高三月考难度数据导入

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WangWeiye 2023-05-17 12:16:03 +08:00
parent 9fac56fa4e
commit c4906d2d4f
4 changed files with 855 additions and 291 deletions

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ans usages
017223
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017224
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017225
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017226
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017227
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017228
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017229
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017230
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017231
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017232
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017233
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017234
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017235
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017236
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017237
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017238
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017239
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017240
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017241
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017242
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20230517 2023届高三03班 1.000 0.603 0.250
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20230517 2023届高三12班 1.000 0.347 0.099
017243
20230517 2023届高三01班 0.934 0.721 0.070
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20230517 2023届高三03班 0.879 0.523 0.017
20230517 2023届高三04班 0.958 0.439 0.004
20230517 2023届高三05班 0.875 0.434 0.007
20230517 2023届高三06班 0.865 0.355 0.035
20230517 2023届高三07班 0.939 0.359 0.000
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20230517 2023届高三12班 0.854 0.319 0.005
021639
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[BEGIN] [BEGIN]
## 20230418 ## 20230418
** 2025届高一01班 ** 2023届高三01班
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[END] [END]
[BEGIN] [BEGIN]
## 20230418 ## 20230418
** 2025届高一02班 ** 2023届高三02班
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[END] [END]
[BEGIN] [BEGIN]
## 20230418 ## 20230418
** 2025届高一03班 ** 2023届高三03班
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[END] [END]
[BEGIN] [BEGIN]
## 20230418 ## 20230418
** 2025届高一04班 ** 2023届高三04班
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[END] [END]
[BEGIN] [BEGIN]
## 20230418 ## 20230418
** 2025届高一05班 ** 2023届高三05班
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015289 0.573 0.455 0.043 017243 0.875 0.434 0.007
[END] [END]
[BEGIN] [BEGIN]
## 20230418 ## 20230418
** 2025届高一06班 ** 2023届高三06班
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[END] [END]
[BEGIN] [BEGIN]
## 20230418 ## 20230418
** 2025届高一07班 ** 2023届高三07班
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015289 0.672 0.517 0.005 017243 0.939 0.359 0.000
[END] [END]
[BEGIN] [BEGIN]
## 20230418 ## 20230418
** 2025届高一08班 ** 2023届高三08班
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015270 1.000 017224 1.000
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015289 0.535 0.398 0.007 017243 0.904 0.423 0.029
[END] [END]
[BEGIN] [BEGIN]
## 20230418 ## 20230418
** 2025届高一09班 ** 2023届高三09班
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015287 0.958 0.612 017241 1.000 0.658 0.783
015288 0.833 0.517 017242 0.967 0.283 0.058
015289 0.662 0.532 0.005 017243 0.858 0.267 0.013
[END] [END]
[BEGIN] [BEGIN]
## 20230418 ## 20230418
** 2025届高一10班 ** 2023届高三10班
015269 0.950 017223 0.973
015270 0.950 017224 1.000
015271 1.000 017225 0.838
015272 0.977 017226 0.973
015273 0.927 017227 0.973
015274 0.877 017228 0.595
015275 0.733 017229 0.892
015276 0.903 017230 0.865
015277 0.513 017231 0.541
015278 0.633 017232 0.243
015279 0.587 017233 0.459
015280 0.487 017234 0.000
015281 0.903 017235 1.000
015282 0.903 017236 0.946
015283 0.682 017237 0.243
015284 0.292 017238 0.216
015285 0.979 017239 0.973 0.939
015286 0.948 0.950 017240 0.973 0.889
015287 0.990 0.676 017241 0.946 0.615 0.761
015288 0.890 0.650 017242 0.973 0.360 0.057
015289 0.750 0.695 0.040 017243 0.858 0.306 0.024
[END] [END]
[BEGIN] [BEGIN]
## 20230418 ## 20230418
** 2025届高一11班 ** 2023届高三11班
015269 1.000 017223 1.000
015270 1.000 017224 0.950
015271 1.000 017225 0.850
015272 0.953 017226 1.000
015273 0.883 017227 0.900
015274 0.907 017228 0.550
015275 0.767 017229 0.900
015276 0.907 017230 0.750
015277 0.533 017231 0.400
015278 0.697 017232 0.250
015279 0.533 017233 0.400
015280 0.533 017234 0.000
015281 0.883 017235 1.000
015282 0.907 017236 1.000
015283 0.767 017237 0.300
015284 0.185 017238 0.000
015285 0.985 017239 0.992 0.950
015286 0.995 0.927 017240 0.958 0.894
015287 0.990 0.724 017241 1.000 0.812 0.758
015288 0.953 0.685 017242 0.950 0.333 0.019
015289 0.780 0.800 0.122 017243 0.863 0.333 0.006
[END] [END]
[BEGIN] [BEGIN]
## 20230418 ## 20230418
** 2025届高一12班 ** 2023届高三12班
015269 0.977 017223 0.958
015270 1.000 017224 1.000
015271 1.000 017225 0.833
015272 0.933 017226 0.958
015273 0.887 017227 0.958
015274 0.933 017228 0.458
015275 0.703 017229 1.000
015276 0.953 017230 0.792
015277 0.637 017231 0.542
015278 0.773 017232 0.250
015279 0.613 017233 0.375
015280 0.477 017234 0.000
015281 0.910 017235 1.000
015282 0.840 017236 1.000
015283 0.660 017237 0.375
015284 0.340 017238 0.208
015285 0.971 017239 1.000 0.943
015286 0.917 0.983 017240 0.938 0.932
015287 0.988 0.632 017241 1.000 0.698 0.812
015288 0.882 0.632 017242 1.000 0.347 0.099
015289 0.770 0.710 0.037 017243 0.854 0.319 0.005
[END] [END]

View File

@ -1,7 +1,7 @@
texfile = r"D:\mathdeptv2\工具\临时文件\高一区统考_学生用_20230421.tex" texfile = r"C:\Users\weiye\Documents\wwy sync\23届\下学期测验卷\高三下学期月考02.tex"
excelfile = r"C:\Users\weiye\Documents\wwy sync\临时工作区\高一区统考.xlsx" excelfile = r"C:\Users\weiye\Documents\wwy sync\23届\统计数据\大型考试\年级_20230516月考.xlsx"
date = "20230418" date = "20230418"
grade = "2025届高一" grade = "2023届高三"
sheetname = "难度统计" sheetname = "难度统计"
max_classnum = 12 max_classnum = 12

View File

@ -441830,7 +441830,20 @@
"ans": "$\\{1\\}$", "ans": "$\\{1\\}$",
"solution": "", "solution": "",
"duration": -1, "duration": -1,
"usages": [], "usages": [
"20230517\t2023届高三01班\t0.971",
"20230517\t2023届高三02班\t0.969",
"20230517\t2023届高三03班\t0.966",
"20230517\t2023届高三04班\t0.967",
"20230517\t2023届高三05班\t1.000",
"20230517\t2023届高三06班\t0.974",
"20230517\t2023届高三07班\t0.970",
"20230517\t2023届高三08班\t1.000",
"20230517\t2023届高三09班\t0.933",
"20230517\t2023届高三10班\t0.973",
"20230517\t2023届高三11班\t1.000",
"20230517\t2023届高三12班\t0.958"
],
"origin": "2023届高三下学期月考2试题1", "origin": "2023届高三下学期月考2试题1",
"edit": [ "edit": [
"20230507\t余利成" "20230507\t余利成"
@ -441850,7 +441863,20 @@
"ans": "$1$", "ans": "$1$",
"solution": "", "solution": "",
"duration": -1, "duration": -1,
"usages": [], "usages": [
"20230517\t2023届高三01班\t0.971",
"20230517\t2023届高三02班\t1.000",
"20230517\t2023届高三03班\t0.966",
"20230517\t2023届高三04班\t0.967",
"20230517\t2023届高三05班\t0.921",
"20230517\t2023届高三06班\t0.949",
"20230517\t2023届高三07班\t0.970",
"20230517\t2023届高三08班\t1.000",
"20230517\t2023届高三09班\t1.000",
"20230517\t2023届高三10班\t1.000",
"20230517\t2023届高三11班\t0.950",
"20230517\t2023届高三12班\t1.000"
],
"origin": "2023届高三下学期月考2试题2", "origin": "2023届高三下学期月考2试题2",
"edit": [ "edit": [
"20230507\t余利成" "20230507\t余利成"
@ -441870,7 +441896,20 @@
"ans": "$\\dfrac{\\pi}{3}$", "ans": "$\\dfrac{\\pi}{3}$",
"solution": "", "solution": "",
"duration": -1, "duration": -1,
"usages": [], "usages": [
"20230517\t2023届高三01班\t0.941",
"20230517\t2023届高三02班\t0.938",
"20230517\t2023届高三03班\t0.897",
"20230517\t2023届高三04班\t0.933",
"20230517\t2023届高三05班\t0.947",
"20230517\t2023届高三06班\t0.949",
"20230517\t2023届高三07班\t0.970",
"20230517\t2023届高三08班\t0.846",
"20230517\t2023届高三09班\t0.767",
"20230517\t2023届高三10班\t0.838",
"20230517\t2023届高三11班\t0.850",
"20230517\t2023届高三12班\t0.833"
],
"origin": "2023届高三下学期月考2试题3", "origin": "2023届高三下学期月考2试题3",
"edit": [ "edit": [
"20230507\t余利成" "20230507\t余利成"
@ -441890,7 +441929,20 @@
"ans": "$70$", "ans": "$70$",
"solution": "", "solution": "",
"duration": -1, "duration": -1,
"usages": [], "usages": [
"20230517\t2023届高三01班\t0.971",
"20230517\t2023届高三02班\t1.000",
"20230517\t2023届高三03班\t0.966",
"20230517\t2023届高三04班\t1.000",
"20230517\t2023届高三05班\t0.974",
"20230517\t2023届高三06班\t0.974",
"20230517\t2023届高三07班\t1.000",
"20230517\t2023届高三08班\t1.000",
"20230517\t2023届高三09班\t0.933",
"20230517\t2023届高三10班\t0.973",
"20230517\t2023届高三11班\t1.000",
"20230517\t2023届高三12班\t0.958"
],
"origin": "2023届高三下学期月考2试题4", "origin": "2023届高三下学期月考2试题4",
"edit": [ "edit": [
"20230507\t余利成" "20230507\t余利成"
@ -441910,7 +441962,20 @@
"ans": "$\\dfrac{1}{8}$", "ans": "$\\dfrac{1}{8}$",
"solution": "", "solution": "",
"duration": -1, "duration": -1,
"usages": [], "usages": [
"20230517\t2023届高三01班\t1.000",
"20230517\t2023届高三02班\t1.000",
"20230517\t2023届高三03班\t1.000",
"20230517\t2023届高三04班\t0.967",
"20230517\t2023届高三05班\t1.000",
"20230517\t2023届高三06班\t0.949",
"20230517\t2023届高三07班\t0.939",
"20230517\t2023届高三08班\t0.923",
"20230517\t2023届高三09班\t0.900",
"20230517\t2023届高三10班\t0.973",
"20230517\t2023届高三11班\t0.900",
"20230517\t2023届高三12班\t0.958"
],
"origin": "2023届高三下学期月考2试题5", "origin": "2023届高三下学期月考2试题5",
"edit": [ "edit": [
"20230507\t余利成" "20230507\t余利成"
@ -441930,7 +441995,20 @@
"ans": "$y=-\\dfrac{1}{32}$", "ans": "$y=-\\dfrac{1}{32}$",
"solution": "", "solution": "",
"duration": -1, "duration": -1,
"usages": [], "usages": [
"20230517\t2023届高三01班\t0.529",
"20230517\t2023届高三02班\t0.594",
"20230517\t2023届高三03班\t0.345",
"20230517\t2023届高三04班\t0.500",
"20230517\t2023届高三05班\t0.526",
"20230517\t2023届高三06班\t0.436",
"20230517\t2023届高三07班\t0.545",
"20230517\t2023届高三08班\t0.577",
"20230517\t2023届高三09班\t0.633",
"20230517\t2023届高三10班\t0.595",
"20230517\t2023届高三11班\t0.550",
"20230517\t2023届高三12班\t0.458"
],
"origin": "2023届高三下学期月考2试题6", "origin": "2023届高三下学期月考2试题6",
"edit": [ "edit": [
"20230507\t余利成" "20230507\t余利成"
@ -441950,7 +442028,20 @@
"ans": "$\\dfrac{2}{3}$", "ans": "$\\dfrac{2}{3}$",
"solution": "", "solution": "",
"duration": -1, "duration": -1,
"usages": [], "usages": [
"20230517\t2023届高三01班\t1.000",
"20230517\t2023届高三02班\t0.938",
"20230517\t2023届高三03班\t0.966",
"20230517\t2023届高三04班\t0.967",
"20230517\t2023届高三05班\t0.947",
"20230517\t2023届高三06班\t0.974",
"20230517\t2023届高三07班\t1.000",
"20230517\t2023届高三08班\t0.962",
"20230517\t2023届高三09班\t0.800",
"20230517\t2023届高三10班\t0.892",
"20230517\t2023届高三11班\t0.900",
"20230517\t2023届高三12班\t1.000"
],
"origin": "2023届高三下学期月考2试题7", "origin": "2023届高三下学期月考2试题7",
"edit": [ "edit": [
"20230507\t余利成" "20230507\t余利成"
@ -441970,7 +442061,20 @@
"ans": "$36\\pi$", "ans": "$36\\pi$",
"solution": "", "solution": "",
"duration": -1, "duration": -1,
"usages": [], "usages": [
"20230517\t2023届高三01班\t0.824",
"20230517\t2023届高三02班\t0.781",
"20230517\t2023届高三03班\t0.897",
"20230517\t2023届高三04班\t0.833",
"20230517\t2023届高三05班\t0.895",
"20230517\t2023届高三06班\t0.846",
"20230517\t2023届高三07班\t0.636",
"20230517\t2023届高三08班\t0.615",
"20230517\t2023届高三09班\t0.667",
"20230517\t2023届高三10班\t0.865",
"20230517\t2023届高三11班\t0.750",
"20230517\t2023届高三12班\t0.792"
],
"origin": "2023届高三下学期月考2试题8", "origin": "2023届高三下学期月考2试题8",
"edit": [ "edit": [
"20230507\t余利成" "20230507\t余利成"
@ -441990,7 +442094,20 @@
"ans": "$0.1$", "ans": "$0.1$",
"solution": "", "solution": "",
"duration": -1, "duration": -1,
"usages": [], "usages": [
"20230517\t2023届高三01班\t0.735",
"20230517\t2023届高三02班\t0.500",
"20230517\t2023届高三03班\t0.655",
"20230517\t2023届高三04班\t0.633",
"20230517\t2023届高三05班\t0.737",
"20230517\t2023届高三06班\t0.667",
"20230517\t2023届高三07班\t0.394",
"20230517\t2023届高三08班\t0.692",
"20230517\t2023届高三09班\t0.567",
"20230517\t2023届高三10班\t0.541",
"20230517\t2023届高三11班\t0.400",
"20230517\t2023届高三12班\t0.542"
],
"origin": "2023届高三下学期月考2试题9", "origin": "2023届高三下学期月考2试题9",
"edit": [ "edit": [
"20230507\t余利成" "20230507\t余利成"
@ -442010,7 +442127,20 @@
"ans": "$\\dfrac{17}{32}$", "ans": "$\\dfrac{17}{32}$",
"solution": "", "solution": "",
"duration": -1, "duration": -1,
"usages": [], "usages": [
"20230517\t2023届高三01班\t0.382",
"20230517\t2023届高三02班\t0.469",
"20230517\t2023届高三03班\t0.586",
"20230517\t2023届高三04班\t0.233",
"20230517\t2023届高三05班\t0.316",
"20230517\t2023届高三06班\t0.231",
"20230517\t2023届高三07班\t0.333",
"20230517\t2023届高三08班\t0.231",
"20230517\t2023届高三09班\t0.267",
"20230517\t2023届高三10班\t0.243",
"20230517\t2023届高三11班\t0.250",
"20230517\t2023届高三12班\t0.250"
],
"origin": "2023届高三下学期月考2试题10", "origin": "2023届高三下学期月考2试题10",
"edit": [ "edit": [
"20230507\t余利成" "20230507\t余利成"
@ -442030,7 +442160,20 @@
"ans": "$[2,2^{2023}]$", "ans": "$[2,2^{2023}]$",
"solution": "", "solution": "",
"duration": -1, "duration": -1,
"usages": [], "usages": [
"20230517\t2023届高三01班\t0.853",
"20230517\t2023届高三02班\t0.719",
"20230517\t2023届高三03班\t0.586",
"20230517\t2023届高三04班\t0.833",
"20230517\t2023届高三05班\t0.737",
"20230517\t2023届高三06班\t0.667",
"20230517\t2023届高三07班\t0.576",
"20230517\t2023届高三08班\t0.731",
"20230517\t2023届高三09班\t0.533",
"20230517\t2023届高三10班\t0.459",
"20230517\t2023届高三11班\t0.400",
"20230517\t2023届高三12班\t0.375"
],
"origin": "2023届高三下学期月考2试题11", "origin": "2023届高三下学期月考2试题11",
"edit": [ "edit": [
"20230507\t余利成" "20230507\t余利成"
@ -442050,7 +442193,20 @@
"ans": "$[\\dfrac{1}{2},+\\infty)$", "ans": "$[\\dfrac{1}{2},+\\infty)$",
"solution": "", "solution": "",
"duration": -1, "duration": -1,
"usages": [], "usages": [
"20230517\t2023届高三01班\t0.147",
"20230517\t2023届高三02班\t0.062",
"20230517\t2023届高三03班\t0.138",
"20230517\t2023届高三04班\t0.167",
"20230517\t2023届高三05班\t0.053",
"20230517\t2023届高三06班\t0.077",
"20230517\t2023届高三07班\t0.000",
"20230517\t2023届高三08班\t0.038",
"20230517\t2023届高三09班\t0.033",
"20230517\t2023届高三10班\t0.000",
"20230517\t2023届高三11班\t0.000",
"20230517\t2023届高三12班\t0.000"
],
"origin": "2023届高三下学期月考2试题12", "origin": "2023届高三下学期月考2试题12",
"edit": [ "edit": [
"20230507\t余利成" "20230507\t余利成"
@ -442070,7 +442226,20 @@
"ans": "D", "ans": "D",
"solution": "", "solution": "",
"duration": -1, "duration": -1,
"usages": [], "usages": [
"20230517\t2023届高三01班\t1.000",
"20230517\t2023届高三02班\t1.000",
"20230517\t2023届高三03班\t1.000",
"20230517\t2023届高三04班\t1.000",
"20230517\t2023届高三05班\t1.000",
"20230517\t2023届高三06班\t1.000",
"20230517\t2023届高三07班\t1.000",
"20230517\t2023届高三08班\t1.000",
"20230517\t2023届高三09班\t1.000",
"20230517\t2023届高三10班\t1.000",
"20230517\t2023届高三11班\t1.000",
"20230517\t2023届高三12班\t1.000"
],
"origin": "2023届高三下学期月考2试题13", "origin": "2023届高三下学期月考2试题13",
"edit": [ "edit": [
"20230507\t余利成" "20230507\t余利成"
@ -442090,7 +442259,20 @@
"ans": "B", "ans": "B",
"solution": "", "solution": "",
"duration": -1, "duration": -1,
"usages": [], "usages": [
"20230517\t2023届高三01班\t0.941",
"20230517\t2023届高三02班\t0.969",
"20230517\t2023届高三03班\t0.966",
"20230517\t2023届高三04班\t0.967",
"20230517\t2023届高三05班\t0.947",
"20230517\t2023届高三06班\t1.000",
"20230517\t2023届高三07班\t1.000",
"20230517\t2023届高三08班\t1.000",
"20230517\t2023届高三09班\t0.967",
"20230517\t2023届高三10班\t0.946",
"20230517\t2023届高三11班\t1.000",
"20230517\t2023届高三12班\t1.000"
],
"origin": "2023届高三下学期月考2试题14", "origin": "2023届高三下学期月考2试题14",
"edit": [ "edit": [
"20230507\t余利成" "20230507\t余利成"
@ -442110,7 +442292,20 @@
"ans": "C", "ans": "C",
"solution": "", "solution": "",
"duration": -1, "duration": -1,
"usages": [], "usages": [
"20230517\t2023届高三01班\t0.765",
"20230517\t2023届高三02班\t0.562",
"20230517\t2023届高三03班\t0.552",
"20230517\t2023届高三04班\t0.433",
"20230517\t2023届高三05班\t0.526",
"20230517\t2023届高三06班\t0.641",
"20230517\t2023届高三07班\t0.424",
"20230517\t2023届高三08班\t0.577",
"20230517\t2023届高三09班\t0.300",
"20230517\t2023届高三10班\t0.243",
"20230517\t2023届高三11班\t0.300",
"20230517\t2023届高三12班\t0.375"
],
"origin": "2023届高三下学期月考2试题15", "origin": "2023届高三下学期月考2试题15",
"edit": [ "edit": [
"20230507\t余利成" "20230507\t余利成"
@ -442130,7 +442325,20 @@
"ans": "D", "ans": "D",
"solution": "", "solution": "",
"duration": -1, "duration": -1,
"usages": [], "usages": [
"20230517\t2023届高三01班\t0.412",
"20230517\t2023届高三02班\t0.562",
"20230517\t2023届高三03班\t0.517",
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"origin": "2023届高三下学期月考2试题16", "origin": "2023届高三下学期月考2试题16",
"edit": [ "edit": [
"20230507\t余利成" "20230507\t余利成"
@ -442150,7 +442358,20 @@
"ans": "(1) $h=1$; (2) $\\arcsin\\dfrac{\\sqrt{10}}{5}$", "ans": "(1) $h=1$; (2) $\\arcsin\\dfrac{\\sqrt{10}}{5}$",
"solution": "(1) 由底面$ABC$为等腰直角三角形且 $AB \\perp AC$知$AB\\perp$平面$ACC_1A_1$, \\\\\n从而$BM$在平面$ACC_1A_1$上的投影为$AM$,\n故由$BM \\perp A_1C$ 知 $AM\\perp A_1C $, \\\\\n结合$AC=2$, $AA_1=4$ 得 $MC=1$, 即$h=1$.\\\\\n(2) 如图建系:以$A$为原点,分别以$\\overrightarrow{AB}$、$\\overrightarrow{AC}$、$\\overrightarrow{AA_1}$方向为$x$轴、 $y$轴、$z$ 轴正方向建立平面直角坐标系.\\\\\n$A(0,0,0),B(2,0,0),C(0,2,0),A_1(0,0,4),M(0,2,2)$,\n$\\overrightarrow{BA_1}=(-2,0,4),\\overrightarrow{AB}=(2,0,0),\\overrightarrow{AM}=(0,2,2),$\\\\\n设平面$ABM$的一个法向量为$\\overrightarrow{n}=(x,y,z)$, 则$\\begin{cases}\n2x=0,\\\\\n2y+2z=0.\n\\end{cases}$ 取$\\overrightarrow{n}=(0,1,-1),$设直线$BA_1$与平面$ABM$所成的角为$\\theta$, 则$\\sin\\theta=|\\cos \\langle\\overrightarrow{BA_1},\\overrightarrow{n}\\rangle|=\\dfrac{|\\overrightarrow{BA_1}\\cdot\\overrightarrow{n}|}{|\\overrightarrow{BA_1}|\\cdot|\\overrightarrow{n}|}=\\dfrac{\\sqrt{10}}{5}$, 故直线$BA_1$与平面$ABM$所成的角为$\\arcsin\\dfrac{\\sqrt{10}}{5}$.", "solution": "(1) 由底面$ABC$为等腰直角三角形且 $AB \\perp AC$知$AB\\perp$平面$ACC_1A_1$, \\\\\n从而$BM$在平面$ACC_1A_1$上的投影为$AM$,\n故由$BM \\perp A_1C$ 知 $AM\\perp A_1C $, \\\\\n结合$AC=2$, $AA_1=4$ 得 $MC=1$, 即$h=1$.\\\\\n(2) 如图建系:以$A$为原点,分别以$\\overrightarrow{AB}$、$\\overrightarrow{AC}$、$\\overrightarrow{AA_1}$方向为$x$轴、 $y$轴、$z$ 轴正方向建立平面直角坐标系.\\\\\n$A(0,0,0),B(2,0,0),C(0,2,0),A_1(0,0,4),M(0,2,2)$,\n$\\overrightarrow{BA_1}=(-2,0,4),\\overrightarrow{AB}=(2,0,0),\\overrightarrow{AM}=(0,2,2),$\\\\\n设平面$ABM$的一个法向量为$\\overrightarrow{n}=(x,y,z)$, 则$\\begin{cases}\n2x=0,\\\\\n2y+2z=0.\n\\end{cases}$ 取$\\overrightarrow{n}=(0,1,-1),$设直线$BA_1$与平面$ABM$所成的角为$\\theta$, 则$\\sin\\theta=|\\cos \\langle\\overrightarrow{BA_1},\\overrightarrow{n}\\rangle|=\\dfrac{|\\overrightarrow{BA_1}\\cdot\\overrightarrow{n}|}{|\\overrightarrow{BA_1}|\\cdot|\\overrightarrow{n}|}=\\dfrac{\\sqrt{10}}{5}$, 故直线$BA_1$与平面$ABM$所成的角为$\\arcsin\\dfrac{\\sqrt{10}}{5}$.",
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"origin": "2023届高三下学期月考2试题17", "origin": "2023届高三下学期月考2试题17",
"edit": [ "edit": [
"20230507\t余利成" "20230507\t余利成"
@ -442170,7 +442391,20 @@
"ans": "(1) $a-c=\\pm 1$; (2) $C=\\dfrac{\\pi}{3}$", "ans": "(1) $a-c=\\pm 1$; (2) $C=\\dfrac{\\pi}{3}$",
"solution": "(1) 由 $S=\\dfrac{1}{2}ac\\sin B=\\dfrac{\\sqrt{3}}{4}ac=\\sqrt{3}$知$ac=4$\\\\\n由$a^2+c^2-b^2=2ac\\cos B$知$a^2+c^2=9$.\\\\\n结合两式得$(a-c)^2=1$, 故$a-c=\\pm 1$\\\\\n(2) 由$2\\cos C (ac\\cos B+cb\\cos A)=c^2$知$2a\\cos B\\cos C+2b \\cos A\\cos C=c$,\\\\\n又由正弦定理知 $2\\cos C(\\sin A\\cos B+\\sin B\\cos A)=\\sin C$\\\\\n$2\\cos C \\sin (A+B)=2\\cos C \\sin C=\\sin C$,其中$C\\in (0,\\pi), \\sin C>0$,\\\\\n故$\\cos C=\\dfrac{1}{2}$, $C\\in(0,\\pi)$, $C=\\dfrac{\\pi}{3}$.", "solution": "(1) 由 $S=\\dfrac{1}{2}ac\\sin B=\\dfrac{\\sqrt{3}}{4}ac=\\sqrt{3}$知$ac=4$\\\\\n由$a^2+c^2-b^2=2ac\\cos B$知$a^2+c^2=9$.\\\\\n结合两式得$(a-c)^2=1$, 故$a-c=\\pm 1$\\\\\n(2) 由$2\\cos C (ac\\cos B+cb\\cos A)=c^2$知$2a\\cos B\\cos C+2b \\cos A\\cos C=c$,\\\\\n又由正弦定理知 $2\\cos C(\\sin A\\cos B+\\sin B\\cos A)=\\sin C$\\\\\n$2\\cos C \\sin (A+B)=2\\cos C \\sin C=\\sin C$,其中$C\\in (0,\\pi), \\sin C>0$,\\\\\n故$\\cos C=\\dfrac{1}{2}$, $C\\in(0,\\pi)$, $C=\\dfrac{\\pi}{3}$.",
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"origin": "2023届高三下学期月考2试题18", "origin": "2023届高三下学期月考2试题18",
"edit": [ "edit": [
"20230507\t余利成" "20230507\t余利成"
@ -442190,7 +442424,20 @@
"ans": "(1) $p=60$, $q=40$, $x=100$, $y=100$; (2) $\\chi^2=8$, 有$95 \\%$的把握认为注射此种疫苗有效; (3) 分布为$\\begin{pmatrix}0&1&2&3&4\\\\\\dfrac{1}{210}&\\dfrac{4}{35}&\\dfrac{3}{7}&\\dfrac{8}{21}&\\dfrac{1}{14}\\end{pmatrix}$, $E[X]=\\dfrac{12}{5}$", "ans": "(1) $p=60$, $q=40$, $x=100$, $y=100$; (2) $\\chi^2=8$, 有$95 \\%$的把握认为注射此种疫苗有效; (3) 分布为$\\begin{pmatrix}0&1&2&3&4\\\\\\dfrac{1}{210}&\\dfrac{4}{35}&\\dfrac{3}{7}&\\dfrac{8}{21}&\\dfrac{1}{14}\\end{pmatrix}$, $E[X]=\\dfrac{12}{5}$",
"solution": "(1) $\\dfrac{p}{p+40}=\\dfrac{3}{5}$得$p=60$, $q=40$, $x=100$, $y=100$.\\\\\n(2) 原假设$H_0:$ 是否注射此种疫苗与是否感染病毒无关.\\\\\n$\\chi^2=\\dfrac{200\\times (40\\times 40-60\\times 60)^2}{100\\times 100\\times 100\\times 100}=8>3.841$\\\\\n故拒绝原假设即有$95 \\%$的把握认为注射此种疫苗有效.\\\\\n(3) 抽取$6$只未注射疫苗、$4$只注射疫苗的小白鼠.\\\\\n$P(X=0)=\\dfrac{C_6^0C_4^4}{C_{10}^4}=\\dfrac{1}{210}$;\\\\\n$P(X=1)=\\dfrac{C_6^1C_4^3}{C_{10}^4}=\\dfrac{4}{35}$;\\\\\n$P(X=2)=\\dfrac{C_6^2C_4^2}{C_{10}^4}=\\dfrac{3}{7}$;\\\\\n$P(X=3)=\\dfrac{C_6^3C_4^1}{C_{10}^4}=\\dfrac{8}{21}$;\\\\\n$P(X=4)=\\dfrac{C_6^4C_4^0}{C_{10}^4}=\\dfrac{1}{14}$.\\\\\n故$X$的分布为$\\begin{pmatrix}\n0&1&2&3&4\\\\\n\\dfrac{1}{210}&\\dfrac{4}{35}&\\dfrac{3}{7}&\\dfrac{8}{21}&\\dfrac{1}{14}\n\\end{pmatrix},$\\\\\n期望$E[X]=0\\times \\dfrac{1}{210}+1\\times \\dfrac{4}{35}+2\\times \\dfrac{3}{7}+3\\times \\dfrac{8}{21}+4\\times \\dfrac{1}{14}=\\dfrac{12}{5}$.", "solution": "(1) $\\dfrac{p}{p+40}=\\dfrac{3}{5}$得$p=60$, $q=40$, $x=100$, $y=100$.\\\\\n(2) 原假设$H_0:$ 是否注射此种疫苗与是否感染病毒无关.\\\\\n$\\chi^2=\\dfrac{200\\times (40\\times 40-60\\times 60)^2}{100\\times 100\\times 100\\times 100}=8>3.841$\\\\\n故拒绝原假设即有$95 \\%$的把握认为注射此种疫苗有效.\\\\\n(3) 抽取$6$只未注射疫苗、$4$只注射疫苗的小白鼠.\\\\\n$P(X=0)=\\dfrac{C_6^0C_4^4}{C_{10}^4}=\\dfrac{1}{210}$;\\\\\n$P(X=1)=\\dfrac{C_6^1C_4^3}{C_{10}^4}=\\dfrac{4}{35}$;\\\\\n$P(X=2)=\\dfrac{C_6^2C_4^2}{C_{10}^4}=\\dfrac{3}{7}$;\\\\\n$P(X=3)=\\dfrac{C_6^3C_4^1}{C_{10}^4}=\\dfrac{8}{21}$;\\\\\n$P(X=4)=\\dfrac{C_6^4C_4^0}{C_{10}^4}=\\dfrac{1}{14}$.\\\\\n故$X$的分布为$\\begin{pmatrix}\n0&1&2&3&4\\\\\n\\dfrac{1}{210}&\\dfrac{4}{35}&\\dfrac{3}{7}&\\dfrac{8}{21}&\\dfrac{1}{14}\n\\end{pmatrix},$\\\\\n期望$E[X]=0\\times \\dfrac{1}{210}+1\\times \\dfrac{4}{35}+2\\times \\dfrac{3}{7}+3\\times \\dfrac{8}{21}+4\\times \\dfrac{1}{14}=\\dfrac{12}{5}$.",
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"origin": "2023届高三下学期月考2试题19", "origin": "2023届高三下学期月考2试题19",
"edit": [ "edit": [
"20230507\t余利成" "20230507\t余利成"
@ -442210,7 +442457,20 @@
"ans": "(1) $\\dfrac 12$; (2) 证明略; (3) $\\dfrac 92$", "ans": "(1) $\\dfrac 12$; (2) 证明略; (3) $\\dfrac 92$",
"solution": "(1) 椭圆的离心率$e=\\dfrac{1}{2}$;\\\\\n(2) 证明: 当$x_0=2$时,$y_0=0,$ 过点$P$的椭圆$C$的切线方程为 $x=2$,符合$\\dfrac{x_0 x}{4}+\\dfrac{y_0 y}{3}=1$\\\\\n同理当$x_0=-2$时,也符合;\\\\\n当$x_0\\neq\\pm 2$时,设过点$P$的椭圆$C$的切线方程为$y-y_0=k(x-x_0)$($k$存在),\\\\\n联立$\\begin{cases}\ny-y_0=k(x-x_0),\\\\\n \\dfrac{x^2}{4}+\\dfrac{y^2}{3}=1\n\\end{cases}$得$(3+4k^2)x^2+8k(y-kx_0)x+4(y_0-kx_0)^2-12=0$,\\\\\n$\\Delta=0$得$(x_0^2-4)k^2-2x_0y_0k+y_0^2-3=0$,解得$k=\\dfrac{x_0y_0}{x_0^2-4}=\\dfrac{x_0y_0}{4(1-\\dfrac{y_0^2}{3})-4}=-\\dfrac{3x_0}{4y_0}.$\\\\\n故$y-y_0=-\\dfrac{3x_0}{4y_0}(x-x_0)$,即$\\dfrac{x_0 x}{4}+\\dfrac{y_0 y}{3}=1$.\\\\\n综上过点$P$的椭圆$C$的切线方程为$\\dfrac{x_0 x}{4}+\\dfrac{y_0 y}{3}=1$.\\\\\n(3) 设$A(x_1,y_1).B(x_2,y_2),x_1\\neq x_2,M(4,t)$.\\\\\n则切线$MA:\\dfrac{x_1x}{4}+\\dfrac{y_1y}{3}=1,$代入$(4,t)$得$x_1+\\dfrac{ty_1}{3}=1$,\\\\\n同理$x_1+\\dfrac{ty_1}{3}=1$\\\\\n故$A(x_1,y_1),B(x_2,y_2)$在直线$x+\\dfrac{ty}{3}=1$上,故直线$AB:x=-\\dfrac{ty}{3}+1$.\\\\\n联立$\\begin{cases}\nx=-\\dfrac{ty}{3}+1,\\\\\n\\dfrac{x^2}{4}+\\dfrac{y^2}{3}=1\n\\end{cases}$\n得\n$(4+\\dfrac{t^2}{3})y^2-2ty-9=0$, $\\Delta=16t^2+144>0$,\\\\\n$|AB|=\\sqrt{1+\\dfrac{t^2}{9}}\\cdot |y_1-y_2|=\\sqrt{1+\\dfrac{t^2}{9}}\\cdot \\dfrac{\\sqrt{16t^2+144}}{4+\\dfrac{t^2}{3}}$,\\\\\n$M$到直线$AB$的距离$d=\\dfrac{|4+\\dfrac{t^2}{3}-1|}{\\sqrt{1+\\dfrac{t^2}{9}}}=\\dfrac{3+\\dfrac{t^2}{3}}{\\sqrt{1+\\dfrac{t^2}{9}}}$,\\\\\n$\\triangle MAB$的面积$S=\\dfrac{1}{2}|AB|\\cdot d=\\dfrac{1}{2}\\cdot \\sqrt{1+\\dfrac{t^2}{9}}\\cdot \\dfrac{\\sqrt{16t^2+144}}{4+\\dfrac{t^2}{3}}\\cdot \\dfrac{3+\\dfrac{t^2}{3}}{\\sqrt{1+\\dfrac{t^2}{9}}}=\\dfrac{2(t^2+9)\\sqrt{t^2+9}}{t^2+12}$,\\\\\n令$\\lambda=\\sqrt{t^2+9}\\geq3, S=f(\\lambda)=\\dfrac{2\\lambda^3}{\\lambda^2+3},$\n则$f^{'} (\\lambda)=\\dfrac{2\\lambda^4+18\\lambda^2}{(\\lambda^2+3)^2}>0$,故$f(\\lambda)$在$[3,+\\infty)$严格增,$f(\\lambda)_{\\min}=f(3)=\\dfrac{9}{2}$,故$\\triangle MAB$的面积的最小值为$\\dfrac{9}{2},$ 此时$M$的坐标为$(4,0)$.", "solution": "(1) 椭圆的离心率$e=\\dfrac{1}{2}$;\\\\\n(2) 证明: 当$x_0=2$时,$y_0=0,$ 过点$P$的椭圆$C$的切线方程为 $x=2$,符合$\\dfrac{x_0 x}{4}+\\dfrac{y_0 y}{3}=1$\\\\\n同理当$x_0=-2$时,也符合;\\\\\n当$x_0\\neq\\pm 2$时,设过点$P$的椭圆$C$的切线方程为$y-y_0=k(x-x_0)$($k$存在),\\\\\n联立$\\begin{cases}\ny-y_0=k(x-x_0),\\\\\n \\dfrac{x^2}{4}+\\dfrac{y^2}{3}=1\n\\end{cases}$得$(3+4k^2)x^2+8k(y-kx_0)x+4(y_0-kx_0)^2-12=0$,\\\\\n$\\Delta=0$得$(x_0^2-4)k^2-2x_0y_0k+y_0^2-3=0$,解得$k=\\dfrac{x_0y_0}{x_0^2-4}=\\dfrac{x_0y_0}{4(1-\\dfrac{y_0^2}{3})-4}=-\\dfrac{3x_0}{4y_0}.$\\\\\n故$y-y_0=-\\dfrac{3x_0}{4y_0}(x-x_0)$,即$\\dfrac{x_0 x}{4}+\\dfrac{y_0 y}{3}=1$.\\\\\n综上过点$P$的椭圆$C$的切线方程为$\\dfrac{x_0 x}{4}+\\dfrac{y_0 y}{3}=1$.\\\\\n(3) 设$A(x_1,y_1).B(x_2,y_2),x_1\\neq x_2,M(4,t)$.\\\\\n则切线$MA:\\dfrac{x_1x}{4}+\\dfrac{y_1y}{3}=1,$代入$(4,t)$得$x_1+\\dfrac{ty_1}{3}=1$,\\\\\n同理$x_1+\\dfrac{ty_1}{3}=1$\\\\\n故$A(x_1,y_1),B(x_2,y_2)$在直线$x+\\dfrac{ty}{3}=1$上,故直线$AB:x=-\\dfrac{ty}{3}+1$.\\\\\n联立$\\begin{cases}\nx=-\\dfrac{ty}{3}+1,\\\\\n\\dfrac{x^2}{4}+\\dfrac{y^2}{3}=1\n\\end{cases}$\n得\n$(4+\\dfrac{t^2}{3})y^2-2ty-9=0$, $\\Delta=16t^2+144>0$,\\\\\n$|AB|=\\sqrt{1+\\dfrac{t^2}{9}}\\cdot |y_1-y_2|=\\sqrt{1+\\dfrac{t^2}{9}}\\cdot \\dfrac{\\sqrt{16t^2+144}}{4+\\dfrac{t^2}{3}}$,\\\\\n$M$到直线$AB$的距离$d=\\dfrac{|4+\\dfrac{t^2}{3}-1|}{\\sqrt{1+\\dfrac{t^2}{9}}}=\\dfrac{3+\\dfrac{t^2}{3}}{\\sqrt{1+\\dfrac{t^2}{9}}}$,\\\\\n$\\triangle MAB$的面积$S=\\dfrac{1}{2}|AB|\\cdot d=\\dfrac{1}{2}\\cdot \\sqrt{1+\\dfrac{t^2}{9}}\\cdot \\dfrac{\\sqrt{16t^2+144}}{4+\\dfrac{t^2}{3}}\\cdot \\dfrac{3+\\dfrac{t^2}{3}}{\\sqrt{1+\\dfrac{t^2}{9}}}=\\dfrac{2(t^2+9)\\sqrt{t^2+9}}{t^2+12}$,\\\\\n令$\\lambda=\\sqrt{t^2+9}\\geq3, S=f(\\lambda)=\\dfrac{2\\lambda^3}{\\lambda^2+3},$\n则$f^{'} (\\lambda)=\\dfrac{2\\lambda^4+18\\lambda^2}{(\\lambda^2+3)^2}>0$,故$f(\\lambda)$在$[3,+\\infty)$严格增,$f(\\lambda)_{\\min}=f(3)=\\dfrac{9}{2}$,故$\\triangle MAB$的面积的最小值为$\\dfrac{9}{2},$ 此时$M$的坐标为$(4,0)$.",
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"origin": "2023届高三下学期月考2试题20", "origin": "2023届高三下学期月考2试题20",
"edit": [ "edit": [
"20230507\t余利成" "20230507\t余利成"
@ -442230,7 +442490,20 @@
"ans": "(1) 证明略; (2) $3$; (3) 不存在, 证明略", "ans": "(1) 证明略; (2) $3$; (3) 不存在, 证明略",
"solution": "(1) $x_1=\\dfrac{1}{2},x_{n+1}=g(x_n)=\\dfrac{x_n}{x_{n+1}},$故$x_n>0,\\dfrac{1}{x_{n+1}}=\\dfrac{x_n+1}{x_n}=\\dfrac{1}{x_n}+1$,即$\\dfrac{1}{x_{n+1}}-\\dfrac{1}{x_{n}}=1$,\\\\因此数列$\\{\\dfrac{1}{x_n}\\}$是以$2$为首项,$1$为公差的等差数列.\\\\\n(2) 对任意$x>0$ 均有$f(x)-mg(x)=\\ln (x+1)-\\dfrac{mx}{x+1}+1>0,$\\\\\n令$h(x)=\\ln (x+1)-\\dfrac{mx}{x+1}+1,x>0,$则$h^{'}(x)=\\dfrac{x+1-m}{(x+1)^2}$.\\\\\n当$m\\geq4$时,$h(1)=\\ln 2-\\dfrac{m}{2}+1<2-\\dfrac{m}{2}<0,$这与$h(x)>0$对$x>0$恒成立矛盾;\\\\\n当$m=3$时,$h(x)=\\ln (x+1)-\\dfrac{3x}{x+1}+1,h^{'}(x)\\dfrac{x-2}{(x+1)^2}.h(x)$在$(0,2]$严格减,在$[2,+\\infty)$严格增,故$h(x)_{\\min}=h(2)=\\ln 3-2+1=\\ln 3-1>0$,符合题意.\\\\\n综上整数$m$的最大值为$3$.\\\\\n(3) 对任意正整数$t$,取$n=100t,$则 \\\\$\\displaystyle f(n-t)=f(99t)=\\ln (1+99t)=\\ln \\dfrac{1+99t}{99t}+\\ln \\dfrac{99t}{99t-1}+\\cdots +\\ln \\dfrac{2}{1}=\\sum_{k=1}^{99t}\\ln (1+\\dfrac{1}{k}).$\\\\\n$\\displaystyle n-\\sum_{k=1}^{100t}g(k)=\\sum_{k=1}^{100t}(1- g(k))=\\sum_{k=1}^{100t} \\dfrac{1}{1+k}=\\sum_{k=1}^{99t} \\dfrac{1}{1+k}+\\sum_{k=99t+1}^{100t} \\dfrac{1}{1+k}.$\\\\\n$\\displaystyle f(n-t)-[ n-\\sum_{k=1}^{100t}g(k)]=\\sum_{k=1}^{99t} (\\ln (1+\\dfrac{1}{k})-\\dfrac{1}{1+k})-\\sum_{k=99t+1}^{100t} \\dfrac{1}{1+k}.$\\\\\n令$H(x)=\\ln (x+1)-\\dfrac{x}{x+1},x>0$,则$H^{'}(x)=\\dfrac{x}{(x+1)^2}>0$恒成立,故$H(x)$在$(0,+\\infty)$严格增,$H(x)>H(0)=0$,故$\\ln (1+x)>\\dfrac{x}{x+1}$对$x>0$恒成立.\\\\\n因此$\\ln (1+\\dfrac{1}{k})>\\dfrac{\\dfrac{1}{k}}{1+\\dfrac{1}{k}}=\\dfrac{1}{1+k}$,\\\\\n$\\displaystyle f(n-t)-[ n-\\sum_{k=1}^{100t}g(k)]=\\sum_{k=1}^{99t} (\\ln (1+\\dfrac{1}{k})-\\dfrac{1}{1+k})-\\sum_{k=99t+1}^{100t} \\dfrac{1}{1+k}$\\\\\n$\\displaystyle>\\sum_{k=2}^{99t} (\\ln (1+\\dfrac{1}{k})-\\dfrac{1}{1+k})+(\\ln 2-\\dfrac{1}{2})- \\dfrac{t}{1+99t+1}>\\ln 2-\\dfrac{1}{2}-\\dfrac{1}{99+\\dfrac{2}{t}}>0.1-\\dfrac{1}{99}>0.$\\\\\n因此不存在正整数$t$使得对任意$n \\in \\mathbf{N}$, $n \\geq t$, 都有$\\displaystyle f(n-t)<n-\\sum_{k=1}^n g(k)$成立.", "solution": "(1) $x_1=\\dfrac{1}{2},x_{n+1}=g(x_n)=\\dfrac{x_n}{x_{n+1}},$故$x_n>0,\\dfrac{1}{x_{n+1}}=\\dfrac{x_n+1}{x_n}=\\dfrac{1}{x_n}+1$,即$\\dfrac{1}{x_{n+1}}-\\dfrac{1}{x_{n}}=1$,\\\\因此数列$\\{\\dfrac{1}{x_n}\\}$是以$2$为首项,$1$为公差的等差数列.\\\\\n(2) 对任意$x>0$ 均有$f(x)-mg(x)=\\ln (x+1)-\\dfrac{mx}{x+1}+1>0,$\\\\\n令$h(x)=\\ln (x+1)-\\dfrac{mx}{x+1}+1,x>0,$则$h^{'}(x)=\\dfrac{x+1-m}{(x+1)^2}$.\\\\\n当$m\\geq4$时,$h(1)=\\ln 2-\\dfrac{m}{2}+1<2-\\dfrac{m}{2}<0,$这与$h(x)>0$对$x>0$恒成立矛盾;\\\\\n当$m=3$时,$h(x)=\\ln (x+1)-\\dfrac{3x}{x+1}+1,h^{'}(x)\\dfrac{x-2}{(x+1)^2}.h(x)$在$(0,2]$严格减,在$[2,+\\infty)$严格增,故$h(x)_{\\min}=h(2)=\\ln 3-2+1=\\ln 3-1>0$,符合题意.\\\\\n综上整数$m$的最大值为$3$.\\\\\n(3) 对任意正整数$t$,取$n=100t,$则 \\\\$\\displaystyle f(n-t)=f(99t)=\\ln (1+99t)=\\ln \\dfrac{1+99t}{99t}+\\ln \\dfrac{99t}{99t-1}+\\cdots +\\ln \\dfrac{2}{1}=\\sum_{k=1}^{99t}\\ln (1+\\dfrac{1}{k}).$\\\\\n$\\displaystyle n-\\sum_{k=1}^{100t}g(k)=\\sum_{k=1}^{100t}(1- g(k))=\\sum_{k=1}^{100t} \\dfrac{1}{1+k}=\\sum_{k=1}^{99t} \\dfrac{1}{1+k}+\\sum_{k=99t+1}^{100t} \\dfrac{1}{1+k}.$\\\\\n$\\displaystyle f(n-t)-[ n-\\sum_{k=1}^{100t}g(k)]=\\sum_{k=1}^{99t} (\\ln (1+\\dfrac{1}{k})-\\dfrac{1}{1+k})-\\sum_{k=99t+1}^{100t} \\dfrac{1}{1+k}.$\\\\\n令$H(x)=\\ln (x+1)-\\dfrac{x}{x+1},x>0$,则$H^{'}(x)=\\dfrac{x}{(x+1)^2}>0$恒成立,故$H(x)$在$(0,+\\infty)$严格增,$H(x)>H(0)=0$,故$\\ln (1+x)>\\dfrac{x}{x+1}$对$x>0$恒成立.\\\\\n因此$\\ln (1+\\dfrac{1}{k})>\\dfrac{\\dfrac{1}{k}}{1+\\dfrac{1}{k}}=\\dfrac{1}{1+k}$,\\\\\n$\\displaystyle f(n-t)-[ n-\\sum_{k=1}^{100t}g(k)]=\\sum_{k=1}^{99t} (\\ln (1+\\dfrac{1}{k})-\\dfrac{1}{1+k})-\\sum_{k=99t+1}^{100t} \\dfrac{1}{1+k}$\\\\\n$\\displaystyle>\\sum_{k=2}^{99t} (\\ln (1+\\dfrac{1}{k})-\\dfrac{1}{1+k})+(\\ln 2-\\dfrac{1}{2})- \\dfrac{t}{1+99t+1}>\\ln 2-\\dfrac{1}{2}-\\dfrac{1}{99+\\dfrac{2}{t}}>0.1-\\dfrac{1}{99}>0.$\\\\\n因此不存在正整数$t$使得对任意$n \\in \\mathbf{N}$, $n \\geq t$, 都有$\\displaystyle f(n-t)<n-\\sum_{k=1}^n g(k)$成立.",
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"origin": "2023届高三下学期月考2试题21", "origin": "2023届高三下学期月考2试题21",
"edit": [ "edit": [
"20230507\t余利成" "20230507\t余利成"