20230315 evening
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"execution_count": 2,
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"cells": [
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"cell_type": "code",
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"execution_count": null,
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"execution_count": 1,
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"metadata": {},
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"outputs": [],
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"ID 031310 已被使用.\n"
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]
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}
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],
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"source": [
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"import os,re,json,time\n",
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"\n",
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"cells": [
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{
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"cell_type": "code",
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"execution_count": 3,
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"metadata": {},
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"outputs": [],
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"source": [
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"#修改起始id,出处,文件名\n",
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"starting_id = 14743\n",
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"starting_id = 31311\n",
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"raworigin = \"\"\n",
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"filename = r\"C:\\Users\\weiye\\Documents\\wwy sync\\临时工作区\\自拟题目9.tex\"\n",
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"editor = \"20230314\\t王伟叶\"\n",
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"editor = \"20230315\\t王伟叶\"\n",
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"indexed = True\n"
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]
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},
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{
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"cell_type": "code",
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"execution_count": 2,
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"execution_count": 4,
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"metadata": {},
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"outputs": [
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{
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"name": "stdout",
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"output_type": "stream",
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"text": [
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"添加题号014743, 来源: 23届宝山区混合式教学适应性练习试题1\n",
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"添加题号014744, 来源: 23届宝山区混合式教学适应性练习试题2\n",
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"添加题号014745, 来源: 23届宝山区混合式教学适应性练习试题3\n",
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"添加题号014746, 来源: 23届宝山区混合式教学适应性练习试题4\n",
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"添加题号014747, 来源: 23届宝山区混合式教学适应性练习试题5\n",
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"添加题号014748, 来源: 23届宝山区混合式教学适应性练习试题6\n",
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"添加题号014749, 来源: 23届宝山区混合式教学适应性练习试题7\n",
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"添加题号014750, 来源: 23届宝山区混合式教学适应性练习试题8\n",
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"添加题号014751, 来源: 23届宝山区混合式教学适应性练习试题9\n",
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"添加题号014752, 来源: 23届宝山区混合式教学适应性练习试题10\n",
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"添加题号014753, 来源: 23届宝山区混合式教学适应性练习试题11\n",
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"添加题号014754, 来源: 23届宝山区混合式教学适应性练习试题12\n",
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"添加题号014755, 来源: 23届宝山区混合式教学适应性练习试题13\n",
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"添加题号014756, 来源: 23届宝山区混合式教学适应性练习试题14\n",
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"添加题号014757, 来源: 23届宝山区混合式教学适应性练习试题15\n",
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"添加题号014758, 来源: 23届宝山区混合式教学适应性练习试题16\n",
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"添加题号014759, 来源: 23届宝山区混合式教学适应性练习试题17\n",
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"添加题号014760, 来源: 23届宝山区混合式教学适应性练习试题18\n",
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"添加题号014761, 来源: 23届宝山区混合式教学适应性练习试题19\n",
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"添加题号014762, 来源: 23届宝山区混合式教学适应性练习试题20\n",
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"添加题号014763, 来源: 23届宝山区混合式教学适应性练习试题21\n"
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"添加题号031311, 来源: 23届四校联考试题1\n",
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"添加题号031312, 来源: 23届四校联考试题2\n",
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"添加题号031313, 来源: 23届四校联考试题3\n",
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"添加题号031314, 来源: 23届四校联考试题4\n",
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"添加题号031315, 来源: 23届四校联考试题5\n",
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"添加题号031316, 来源: 23届四校联考试题6\n",
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"添加题号031317, 来源: 23届四校联考试题7\n",
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"添加题号031318, 来源: 23届四校联考试题8\n",
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"添加题号031319, 来源: 23届四校联考试题9\n",
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"添加题号031320, 来源: 23届四校联考试题10\n",
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"添加题号031321, 来源: 23届四校联考试题11\n",
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"添加题号031322, 来源: 23届四校联考试题12\n",
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"添加题号031323, 来源: 23届四校联考试题13\n",
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"添加题号031324, 来源: 23届四校联考试题14\n",
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"添加题号031325, 来源: 23届四校联考试题15\n",
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"添加题号031326, 来源: 23届四校联考试题16\n",
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"添加题号031327, 来源: 23届四校联考试题17\n",
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"添加题号031328, 来源: 23届四校联考试题18\n",
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"添加题号031329, 来源: 23届四校联考试题19\n",
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"添加题号031330, 来源: 23届四校联考试题20\n",
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"添加题号031331, 来源: 23届四校联考试题21\n"
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]
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}
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],
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@ -148,7 +148,7 @@
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],
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"metadata": {
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"kernelspec": {
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"display_name": "mathdept",
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"display_name": "pythontest",
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"language": "python",
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"name": "python3"
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},
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@ -162,12 +162,12 @@
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.8.15"
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"version": "3.9.15"
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"orig_nbformat": 4,
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"interpreter": {
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"hash": "42dd566da87765ddbe9b5c5b483063747fec4aacc5469ad554706e4b742e67b2"
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"hash": "91219a98e0e9be72efb992f647fe78b593124968b75db0b865552d6787c8db93"
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@ -440397,6 +440397,431 @@
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"remark": "",
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"space": ""
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},
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"031310": {
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"id": "031310",
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"content": "命题$p$: 存在$a\\in \\mathbf{R}$且$a\\ne 0$, 对任意的$x\\in \\mathbf{R}$, 均有$f(x+a)<f(x)+f(a)$恒成立. 已知命题$q_1$: $f(x)$单调递减, 且$f(x)>0$恒成立; 命题$q_2$: $f(x)$单调递增, 且存在${x_0}<0$使得$f({x_0})=0$. 则下列说法正确的是\\bracket{20}.\n\\twoch{$q_1$、$q_2$都是$p$的充分条件}{只有$q_1$是$p$的充分条件}{只有$q_2$是$p$的充分条件}{$q_1$、$q_2$都不是$p$的充分条件}",
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"objs": [
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"K0219001B"
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],
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"tags": [
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"第二单元"
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],
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"genre": "选择题",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"origin": "上海2020年秋季高考试题16-20230314修改",
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"edit": [
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"20220701\t王伟叶",
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"20230314\t"
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],
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"same": [],
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"related": [
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"003625"
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],
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"remark": "",
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"space": ""
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},
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"031311": {
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"id": "031311",
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"content": "已知集合$A=\\{x | x^2-4 x<0\\}$, $B=\\{1,2,3,4,5\\}$, 则$\\overline {A} \\cap B=$\\blank{50}.",
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"objs": [],
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"tags": [],
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"genre": "",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"origin": "23届四校联考试题1",
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"edit": [
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"20230315\t王伟叶"
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],
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"same": [],
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"related": [],
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"remark": "",
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"space": ""
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},
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"031312": {
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"id": "031312",
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"content": "不等式$\\lg (x-1)<1$的解集为\\blank{50}.",
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"objs": [],
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"tags": [],
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"genre": "",
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"ans": "",
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"solution": "",
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"duration": -1,
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||||
"usages": [],
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||||
"origin": "23届四校联考试题2",
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"edit": [
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"20230315\t王伟叶"
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],
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"same": [],
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"related": [],
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"remark": "",
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"space": ""
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},
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"031313": {
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"id": "031313",
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"content": "已知复数$z=1+2 \\mathrm{i}$, 则$z\\cdot (\\overline{z})^{-1}=$\\blank{50}.",
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"objs": [],
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"tags": [],
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"genre": "",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"origin": "23届四校联考试题3",
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"edit": [
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"20230315\t王伟叶"
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],
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"same": [],
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"related": [],
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"remark": "",
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"space": ""
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},
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"031314": {
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"id": "031314",
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"content": "函数$y=\\sin ^2(\\pi x)$的最小正周期为\\blank{50}.",
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"objs": [],
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"tags": [],
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"genre": "",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"origin": "23届四校联考试题4",
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"edit": [
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"20230315\t王伟叶"
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],
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"same": [],
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"related": [],
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"remark": "",
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"space": ""
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},
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"031315": {
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"id": "031315",
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"content": "平行直线$x+\\sqrt{3} y+\\sqrt{3}=0$与$\\sqrt{3} x+3 y-9=0$之间的距离为\\blank{50}.",
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"objs": [],
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"tags": [],
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"genre": "",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"origin": "23届四校联考试题5",
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"edit": [
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"20230315\t王伟叶"
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],
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"same": [],
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"related": [],
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"remark": "",
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"space": ""
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},
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"031316": {
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"id": "031316",
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"content": "若$12^a=3^b=m$, 且$\\dfrac{1}{a}-\\dfrac{1}{b}=2$, 则$m=$\\blank{50}.",
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"objs": [],
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"tags": [],
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"genre": "",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"origin": "23届四校联考试题6",
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"edit": [
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"20230315\t王伟叶"
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],
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"same": [],
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"related": [],
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"remark": "",
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"space": ""
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},
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"031317": {
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"id": "031317",
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"content": "向量$\\overrightarrow {n}=(a, 2)$为直线$x-2 y+2=0$的法向量, 则向量$(1,1)$在$\\overrightarrow {n}$方向上的投影为\\blank{50}.",
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"objs": [],
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"tags": [],
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"genre": "",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"origin": "23届四校联考试题7",
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"edit": [
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"20230315\t王伟叶"
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],
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"same": [],
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"related": [],
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"remark": "",
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"space": ""
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},
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"031318": {
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"id": "031318",
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"content": "长方体$ABCD-A_1B_1C_1D_1$为不计容器壁厚度的密封容器, 里面盛有体积为$V$的水, 已知$AB=3$, $AA_1=2$, $AD=1$, 如果将该密封容器任意摆放均不能使水面呈三角形, 则$V$的取值范围为\\blank{50}.\n\\begin{center}\n\\begin{tikzpicture}[>=latex,scale= 0.6]\n\\def\\l{3}\n\\def\\m{2}\n\\def\\n{2}\n\\begin{scope}[x = {(10:1cm)}, y = {(100:1cm)}]\n\\draw (0,0,0) node [below left] {$A$} coordinate (A);\n\\draw (A) ++ (\\l,0,0) node [below right] {$B$} coordinate (B);\n\\draw (A) ++ (\\l,0,-\\m) node [right] {$C$} coordinate (C);\n\\draw (A) ++ (0,0,-\\m) node [left] {$D$} coordinate (D);\n\\draw (A) ++ (0,\\n,0) node [left] {$A_1$} coordinate (A1);\n\\draw (B) ++ (0,\\n,0) node [right] {$B_1$} coordinate (B1);\n\\draw (C) ++ (0,\\n,0) node [above right] {$C_1$} coordinate (C1);\n\\draw (D) ++ (0,\\n,0) node [above left] {$D_1$} coordinate (D1);\n\\end{scope}\n\\draw ($(A)!0.6!(A1)$) coordinate (P1);\n\\draw (P1) ++ (0,0,-\\m) coordinate (P4);\n\\draw (P1) ++ ({\\l/cos(10)},0) coordinate (P2);\n\\draw (P2) ++ (0,0,-\\m) coordinate (P3);\n\\fill [gray!30] (P1)--(P4)--(P3)--(C)--(B)--(A)--cycle;\n\\draw [thick] (A1) -- (B1) -- (C1) -- (D1) -- cycle;\n\\draw [thick] (A) -- (A1) (B) -- (B1) (C) -- (C1);\n\\draw [thick, dashed] (D) -- (D1);\n\\draw [thick] (A) -- (B) -- (C);\n\\draw [thick, dashed] (A) -- (D) -- (C);\n\\draw (P1)--(P2)--(P3);\n\\draw [dashed] (P1)--(P4)--(P3);\n\\end{tikzpicture}\n\\end{center}",
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"objs": [],
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"tags": [],
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"genre": "",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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||||
"origin": "23届四校联考试题8",
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"edit": [
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"20230315\t王伟叶"
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],
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"same": [],
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"related": [],
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"remark": "",
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"space": ""
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},
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"031319": {
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"id": "031319",
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"content": "在$\\triangle ABC$中, $\\angle A=150^{\\circ}, D_1, D_2, \\cdots, D_{2022}$依次为边$BC$上的点, \n且$BD_1=D_1D_2=D_2D_3=\\cdots=D_{2021} D_{2022}=D_{2022} C$, 设$\\angle BAD_1=\\alpha_1$, $\\angle D_1AD_2=\\alpha_2$, $\\cdots$, \n$\\angle D_{2021} AD_{2022}=\\alpha_{2022}$, $\\angle D_{2022} AC=\\alpha_{2023}$, 则$\\dfrac{\\sin \\alpha_1 \\sin \\alpha_3 \\cdots \\sin \\alpha_{2023}}{\\sin \\alpha_2 \\sin \\alpha_4 \\cdots \\sin \\alpha_{2022}}$的值为\\blank{50}.",
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"objs": [],
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"tags": [],
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"genre": "",
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"ans": "",
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"solution": "",
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"duration": -1,
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"usages": [],
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"origin": "23届四校联考试题9",
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||||
"edit": [
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"20230315\t王伟叶"
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],
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"same": [],
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"related": [],
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"remark": "",
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"space": ""
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},
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"031320": {
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"id": "031320",
|
||||
"content": "上海电视台五星体育频道有一档四人扑克牌竞技节目``上海三打一'', 在打法中有一种``三带二''的牌型, 即点数相同的三张牌外加一对牌, (三张牌的点数必须和对牌的点数不同). 在一副不含大小王的$52$张扑克牌中不放回的抽取五次, 已知前三次抽到两张A, 一张K, 则接下来两次抽取能抽到``三带二''的牌型(AAAKK或KKKAA)的概率为\\blank{50}.",
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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||||
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||||
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|
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|
||||
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|
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|
||||
"031321": {
|
||||
"id": "031321",
|
||||
"content": "函数$y=f(x)$的表达式为$f(x)=2 x^3-5 x^2-4 x$, 如果$f(a)=f(b)=f(c)$且$a<b<c$, 则$a b c$的取值范围为\\blank{50}.",
|
||||
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|
||||
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||||
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||||
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|
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|
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|
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"031322": {
|
||||
"id": "031322",
|
||||
"content": "已知抛物线$y^2=2 p x$($p>0$), $P(2,1)$为抛物线内一点, 不经过点$P$的直线$l: y=2 x+m$与抛物线相交于$A, B$两点, 直线$AP, BP$分别交抛物线于$C, D$两点, 若对任意直线$l$, 总存在$\\lambda$, 使得$\\overrightarrow{AP}=\\lambda \\overrightarrow{PC}$, $\\overrightarrow{BP}=\\lambda \\overrightarrow{PD}$($\\lambda>0$, $\\lambda \\neq 1$)成立, 则$p=$\\blank{50}.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, scale = 0.3]\n\\draw [->] (-2,0) -- (10,0) node [below] {$x$};\n\\draw [->] (0,-6.5) -- (0,6.5) node [left] {$y$};\n\\draw (0,0) node [above left] {$O$};\n\\draw [domain = -6.3:6.3] plot ({\\x*\\x/4},\\x);\n\\draw (2,1) node [right] {$P$} coordinate (P);\n\\draw (4,-4) node [below] {$C$} coordinate (C);\n\\draw (9,6) node [above] {$D$} coordinate (D);\n\\draw (1.44,2.4) node [above left] {$A$} coordinate (A);\n\\draw (0.04,-0.4) node [below left] {$B$} coordinate (B);\n\\draw ($(A)!-0.5!(B)$)--($(A)!1.5!(B)$) (C)--(D) (B)--(D)(A)--(C);\n\\end{tikzpicture}\n\\end{center}",
|
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|
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|
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|
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|
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"031323": {
|
||||
"id": "031323",
|
||||
"content": "已知直线$l_1: x+a y-2=0$, $l_2: (a+1) x-a y+1=0$, 则$a=-2$是$l_1\\parallel l_2$的\\bracket{20}.\n\\twoch{充分不必要条件}{必要不充分条件}{充要条件}{既不充分也不必要条件}",
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|
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"031324": {
|
||||
"id": "031324",
|
||||
"content": "已知函数$y=\\dfrac{\\mathrm{e}^x}{\\mathrm{e}^x-1}$, 则其图像大致是\\bracket{20}.\n\\fourch{\\begin{tikzpicture}[>=latex, scale = 0.5]\n\\begin{scope}\n\\clip (-3,-3) rectangle (3,3);\n\\draw [domain = -3.5:-0.1, samples = 100] plot (\\x+0.5,{exp(\\x)/(exp(\\x)-1)+0.3});\n\\draw [domain = 0.1:3, samples = 100] plot (\\x+0.5,{exp(\\x)/(exp(\\x)-1)+0.3});\n\\end{scope}\n\\draw [->] (-3,0) -- (3,0) node [below] {$x$};\n\\draw [->] (0,-3) -- (0,3) node [left] {$y$};\n\\draw (0,0) node [below right] {$O$};\n\\end{tikzpicture}\n}{\\begin{tikzpicture}[>=latex, scale = 0.5]\n\\begin{scope}\n\\clip (-3,-3) rectangle (3,3);\n\\draw [domain = -3:-0.1, samples = 100] plot (\\x,{exp(\\x)/(exp(\\x)-1)});\n\\draw [domain = 0.1:3, samples = 100] plot (\\x,{exp(\\x)/(exp(\\x)-1)});\n\\end{scope}\n\\draw [->] (-3,0) -- (3,0) node [below] {$x$};\n\\draw [->] (0,-3) -- (0,3) node [left] {$y$};\n\\draw (0,0) node [below right] {$O$};\n\\end{tikzpicture}}{\\begin{tikzpicture}[>=latex, scale = 0.5]\n\\begin{scope}\n\\clip (-3,-3) rectangle (3,3);\n\\draw [domain = -3.5:-0.1, samples = 100] plot (\\x+0.6,{exp(\\x)/(exp(\\x)-1)+1.3});\n\\draw [domain = 0.1:3, samples = 100] plot (\\x+0.5,{exp(\\x)/(exp(\\x)-1)+0.3});\n\\end{scope}\n\\draw [->] (-3,0) -- (3,0) node [below] {$x$};\n\\draw [->] (0,-3) -- (0,3) node [left] {$y$};\n\\draw (0,0) node [below left] {$O$};\n\\end{tikzpicture}}{\\begin{tikzpicture}[>=latex, scale = 0.5]\n\\begin{scope}\n\\clip (-3,-3) rectangle (3,3);\n\\draw [domain = -3:-0.1, samples = 100] plot (\\x,{exp(\\x)/(exp(\\x)-1)-1});\n\\draw [domain = 0.1:3, samples = 100] plot (\\x,{exp(\\x)/(exp(\\x)-1)-1});\n\\end{scope}\n\\draw [->] (-3,0) -- (3,0) node [below] {$x$};\n\\draw [->] (0,-3) -- (0,3) node [left] {$y$};\n\\draw (0,0) node [below right] {$O$};\n\\end{tikzpicture}}",
|
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|
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|
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|
||||
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|
||||
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|
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"031325": {
|
||||
"id": "031325",
|
||||
"content": "如图所示, 正三棱柱$ABC-A_1B_1C_1$的所有棱长均为$1$, 点$P$、$M$、$N$分别为棱$AA_1$、$AB$、$A_1B_1$的中点, 点$Q$为线段$MN$上的动点. 当点$Q$由点$N$出发向点$M$运动的过程中, 以下结论中正确的是\\bracket{20}.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw (0,0,0) node [below] {$A$} coordinate (A);\n\\draw (2,0,0) node [below] {$B$} coordinate (B);\n\\draw (1,0,{-sqrt(3)}) node [below] {$C$} coordinate (C);\n\\draw (A) ++ (0,2,0) node [left] {$A_1$} coordinate (A_1);\n\\draw (B) ++ (0,2,0) node [right] {$B_1$} coordinate (B_1);\n\\draw (C) ++ (0,2,0) node [above] {$C_1$} coordinate (C_1);\n\\draw ($(A)!0.5!(B)$) node [below] {$M$} coordinate (M);\n\\draw ($(A_1)!0.5!(B_1)$) node [above] {$N$} coordinate (N);\n\\draw ($(A)!0.5!(A_1)$) node [left] {$P$} coordinate (P);\n\\draw ($(N)!0.3!(M)$) node [left] {$Q$} coordinate (Q);\n\\draw (A)--(B)--(B_1)--(A_1)--cycle(A_1)--(C_1)--(B_1)(M)--(N)(P)--(B);\n\\draw [dashed] (A)--(C)--(B)(C)--(P)(C)--(C_1)--(Q);\n\\end{tikzpicture}\n\\end{center}\n\\twoch{直线$C_1Q$与直线$CP$可能相交}{直线$C_1Q$与直线$CP$始终异面}{直线$C_1Q$与直线$CP$可能垂直}{直线$C_1Q$与直线$BP$不可能垂直}",
|
||||
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|
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|
||||
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|
||||
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|
||||
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|
||||
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|
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|
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||||
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|
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|
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},
|
||||
"031326": {
|
||||
"id": "031326",
|
||||
"content": "下列用递推公式表示的数列中, 使得$\\displaystyle\\lim _{n \\to+\\infty} a_n=\\sqrt{2}$成立的是\\bracket{20}.\n\\twoch{$\\begin{cases}a_n=\\dfrac{1}{2}(a_{n-1}+\\dfrac{2}{a_{n-1}})(n \\geq 2), \\\\ a_1=-1\\end{cases}$}{$\\begin{cases}a_n=\\dfrac{2-3 a_{n-1}}{a_{n-1}-3}(n \\geq 2), \\\\ a_1=1\\end{cases}$}{$\\begin{cases}a_n=\\dfrac{a_{n-1}+99}{49 a_{n-1}+1}(n \\geq 2), \\\\ a_1=1\\end{cases}$}{$\\begin{cases}a_n=\\dfrac{2+a_{n-1} \\ln a_{n-1}}{a_{n-1}+\\ln a_{n-1}}(n \\geq 2), \\\\ a_1=1\\end{cases}$}",
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
"20230315\t王伟叶"
|
||||
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|
||||
},
|
||||
"031327": {
|
||||
"id": "031327",
|
||||
"content": "如图, 四棱锥$P-ABCD$中, 等腰$\\triangle PAB$的边长分别为$PA=PB=5$, $AB=6$, 矩形$ABCD$所在的平面与平面$PAB$垂直.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, scale = 0.5]\n\\draw (0,0,0) node [left] {$A$} coordinate (A);\n\\draw (6,0,0) node [right] {$B$} coordinate (B);\n\\draw (A) ++ (0,3,0) node [left] {$D$} coordinate (D);\n\\draw (B) ++ (0,3,0) node [right] {$C$} coordinate (C);\n\\draw (3,0,5) node [below] {$P$} coordinate (P);\n\\draw (A)--(P)--(B)--(C)--(D)--cycle(D)--(P)--(C);\n\\draw [dashed] (A)--(B);\n\\end{tikzpicture}\n\\end{center}\n(1) 如果$BC=3$, 求直线$PC$与平面$PAB$所成的角的大小;\\\\\n(2) 如果$PC \\perp BD$, 求$BC$的长.",
|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
||||
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|
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|
||||
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|
||||
"20230315\t王伟叶"
|
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||||
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|
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},
|
||||
"031328": {
|
||||
"id": "031328",
|
||||
"content": "自$2015$年上海启动 《上海绿道专项规划(2035)》至今上海已建成绿道总长度近$1600$公里. 根据 《上海市生态空间专项规划(2021-2035)》, 到$2035$年, 上海绿道总长度将超过$2000$公里. 届时, 绿道会像城市的毛细血管一样, 延伸到市民生活的各个角落. 绿荫下的绿道 (步道、骑行道) 给市民提供了散步休憩、跑步骑行运动的生态空间. 某一线品牌自行车制造商在布局线下自行车体验与销售店时随机调研了$1000$位市民, 调研数据如左下表所示. $166$位有意愿购买万元级运动自行车的受访者的年龄(单位: 岁), 在各区间内的频数记录如右下表所示.\\\\\n(1) 试估计有意愿购买万元级运动自行车人群的平均年龄 (结果精确到$0.1$岁).\\\\\n(2) 将表$1$的$2 \\times 2$列联表中的数据补充完整, 并判断是否有$95\\%$的把握认为``离家附近($2$千米内)有骑行绿道与万元级运动自行车消费有关''? \\\\\n附: $\\chi^2=\\dfrac{n(a d-b c)^2}{(a+b)(c+d)(a+c)(b+d)}$, 其中$n=a+b+c+d$.\n\\begin{center}\n\\begin{tabular}{|c|c|c|c|c|}\n\\hline$P(\\chi^2 \\geq k)$& 0.10 & 0.05 & 0.01 & 0.005 \\\\\n\\hline$k$& 2.706 & 3.841 & 6.635 & 7.879 \\\\\n\\hline\n\\end{tabular}\n\\end{center}\n\\begin{center}\n\\begin{tabular}{|c|c|c|c|}\n\\hline &\\makecell{有意愿购买 \\\\ 万元级 \\\\ 运动自行车 }& \\makecell{没有意愿购买 \\\\ 万元级 \\\\ 运动自行车 }& 总计 \\\\ \\hline\n\\makecell{距家$2$千米内\\\\有骑行绿道}& 118 & 270 & \\\\ \\hline\n\\makecell{距家$2$千米内\\\\无骑行绿道}& & & \\\\\\hline \n总计 & 166 & & 1000 \\\\\\hline\n\\end{tabular}\n\\begin{tabular}{|c|c|}\n\\hline\n\\makecell{ 年龄分 \\\\ 组区间 } & 频数 \\\\\n\\hline$[12,18)$& 16 \\\\\n\\hline$[18,24)$& 24 \\\\\n\\hline$[24,30)$& 35 \\\\\n\\hline$[30,36)$& 30 \\\\\n\\hline$[36,42)$& 21 \\\\\n\\hline$[42,48)$& 15 \\\\\n\\hline$[48,54)$& 11 \\\\\n\\hline$[54,60)$& 6 \\\\\n\\hline$[60,66)$& 5 \\\\\n\\hline$[66,72)$& 3 \\\\\n\\hline\n\\end{tabular}\n\\end{center}",
|
||||
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|
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|
||||
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||||
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|
||||
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|
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||||
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|
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},
|
||||
"031329": {
|
||||
"id": "031329",
|
||||
"content": "雨天外出虽然有撑雨伞, 时常却总免不了淋湿衣袖、裤脚、背包等, 小明想通过数学建模的方法研究如何撑伞可以让淋湿的面积尽量小. 为了简化问题小明做出下列假设:\\\\\n假设 1: 在网上查阅了人均身高和肩宽的数据后, 小明把人假设为身高、肩宽分别为$170 \\text{cm}$、$40 \\text{cm}$的矩形``纸片人'';\\\\\n假设 2: 受风的影响, 雨滴下落轨迹视为与水平地面所成角为$60^{\\circ}$的直线;\\\\\n假设 3: 伞柄$OT$长为$60 \\text{cm}$, 可绕矩形``纸片人''上点$O$旋转;\\\\\n假设 4: 伞面为被伞柄$OT$垂直平分的线段$AB$, $AB=120 \\text{cm}$.\\\\\n以如图$1$方式撑伞矩形``纸片人''将淋湿``裤脚''; 以如图$2$方式撑伞矩形``纸片人''将淋湿``头和肩膀''(``裤脚''也会有一小部分被淋湿).\n\\begin{center}\n\\begin{tikzpicture}[>=latex,scale = 0.7]\n\\begin{scope}\n\\clip (-3,0) rectangle (3,6);\n\\draw (0,0) -- (0,3.4) -- (0.8,3.4) node [below right] {$O$} coordinate (O) -- (0.8,0);\n\\draw (O) --++ (60:1.2) node [above right] {$T$} coordinate (T);\n\\draw (T) --++ (150:1.2) node [left] {$A$} coordinate (A);\n\\draw (T) --++ (-30:1.2) node [right] {$B$} coordinate (B);\n\\draw (A) ++ (60:1) --++ (60:-15);\n\\draw (B) ++ (60:1) --++ (60:-15);\n\\end{scope}\n\\draw (-3,0) -- (3,0);\n\\end{tikzpicture}\n\\begin{tikzpicture}[>=latex,scale = 0.7]\n\\begin{scope}\n\\clip (-3,0) rectangle (3,6);\n\\draw (0,0) -- (0,3.4) -- (0.8,3.4) node [below right] {$O$} coordinate (O) -- (0.8,0);\n\\draw (O) --++ (20:1.2) node [above right] {$T$} coordinate (T);\n\\draw (T) --++ (110:1.2) node [left] {$A$} coordinate (A);\n\\draw (T) --++ (-70:1.2) node [right] {$B$} coordinate (B);\n\\draw (A) ++ (60:1) --++ (60:-15);\n\\draw (B) ++ (60:1) --++ (60:-15);\n\\end{scope}\n\\draw (-3,0) -- (3,0);\n\\end{tikzpicture}\n\\begin{tikzpicture}[>=latex,scale = 0.7]\n\\begin{scope}\n\\clip (-3,0) rectangle (3.5,6);\n\\draw (0,0) -- (0,3.4) -- (0.8,3.4) node [below right] {$O$} coordinate (O) -- (0.8,0);\n\\draw (O) --++ (39.09484:1.2) node [above right] {$T$} coordinate (T);\n\\draw (T) --++ (129.09484:1.2) node [left] {$A$} coordinate (A);\n\\draw (T) --++ ({39.09484-90}:1.2) node [right] {$B$} coordinate (B);\n\\draw (A) ++ (60:1) --++ (60:-15);\n\\draw (B) ++ (60:1) --++ (60:-15);\n\\end{scope}\n\\draw (-3,0) -- (3,0);\n\\end{tikzpicture}\n\\end{center}\n(1) 如图$3$在矩形``纸片人''上身恰好不被淋湿时, 求其``裤脚''被淋湿 (阴影) 部分的面积(结果精确到$0.1 \\text{cm}^2)$;\\\\\n(2) 请根据你的生活经验对小明建立的数学模型提两条改进建议. (无需求解改进后的模型, 如果建议超过两条仅对前两条评分)",
|
||||
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|
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|
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|
||||
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|
||||
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|
||||
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|
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|
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|
||||
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|
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|
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},
|
||||
"031330": {
|
||||
"id": "031330",
|
||||
"content": "已知椭圆$c: \\dfrac{x^2}{4}+\\dfrac{y^2}{b^2}=1$($0<b<2$)经过点$M(1,-\\dfrac{3}{2})$, $F_1$, $F_2$为椭圆$c$的左右焦点, $Q(x_0, y_0)$为平面内一个动点, 其中$y_0>0$, 记直线$QF_1$与椭圆$C$在$x$轴上方的交点为$A(x_1, y_1)$, 直线$QF_2$与椭圆$c$在$x$轴上方的交点为$B(x_2, y_2)$.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw [->] (-3,0) -- (3,0) node [below] {$x$};\n\\draw [->] (0,-2.5) -- (0,2.5) node [left] {$y$};\n\\draw (0,0) node [below left] {$O$};\n\\draw [name path = elli] (0,0) ellipse (2 and {sqrt(3)});\n\\filldraw (-1,0) circle (0.03) node [below] {$F_1$} coordinate (F_1);\n\\filldraw (1,0) circle (0.03) node [below] {$F_2$} coordinate (F_2);\n\\path [name path = line1] (F_1) --++ (50:2.5);\n\\path [name path = line2] (F_2) --++ (50:1.5);\n\\path [name intersections = {of = line1 and elli, by = B}];\n\\path [name intersections = {of = line2 and elli, by = A}];\n\\path [draw, name path = line3] (F_1) -- (A) node [above] {$A$};\n\\path [draw, name path = line4] (F_2) -- (B) node [above] {$B$};\n\\path [name intersections = {of = line3 and line4, by = Q}];\n\\draw (Q) node [below] {$Q$};\n\\draw (F_1)--(B) (F_2)--(A);\n\\end{tikzpicture}\n\\end{center}\n(1) 求椭圆$c$的的离心率;\\\\\n(2) 若$AF_2\\parallel BF_1$, 证明: $\\dfrac{1}{y_1}+\\dfrac{1}{y_2}=\\dfrac{1}{y_0}$;\\\\\n(3) 若$\\dfrac{1}{y_1}+\\dfrac{1}{y_2}=\\dfrac{4}{3 y_0}$, 求点$Q$的轨迹方程.",
|
||||
"objs": [],
|
||||
"tags": [],
|
||||
"genre": "",
|
||||
"ans": "",
|
||||
"solution": "",
|
||||
"duration": -1,
|
||||
"usages": [],
|
||||
"origin": "23届四校联考试题20",
|
||||
"edit": [
|
||||
"20230315\t王伟叶"
|
||||
],
|
||||
"same": [],
|
||||
"related": [],
|
||||
"remark": "",
|
||||
"space": ""
|
||||
},
|
||||
"031331": {
|
||||
"id": "031331",
|
||||
"content": "已知函数$y=f(x)$的表达式为$f(x)=\\dfrac{1}{2} a x^2+(a+1) x+\\ln x$($a \\in \\mathbf{R}$).\\\\\n(1) 若$1$是$f(x)$的极值点, 求$a$的值;\\\\\n(2) 求$f(x)$的单调区间;\\\\\n(3) 若$f(x)=\\dfrac{1}{2} a x^2+x$有两个实数解$x_1, x_2$($x_1<x_2$),\\\\ \n(i) 直接写出$a$的取值范围;\\\\\n(ii) $\\lambda$为正实数, 若对于符合题意的任意$x_1, x_2$, 当$s=\\lambda(x_1+x_2)$时都有$f'(s)<0$, 求$\\lambda$的取值范围.",
|
||||
"objs": [],
|
||||
"tags": [],
|
||||
"genre": "",
|
||||
"ans": "",
|
||||
"solution": "",
|
||||
"duration": -1,
|
||||
"usages": [],
|
||||
"origin": "23届四校联考试题21",
|
||||
"edit": [
|
||||
"20230315\t王伟叶"
|
||||
],
|
||||
"same": [],
|
||||
"related": [],
|
||||
"remark": "",
|
||||
"space": ""
|
||||
},
|
||||
"040001": {
|
||||
"id": "040001",
|
||||
"content": "参数方程$\\begin{cases}x=3 t^2+4, \\\\ y=t^2-2\\end{cases}$($0 \\leq t \\leq 3$)所表示的曲线是\\bracket{20}.\n\\fourch{一支双曲线}{线段}{圆弧}{射线}",
|
||||
|
|
@ -444343,31 +444768,5 @@
|
|||
"related": [],
|
||||
"remark": "",
|
||||
"space": ""
|
||||
},
|
||||
"031310": {
|
||||
"id": "031310",
|
||||
"content": "命题$p$: 存在$a\\in \\mathbf{R}$且$a\\ne 0$, 对任意的$x\\in \\mathbf{R}$, 均有$f(x+a)<f(x)+f(a)$恒成立. 已知命题$q_1$: $f(x)$单调递减, 且$f(x)>0$恒成立; 命题$q_2$: $f(x)$单调递增, 且存在${x_0}<0$使得$f({x_0})=0$. 则下列说法正确的是\\bracket{20}.\n\\twoch{$q_1$、$q_2$都是$p$的充分条件}{只有$q_1$是$p$的充分条件}{只有$q_2$是$p$的充分条件}{$q_1$、$q_2$都不是$p$的充分条件}",
|
||||
"objs": [
|
||||
"K0219001B"
|
||||
],
|
||||
"tags": [
|
||||
"第二单元"
|
||||
],
|
||||
"genre": "选择题",
|
||||
"ans": "",
|
||||
"solution": "",
|
||||
"duration": -1,
|
||||
"usages": [],
|
||||
"origin": "上海2020年秋季高考试题16-20230314修改",
|
||||
"edit": [
|
||||
"20220701\t王伟叶",
|
||||
"20230314\t"
|
||||
],
|
||||
"same": [],
|
||||
"related": [
|
||||
"003625"
|
||||
],
|
||||
"remark": "",
|
||||
"space": ""
|
||||
}
|
||||
}
|
||||
Reference in New Issue