20230227 evening
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"cells": [
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{
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"cell_type": "code",
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"execution_count": 5,
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"execution_count": 50,
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"metadata": {},
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"outputs": [
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{
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"0"
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]
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},
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"execution_count": 5,
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"execution_count": 50,
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"metadata": {},
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"output_type": "execute_result"
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}
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@ -19,7 +19,7 @@
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"source": [
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"import os,re,json\n",
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"\"\"\"这里编辑题号(列表)后将在vscode中打开窗口, 编辑后保存关闭, 随后运行第二个代码块\"\"\"\n",
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"problems = \"4129\"\n",
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"problems = \"31237\"\n",
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"\n",
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"def generate_number_set(string,dict):\n",
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" string = re.sub(r\"[\\n\\s]\",\"\",string)\n",
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@ -51,7 +51,7 @@
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},
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{
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"cell_type": "code",
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"execution_count": 4,
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"execution_count": 49,
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"metadata": {},
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"outputs": [],
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"source": [
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@ -21,7 +21,9 @@
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"\n",
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"\"\"\"---设置关键字, 同一field下不同选项为or关系, 同一字典中不同字段间为and关系, 不同字典间为or关系, _not表示列表中的关键字都不含, 同一字典中的数字用来供应同一字段不同的条件之间的and---\"\"\"\n",
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"keywords_dict_table = [\n",
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" {\"usages\":[\"2025届高一\"],\"usages2\":[\"202302\"]}\n",
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" {\"origin\":[\"教材复习题\"],\"tags\":[\"第三单元\"]},\n",
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" {\"origin\":[\"新教材\"],\"origin2\":[\"课堂练习\"],\"tags\":[\"第三单元\"]},\n",
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" {\"origin\":[\"新教材\"],\"origin2\":[\"册习题\"],\"tags\":[\"第三单元\"]},\n",
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"]\n",
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"\"\"\"---关键字设置完毕---\"\"\"\n",
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"# 示例： keywords_dict_table = [\n",
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@ -96,7 +98,7 @@
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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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@ -115,7 +117,7 @@
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"orig_nbformat": 4,
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"vscode": {
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"interpreter": {
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"hash": "ff3c292c316ba85de6f1ad75f19c731e79d694e741b6f515ec18f14996fe48dc"
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"hash": "91219a98e0e9be72efb992f647fe78b593124968b75db0b865552d6787c8db93"
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}
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}
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},
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@ -2,125 +2,26 @@
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"cells": [
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{
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"cell_type": "code",
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"execution_count": 7,
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"execution_count": 6,
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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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"题号: 040001 , 字段: usages 中已添加数据: 202302217\t2024届高二10班\t0.962\n",
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"题号: 040002 , 字段: usages 中已添加数据: 202302217\t2024届高二10班\t1.000\n",
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"题号: 040003 , 字段: usages 中已添加数据: 202302217\t2024届高二10班\t0.962\n",
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"题号: 040004 , 字段: usages 中已添加数据: 202302217\t2024届高二10班\t1.000\n",
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"题号: 040005 , 字段: usages 中已添加数据: 202302217\t2024届高二10班\t1.000\n",
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"题号: 040006 , 字段: usages 中已添加数据: 202302217\t2024届高二10班\t0.231\n",
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"题号: 040007 , 字段: usages 中已添加数据: 202302217\t2024届高二10班\t0.885\n",
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"题号: 040008 , 字段: usages 中已添加数据: 202302217\t2024届高二10班\t0.885\n",
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"题号: 040009 , 字段: usages 中已添加数据: 202302217\t2024届高二10班\t0.923\n",
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"题号: 040011 , 字段: usages 中已添加数据: 202302217\t2024届高二10班\t0.923\n",
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"题号: 040012 , 字段: usages 中已添加数据: 202302217\t2024届高二10班\t0.000\n",
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"题号: 040013 , 字段: usages 中已添加数据: 202302217\t2024届高二10班\t0.615\n",
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"题号: 040014 , 字段: usages 中已添加数据: 202302217\t2024届高二10班\t0.385\n",
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"题号: 040015 , 字段: usages 中已添加数据: 202302217\t2024届高二10班\t0.712\t0.808\n",
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"题号: 040016 , 字段: usages 中已添加数据: 202302217\t2024届高二10班\t0.885\t0.308\n",
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"题号: 040017 , 字段: usages 中已添加数据: 202302217\t2024届高二10班\t0.962\t0.769\t0.462\n",
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"题号: 040001 , 字段: usages 中已添加数据: 202302217\t2024届高二01班\t0.944\n",
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"题号: 040002 , 字段: usages 中已添加数据: 202302217\t2024届高二01班\t0.972\n",
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"题号: 040003 , 字段: usages 中已添加数据: 202302217\t2024届高二01班\t1.000\n",
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"题号: 040004 , 字段: usages 中已添加数据: 202302217\t2024届高二01班\t0.722\n",
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"题号: 040005 , 字段: usages 中已添加数据: 202302217\t2024届高二01班\t0.889\n",
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"题号: 040006 , 字段: usages 中已添加数据: 202302217\t2024届高二01班\t0.528\n",
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"题号: 040007 , 字段: usages 中已添加数据: 202302217\t2024届高二01班\t0.694\n",
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"题号: 040008 , 字段: usages 中已添加数据: 202302217\t2024届高二01班\t1.000\n",
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"题号: 040009 , 字段: usages 中已添加数据: 202302217\t2024届高二01班\t1.000\n",
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"题号: 040011 , 字段: usages 中已添加数据: 202302217\t2024届高二01班\t0.889\n",
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"题号: 040012 , 字段: usages 中已添加数据: 202302217\t2024届高二01班\t0.111\n",
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"题号: 040013 , 字段: usages 中已添加数据: 202302217\t2024届高二01班\t0.833\n",
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"题号: 040014 , 字段: usages 中已添加数据: 202302217\t2024届高二01班\t0.583\n",
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"题号: 040015 , 字段: usages 中已添加数据: 202302217\t2024届高二01班\t1.000\t0.833\n",
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"题号: 040016 , 字段: usages 中已添加数据: 202302217\t2024届高二01班\t0.958\t0.889\n",
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"题号: 040017 , 字段: usages 中已添加数据: 202302217\t2024届高二01班\t0.944\t0.889\t0.708\n",
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"题号: 040001 , 字段: usages 中已添加数据: 202302217\t2024届高二12班\t0.655\n",
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"题号: 040002 , 字段: usages 中已添加数据: 202302217\t2024届高二12班\t0.828\n",
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"题号: 040003 , 字段: usages 中已添加数据: 202302217\t2024届高二12班\t0.655\n",
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"题号: 040004 , 字段: usages 中已添加数据: 202302217\t2024届高二12班\t0.483\n",
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"题号: 040005 , 字段: usages 中已添加数据: 202302217\t2024届高二12班\t0.966\n",
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"题号: 040006 , 字段: usages 中已添加数据: 202302217\t2024届高二12班\t0.310\n",
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"题号: 040007 , 字段: usages 中已添加数据: 202302217\t2024届高二12班\t0.828\n",
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"题号: 040008 , 字段: usages 中已添加数据: 202302217\t2024届高二12班\t0.966\n",
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"题号: 040009 , 字段: usages 中已添加数据: 202302217\t2024届高二12班\t0.897\n",
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"题号: 040011 , 字段: usages 中已添加数据: 202302217\t2024届高二12班\t0.793\n",
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"题号: 040012 , 字段: usages 中已添加数据: 202302217\t2024届高二12班\t0.000\n",
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"题号: 040013 , 字段: usages 中已添加数据: 202302217\t2024届高二12班\t0.690\n",
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"题号: 040014 , 字段: usages 中已添加数据: 202302217\t2024届高二12班\t0.345\n",
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"题号: 040015 , 字段: usages 中已添加数据: 202302217\t2024届高二12班\t0.983\t0.741\n",
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"题号: 040016 , 字段: usages 中已添加数据: 202302217\t2024届高二12班\t0.931\t0.293\n",
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"题号: 040017 , 字段: usages 中已添加数据: 202302217\t2024届高二12班\t1.000\t0.483\t0.534\n",
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"题号: 040001 , 字段: usages 中已添加数据: 202302217\t2024届高二02班\t0.974\n",
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"题号: 040002 , 字段: usages 中已添加数据: 202302217\t2024届高二02班\t1.000\n",
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"题号: 040003 , 字段: usages 中已添加数据: 202302217\t2024届高二02班\t0.974\n",
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"题号: 040004 , 字段: usages 中已添加数据: 202302217\t2024届高二02班\t0.795\n",
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"题号: 040005 , 字段: usages 中已添加数据: 202302217\t2024届高二02班\t0.872\n",
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"题号: 040006 , 字段: usages 中已添加数据: 202302217\t2024届高二02班\t0.615\n",
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"题号: 040007 , 字段: usages 中已添加数据: 202302217\t2024届高二02班\t0.769\n",
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"题号: 040008 , 字段: usages 中已添加数据: 202302217\t2024届高二02班\t1.000\n",
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"题号: 040009 , 字段: usages 中已添加数据: 202302217\t2024届高二02班\t0.974\n",
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"题号: 040011 , 字段: usages 中已添加数据: 202302217\t2024届高二02班\t0.821\n",
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"题号: 040012 , 字段: usages 中已添加数据: 202302217\t2024届高二02班\t0.154\n",
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"题号: 040013 , 字段: usages 中已添加数据: 202302217\t2024届高二02班\t0.974\n",
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"题号: 040014 , 字段: usages 中已添加数据: 202302217\t2024届高二02班\t0.718\n",
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"题号: 040015 , 字段: usages 中已添加数据: 202302217\t2024届高二02班\t0.872\t0.718\n",
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"题号: 040016 , 字段: usages 中已添加数据: 202302217\t2024届高二02班\t0.949\t0.590\n",
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"题号: 040017 , 字段: usages 中已添加数据: 202302217\t2024届高二02班\t0.923\t0.782\t0.590\n",
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"题号: 040001 , 字段: usages 中已添加数据: 202302217\t2024届高二06班\t0.950\n",
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"题号: 040002 , 字段: usages 中已添加数据: 202302217\t2024届高二06班\t0.975\n",
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"题号: 040003 , 字段: usages 中已添加数据: 202302217\t2024届高二06班\t0.950\n",
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"题号: 040004 , 字段: usages 中已添加数据: 202302217\t2024届高二06班\t0.950\n",
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"题号: 040005 , 字段: usages 中已添加数据: 202302217\t2024届高二06班\t0.875\n",
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"题号: 040006 , 字段: usages 中已添加数据: 202302217\t2024届高二06班\t0.250\n",
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"题号: 040007 , 字段: usages 中已添加数据: 202302217\t2024届高二06班\t0.925\n",
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"题号: 040008 , 字段: usages 中已添加数据: 202302217\t2024届高二06班\t1.000\n",
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"题号: 040009 , 字段: usages 中已添加数据: 202302217\t2024届高二06班\t0.950\n",
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"题号: 040011 , 字段: usages 中已添加数据: 202302217\t2024届高二06班\t0.950\n",
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"题号: 040012 , 字段: usages 中已添加数据: 202302217\t2024届高二06班\t0.050\n",
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"题号: 040013 , 字段: usages 中已添加数据: 202302217\t2024届高二06班\t0.675\n",
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"题号: 040014 , 字段: usages 中已添加数据: 202302217\t2024届高二06班\t0.500\n",
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"题号: 040015 , 字段: usages 中已添加数据: 202302217\t2024届高二06班\t0.725\t0.938\n",
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"题号: 040016 , 字段: usages 中已添加数据: 202302217\t2024届高二06班\t0.975\t0.388\n",
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"题号: 040017 , 字段: usages 中已添加数据: 202302217\t2024届高二06班\t0.925\t0.825\t0.588\n",
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"题号: 040001 , 字段: usages 中已添加数据: 202302217\t2024届高二07班\t0.975\n",
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"题号: 040002 , 字段: usages 中已添加数据: 202302217\t2024届高二07班\t0.925\n",
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"题号: 040003 , 字段: usages 中已添加数据: 202302217\t2024届高二07班\t0.925\n",
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"题号: 040004 , 字段: usages 中已添加数据: 202302217\t2024届高二07班\t0.075\n",
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"题号: 040005 , 字段: usages 中已添加数据: 202302217\t2024届高二07班\t0.750\n",
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"题号: 040006 , 字段: usages 中已添加数据: 202302217\t2024届高二07班\t0.425\n",
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"题号: 040007 , 字段: usages 中已添加数据: 202302217\t2024届高二07班\t0.775\n",
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"题号: 040008 , 字段: usages 中已添加数据: 202302217\t2024届高二07班\t0.950\n",
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"题号: 040009 , 字段: usages 中已添加数据: 202302217\t2024届高二07班\t0.775\n",
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"题号: 040011 , 字段: usages 中已添加数据: 202302217\t2024届高二07班\t0.575\n",
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"题号: 040012 , 字段: usages 中已添加数据: 202302217\t2024届高二07班\t0.075\n",
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"题号: 040013 , 字段: usages 中已添加数据: 202302217\t2024届高二07班\t0.625\n",
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"题号: 040014 , 字段: usages 中已添加数据: 202302217\t2024届高二07班\t0.425\n",
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"题号: 040015 , 字段: usages 中已添加数据: 202302217\t2024届高二07班\t0.662\t0.787\n",
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"题号: 040016 , 字段: usages 中已添加数据: 202302217\t2024届高二07班\t0.762\t0.438\n",
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"题号: 040017 , 字段: usages 中已添加数据: 202302217\t2024届高二07班\t0.762\t0.537\t0.388\n",
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"题号: 040001 , 字段: usages 中已添加数据: 202302217\t2024届高二08班\t0.652\n",
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"题号: 040002 , 字段: usages 中已添加数据: 202302217\t2024届高二08班\t1.000\n",
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"题号: 040003 , 字段: usages 中已添加数据: 202302217\t2024届高二08班\t0.609\n",
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"题号: 040004 , 字段: usages 中已添加数据: 202302217\t2024届高二08班\t0.304\n",
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"题号: 040005 , 字段: usages 中已添加数据: 202302217\t2024届高二08班\t0.870\n",
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"题号: 040006 , 字段: usages 中已添加数据: 202302217\t2024届高二08班\t0.391\n",
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"题号: 040007 , 字段: usages 中已添加数据: 202302217\t2024届高二08班\t0.783\n",
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"题号: 040008 , 字段: usages 中已添加数据: 202302217\t2024届高二08班\t0.957\n",
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"题号: 040009 , 字段: usages 中已添加数据: 202302217\t2024届高二08班\t0.957\n",
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"题号: 040011 , 字段: usages 中已添加数据: 202302217\t2024届高二08班\t0.913\n",
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"题号: 040012 , 字段: usages 中已添加数据: 202302217\t2024届高二08班\t0.087\n",
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"题号: 040013 , 字段: usages 中已添加数据: 202302217\t2024届高二08班\t0.696\n",
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"题号: 040014 , 字段: usages 中已添加数据: 202302217\t2024届高二08班\t0.522\n",
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"题号: 040015 , 字段: usages 中已添加数据: 202302217\t2024届高二08班\t0.848\t0.652\n",
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"题号: 040016 , 字段: usages 中已添加数据: 202302217\t2024届高二08班\t0.891\t0.348\n",
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"题号: 040017 , 字段: usages 中已添加数据: 202302217\t2024届高二08班\t0.957\t0.391\t0.435\n"
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"题号: 010000 , 字段: solution 中已添加数据: (1) $S_{\\triangle ABC}=\\dfrac 12 \\cdot 2\\times 2\\times \\sin 60^\\circ=\\sqrt{3}$, 三棱锥的高$PO=2\\sin 60^\\circ = \\sqrt{3}$, 故$V_{P-ABC}=\\dfrac 13\\cdot S_{\\triangle ABC}\\cdot PO=\\dfrac 13\\times \\sqrt{3}\\times \\sqrt{3}=1$.\\\\\n",
|
||||
"(2) 取$OC$中点$N$, 由于$MN\\parallel BO$, 且$BO\\perp AC$, 故$MN\\perp AC$. 又因为$PO\\perp$平面$ABC$, 故$MN\\perp PO$. 注意到$PO\\cap AC=O$, 于是$MN\\perp$平面$PAC$, 从而$PN$是$PM$在平面$PAC$上的射影. $MN=\\dfrac{\\sqrt{3}}{2}$, $PN=\\dfrac{\\sqrt{13}}{2}$, 故$\\angle MPN=\\arctan \\dfrac{MN}{PN}=\\arctan \\dfrac{\\sqrt{39}}{13}$. 综上, $PM$与平面$PAC$所成角的大小为$\\arctan \\dfrac{\\sqrt{39}}{13}$.\n",
|
||||
"题号: 011301 , 字段: solution 中已添加数据: (1) 由$a^2+c^2-2ac\\cos B=b^2$, 即$c^2-5c-24=0$解得$c=8$或$-3$(负数舍去). 故$c=8$.\\\\\n",
|
||||
"(2) 延长$CM$至$N$, 使得$CM=MN$, 则四边形$ACBN$是平行四边形, 故$S_{\\triangle ABC}=S_{\\triangle ACN}$. $\\triangle ACN$的三边长分别为$5,6,7$, 故$\\cos \\angle CAN=\\dfrac{5^2+7^2-6^2}{2\\times 5\\times 7}=\\dfrac{19}{35}$, 所以$\\sin \\angle CAN=\\dfrac{12\\sqrt{6}}{35}$. 由此, $S_{\\triangle ACN}=\\dfrac 12\\cdot AC\\cdot AN\\cdot \\sin \\angle CAN=6\\sqrt{6}$. 综上, 三角形$ABC$的面积为$6\\sqrt{6}$.\n",
|
||||
"题号: 003691 , 字段: solution 中已添加数据: (1) 前$4$个月的累计投放量为$a_1+a_2+a_3+a_4=20+95+420+430=965$. 前$4$个月的累计损失量为$b_1+b_2+b_3+b_4=6+7+8+9=30$, 因此该地区第$4$个月底的共享单车的保有量为$965-30=935$.\\\\\n",
|
||||
"(2) 考察不等式$a_n \\geq b_n$在整数范围内的解集. 当$n \\leq 3$时, $5 n^4+15 \\geq n+5$确实成立, 当$n \\geq 4$时, 由$-10 n+470 \\geq n+5$, 解得$n \\leq \\dfrac{465}{11}\\approx 42.27$, 因此第$42$个月底, 保有量达到最大. 当$n \\geq 4$时, $\\{a_n\\}$是公差为$-10$的等差数列, 而$\\{b_n\\}$是公差为$1$的等差数列, 于是到第$42$个月底, 单车保有量为$\\dfrac{38(a_5+a_{42})}{2}+965-\\dfrac{42(b_1+b_{42})}{2}=\\dfrac{38(420+50)}{2}+965-\\dfrac{42(6+47)}{2}=8782$. 而$S_{42}=-4\\times 16+8800=8736<8782$. 因此该月底单车保有量超过了容纳量.\n",
|
||||
"题号: 011450 , 字段: solution 中已添加数据: (1) 设直线$l$的方程为$x=m$, 由$S_{\\triangle OPQ}=\\dfrac 12\\cdot |m|\\cdot |PQ|=\\dfrac{\\sqrt{2}}{2}$得$|PQ|=\\dfrac{\\sqrt{2}}{|m|}$, 于是点$(m,\\dfrac{\\sqrt{2}}{2|m|})$在椭圆上. 代入椭圆$C$的方程, 得$m^2+\\dfrac{2}{8m^2}=1$, 解得$m^2=\\dfrac 12$. 故直线$l$的方程为$x=\\dfrac{\\sqrt{2}}2$或$x=-\\dfrac{\\sqrt{2}}2$.\\\\\n",
|
||||
"(2) 当直线$l$的斜率不存在时, 点$P,Q$关于$x$轴对称, 设$P(x_1,y_1)$, $Q(x_1,-y_1)$, 由$x_1^2+\\dfrac{y_1^2}{2}=1$及$|x_1y_1|=\\dfrac{\\sqrt{2}}{2}$解得$|x_1|=\\dfrac{\\sqrt{2}}2$, $|y_1|=1$, 故此时$x_1^2+x_2^2=1$, $y_1^2+y_2^2=2$.\\\\\n",
|
||||
"当直线$l$的斜率存在时, 设直线$l$的方程为$y=kx+m$, 与椭圆$C$的方程联立, 整理得$(2+k^2)x^2+2kmx+m^2-2=0$, 由$\\dfrac 12|x_1y_2-x_2y_1|=\\dfrac 12|x_1(kx_2+m)-x_2(kx_1+m)|=\\dfrac 12|m||x_1-x_2|=\\dfrac{\\sqrt{2}}{2}$得$\\sqrt{2}(2+k^2)=|m|\\sqrt{2+k^2-m^2}$. 于是$2m^2=2+k^2$, $x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=(\\dfrac{-2km}{2+k^2})^2-\\dfrac{2m^2-4}{2+k^2}=1$, $y_1^2+y_2^2=2(1-x_1^2+1-x_2^2)=2$.\\\\\n",
|
||||
"(3) 假设这样的三角形存在, 则三点$D(x_1,y_1)$, $E(x_2,y_2)$, $G(x_3,y_3)$应满足$x_1^2+x_2^2=x_2^2+x_3^2=x_3^2+x_1^2=1$, 从而$x_1^2=x_2^2=x_3^2=\\dfrac 12$, 但椭圆$C$上任取横坐标为$\\pm \\dfrac{\\sqrt{2}}{2}$的三点, 总有两个点与原点共线, 此时三角形不存在, 矛盾. 因此假设不成立, 满足条件的三角形不存在.\n",
|
||||
"题号: 031236 , 字段: solution 中已添加数据: (1) $f'(0)=\\mathrm{e}^0(\\dfrac{1}{1+0}+\\ln (1+0))=1$, 又$f(0)=0$, 故所求切线的方程为$y=x$.\\\\\n",
|
||||
"(2) $g(x)=f'(x)=\\mathrm{e}^x(\\dfrac{1}{1+x}+\\ln (1+x))$. $g'(x)=\\mathrm{e}^x(\\ln (1+x)+\\dfrac{2}{1+x}-\\dfrac{1}{(1+x)^2})$, 当$x>0$时, $\\mathrm{e}^x>1>0$, $\\ln(1+x)>0$, 且$\\dfrac{2}{1+x}-\\dfrac{1}{(1+x)^2}=\\dfrac{2x+1}{(1+x)^2}>0$, 所以$g'(x)$在$(0,+\\infty)$上恒正. 因此$g(x)$在$[0,+\\infty)$上是严格增函数.\\\\\n",
|
||||
"(3) 对于固定的常数$t$, 令$h(x)=f(x+t)-f(x)-f(t)$, 则$h(t)=0$, 而$h'(x)=f'(x+t)-f'(x)=g(x+t)-g(x)$. 由(2)可知$g(x+t)>g(x)$, 故$h'(x)$在$(0,+\\infty)$上恒正, 于是有$h(s)>h(0)=0$, 即$f(s+t)>f(s)+f(t)$.\n"
|
||||
]
|
||||
}
|
||||
],
|
||||
|
|
|
|||
|
|
@ -1,338 +1,26 @@
|
|||
usages
|
||||
solution
|
||||
|
||||
021492
|
||||
20230221 2025届高一11班 0.976
|
||||
|
||||
021493
|
||||
20230221 2025届高一11班 0.976
|
||||
010000
|
||||
(1) $S_{\triangle ABC}=\dfrac 12 \cdot 2\times 2\times \sin 60^\circ=\sqrt{3}$, 三棱锥的高$PO=2\sin 60^\circ = \sqrt{3}$, 故$V_{P-ABC}=\dfrac 13\cdot S_{\triangle ABC}\cdot PO=\dfrac 13\times \sqrt{3}\times \sqrt{3}=1$.\\
|
||||
(2) 取$OC$中点$N$, 由于$MN\parallel BO$, 且$BO\perp AC$, 故$MN\perp AC$. 又因为$PO\perp$平面$ABC$, 故$MN\perp PO$. 注意到$PO\cap AC=O$, 于是$MN\perp$平面$PAC$, 从而$PN$是$PM$在平面$PAC$上的射影. $MN=\dfrac{\sqrt{3}}{2}$, $PN=\dfrac{\sqrt{13}}{2}$, 故$\angle MPN=\arctan \dfrac{MN}{PN}=\arctan \dfrac{\sqrt{39}}{13}$. 综上, $PM$与平面$PAC$所成角的大小为$\arctan \dfrac{\sqrt{39}}{13}$.
|
||||
|
||||
021494
|
||||
20230221 2025届高一11班 1.000
|
||||
011301
|
||||
(1) 由$a^2+c^2-2ac\cos B=b^2$, 即$c^2-5c-24=0$解得$c=8$或$-3$(负数舍去). 故$c=8$.\\
|
||||
(2) 延长$CM$至$N$, 使得$CM=MN$, 则四边形$ACBN$是平行四边形, 故$S_{\triangle ABC}=S_{\triangle ACN}$. $\triangle ACN$的三边长分别为$5,6,7$, 故$\cos \angle CAN=\dfrac{5^2+7^2-6^2}{2\times 5\times 7}=\dfrac{19}{35}$, 所以$\sin \angle CAN=\dfrac{12\sqrt{6}}{35}$. 由此, $S_{\triangle ACN}=\dfrac 12\cdot AC\cdot AN\cdot \sin \angle CAN=6\sqrt{6}$. 综上, 三角形$ABC$的面积为$6\sqrt{6}$.
|
||||
|
||||
021495
|
||||
20230221 2025届高一11班 0.762
|
||||
003691
|
||||
(1) 前$4$个月的累计投放量为$a_1+a_2+a_3+a_4=20+95+420+430=965$. 前$4$个月的累计损失量为$b_1+b_2+b_3+b_4=6+7+8+9=30$, 因此该地区第$4$个月底的共享单车的保有量为$965-30=935$.\\
|
||||
(2) 考察不等式$a_n \geq b_n$在整数范围内的解集. 当$n \leq 3$时, $5 n^4+15 \geq n+5$确实成立, 当$n \geq 4$时, 由$-10 n+470 \geq n+5$, 解得$n \leq \dfrac{465}{11}\approx 42.27$, 因此第$42$个月底, 保有量达到最大. 当$n \geq 4$时, $\{a_n\}$是公差为$-10$的等差数列, 而$\{b_n\}$是公差为$1$的等差数列, 于是到第$42$个月底, 单车保有量为$\dfrac{38(a_5+a_{42})}{2}+965-\dfrac{42(b_1+b_{42})}{2}=\dfrac{38(420+50)}{2}+965-\dfrac{42(6+47)}{2}=8782$. 而$S_{42}=-4\times 16+8800=8736<8782$. 因此该月底单车保有量超过了容纳量.
|
||||
|
||||
021496
|
||||
20230221 2025届高一11班 0.952
|
||||
|
||||
021497
|
||||
20230221 2025届高一11班 0.976 0.762 0.905
|
||||
|
||||
021498
|
||||
20230221 2025届高一11班 0.964
|
||||
|
||||
021499
|
||||
20230221 2025届高一11班 0.952
|
||||
|
||||
021500
|
||||
20230221 2025届高一11班 0.988
|
||||
|
||||
021501
|
||||
20230221 2025届高一11班 1.000
|
||||
|
||||
021502
|
||||
20230221 2025届高一11班 0.762
|
||||
|
||||
021503
|
||||
20230221 2025届高一11班 0.929
|
||||
|
||||
021504
|
||||
20230221 2025届高一11班 0.869 0.869
|
||||
|
||||
021505
|
||||
20230221 2025届高一11班 0.643
|
||||
|
||||
021506
|
||||
20230221 2025届高一11班 0.857
|
||||
|
||||
021507
|
||||
20230221 2025届高一11班 0.702
|
||||
|
||||
021492
|
||||
20230221 2025届高一01班 0.951
|
||||
|
||||
021493
|
||||
20230221 2025届高一01班 0.829
|
||||
|
||||
021494
|
||||
20230221 2025届高一01班 0.732
|
||||
|
||||
021495
|
||||
20230221 2025届高一01班 0.610
|
||||
|
||||
021496
|
||||
20230221 2025届高一01班 0.829
|
||||
|
||||
021497
|
||||
20230221 2025届高一01班 0.829 0.732 0.707
|
||||
|
||||
021498
|
||||
20230221 2025届高一01班 0.902
|
||||
|
||||
021499
|
||||
20230221 2025届高一01班 0.842
|
||||
|
||||
021500
|
||||
20230221 2025届高一01班 0.842
|
||||
|
||||
021501
|
||||
20230221 2025届高一01班 0.890
|
||||
|
||||
021502
|
||||
20230221 2025届高一01班 0.756
|
||||
|
||||
021503
|
||||
20230221 2025届高一01班 0.780
|
||||
|
||||
021504
|
||||
20230221 2025届高一01班 0.695 0.671
|
||||
|
||||
021505
|
||||
20230221 2025届高一01班 0.658
|
||||
|
||||
021506
|
||||
20230221 2025届高一01班 0.707
|
||||
|
||||
021507
|
||||
20230221 2025届高一01班 0.732
|
||||
|
||||
021492
|
||||
20230221 2025届高一12班 1.000
|
||||
|
||||
021493
|
||||
20230221 2025届高一12班 0.905
|
||||
|
||||
021494
|
||||
20230221 2025届高一12班 0.905
|
||||
|
||||
021495
|
||||
20230221 2025届高一12班 0.571
|
||||
|
||||
021496
|
||||
20230221 2025届高一12班 0.952
|
||||
|
||||
021497
|
||||
20230221 2025届高一12班 1.000 0.833 0.952
|
||||
|
||||
021498
|
||||
20230221 2025届高一12班 0.964
|
||||
|
||||
021499
|
||||
20230221 2025届高一12班 0.976
|
||||
|
||||
021500
|
||||
20230221 2025届高一12班 1.000
|
||||
|
||||
021501
|
||||
20230221 2025届高一12班 0.976
|
||||
|
||||
021502
|
||||
20230221 2025届高一12班 0.845
|
||||
|
||||
021503
|
||||
20230221 2025届高一12班 0.881
|
||||
|
||||
021504
|
||||
20230221 2025届高一12班 0.774 0.905
|
||||
|
||||
021505
|
||||
20230221 2025届高一12班 0.774
|
||||
|
||||
021506
|
||||
20230221 2025届高一12班 0.702
|
||||
|
||||
021507
|
||||
20230221 2025届高一12班 0.679
|
||||
|
||||
021492
|
||||
20230221 2025届高一02班 0.943
|
||||
|
||||
021493
|
||||
20230221 2025届高一02班 0.886
|
||||
|
||||
021494
|
||||
20230221 2025届高一02班 0.857
|
||||
|
||||
021495
|
||||
20230221 2025届高一02班 0.571
|
||||
|
||||
021496
|
||||
20230221 2025届高一02班 0.971
|
||||
|
||||
021497
|
||||
20230221 2025届高一02班 0.971 0.857 0.971
|
||||
|
||||
021498
|
||||
20230221 2025届高一02班 0.929
|
||||
|
||||
021499
|
||||
20230221 2025届高一02班 0.871
|
||||
|
||||
021500
|
||||
20230221 2025届高一02班 0.943
|
||||
|
||||
021501
|
||||
20230221 2025届高一02班 1.000
|
||||
|
||||
021502
|
||||
20230221 2025届高一02班 0.743
|
||||
|
||||
021503
|
||||
20230221 2025届高一02班 0.886
|
||||
|
||||
021504
|
||||
20230221 2025届高一02班 0.700 0.729
|
||||
|
||||
021505
|
||||
20230221 2025届高一02班 0.729
|
||||
|
||||
021506
|
||||
20230221 2025届高一02班 0.614
|
||||
|
||||
021507
|
||||
20230221 2025届高一02班 0.686
|
||||
|
||||
021492
|
||||
20230221 2025届高一06班 0.923
|
||||
|
||||
021493
|
||||
20230221 2025届高一06班 0.897
|
||||
|
||||
021494
|
||||
20230221 2025届高一06班 0.846
|
||||
|
||||
021495
|
||||
20230221 2025届高一06班 0.385
|
||||
|
||||
021496
|
||||
20230221 2025届高一06班 0.923
|
||||
|
||||
021497
|
||||
20230221 2025届高一06班 0.949 0.769 0.821
|
||||
|
||||
021498
|
||||
20230221 2025届高一06班 0.846
|
||||
|
||||
021499
|
||||
20230221 2025届高一06班 0.808
|
||||
|
||||
021500
|
||||
20230221 2025届高一06班 0.974
|
||||
|
||||
021501
|
||||
20230221 2025届高一06班 0.962
|
||||
|
||||
021502
|
||||
20230221 2025届高一06班 0.897
|
||||
|
||||
021503
|
||||
20230221 2025届高一06班 0.718
|
||||
|
||||
021504
|
||||
20230221 2025届高一06班 0.500 0.667
|
||||
|
||||
021505
|
||||
20230221 2025届高一06班 0.564
|
||||
|
||||
021506
|
||||
20230221 2025届高一06班 0.718
|
||||
|
||||
021507
|
||||
20230221 2025届高一06班 0.500
|
||||
|
||||
021492
|
||||
20230221 2025届高一08班 0.946
|
||||
|
||||
021493
|
||||
20230221 2025届高一08班 0.865
|
||||
|
||||
021494
|
||||
20230221 2025届高一08班 0.973
|
||||
|
||||
021495
|
||||
20230221 2025届高一08班 0.460
|
||||
|
||||
021496
|
||||
20230221 2025届高一08班 0.784
|
||||
|
||||
021497
|
||||
20230221 2025届高一08班 0.946 0.811 0.811
|
||||
|
||||
021498
|
||||
20230221 2025届高一08班 0.838
|
||||
|
||||
021499
|
||||
20230221 2025届高一08班 0.878
|
||||
|
||||
021500
|
||||
20230221 2025届高一08班 0.838
|
||||
|
||||
021501
|
||||
20230221 2025届高一08班 0.946
|
||||
|
||||
021502
|
||||
20230221 2025届高一08班 0.703
|
||||
|
||||
021503
|
||||
20230221 2025届高一08班 0.838
|
||||
|
||||
021504
|
||||
20230221 2025届高一08班 0.797 0.635
|
||||
|
||||
021505
|
||||
20230221 2025届高一08班 0.486
|
||||
|
||||
021506
|
||||
20230221 2025届高一08班 0.676
|
||||
|
||||
021507
|
||||
20230221 2025届高一08班 0.689
|
||||
|
||||
021492
|
||||
20230221 2025届高一09班 0.952
|
||||
|
||||
021493
|
||||
20230221 2025届高一09班 0.929
|
||||
|
||||
021494
|
||||
20230221 2025届高一09班 0.952
|
||||
|
||||
021495
|
||||
20230221 2025届高一09班 0.691
|
||||
|
||||
021496
|
||||
20230221 2025届高一09班 0.905
|
||||
|
||||
021497
|
||||
20230221 2025届高一09班 0.976 0.881 0.857
|
||||
|
||||
021498
|
||||
20230221 2025届高一09班 0.869
|
||||
|
||||
021499
|
||||
20230221 2025届高一09班 0.786
|
||||
|
||||
021500
|
||||
20230221 2025届高一09班 0.941
|
||||
|
||||
021501
|
||||
20230221 2025届高一09班 0.941
|
||||
|
||||
021502
|
||||
20230221 2025届高一09班 0.833
|
||||
|
||||
021503
|
||||
20230221 2025届高一09班 0.905
|
||||
|
||||
021504
|
||||
20230221 2025届高一09班 0.667 0.571
|
||||
|
||||
021505
|
||||
20230221 2025届高一09班 0.559
|
||||
|
||||
021506
|
||||
20230221 2025届高一09班 0.738
|
||||
|
||||
021507
|
||||
20230221 2025届高一09班 0.702
|
||||
011450
|
||||
(1) 设直线$l$的方程为$x=m$, 由$S_{\triangle OPQ}=\dfrac 12\cdot |m|\cdot |PQ|=\dfrac{\sqrt{2}}{2}$得$|PQ|=\dfrac{\sqrt{2}}{|m|}$, 于是点$(m,\dfrac{\sqrt{2}}{2|m|})$在椭圆上. 代入椭圆$C$的方程, 得$m^2+\dfrac{2}{8m^2}=1$, 解得$m^2=\dfrac 12$. 故直线$l$的方程为$x=\dfrac{\sqrt{2}}2$或$x=-\dfrac{\sqrt{2}}2$.\\
|
||||
(2) 当直线$l$的斜率不存在时, 点$P,Q$关于$x$轴对称, 设$P(x_1,y_1)$, $Q(x_1,-y_1)$, 由$x_1^2+\dfrac{y_1^2}{2}=1$及$|x_1y_1|=\dfrac{\sqrt{2}}{2}$解得$|x_1|=\dfrac{\sqrt{2}}2$, $|y_1|=1$, 故此时$x_1^2+x_2^2=1$, $y_1^2+y_2^2=2$.\\
|
||||
当直线$l$的斜率存在时, 设直线$l$的方程为$y=kx+m$, 与椭圆$C$的方程联立, 整理得$(2+k^2)x^2+2kmx+m^2-2=0$, 由$\dfrac 12|x_1y_2-x_2y_1|=\dfrac 12|x_1(kx_2+m)-x_2(kx_1+m)|=\dfrac 12|m||x_1-x_2|=\dfrac{\sqrt{2}}{2}$得$\sqrt{2}(2+k^2)=|m|\sqrt{2+k^2-m^2}$. 于是$2m^2=2+k^2$, $x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=(\dfrac{-2km}{2+k^2})^2-\dfrac{2m^2-4}{2+k^2}=1$, $y_1^2+y_2^2=2(1-x_1^2+1-x_2^2)=2$.\\
|
||||
(3) 假设这样的三角形存在, 则三点$D(x_1,y_1)$, $E(x_2,y_2)$, $G(x_3,y_3)$应满足$x_1^2+x_2^2=x_2^2+x_3^2=x_3^2+x_1^2=1$, 从而$x_1^2=x_2^2=x_3^2=\dfrac 12$, 但椭圆$C$上任取横坐标为$\pm \dfrac{\sqrt{2}}{2}$的三点, 总有两个点与原点共线, 此时三角形不存在, 矛盾. 因此假设不成立, 满足条件的三角形不存在.
|
||||
|
||||
031236
|
||||
(1) $f'(0)=\mathrm{e}^0(\dfrac{1}{1+0}+\ln (1+0))=1$, 又$f(0)=0$, 故所求切线的方程为$y=x$.\\
|
||||
(2) $g(x)=f'(x)=\mathrm{e}^x(\dfrac{1}{1+x}+\ln (1+x))$. $g'(x)=\mathrm{e}^x(\ln (1+x)+\dfrac{2}{1+x}-\dfrac{1}{(1+x)^2})$, 当$x>0$时, $\mathrm{e}^x>1>0$, $\ln(1+x)>0$, 且$\dfrac{2}{1+x}-\dfrac{1}{(1+x)^2}=\dfrac{2x+1}{(1+x)^2}>0$, 所以$g'(x)$在$(0,+\infty)$上恒正. 因此$g(x)$在$[0,+\infty)$上是严格增函数.\\
|
||||
(3) 对于固定的常数$t$, 令$h(x)=f(x+t)-f(x)-f(t)$, 则$h(t)=0$, 而$h'(x)=f'(x+t)-f'(x)=g(x+t)-g(x)$. 由(2)可知$g(x+t)>g(x)$, 故$h'(x)$在$(0,+\infty)$上恒正, 于是有$h(s)>h(0)=0$, 即$f(s+t)>f(s)+f(t)$.
|
||||
|
|
@ -1 +1 @@
|
|||
021441,021442,021443,021444,021445,021446,021447,021448,021449,021450,021451,021452,021453,021454,021455,021456,021457,021458,021459,021460,021461,021462,021463,021464,021465,021466,021467,021468,021469,021470,021471,021472,021473,021474,021475,021476,021477,021478,021479,021480,021481,021482,021483,021484,021485,021486,021487,021488,021489,021490,021491,021492,021493,021494,021495,021496,021497,021498,021499,021500,021501,021502,021503,021504,021505,021506,021507
|
||||
003675,012165,004252,031239,031238,031266,031240,031237,004256,004499,004132,004217,008972,009082,012445,030691,010000,011301,003691,011450,031236
|
||||
|
|
@ -2,7 +2,7 @@
|
|||
"cells": [
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 1,
|
||||
"execution_count": 2,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
|
|
@ -13,11 +13,9 @@
|
|||
"题块 1 处理完毕.\n",
|
||||
"正在处理题块 2 .\n",
|
||||
"题块 2 处理完毕.\n",
|
||||
"正在处理题块 3 .\n",
|
||||
"题块 3 处理完毕.\n",
|
||||
"开始编译教师版本pdf文件: 临时文件/高三下学期月考01_教师_20230224.tex\n",
|
||||
"开始编译教师版本pdf文件: 临时文件/15_数列综合_教师_20230227.tex\n",
|
||||
"0\n",
|
||||
"开始编译学生版本pdf文件: 临时文件/高三下学期月考01_学生_20230224.tex\n",
|
||||
"开始编译学生版本pdf文件: 临时文件/15_数列综合_学生_20230227.tex\n",
|
||||
"0\n"
|
||||
]
|
||||
}
|
||||
|
|
@ -30,12 +28,12 @@
|
|||
"\"\"\"2: 测验卷与周末卷(填空题, 选择题, 解答题)\"\"\"\n",
|
||||
"\"\"\"3: 日常选题讲义(一个section)\"\"\"\n",
|
||||
"\n",
|
||||
"paper_type = 2 # 随后设置一下后续的讲义标题\n",
|
||||
"paper_type = 1 # 随后设置一下后续的讲义标题\n",
|
||||
"\n",
|
||||
"\"\"\"---设置题块编号---\"\"\"\n",
|
||||
"\n",
|
||||
"problems = [\n",
|
||||
"\"3675,12165,4252,31239,31238,31266,31240,31237,4256,4499,4132,4217\",\"8972,9082,12445,30691\",\"10000,11301,3691,11450,31236\"\n",
|
||||
"\"012942,012941,012976,012923,013002,014550,012966,013005,013935,012978,012932,012947,012971,013960\",\"013000,012956,014553,012959,012977,012982,013955,013944,012936,014560,013006,012983,012938\"\n",
|
||||
"]\n",
|
||||
"\n",
|
||||
"\"\"\"---设置结束---\"\"\"\n",
|
||||
|
|
@ -44,7 +42,7 @@
|
|||
"if paper_type == 1:\n",
|
||||
" enumi_mode = 0 #设置模式(1为整卷统一编号, 0为每一部分从1开始编号)\n",
|
||||
" template_file = \"模板文件/复习讲义模板.txt\" #设置模板文件名\n",
|
||||
" exec_list = [(\"标题数字待处理\",\"10\"),(\"标题文字待处理\",\"空间向量与应用\")] #设置讲义标题\n",
|
||||
" exec_list = [(\"标题数字待处理\",\"15\"),(\"标题文字待处理\",\"数列综合\")] #设置讲义标题\n",
|
||||
" destination_file = \"临时文件/\"+exec_list[0][1]+\"_\"+exec_list[1][1] # 设置输出文件名\n",
|
||||
"elif paper_type == 2:\n",
|
||||
" enumi_mode = 1 #设置模式(1为整卷统一编号, 0为每一部分从1开始编号)\n",
|
||||
|
|
@ -204,7 +202,7 @@
|
|||
],
|
||||
"metadata": {
|
||||
"kernelspec": {
|
||||
"display_name": "mathdept",
|
||||
"display_name": "pythontest",
|
||||
"language": "python",
|
||||
"name": "python3"
|
||||
},
|
||||
|
|
@ -223,7 +221,7 @@
|
|||
"orig_nbformat": 4,
|
||||
"vscode": {
|
||||
"interpreter": {
|
||||
"hash": "ff3c292c316ba85de6f1ad75f19c731e79d694e741b6f515ec18f14996fe48dc"
|
||||
"hash": "91219a98e0e9be72efb992f647fe78b593124968b75db0b865552d6787c8db93"
|
||||
}
|
||||
}
|
||||
},
|
||||
|
|
|
|||
|
|
@ -2,16 +2,16 @@
|
|||
"cells": [
|
||||
{
|
||||
"cell_type": "code",
|
||||
"execution_count": 2,
|
||||
"execution_count": 7,
|
||||
"metadata": {},
|
||||
"outputs": [
|
||||
{
|
||||
"name": "stdout",
|
||||
"output_type": "stream",
|
||||
"text": [
|
||||
"开始编译教师版本pdf文件: 临时文件/2025届高一2月作业情况_教师用_20230226.tex\n",
|
||||
"开始编译教师版本pdf文件: 临时文件/月考解答_教师用_20230227.tex\n",
|
||||
"0\n",
|
||||
"开始编译学生版本pdf文件: 临时文件/2025届高一2月作业情况_学生用_20230226.tex\n",
|
||||
"开始编译学生版本pdf文件: 临时文件/月考解答_学生用_20230227.tex\n",
|
||||
"0\n"
|
||||
]
|
||||
}
|
||||
|
|
@ -26,7 +26,7 @@
|
|||
"\"\"\"---设置题目列表---\"\"\"\n",
|
||||
"#留空为编译全题库, a为读取临时文件中的题号筛选.txt文件生成题库\n",
|
||||
"problems = r\"\"\"\n",
|
||||
"a\n",
|
||||
"003675,012165,004252,031239,031238,031266,031240,031237,004256,004499,004132,004217,008972,009082,012445,030691,010000,011301,003691,011450,031236\n",
|
||||
"\n",
|
||||
"\n",
|
||||
"\"\"\"\n",
|
||||
|
|
@ -34,7 +34,7 @@
|
|||
"\n",
|
||||
"\"\"\"---设置文件名---\"\"\"\n",
|
||||
"#目录和文件的分隔务必用/\n",
|
||||
"filename = \"临时文件/2025届高一2月作业情况\"\n",
|
||||
"filename = \"临时文件/月考解答\"\n",
|
||||
"\"\"\"---设置文件名结束---\"\"\"\n",
|
||||
"\n",
|
||||
"\n",
|
||||
|
|
@ -189,12 +189,12 @@
|
|||
"name": "python",
|
||||
"nbconvert_exporter": "python",
|
||||
"pygments_lexer": "ipython3",
|
||||
"version": "3.9.15"
|
||||
"version": "3.8.15"
|
||||
},
|
||||
"orig_nbformat": 4,
|
||||
"vscode": {
|
||||
"interpreter": {
|
||||
"hash": "ff3c292c316ba85de6f1ad75f19c731e79d694e741b6f515ec18f14996fe48dc"
|
||||
"hash": "42dd566da87765ddbe9b5c5b483063747fec4aacc5469ad554706e4b742e67b2"
|
||||
}
|
||||
}
|
||||
},
|
||||
|
|
|
|||
|
|
@ -99392,7 +99392,7 @@
|
|||
],
|
||||
"genre": "填空题",
|
||||
"ans": "$(-\\infty,0)$",
|
||||
"solution": "",
|
||||
"solution": "即$\\dfrac 1x<0$",
|
||||
"duration": -1,
|
||||
"usages": [],
|
||||
"origin": "上海2017年秋季高考试题3",
|
||||
|
|
@ -99818,7 +99818,7 @@
|
|||
],
|
||||
"genre": "解答题",
|
||||
"ans": "(1) $935$辆; (2) 因$8782>8736$, 故第$42$个月底单车保有量超过了容纳量",
|
||||
"solution": "",
|
||||
"solution": "(1) 前$4$个月的累计投放量为$a_1+a_2+a_3+a_4=20+95+420+430=965$. 前$4$个月的累计损失量为$b_1+b_2+b_3+b_4=6+7+8+9=30$, 因此该地区第$4$个月底的共享单车的保有量为$965-30=935$.\\\\\n(2) 考察不等式$a_n \\geq b_n$在整数范围内的解集. 当$n \\leq 3$时, $5 n^4+15 \\geq n+5$确实成立, 当$n \\geq 4$时, 由$-10 n+470 \\geq n+5$, 解得$n \\leq \\dfrac{465}{11}\\approx 42.27$, 因此第$42$个月底, 保有量达到最大. 当$n \\geq 4$时, $\\{a_n\\}$是公差为$-10$的等差数列, 而$\\{b_n\\}$是公差为$1$的等差数列, 于是到第$42$个月底, 单车保有量为$\\dfrac{38(a_5+a_{42})}{2}+965-\\dfrac{42(b_1+b_{42})}{2}=\\dfrac{38(420+50)}{2}+965-\\dfrac{42(6+47)}{2}=8782$. 而$S_{42}=-4\\times 16+8800=8736<8782$. 因此该月底单车保有量超过了容纳量.",
|
||||
"duration": -1,
|
||||
"usages": [],
|
||||
"origin": "上海2017年秋季高考试题19",
|
||||
|
|
@ -110677,7 +110677,7 @@
|
|||
],
|
||||
"genre": "填空题",
|
||||
"ans": "$[-3,1]$",
|
||||
"solution": "",
|
||||
"solution": "设$MN$的中点为$T$, 则$|OT|=1$且$T$能取遍单位圆上的每一个点. 而$\\overrightarrow{AM}\\cdot \\overrightarrow{AN}=(\\overrightarrow{AT}+\\overrightarrow{TM})\\cdot (\\overrightarrow{AT}-\\overrightarrow{TM})=\\overrightarrow{AT}^2-3$, 而$|AT|$的范围为$[0,2]$, 所以$\\overrightarrow{AM}\\cdot \\overrightarrow{AN}$的取值范围为$[-3,1]$",
|
||||
"duration": -1,
|
||||
"usages": [
|
||||
"20220331\t2022届高三01班\t0.744"
|
||||
|
|
@ -112937,7 +112937,7 @@
|
|||
],
|
||||
"genre": "填空题",
|
||||
"ans": "$36$",
|
||||
"solution": "",
|
||||
"solution": "$f(2020)=f(2^{10}\\cdot \\dfrac{2020}{1024})=2^{10}(2-\\dfrac{2020}{1024})=28$. 而当$x\\in (2^i,2^{i+1}]$($i\\in \\mathbf{Z}$)时, $f(x)$的取值范围为$[0,2^i)$. 所以$x>32$, 设$x\\in (32,64]$, 则$f(x)=32f(\\dfrac{x}{32})=32\\cdot (2-\\dfrac{x}{32})=64-x$, 解$64-x=28$得$x=36$",
|
||||
"duration": -1,
|
||||
"usages": [
|
||||
"20220505\t2022届高三01班\t0.907"
|
||||
|
|
@ -113901,7 +113901,7 @@
|
|||
],
|
||||
"genre": "填空题",
|
||||
"ans": "$3$",
|
||||
"solution": "",
|
||||
"solution": "圆心坐标为$(1,2)$,. 由点到直线的距离公式得$d=\\dfrac{|3\\cdot 1+4\\cdot 2+4|}{\\sqrt{3^2+4^2}}=\\dfrac{15}{5}=3$",
|
||||
"duration": -1,
|
||||
"usages": [
|
||||
"20220517\t2022届高三01班\t0.977",
|
||||
|
|
@ -114013,7 +114013,7 @@
|
|||
],
|
||||
"genre": "填空题",
|
||||
"ans": "$(-\\infty,-1]\\cup [0,+\\infty)$",
|
||||
"solution": "",
|
||||
"solution": "该函数在$x>0$时的取值范围为$(b,1+b)$, 所以该函数存在反函数的一个充要条件为$(-b-1,-b)$, $\\{0\\}$, $(b,b+1)$两两的交集为$\\varnothing$, 解得$b\\le -1$或$b\\ge 0$",
|
||||
"duration": -1,
|
||||
"usages": [
|
||||
"20220517\t2022届高三01班\t0.698"
|
||||
|
|
@ -120893,8 +120893,8 @@
|
|||
"第八单元"
|
||||
],
|
||||
"genre": "填空题",
|
||||
"ans": "$0217$",
|
||||
"solution": "",
|
||||
"ans": "$9217$",
|
||||
"solution": "最大数为$1$的集合有$2^0$个, 最大数为$2$的集合有$2^1$个, 以此类推. 所以$S_{10}=\\displaystyle\\sum_{i=1}^{10}(i\\cdot 2^{i-1})=9217$",
|
||||
"duration": -1,
|
||||
"usages": [
|
||||
"20211123\t2022届高三01班\t0.810"
|
||||
|
|
@ -226727,7 +226727,7 @@
|
|||
],
|
||||
"genre": "选择题",
|
||||
"ans": "B",
|
||||
"solution": "",
|
||||
"solution": "根据复数相等的定义, $z_1=z_2$时必有$a=c$, 反之如$z_1=1$, $z_2=1+\\mathrm{i}$, 则有$a=c$且$z_1\\ne z_2$",
|
||||
"duration": -1,
|
||||
"usages": [],
|
||||
"origin": "二期课改练习册高二第二学期",
|
||||
|
|
@ -229135,7 +229135,7 @@
|
|||
],
|
||||
"genre": "选择题",
|
||||
"ans": "D",
|
||||
"solution": "",
|
||||
"solution": "如图, 点$(x,y)$的轨迹为图中的圆, 圆上动点与原点连线斜率的最大值是如图切线的斜率\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw [->] (-2,0) -- (4,0) node [below] {$x$};\n\\draw [->] (0,-2) -- (0,2) node [left] {$y$};\n\\draw (0,0) node [below left] {$O$};\n\\draw (2,0) circle ({sqrt(3)});\n\\draw (0,0) --++ (60:2);\n\\filldraw (2,0) node [below] {$C$} coordinate (C) circle (0.03);\n\\end{tikzpicture}\n\\end{center}",
|
||||
"duration": -1,
|
||||
"usages": [],
|
||||
"origin": "二期课改练习册高二第二学期",
|
||||
|
|
@ -250544,7 +250544,7 @@
|
|||
],
|
||||
"genre": "解答题",
|
||||
"ans": "(1) $1$; (2) $\\arctan\\dfrac{\\sqrt{39}}{13}$",
|
||||
"solution": "",
|
||||
"solution": "(1) $S_{\\triangle ABC}=\\dfrac 12 \\cdot 2\\times 2\\times \\sin 60^\\circ=\\sqrt{3}$, 三棱锥的高$PO=2\\sin 60^\\circ = \\sqrt{3}$, 故$V_{P-ABC}=\\dfrac 13\\cdot S_{\\triangle ABC}\\cdot PO=\\dfrac 13\\times \\sqrt{3}\\times \\sqrt{3}=1$.\\\\\n(2) 取$OC$中点$N$, 由于$MN\\parallel BO$, 且$BO\\perp AC$, 故$MN\\perp AC$. 又因为$PO\\perp$平面$ABC$, 故$MN\\perp PO$. 注意到$PO\\cap AC=O$, 于是$MN\\perp$平面$PAC$, 从而$PN$是$PM$在平面$PAC$上的射影. $MN=\\dfrac{\\sqrt{3}}{2}$, $PN=\\dfrac{\\sqrt{13}}{2}$, 故$\\angle MPN=\\arctan \\dfrac{MN}{PN}=\\arctan \\dfrac{\\sqrt{39}}{13}$. 综上, $PM$与平面$PAC$所成角的大小为$\\arctan \\dfrac{\\sqrt{39}}{13}$.",
|
||||
"duration": -1,
|
||||
"usages": [
|
||||
"20221101\t2023届高三10班\t0.886\t0.686",
|
||||
|
|
@ -282133,7 +282133,7 @@
|
|||
],
|
||||
"genre": "解答题",
|
||||
"ans": "(1) $c=8$; (2) $6\\sqrt{6}$",
|
||||
"solution": "",
|
||||
"solution": "(1) 由$a^2+c^2-2ac\\cos B=b^2$, 即$c^2-5c-24=0$解得$c=8$或$-3$(负数舍去). 故$c=8$.\\\\\n(2) 延长$CM$至$N$, 使得$CM=MN$, 则四边形$ACBN$是平行四边形, 故$S_{\\triangle ABC}=S_{\\triangle ACN}$. $\\triangle ACN$的三边长分别为$5,6,7$, 故$\\cos \\angle CAN=\\dfrac{5^2+7^2-6^2}{2\\times 5\\times 7}=\\dfrac{19}{35}$, 所以$\\sin \\angle CAN=\\dfrac{12\\sqrt{6}}{35}$. 由此, $S_{\\triangle ACN}=\\dfrac 12\\cdot AC\\cdot AN\\cdot \\sin \\angle CAN=6\\sqrt{6}$. 综上, 三角形$ABC$的面积为$6\\sqrt{6}$.",
|
||||
"duration": -1,
|
||||
"usages": [],
|
||||
"origin": "2022届高三下学期周末卷3试题18",
|
||||
|
|
@ -285466,7 +285466,7 @@
|
|||
],
|
||||
"genre": "解答题",
|
||||
"ans": "(1) $x=\\dfrac{\\sqrt{2}}2$或$x=-\\dfrac{\\sqrt{2}}2$; (2) $x_1^2+x_2^2=1$, $y_1^2+y_2^2=2$; (3) 不存在, 证明略.",
|
||||
"solution": "",
|
||||
"solution": "(1) 设直线$l$的方程为$x=m$, 由$S_{\\triangle OPQ}=\\dfrac 12\\cdot |m|\\cdot |PQ|=\\dfrac{\\sqrt{2}}{2}$得$|PQ|=\\dfrac{\\sqrt{2}}{|m|}$, 于是点$(m,\\dfrac{\\sqrt{2}}{2|m|})$在椭圆上. 代入椭圆$C$的方程, 得$m^2+\\dfrac{2}{8m^2}=1$, 解得$m^2=\\dfrac 12$. 故直线$l$的方程为$x=\\dfrac{\\sqrt{2}}2$或$x=-\\dfrac{\\sqrt{2}}2$.\\\\\n(2) 当直线$l$的斜率不存在时, 点$P,Q$关于$x$轴对称, 设$P(x_1,y_1)$, $Q(x_1,-y_1)$, 由$x_1^2+\\dfrac{y_1^2}{2}=1$及$|x_1y_1|=\\dfrac{\\sqrt{2}}{2}$解得$|x_1|=\\dfrac{\\sqrt{2}}2$, $|y_1|=1$, 故此时$x_1^2+x_2^2=1$, $y_1^2+y_2^2=2$.\\\\\n当直线$l$的斜率存在时, 设直线$l$的方程为$y=kx+m$, 与椭圆$C$的方程联立, 整理得$(2+k^2)x^2+2kmx+m^2-2=0$, 由$\\dfrac 12|x_1y_2-x_2y_1|=\\dfrac 12|x_1(kx_2+m)-x_2(kx_1+m)|=\\dfrac 12|m||x_1-x_2|=\\dfrac{\\sqrt{2}}{2}$得$\\sqrt{2}(2+k^2)=|m|\\sqrt{2+k^2-m^2}$. 于是$2m^2=2+k^2$, $x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=(\\dfrac{-2km}{2+k^2})^2-\\dfrac{2m^2-4}{2+k^2}=1$, $y_1^2+y_2^2=2(1-x_1^2+1-x_2^2)=2$.\\\\\n(3) 假设这样的三角形存在, 则三点$D(x_1,y_1)$, $E(x_2,y_2)$, $G(x_3,y_3)$应满足$x_1^2+x_2^2=x_2^2+x_3^2=x_3^2+x_1^2=1$, 从而$x_1^2=x_2^2=x_3^2=\\dfrac 12$, 但椭圆$C$上任取横坐标为$\\pm \\dfrac{\\sqrt{2}}{2}$的三点, 总有两个点与原点共线, 此时三角形不存在, 矛盾. 因此假设不成立, 满足条件的三角形不存在.",
|
||||
"duration": -1,
|
||||
"usages": [],
|
||||
"origin": "2022届高三下学期周末卷10试题20",
|
||||
|
|
@ -302089,7 +302089,7 @@
|
|||
],
|
||||
"genre": "填空题",
|
||||
"ans": "$4$",
|
||||
"solution": "",
|
||||
"solution": "解$1\\cdot a-2\\cdot 3=0$得$a=6$, 此时方程组确实无解",
|
||||
"duration": -1,
|
||||
"usages": [
|
||||
"20230103\t2023届高三10班\t1.000",
|
||||
|
|
@ -308885,7 +308885,7 @@
|
|||
],
|
||||
"genre": "选择题",
|
||||
"ans": "B",
|
||||
"solution": "",
|
||||
"solution": "要求$a^2-(k+2)ab+b^2\\ge 0$恒成立, 即要求$(k+2)^2b^2-4b^2\\le 0$对一切$b\\in \\mathbf{R}$恒成立, 即要求$(k^2+4k)\\le 0$, 解得$k$的范围为$[-4,0]$",
|
||||
"duration": -1,
|
||||
"usages": [],
|
||||
"origin": "2015届春季高考附加卷试题2",
|
||||
|
|
@ -322188,7 +322188,7 @@
|
|||
},
|
||||
"013040": {
|
||||
"id": "013040",
|
||||
"content": "已知函数$y=f(x)$的周期为$2 \\pi$, 当$x \\in[0,2 \\pi)$时, $f(x)=\\sin \\dfrac{x}{2}$, 那么方程$f(x)=\\dfrac{1}{2}$的解集是\\blank{50}.",
|
||||
"content": "已知定义在$\\mathbf{R}$上的函数$y=f(x)$的周期为$2 \\pi$, 当$x \\in[0,2 \\pi)$时, $f(x)=\\sin \\dfrac{x}{2}$, 那么方程$f(x)=\\dfrac{1}{2}$的解集是\\blank{50}.",
|
||||
"objs": [],
|
||||
"tags": [
|
||||
"第三单元"
|
||||
|
|
@ -349640,7 +349640,7 @@
|
|||
},
|
||||
"014331": {
|
||||
"id": "014331",
|
||||
"content": "如图, 在直角三角形$ABC$中, $|CA|=|CB|=2, M$、$N$是斜边$AB$上的两个动点(点$M$靠在线段$BN$上), 且$|MN|=\\sqrt{2}$.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw (0,0) node [left] {$C$} coordinate (C);\n\\draw (2,0) node [right] {$A$} coordinate (A);\n\\draw (0,2) node [above] {$B$} coordinate (B);\n\\draw ($(A)!0.2!(B)$) node [above right] {$M$} coordinate (M);\n\\draw ($(A)!0.7!(B)$) node [above right] {$N$} coordinate (N);\n\\draw (A)--(B)--(C)--cycle;\n\\draw (C)--(M)(C)--(N);\n\\end{tikzpicture}\n\\end{center}\n(1) 求向量$\\overrightarrow{MN}$在$\\overrightarrow{CB}$方向上的投影与数量投影;\\\\\n(2) 求$\\overrightarrow{CM} \\cdot \\overrightarrow{CN}$的取值范围.",
|
||||
"content": "如图, 在直角三角形$ABC$中, $|CA|=|CB|=2, M$、$N$是斜边$AB$上的两个动点(点$N$在线段$BM$上), 且$|MN|=\\sqrt{2}$.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw (0,0) node [left] {$C$} coordinate (C);\n\\draw (2,0) node [right] {$A$} coordinate (A);\n\\draw (0,2) node [above] {$B$} coordinate (B);\n\\draw ($(A)!0.2!(B)$) node [above right] {$M$} coordinate (M);\n\\draw ($(A)!0.7!(B)$) node [above right] {$N$} coordinate (N);\n\\draw (A)--(B)--(C)--cycle;\n\\draw (C)--(M)(C)--(N);\n\\end{tikzpicture}\n\\end{center}\n(1) 求向量$\\overrightarrow{MN}$在$\\overrightarrow{CB}$方向上的投影与数量投影;\\\\\n(2) 求$\\overrightarrow{CM} \\cdot \\overrightarrow{CN}$的取值范围.",
|
||||
"objs": [],
|
||||
"tags": [],
|
||||
"genre": "解答题",
|
||||
|
|
@ -350839,7 +350839,7 @@
|
|||
},
|
||||
"014394": {
|
||||
"id": "014394",
|
||||
"content": "已知复数$z=\\dfrac{\\mathrm{i}+\\mathrm{i}^2++\\mathrm{i}^3+\\cdots+\\mathrm{i}^{2003}}{1+\\mathrm{i}}$, 则复数$z=$\\blank{50}.",
|
||||
"content": "已知复数$z=\\dfrac{\\mathrm{i}+\\mathrm{i}^2+\\mathrm{i}^3+\\cdots+\\mathrm{i}^{2003}}{1+\\mathrm{i}}$, 则复数$z=$\\blank{50}.",
|
||||
"objs": [],
|
||||
"tags": [],
|
||||
"genre": "填空题",
|
||||
|
|
@ -353499,10 +353499,10 @@
|
|||
},
|
||||
"014534": {
|
||||
"id": "014534",
|
||||
"content": "已知数列$\\{a_n\\}$的前$n$项和为$S_n=2 \\times 3^n+a$. 当常数$a=$时, 数列$\\{a_n\\}$为等比数列.",
|
||||
"content": "已知数列$\\{a_n\\}$的前$n$项和为$S_n=2 \\times 3^n+a$. 当常数$a=$\\blank{50}时, 数列$\\{a_n\\}$为等比数列.",
|
||||
"objs": [],
|
||||
"tags": [],
|
||||
"genre": "解答题",
|
||||
"genre": "填空题",
|
||||
"ans": "",
|
||||
"solution": "",
|
||||
"duration": -1,
|
||||
|
|
@ -353514,7 +353514,7 @@
|
|||
"same": [],
|
||||
"related": [],
|
||||
"remark": "",
|
||||
"space": "12ex"
|
||||
"space": ""
|
||||
},
|
||||
"014535": {
|
||||
"id": "014535",
|
||||
|
|
@ -353727,7 +353727,7 @@
|
|||
},
|
||||
"014546": {
|
||||
"id": "014546",
|
||||
"content": "已知无穷等比数列$\\{a_n\\}$的公比为$q$, 前$n$项和为$S_n$, 且$\\displaystyle\\lim _{n \\to+\\infty} S_n=S$. 下列条件中, 使得$2S_n<S$桓成立的是\\bracket{20}.\n\\twoch{$a_1>0$, $0.6<q<0.7$}{$a_1<0$, $-0.7<q<-0.6$}{$a_1>0$, $0.7<q<0.8$}{$a_1<0$, $-0.8<q<-0.7$}",
|
||||
"content": "已知无穷等比数列$\\{a_n\\}$的公比为$q$, 前$n$项和为$S_n$, 且$\\displaystyle\\lim _{n \\to+\\infty} S_n=S$. 下列条件中, 使得$2S_n<S$恒成立的是\\bracket{20}.\n\\twoch{$a_1>0$, $0.6<q<0.7$}{$a_1<0$, $-0.7<q<-0.6$}{$a_1>0$, $0.7<q<0.8$}{$a_1<0$, $-0.8<q<-0.7$}",
|
||||
"objs": [],
|
||||
"tags": [],
|
||||
"genre": "选择题",
|
||||
|
|
@ -353993,7 +353993,7 @@
|
|||
},
|
||||
"014560": {
|
||||
"id": "014560",
|
||||
"content": "已知数列$\\{a_n\\}$的前$n$项的和$S_n$满足$S_{n+1}+S_n=n$. 对于以下两个命题: \\textcircled{1} 若$a_1=-1$, 则$S_{203}=1010$; \\textcircled{2} 数列$\\{a_{n+1}+a_n\\}$是常数列, 下列说法正确的是\\bracket{20}.\n\\fourch{\\textcircled{1}\\textcircled{2}都正确}{\\textcircled{1}正确, \\textcircled{2}不正确}{\\textcircled{1}\\textcircled{2}都不正确}{\\textcircled{1}不正确, \\textcircled{2}正确}",
|
||||
"content": "已知数列$\\{a_n\\}$的前$n$项的和$S_n$满足$S_{n+1}+S_n=n$. 对于以下两个命题: \\textcircled{1} 若$a_1=-1$, 则$S_{2023}=1010$; \\textcircled{2} 数列$\\{a_{n+1}+a_n\\}$是常数列, 下列说法正确的是\\bracket{20}.\n\\fourch{\\textcircled{1}\\textcircled{2}都正确}{\\textcircled{1}正确, \\textcircled{2}不正确}{\\textcircled{1}\\textcircled{2}都不正确}{\\textcircled{1}不正确, \\textcircled{2}正确}",
|
||||
"objs": [],
|
||||
"tags": [],
|
||||
"genre": "选择题",
|
||||
|
|
@ -420452,7 +420452,7 @@
|
|||
],
|
||||
"genre": "选择题",
|
||||
"ans": "B",
|
||||
"solution": "",
|
||||
"solution": "注意到$1^2=2\\cdot 1-1$, 因此$A=\\{x|x\\in \\mathbf{R},\\ x\\ne 1\\}$与$B=\\{1\\}$时$f(x)$是偶函数, 而$A=\\{x|x\\in \\mathbf{R},\\ x\\ne 1, \\ x\\ne 0\\}$与$B=\\{0,1\\}$时$f(x)$也是偶函数, 故\\textcircled{1}错误.\\\\\n任何一组满足$\\dfrac 32\\in A$, $\\pm \\sqrt{2}\\in B$的满足要求的集合都能使得$f(x)=2$无解, 注意到$3,4,5,6,\\cdot$都可以随意放置在$A$或$B$之一内, 因此这样的集合确实有无穷多对, \\textcircled{2}正确.",
|
||||
"duration": -1,
|
||||
"usages": [],
|
||||
"origin": "2022届杨浦一模试题16",
|
||||
|
|
@ -432419,7 +432419,7 @@
|
|||
"tags": [],
|
||||
"genre": "解答题",
|
||||
"ans": "(1) $y=x$; (2) $g'(x)=\\mathrm{e}^x(\\ln (x+1)+\\dfrac 1{x+1}+\\dfrac{x}{(x+1)^2})$, 故$g(x)$在$[1,+\\infty)$上是严格增函数; (3) 证明略.",
|
||||
"solution": "",
|
||||
"solution": "(1) $f'(0)=\\mathrm{e}^0(\\dfrac{1}{1+0}+\\ln (1+0))=1$, 又$f(0)=0$, 故所求切线的方程为$y=x$.\\\\\n(2) $g(x)=f'(x)=\\mathrm{e}^x(\\dfrac{1}{1+x}+\\ln (1+x))$. $g'(x)=\\mathrm{e}^x(\\ln (1+x)+\\dfrac{2}{1+x}-\\dfrac{1}{(1+x)^2})$, 当$x>0$时, $\\mathrm{e}^x>1>0$, $\\ln(1+x)>0$, 且$\\dfrac{2}{1+x}-\\dfrac{1}{(1+x)^2}=\\dfrac{2x+1}{(1+x)^2}>0$, 所以$g'(x)$在$(0,+\\infty)$上恒正. 因此$g(x)$在$[0,+\\infty)$上是严格增函数.\\\\\n(3) 对于固定的常数$t$, 令$h(x)=f(x+t)-f(x)-f(t)$, 则$h(t)=0$, 而$h'(x)=f'(x+t)-f'(x)=g(x+t)-g(x)$. 由(2)可知$g(x+t)>g(x)$, 故$h'(x)$在$(0,+\\infty)$上恒正, 于是有$h(s)>h(0)=0$, 即$f(s+t)>f(s)+f(t)$.",
|
||||
"duration": -1,
|
||||
"usages": [],
|
||||
"origin": "2022年北京高考",
|
||||
|
|
@ -432444,7 +432444,7 @@
|
|||
],
|
||||
"genre": "填空题",
|
||||
"ans": "$\\dfrac 35$",
|
||||
"solution": "",
|
||||
"solution": "所有取法共有$\\mathrm{C}_6^3=20$种, 其中恰有两个小球编号相邻的取法有$124$, $125$, $126$, $235$, $236$, $134$, $346$, $145$, $245$, $156$, $256$, $356$共$12$种, 所以所求概率为$\\dfrac{12}{20}=\\dfrac 35$",
|
||||
"duration": -1,
|
||||
"usages": [],
|
||||
"origin": "教材复习题-20230220修改",
|
||||
|
|
@ -432470,7 +432470,7 @@
|
|||
"二项式定理"
|
||||
],
|
||||
"genre": "填空题",
|
||||
"ans": "有$17$项是有理项",
|
||||
"ans": "$17$",
|
||||
"solution": "考虑$(x^{\\frac 12}+x^{-\\frac 13})^{100}$展开式的通项$T_{r+1}=\\mathrm{C}_{100}^rx^{\\frac{100-r}2}\\cdot (x^{-\\frac 13})^r=\\mathrm{C}_{100}^rx^{50-\\dfrac{3r}6}$.\n令$r=6k$($k\\in \\mathbf{Z}$), 则$0\\le 6k\\le 100$, 即$r=0,6,12,\\cdots ,96$.\n因此共有$17$个有理项.",
|
||||
"duration": -1,
|
||||
"usages": [],
|
||||
|
|
@ -432495,7 +432495,7 @@
|
|||
],
|
||||
"genre": "填空题",
|
||||
"ans": "$\\dfrac 23$",
|
||||
"solution": "",
|
||||
"solution": "$\\dfrac{1}{\\sin^2\\alpha+2\\sin\\alpha\\cos\\alpha}=\\dfrac{\\sin^2\\alpha+\\cos^2\\alpha}{\\sin^2\\alpha+2\\sin\\alpha\\cos\\alpha}=\\dfrac{\\tan^2\\alpha+1}{\\tan^2\\alpha+2\\tan\\alpha}=\\dfrac 2 3$",
|
||||
"duration": -1,
|
||||
"usages": [],
|
||||
"origin": "二期课改练习册高一第二学期-20230220修改",
|
||||
|
|
@ -432520,7 +432520,7 @@
|
|||
],
|
||||
"genre": "填空题",
|
||||
"ans": "$\\dfrac{2\\pi}3$",
|
||||
"solution": "",
|
||||
"solution": "因$\\overrightarrow{b}\\cdot \\overrightarrow{b}=36$, 故$\\overrightarrow{a}\\cdot \\overrightarrow{b}=-15$, 进而$\\cos\\langle \\overrightarrow a,\\overrightarrow b\\rangle=\\dfrac{-15}{30}=-\\dfrac 12$",
|
||||
"duration": -1,
|
||||
"usages": [],
|
||||
"origin": "新教材必修第二册课堂练习-20230220修改",
|
||||
|
|
@ -433034,7 +433034,7 @@
|
|||
"tags": [],
|
||||
"genre": "填空题",
|
||||
"ans": "$6$",
|
||||
"solution": "",
|
||||
"solution": "从小到大排列后, 第$9$和第$10$个数的算术平均数为$5$. $18\\cdot 0.75=13.5$, 故第$14$个数也为$5$, 从而第$10$个数小于等于$5$, 因此它等于$5$, 第九个数也必须等于$5$, 这样, 第$9,10,11,12,13,14$这六个数都等于$5$. 另一方面, $8$个$4$, $6$个$5$, $4$个$6$的数据满足要求.",
|
||||
"duration": -1,
|
||||
"usages": [],
|
||||
"origin": "自拟题目",
|
||||
|
|
|
|||
Reference in New Issue