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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4060

100··········································································································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＋lnxx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：00

01()()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=ïïíïïï-=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23452155111111222222CC

，························3分设X为该电子产品需要维修的系统个数，则12XB，，200X.221120001222kkkPkPXkCk

，，，··································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2

543226151233121BAfpPPppppppp.∵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