad\gt \ 0$.
+
+$\begin{cases}bc-ad\gt \ 0,&\cdots(2)\\
+\dfrac{c}{a}-\dfrac{d}{b}\gt \ 0,&\cdots(3)
+\end{cases}$
+
+用反证法,显然$ab\neq{}0$,
+
+设$ab\lt 0$,$(2)$式同除$ab$得$\dfrac{c}{a}-\dfrac{d}{b}\lt 0$,矛盾,
+
+故$ab\gt \ 0$.
+故选D.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 设$a\gt \ 0$,$b\gt \ 0$,则以下不等式中不恒成立的是\key{B}.
+\fourchoices{$(a+b)(\dfrac{1}{a}+\dfrac{1}{b})\geqslant{}4$}{$a^{3}+b^{3}\geqslant{}2ab^{2}$}{$a^{2}+b^{2}+2\geqslant{}2a+2b$}{$\sqrt{|{a-b}|}\geqslant{}\sqrt{a}-\sqrt{b}$}
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$a\gt \ 0$,$b\gt \ 0$,
+
+$(a+b)(\dfrac{1}{a}+\dfrac{1}{b})\geqslant{}2\sqrt{ab}\cdot{}2\sqrt{\dfrac{1}{ab}}\geqslant{}4$,
+
+故A恒成立;
+
+$a^{3}+b^{3}\geqslant{}2ab^{2}$,
+
+取$a=\dfrac{1}{2}$,$b=\dfrac{2}{3}$,则B不成立;
+
+$a^{2}+b^{2}+2-(2a+2b)=(a-1)^{2}+(b-1)^{2}\geqslant{}0$,
+
+故C恒成立;
+
+若$a\lt b$,则$\sqrt{|{a-b}|}\geqslant{}\sqrt{a}-\sqrt{b}$恒成立,
+
+若$a\geqslant{}b$,则${(\sqrt{|{a-b}|})^2}-{(\sqrt{a}-\sqrt{b})^2}=2\sqrt{ab}\geqslant{}0$,
+
+故$\sqrt{|{a-b}|}\geqslant{}\sqrt{a}-\sqrt{b}$,
+即D恒成立.
+
+\end{solution}
+\end{questions}
+\group{简答}{第17-19题每题14分，第20题16分，第21题18分.}
+
+\begin{questions}[stp]
+\begin{minipage}{\linewidth}
+\question [14] 已知$\triangle ABC$为直角三角形, 记其两条直角边长分别为$a,b\in\symbf{R}^+$, 记面积为$S$, 周长为$C$. 若三角形面积为定值，其周长是否有最值，最大值还是最小值，何时取到，为多少（结果用$S$表示）？
+
+\end{minipage}
+\begin{solution}{6cm}
+\method
+$C=a+b+\sqrt{a^2+b^2}$,
+\score{2}{2}由$\sqrt{a^2+b^2} \ge \sqrt{2ab}$，$a+b \ge 2\sqrt{ab}$,
+
+得$C \ge 2\sqrt{ab}+\sqrt{2ab}$
+\score{6}{8}由$S=\frac{1}{2}ab$，
+
+得$C \ge (2+2\sqrt{2})\sqrt{S}$，
+\score{4}{12}当且仅当$a=b=\sqrt{2S}$时取到最小值.
+\score{2}{14}
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [14] 已知$a\in\textbf{R}$,若关于$x$的方程$x^2+x+|a-\dfrac{1}{4}|+|a|=0$有实根，求$a$的取值范围.
+
+\end{minipage}
+\begin{solution}{6cm}
+\method
+方程即$|a-\frac{1}{4}|+|a|=-x^2-x\in[0,\frac{1}{4}]$,
+
+即解不等式$|a-\frac{1}{4}|+|a| \le \frac{1}{4}$.
+
+当$a\le 0$，化简得$\frac{1}{4}-2a \le \frac{1}{4}$，解得$a=0$;
+
+当$0 \lt a\le \frac{1}{4}$, 化简得$\frac{1}{4} \le \frac{1}{4}$，解得$a\in(0,\frac{1}{4}]$;
+
+当$a \gt \frac{1}{4}$, 化简得$2a-\frac{1}{4}\le\frac{1}{4}$，无解.
+
+综上所述，$a\in [0,\frac{1}{4}]$
+\score{14}{14}
+\end{solution}
+\vfill
+\newpage
+\begin{minipage}{\linewidth}
+\question [14] 先阅读下列不等式的证法，再解决后面的问题:
+\[
+\text{证明:} (a_1b_1+a_2b_2)^2 \le (a_1^2+a_2^2)(b_1^2+b_2^2)\]
+
+证: 令$A=\sqrt{a_1^2+a_2^2}$, $B=\sqrt{b_1^2+b_2^2}$
+
+\[
+\begin{array}{rl}
+\frac{a_1b_1+a_2b_2}{\sqrt{a_1^2+a_2^2}\sqrt{b_1^2+b_2^2}}& =\frac{a_1b_1}{AB}+\frac{a_2b_2}{AB}\\
+& = \frac{a_1}{A}\cdot\frac{b_1}{B}+\frac{a_2}{A}\cdot\frac{b_2}{B} \\
+& \le \frac{1}{2}(\frac{a_1^2}{A^2}+\frac{b_1^2}{B^2})+ \frac{1}{2}(\frac{a_2^2}{A^2}+\frac{b_2^2}{B^2})\\
+& =\frac{1}{2}(\frac{a_1^2+a_2^2}{A^2}+\frac{b_1^2+b_2^2}{B^2})\\
+& =1
+\end{array}
+\]
+故$ (a_1b_1+a_2b_2)^2 \le (a_1^2+a_2^2)(b_1^2+b_2^2)$.
+\begin{subquestions}
+ \subquestion 若$x_1,y_1,x_2,y_2\in\symbf{R^+}$，利用上述结论，证明:\[
+(x_1+x_2)(y_1+y_2) \ge (\sqrt{x_1y_1}+\sqrt{x_2y_2})^2
+\]
+ \subquestion 若$x_1,y_1,x_2,y_2,z_1,z_2\in\symbf{R^+}$，模仿上述证法并结合(1)的证法，证明:\[
+(x_1+x_2)(y_1+y_2)(z_1+z_2) \ge (\sqrt[3]{x_1y_1z_1}+\sqrt[3]{x_2y_2z_2})^3
+\]
+(提示：若$a,b,c\in\symbf{R^+}$，有$\frac{a^3+b^3+c^3}{3}\ge abc$,)
+\end{subquestions}
+\end{minipage}
+\begin{solution}{3cm}
+\method
+(1)设$a_1=\sqrt{x_1}$，$a_2=\sqrt{x_2}$，$b_1=\sqrt{y_1}$，$b_2=\sqrt{y_2}$.
+
+由\[ (a_1b_1+a_2b_2)^2 \le (a_1^2+a_2^2)(b_1^2+b_2^2)\]
+
+得\[ (\sqrt{x_1}\sqrt{y_1}+\sqrt{x_2}\sqrt{y_2})^2 \le [((\sqrt{x_1})^2+(\sqrt{x_2})^2)((\sqrt{y_1})^2+(\sqrt{y_2})^2)]\]
+
+即\[
+(x_1+x_2)(y_1+y_2) \ge (\sqrt{x_1y_1}+\sqrt{x_2y_2})^2
+\]
+\score{6}{6}(2) 设$a_1=\sqrt[3]{x_1}, b_1=\sqrt[3]{y_1}, c_1=\sqrt[3]{z_1}$, $a_2=\sqrt[3]{x_2}, b_2=\sqrt[3]{y_2}, c_2=\sqrt[3]{z_2}$.
+
+要证
+\[
+(x_1+x_2)(y_1+y_2)(z_1+z_2) \ge (\sqrt[3]{x_1y_1z_1}+\sqrt[3]{x_2y_2z_2})^3
+\]
+
+即证
+\[
+(a_1b_1c_1+a_2b_2c_2)^3 \le (a_1^3+a_2^3)(b_1^3+b_2^3)(c_1^3+c_2^3)
+\]
+
+(不换元，直接证不扣分)
+\score{2}{8}令$A=\sqrt[3]{a_1^3+a_2^3}$，$B=\sqrt[3]{b_1^3+b_2^3}$, $C=\sqrt[3]{c_1^3+c_2^3}$.
+
+\[
+\begin{array}{rl}
+& \frac{a_1b_1c_1+a_2b_2c_2}{\sqrt[3]{a_1^3+a_2^3}\sqrt[3]{b_1^3+b_2^3}\sqrt[3]{c_1^3+c_2^3}}\\
+= & \frac{a_1b_1c_1}{ABC}+\frac{a_2b_2c_2}{ABC}\\
+= & \frac{a_1}{A}\cdot\frac{b_1}{B}\cdot\frac{c_1}{C}+\frac{a_2}{A}\cdot\frac{b_2}{B}\cdot\frac{c_2}{C}\\
+\le & \frac{1}{3}(\frac{a_1^3}{A^3}+\frac{b_1^3}{B^3}+\frac{c_1^3}{C^3})+ \frac{1}{3}(\frac{a_2^3}{A^3}+\frac{b_2^3}{B^3}+\frac{c_2^3}{C^3})\\
+= & \frac{1}{3}(\frac{a_1^3+a_2^3}{A^3}+\frac{b_1^3+b_2^3}{B^3}++\frac{c_1^3+c_2^3}{C^3})\\
+= & 1
+\end{array}
+\]
+
+故\[(a_1b_1c_1+a_2b_2c_2)^3 \le (a_1^3+a_2^3)(b_1^3+b_2^3)(c_1^3+c_2^3)\]
+
+得证.
+\score{6}{14}
+\end{solution}
+\vfill
+\newpage
+\begin{minipage}{\linewidth}
+\question [16] 公元2222年，有一种高危传染疾病在全球范围内蔓延，被感染者的潜伏期可以长达10年，期间会有约$0.05\%$的概率传染给他人，一旦发病三天内即死亡。某城市总人口约200万人，专家分析其中约有$1000$名感染者，为了防止疾病继续扩散，疾病预防控制中心现决定对全市人口进行血液检测以筛选出被感染者。由于检测试剂十分昂贵且数量有限，需要将血样混合后一起检测以节约试剂。已知感染者的检测结果为阳性, 未被感染者则为阴性. 阳性血样与阴性血样混合后的检测结果为阳性, 同为阴性或阳性的血样混合后结果不发生改变.
+\begin{subquestions}
+ \subquestion 若对全市人口进行平均分组，同一分组的血样将被混合到一起检测，若发现结果为阳性，则再在该分组内逐个检测排查. 设每个组$x$个人，那么最坏情况下，需要进行约多少次检测可以找到所有的被感染者? 在当前方案下，若要使检测的次数尽可能少，每个分组的最优人数是？
+ \subquestion 在(1)的检测方案中，对于检测结果为阳性的组采取逐一检测排查的方法并不是很好，或可将这些组的血样再进行一次分组混合血样检测，然后再进行逐一排查。仍然考虑最坏的情况，请问两次要如何分组，使检测总次数尽可能少？
+ \subquestion 在(2)的检测方案中，进行了两次分组混合血样检测。仍然考虑最坏情况，若再进行若干次分组混合血样检测，是否会使检测次数更少？请给出最优的检测方案.
+\end{subquestions}
+\end{minipage}
+\begin{solution}{6cm}
+\method
+设需要进行$y$次检测. 由每个组$x$个人，则有$\frac{2000000}{x}$个小组, 最坏情况是1000人被分配到了不同的组.
+
+\[
+y=\begin{cases}
+\frac{2\cdot 10^6}{x}+2\cdot 10^6, & x \ge 2000 \\
+\frac{2\cdot 10^6}{x}+1000\cdot x, & x \lt 2000 \\
+\end{cases}
+\]
+
+(题目要求近似计算，可以不考虑整除，考虑整除不扣分)
+\score{2}{2}分组数小于1000的情况比逐个检测还要多，可以不用考虑，
+
+$y=\frac{2\times 10^6}{x}+1000\cdot x \ge 2\sqrt{2\cdot 10^9}$.
+
+当$x=\sqrt{2\times10^3}\approx45$人，检测次数最少.
+
+(结果为44-45都给分)
+\score{2}{4}(2) 设第二次分组每个组$w$人.
+
+\[
+\begin{array}{rl}
+y & = \frac{2\times10^6}{x}+\frac{1000x}{w}+1000w\\
+& \ge 3\sqrt[3]{2\times10^{12}}\\
+\end{array}
+\]
+
+当$\frac{2\times10^6}{x}=\frac{1000x}{w}=1000w$时，
+即$x=2000^\frac{2}{3}\approx159$,$w=2000^\frac{1}{3}\approx13$，
+检测次数最少.
+
+所以第一次分组159人一组，第二次分组13人一组.
+
+(也可以用两次基本不等式求得，允许$10\%$的误差.)
+\score{6}{10}(3) 若进行$k$次分组，设第$i$次分组时每组人数为$x_i$. 则总的检测次数为:
+\[
+\begin{array}{rl}
+y=&\dfrac{2\times10^6}{x_1}+\dfrac{1000x_1}{x_2}+\cdots\\
+&+\dfrac{1000x_{k-1}}{x_k}+1000x_k\\
+\ge & (k+1)\sqrt[k+1]{2\times10^{3k+2}}
+\end{array}
+\]
+
+利用计算器可得，
+
+1次分组， $y \ge 2\sqrt{2\cdot 10^9}\approx8.94\times10^4$;
+
+2次分组， $y \ge 3\sqrt[3]{2\cdot 10^{12}}\approx3.78\times10^4$;
+
+3次分组， $y \ge 4\sqrt[4]{2\cdot 10^{15}}\approx2.68\times10^4$;
+
+4次分组， $y \ge 5\sqrt[5]{2\cdot 10^{18}}\approx2.29\times10^4$;
+
+5次分组， $y \ge 6\sqrt[6]{2\cdot 10^{21}}\approx2.13\times10^4$;
+
+6次分组， $y \ge 7\sqrt[7]{2\cdot 10^{24}}\approx2.073\times10^4$;
+
+7次分组， $y \ge 8\sqrt[8]{2\cdot 10^{27}}\approx2.068\times10^4$;
+
+8次分组， $y \ge 9\sqrt[9]{2\cdot 10^{30}}\approx2.094\times10^4$;
+
+9次分组， $y \ge 10\sqrt[10]{2\cdot 10^{33}}\approx2.138\times10^4$;
+
+可见进行7次分组混合血样检测最优.
+
+其中第k次分组时$(2000)^{\frac{8-k}{8}}$人一组.
+\score{6}{16}
+\end{solution}
+\vfill
+\newpage
+\begin{minipage}{\linewidth}
+\question [18] 已知函数$f(x)=a{{x}^{2}}-\frac{1}{2}x+c$（$a$、$c\in R$），满足$f(1)=0$，且$f(x)\ge 0$在$x\in R$时恒成立．
+\begin{subquestions}
+ \subquestion 求$a$、$c$的值；
+ \subquestion 若$h(x)=\frac{3}{4}{{x}^{2}}-bx+\frac{b}{2}-\frac{1}{4}$，解不等式$f(x)+h(x) \lt 0$；
+ \subquestion 是否存在实数$m$，使函数$g(x)=f(x)-mx$在区间$[m,m+2]$上有最小值$-5$？若存在，请求出$m$的值；若不存在，请说明理由．
+\end{subquestions}
+\end{minipage}
+\begin{solution}{6cm}
+\method
+由$f(1)=0$，得$a+c=\frac{1}{2}$,
+\score{1}{1}因为$f(x)\ge 0$在$x\in R$时恒成立，所以$a \gt 0$且△$=\frac{1}{4}-4ac\le 0$，$ac\ge \frac{1}{16}$，
+\score{1}{2}即$a\left( \frac{1}{2}-a \right)\ge \frac{1}{16}$，${{a}^{2}}-\frac{1}{2}a+\frac{1}{16}\le 0$，${{\left( a-\frac{1}{4} \right)}^{2}}\le 0$，所以$a=c=\frac{1}{4}$．
+\score{2}{4}(2) 由（1）得$f(x)=\frac{1}{4}{{x}^{2}}-\frac{1}{2}x+\frac{1}{4}$，由$f(x)+h(x) \lt 0$，得
+${{x}^{2}}-\left( b+\frac{1}{2} \right)x+\frac{b}{2} \lt 0$，即$(x-b)\left( x-\frac{1}{2} \right) \lt 0$，
+\score{3}{7}所以，当$b \lt \frac{1}{2}$时，原不等式解集为$(b,\frac{1}{2})$；
+当$b \gt \frac{1}{2}$时，原不等式解集为$(\frac{1}{2},b)$；
+当$b=\frac{1}{2}$时，原不等式解集为空集 ．
+\score{3}{10}(3) $g(x)=\frac{1}{4}{{x}^{2}}-\left( \frac{1}{2}+m \right)x+\frac{1}{4}$，
+\score{1}{11}$g(x)$的图像是开口向上的抛物线，对称轴为直线$x=2m+1$．
+假设存在实数$m$，使函数$g(x)$在区间$[m,m+2]$上有最小值$-5$．
+\ding{192} 当$2m+1 \lt m$，即$m \lt -1$时，函数$g(x)$在区间$[m,m+2]$上是增函数，所以$g(m)=-5$，即$\frac{1}{4}{{m}^{2}}-\left( \frac{1}{2}+m \right)m+\frac{1}{4}=-5$，解得$m=-3$或$m=\frac{7}{3}$，
+因为$m \lt -1$，所以$m=-3$；
+\score{2}{13}\ding{193}当$m\le 2m+1\le m+2$，即$-1\le m\le 1$时，函数$g(x)$的最小值为$g(2m+1)=-5$，即
+$\frac{1}{4}{{(2m+1)}^{2}}-\left( \frac{1}{2}+m \right)(2m+1)+\frac{1}{4}=-5$，解得$m=-\frac{1}{2}-\frac{\sqrt{21}}{2}$或$m=-\frac{1}{2}+\frac{\sqrt{21}}{2}$，均舍去；
+\score{2}{15}\ding{194}当$2m+1 \gt m+2$，即$m \gt 1$时，$g(x)$在区间$[m,m+2]$上是减函数，所以$g(m+2)=-5$，即$\frac{1}{4}{{(m+2)}^{2}}-\left( \frac{1}{2}+m \right)(m+2)+\frac{1}{4}=-5$，解得$m=-1-2\sqrt{2}$或$m=-1+2\sqrt{2}$，因$m \gt 1$，所以$m=-1+2\sqrt{2}$．
+\score{2}{17}综上，存在实数$m$，$m=-3$或$m=-1+2\sqrt{2}$时，函数$g(x)$在区间$[m,m+2]$上有最小值$-5$．
+\score{1}{18}
+\end{solution}
+\end{questions}
+\end{groups}
+\label{lastpage}
+\end{document}
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+\documentclass[list, windows, answers]{BHCexam}
+\pagestyle{fancy}
+\fancyfoot[C]{\kaishu \small 第 \thepage 页 共 \pageref{lastpage} 页}
+\fancyhead[L]{\includegraphics[width=2cm]{qrcode.png}}
+\fancyhead[R]{\raisebox{0.5\height}{\includegraphics[height=1cm]{logo.png}}}
+\begin{document}
+\title{上海某高中 2017-2018 学年度第一学期}
+\subtitle{高一数学期中试卷}
+\notice{满分150分，120分钟完成，允许使用计算器，答案一律写在答题纸上}
+\author{微信关注公众号：橘子数学}
+\date{2017.11}
+\maketitle
+\begin{groups}
+\group{填空}{第1-6题每题4分，第7-12题每题5分.}
+
+\begin{questions}[p]
+\begin{minipage}{\linewidth}
+\question [4] 已知集合$U=\{1,2,3,4\}$,集合$A=\{1,2\}$,$B=\{2,3\}$,则$(A\cap\complement_UB) \cup (\complement_UA\cap B)=$\key{$\{1,3\}$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$A\cap\complement_UB=\{1\}$, $B\cap\complement_UA=\{3\}$.
+
+$(A\cap\complement_UB) \cup (\complement_UA\cap B)=\{1,3\}$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [4] 设集合$M=\{x|0\lt x\leqslant{}3\}$,$N=\{x|0\lt x\leqslant{}2\}$,那么 “ $a\in{M}$ ” 是 “ $a\in{N}$ ” 的\key{必要非充分}条件.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+由$N \subsetneqq M$，可知 “ $a\in{M}$ ” 是 “ $a\in{N}$ ” 的必要非充分条件.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [4] 函数$f(x)=\sqrt{x+1}+\dfrac{1}{2-x}$的定义域为\key{$[-1,2)\cup{}(2,+\infty)$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+根据题意:$\left\{\begin{array}{l}{x+1\geqslant{}0}\\{2-x\neq{}0}\end{array}\right.$
+
+解得:$x\geqslant{}-1$且$x\neq{}2$
+
+定义域是:$[-1,2)\cup{}(2,+\infty)$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [4] 已知集合$A=\{x \big| |x-a| \lt 1,x\in\symbf{R}\}$,
+$B=\{x\Big|\dfrac{2x-a}{x+1} \lt 1,x\in\symbf{R}\}$,
+且$A\cap B=\varnothing$,
+ 则实数$a$的取值范围是\key{$a \in (-\infty,-2]$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$A=(a-1,a+1)$，
+
+由$\dfrac{2x-a}{x+1} \lt 1$化简得$\dfrac{x-a-1}{x+1} \lt 0$.
+
+若$a+1 \gt -1$，不满足$A \cap B=\varnothing$.
+
+若$a+1 \le -1$，满足$A \cap B=\varnothing$.
+
+故$a\in(-\infty,-2]$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [4] 已知$y=f(x), y=g(x)$是两个定义在$\symbf{R}$上的二次函数，其$x, y$的取值如下表所示:
+\[\begin{array}{|c|c|c|c|c|}
+\hline
+x & 1 & 2 & 3 & 4 \\
+\hline
+f(x) & -3 & -4 & -3 & 0\\
+\hline
+g(x) & 0 & 1 & 0 & -3\\
+\hline
+\end{array}\]
+则不等式$f(g(x))\ge0$的解集为\key{$(-\infty,1]\cup[3,+\infty)$.}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+由表格可得$y=f(x)$图像开口向上且关于$x=2$对称，其零点为$x=4$和$x=0$.
+
+故不等式$f(g(x))\ge0$, 即$g(x)\in(-\infty,0]\cup[4,+\infty)$.
+
+又$y=g(x)$的图像开口向下且关于$x=2$对称，其最大值为$1$.
+
+故$g(x)\in(-\infty,0]$，解得$x\in(-\infty,1]\cup[3,+\infty)$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [4] 关于$x$的不等式$2kx^2+kx+\frac{3}{8} \lt 0$的解集不为空集，则$k$的取值范围为\key{$k \in (-\infty,0)\cup(3,+\infty)$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+若$k \lt 0 $, 二次函数$y=2kx^2+kx+\frac{3}{8}$图像开口向下，解集恒不为空集. 满足.
+
+若$k=0$, 不等式解集为$\varnothing$，不满足.
+
+若$k \gt 0$，二次函数$y=2kx^2+kx+\frac{3}{8}$图像开口向上，要求$\Delta=k^2-4 \cdot (2k) \cdot \frac{3}{8}=k^2-3k \gt 0$，解得$k\in(3,+\infty)$.
+
+综上所属，$k \in (-\infty,0)\cup(3,+\infty)$
+
+\end{solution}
+\vfill
+\newpage
+\begin{minipage}{\linewidth}
+\question [5] 已知本张试卷的出卷人在公元$x^2$年时年龄为$x-8$岁，则出卷人的出生年份是\key{$1989$}.(假设出生当年的年龄为$1$岁)
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+设出卷人的出生年份是$y$.
+
+则有$x^2-y+1=x-8$.
+
+化简得$y=x^2-x+9$.
+
+由生活常识，
+
+$1952 \lt x^2-x+9 \lt 2000$, $x\in\symbf{N}$.
+
+解得$x=45$, 故$y=1989$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 若对任意$x\in\symbf{R}$,不等式$|x|\geqslant{}ax$恒成立,则实数$a$的取值范围是\key{$a \in [-1,1]$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+若$x \gt 0$，得$x \ge ax$， 解得$a \le 1$.
+
+若$x = 0$，得$0 \ge 0$成立.
+
+若$x \lt 0$，得$-x \ge ax$，解得$a \ge -1$.
+
+综上所述，$a \in [-1,1]$.
+\method
+画出$y=|x|$与$y=ax$的图像,
+
+要使$y=ax$的图像恒在$y=|x|$下方，
+
+则$a\in[-1,1]$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 设常数$a\gt \ 0$,若$9x+\dfrac{a^2}{x}\geqslant{}a+1$对一切正实数$x$成立,则$a$的取值范围为\key{$\left[\dfrac{1}{5}, + \infty \right)$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$\forall x\in{R}^{+},9x+\dfrac{a^2}{x}\geqslant{}2\sqrt{9x\cdot{}\dfrac{a^2}{x}}=6a$,
+
+$\therefore{}6a\geqslant{}a+1$,
+
+$\therefore{}a\geqslant{}\dfrac{1}{5}$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 设函数$f(x)=\begin{cases}x^2+2x+2,x\leqslant{}0\\-x^2,x\gt \ 0\end{cases}$,若$f(f(a))=2$,则$a=$\key{}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+若$a\leqslant{}0$,
+则$f(a)=a^2+2a+2=(a+1)^2+1 \gt 0$,
+而$f(f(a))=-(a^2+2a+2)^2=2$无解，舍去;
+
+若$a\gt \ 0$,
+则$f(a)=-a^2\lt 0$,
+故$f(f(a))=(-a^2)^2+2(-a^2)+2=2$,
+解得$a=\sqrt{2}$.
+
+综上所述, $a=\sqrt{2}$.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 若二次函数$y=f(x)$对一切$x\in \text{R}$恒有${{x}^{2}}-2x+4\le f(x)\le 2{{x}^{2}}-4x+5$成立，且$f(5)=27$，则$f(11)=$\key{$153$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+画出不等式两边的两个二次函数图像，如图，可得未知的二次函数$y=f(x)$开口向上，以$(1,3)$为顶点，故可设函数解析式为$f(x)=a(x-1)^2+3$，将$f(5)=27$代入可得$a=\frac{3}{2}$，则$f(x)=\frac{3}{2}(x-1)^2+3$，故$f(11)=153$.
+\begin{center}
+\includegraphics[height=8cm]{./To7VxKXpZaOWO5Qq9qzeZGtlwYTwJ5KF.jpg}\\
+第11题
+\vspace{0.5cm}
+\end{center}
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 已知$f(x)=(a^2-5)x^2+2x+2$. 若不等式$f(x) \gt x$的解集为$A$. 已知$(0,1)\subseteq A$，则$a$的取值范围为\key{$a \in (-\infty,-\sqrt{2}]\cup[\sqrt{2},+\infty)$}.
+
+\end{minipage}
+\begin{solution}{4cm}
+\method
+由题意得，$(a^2-5)x^2+x+2 \gt 0$在$(0,1)$上恒成立.
+
+即$a^2-5 \gt -\frac{2}{x^2}-\frac{1}{x}$在$(0,1)$上恒成立.
+
+设$f(x)=-\frac{2}{x^2}-\frac{1}{x}=-2(\frac{1}{x}+\frac{1}{4})^2+\frac{1}{8}$,
+
+又$\frac{1}{x}\in(1,+\infty)$, 故$f(x) \lt -3$.
+
+即$a^2-5 \ge -3$, 解得$a \in (-\infty,-\sqrt{2}]\cup[\sqrt{2},+\infty)$.
+
+\end{solution}
+\end{questions}
+
+\group{选择}{第13-16题每题5分.}
+
+\begin{questions}[p]
+\begin{minipage}{\linewidth}
+\question [5] 设$P,Q$为两个非空实数集,定义集合$P+Q=\{a+b|a\in{P},b\in{Q}\}$.若$P=\{0,2,5\}$,$Q=\{1,2,6\}$,则$P+Q$中元素的个数是\key{B}.
+\fourchoices{$9$}{$8$}{$7$}{$6$}
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$\because P=\{0,2,5\}$,$Q=\{1,2,6\}$,
+
+$\therefore P+Q=\{1,2,3,4,6,7,8,11\}$.
+
+故选B
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 不等式$(1+x)(1-|x|)\gt \ 0$的解集是\key{D}.
+\fourchoices{$\{x|0\leqslant{}x\lt 1\}$}{$\{x|x\lt 0$且$x\neq{}-1\}$}{$\{x|-1\lt x\lt 1\}$}{$\{x|x\lt 1$且$x\neq{}-1\}$}
+\end{minipage}
+\begin{solution}{4cm}
+\method
+求不等式$(1+x)(1-|x|)\gt \ 0$的解集,则分两种情况讨论:
+
+情况$1:\left\{\begin{array}{l}{1+x\gt \ 0\;\;}\\{1-|x|\gt \ 0}\end{array}\right.$
+
+即$\left\{\begin{array}{l}{x\gt \ -1}\\{-1\lt x\lt 1}\end{array}\right.$
+则$-1\lt x\lt 1$.
+
+情况$2:\left\{\begin{array}{l}{1+x\lt 0}\\{1-|x|\lt 0}\end{array}\right.$
+
+即$\left\{\begin{array}{l}{x\lt -1}\\{x\gt 1\text{或}x\lt -1}\end{array}\right.$
+则$x\lt -1$.
+
+两种情况取并集得$\{x|x\lt 1$且$x\neq{}-1\}$.
+
+故选D.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 已知三个不等式:$ab\gt \ 0$,$bc-ad\gt \ 0$,$\dfrac{c}{a}-\dfrac{d}{b}\gt \ 0$(其中$a,b,c,d$均为实数),用其中两个不等式作为条件,余下的一个不等式作为结论组成一个命题,可组成的正确命题的个数是\key{D}.
+\fourchoices{$0$}{$1$}{$2$}{$3$}
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$\begin{cases}ab\gt \ 0, &\cdots(1)\\
+bc-ad\gt \ 0,&\cdots(2)
+\end{cases}$
+
+$(2)\div(1)$得$\dfrac{bc-ad}{ab}=\dfrac{c}{a}-\dfrac{d}{b}\gt \ 0$.
+
+$\begin{cases}
+ab\gt \ 0,&\cdots(1)\\
+\dfrac{c}{a}-\dfrac{d}{b}\gt \ 0,&\cdots(3)
+\end{cases}$
+
+$(1)\times{}(3)$得$ab(\dfrac{c}{a}-\dfrac{d}{b})=bc-ad\gt \ 0$.
+
+$\begin{cases}bc-ad\gt \ 0,&\cdots(2)\\
+\dfrac{c}{a}-\dfrac{d}{b}\gt \ 0,&\cdots(3)
+\end{cases}$
+
+用反证法,显然$ab\neq{}0$,
+
+设$ab\lt 0$,$(2)$式同除$ab$得$\dfrac{c}{a}-\dfrac{d}{b}\lt 0$,矛盾,
+
+故$ab\gt \ 0$.
+故选D.
+
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [5] 设$a\gt \ 0$,$b\gt \ 0$,则以下不等式中不恒成立的是\key{B}.
+\fourchoices{$(a+b)(\dfrac{1}{a}+\dfrac{1}{b})\geqslant{}4$}{$a^{3}+b^{3}\geqslant{}2ab^{2}$}{$a^{2}+b^{2}+2\geqslant{}2a+2b$}{$\sqrt{|{a-b}|}\geqslant{}\sqrt{a}-\sqrt{b}$}
+\end{minipage}
+\begin{solution}{4cm}
+\method
+$a\gt \ 0$,$b\gt \ 0$,
+
+$(a+b)(\dfrac{1}{a}+\dfrac{1}{b})\geqslant{}2\sqrt{ab}\cdot{}2\sqrt{\dfrac{1}{ab}}\geqslant{}4$,
+
+故A恒成立;
+
+$a^{3}+b^{3}\geqslant{}2ab^{2}$,
+
+取$a=\dfrac{1}{2}$,$b=\dfrac{2}{3}$,则B不成立;
+
+$a^{2}+b^{2}+2-(2a+2b)=(a-1)^{2}+(b-1)^{2}\geqslant{}0$,
+
+故C恒成立;
+
+若$a\lt b$,则$\sqrt{|{a-b}|}\geqslant{}\sqrt{a}-\sqrt{b}$恒成立,
+
+若$a\geqslant{}b$,则${(\sqrt{|{a-b}|})^2}-{(\sqrt{a}-\sqrt{b})^2}=2\sqrt{ab}\geqslant{}0$,
+
+故$\sqrt{|{a-b}|}\geqslant{}\sqrt{a}-\sqrt{b}$,
+即D恒成立.
+
+\end{solution}
+\end{questions}
+\group{简答}{第17-19题每题14分，第20题16分，第21题18分.}
+
+\begin{questions}[stp]
+\begin{minipage}{\linewidth}
+\question [14] 已知$\triangle ABC$为直角三角形, 记其两条直角边长分别为$a,b\in\symbf{R}^+$, 记面积为$S$, 周长为$C$. 若三角形面积为定值，其周长是否有最值，最大值还是最小值，何时取到，为多少（结果用$S$表示）？
+
+\end{minipage}
+\begin{solution}{6cm}
+\method
+$C=a+b+\sqrt{a^2+b^2}$,
+\score{2}{2}由$\sqrt{a^2+b^2} \ge \sqrt{2ab}$，$a+b \ge 2\sqrt{ab}$,
+
+得$C \ge 2\sqrt{ab}+\sqrt{2ab}$
+\score{6}{8}由$S=\frac{1}{2}ab$，
+
+得$C \ge (2+2\sqrt{2})\sqrt{S}$，
+\score{4}{12}当且仅当$a=b=\sqrt{2S}$时取到最小值.
+\score{2}{14}
+\end{solution}
+\vfill
+\begin{minipage}{\linewidth}
+\question [14] 已知$a\in\textbf{R}$,若关于$x$的方程$x^2+x+|a-\dfrac{1}{4}|+|a|=0$有实根，求$a$的取值范围.
+
+\end{minipage}
+\begin{solution}{6cm}
+\method
+方程即$|a-\frac{1}{4}|+|a|=-x^2-x\in[0,\frac{1}{4}]$,
+
+即解不等式$|a-\frac{1}{4}|+|a| \le \frac{1}{4}$.
+
+当$a\le 0$，化简得$\frac{1}{4}-2a \le \frac{1}{4}$，解得$a=0$;
+
+当$0 \lt a\le \frac{1}{4}$, 化简得$\frac{1}{4} \le \frac{1}{4}$，解得$a\in(0,\frac{1}{4}]$;
+
+当$a \gt \frac{1}{4}$, 化简得$2a-\frac{1}{4}\le\frac{1}{4}$，无解.
+
+综上所述，$a\in [0,\frac{1}{4}]$
+\score{14}{14}
+\end{solution}
+\vfill
+\newpage
+\begin{minipage}{\linewidth}
+\question [14] 先阅读下列不等式的证法，再解决后面的问题:
+\[
+\text{证明:} (a_1b_1+a_2b_2)^2 \le (a_1^2+a_2^2)(b_1^2+b_2^2)\]
+
+证: 令$A=\sqrt{a_1^2+a_2^2}$, $B=\sqrt{b_1^2+b_2^2}$
+
+\[
+\begin{array}{rl}
+\frac{a_1b_1+a_2b_2}{\sqrt{a_1^2+a_2^2}\sqrt{b_1^2+b_2^2}}& =\frac{a_1b_1}{AB}+\frac{a_2b_2}{AB}\\
+& = \frac{a_1}{A}\cdot\frac{b_1}{B}+\frac{a_2}{A}\cdot\frac{b_2}{B} \\
+& \le \frac{1}{2}(\frac{a_1^2}{A^2}+\frac{b_1^2}{B^2})+ \frac{1}{2}(\frac{a_2^2}{A^2}+\frac{b_2^2}{B^2})\\
+& =\frac{1}{2}(\frac{a_1^2+a_2^2}{A^2}+\frac{b_1^2+b_2^2}{B^2})\\
+& =1
+\end{array}
+\]
+故$ (a_1b_1+a_2b_2)^2 \le (a_1^2+a_2^2)(b_1^2+b_2^2)$.
+\begin{subquestions}
+ \subquestion 若$x_1,y_1,x_2,y_2\in\symbf{R^+}$，利用上述结论，证明:\[
+(x_1+x_2)(y_1+y_2) \ge (\sqrt{x_1y_1}+\sqrt{x_2y_2})^2
+\]
+ \subquestion 若$x_1,y_1,x_2,y_2,z_1,z_2\in\symbf{R^+}$，模仿上述证法并结合(1)的证法，证明:\[
+(x_1+x_2)(y_1+y_2)(z_1+z_2) \ge (\sqrt[3]{x_1y_1z_1}+\sqrt[3]{x_2y_2z_2})^3
+\]
+(提示：若$a,b,c\in\symbf{R^+}$，有$\frac{a^3+b^3+c^3}{3}\ge abc$,)
+\end{subquestions}
+\end{minipage}
+\begin{solution}{3cm}
+\method
+(1)设$a_1=\sqrt{x_1}$，$a_2=\sqrt{x_2}$，$b_1=\sqrt{y_1}$，$b_2=\sqrt{y_2}$.
+
+由\[ (a_1b_1+a_2b_2)^2 \le (a_1^2+a_2^2)(b_1^2+b_2^2)\]
+
+得\[ (\sqrt{x_1}\sqrt{y_1}+\sqrt{x_2}\sqrt{y_2})^2 \le [((\sqrt{x_1})^2+(\sqrt{x_2})^2)((\sqrt{y_1})^2+(\sqrt{y_2})^2)]\]
+
+即\[
+(x_1+x_2)(y_1+y_2) \ge (\sqrt{x_1y_1}+\sqrt{x_2y_2})^2
+\]
+\score{6}{6}(2) 设$a_1=\sqrt[3]{x_1}, b_1=\sqrt[3]{y_1}, c_1=\sqrt[3]{z_1}$, $a_2=\sqrt[3]{x_2}, b_2=\sqrt[3]{y_2}, c_2=\sqrt[3]{z_2}$.
+
+要证
+\[
+(x_1+x_2)(y_1+y_2)(z_1+z_2) \ge (\sqrt[3]{x_1y_1z_1}+\sqrt[3]{x_2y_2z_2})^3
+\]
+
+即证
+\[
+(a_1b_1c_1+a_2b_2c_2)^3 \le (a_1^3+a_2^3)(b_1^3+b_2^3)(c_1^3+c_2^3)
+\]
+
+(不换元，直接证不扣分)
+\score{2}{8}令$A=\sqrt[3]{a_1^3+a_2^3}$，$B=\sqrt[3]{b_1^3+b_2^3}$, $C=\sqrt[3]{c_1^3+c_2^3}$.
+
+\[
+\begin{array}{rl}
+& \frac{a_1b_1c_1+a_2b_2c_2}{\sqrt[3]{a_1^3+a_2^3}\sqrt[3]{b_1^3+b_2^3}\sqrt[3]{c_1^3+c_2^3}}\\
+= & \frac{a_1b_1c_1}{ABC}+\frac{a_2b_2c_2}{ABC}\\
+= & \frac{a_1}{A}\cdot\frac{b_1}{B}\cdot\frac{c_1}{C}+\frac{a_2}{A}\cdot\frac{b_2}{B}\cdot\frac{c_2}{C}\\
+\le & \frac{1}{3}(\frac{a_1^3}{A^3}+\frac{b_1^3}{B^3}+\frac{c_1^3}{C^3})+ \frac{1}{3}(\frac{a_2^3}{A^3}+\frac{b_2^3}{B^3}+\frac{c_2^3}{C^3})\\
+= & \frac{1}{3}(\frac{a_1^3+a_2^3}{A^3}+\frac{b_1^3+b_2^3}{B^3}++\frac{c_1^3+c_2^3}{C^3})\\
+= & 1
+\end{array}
+\]
+
+故\[(a_1b_1c_1+a_2b_2c_2)^3 \le (a_1^3+a_2^3)(b_1^3+b_2^3)(c_1^3+c_2^3)\]
+
+得证.
+\score{6}{14}
+\end{solution}
+\vfill
+\newpage
+\begin{minipage}{\linewidth}
+\question [16] 公元2222年，有一种高危传染疾病在全球范围内蔓延，被感染者的潜伏期可以长达10年，期间会有约$0.05\%$的概率传染给他人，一旦发病三天内即死亡。某城市总人口约200万人，专家分析其中约有$1000$名感染者，为了防止疾病继续扩散，疾病预防控制中心现决定对全市人口进行血液检测以筛选出被感染者。由于检测试剂十分昂贵且数量有限，需要将血样混合后一起检测以节约试剂。已知感染者的检测结果为阳性, 未被感染者则为阴性. 阳性血样与阴性血样混合后的检测结果为阳性, 同为阴性或阳性的血样混合后结果不发生改变.
+\begin{subquestions}
+ \subquestion 若对全市人口进行平均分组，同一分组的血样将被混合到一起检测，若发现结果为阳性，则再在该分组内逐个检测排查. 设每个组$x$个人，那么最坏情况下，需要进行约多少次检测可以找到所有的被感染者? 在当前方案下，若要使检测的次数尽可能少，每个分组的最优人数是？
+ \subquestion 在(1)的检测方案中，对于检测结果为阳性的组采取逐一检测排查的方法并不是很好，或可将这些组的血样再进行一次分组混合血样检测，然后再进行逐一排查。仍然考虑最坏的情况，请问两次要如何分组，使检测总次数尽可能少？
+ \subquestion 在(2)的检测方案中，进行了两次分组混合血样检测。仍然考虑最坏情况，若再进行若干次分组混合血样检测，是否会使检测次数更少？请给出最优的检测方案.
+\end{subquestions}
+\end{minipage}
+\begin{solution}{6cm}
+\method
+设需要进行$y$次检测. 由每个组$x$个人，则有$\frac{2000000}{x}$个小组, 最坏情况是1000人被分配到了不同的组.
+
+\[
+y=\begin{cases}
+\frac{2\cdot 10^6}{x}+2\cdot 10^6, & x \ge 2000 \\
+\frac{2\cdot 10^6}{x}+1000\cdot x, & x \lt 2000 \\
+\end{cases}
+\]
+
+(题目要求近似计算，可以不考虑整除，考虑整除不扣分)
+\score{2}{2}分组数小于1000的情况比逐个检测还要多，可以不用考虑，
+
+$y=\frac{2\times 10^6}{x}+1000\cdot x \ge 2\sqrt{2\cdot 10^9}$.
+
+当$x=\sqrt{2\times10^3}\approx45$人，检测次数最少.
+
+(结果为44-45都给分)
+\score{2}{4}(2) 设第二次分组每个组$w$人.
+
+\[
+\begin{array}{rl}
+y & = \frac{2\times10^6}{x}+\frac{1000x}{w}+1000w\\
+& \ge 3\sqrt[3]{2\times10^{12}}\\
+\end{array}
+\]
+
+当$\frac{2\times10^6}{x}=\frac{1000x}{w}=1000w$时，
+即$x=2000^\frac{2}{3}\approx159$,$w=2000^\frac{1}{3}\approx13$，
+检测次数最少.
+
+所以第一次分组159人一组，第二次分组13人一组.
+
+(也可以用两次基本不等式求得，允许$10\%$的误差.)
+\score{6}{10}(3) 若进行$k$次分组，设第$i$次分组时每组人数为$x_i$. 则总的检测次数为:
+\[
+\begin{array}{rl}
+y=&\dfrac{2\times10^6}{x_1}+\dfrac{1000x_1}{x_2}+\cdots\\
+&+\dfrac{1000x_{k-1}}{x_k}+1000x_k\\
+\ge & (k+1)\sqrt[k+1]{2\times10^{3k+2}}
+\end{array}
+\]
+
+利用计算器可得，
+
+1次分组， $y \ge 2\sqrt{2\cdot 10^9}\approx8.94\times10^4$;
+
+2次分组， $y \ge 3\sqrt[3]{2\cdot 10^{12}}\approx3.78\times10^4$;
+
+3次分组， $y \ge 4\sqrt[4]{2\cdot 10^{15}}\approx2.68\times10^4$;
+
+4次分组， $y \ge 5\sqrt[5]{2\cdot 10^{18}}\approx2.29\times10^4$;
+
+5次分组， $y \ge 6\sqrt[6]{2\cdot 10^{21}}\approx2.13\times10^4$;
+
+6次分组， $y \ge 7\sqrt[7]{2\cdot 10^{24}}\approx2.073\times10^4$;
+
+7次分组， $y \ge 8\sqrt[8]{2\cdot 10^{27}}\approx2.068\times10^4$;
+
+8次分组， $y \ge 9\sqrt[9]{2\cdot 10^{30}}\approx2.094\times10^4$;
+
+9次分组， $y \ge 10\sqrt[10]{2\cdot 10^{33}}\approx2.138\times10^4$;
+
+可见进行7次分组混合血样检测最优.
+
+其中第k次分组时$(2000)^{\frac{8-k}{8}}$人一组.
+\score{6}{16}
+\end{solution}
+\vfill
+\newpage
+\begin{minipage}{\linewidth}
+\question [18] 已知函数$f(x)=a{{x}^{2}}-\frac{1}{2}x+c$（$a$、$c\in R$），满足$f(1)=0$，且$f(x)\ge 0$在$x\in R$时恒成立．
+\begin{subquestions}
+ \subquestion 求$a$、$c$的值；
+ \subquestion 若$h(x)=\frac{3}{4}{{x}^{2}}-bx+\frac{b}{2}-\frac{1}{4}$，解不等式$f(x)+h(x) \lt 0$；
+ \subquestion 是否存在实数$m$，使函数$g(x)=f(x)-mx$在区间$[m,m+2]$上有最小值$-5$？若存在，请求出$m$的值；若不存在，请说明理由．
+\end{subquestions}
+\end{minipage}
+\begin{solution}{6cm}
+\method
+由$f(1)=0$，得$a+c=\frac{1}{2}$,
+\score{1}{1}因为$f(x)\ge 0$在$x\in R$时恒成立，所以$a \gt 0$且△$=\frac{1}{4}-4ac\le 0$，$ac\ge \frac{1}{16}$，
+\score{1}{2}即$a\left( \frac{1}{2}-a \right)\ge \frac{1}{16}$，${{a}^{2}}-\frac{1}{2}a+\frac{1}{16}\le 0$，${{\left( a-\frac{1}{4} \right)}^{2}}\le 0$，所以$a=c=\frac{1}{4}$．
+\score{2}{4}(2) 由（1）得$f(x)=\frac{1}{4}{{x}^{2}}-\frac{1}{2}x+\frac{1}{4}$，由$f(x)+h(x) \lt 0$，得
+${{x}^{2}}-\left( b+\frac{1}{2} \right)x+\frac{b}{2} \lt 0$，即$(x-b)\left( x-\frac{1}{2} \right) \lt 0$，
+\score{3}{7}所以，当$b \lt \frac{1}{2}$时，原不等式解集为$(b,\frac{1}{2})$；
+当$b \gt \frac{1}{2}$时，原不等式解集为$(\frac{1}{2},b)$；
+当$b=\frac{1}{2}$时，原不等式解集为空集 ．
+\score{3}{10}(3) $g(x)=\frac{1}{4}{{x}^{2}}-\left( \frac{1}{2}+m \right)x+\frac{1}{4}$，
+\score{1}{11}$g(x)$的图像是开口向上的抛物线，对称轴为直线$x=2m+1$．
+假设存在实数$m$，使函数$g(x)$在区间$[m,m+2]$上有最小值$-5$．
+\ding{192} 当$2m+1 \lt m$，即$m \lt -1$时，函数$g(x)$在区间$[m,m+2]$上是增函数，所以$g(m)=-5$，即$\frac{1}{4}{{m}^{2}}-\left( \frac{1}{2}+m \right)m+\frac{1}{4}=-5$，解得$m=-3$或$m=\frac{7}{3}$，
+因为$m \lt -1$，所以$m=-3$；
+\score{2}{13}\ding{193}当$m\le 2m+1\le m+2$，即$-1\le m\le 1$时，函数$g(x)$的最小值为$g(2m+1)=-5$，即
+$\frac{1}{4}{{(2m+1)}^{2}}-\left( \frac{1}{2}+m \right)(2m+1)+\frac{1}{4}=-5$，解得$m=-\frac{1}{2}-\frac{\sqrt{21}}{2}$或$m=-\frac{1}{2}+\frac{\sqrt{21}}{2}$，均舍去；
+\score{2}{15}\ding{194}当$2m+1 \gt m+2$，即$m \gt 1$时，$g(x)$在区间$[m,m+2]$上是减函数，所以$g(m+2)=-5$，即$\frac{1}{4}{{(m+2)}^{2}}-\left( \frac{1}{2}+m \right)(m+2)+\frac{1}{4}=-5$，解得$m=-1-2\sqrt{2}$或$m=-1+2\sqrt{2}$，因$m \gt 1$，所以$m=-1+2\sqrt{2}$．
+\score{2}{17}综上，存在实数$m$，$m=-3$或$m=-1+2\sqrt{2}$时，函数$g(x)$在区间$[m,m+2]$上有最小值$-5$．
+\score{1}{18}
+\end{solution}
+\end{questions}
+\end{groups}
+\label{lastpage}
+\end{document}
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+++ trunk/Master/texmf-dist/doc/xelatex/bhcexam/examples/naive.tex 2022-08-10 20:38:07 UTC (rev 64093)
@@ -0,0 +1,86 @@
+% 使用 BHCexam 文档类，并传递选项
+\documentclass[answers]{BHCexam}
+\usepackage{hyperref}
+
+\begin{document}
+
+% 第一行主标题
+\title{BHCexam试卷排版宏包}
+
+% 第二行主标题
+\subtitle{样例}
+
+% 考试说明
+\notice{满分100分, 10分钟完成.}
+
+% 命题人信息
+\author{微信关注公众号：橘子数学}
+
+% 考试日期
+\date{2019.12.1}
+
+% 生成试卷头
+\maketitle
+
+\begin{groups}
+
+% 第一个题组，显示分值，不预留空间
+\group{填空}{本题组共1小题，共30.0分}
+\begin{questions}[s]
+
+% 填空题，两个空
+\question[30] 橘子数学的网址是\key{www.mathcrowd.cn}, 微信服务号\key{橘子数学}.
+\question[30] 橘子数学趣味挑战的网址是\key{qa.mathcrowd.cn},微信订阅号是\key{试题工坊}, \key{橘子数学题库}.
+
+\end{questions}
+
+% 第二个题组，显示分值，不预留空间
+\group{选择}{本题组共2小题，共40.0分}
+\begin{questions}[ps]
+
+% 选择题，四个选项
+\question[30] 以下哪一项不是橘子数学社区的宗旨\key{C}.
+\fourchoices{开放}{高效}{无视版权}{合作}
+
+% 解答，4cm 参数被忽略
+\begin{solution}{4cm}
+\method 橘子数学社区的宗旨是开放、高效、合作、变革.
+\method 见 \url{http://docs.mathcrowd.cn/zh_CN/latest/community/principles.html}
+\end{solution}
+
+% 选择题，五个选项
+\question[40] 以下数学公式显示有明显瑕疵的是\key{D}.
+\fivechoices{$\sin A$}{$2+3\mathrm{i}$}{$x^2$}{$\ln x$}{$\mathrm{e}^{\mathrm{i}\theta}$}
+
+\begin{solution}{4cm}
+\methodonly D 中正确的公式显示效果为$\ln{x}$.
+\end{solution}
+\end{questions}
+
+% 第三个题组，显示分值，预留空间
+\group{主观题}{本题组共1小题，共30.0分}
+\begin{questions}[st]
+% 简答题，两个小问
+\question[30] 请回答以下问题：
+\begin{subquestions}
+ \subquestion 你觉得有必要创建这样一个试题社区吗? 为什么?
+ \subquestion 你对社区的建设有什么建议.
+\end{subquestions}
+
+% 解答，学生版会预留8cm的答题空间.
+\begin{solution}{4cm}
+ \methodonly 欢迎加入用户群组发言讨论.
+
+telegram 交流群组: https://t.me/mathcrowd
+
+QQ 群: 319701002
+
+Github项目页: \url{https://github.com/mathedu4all/bhcexam}
+
+\score{30}{30}
+
+\end{solution}
+\end{questions}
+
+\end{groups}
+\end{document}
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--- trunk/Master/texmf-dist/tex/xelatex/bhcexam/BHCexam.cls (rev 0)
+++ trunk/Master/texmf-dist/tex/xelatex/bhcexam/BHCexam.cls 2022-08-10 20:38:07 UTC (rev 64093)
@@ -0,0 +1,550 @@
+% BHCexam.cls
+%
+% An exam class designed for mathematics teacher in China.
+% Since 2016, it was used by mathcrowd.cn ( an opensource math exam database) as the default class to export exam papers.
+%
+
+%% BHCexam.cls
+%% Copyright (c) 2011-2022 BAO HONG CHANG
+%
+% This work may be distributed and/or modified under the
+% conditions of the LaTeX Project Public License, either version 1.3
+% of this license or (at your option) any later version.
+% The latest version of this license is in
+% http://www.latex-project.org/lppl.txt
+% and version 1.3 or later is part of all distributions of LaTeX
+% version 2003/12/01 or later.
+%
+% This work has the LPPL maintenance status "author-maintained".
+%
+% This work consists of the files BHCexam.cls.
+% Documation on https://docs.mathcrowd.cn/advances/bhcexam.html
+% Support on https://github.com/mathedu4all/bhcexam/issues
+%
+
+%%% BAO HONG CHANG
+%%% Mathcrowd Inc.
+%%% Shanghai
+%%% charles at mathcrowd.cn
+
+% The newest version of this documentclass should always be available
+% from my web page: https://github.com/mathedu4all/bhcexam
+
+\def\fileversion{1.7}
+\def\filedate{2022/07/29}
+
+\NeedsTeXFormat{LaTeX2e}
+\ProvidesClass{BHCexam}[\filedate\space Version \fileversion\space by
+ BAO HONG CHANG]
+
+ % *****************
+ % ** 宏包选项 **
+ % *****************
+
+
+ % 文档类支持以下宏包选项:
+
+ % answers 显示解析
+ % adobe 使用adobe字体
+ % ubuntu 使用ubuntu字体
+ % windows 使用windows字体
+ % fandol 使用fandol字体，随texlive默认安装
+ % mac 使用mac字体
+ % list 以列表项目格式显示试题
+ % twocolumn 使用A3纸张并分栏
+
+
+\newif\if at printanswers \@printanswersfalse
+\DeclareOption{answers}{\@printanswerstrue}
+
+\newcommand\@fontset{windows}
+\DeclareOption{adobe}{\renewcommand\@fontset{adobe}}
+\DeclareOption{ubuntu}{\renewcommand\@fontset{ubuntu}}
+\DeclareOption{mac}{\renewcommand\@fontset{mac}}
+\DeclareOption{windows}{\renewcommand\@fontset{windows}}
+\DeclareOption{fandol}{\renewcommand\@fontset{fandol}}
+
+\newif\if at twocolumn \@twocolumnfalse
+\DeclareOption{twocolumn}{\@twocolumntrue}
+
+\newif\if at list \@listfalse
+\DeclareOption{list}{\@listtrue}
+
+
+\DeclareOption*{\PassOptionsToClass{\CurrentOption}{article}}
+\ProcessOptions\relax
+\LoadClass{article}
+
+\RequirePackage[fontset = \@fontset, punct=kaiming]{ctex}
+\ctexset{linestretch = 4, autoindent = 0pt}
+
+
+% *****************************
+% ** 加载其他宏包 **
+% *****************************
+
+\RequirePackage{tabularx, ifthen} % 排选项用
+\RequirePackage{xcolor,graphicx,caption} % 彩色、图片、图释
+\RequirePackage{geometry,fancyhdr} % 纸张、边距、页眉、页脚
+\RequirePackage{etoolbox}
+\RequirePackage{amsmath,amsmath,amssymb} % ams数学相关
+\RequirePackage{unicode-math}
+\RequirePackage{pifont} % 带圈数字\ding
+\RequirePackage{bbding} % 图案
+\RequirePackage{romannum} % 罗马数字
+\RequirePackage{enumitem}
+
+
+
+% *****************
+% ** 页面设置 **
+% *****************
+
+
+\AtBeginDocument{
+ \pagenumbering{arabic} % 使用阿拉伯数字页码
+ \renewcommand{\parallel}{\mathrel{\mathpalette\new at parallel\relax}} % 重定义平行符号
+ \newcommand{\new at parallel}[2]{%
+ \begingroup
+ \sbox\z@{高度}% get the height of an uppercase letter
+ \resizebox{!}{\ht\z@}{\raisebox{\depth}{$\m at th#1/\mkern-10mu/$}}%
+ \endgroup
+ }
+}
+
+
+%
+% 双栏显示
+%
+\if at twocolumn
+\geometry{a3paper,landscape, twocolumn,columnsep=40mm,left=50mm,right=30mm,top=35mm,bottom=25mm,headheight=20mm}
+\else
+\geometry{a4paper,left=30mm,right=30mm,top=35mm,bottom=25mm, headheight=20mm}
+\fi
+
+%
+% 图释
+%
+\captionsetup[figure]{font=small,belowskip=0pt}
+
+
+
+% *****************
+% ** 试卷头 **
+% *****************
+
+
+
+\gdef\@subtitle{}
+\gdef\@notice{}
+\gdef\@author{}
+\gdef\@date{}
+
+
+
+\renewcommand{\headrulewidth}{0pt}
+\renewcommand{\title}[1]{\gdef\@title{#1}}
+\newcommand{\subtitle}[1]{\gdef\@subtitle{#1}}
+\newcommand{\notice}[1]{\def\@notice{#1}}
+\renewcommand{\author}[1]{\gdef\@author{#1}}
+\renewcommand{\date}[1]{\gdef\@date{#1}}
+
+\renewcommand\maketitle{\begingroup
+ \renewcommand{\baselinestretch}{2}
+ \newpage
+ \begin{center}
+ \heiti \Large
+ \@title \par
+ \ifdefempty{\@subtitle}{}{
+ \@subtitle \par
+ }
+
+ \ifdefempty{\@author}{}{
+ \songti \normalsize
+ \@author \par
+ }
+
+ \ifdefempty{\@notice}{}{
+ \setlength{\fboxsep}{1em}
+ \vspace{0.5\baselineskip}
+ \fbox{
+ \parbox{0.5\linewidth}{
+ \kaishu \normalsize
+ \@notice \par
+ }
+ }
+ \vspace{0.5\baselineskip}
+ }
+
+
+ \end{center}
+ \renewcommand{\baselinestretch}{1}
+ \ifdefempty{\@date}{}{
+ \begin{flushright}
+ \songti \small
+ \@date \par
+ \end{flushright}
+ }
+ \songti \normalsize
+\endgroup}
+
+\renewcommand\arraystretch{1.5}
+\renewcommand{\baselinestretch}{1.5}
+
+% ***************************
+% ** 试题、解答环境定义 **
+% ***************************
+
+\newcounter{Group}
+\newcounter{Question}
+\newcounter{Example}
+\newcounter{Exercise}
+\newcounter{Method}[Question]
+
+\newif\if at showscore
+\@showscorefalse
+\newif\if at showskip
+\@showskipfalse
+\newif\if at showparen
+\@showparenfalse
+\newif\if at resetcounter
+\@resetcounterfalse
+\newlength{\myvertspace}
+
+
+% 定义题组环境
+\newenvironment{groups}{
+ \if at list
+ \list{\heiti\chinese{Group}.}{\usecounter{Group}}
+ \else
+ \par \begingroup \par
+ \fi
+}{
+ \if at list
+ \endlist
+ \else
+ \par \endgroup \par
+ \fi
+}
+
+% 定义新增题组命令
+\newcommand{\group}[2]{
+ \if at list
+ \item \heiti{#1} \quad \kaishu \small #2 \songti \normalsize \par
+ \else
+ \stepcounter{Group}
+ \par \heiti{\par \chinese{Group} 、#1} \quad \small{\kaishu #2} \songti \normalsize \par
+ \fi
+
+}
+
+% 定义试题环境
+\newenvironment{questions}[1][]{
+ \@tfor \@opt :=#1\do
+ {\if\@opt s\global\@showscoretrue\fi
+ \if\@opt t\global\@showskiptrue\fi
+ \if\@opt p\global\@showparentrue\fi
+ \if\@opt r\global\@resetcountertrue\fi}
+ \if at resetcounter
+ \setcounter{Question}{0}
+ \fi
+ \if at list
+ \list{\arabic{Question}.}{\setlength{\leftmargin}{0pt}}
+ \else
+ \par \begingroup \par
+ \fi
+}{
+ \if at list
+ \endlist
+ \else
+ \par \endgroup \par
+ \fi
+
+ \global\@showscorefalse
+ \global\@showskipfalse
+ \global\@showparenfalse
+}
+
+% 定义新增试题命令
+\newcommand{\question}[1][0]{
+ \stepcounter{Question}
+ \if at list
+ \item \if at showscore \kaishu ( #1 分) \songti \fi
+ \else
+ \vspace{2mm}
+
+ \arabic{Question}.
+ \if at showscore \kaishu ( #1 分) \songti \fi
+ \fi
+}
+
+% 定义新增例题命令
+\newcommand{\example}[1][0]{
+ \stepcounter{Example}
+ \if at list
+ \item[例题\arabic{Example}. ]
+ \if at showscore \kaishu ( #1 分) \songti \fi
+ \else
+ \vspace{2mm}
+ \par 例题\arabic{Example}.
+ \if at showscore \kaishu ( #1 分) \songti \fi
+ \fi
+}
+
+% 定义新增习题命令
+\newcommand{\exercise}[1][0]{
+ \stepcounter{Exercise}
+ \if at list
+ \item[习题\arabic{Exercise}. ]
+ \if at showscore \kaishu ( #1 分) \songti \fi
+ \else
+ \vspace{2mm}
+ \par 习题\arabic{Exercise}.
+ \if at showscore \kaishu ( #1 分) \songti \fi
+ \fi
+}
+
+% 定义小问环境
+\newlist{subquestions}{enumerate}{2}
+\setlist[subquestions,1]{label=(\arabic*)}
+\setlist[subquestions,2]{label=(\roman*)}
+
+% 定义新增小问命令
+\newcommand{\subquestion}{\item}
+
+\newenvironment{solution}[1]{
+ \setlength{\myvertspace}{#1}
+ \par \if at printanswers \par \color{red} \begingroup \else \if at showskip \vspace*{\myvertspace} \fi \setbox\z@\vbox\bgroup\fi \songti
+}{
+ \par \if at printanswers \endgroup \color{black} \else \egroup \fi \par
+}
+
+\newcommand{\hint}{
+ \par \fbox{\heiti{提示}} \par \songti
+}
+\newcommand{\method}{
+ \stepcounter{Method}
+ \vspace{2mm}
+ \par \fbox{\heiti{解法\chinese{Method}}} \par \songti
+}
+
+\newcommand{\methodonly}{
+ \par \fbox{\heiti{解答}} \par \songti
+}
+
+\newcommand{\score}[2]{
+ \par
+ \dotfill 本步骤 #1 分， 累计 #2 分
+ \par}
+
+\makeatletter
+\newcommand{\build}[2]{\leavevmode
+ \count@=\z@ \toks@={}%
+ \loop\ifnum\count@<\numexpr#1\relax
+ \toks@=\expandafter{\the\toks@#2}%
+ \advance\count@\@ne
+ \repeat
+ \the\toks@}
+\makeatletter
+
+\newlength{\keylength}
+\newcommand{\key}[1]{
+ \if at printanswers
+ \underline{~~#1~~}
+ \else
+ \settowidth{\keylength}{~~#1~~}
+ \build{13}{\hskip1sp\kern-1sp\hbox to 0.1\keylength{\hrulefill}}
+ \fi}
+
+\newlength{\choicelengtha}
+\newlength{\choicelengthb}
+\newlength{\choicelengthc}
+\newlength{\choicelengthd}
+\newlength{\choicelengthe}
+\newlength{\maxlength}
+
+% 三个选项
+\newcommand{\threechoices}[3]{
+ \if at showparen \dotfill (\qquad) \fi
+ \par
+ \settowidth{\choicelengtha}{A.~#1~~~}
+ \settowidth{\choicelengthb}{B.~#2~~~}
+ \settowidth{\choicelengthc}{C.~#3~~~}
+
+ \ifthenelse{\lengthtest{\choicelengtha>\choicelengthb}}{\setlength{\maxlength}{\choicelengtha}}{\setlength{\maxlength}{\choicelengthb}}
+ \ifthenelse{\lengthtest{\choicelengthc>\maxlength}}{\setlength{\maxlength}{\choicelengthc}}{}
+ \ifthenelse{\lengthtest{\maxlength>0.4\linewidth}}
+ {
+ \begin{tabularx}{\linewidth}{X}
+ \setlength\tabcolsep{0pt}
+ A.~#1~~~\\
+ B.~#2~~~\\
+ C.~#3~~~\\
+ \end{tabularx}
+ }%
+ {
+ \ifthenelse{\lengthtest{\maxlength>0.2\linewidth}}
+ {
+ \begin{tabularx}{\linewidth}{XX}
+ \setlength\tabcolsep{0pt}
+ A.~#1~~~ & B.~#2~~~\\
+ C.~#3~~~ & \\
+ \end{tabularx}
+ }%
+ {
+ \begin{tabularx}{\linewidth}{XXXX}
+ \setlength\tabcolsep{0pt}
+ A.~#1~~~ & B.~#2~~~ & C.~#3~~~ &\\
+ \end{tabularx}
+ }
+ }
+ \unskip \unskip
+}
+
+% 四个选项
+\newcommand{\fourchoices}[4]{
+ \if at showparen \dotfill (\qquad) \fi
+ \par
+ \settowidth{\choicelengtha}{A.~#1~~~}
+ \settowidth{\choicelengthb}{B.~#2~~~}
+ \settowidth{\choicelengthc}{C.~#3~~~}
+ \settowidth{\choicelengthd}{D.~#4~~~}
+ \ifthenelse{\lengthtest{\choicelengtha>\choicelengthb}}{\setlength{\maxlength}{\choicelengtha}}{\setlength{\maxlength}{\choicelengthb}}
+ \ifthenelse{\lengthtest{\choicelengthc>\maxlength}}{\setlength{\maxlength}{\choicelengthc}}{}
+ \ifthenelse{\lengthtest{\choicelengthd>\maxlength}}{\setlength{\maxlength}{\choicelengthd}}{}
+ \ifthenelse{\lengthtest{\maxlength>0.4\linewidth}}
+ {
+ \begin{tabularx}{\linewidth}{X}
+ \setlength\tabcolsep{0pt}
+ A.~#1~~~\\
+ B.~#2~~~\\
+ C.~#3~~~\\
+ D.~#4~~~\\
+ \end{tabularx}
+ }%
+ {
+ \ifthenelse{\lengthtest{\maxlength>0.2\linewidth}}
+ {
+ \begin{tabularx}{\linewidth}{XX}
+ \setlength\tabcolsep{0pt}
+ A.~#1~~~ & B.~#2~~~\\
+ C.~#3~~~ & D.~#4~~~\\
+ \end{tabularx}
+ }%
+ {
+ \begin{tabularx}{\linewidth}{XXXX}
+ \setlength\tabcolsep{0pt}
+ A.~#1~~~ & B.~#2~~~ & C.~#3~~~ & D.~#4~~~\\
+ \end{tabularx}
+ }%
+ }
+ \unskip \unskip
+}
+
+% 五个选项
+\newcommand{\fivechoices}[5]{
+ \if at showparen \hfill \dotfill (\quad) \fi
+ \par
+ \settowidth{\choicelengtha}{A.~#1~~~}
+ \settowidth{\choicelengthb}{B.~#2~~~}
+ \settowidth{\choicelengthc}{C.~#3~~~}
+ \settowidth{\choicelengthd}{D.~#4~~~}
+ \settowidth{\choicelengthe}{E.~#5~~~}
+ \ifthenelse{\lengthtest{\choicelengtha>\choicelengthb}}{\setlength{\maxlength}{\choicelengtha}}{\setlength{\maxlength}{\choicelengthb}}
+ \ifthenelse{\lengthtest{\choicelengthc>\maxlength}}{\setlength{\maxlength}{\choicelengthc}}{}
+ \ifthenelse{\lengthtest{\choicelengthd>\maxlength}}{\setlength{\maxlength}{\choicelengthd}}{}
+ \ifthenelse{\lengthtest{\choicelengthe>\maxlength}}{\setlength{\maxlength}{\choicelengthe}}{}
+ \ifthenelse{\lengthtest{\maxlength>0.4\linewidth}}{
+ \begin{tabularx}{\linewidth}{X}
+ \setlength\tabcolsep{0pt}
+ A.~#1~~~\\
+ B.~#2~~~\\
+ C.~#3~~~\\
+ D.~#4~~~\\
+ E.~#5~~~\\
+ \end{tabularx}
+ }%
+ {
+ \ifthenelse{\lengthtest{\maxlength>0.2\linewidth}}
+ {
+ \begin{tabularx}{\linewidth}{XX}
+ \setlength\tabcolsep{0pt}
+ A.~#1~~~ & B.~#2~~~\\
+ C.~#3~~~ & D.~#4~~~\\
+ E.~#5~~~ & \\
+ \end{tabularx}
+ }%
+ {
+ \begin{tabularx}{\linewidth}{XXXX}
+ \setlength\tabcolsep{0pt}
+ A.~#1~~~ & B.~#2~~~ & C.~#3~~~ & D.~#4~~~ \\
+ E.~#5~~~ & & & \\
+ \end{tabularx}
+ }
+ }%
+ \unskip \unskip
+}
+
+% 六个选项
+\newcommand{\sixchoices}[6]{
+ \if at showparen \hfill \dotfill (\quad) \fi
+ \par
+ \settowidth{\choicelengtha}{A.~#1~~~}
+ \settowidth{\choicelengthb}{B.~#2~~~}
+ \settowidth{\choicelengthc}{C.~#3~~~}
+ \settowidth{\choicelengthd}{D.~#4~~~}
+ \settowidth{\choicelengthe}{D.~#5~~~}
+ \ifthenelse{\lengthtest{\choicelengtha>\choicelengthb}}{\setlength{\maxlength}{\choicelengtha}}{\setlength{\maxlength}{\choicelengthb}}
+ \ifthenelse{\lengthtest{\choicelengthc>\maxlength}}{\setlength{\maxlength}{\choicelengthc}}{}
+ \ifthenelse{\lengthtest{\choicelengthd>\maxlength}}{\setlength{\maxlength}{\choicelengthd}}{} \ifthenelse{\lengthtest{\choicelengthe>\maxlength}}{\setlength{\maxlength}{\choicelengthe}}{}
+
+ \ifthenelse{\lengthtest{\maxlength>0.4\linewidth}}
+ {
+ \begin{tabularx}{\linewidth}{X}
+ \setlength\tabcolsep{0pt}
+ A.~#1~~~\\
+ B.~#2~~~\\
+ C.~#3~~~\\
+ D.~#4~~~\\
+ E.~#5~~~\\
+ F.~#6~~~\\
+ \end{tabularx}
+ }%
+ {
+ \ifthenelse{\lengthtest{\maxlength>0.2\linewidth}}
+ {
+ \begin{tabularx}{\linewidth}{XX}
+ \setlength\tabcolsep{0pt}
+ A.~#1~~~ & B.~#2~~~\\
+ C.~#3~~~ & D.~#4~~~\\
+ E.~#5~~~ & F.~#6~~~\\
+ \end{tabularx}
+ }%
+ {
+ \begin{tabularx}{\linewidth}{XXXX}
+ \setlength\tabcolsep{0pt}
+ A.~#1~~~ & B.~#2~~~ & C.~#3~~~ & D.~#4~~~ \\
+ E.~#2~~~ & F.~#6~~~ & & \\
+ \end{tabularx}
+ }%
+ }
+ \unskip \unskip
+}
+
+
+% ***************************
+% ** CUSTOMIZED COMMAND **
+% ***************************
+
+\newcommand\abs[1]{\left|#1\right|}
+\newcommand{\gt}{>}
+\newcommand{\lt}{<}
+\renewcommand{\geq}{\geqslant}
+\renewcommand{\ge}{\geqslant}
+\renewcommand{\leq}{\leqslant}
+\renewcommand{\le}{\leqslant}
+\renewenvironment{split}{\begin{aligned}}{\end{aligned}}
+
+
+
+\endinput
Property changes on: trunk/Master/texmf-dist/tex/xelatex/bhcexam/BHCexam.cls
___________________________________________________________________
Added: svn:eol-style
## -0,0 +1 ##
+native
\ No newline at end of property
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