From c2e431d1b1f975d5e3da2c1bf3cf6afe8a1c751d Mon Sep 17 00:00:00 2001 From: "weiye.wang" Date: Sat, 24 Jun 2023 20:46:00 +0800 Subject: [PATCH] =?UTF-8?q?=E6=94=B6=E5=BD=952023=E5=B1=8A=E5=8C=97?= =?UTF-8?q?=E4=BA=AC=E9=AB=98=E8=80=83=E8=AF=95=E5=8D=B7?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- 工具/文本文件/批量题目分类号记录.txt | 14 + 工具v2/批量收录题目.py | 4 +- 题库0.3/Problems.json | 420 +++++++++++++++++++++++++++ 3 files changed, 436 insertions(+), 2 deletions(-) diff --git a/工具/文本文件/批量题目分类号记录.txt b/工具/文本文件/批量题目分类号记录.txt index eeb3b719..0be6ccc9 100644 --- a/工具/文本文件/批量题目分类号记录.txt +++ b/工具/文本文件/批量题目分类号记录.txt @@ -1,3 +1,17 @@ +20230624 2023届全国高考试卷 + +problems_dict = { +"2023届全国高考新高考I卷": "018039:018060", +"2023年全国高考上海卷": "018061:018081", +"2023届全国高考新高考II卷": "018082:018103", +"2023届全国高考甲卷理科": "018104:018126", +"2023届全国高考乙卷理科": "018127:018149", +"2023届全国高考天津卷": "018150:018169", +"2023届全国高考甲卷文科": "018170:018192", +"2023届全国高考乙卷文科": "018193:018215", +"2023届全国高考北京卷": "018237:018257" +} + 20230621 2025届高一第一学期材料 problems_dict = { "空中课堂必修第一册例题与习题": "011739:011987", diff --git a/工具v2/批量收录题目.py b/工具v2/批量收录题目.py index 8114b0e6..9f2e81c0 100644 --- a/工具v2/批量收录题目.py +++ b/工具v2/批量收录题目.py @@ -1,9 +1,9 @@ #修改起始id,出处,文件名 starting_id = 18237 #起始id设置, 来自"寻找空闲题号"功能 -raworigin = "测试一下" #题目来源的前缀(中缀在.tex文件中) +raworigin = "" #题目来源的前缀(中缀在.tex文件中) filename = r"C:\Users\weiye\Documents\wwy sync\临时工作区\自拟题目16.tex" #题目的来源.tex文件 editor = "王伟叶" #编辑者姓名 -IndexDescription = " " #设置是否使用后缀, 留空("")则不用后缀, 不留空则以所设字符串作为后缀起始词, 按.tex文件中的顺序编号 +IndexDescription = "试题" #设置是否使用后缀, 留空("")则不用后缀, 不留空则以所设字符串作为后缀起始词, 按.tex文件中的顺序编号 from database_tools import * diff --git a/题库0.3/Problems.json b/题库0.3/Problems.json index 87de915b..36f03ad3 100644 --- a/题库0.3/Problems.json +++ b/题库0.3/Problems.json @@ -467023,6 +467023,426 @@ "space": "4em", "unrelated": [] }, + "018237": { + "id": "018237", + "content": "已知集合$M=\\{x | x+2 \\geq 0\\}$, $N=\\{x | x-1<0\\}$, 则$M \\cap N=$\\bracket{20}.\n\\fourch{$\\{x |-2 \\leq x<1\\}$}{$\\{x |-2=latex, scale = 0.25]\n\\draw (-12.5,0,5) node [left] {$A$} coordinate (A);\n\\draw (12.5,0,5) node [right] {$B$} coordinate (B);\n\\draw (12.5,0,-5) node [right] {$C$} coordinate (C);\n\\draw (-12.5,0,-5) node [left] {$D$} coordinate (D);\n\\draw (A) ++ (5,{sqrt(14)},-5) node [above] {$F$} coordinate (F);\n\\draw (B) ++ (-5,{sqrt(14)},-5) node [above] {$E$} coordinate (E);\n\\draw (D)--(A)--(B)--(C)--(E)--(F)--cycle(A)--(F)(B)--(E);\n\\draw [dashed] (D)--(C);\n\\end{tikzpicture}\n\\end{center}\n\\fourch{$102 \\mathrm{m}$}{$112 \\mathrm{m}$}{$117 \\mathrm{m}$}{$125 \\mathrm{m}$}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2023届全国高考北京卷试题9", + "edit": [ + "20230624\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "018246": { + "id": "018246", + "content": "已知数列$\\{a_n\\}$满足$a_{n+1}=\\dfrac{1}{4}(a_n-6)^3+6$($n=1,2,3, \\cdots$), 则\\bracket{20}.\n\\onech{当$a_1=3$时, $\\{a_n\\}$为递减数列, 且存在常数$M \\leq 0$, 使得$a_n>M$恒成立}{当$a_1=5$时, $\\{a_n\\}$为递增数列, 且存在常数$M \\leq 6$, 使得$a_n6$, 使得$a_n>M$恒成立}{当$a_1=9$时, $\\{a_n\\}$为递增数列, 且存在常数$M>0$, 使得$a_n\\beta$, 则$\\tan \\alpha>\\tan \\beta$. 能说明$p$为假命题的一组$\\alpha, \\beta$的值为$\\alpha=$\\blank{50}, $\\beta=$\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2023届全国高考北京卷试题13", + "edit": [ + "20230624\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "018250": { + "id": "018250", + "content": "我国度量衡的发展有着悠久的历史, 战国时期就已经出现了类似于砝码的、用来测量物体质量的``环权''. 已知$9$枚环权的质量 (单位: 铢) 从小到大构成项数为$9$的数列$\\{a_n\\}$, 该数列的前$3$项成等差数列, 后$7$项成等比数列, 且$a_1=1$, $a_5=12$, $a_9=192$, 则$a_7=$\\blank{50}; 数列$\\{a_n\\}$所有项的和为\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2023届全国高考北京卷试题14", + "edit": [ + "20230624\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "018251": { + "id": "018251", + "content": "设$a>0$, 函数$f(x)=\\begin{cases}x+2, & x<-a, \\\\ \\sqrt{a^2-x^2}, & -a \\leq x \\leq a,\\\\ -\\sqrt{x}-1, & x>a.\\end{cases}$ 给出下列四个结论:\\\\\n\\textcircled{1} $f(x)$在区间$(a-1,+\\infty)$上单调递减;\\\\\n\\textcircled{2} 当$a \\geq 1$时, $f(x)$存在最大值;\\\\\n\\textcircled{3} 设$M(x_1, f(x_1))$($x_1 \\leq a$), $N(x_2, f(x_2))$($x_2>a$), 则$|MN|>1$;\\\\\n\\textcircled{4} 设$P(x_3, f(x_3))$($x_3<-a$), $Q(x_4, f(x_4))$($x_4 \\geq -a$). 若$|PQ|$存在最小值, 则$a$的取值范围是$(0, \\dfrac{1}{2}]$.\\\\\n其中所有正确结论的序号是\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2023届全国高考北京卷试题15", + "edit": [ + "20230624\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "", + "unrelated": [] + }, + "018252": { + "id": "018252", + "content": "如图, 在三棱锥$P-ABC$中, $PA \\perp$平面$ABC$, $PA=AB=BC=1$, $PC=\\sqrt{3}$.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, scale = 2.5]\n\\draw (0,0,0) node [left] {$A$} coordinate (A);\n\\draw (0,1,0) node [above] {$P$} coordinate (P);\n\\draw ({sqrt(2)},0,0) node [right] {$C$} coordinate (C);\n\\draw ({sqrt(2)/2},0,{sqrt(2)/2}) node [below] {$B$} coordinate (B);\n\\draw (A)--(B)--(C)--(P)--cycle(P)--(B);\n\\draw [dashed] (A)--(C);\n\\end{tikzpicture}\n\\end{center}\n(1) 求证: $BC \\perp$平面$PAB$;\\\\\n(2) 求二面角$A-PC-B$的大小.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2023届全国高考北京卷试题16", + "edit": [ + "20230624\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "018253": { + "id": "018253", + "content": "设函数$f(x)=\\sin \\omega x \\cos \\varphi+\\cos \\omega x \\sin \\varphi$($\\omega>0$, $|\\varphi|<\\dfrac{\\pi}{2}$).\\\\\n(1) 若$f(0)=-\\dfrac{\\sqrt{3}}{2}$, 求$\\varphi$的值;\\\\\n(2) 已知$f(x)$在区间$[-\\dfrac{\\pi}{3}, \\dfrac{2 \\pi}{3}]$上单调递增, $f(\\dfrac{2 \\pi}{3})=1$, 再从条件\\textcircled{1}、条件\\textcircled{2}、条件\\textcircled{3}这三个条件中选择一个作为已知, 使函数$f(x)$存在, 求$\\omega, \\varphi$的值.\\\\\n条件\\textcircled{1}: $f(\\dfrac{\\pi}{3})=\\sqrt{2}$;\\\\\n条件\\textcircled{2}: $f(-\\dfrac{\\pi}{3})=-1$;\\\\\n条件\\textcircled{3}: $f(x)$在区间$[-\\dfrac{\\pi}{2},-\\dfrac{\\pi}{3}]$上单调递减.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2023届全国高考北京卷试题17", + "edit": [ + "20230624\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "018254": { + "id": "018254", + "content": "为研究某种农产品价格变化的规律, 收集得到了该农产品连续$40$天的价格变化数据, 如下表所示. 在描述价格变化时, 用``$+$''表示``上涨'', 即当天价格比前一天价格高; 用``$-$''表示``下跌'', 即当天价格比前一天价格低; 用``$0$''表示``不变'', 即当天价格与前一天价格相同. \n\\begin{center}\n\\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|}\n\\hline 时段 & \\multicolumn{10}{c|}{ 价格变化 } \\\\\n\\hline 第 1 天到第 10 天 &$-$&$+$&$+$&$0$&$-$&$-$&$-$&$+$&$+$&$0$ \\\\\n\\hline 第 11 天到第 20 天 & $+$&$0$&$-$&$-$&$+$&$-$&$+$&$0$&$0$&$+$\\\\\n\\hline 第 21 天到第 30 天 &$0$&$+$&$+$&$0$&$-$&$-$&$-$&$+$&$+$&$0$ \\\\\n\\hline 第31天到第40天 & $+$&$0$&$+$&$-$&$-$&$-$&$+$&$0$&$-$&$+$\\\\\n\\hline\n\\end{tabular}\n\\end{center}\n用频率估计概率.\\\\\n(1) 试估计该农产品价格``上涨''的概率;\\\\\n(2) 假设该农产品每天的价格变化是相互独立的. 在未来的日子里任取$4$天, 试估计该农产品价格在这$4$天中$2$天``上涨''、 $1$天``下跌''、 $1$天``不变''的概率;\\\\\n(3) 假设该农产品每天的价格变化只受前一天价格变化的影响. 判断第$41$天该农产品价格``上涨''``下跌''和``不变''的概率估计值哪个最大. (结论不要求证明)", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2023届全国高考北京卷试题18", + "edit": [ + "20230624\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "018255": { + "id": "018255", + "content": "已知椭圆$E: \\dfrac{x^2}{a^2}+\\dfrac{y^2}{b^2}=1$($a>b>0$)的离心率为$\\dfrac{\\sqrt{5}}{3}$, $A$、$C$分别是$E$的上、下顶点, $B$、$D$分别是$E$的左、右顶点, $|AC|=4$.\\\\\n(1) 求$E$的方程;\\\\\n(2) 设$P$为第一象限内$E$上的动点, 直线$PD$与直线$B C$交于点$M$, 直线$P A$与直线$y=-2$交于点$N$. 求证: $MN\\parallel CD$.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2023届全国高考北京卷试题19", + "edit": [ + "20230624\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "018256": { + "id": "018256", + "content": "设函数$f(x)=x-x^3 \\mathrm{e}^{a x+b}$, 曲线$y=f(x)$在点$(1, f(1))$处的切线方程为$y=-x+1$.\\\\\n(1) 求$a, b$的值;\\\\\n(2) 设函数$g(x)=f'(x)$, 求$g(x)$的单调区间;\\\\\n(3) 求$f(x)$的极值点个数.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2023届全国高考北京卷试题20", + "edit": [ + "20230624\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, + "018257": { + "id": "018257", + "content": "已知数列$\\{a_n\\}, \\{b_n\\}$的项数均为$m$($m>2$), 且$a_n, b_n \\in\\{1, 2, \\cdots, m\\}$, $\\{a_n\\}, \\{b_n\\}$的前$n$项和分别为$A_n, B_n$, 并规定$A_0=B_0=0$. 对于$k \\in\\{0, 1, 2, \\cdots, m\\}$, 定义$r_k=\\max \\{i | B_i \\leq A_k,\\ i \\in\\{0, 1, 2, \\cdots, m\\}\\}$, 其中, $\\max M$表示数集$M$中最大的数.\\\\\n(1) 若$a_1=2$, $a_2=1$, $a_3=3$, $b_1=1$, $b_2=3$, $b_3=3$, 求$r_0, r_1, r_2, r_3$的值;\\\\\n(2) 若$a_1 \\geq b_1$, 且$2 r_j \\leq r_{j+1}+r_{j-1}$, $j=1, 2, \\cdots, m-1$, 求$r_n$;\\\\\n(3) 证明: 存在$p, q, s, t \\in\\{0, 1, 2, \\cdots, m\\}$, 满足$p>q$, $s>t$, 使得$A_p+B_t=A_q+B_s$.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2023届全国高考北京卷试题21", + "edit": [ + "20230624\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "4em", + "unrelated": [] + }, "020001": { "id": "020001", "content": "判断下列各组对象能否组成集合, 若能组成集合, 指出是有限集还是无限集.\\\\\n(1) 上海市控江中学$2022$年入学的全体高一年级新生;\\\\\n(2) 中国现有各省的名称;\\\\\n(3) 太阳、$2$、上海市;\\\\\n(4) 大于$10$且小于$15$的有理数;\\\\\n(5) 末位是$3$的自然数;\\\\\n(6) 影响力比较大的中国数学家;\\\\\n(7) 方程$x^2+x-3=0$的所有实数解;\\\\ \n(8) 函数$y=\\dfrac 1x$图像上所有的点;\\\\ \n(9) 在平面直角坐标系中, 到定点$(0, 0)$的距离等于$1$的所有点;\\\\\n(10) 不等式$3x-10<0$的所有正整数解;\\\\\n(11) 所有的平面四边形.",