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高二数学参考答案第1页（共5页）2021年春期高中二年级期末质量评估数学(理)参考答案一、选择题CABDCABCCBCD二、填空题13．14714．200315.2316．(1,2)e三、解答题17．【解析】证明：①当1n=时，左边1a=+，右边1a=+

.所以，当1n=时，命题成立.··································································1分②假设当(1)nkk=³时，命题成立，即(1)1kkaa+³+.··············

·············2分那么，当1nk=+时，因为1a>-，所以10a+>.···································3分根据假设知，1(1)(1)(1)kkaaa++=++····

··············································4分(1)(1)kaa³++21(1)kkaa=+++·············································6分由于20ka³，所以21(

1)1(1)kkkaaa+++³++.从而1(1)1(1)kkaa++³++.·································································8分这表明，当1nk=+时，命题成立.·······················

···································9分根据①②知，该命题成立.·······································································10分18．【解析】(1)由题351,

11,1xxfxxx,所以当01x时,3()11fxx,····················································2分故533(1)(2)ffxfxx,··················

··································4分而53(2)x的展开式共有6项,····························································5分故二项式系数的最大值为235510CC.·

···················································6分（2）当1x时，102()1fxx，··············································

·········7分高二数学参考答案第2页（共5页）即101021001210(1)(11())112)(aaxaxaxxx，······9分由12((1))rnrrrnTCx··

···································································10分可知，7377102(1)960aC.··································

···························12分19．【解析】（1）由表格数据可得2×2列联表如下：非移动支付活跃用户移动支付活跃用户总计男252045女154055总计4060100··················

························································································2分将列联表中的数据代入公式计算，得χ2＝nad－bc2a＋bc＋da＋cb＋d＝100×

25×40－15×20240×60×55×45·····································································3分＝2450297≈8

.249>6.635.··········································································5分所以有99%的把握认为是否为“移动支付活跃用户”与性别有关．··········

·········6分（2）视频率为概率，在我市“移动支付达人”中，随机抽取1名用户，该用户为男“移动支付达人”的概率为13，女“移动支付达人”的概率为23.·····················

····················8分于是，抽取的4名用户中，既有男“移动支付达人”，又有女“移动支付达人”的概率为：P(A)＝1－134－234＝6481.·····························································

········12分20.【解析】(1)证明：由题知证lnx≥-1x＋1成立即可，············································1分令g(x)＝lnx＋1x－1(x>0)，则g′(x)＝1x－1x2＝x－1x2.·············

···························································2分高二数学参考答案第3页（共5页）当0<x<1时，g′(x)<0；当x>1时，g′(x)>

0，故g(x)在(0,1)上单调递减，g(x)在(1，＋∞)上单调递增，·····························3分所以g(x)≥g(1)＝0，即lnx≥-1x＋1.···············································

········4分(当且仅当1x时取得等号)································································5分(2)由(1)知：h(x)=f(x)＋(x2－1)lnx＝x2lnx＋a(1－x2)≥

x2－1x＋1＋a(1－x2)(※)··································································7分=(x－1)[(1－a)x－

a]=(1-)(-1)()1-aaxxa··········································································8分由于x∈0，a1－a，所以01a，即(1-)0a

，数形结合只需011-aa成立即可.解得102a.···············································································

10分又当11-aa，即12a时，(※)式取“”，结合(1)0h，可知12a符合题意.···················································11分综上所述：102a.········

······························································12分21．【解析】(1)令f(x)＝lnx－mx＋1＝0，得m＝1＋lnxx，即函数g(x)＝1＋ln

xx与直线y＝m在(0，＋∞)上有两个不同交点，················1分因为g′(x)＝－lnxx2(x>0)，·························································

·············2分高二数学参考答案第4页（共5页）当x∈(0,1)时，g′(x)>0；当x∈(1，＋∞)时，g′(x)<0，故g(x)在(0,1)上单调递增，在(1，＋∞)上单调递减，所以g(x)max＝g(1)＝1.又g1e＝0，故当x∈0，

1e时，g(x)<0；当x∈1e，＋∞时，g(x)>0.·····································································4分画出图象，如图所示，······

·····································································5分可得m∈(0,1).·························

·····························································6分(2)由题，当0m时，()1lfxnx＋，由1()fxx，知001()fxx，····················

··········································7分故曲线()yfx在点00,())xfx(的切线为：0001()()yfxxxx，即：0001ln1()yxxxx，所以001lnyxxx.·········

···································································8分又设该切线与xye相切于点11(,)xy，则由xye，易知111()xxyeexx

，即：111(1)xxyexex.·····································································9分于是：110101(1)lnxxexexxìïï=ïï

íïïï-=ïî····································································10分从而有10lnxx，···········································

································11分整理可得：00(1)ln1xx-=.又01x=显然不满足，高二数学参考答案第5页（共5页）因此001ln1xx=-成立.······································

································12分22．【解析】(1)A系统需要维修的概率为231311112222C，·······································1分B系统需要维修的概率为234

52155111111222222CC，························3分设X为该电子产品需要维修的系统个数，则12XB，，200X.2211200012

22kkkPkPXkCk，，，··································4分∴的分布列为:0200400P141214∴120022002E.·····································

·····································6分(2)A系统3个元件至少有2个正常工作的概率为:223323123APCppppp，·················································

···········7分B系统5个元件至少有3个正常工作的概率为:2334455511BPCppCppp54361510ppp，··································9分则2543226151233121BAfp

PPppppppp.∵01p.令0fp，解得112p.····················································10分所以:当112p

时，B系统比A系统正常工作的概率大，当该产品出现故障时，优先检测A系统；当102p时，A系统比B系统正常工作的概率大，当该产品出现故障时，优先检测B系统；当12p时，A系统与B系统正常工作的概率相等，当该产品出现故

障时，A，B系统检测不分次序.···························································································12分