diff --git a/工具/latex界面修改题目内容.py b/工具/latex界面修改题目内容.py index c3798bdc..26ac0501 100644 --- a/工具/latex界面修改题目内容.py +++ b/工具/latex界面修改题目内容.py @@ -1,6 +1,6 @@ import os,re,json """这里编辑题号(列表)后将在vscode中打开窗口, 编辑后保存关闭""" -problems = "014805,014806,014807,014808,014809,014810,014811,014812,014813,014814,014815,014816,014817,014818,014819,014820,014821,014822,014823,014824,014825,030608,030632,030636,030679,030715,030757,030836,030856,030895,030927,030955,030977,031022,031040,031115,031136,031149" +problems = "013535" editor = "王伟叶" def generate_number_set(string,dict): diff --git a/工具/关键字筛选题号.py b/工具/关键字筛选题号.py index d3d43565..56070979 100644 --- a/工具/关键字筛选题号.py +++ b/工具/关键字筛选题号.py @@ -2,7 +2,7 @@ import os,re,json """---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---""" keywords_dict_table = [ - {"origin":["2025"],"origin2":["校本"],"origin3":["高一下"]} + {"content":["阿基米德"]} ] """---关键字设置完毕---""" # 示例： keywords_dict_table = [ diff --git a/工具/工具面板.py b/工具/工具面板.py index 559d0eef..4416e230 100644 --- a/工具/工具面板.py +++ b/工具/工具面板.py @@ -101,6 +101,8 @@ def run_command1(): call(["python","latex界面修改题目内容.py"]) elif selectedtool == "答案metadata生成": call(["python","答案metadata生成.py"]) + elif selectedtool == "错题重做来源清点": + call(["python","错题重做来源清点.py"]) LabelTool.config(text = selectedtool+"STEP1命令执行完毕") button1.place_forget() @@ -168,6 +170,7 @@ BKZMenu.add_command(label = "试卷答案生成", command = lambda: SetButton(" BKZMenu.add_command(label = "题号清单生成", command = lambda: SetButton("题号清单生成",1,["题号清单生成.py"])) BKZMenu.add_command(label = "已用题号剔除", command = lambda: SetButton("已用题号剔除",1,["已用题号剔除.py"])) BKZMenu.add_command(label = "寻找未赋答案题目", command = lambda: SetButton("寻找未赋答案题目",1,["寻找未赋答案题目.py"])) +BKZMenu.add_command(label = "错题重做来源清点", command = lambda: SetButton("错题重做来源清点",1,["错题重做来源清点.py"])) # 设置 目标及标签 菜单项 ObjTagMenu = Menu(menubar, tearoff = False) diff --git a/工具/批量收录题目.py b/工具/批量收录题目.py index f17c5ecf..728cf811 100644 --- a/工具/批量收录题目.py +++ b/工具/批量收录题目.py @@ -1,8 +1,8 @@ #修改起始id,出处,文件名 -starting_id = 14847 -raworigin = "" +starting_id = 14920 +raworigin = "2022届空中课堂学科精要名师点拨-" filename = r"C:\Users\weiye\Documents\wwy sync\临时工作区\自拟题目6.tex" -editor = "202304011\t王伟叶" +editor = "202304012\t王伟叶" indexed = False import os,re,json diff --git a/工具/文本文件/题号筛选.txt b/工具/文本文件/题号筛选.txt index c06c70e8..d094bd2f 100644 --- a/工具/文本文件/题号筛选.txt +++ b/工具/文本文件/题号筛选.txt @@ -1 +1 @@ -021441,021442,021443,021444,021445,021446,021447,021448,021449,021450,021451,021452,021453,021454,021455,021456,021457,021458,021459,021460,021461,021462,021463,021464,021465,021466,021467,021468,021469,021470,021471,021472,021473,021474,021475,021476,021477,021478,021479,021480,021481,021482,021483,021484,021485,021486,021487,021488,021489,021490,021491,021492,021493,021494,021495,021496,021497,021498,021499,021500,021501,021502,021503,021504,021505,021506,021507,021508,021509,021510,021511,021512,021513,021514,021515,021516,021517,021518,021519,021520,021521,021522,021523,021524,021525,021526,021527,021528,021529,021530,021531,021532,021533,021534,021535,021536,021537,021538,021539,021540,021541,021542,021543,021544,021545,021546,021547,021548,021549,021550,021551,021552,021553,021554,021555,021556,021557,021558,021559,021560,021561,021562,021563,021564,021565,021566,021567,021568,021569,021570,021571,021572,021573,021574,021575,021576,021577,021578,021579,021580,021581,021582,021583,021584,021585,021586,021587,021588,021589,021590,021591,021592,021593,021594,021595,021596,021597,021598,021599,021600,021601,021602,021603,021604,021605,021606,021607,021608,021609,021610,021611,021612,021613,021614,021615,021616,021617,021618,021619,021620,021621,021622,021623,021624,021625,021626,021627,021628,021629,021630,021631,021632,021633,021634,021635,021636,021637,021638,021639,021640,021641,021642,021643,021644,021645,021646,021647,021648,021649,021650,021651,021652,021653,021654,021655,021656,021657,021658,021659,021660,021661,021662,021663,021664,021665,021666,021667,021668,021669,021670,021671,021672,021673,021674,021675,021676,021677,021678,021679,021680,021681,021682,021683,021684,021685,021686,021687,021688,021689,021690,021691,021692,021693,021694,021695,021696,021697,021698,021699,021700,021701,021702,021703,021704,021705,021706,021707,021708,021709,021710,021711,021712,021713,021714,021715,021716,021717,021718,021719,021720,021721,021722,021723,021724,021725,021726,021727,021728,021729,021730,021731,021732,021733,021734,021735,021736,021737,021738,021739,021740,021741,021742,021743,021744,021745,021746,021747,021748,021749,021750,021751,021752,021753,021754,021755,021756,021757,021758,021759,021760,021761,021762,021763,021764,021765,021766,021767,021768,021769,021770,021771,021772,021773,021774,021775,021776,021777,021778,021779,021780,021781,021782,021783,021784,021785,021786,021787,021788,021789,021790,021791,021792,021793,021794,021795,021796,021797,021798,021799,021800,021801,021802,021803,021804,021805,021806,021807,021808,021809,021810,021811,021812,021813,021814,021815,021816,021817,021818,021819,021820,021821,021822,021823,021824,021825,021826,021827,021828,021829,021830,021831,021832,021833,021834,021835,021836,021837,021838,021839,021840,021841,021842,021843,021844,021845,021846,021847,021848,021849,021850,021851,021852,021853,021854,021855,021856,021857,021858,021859,021860,021861,021862,021863,021864,021865,021866,021867,021868,021869,021870,021871,021872,021873,021874,021875,021876,021877,021878,021879,021880,021881,021882,021883,021884,021885,021886,021887,021888,021889,021890,021891,021892,021893,021894,021895,021896,021897,021898,021899,021900,021901,021902,021903,021904,021905,021906,021907,021908,021909,021910,021911,021912,021913,021914,021915,021916,021917,021918,021919,021920,021921,021922,021923,021924,021925,021926,021927,021928,021929,021930,021931,021932,021933,021934,021935,021936,021937,021938,021939,021940,021941,021942,021943,021944,021945,021946,021947,021948,021949,021950,021951,021952,021953,021954,021955,021956,021957,021958,021959,021960,021961,021962,021963,021964,021965,021966,021967,021968,021969,021970,021971,021972,021973,021974,021975,021976,021977,021978,021979,021980,021981,021982,021983,021984,021985,021986,021987,021988,021989,021990,021991,021992,021993,021994,021995,021996,021997,021998,021999,022000,022001,022002,022003,022004,022005,022006,022007,022008,022009,022010,022011,022012,022013,022014,022015,022016,022017,022018,022019,022020,022021,022022,022023,022024,022025,022026,022027,022028,022029,022030,022031,022032,022033,022034,022035,022036,022037,022038,022039,022040,022041,022042,022043,022044,022045,022046,022047 \ No newline at end of file +013535 \ No newline at end of file diff --git a/工具/讲义生成.py b/工具/讲义生成.py index 6964c1e7..4e72fe28 100644 --- a/工具/讲义生成.py +++ b/工具/讲义生成.py @@ -5,12 +5,12 @@ import os,re,json,time,sys """2: 测验卷与周末卷(填空题, 选择题, 解答题)""" """3: 日常选题讲义(一个section)""" -paper_type = 1 # 随后设置一下后续的讲义标题 +paper_type = 3 # 随后设置一下后续的讲义标题 """---设置题块编号---""" problems = [ -"332,401,654,2605,2664,3574,3640,31196,31320,4584,7361,7423,30227,30275,30540","340,412,2586,3585,4575,11993,7630,14091,31158,7502,10868,30495,30520" +"4612:4618" ] @@ -30,7 +30,7 @@ elif paper_type == 2: elif paper_type == 3: 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) template_file = "模板文件/日常选题讲义模板.txt" #设置模板文件名 - exec_list = [("标题文字待处理","第四讲")] #设置讲义标题 + exec_list = [("标题文字待处理","07超几何分布")] #设置讲义标题 destination_file = "临时文件/"+exec_list[0][1] # 设置输出文件名 diff --git a/工具/错题重做来源清点.py b/工具/错题重做来源清点.py new file mode 100644 index 00000000..715dee54 --- /dev/null +++ b/工具/错题重做来源清点.py @@ -0,0 +1,78 @@ +import os,re,json + +outputfile = "临时文件/已做题目.txt" +new_dir = r"C:\Users\weiye\Documents\wwy sync\23届\四月错题重做" + +old_dirs = [ +r"C:\Users\weiye\Documents\wwy sync\23届\暑假概率初步续", +r"C:\Users\weiye\Documents\wwy sync\23届\上学期测验卷", +r"C:\Users\weiye\Documents\wwy sync\23届\上学期周末卷", +r"C:\Users\weiye\Documents\wwy sync\23届\第一轮复习讲义", +r"C:\Users\weiye\Documents\wwy sync\23届\其他讲义与试卷", +r"C:\Users\weiye\Documents\wwy sync\23届\赋能", +r"C:\Users\weiye\Documents\wwy sync\23届\正态分布及成对数据新课", +r"C:\Users\weiye\Documents\wwy sync\23届\下学期测验卷", +r"C:\Users\weiye\Documents\wwy sync\23届\下学期周末卷", +r"C:\Users\weiye\Documents\wwy sync\23届\第二轮复习讲义" + +] + +def id_related(id,prodict,idrespdict): + for rid in prodict[id]["related"]: + if rid in idrespdict: + return True + return False + +with open(r"../题库0.3/Problems.json","r",encoding="u8") as f: + data = f.read() +pro_dict = json.loads(data) + +id_resp_dict = {} + +for dir in old_dirs: + for texfile in os.listdir(dir): + if ".tex" in texfile: + with open(os.path.join(dir,texfile),"r",encoding="u8") as f: + data = f.read() + ids = re.findall(r"\((\d{6})\)",data) + for id in ids: + if not id in id_resp_dict: + id_resp_dict[id] = [os.path.join(os.path.split(dir)[1],texfile[:-4])+"--题序"+str(ids.index(id)+1)] + else: + id_resp_dict[id].append(os.path.join(os.path.split(dir)[1],texfile[:-4])+"--题序"+str(ids.index(id)+1)) + +new_texfiles = [f for f in os.listdir(new_dir) if ".tex" in f] + +output = "" + +for texfile in new_texfiles: + with open(os.path.join(new_dir,texfile),"r",encoding="u8") as f: + data = f.read() + ids = re.findall(r"\((\d{6})\)",data) + output += texfile[:-4] + "\n" + print(texfile[:-4]) + output += ",".join(ids) + "\n" + print(",".join(ids)) + for id in ids: + if id in id_resp_dict: + print("%s\t(已做题)\t%s"%(id,",".join(id_resp_dict[id]))) + output += "%s\t(已做题)\t%s\n"%(id,",".join(id_resp_dict[id])) + elif id_related: + rel_ids = pro_dict[id]["related"] + origins = [] + for id in rel_ids: + if id in id_resp_dict: + origins += id_resp_dict[id] + print("%s\t(关联题)\t%s"%(id,",".join(origins))) + output += "%s\t(关联题)\t%s\n"%(id,",".join(origins)) + else: + print("%s\t(无关题)") + output += "%s\t(无关题)\n" + print("") + output += "\n" + +with open(outputfile,"w",encoding="u8") as f: + f.write(output) + + + diff --git a/工具/题号清单生成.py b/工具/题号清单生成.py index d0bcdf9e..ee3a3716 100644 --- a/工具/题号清单生成.py +++ b/工具/题号清单生成.py @@ -1,6 +1,8 @@ import os,re "---此处输入文件夹列表---" -directories = [r"C:\Users\weiye\Documents\wwy sync\23届\上学期测验卷", +directories = [ +r"C:\Users\weiye\Documents\wwy sync\23届\暑假概率初步续", +r"C:\Users\weiye\Documents\wwy sync\23届\上学期测验卷", r"C:\Users\weiye\Documents\wwy sync\23届\上学期周末卷", r"C:\Users\weiye\Documents\wwy sync\23届\第一轮复习讲义", r"C:\Users\weiye\Documents\wwy sync\23届\赋能", @@ -11,7 +13,7 @@ r"C:\Users\weiye\Documents\wwy sync\23届\下学期测验卷", r"C:\Users\weiye\Documents\wwy sync\23届\下学期周末卷", r"C:\Users\weiye\Documents\wwy sync\23届\第二轮复习讲义", r"C:\Users\weiye\Documents\wwy sync\23届\四月错题重做", -r"d:\mathdeptv2\工具\临时文件" + ] "---文件夹列表输入结束---" diff --git a/工具/题号选题pdf生成.py b/工具/题号选题pdf生成.py index 903a5a90..becd9fa2 100644 --- a/工具/题号选题pdf生成.py +++ b/工具/题号选题pdf生成.py @@ -7,13 +7,13 @@ import os,re,time,json,sys """---设置题目列表---""" #留空为编译全题库, a为读取文本文件中的题号筛选.txt文件生成题库 problems = r""" -000778,001253,001325,003747,013721,031392,010938,011148,004157,010197,030060,001339,002918,030327,004009,001262,004435,030398,030051,012859,031321,012543,001352,005650,005123,030337,007939,002968,012902,030356 +4572:4618 """ """---设置题目列表结束---""" """---设置文件名---""" #目录和文件的分隔务必用/ -filename = "临时文件/易错题1原结果" +filename = "临时文件/临时" """---设置文件名结束---""" """---设置是否需要解答题的空格---""" diff --git a/题库0.3/Problems.json b/题库0.3/Problems.json index 64b9efdc..5b130351 100644 --- a/题库0.3/Problems.json +++ b/题库0.3/Problems.json @@ -367361,6 +367361,1450 @@ "remark": "", "space": "12ex" }, + "014920": { + "id": "014920", + "content": "已知点$P(2,1)$和圆$C: x^2+y^2=4$, 则过点$P$的圆$C$的切线方程为\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-分类讨论化繁为简", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014921": { + "id": "014921", + "content": "已知数列$\\{a_n\\}$的前$n$项和为$S_n$, 且$S_n=n^2-n+1$, 则数列$\\{a_n\\}$的通项公式可以为\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-分类讨论化繁为简", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014922": { + "id": "014922", + "content": "设常数$a \\geq 0$, 函数$f(x)=\\dfrac{2^x+a}{2^x-a}$, 根据$a$的不同取值, 讨论函数$y=f(x)$的奇偶性, 并说明理由.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-分类讨论化繁为简", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014923": { + "id": "014923", + "content": "直线$l$过点$(1,-2)$, 且与抛物线$y^2=2 x$只有一个公共点, 求直线$l$的方程.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-分类讨论化繁为简", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014924": { + "id": "014924", + "content": "如果一条直线与一个平面垂直, 那么称此直线与平面构成一个``正交线面对''. 在一个正方体中, 由两个顶点确定的直线与含有四个顶点的平面构成的``正交线面对''的个数是\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-分类讨论化繁为简", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014925": { + "id": "014925", + "content": "规定$\\mathrm{C}_x^m=\\dfrac{x(x-1) \\cdots(x-m+1)}{m !}$, 其中$x \\in \\mathbf{R}$, $m$是正整数, 且$\\mathrm{C}_x^0=1$, 这是组合数$\\mathrm{C}_n^m$($n, m$是正整数, 且$m \\leq n$)的一种推广. 已知组合数$\\mathrm{C}_n^m$是正整数, 证明: 当$x \\in \\mathbf{Z}$, $m$是正整数时, $\\mathrm{C}_x^m \\in \\mathbf{Z}$.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-分类讨论化繁为简", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014926": { + "id": "014926", + "content": "实数$a, b$满足$a b>0$且$a \\neq b$, 由$a, b, \\dfrac{a+b}{2}, \\sqrt{a b}$按一定顺序构成的数列\\bracket{20}.\n\\twoch{可能是等差数列, 也可能是等比数列}{可能是等差数列, 但不可能是等比数列}{不可能是等差数列, 但可能是等比数列}{不可能是等差数列, 也不可能是等比数列}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-分类讨论化繁为简", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014927": { + "id": "014927", + "content": "设函数$f(x)=a x^2-2 x+2$, 对于满足$10$, 求实数$a$的取值范围.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-分类讨论化繁为简", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014928": { + "id": "014928", + "content": "已知数列$\\{a_n\\}$的前$n$项和为$S_n=-\\dfrac{3}{2} n^2+\\dfrac{27}{2} n$, 设$T_n=|a_1|+|a_2|+\\cdots+|a_n|$, 求$T_n$.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-分类讨论化繁为简", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014929": { + "id": "014929", + "content": "若函数$f(x)=\\begin{cases}2^x-a, & x \\leq 1, \\\\ (x-a)(x-2 a), & x>1\\end{cases}$恰有两个零点, 则实数$a$的取值范围是\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-分类讨论化繁为简", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014930": { + "id": "014930", + "content": "我们称点$P$到图形$C$上任意一点距离的最小值为点$P$到图形$C$的距离, 那么平面内到定圆$C$的距离与到定点$A$的距离相等的点的轨迹不可能是\\bracket{20}.\n\\fourch{圆}{椭圆}{双曲线的一支}{直线}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-分类讨论化繁为简", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014931": { + "id": "014931", + "content": "在集合$U=\\{a, b, c, d\\}$的子集中选出$4$个不同的子集, 需同时满足以下两个条件: \\textcircled{1} $\\varnothing, U$都要选出; \\textcircled{2} 对选出的任意两个子集$A$和$B$, 必有$A \\subseteq B$或$B \\subseteq A$. 那么共有\\blank{50}种不同的选法.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-分类讨论化繁为简", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014932": { + "id": "014932", + "content": "给定常数$c>0$, 定义函数$f(x)=2|x+c+4|-|x+c|$, 数列$a_1, a_2, a_3, \\cdots$满足$a_{n+1}=f(a_n)$, $n \\in \\mathbf{N}$, $n \\ge 1$.\\\\\n(1) 求证: 对任意正整数$n$, $a_{n+1}-a_n \\geq c$;\\\\ \n(2) 是否存在$a_1$, 使得$a_1, a_2, a_3, \\cdots, a_n, \\cdots$成等差数列? 若存在, 求出所有这样的$a_1$; 若不存在, 说明理由.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-分类讨论化繁为简", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014933": { + "id": "014933", + "content": "是否存在第一象限的角$\\alpha$和第三象限的角$\\beta$, 使得$\\tan \\alpha \\tan \\beta=\\tan (\\alpha-\\beta)$? 请说明理由.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-向数学家曹冲学转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014934": { + "id": "014934", + "content": "是否存在第二象限的角$\\alpha$和第四象限的角$\\beta$, 使得$\\tan \\alpha \\tan \\beta=\\tan (\\alpha-\\beta)$? 请说明理由.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-向数学家曹冲学转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014935": { + "id": "014935", + "content": "是否存在第一象限的角$\\alpha$和第三象限的角$\\beta$, 使得$\\sin \\alpha \\sin \\beta=\\sin (\\alpha-\\beta)$? 请说明理由.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-向数学家曹冲学转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014936": { + "id": "014936", + "content": "设$n \\in \\mathbf{N}$, $n \\ge 1$, 数列$\\{a_n\\}$、$\\{b_n\\}$满足: $a_n$为$(x+4)^n-(x+1)^n$的展开式中各项系数之和, $b_n=[\\dfrac{a_1}{5}]+[\\dfrac{2 a_2}{5^2}]+\\cdots+[\\dfrac{n a_n}{5^n}]$($[x]$表示不超过实数$x$的最大整数), 则当$n$变化时, $b_n-5 n$的最小值为\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-向数学家曹冲学转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014937": { + "id": "014937", + "content": "如图, 棱长为$2$的正方体$ABCD-A_1B_1C_1D_1$中, $E$为棱$CC_1$的中点, 点$P$、$Q$分别为面$A_1B_1C_1D_1$和线段$B_1C$上的动点, 则$\\triangle PEQ$周长的最小值为\\bracket{20}.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\def\\l{2}\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,-\\l) node [right] {$C$} coordinate (C);\n\\draw (A) ++ (0,0,-\\l) node [left] {$D$} coordinate (D);\n\\draw (A) -- (B) -- (C);\n\\draw [dashed] (A) -- (D) -- (C);\n\\draw (A) ++ (0,\\l,0) node [left] {$A_1$} coordinate (A1);\n\\draw (B) ++ (0,\\l,0) node [right] {$B_1$} coordinate (B1);\n\\draw (C) ++ (0,\\l,0) node [above right] {$C_1$} coordinate (C1);\n\\draw (D) ++ (0,\\l,0) node [above left] {$D_1$} coordinate (D1);\n\\draw (A1) -- (B1) -- (C1) -- (D1) -- cycle;\n\\draw (A) -- (A1) (B) -- (B1) (C) -- (C1);\n\\draw [dashed] (D) -- (D1);\n\\draw ($(C)!0.5!(C1)$) node [right] {$E$} coordinate (E);\n\\draw ($(B1)!0.3!(C)$) node [below] {$Q$} coordinate (Q);\n\\draw ($1/3*(A1)+1/3*(B1)+1/3*(C1)$) node [left] {$P$} coordinate (P);\n\\draw (B1)--(C)(E)--(Q);;\n\\draw [dashed] (Q)--(P)--(E);\n\\end{tikzpicture}\n\\end{center}\n\\fourch{$2 \\sqrt{2}$}{$\\sqrt{10}$}{$\\sqrt{11}$}{$2 \\sqrt{3}$}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-向数学家曹冲学转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014938": { + "id": "014938", + "content": "已知椭圆$\\Gamma: \\dfrac{x^2}{2}+y^2=1$的右焦点为$F_2$, 过$F_2$作两条不重合的动直线$l_1$、$l_2$, 其中$l_1$与$\\Gamma$交于$A$、$B$两点, $l_2$与$\\Gamma$交于$C$、$D$两点, $M$、$N$分别是线段$AB$、$CD$的中点. 若直线$MN$过定点$(\\dfrac{2}{3}, 0)$, 试问$l_1$与$l_2$的夹角是否为定值? 如果是, 求出该定值; 如果不是, 请说明理由.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-向数学家曹冲学转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014939": { + "id": "014939", + "content": "如图, 已知椭圆$\\Gamma: \\dfrac{x^2}{2}+y^2=1$的右焦点为$F_2$. 过点$F_2$作互相垂直且与$x$轴不重合的两条直线, 分别交$\\Gamma$于$A$、$B$两点和$C$、$D$两点, $M$、$N$分别是线段$AB$、$CD$的中点. 求证: 直线$MN$过定点.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw [->] (-2,0) -- (2,0) node [below] {$x$};\n\\draw [->] (0,-1.5) -- (0,1.5) node [left] {$y$};\n\\draw (0,0) node [below left] {$O$};\n\\path [draw, name path = elli] (0,0) ellipse ({sqrt(2)} and 1);\n\\path [domain = 0:2, name path = AB] plot (\\x,{2*(\\x-1)});\n\\path [domain = -2:2, name path = CD] plot (\\x,{-0.5*(\\x-1)});\n\\path [name intersections = {of = AB and elli, by = {A,B}}];\n\\path [name intersections = {of = CD and elli, by = {D,C}}];\n\\draw (A) node [below] {$A$}-- (B) node [above] {$B$};\n\\draw (C) node [below] {$C$} -- (D) node [above] {$D$};\n\\draw ($(A)!0.5!(B)$) node [below] {$M$} coordinate (M);\n\\draw ($(C)!0.5!(D)$) node [above] {$N$} coordinate (N);\n\\draw (M)--(N);\n\\draw (1,0) node [above] {$F_2$} coordinate (F_2);\n\\foreach \\i in {A,B,C,D,M,N,F_2}\n{\\filldraw (\\i) circle (0.03);};\n\\end{tikzpicture}\n\\end{center}", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-向数学家曹冲学转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014940": { + "id": "014940", + "content": "设定义在$\\mathbf{R}$上的函数$f(x)$满足: 对于任意的$x_1$、$x_2 \\in \\mathbf{R}$, 当$x_1 \\neq x_2$时, 都有$\\dfrac{f(x_1)-f(x_2)}{x_1-x_2} \\geq 0$, 且存在$u \\neq v$, 使得$f(u) \\neq f(v)$. 求证: $f(x)$不是周期函数.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-向数学家曹冲学转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014941": { + "id": "014941", + "content": "已知函数$f(x)=2^x+x$的反函数为$y=f^{-1}(x)$, 若$a, b, c$依次成公差大于零的等差数列, 且各项均在函数$f(x)$的值域中, 求证: $2 f^{-1}(b)>f^{-1}(a)+f^{-1}(c)$.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-向数学家曹冲学转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014942": { + "id": "014942", + "content": "已知函数$f(x)=3 \\cos x$, 则函数$y=f(2 x)-f(x)$的值域是\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-向数学家曹冲学转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014943": { + "id": "014943", + "content": "若关于$x$的不等式$|2^x-m|-\\dfrac{1}{2^x}<0$在区间$[0,1]$内恒成立, 则实数$m$的取值范围为\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-向数学家曹冲学转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014944": { + "id": "014944", + "content": "在$xOy$平面上, 将曲线$\\dfrac{x^2}{9}-\\dfrac{y^2}{16}=1$($x>0$)、直线$y=\\dfrac{4}{3} x$、直线$y=0$和直线$y=4$围成的封闭图形记为$D$, 记$D$绕$y$轴旋转一周所得的几何体为$\\Omega$, 利用祖暅原理得出$\\Omega$的体积为\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-向数学家曹冲学转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014945": { + "id": "014945", + "content": "已知数列$\\{a_n\\}$满足$a_1=1$, $a_2=3$, 若$|a_{n+1}-a_n|=2^n$($n \\in \\mathbf{N}$, $n\\ge 1$), 且$\\{a_{2 n-1}\\}$是递增数列、$\\{a_{2 n}\\}$是递减数列, 求$\\displaystyle\\lim_{n\\to\\infty} \\dfrac{a_{2 n-1}}{a_{2 n}}$.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-向数学家曹冲学转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014946": { + "id": "014946", + "content": "已知曲线$C_1: \\dfrac{x^2}{4}-\\dfrac{y^2}{3}=1$和曲线$C_2: |y|=|x|+2$, 求证: 不存在过点$(1,0)$的直线, 同时与$C_1, C_2$都有公共点.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-向数学家曹冲学转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014947": { + "id": "014947", + "content": "若关于$x$的方程$|x^2-2 a x-3|=4 a$有且仅有两个实数解, 则实数$a$的取值范围是\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-活用函数与方程阡陌变通途", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014948": { + "id": "014948", + "content": "设$a_1, d$为实数, 且首项为$a_1$、公差为$d$的等差数列$\\{a_n\\}$的前$n$项和$S_n$满足$S_5S_6+15=0$, 则$d$的取值范围是\\blank{50}", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-活用函数与方程阡陌变通途", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014949": { + "id": "014949", + "content": "若关于$x$的方程$\\cos 2 x-2 \\cos x+m=0$有实数解, 则实数$m$的取值范围是\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-活用函数与方程阡陌变通途", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014950": { + "id": "014950", + "content": "若关于$x$的方程$\\cos 2 x-2 \\cos x+m=0$在$[0, \\pi]$内有且仅有两个解, 则实数$m$的取值范围是\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-活用函数与方程阡陌变通途", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014951": { + "id": "014951", + "content": "若关于$x$的不等式$\\cos 2 x-2 \\cos x+m>0$恒成立, 则实数$m$的取值范围是\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-活用函数与方程阡陌变通途", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014952": { + "id": "014952", + "content": "若$\\{x | \\cos 2 x-2 \\cos x+m>0, \\ x \\in \\mathbf{R}\\} \\neq \\varnothing$, 则实数$m$的取值范围是\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-活用函数与方程阡陌变通途", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014953": { + "id": "014953", + "content": "已知函数$f(x)=\\dfrac{4 x+a}{x^2+1}$的值域为$[-1, b]$, 求实数$a, b$的值.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-活用函数与方程阡陌变通途", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014954": { + "id": "014954", + "content": "已知函数$f(x)=\\sqrt{x^2-1}-a x+1$有且仅有两个零点, 则实数$a$的取值范围是\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-活用函数与方程阡陌变通途", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014955": { + "id": "014955", + "content": "在正四棱柱$ABCD-A_1B_1C_1D_1$中, 顶点$B_1$到对角线$BD_1$、平面$A_1BCD_1$的距离分别为$h$、$d$, 则下列命题中正确的是\\bracket{20}.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\def\\l{2}\n\\def\\m{2}\n\\def\\n{2.5}\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) -- (B) -- (C);\n\\draw [dashed] (A) -- (D) -- (C);\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\\draw (A1) -- (B1) -- (C1) -- (D1) -- cycle;\n\\draw (A) -- (A1) (B) -- (B1) (C) -- (C1);\n\\draw [dashed] (D) -- (D1);\n\\draw (A1)--(B);\n\\draw [dashed] (B)--(D1)--(C);\n\\end{tikzpicture}\n\\end{center}\n\\onech{若侧棱的长小于底面的边长, 则$\\dfrac{h}{d}$的取值范围为$(0,1)$}{若侧棱的长小于底面的边长, 则$\\dfrac{h}{d}$的取值范围为$(\\dfrac{\\sqrt{2}}{2}, \\dfrac{2 \\sqrt{3}}{3})$}{若侧棱的长大于底面的边长, 则$\\dfrac{h}{d}$的取值范围为$(\\dfrac{2 \\sqrt{3}}{3}, \\sqrt{2})$}{若侧棱的长大于底面的边长, 则$\\dfrac{h}{d}$的取值范围为$(\\dfrac{2 \\sqrt{3}}{3},+\\infty)$}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-活用函数与方程阡陌变通途", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014956": { + "id": "014956", + "content": "已知点$P(0,1)$, 椭圆$\\dfrac{x^2}{4}+y^2=m$($m>1$)上两点$A, B$满足$\\overrightarrow{AP}=2 \\overrightarrow{PB}$, 则当$m=$\\blank{50}时, 点$B$横坐标的绝对值最大.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-活用函数与方程阡陌变通途", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014957": { + "id": "014957", + "content": "已知实数$a, b$满足$\\dfrac{8}{(a+1)^3}+\\dfrac{10}{a+1}-b^3-5 b>0$, 对于命题: \\textcircled{1} 若$a, b$两数中有一个大于$1$, 则另一个必小于$1$; \\textcircled{2} 若$a \\in(-2,-1)$, 则$a>b$, 下列判断正确的是\\bracket{20}.\n\\twoch{\\textcircled{1}和\\textcircled{2}均为真命题}{\\textcircled{1}和\\textcircled{2}均为假命题}{\\textcircled{1}为真命题, \n\\textcircled{2}为假命题}{\\textcircled{1}为假命题, \\textcircled{2}为真命题}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-活用函数与方程阡陌变通途", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014958": { + "id": "014958", + "content": "若关于$x$的方程$m+\\sqrt{1-x}=x$有实数解, 则实数$m$的最大值为\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-活用函数与方程阡陌变通途", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014959": { + "id": "014959", + "content": "设常数$a$使方程$\\sin x+\\sqrt{3} \\cos x=a$在闭区间$[0,2 \\pi]$上恰有三个解$x_1, x_2, x_3$, 则$x_1+x_2+x_3=$\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-活用函数与方程阡陌变通途", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014960": { + "id": "014960", + "content": "已知数列$\\{a_n\\}$满足: $a_{n+1}(a_n+1)=1$($n \\in \\mathbf{N}$, $n\\ge 1$), 且$a_1=1$, 若$\\displaystyle\\lim_{n\\to\\infty} a_n=A$, 则$A=$\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-活用函数与方程阡陌变通途", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014961": { + "id": "014961", + "content": "已知函数$f(x)=\\dfrac{(1-t) x-t^2}{x}$($t \\in \\mathbf{R}$)的定义域为$D$, 若存在区间$[a, b] \\subseteq D$, 使得当$x \\in[a, b]$时, $f(x)$的取值范围也是$[a, b]$, 则当$t$变化时, $b-a$的最大值为\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-活用函数与方程阡陌变通途", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014962": { + "id": "014962", + "content": "(1) 求证: 关于$x$的方程$x^n+x-1=0$($n \\in \\mathbf{N}$, $n \\geq 2$)在区间$(\\dfrac{1}{2}, 1)$内存在唯一解;\\\\\n(2) 设(1) 中方程的唯一解为$x_n$, 判断数列$\\{x_{n+1}\\}$的单调性并说明理由.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-活用函数与方程阡陌变通途", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014963": { + "id": "014963", + "content": "泳池长$90$米, 甲乙两人分别在泳池两边, 同时相向游泳, 甲的速度是$3$米/秒, 乙的速度是$2$米/秒, 若不计转向时间, $3$分钟后他们共相遇多少次?", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-代数几何统一体数形结合莫分离", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014964": { + "id": "014964", + "content": "函数$y=f(x)$的图像如图所示, 在区间$[a, b]$上可找到$n$($n \\geq 2$, $n \\in \\mathbf{N}$)个不同的实数$x_1,x_2, \\cdots, x_n$, 使得$\\dfrac{f(x_1)}{x_1}=\\dfrac{f(x_2)}{x_2}=\\cdots=\\dfrac{f(x_n)}{x_n}$, 则$n$的取值集合为\\bracket{20}.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw [->] (-1,0) -- (3,0) node [below] {$x$};\n\\draw [->] (0,-1) -- (0,2) node [left] {$y$};\n\\draw (0,0) node [below left] {$O$};\n\\draw (0.5,0) node [below] {$a$} -- (1,1.5) -- (1.5,0.2) -- (2,1) -- (2.5,0) node [below] {$b$};\n\\end{tikzpicture}\n\\end{center}\n\\fourch{$\\{2,3\\}$}{$\\{3,4\\}$}{$\\{2,3,4\\}$}{$\\{3,4,5\\}$}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-代数几何统一体数形结合莫分离", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014965": { + "id": "014965", + "content": "对任意$s \\in(-\\infty, 0) \\cup(0,+\\infty)$和$t \\in[-1,1]$, 不等式$s^2+\\dfrac{16}{s^2}-2 s t-\\dfrac{8}{s} \\sqrt{1-t^2}-a \\geq 0$恒成立, 则实数$a$的取值范围是\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-代数几何统一体数形结合莫分离", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014966": { + "id": "014966", + "content": "若四面体$S-ABC$的一条棱长为$x$, 其余棱长均为$1$, 它的体积是$V(x)$, 则函数$V(x)$在其定义域上\\bracket{20}.\n\\twoch{是增函数但无最大值}{是增函数且有最大值}{不是增函数且无最大值}{不是增函数但有最大值}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-代数几何统一体数形结合莫分离", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014967": { + "id": "014967", + "content": "若两函数$y=x+a$与$y=\\sqrt{1-2 x^2}$的图像有两个公共点$A$、$B$, $O$是坐标原点, $\\triangle OAB$是锐角三角形, 则实数$a$的取值范围是\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-代数几何统一体数形结合莫分离", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014968": { + "id": "014968", + "content": "如图, 在平面直角坐标系$xOy$中, 菱形$ABCD$的边长为$4$, 且$|OB|=|OD|=6$.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, scale = 0.5]\n\\draw [->] (-1,0) -- (7,0) node [below] {$x$};\n\\draw [->] (0,-3) -- (0,5) node [left] {$y$};\n\\draw (0,0) node [below left] {$O$} coordinate (O);\n\\draw (5,2) node [right] {$C$} coordinate (C);\n\\path [draw, name path = arc] (2,-2) arc (-90:90:2);\n\\path [name path = OC] (O) -- (C);\n\\path [name intersections = {of = OC and arc, by = A}];\n\\draw (A) node [below left] {$A$} coordinate (A);\n\\path [name path = circle1] (A) circle (4);\n\\path [name path = circle2] (C) circle (4);\n\\path [name intersections = {of = circle1 and circle2, by = {B,D}}];\n\\draw (A)--(B)--(C)--(D)--cycle(B) node [above] {$B$} --(O)--(D) node [below] {$D$};\n\\draw [dashed] (2,-2)--(2,2);\n\\end{tikzpicture}\n\\end{center}\n(1) 求证: $|OA| \\cdot|OC|$为定值;\\\\\n(2) 当点$A$在半圆$(x-2)^2+y^2=4$($2 \\leq x \\leq 4$)上运动时, 求点$C$的轨迹方程.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-代数几何统一体数形结合莫分离", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014969": { + "id": "014969", + "content": "已知圆$O$的半径为$1$, $PA$、$PB$为该圆的两条切线, $A$、$B$为两个切点, 则$\\overrightarrow{PA} \\cdot \\overrightarrow{PB}$的最小值为\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-代数几何统一体数形结合莫分离", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014970": { + "id": "014970", + "content": "已知$f(x)$是定义在$\\mathbf{R}$上的偶函数, 且$f(x)$在$[0,+\\infty)$上是增函数, 如果对于任意$x \\in[1,2]$, 不等式$f(a x+1) \\leq f(x-3)$恒成立, 则实数$a$的取值范围是\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-代数几何统一体数形结合莫分离", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014971": { + "id": "014971", + "content": "已知$f(x)$是定义在$[1,+\\infty)$上的函数, 且$f(x)=\\begin{cases}1-|2 x-3|,& 1 \\leq x<2, \\\\ \\dfrac{1}{2} f(\\dfrac{1}{2} x), & x \\geq 2,\\end{cases}$则函数$y=2 x f(x)-3$在区间$(1,2022)$上的零点个数为\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-代数几何统一体数形结合莫分离", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014972": { + "id": "014972", + "content": "设$\\theta \\in(0, \\dfrac{\\pi}{2}]$, 函数$f(x)=5 \\sin (2 x-\\theta)$, $x \\in[0,5 \\pi]$, 若函数$F(x)=f(x)-3$的所有零点依次记为$x_1, x_2, x_3, \\cdots, x_n$, 且$x_1=latex]\n\\fill [pattern = north east lines] (0,0.2) rectangle (4,0);\n\\draw (0,0) -- (4,0);\n\\draw (0.4,0) -- (0.4,-1) -- (3.6,-1) -- (3.6,0) (2,0)--(2,-1);\n\\draw (0.4,-1.1) -- (0.4,-1.7) (3.6,-1.1)-- (3.6,-1.7);\n\\draw [<->] (0.4,-1.4) -- (3.6,-1.4) node [midway, fill = white] {$30-3x$};\n\\draw [<->] (3.9,-1) -- (3.9,0) node [midway, fill = white, rotate = 90] {$x$};\n\\draw (3.7,-1) -- (4,-1);\n\\end{tikzpicture}\n\\end{center}", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-运用模型思想解决实际问题", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014975": { + "id": "014975", + "content": "动物园要建造一面靠墙(该墙的长度为$12$米)的$2$间面积相同的长方形熊猫居室. 如果可供建造围墙的材料长是$30$米, 那么宽$x$为多少米时才能使所建造的熊猫居室面积最大? 熊猫居室的最大面积是多少平方米?\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\fill [pattern = north east lines] (0,0.2) rectangle (4,0);\n\\draw (0,0) -- (4,0);\n\\draw (0.4,0) -- (0.4,-1) -- (3.6,-1) -- (3.6,0) (2,0)--(2,-1);\n\\draw (0.4,-1.1) -- (0.4,-1.7) (3.6,-1.1)-- (3.6,-1.7);\n\\draw [<->] (0.4,-1.4) -- (3.6,-1.4) node [midway, fill = white] {$30-3x$};\n\\draw [<->] (3.9,-1) -- (3.9,0) node [midway, fill = white, rotate = 90] {$x$};\n\\draw (3.7,-1) -- (4,-1);\n\\end{tikzpicture}\n\\end{center}", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-运用模型思想解决实际问题", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014976": { + "id": "014976", + "content": "北京天坛的圜丘坛为古代祭天的场所, 分上、中、下三层. 上层中心有一块圆形石板(称为天心石), 环绕天心石砌$9$块扇面形石板构成第一环, 向外每环依次增加$9$块. 中层起, 每层的第一环比上一层的最后一环多$9$块, 向外每环依次也增加$9$块. 已知每层环数相同, 且下层比中层多$729$块, 则三层共有扇面形石板(不含天心石)\\blank{50}块.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-运用模型思想解决实际问题", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014977": { + "id": "014977", + "content": "小明将上周每天骑车上学路上的情况用图像表示. 很遗憾图像的先后次序不小心被打乱了. 还好小明同时用文字进行了记录:\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw [->] (0,0) -- (2,0) node [below] {时间};\n\\draw [->] (0,0) -- (0,2) node [right] {离开家的距离};\n\\draw (0,0) node [below left] {$O$};\n\\draw [thick] (0,0) -- (0.8,0.8) -- (1.2,0.8) -- (2,1.6);\n\\draw (1,-0.5) node {A};\n\\end{tikzpicture}\n\\begin{tikzpicture}[>=latex]\n\\draw [->] (0,0) -- (2,0) node [below] {时间};\n\\draw [->] (0,0) -- (0,2) node [right] {离开家的距离};\n\\draw (0,0) node [below left] {$O$};\n\\draw [thick] (0,0) -- (0.5,0.5) -- (0.7,0) -- (1.2,0) -- (2,1.6);\n\\draw (1,-0.5) node {B};\n\\end{tikzpicture}\n\\begin{tikzpicture}[>=latex]\n\\draw [->] (0,0) -- (2,0) node [below] {时间};\n\\draw [->] (0,0) -- (0,2) node [right] {离开家的距离};\n\\draw (0,0) node [below left] {$O$};\n\\draw [thick,domain = 0:1.6, samples = 50] plot (\\x,{\\x*\\x*\\x/1.6/1.6}); \n\\draw (1,-0.5) node {C};\n\\end{tikzpicture}\n\\begin{tikzpicture}[>=latex]\n\\draw [->] (0,0) -- (2,0) node [below] {时间};\n\\draw [->] (0,0) -- (0,2) node [right] {离开家的距离};\n\\draw (0,0) node [below left] {$O$};\n\\draw [thick] (0,0) -- (1,1.3) -- (1.3,1.3) -- (2,1.6);\n\\draw (1,-0.5) node {D};\n\\end{tikzpicture}\n\\begin{tikzpicture}[>=latex]\n\\draw [->] (0,0) -- (2,0) node [below] {时间};\n\\draw [->] (0,0) -- (0,2) node [right] {离开家的距离};\n\\draw (0,0) node [below left] {$O$};\n\\draw [thick] (0,0) -- (1.6,1.6);\n\\draw (1,-0.5) node {E};\n\\end{tikzpicture}\n\\end{center}\n周一: 匀速骑车前进;\\\\\n周二: 匀速骑车前进, 中间遇到红灯停了一次, 然后用与之前同样的速度匀速前进;\\\\\n周三: 骑车出门晩了, 越骑越快;\\\\\n周四: 匀速骑车出门后一会儿想起忘带东西又回去拿, 然后再赶回学校;\\\\\n周五: $\\cdots$.\\\\\n请将图像的编号填入表格中对应日期的下方:\n\\begin{center}\n\\begin{tabular}{|c|c|c|c|c|c|}\n\\hline 日期 & 周一 & 周二 & 周三 & 周四 & 周五 \\\\\n\\hline 图像编号 &&&&&\\\\\n\\hline\n\\end{tabular} \n\\end{center}\n并描述周五小明上学途中可能发生的情况, 填在下面的空格中:\\\\\n周五: \\blank{200}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-运用模型思想解决实际问题", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014978": { + "id": "014978", + "content": "为防止和严惩酒后驾车, 国家质量监督检验检疫局于$2004$年$5$月$31$日发布了《车辆驾驶人员血液、呼气酒精含量阈值与检验》国家标准. 标准规定, 车辆驾驶人员血液中的酒精含量大于或等于$20$毫克/百毫升, 小于$80$毫克/百毫升为饮酒驾车, 血液中的酒精含量大于或等于$80$毫克/百毫升为醉酒驾车, 经过大样本容量的试验分析得到一个函数模型$y=f(x)$近似地表达$y$与$x$的关系: $f(x)=\\begin{cases}a(x-1.3)^2+46.83, & 0=latex]\n\\draw (0,0) node [below] {$O$} coordinate (O);\n\\draw (2,0) node [below] {$A$} coordinate (A);\n\\draw (120:2) node [left] {$B$} coordinate (B);\n\\draw ($(O)!0.5!(B)$) node [left] {$Q$} coordinate (Q);\n\\path [name path = QP] (Q) --++ (2.5,0);\n\\path [name path = arc,draw] (A) arc (0:120:2);\n\\path [name intersections = {of = QP and arc, by = P}];\n\\draw (Q) -- (P) node [right] {$P$};\n\\draw (B)--(O)--(A)(O)--(P);\n\\draw ($(Q)!0.4!(P)$) ++ (0,0.5) node {桃花区};\n\\draw ($(Q)!0.4!(P)$) ++ (0,-0.2) node {郁金香区};\n\\draw ($(O)!0.7!(A)$) ++ (0,0.3) node {海棠区};\n\\end{tikzpicture}\n\\end{center}\n(1) 当$Q$是$OB$的中点时, 求$PQ$的长; (结果精确到$1$米)\\\\\n(2) 郁金香因其花朵饱满、颜色鲜㓞而受到游客欢迎. 请规划观赏区划分方案, 使郁金香种植区$\\triangle OPQ$的面积尽可能地大.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-运用模型思想解决实际问题", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014980": { + "id": "014980", + "content": "某项目组决定开发一款``猫捉老鼠''的游戏. 如图, 两个信号源$A$、$B$相距$10$米, $O$是$AB$的中点, 过$O$点的直线$l$与直线$AB$的夹角为$45^{\\circ}$. 机器猫在直线$l$上运动, 机器鼠的运动轨迹始终满足: 接收到$A$点的信号比接收到$B$点的信号晩$\\dfrac{8}{v_0}$秒 (注: 信号每秒传播$v_0$米). 在时刻$t_0$时, 测得机器鼠距离$O$点为$4$米. 以$O$为原点, 直线$AB$为$x$轴建立平面直角坐标系.\n\\begin{center}\n\\begin{tikzpicture}[>=latex]\n\\draw [->] (-2,0) -- (2,0) node [below] {$x$};\n\\draw [->] (0,-2) -- (0,2) node [left] {$y$};\n\\draw (0,0) node [below right] {$O$};\n\\filldraw (-1,0) circle (0.03) node [below] {$A$} coordinate (A);\n\\filldraw (1,0) circle (0.03) node [below] {$B$} coordinate (B);\n\\draw (-1.6,-1.6) -- (1.6,1.6) node [right] {$l$};\n\\draw (0.8,0.8) node [fill = white] {\\rotatebox{45}{猫}};\n\\draw ({4/3},0.8) node {鼠};\n\\end{tikzpicture}\n\\end{center}\n(1) 求时刻$t_0$时机器鼠所在位置的坐标;\\\\\n(2) 游戏设定: 机器鼠在距离直线$l$不超过$1.5$米的区域运动时, 有``被抓''的风险. 如果机器鼠保持目前的运动轨迹不变, 是否有``被抓''的风险?", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-运用模型思想解决实际问题", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014981": { + "id": "014981", + "content": "有一块正方形菜地$EFGH$, $EH$所在直线是一条小河, 收获的蔬菜可以送到$F$点或河边运走. 于是, 菜地分为两个区域$S_1$和$S_2$, 其中$S_1$中的蔬菜运到河边较近, $S_2$中的蔬菜运到$F$点较近, 而菜地内$S_1$和$S_2$的分界线$C$上的点到河边与到$F$点的距离相等. 现建立平面直角坐标系, 如图, 其中原点$O$为线段$EF$的中点, 点$F$的坐标为$(1,0)$, 则菜地内分界线$C$的方程为\\blank{50}.\n\\begin{center}\n\\begin{tikzpicture}[>=latex,scale = 0.6]\n\\draw [->] (-3,0) -- (3,0) node [below] {$x$};\n\\draw [->] (0,0) -- (0,5) node [left] {$y$};\n\\draw (0,0) node [below left] {$O$};\n\\draw (-2,0) node [below] {$E$} coordinate (E) rectangle (2,4) node [above] {$G$} coordinate (G);\n\\draw (2,0) node [below] {$F$} coordinate (F);\n\\draw (-2,4) node [above] {$H$} coordinate (H);\n\\draw (-1,2.5) node {$S_1$};\n\\draw (1,1) node {$S_2$};\n\\draw [very thick, domain = 0:4, samples = 100] plot ({pow(\\x,2)/8},\\x);\n\\filldraw ({1/2},2) circle (0.05) node [right] {$M$};\n\\end{tikzpicture}\n\\end{center}", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-运用模型思想解决实际问题", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014982": { + "id": "014982", + "content": "产能利用率是指实际产出与生产能力的比率, 工业产能利用率是衡量工业生产经营状况的重要指标. 下图为某地区统计局发布的$2015$年至$2018$年第$2$季度该地区工业产能利用率的折线图.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, yscale = 0.5]\n\\draw (1,4.2) coordinate (P);\n\\foreach \\i/\\j in {1/74.2,2/74.3,3/74.0,4/74.6,5/72.9,6/73.1,7/73.2,8/73.8,9/75.8,10/76.8,11/76.8,12/78.0,13/76.5,14/76.8}\n{\\filldraw (\\i,{\\j-70}) circle (0.05 and 0.1) coordinate (Q) node [above = 0.2] {$\\j$};\n\\draw (P)--(Q);\n\\draw (\\i,{\\j-70}) coordinate (P);};\n\\draw (0,0) rectangle (15,10);\n\\draw (7.5,10) node [above] {分季度工业产能利用率};\n\\foreach \\i in {70,72,74,76,78,80}\n{\\draw (0,{\\i-70}) node [left] {$\\i$};};\n\\draw (0,10) node [above left] {$(\\%)$};\n\\foreach \\i/\\j in {1/1,2/2,3/3,4/4,5/1,6/2,7/3,8/4,9/1,10/2,11/3,12/4,13/1,14/2}\n{\\draw (\\i,0) node [below] {\\small$\\j$季度};};\n\\draw (0.6,-1) -- (4.4,-1) node [below, midway] {$2015$年};\n\\draw (4.6,-1) -- (8.4,-1) node [below, midway] {$2016$年};\n\\draw (8.6,-1) -- (12.4,-1) node [below, midway] {$2017$年};\n\\draw (12.6,-1) -- (14.4,-1) node [below, midway] {$2018$年};\n\\end{tikzpicture}\n\\end{center}\n在统计学中, 同比是指本期统计数据与上一年同期统计数据相比较, 例如$2016$年第二季度与$2015$年第二季度相比较; 环比是指本期统计数据与上期统计数据相比较, 例如$2015$年第二季度与$2015$年第一季度相比较.\n根据上述信息, 下列结论中正确的是\\bracket{20}.\n\\twoch{2015 年第三季度环比有所提高}{2016 年第一季度同比有所提高}{2017 年第三季度同比有所提高}{2018 年第一季度环比有所提高}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-运用模型思想解决实际问题", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014983": { + "id": "014983", + "content": "某居民小区为缓解业主停车难的问题, 拟对小区内一块扇形空地$AOB$进行改造. 如图所示, 平行四边形$OMPN$区域为停车场, 其余部分建成绿地, 点$P$在围墙$AB$弧上, 点$M$和点$N$分别在道路$OA$和道路$OB$上, 且$OA=90$米, $\\angle AOB=\\dfrac{\\pi}{3}$, 设$\\angle POB=\\theta$, 停车场$OMPN$的面积为$S$.\n\\begin{center}\n\\begin{tikzpicture}\n\\draw (0,0) node [left] {$B$} arc (180:140:3) node [left] {$P$} coordinate (P) arc (140:120:3) node [above] {$A$} -- (3,0) node [right] {$O$}-- cycle;\n\\draw [dashed] (3,0) -- (P);\n\\draw (P) --++ ({3/sin(60)*sin(20)},0) node [right] {$M$};\n\\draw (P) --++ (-60:{3/sin(60)*sin(40)}) node [below] {$N$};\n\\end{tikzpicture}\n\\end{center}\n(1) 当$\\theta=\\dfrac{\\pi}{12}$时, 求$S$的值;(结果精确到$0.1$平方米)\\\\\n(2) 写出$S$关于$\\theta$的函数关系式, 并求当$\\theta$为何值时, $S$取到最大值.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-运用模型思想解决实际问题", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014984": { + "id": "014984", + "content": "据相关数据统计, 至$2021$年底全国已开通5G基站$140$万个, 部分省市的政府工作报告将``推进5G通信网络建设''列入$2022$年的重点工作, $2022$年一月份全国开通5G基站$4$万个.\\\\\n(1) 如果从$2022$年$2$月份起, 每个月比上一个月多开通$2000$个, 那么, 到$2022$年底全国共开通5G基站多少万个;(结果精确到$0.1$万个)\\\\\n(2) 如果$2022$年计划开通5G基站$60$万个, 并且自$2023$年起每年新开通的基站数量比上一年增加$x \\%$, 若到$2024$年底全国开通的5G基站总数至少达到$500$万个, 求$x$的最小值. (结果精确到$0.01$)", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-运用模型思想解决实际问题", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014985": { + "id": "014985", + "content": "某民营企业开发出了一种新产品, 预计能获得$50$万元到$1500$万元的经济收益. 企业财务部门研究对开发该新产品的团队进行奖励, 并讨论了一个奖励方案: 奖金金额$y$(单位: 万元) 随经济收益$x$(单位: 万元) 的增加而严格增加, 且$y>0$, 奖金金额不超过$20$万元.\\\\\n(1) 请你为该企业构建一个$y$关于$x$的函数模型, 并说明你的函数模型符合企业奖励要求的理由;\\\\\n(2) 若该企业采用函数$y=\\begin{cases}\\dfrac{1}{50} x+1, & 50 \\leq x \\leq 500, \\\\ 19+\\dfrac{1-a}{x}, & 500=latex]\n\\draw [->] (-0.5,0) -- (3,0) node [below] {$x$};\n\\draw [->] (0,-2.5) -- (0,2.5) node [left] {$y$};\n\\draw (0,0) node [below left] {$O$};\n\\filldraw [domain = -2:2, pattern = horizontal lines] plot ({sqrt(\\x*\\x+1)},\\x) -- (2,2) -- (0,0) -- (2,-2) -- cycle;\n\\end{tikzpicture}\n\\end{center}", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-探究学习型问题建立联系巧转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014988": { + "id": "014988", + "content": "我国南宋时期的数学家杨辉, 在他$1261$年所著的《详解九章算法》一书中, 以如下图所示的三角形图表的方式呈现了一些二项展开式中的二项式系数, 此图称为``杨辉三角'', 也称为``贾宪三角''. 在此图中, 从第三行开始, 首尾两数为$1$, 其他各数均为它``肩上''两数之和.\n\\begin{center}\n\\begin{tikzpicture}[>=latex,scale = 0.7]\n\\foreach \\i in {0,1,...,6}\n{\\draw (\\i,0) --++ (60:{6-\\i});\n\\draw (\\i,0) --++ (120:\\i);\n\\draw (60:\\i) --++ ({6-\\i},0);};\n\\foreach \\i/\\j in {0/1,1/6,2/15,3/20,4/15,5/6,6/1}\n{\\draw [fill=white] (\\i,0) circle (0.4) node {\\tiny\\zhnumber{\\j}};};\n\\foreach \\i/\\j in {0/1,1/5,2/10,3/10,4/5,5/1}\n{\\draw [fill=white] (60:1) ++ (\\i,0) circle (0.4) node {\\tiny\\zhnumber{\\j}};};\n\\foreach \\i/\\j in {0/1,1/4,2/6,3/4,4/1}\n{\\draw [fill=white] (60:2) ++ (\\i,0) circle (0.4) node {\\tiny\\zhnumber{\\j}};};\n\\foreach \\i/\\j in {0/1,1/3,2/3,3/1}\n{\\draw [fill=white] (60:3) ++ (\\i,0) circle (0.4) node {\\tiny\\zhnumber{\\j}};};\n\\foreach \\i/\\j in {0/1,1/2,2/1}\n{\\draw [fill=white] (60:4) ++ (\\i,0) circle (0.4) node {\\tiny\\zhnumber{\\j}};};\n\\foreach \\i/\\j in {0/1,1/1}\n{\\draw [fill=white] (60:5) ++ (\\i,0) circle (0.4) node {\\tiny\\zhnumber{\\j}};};\n\\foreach \\i/\\j in {0/1}\n{\\draw [fill=white] (60:6) ++ (\\i,0) circle (0.4) node {\\tiny\\zhnumber{\\j}};};\n\\end{tikzpicture}\n\\end{center}\n(1) 如图, 以``杨辉三角''第三行第一个数为首项, 从左上至右下, 将第三斜列各数取出按原来的顺序排列得一数列: $1,3,6,10,15, \\cdots$, 记作数列$\\{a_n\\}$, 写出$a_n$与$a_{n-1}$($n \\in \\mathbf{N}$, $n \\geq 2$)的递推关系, 并求其通项公式;\\\\\n(2) 对于 (1) 中的数列$\\{a_n\\}$, 若数列$\\{b_n\\}$满足$b_1+\\dfrac{1}{2} b_2+\\dfrac{1}{3} b_3+\\cdots+\\dfrac{1}{n} b_n=2 a_n$($n \\in \\mathbf{N}$, $n\\ge 1$), 求数列$\\{b_n\\}$的通项公式.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-探究学习型问题建立联系巧转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014989": { + "id": "014989", + "content": "数学中有许多形状优美的曲线, 如星形线. 设定大圆的半径是小圆半径的$4$倍, 让小圆在大圆内部沿着大圆的圆周滚动, 小圆圆周上的任一点形成的轨迹即为星形线. 已知小圆圆周上某一点$P$形成的星形线$C$的方程为$x^{\\frac{2}{3}}+y^{\\frac{2}{3}}=a^{\\frac{2}{3}}$($a>0$), 有如下结论:\\\\\n\\textcircled{1} 曲线$D: |x|+|y|=a$的周长大于星形线$C$的周长;\\\\\n\\textcircled{2} 星形线$C$上任意两点间距离的最大值为$2 a$;\\\\\n\\textcircled{3} 星形线$C$与圆$E: x^2+y^2=\\dfrac{a^2}{4}$有且仅有$4$个公共点.\\\\\n其中所有正确结论的序号是\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-探究学习型问题建立联系巧转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014990": { + "id": "014990", + "content": "对于函数$f(x)$和实数$t$, 若在其定义域内存在实数$x_0$, 使得$f(x_0+t)=f(x_0)+f(t)$成立, 则称$f(x)$是``$t$跃点''函数, 并称$x_0$是函数$f(x)$的``$t$跃点''.\\\\\n(1) 求证: 函数$f(x)=2^x+3 x^2$, $x \\in[0,3]$是``$2$跃点''函数;\\\\\n(2) 若函数$g(x)=x^3-\\dfrac{a}{2} x+3 a$, $x \\in(-3,+\\infty)$是``$1$跃点''函数, 求实数$a$的取值范围.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-探究学习型问题建立联系巧转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, + "014991": { + "id": "014991", + "content": "若实系数一元二次方程$a x^2+b x+c=0$在复数集$\\mathbf{C}$内的根为$x_1$、$x_2$, 则有$a(x-x_1)(x-x_2)=a x^2-a(x_1+x_2) x+a x_1 x_2=0$, 所以$x_1+x_2=-\\dfrac{b}{a}$, $x_1 x_2=\\dfrac{c}{a}$(韦达定理), 类比此方法求解如下问题: 设实系数一元三次方程$a x^3+b x^2+c x+d=0$在复数集$\\mathbf{C}$内的根为$x_1$、$x_2$、$x_3$, 则$\\dfrac{1}{x_1}+\\dfrac{1}{x_2}+\\dfrac{1}{x_3}$的值为\\bracket{20}.\n\\fourch{$-\\dfrac{c}{d}$}{$\\dfrac{b}{d}$}{$\\dfrac{c}{a}$}{$\\dfrac{d}{a}$}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-探究学习型问题建立联系巧转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014992": { + "id": "014992", + "content": "``阿基米德多面体''也称为半正多面体, 是由边数不全相同的正多边形为面围成的多面体, 它体现了数学的对称美. 如图, 将正方体沿交于一顶点的三条棱的中点截去一个三棱锥, 共可截去八个三棱锥, 得到八个面为正三角形、六个面为正方形的``阿基米德多面体'', 则异面直线$AB$与$CD$所成角的大小是\\bracket{20}.\n\\begin{center}\n\\begin{tikzpicture}[>=latex, z = {(220:0.5cm)}]\n\\draw (-1,0,0) coordinate (A);\n\\draw (0,0,1) node [below] {$C$} coordinate (B);\n\\draw (1,0,0) coordinate (C);\n\\draw (0,0,-1) coordinate (D);\n\\draw (-1,1,1) coordinate (E);\n\\draw (1,1,1) node [left] {$D$} coordinate (F);\n\\draw (1,1,-1) coordinate (G);\n\\draw (-1,1,-1) coordinate (H);\n\\draw (-1,2,0) node [left] {$A$} coordinate (M);\n\\draw (0,2,1) coordinate (N);\n\\draw (1,2,0) coordinate (P);\n\\draw (0,2,-1) node [above] {$B$} coordinate (Q);\n\\draw (A)--(B)--(E)--cycle (B)--(C)--(F)--cycle (C)--(G) (G)--(P) (F)--(P)--(N) --cycle (M)--(N)--(E)--cycle (M)--(Q)--(P);\n\\draw [dashed] (A)--(D)--(H)--cycle (C)--(D)--(G) (M)--(H)--(Q)--(G);\n\\end{tikzpicture}\n\\end{center}\n\\fourch{$30^{\\circ}$}{$45^{\\circ}$}{$60^{\\circ}$}{$120^{\\circ}$}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-探究学习型问题建立联系巧转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014993": { + "id": "014993", + "content": "对于非零向量$\\overrightarrow {a}$、$\\overrightarrow {b}$, 定义一种向量的运算: $\\overrightarrow {a} \\otimes \\overrightarrow {b}=\\dfrac{\\overrightarrow {a} \\cdot \\overrightarrow {b}}{\\overrightarrow {b} \\cdot \\overrightarrow {b}}$. 设集合$P=\\{\\dfrac{n}{2} | n \\in \\mathbf{N}\\}$, 若非零向量$\\overrightarrow {a}$、$\\overrightarrow {b}$满足$\\overrightarrow {a} \\otimes \\overrightarrow {b} \\in P$, $\\overrightarrow {b} \\otimes \\overrightarrow {a} \\in P$, 且其夹角$\\theta \\in(\\dfrac{\\pi}{4}, \\dfrac{\\pi}{2})$, 则$\\overrightarrow {a} \\otimes \\overrightarrow {b}$的所有可能的值组成的集合为\\bracket{20}.\n\\fourch{$\\{\\dfrac{3}{2}, \\dfrac{5}{2}\\}$}{$\\{\\dfrac{1}{2}, \\dfrac{3}{2}\\}$}{$\\{1\\}$}{$\\{\\dfrac{1}{2}\\}$}", + "objs": [], + "tags": [], + "genre": "选择题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-探究学习型问题建立联系巧转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014994": { + "id": "014994", + "content": "在平面直角坐标系$xOy$中, 已知任意角$\\theta$的顶点与原点$O$重合, 始边与$x$轴的正半轴重合. 若角$\\theta$的终边经过点$P(x_0, y_0)$且$|OP|=r$($r>0$), 定义$\\mathrm{sicos} \\theta=\\dfrac{x_0+y_0}{r}$, 称``函数$y=\\mathrm{sicos} x$, $x \\in \\mathbf{R}$''为``正余弦函数''. 对于正余弦函数, 有如下结论:\\\\\n\\textcircled{1} 该函数是偶函数;\\\\\n\\textcircled{2} 该函数图像的一个对称中心是点$(\\dfrac{3 \\pi}{4}, 0)$;\\\\\n\\textcircled{3} 该函数的单调递减区间是$[2 k \\pi-\\dfrac{3 \\pi}{4}, 2 k \\pi+\\dfrac{\\pi}{4}]$($k \\in \\mathbf{Z}$);\\\\\n\\textcircled{4} 该函数的图像与直线$y=\\dfrac{3}{2}$没有公共点.\\\\\n其中所有正确结论的序号是\\blank{50}.", + "objs": [], + "tags": [], + "genre": "填空题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-探究学习型问题建立联系巧转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "" + }, + "014995": { + "id": "014995", + "content": "如果有穷数列$a_1, a_2, a_3, \\cdots, a_n$($n$为正整数) 满足条件$a_1=a_n$, $a_2=a_{n-1}$, $\\cdots$, $a_n=a_1$, 即$a_i=a_{n-i+1}$($i=1,2, \\cdots, n$), 我们称其为``对称数列''. 例如, 由组合数组成的数列$\\mathrm{C}_m^0, \\mathrm{C}_m^1, \\cdots, \\mathrm{C}_m^m$就是``对称数列''.\\\\\n(1) 设$\\{b_n\\}$是项数为$7$的``对称数列'', 其中$b_1, b_2, b_3, b_4$是等差数列, 且$b_1=2$, $b_4=11$. 依次写出$\\{b_n\\}$的每一项;\\\\\n(2) 设$\\{c_n\\}$是项数为$2 k-1$的``对称数列''($k$是大于$1$的正整数), 其中$c_k, c_{k+1}, \\cdots, c_{2 k-1}$是首项为$50$, 公差为$-4$的等差数列. 记$\\{c_n\\}$各项的和为$S_{2 k-1}$. 当$k$为何值时, $S_{2 k-1}$取得最大值? 并求出$S_{2 k-1}$的最大值;\\\\\n(3) 对于确定的$m$($m$是大于$1$的正整数), 写出所有项数不超过$2 m$的``对称数列'', 使得$1,2,2^2, \\cdots, 2^{m-1}$依次是该数列中连续的项; 当$m>1500$时, 求其中一个``对称数列''前$2022$项的和$S_{2022}$.", + "objs": [], + "tags": [], + "genre": "解答题", + "ans": "", + "solution": "", + "duration": -1, + "usages": [], + "origin": "2022届空中课堂学科精要名师点拨-探究学习型问题建立联系巧转化", + "edit": [ + "202304012\t王伟叶" + ], + "same": [], + "related": [], + "remark": "", + "space": "12ex" + }, "020001": { "id": "020001", "content": "判断下列各组对象能否组成集合, 若能组成集合, 指出是有限集还是无限集.\\\\\n(1) 上海市控江中学$2022$年入学的全体高一年级新生;\\\\\n(2) 中国现有各省的名称;\\\\\n(3) 太阳、$2$、上海市;\\\\\n(4) 大于$10$且小于$15$的有理数;\\\\\n(5) 末位是$3$的自然数;\\\\\n(6) 影响力比较大的中国数学家;\\\\\n(7) 方程$x^2+x-3=0$的所有实数解;\\\\ \n(8) 函数$y=\\dfrac 1x$图像上所有的点;\\\\ \n(9) 在平面直角坐标系中, 到定点$(0, 0)$的距离等于$1$的所有点;\\\\\n(10) 不等式$3x-10<0$的所有正整数解;\\\\\n(11) 所有的平面四边形.",