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author | Karl Berry <karl@freefriends.org> | 2019-11-08 21:40:58 +0000 |
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committer | Karl Berry <karl@freefriends.org> | 2019-11-08 21:40:58 +0000 |
commit | cee1b7f2580148a33ae6cc0631c56c7e35ce0bff (patch) | |
tree | f1f4b8c3cadb58d72d04bf20dcc136eae4f57a23 /Master/texmf-dist/doc/latex | |
parent | 69e924ee6c8d31ad1ad84dfe14dd724fe50553df (diff) |
jnuexam (8nov19)
git-svn-id: svn://tug.org/texlive/trunk@52696 c570f23f-e606-0410-a88d-b1316a301751
Diffstat (limited to 'Master/texmf-dist/doc/latex')
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-rw-r--r-- | Master/texmf-dist/doc/latex/jnuexam/exam-a-a3split.pdf | bin | 103007 -> 94090 bytes | |||
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-rw-r--r-- | Master/texmf-dist/doc/latex/jnuexam/jnuexam.tex | 107 |
11 files changed, 120 insertions, 55 deletions
diff --git a/Master/texmf-dist/doc/latex/jnuexam/exam-a-a3input.pdf b/Master/texmf-dist/doc/latex/jnuexam/exam-a-a3input.pdf Binary files differindex 5c185d3fc7d..c8ade766802 100644 --- a/Master/texmf-dist/doc/latex/jnuexam/exam-a-a3input.pdf +++ b/Master/texmf-dist/doc/latex/jnuexam/exam-a-a3input.pdf diff --git a/Master/texmf-dist/doc/latex/jnuexam/exam-a-a3split.pdf b/Master/texmf-dist/doc/latex/jnuexam/exam-a-a3split.pdf Binary files differindex f17c0ce2e38..de83638ce7d 100644 --- a/Master/texmf-dist/doc/latex/jnuexam/exam-a-a3split.pdf +++ b/Master/texmf-dist/doc/latex/jnuexam/exam-a-a3split.pdf diff --git a/Master/texmf-dist/doc/latex/jnuexam/exam-a-empty.pdf b/Master/texmf-dist/doc/latex/jnuexam/exam-a-empty.pdf Binary files differindex 846854d86dc..653325a6525 100644 --- a/Master/texmf-dist/doc/latex/jnuexam/exam-a-empty.pdf +++ b/Master/texmf-dist/doc/latex/jnuexam/exam-a-empty.pdf diff --git a/Master/texmf-dist/doc/latex/jnuexam/exam-a.pdf b/Master/texmf-dist/doc/latex/jnuexam/exam-a.pdf Binary files differindex c57592e6754..5eb115847a8 100644 --- a/Master/texmf-dist/doc/latex/jnuexam/exam-a.pdf +++ b/Master/texmf-dist/doc/latex/jnuexam/exam-a.pdf diff --git a/Master/texmf-dist/doc/latex/jnuexam/exam-a.tex b/Master/texmf-dist/doc/latex/jnuexam/exam-a.tex index 44bc600fae0..54b2d1ac404 100644 --- a/Master/texmf-dist/doc/latex/jnuexam/exam-a.tex +++ b/Master/texmf-dist/doc/latex/jnuexam/exam-a.tex @@ -39,10 +39,10 @@ \vfill \begin{problem} -已知二阶行列式 $\text{$\left|\begin{array}{cc} +已知二阶行列式 $\left|\begin{array}{cc} 1 & 2\\ - 3 & x -\end{array}\right|$=0}$,则 $x=$ \fillout{$-6$}. +\end{array}\right|=0$,则 $x=$ \fillout{$-6$}. \end{problem} \vfill @@ -178,10 +178,10 @@ \begin{solution} \everymath{\displaystyle}% -原式$=\int\e^{2x}\,\sec^2 x\dx+2\int\e^{2x}\,\tan x\dx$ \score{2} -\hspace{5em}${}=\int\e^{2x}\,\d(\tan x)+ 2\int\e^{2x}\,\tan x\dx$ \score{4} -\hspace{5em}${}=\e^{2x}\,\tan x - 2\int\e^{2x}\,\tan x\dx+ 2\int\e^{2x}\,\tan x\dx$ \score{6} -\hspace{5em}${}=\e^{2x}\,\tan x + C$ \score{8} +原式 \? $=\int\e^{2x}\,\sec^2 x\dx+2\int\e^{2x}\,\tan x\dx$ \score{2} +\+ $=\int\e^{2x}\,\d(\tan x)+ 2\int\e^{2x}\,\tan x\dx$ \score{4} +\+ $=\e^{2x}\,\tan x - 2\int\e^{2x}\,\tan x\dx+ 2\int\e^{2x}\,\tan x\dx$ \score{6} +\+ $=\e^{2x}\,\tan x + C$ \score{8} \end{solution} \vfill @@ -220,7 +220,7 @@ \bigskip \begin{solution} -$A = \left|\begin{array}{cccc} +$A \? = \left|\begin{array}{cccc} 0 & 1 & 2 & 3\\ 1 & 2 & 3 & 0\\ 2 & 3 & 0 & 1\\ @@ -235,7 +235,7 @@ $A = \left|\begin{array}{cccc} - 1 & - 6 & 1\\ - 6 & - 8 & 2 \end{array}\right|$ \score{4} -\qquad $= -\left|\begin{array}{ccc} +\+ $= -\left|\begin{array}{ccc} 1 & 2 & 3\\ 0 & - 4 & 4\\ 0 & 4 & 20 @@ -255,11 +255,11 @@ x_2 x_3 + 9 x^2_3$ 化为标准形 $f = d_1 y^2_1 + d_2 y^2_2 + d_3 y^2_3$ . \bigskip \begin{solution} -$f = x_1^2 + 2 x_1 x_2 - 6 x_1 x_3 + 2 x_2^2 - 12 x_2 x_3 + 9 x^2_3$ \par -\qquad$= x_1^2 + 2 x_1 (x_2 - 3 x_3) + (x_2 - 3 x_3)^2 + x_2^2 - 6 x_2 x_3 $ \par -\qquad$= (x_1 + x_2 - 3 x_3)^2 + x_2^2 - 6 x_2 x_3$ \score{3} -\qquad$= (x_1 + x_2 - 3 x_3)^2 + x_2^2 - 2 x_2 \cdot 3 x_3 + (3 x_3)^2 - 9x_3^2$ \par -\qquad$= (x_1 + x_2 - 3 x_3)^2 + (x_2 - 3 x_3)^2 - 9 x_3^2$ \score{6} +$f \? = x_1^2 + 2 x_1 x_2 - 6 x_1 x_3 + 2 x_2^2 - 12 x_2 x_3 + 9 x^2_3$ \par + \+ $= x_1^2 + 2 x_1 (x_2 - 3 x_3) + (x_2 - 3 x_3)^2 + x_2^2 - 6 x_2 x_3 $ \par + \+ $= (x_1 + x_2 - 3 x_3)^2 + x_2^2 - 6 x_2 x_3$ \score{3} + \+ $= (x_1 + x_2 - 3 x_3)^2 + x_2^2 - 2 x_2 \cdot 3 x_3 + (3 x_3)^2 - 9x_3^2$ \par + \+ $= (x_1 + x_2 - 3 x_3)^2 + (x_2 - 3 x_3)^2 - 9 x_3^2$ \score{6} 令$y_1 = x_1 + x_2 - 3 x_3, y_2 = x_2 - 3 x_3, y_3 = x_3$, \newline 则$f = y_1^2 + y_2^2 - 9y_3^2$为标准形.\score{8} \end{solution} @@ -270,19 +270,19 @@ $f = x_1^2 + 2 x_1 x_2 - 6 x_1 x_3 + 2 x_2^2 - 12 x_2 x_3 + 9 x^2_3$ \par \begin{problem} 设每发炮弹命中飞机的概率是0.2且相互独立,现在发射100发炮弹.\par -\step 用切贝谢夫不等式估计命中数目$\xi$在10发到30发之间的概率.\par -\step 用中心极限定理估计命中数目$\xi$在10发到30发之间的概率. +(1) 用切贝谢夫不等式估计命中数目$\xi$在10发到30发之间的概率.\par +(2) 用中心极限定理估计命中数目$\xi$在10发到30发之间的概率. \end{problem} \bigskip \begin{solution} $E\xi = n p = 100 \cdot 0.2 = 20, D\xi = n p q = 100 \cdot 0.2 \cdot 0.8 = 16$. \score{2} -\step $P (10 < \xi < 30) = P (| \xi - E \xi | < 10) \ge 1 - \frac{D\xi}{10^2} +(1) $P (10 < \xi < 30) = P (|\xi - E\xi| < 10) \ge 1 - \frac{D\xi}{10^2} = 1 - \frac{16}{100} = 0.84$. \score{4} -\step $P (10 < \xi < 30) \approx \Phi_0 \left( \frac{30 - 20}{\sqrt{16}}\right) - - \Phi_0 \left( \frac{10 - 20}{\sqrt{16}} \right)$ \score{6} -\qquad $= 2 \Phi_0 (2.5) - 1 = 2 \cdot 0.9938 - 1 =0.9876$ \score{8} +(2) $P (10 < \xi < 30) \? \approx \Phi_0\left(\frac{30 - 20}{\sqrt{16}}\right) + - \Phi_0\left(\frac{10 - 20}{\sqrt{16}}\right)$ \score{6} + \+ $= 2 \Phi_0(2.5) - 1 = 2 \cdot 0.9938 - 1 =0.9876$ \score{8} \end{solution} \vfill @@ -295,11 +295,11 @@ $E\xi = n p = 100 \cdot 0.2 = 20, D\xi = n p q = 100 \cdot 0.2 \cdot 0.8 = 16$. \bigskip \begin{solution} -\step 待检假设 $H_0 : \mu = 3140$. \score{1} -\step 选取统计量 $T = \frac{\bar{X}-\mu}{S / \sqrt{n}} \sim t(n-1)$. \score{3} -\step 查表得到 $t_{\alpha} = t_{\alpha} (n - 1) = t_{0.01} (15) =2.947$. \score{5} -\step 计算统计值 $t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}} =\frac{3160-3140}{100/4} = 0.8$.\score{7} -\step 由于 $| t | < t_{\alpha}$, 故接受 $H_0$, 即假设成立. \score{8} +(1) 待检假设 $H_0 : \mu = 3140$. \score{1} +(2) 选取统计量 $T = \frac{\widebar{X}-\mu}{S / \sqrt{n}} \sim t(n-1)$. \score{3} +(3) 查表得到 $t_{\alpha} = t_{\alpha} (n - 1) = t_{0.01} (15) =2.947$. \score{5} +(4) 计算统计值 $t = \frac{\widebar{x} - \mu_0}{s/\sqrt{n}} =\frac{3160-3140}{100/4} = 0.8$.\score{7} +(5) 由于 $| t | < t_{\alpha}$, 故接受 $H_0$, 即假设成立. \score{8} \end{solution} \vfill @@ -308,6 +308,8 @@ $E\xi = n p = 100 \cdot 0.2 = 20, D\xi = n p q = 100 \cdot 0.2 \cdot 0.8 = 16$. \makepart{证明题}{共~2~小题,每小题~8~分,共~16~分} +\renewcommand{\solutionname}{证} % 将“解”字改为“证”字 + \begin{problem} 设数列$\{x_n\}$满足$x_1=\sqrt2$,$x_{n+1}=\sqrt{2+x_n}$.证明数列收敛,并求出极限. \end{problem} @@ -315,13 +317,13 @@ $E\xi = n p = 100 \cdot 0.2 = 20, D\xi = n p q = 100 \cdot 0.2 \cdot 0.8 = 16$. \bigskip \begin{solution} -\step 事实上,由于$x_1<2$,且$x_k<2$时 +(1) 事实上,由于$x_1<2$,且$x_k<2$时 $$x_{k+1}=\sqrt{2+x_k}<\sqrt{2+2}=2,$$ 由数学归纳法知对所有$n$都有$x_n<2$,即数列有上界. 又由于 $$\frac{x_{n+1}}{x_n}=\sqrt{\frac{2}{x_n^2}+\frac{1}{x_n}}>\sqrt{\frac{2}{2^2}+\frac{1}{2}}=1,$$ 所以数列单调增加.由极限存在准则II,数列必定收敛.\score{4} -\step 设数列的极限为$A$,对递推公式两边同时取极限得到 +(2) 设数列的极限为$A$,对递推公式两边同时取极限得到 $$A=\sqrt{2+A}.$$ 解得$A=2$,即数列$\{x_n\}$的极限为$2$.\score{8} \end{solution} @@ -329,29 +331,29 @@ $$A=\sqrt{2+A}.$$ \vfill \begin{problem} -设事件$A$和$B$相互独立,证明$A$和$\bar{B}$相互独立. +设事件$A$和$B$相互独立,证明$A$和$\widebar{B}$相互独立. \end{problem} \bigskip \begin{solution} -$P (A \cdot \bar{B}) = P (A - B) = P (A - A B)$ \score{2} -\qquad $= P (A) - P (A B) = P (A) - P (A) P (B)$ \score{4} -\qquad $= P (A) (1 - P (B)) = P (A) P (\bar{B})$ \score{6} -所以$A$和$\bar{B}$相互独立.\score{8} +\? $P (A \cdot \widebar{B}) = P (A - B) = P (A - A B)$ \score{2} +\< $= P (A) - P (A B) = P (A) - P (A) P (B)$ \score{4} +\< $= P (A) (1 - P (B)) = P (A) P (\widebar{B})$ \score{6} +所以$A$和$\widebar{B}$相互独立.\score{8} \end{solution} \vfill \makedata{一些可能用到的数据} %附录数据 -\begin{tabu}{*{4}{X[l,$]}} +\begin{tabularx}{\linewidth}{*{4}{>{$}X<{$}}} \hline \Phi_0(0.5)=0.6915 & \Phi_0(1)=0.8413 & \Phi_0(2)=0.9773 & \Phi_0(2.5)=0.9938 \\ t_{0.01}(8)=3.355 & t_{0.01}(9)=3.250 & t_{0.01}(15)=2.947 & t_{0.01}(16)=2.921 \\ \chi_{0.005}^2(8)=22.0 & \chi_{0.005}^2(9)=23.6 & \chi_{0.005}^2(15)=32.8 & \chi_{0.005}^2(16)=34.3 \\ \chi_{0.995}^2(8)=1.34 & \chi_{0.995}^2(9)=1.73 & \chi_{0.995}^2(15)=4.60 & \chi_{0.995}^2(16)=5.14 \\ \hline -\end{tabu} +\end{tabularx} \end{document} diff --git a/Master/texmf-dist/doc/latex/jnuexam/exam-b-a3input.pdf b/Master/texmf-dist/doc/latex/jnuexam/exam-b-a3input.pdf Binary files differindex 9b0fd8d18df..7351872d942 100644 --- a/Master/texmf-dist/doc/latex/jnuexam/exam-b-a3input.pdf +++ b/Master/texmf-dist/doc/latex/jnuexam/exam-b-a3input.pdf diff --git a/Master/texmf-dist/doc/latex/jnuexam/exam-b-a3split.pdf b/Master/texmf-dist/doc/latex/jnuexam/exam-b-a3split.pdf Binary files differindex c224b9ab33d..ff8a2e40bab 100644 --- a/Master/texmf-dist/doc/latex/jnuexam/exam-b-a3split.pdf +++ b/Master/texmf-dist/doc/latex/jnuexam/exam-b-a3split.pdf diff --git a/Master/texmf-dist/doc/latex/jnuexam/exam-b-empty.pdf b/Master/texmf-dist/doc/latex/jnuexam/exam-b-empty.pdf Binary files differindex 1b27aa32c81..9809fb34cc5 100644 --- a/Master/texmf-dist/doc/latex/jnuexam/exam-b-empty.pdf +++ b/Master/texmf-dist/doc/latex/jnuexam/exam-b-empty.pdf diff --git a/Master/texmf-dist/doc/latex/jnuexam/exam-b.pdf b/Master/texmf-dist/doc/latex/jnuexam/exam-b.pdf Binary files differindex 8a378a7e3d8..69dd16f1879 100644 --- a/Master/texmf-dist/doc/latex/jnuexam/exam-b.pdf +++ b/Master/texmf-dist/doc/latex/jnuexam/exam-b.pdf diff --git a/Master/texmf-dist/doc/latex/jnuexam/jnuexam.pdf b/Master/texmf-dist/doc/latex/jnuexam/jnuexam.pdf Binary files differindex 79517e4623c..dac435686dd 100644 --- a/Master/texmf-dist/doc/latex/jnuexam/jnuexam.pdf +++ b/Master/texmf-dist/doc/latex/jnuexam/jnuexam.pdf diff --git a/Master/texmf-dist/doc/latex/jnuexam/jnuexam.tex b/Master/texmf-dist/doc/latex/jnuexam/jnuexam.tex index f41fb784f6c..bfc98e08349 100644 --- a/Master/texmf-dist/doc/latex/jnuexam/jnuexam.tex +++ b/Master/texmf-dist/doc/latex/jnuexam/jnuexam.tex @@ -5,9 +5,7 @@ \usepackage[UTF8,noindent]{ctex} \usepackage{arev} -\usepackage{ragged2e} -\usepackage{listings} -\usepackage{tabu} +\usefonttheme{professionalfonts} \makeatletter @@ -28,39 +26,45 @@ \setlength{\parskip}{7pt plus 1pt minus 1pt} -\justifying -\let\oldraggedright\raggedright -\let\raggedright\justifying - -\lstset{ - basicstyle=\ttfamily\color{blue!50!red} -} - -\lstnewenvironment{code}{}{} - \setbeamersize{text margin left=8mm,text margin right=8mm} -\newenvironment{framex}{\begin{frame}[fragile=singleslide,environment=framex]}{\end{frame}} +\setbeamercolor{normal text}{bg=gray!20} \setbeamertemplate{frametitle}{\strut\insertframetitle\strut\par} \setbeamertemplate{navigation symbols}{} +\newcommand{\cdotfill}{\leavevmode\xleaders\hbox to 0.5em{\hss$\cdot$\hss}\hfill\kern0pt\relax} + +\usepackage{tabularx} + \newcommand{\ulinefill}[1]{\xleaders\hbox{\underline{\vphantom{#1}\kern1pt}}\hfill\kern0pt} \newcommand{\fillbox}[1]{\ulinefill{#1}\underline{#1}\ulinefill{#1}} -\setbeamercolor{normal text}{bg=gray!20} - \setbeamertemplate{title page}{% \renewcommand{\arraystretch}{2}% \usebeamerfont{title} - \begin{tabu}{|X|} + \begin{tabularx}{\linewidth}{|X|} \hline 模板名称:\fillbox{\usebeamercolor[fg]{title}\inserttitle} \\ 模板作者:\fillbox{\insertauthor} \\ 所在单位:\fillbox{\insertinstitute} \\ 更新日期:\fillbox{\the\year}年\fillbox{\the\month}月\fillbox{\the\day}日\\ \hline - \end{tabu}% + \end{tabularx}% +} + +\usepackage{ragged2e} + +\justifying +\let\oldraggedright\raggedright +\let\raggedright\justifying + +\usepackage{fancyvrb} + +\newenvironment{framex}{\begin{frame}[fragile=singleslide,environment=framex]}{\end{frame}} + +\DefineVerbatimEnvironment{code}{Verbatim}{% + formatcom=\color{blue!50!red}% } \begin{document} @@ -136,6 +140,22 @@ \end{framex} \begin{framex} +\frametitle{判断题目} +\begin{code} +\makepart{判断题}{题数分值} + +\begin{problem} +第一道判断题描述。\true +\end{problem} + +\begin{problem} +第二道判断题描述。\false +\end{problem} +\end{code} +其中 \verb!\true! 和 \verb!\false! 命令分别表示正确和错误。 +\end{framex} + +\begin{framex} \frametitle{填空题目} \begin{code} \makepart{填空题}{题数分值} @@ -243,6 +263,15 @@ \end{framex} \begin{framex} +\frametitle{解答名称} +通过重新定义 \verb!\solutionname! 命令,可以改变 \verb!solution! 环境的名称。 +比如下面例子将“解答”二字改为“证明”: +\begin{code} +\renewcommand{\solutionname}{证明} +\end{code} +\end{framex} + +\begin{framex} \frametitle{评分命令} 计算题和证明题等主观题的排版方法是完全一样的。在编写这些主观题的解答时, 可以用 \verb!\score! 命令给出各步骤得分。比如: @@ -256,6 +285,40 @@ $2+2=4$ \score{8} \end{framex} \begin{framex} +\frametitle{对齐命令} +此文档类提供几个对齐命令,用于在不同行之间对齐。比如 +\vskip1em\hrule +我们有$(a+b)^2 = (a+b)(a+b)$ \par +\leavevmode\phantom{我们有$(a+b)^2$}${}= a^2 + 2ab + b^2$ \cdotfill 2分 +\vskip0.6em\hrule\vskip1em +\begin{code} +我们有$(a+b)^2 \? = (a+b)(a+b)$ \\ + \+$= a^2+2ab+b^2$ \score{2} +\end{code} +第一个公式内部的 \verb!\?! 保存当前水平位置, +而第二个公式前面的 \verb!\+! 表示跳到之前保存的位置。 +\par +这两个对齐命令 \verb!\?! 和 \verb!\+! 需要编译两次才能生效。 +\end{framex} + +\begin{framex} +\frametitle{对齐命令} +此文档类提供几个对齐命令,用于在不同行的对齐。比如 +\vskip1em\hrule +我们有$(a+b)^2 = (a+b)(a+b)$ \par +\leavevmode\phantom{我们\,}${}= a^2 + 2ab + b^2$ \cdotfill 2分 +\vskip0.6em\hrule\vskip1em +\begin{code} +我们有 \? $(a+b)^2 = (a+b)(a+b)$ \\ + \< $= a^2+2ab+b^2$ \score{2} +\end{code} +第一行公式前面的 \verb!\?! 保存当前水平位置, +而第二行公式前面的 \verb!\<! 表示跳到之前保存位置的左侧(左移一个等号的宽度)。 +\par +这两个对齐命令 \verb!\?! 和 \verb!\<! 需要编译两次才能生效。 +\end{framex} + +\begin{framex} \frametitle{其它题型} 除了上述四种题型之外,其它题型可以用下面方式编写: \begin{code} @@ -308,7 +371,7 @@ $2+2=4$ \score{8} \frametitle{竖直空白} 在试卷的各个小题后面,可以留下一些竖直空白。本文档类支持下列这些竖直空白命令:\par \renewcommand{\arraystretch}{1.3}% -\begin{tabu}{XX} +\begin{tabularx}{\linewidth}{l<{\qquad}X} \hline \texttt{\string\smallskip} & 竖直小空白 \\ \hline @@ -318,7 +381,7 @@ $2+2=4$ \score{8} \hline \texttt{\string\vfill} & 竖直填充 \\ \hline -\end{tabu} +\end{tabularx}\par 当然,竖直空白命令可以连续使用多个,以得到所需的空白。 \end{framex} @@ -329,7 +392,7 @@ $2+2=4$ \score{8} 分别只对 A 卷和 B 卷有效。 \par \renewcommand{\arraystretch}{1.3}% -\begin{tabu}{XX[2]} +\begin{tabularx}{\linewidth}{l<{\qquad}X} \hline \texttt{\string\newpage} & 分页,对A卷和B卷均有效 \\ \hline @@ -337,7 +400,7 @@ $2+2=4$ \score{8} \hline \texttt{\string\newpageb} & 分页,仅对B卷有效 \\ \hline -\end{tabu} +\end{tabularx}\par 在试卷中\alert{不要}使用其他分页命令,比如 \verb!\clearpage! 等。 \end{framex} |