江苏省南京市、盐城市2022届高三年级第二次模拟考试数学试题

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南京市、盐城市2022届高三年级第二次模拟考试数学参考答案一、单项选择题(本大题共8小题，每小题5分，共40分．在每小题给出的四个选项中，只有一项是符合题目要求的)1．C 2．A 3．B 4．B 5．C 6．D 7．A 8．D二、多项选择题(本大题共4小题，每小题5分，共20分．在每小题给出的四个选项中，有多项符合题目要求．全部选对的得5分，部分选对的得2分，有选错的得0分)9．BCD  10．AD  11．AC  12．ABD三､填空题(本大题共4小题，每小题5分，共20分)13．8  14．144  15．－  16．120四、解答题(本大题共6小题，共70分．解答时应写出文字说明、证明过程或演算步骤)17．(本题满分10分)解：（1）因为∠BAD＝，AC平分∠BAD，所以∠BAC＝∠CAD＝．在△ABC中，因为∠ABC＝，所以∠ACB＝，又因为AC＝2，由＝，得AB＝，················································2分所以S△ABC＝AB·ACsin∠BAC＝．在△ACD中，因为∠ADC＝∠CAD＝，所以CA＝CD＝2，所以S△ACD＝CA·CDsin∠ACD＝，所以S四边形ABCD＝S△ABC＋S△ACD＝．·······································4分（2）因为AC平分∠BAD，所以∠BAC＝∠CAD，在△ACD中，由∠ADC＝， ＝，得AC＝· ．   ①在△ABC中，由∠ABC＝， ＝，得AC＝·．  ②·····································6分由①②得＝．又因为CD＝2AB，所以2sin∠ACB＝sin∠CAD． 设∠BAC＝θ，则sinθ＝2sin(－θ)，··············································8分所以sinθ＝2×(cosθ－sinθ)，即2sinθ＝cosθ．因为θ∈(0，)，所以cosθ≠0，所以tanθ＝，即tan∠BAC＝．·················································10分 18．(本题满分12分) 解：（1）因为2∈[21，22)，所以a2＝22＝4，·········································2分因为20∈[24，25)，所以a20＝25＝32．···········································4分（2）an＝2k的项数为2k－2k－1＝2k－1．·········································6分又因为20＋21＋22＋…＋2k－1＝2k－1，所以数列{an}的前2k－1项和为S＝21×20＋22×21＋23×22＋…＋2k×2k－1＝21＋23＋25＋…＋22k－1＝(4k－1)．····························································8分当k＝5时，S31＝(45－1)＝682＜2022，S51＝S31＋26×20＝682＋1280＝1962＜2022，····································10分S52＝S51＋26＝1962＋64＝2026＞2022．又因为Sn＋1＞Sn，所以使得Sn＜2022成立的正整数n的最大值为51．·································12分19．(本题满分12分)解：（1）取AB中点E，连接PE，DE．因为△PAB是边长为2的等边三角形，所以AB⊥PE，PE＝，AE＝1．又因为PD⊥AB，PD∩PE＝P，PD，PE平面PDE，所以AB⊥平面PDE．························································2分因为DE面PDE，所以AB⊥DE．在Rt△AED中，AD＝2，AE＝1，所以DE＝． 在△PDE中，PD＝，DE＝，PE＝，所以PE2＋DE2＝PD2，所以DE⊥PE．···············4分又因为AB∩PE＝E，AB，PE平面PAB，所以DE⊥平面PAB．又因为DE平面ABCD，所以平面PAB⊥平面ABCD．··················································6分（2）由（1）知，以{，，}为正交基底，建立如图所示的空间直角坐标系E－xyz，则E(0，0，0)，D(0，0，)，C(－2，0，)，P(0，，0)．则＝(－2，0，0)，＝(0，－，)．······································8分设平面PCD的法向量为n＝(x，y，z)，则即取x＝0，y＝1，z＝1．所以n＝(0，1，1)是平面PCD的一个法向量．……………10分因为DE⊥平面PAB，所以＝(0，0，)为平面PAB的一个法向量．所以cos＜n，＞＝＝，所以平面PAB和平面PCD所成锐二面角的大小为．·································12分 20．(本题满分12分) 解：（1）①当1≤X≤9时，P(X＝i)＝(1－p)i－1p，i＝1，2，…，9．当X ＝10时，P(X＝10)＝(1－p)9．所以P(X＝i)＝·····························································4分②E(X)＝i(1－p)i－1p＋10(1－p)9＝pi(1－p)i－1＋10(1－p)9．令S＝i(1－p)i－1，则E(X)＝pS＋10(1－p)9．则S＝1＋2(1－p)＋3(1－p)2＋…＋8(1－p)7＋9(1－p)8，(1－p)S＝(1－p)＋2(1－p)2＋…＋7(1－p)7＋8(1－p)8＋9(1－p)9，两式相减，得pS＝1＋(1－p)＋(1－p)2＋…＋(1－p)7＋(1－p)8－9(1－p)9·················6分＝－9(1－p)9，所以E(X)＝＋(1－p)9＝[1－(1－p)10]． 因为0＜p＜1，所以0＜1－(1－p)10＜1，所以E(X)＜．······························································9分（2）当p＝0.25时，由（1）得E(X)＜4， 则a×E(X) ＜4a＜5a，即试验结束后的平均成本小于试验成功的获利，所以该公司可以考虑投资该产品．····································12分 21．(本题满分12分)解：（1）因为双曲线C渐近线方程为y＝±x，所以＝1．又因为双曲线C经过点(，1)，所以－＝1．····································2分解得a＝b＝．··························································4分（2）方法1当AB斜率不存在时，由双曲线对称性知AD经过原点，此时与题意不符．设AB方程为y＝kx＋m(k≠0)，A(x1，y1)，B(x2，y2)，AB中点E(x3，y3)，则D(－x2，y2)．由消去x，得 (1－k2)x2－2kmx－m2－2＝0，所以x1＋x2＝，x1x2＝－，···················································6分则x3＝＝，y3＝kx3＋m＝，则AB的中垂线方程为y－＝－(x－)，当x＝0时，y＝． 因为B，D两点关于y轴对称，则△ABD的外接圆圆心在y轴上，记圆心为点F，则F(0，)．····················································8分因为△ABD的外接圆经过原点，则OF＝FA，即||＝．又因为－＝1，所以y12－ y1＋1＝0．同理，由OF＝FB，得y22－ y2＋1＝0，所以y1，y2是方程y2－y＋1＝0的两个根，所以y1y2＝1．····························10分则(kx1＋m)(kx2＋m)＝1，即k2x1x2＋km(x1＋x2)＋m2＝1，所以k2×(－)＋km×＋m2＝1，化简得k2＋1＝m2，所以原点O到直线AB距离d＝＝1，所以直线AB与圆x2＋y2＝1相切．·············································12分方法2设直线AB方程为x＝my＋n，A(x1，y1)，B(x2，y2)，则D(－x2，y2)．又因为B，D两点关于y轴对称，则△ABD的外接圆的圆心在y轴上，设为P(0，t)，则PA＝PB，即＝．由－＝1，－＝1，化简得t＝y1＋y2．············································6分因为△ABD的外接圆经过原点O，所以PA＝PO＝|t|，即＝|y1＋y2|，化简得y1y2＝1．···························································8分联立直线AB及双曲线方程消去x，得 (m2－1)y2＋2mny＋n2－2＝0，所以y1y2＝．····················································10分又因为y1y2＝1，所以＝1，即m2＋1＝n2，所以原点O到直线AB距离d＝＝1，所以直线AB与圆x2＋y2＝1相切．·············································12分 22．(本题满分12分)解：（1）由f(x)＝aex＋sinx－3x－2，得f'(x)＝aex＋cosx－3．因为a≤0，所以f'(x)＝aex＋cosx－3≤cosx－3＜0，所以f(x)在(－∞，＋∞)单调递减．·········2分又因为f(0)＝a－2＜0，f(a－2)＝aea－2＋sin(a－2)－3a＋4＞a(ea－2－3)≥0，因此f(x)有唯一的零点．························································4分（2）由（1）知，a≤0符合题意．（i）当a＝2时，由f(x)＝2ex＋sinx－3x－2，得f'(x)＝2ex＋cosx－3．当x＜0时，f'(x)≤2ex－2＜0，所以f(x)单调递减；···································6分当x＞0时，f''(x)＝2ex－sinx≥2ex－1＞0，所以f'(x)在(0，＋∞)上单调递增，从而，当x＞0时，f'(x)＞f'(0)＝0，所以f(x)单调递增， 于是f(x)≥f(0)＝0，当且仅当x＝0时取等号，故此时f(x)有唯一的零点x＝0．·················································8分（ii）当a＞2时，f(x)＞2ex＋sinx－3x－2≥0，此时f(x)无零点；···························9分（iii）当0＜a＜2时，首先证明：当x≥0时，ex＞．设g(x)＝ex－，x≥0，则g'(x)＝ex－x，g''(x)＝ex－1≥0，所以g'(x)在[0，＋∞)上单调递增，故g'(x)≥g'(0)＝1＞0，所以g(x)在[0，＋∞)上单调递增，因此g(x)≥g(0)＝1＞0，即当x≥0时，ex＞．········································10分当x＞0时，f(x)≥aex－3x－3＞x2－3x－3，令x2－3x－3＝0，得x＝．取x0＝＞0，则f(x0)＞0．又f(0)＝a－2＜0，f(－1)＝ae－1＋1－sin1＞0，因此，当0＜a＜2时，f(x)至少有两个零点，不合题意．综上，a＝2或a≤0．························································12分