jnuexam (8nov19)

git-svn-id: svn://tug.org/texlive/trunk@52696 c570f23f-e606-0410-a88d-b1316a301751

1 files changed, 35 insertions, 33 deletions

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}