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高二数学答案（第1页，共4页）2023～2024学年度第二学期期末学业水平诊断高二数学参考答案及评分标准一、选择题CCADBDCA二、选择题9.ABD10.BCD11.AC三、填空题12.80−13.1(

,]e−∞14.14()3nL−，22345L四、解答题15.解：（1）根据已知条件，可得：······················································3分零假

设为0H：创新作文比赛获奖与选修阅读课程无关联，根据列联表中数据计算得到，2250(828212)25==8.3337.879203010403χ××−×≈>×××.·················

··············6分根据小概率值0.005α=的独立性检验，推断0H不成立，即认为创新作文比赛获奖与选修阅读课程有关联，此推断犯错误的概率不大于0.005.·························

···7分（2）由题意可知X的可能取值为1,2,3,则···································8分12823101(1)15CCPXC===，21823107(2)15CCPXC===，383107(3)15CPXC===，·

·······································11分所以,随机变量X的分布列为:所以17712()1231515155EX=×+×+×=.··························13分16.解：（1）当2a=−时，2()(21)exfxxx=−+，所以

2()(1)exfxx′=−.·········1分设切点为00(,)xy，则02000(21)exyxx=−+，020(1)exkx=−，获奖没有获奖合计选修阅读课程81220不选阅读课程22830合计104050X123P115715715{#{QQABRYYQoggoAIB

AAAhCAQXKCkOQkACACQgOBEAIMAAAARNABAA=}#}高二数学答案（第2页，共4页）所以，切线方程为00220000(21)e(1)e()xxyxxxxx−−+=−−.··········

··············3分将(1,0)代入得200(1)0xx−=，解得00x=或01x=.·····························5分故过(1,)0的切线方程为0y=或10xy+−=.·························

·······················7分（2）2()(2)e(1)e(1)(1)exxxfxxaxaxxax′=++++=+++.·····················8分当0a=时，2()(1)exfxx′=+，恒有

()0fx′≥，函数()fx单调递增.·········10分当0a>时，11a−−<−，当(,1)xa∈−∞−−，或(1,)x∈−+∞时，()0fx′>，函数()fx单调递增，当(1,1)xa∈−−

−时，()0fx′<，函数()fx单调递减.····12分当0a<时，11a−−>−，当(,1)x∈−∞−，或(1,)xa∈−−+∞时，()0fx′>，函数()fx单调递增，当(1,1)xa∈−−−时，()0fx′<，函数()fx单调递减.·······14分综上，当0a=时，()fx

在R上单调递增，当0a>时，()fx在(,1)a−∞−−，(1,)−+∞上单调递增，在(1,1)a−−−上单调递减，当0a<时，()fx在(,1)−∞−，(1,)a−−+∞上单调递增，在(1,1)a−−−上单调递减.·······················

·······15分17.解：（1）由题意可知，212bba−=，即211b−=−，故20b=.························1分由323bba−=，可得31a=.··································

····················2分所以数列{}na的公差2d=，所以12(2)25nann=−+−=−.······················3分由1nnnbba−−=，121nnnbba−−−−=，，212bba−=，叠加可得123(1)(125)2nnnnb

baaa−−+−−=+++=，整理可得244(2)nbnnn=−+≥；当1n=时，满足上式，所以244nbnn=−+················································

································5分（2）不妨设(,)mnabmn∗=∈N，即225(2)mn−=−，可得2(2)52nm−+=，········6分当2nk=时，29242mkk=−+，不合题意，当21

nk=−时，22672(3)7mkkkk∗=−+=−+∈N，································7分所以21kb−在数列{}na中均存在公共项，又因为1357bbbb=<<<，所以nc=221(21)nbn+=−.···

······························9分（3）当1n=时，1514T=<，结论成立，············································10分当2n≥时，2111111()(21)(22)241ncnnn

nn=<=−−−×−，·····················12分{#{QQABRYYQoggoAIBAAAhCAQXKCkOQkACACQgOBEAIMAAAARNABAA=}#}高二数学答案（第3页，共4页）所以1111111(1)43351nTnn<+−+−++−−111(1)4n=+

−515444n=−<，综上，54nT<.··················································15分18.解：（1）记事件A=“第2次取出的小球为黑球”；

事件B=“第1次取出的小球为白球”，则333311()666520PA=×+×=，············································2分333()=6510PAB=×，所以(

)6(|)()11PABPBAPA==；··································4分（2）由题意，X的可能取值为0,1,2,3，则··············································5分3331(0)6668PX

==××=，33333333391(1)++655665666200PX==××××××=,32333233237(2)++654655665100PX==××××××=,3211(3)65420PX==××=，所以，随机变量X的分布列为：X0123P1891200371

00120···········································10分（3）由题意可知，前1n−次取了一个白球，第n次取了第二个白球，则：23233333332[()()()]65665665nnn

nP−−−=×+××++××···························12分233232333333=[()()()()]65565656nnnn−−−−××+×+×+=22213555()[1()()]55666nn−−×+++121151()13316()2[()()]

5555216nnnn−−−−−=×=×−−*(2,)nn≥∈N.····················16分所以11312[()()]52nnnP−−=×−*(2,)nn≥∈N.··································17分

19.解：（1）函数()fx定义域为(0,)+∞，11()ln(1)1(ln)1xfxaxaxaxxx+′=++⋅+=++，····································1分

{#{QQABRYYQoggoAIBAAAhCAQXKCkOQkACACQgOBEAIMAAAARNABAA=}#}高二数学答案（第4页，共4页）显然0a≠，令()0fx′=，可得11lnxxxa++=−，令1(

)lnxtxxx+=+，由()fx有两个不同极值点得1()txa=−有两个不同的正根.··3分因为22111()xtxxxx−′=−=.当(0,1)x∈时，()0tx′<，()tx单减，(1,)x∈+∞时，()0tx′>，()tx单

增.················································································5分所以()tx的极小值即最小值(1)2t=，又当0x→时，()tx→+∞，且x→+∞时，()tx→

+∞，所以12a−>，即102a−<<.···········································6分（2）设12,xx为函数()fx的极值点，由（1），不妨设121xx<<，下证122xx+>.要证：2121xx>−>，只要证21()(2)t

xtx>−.令()()(2)(01)gxtxtxx=−−<<.····························8分因为22222114(1)()()(2)0(2)(2)xxxgxtxtxxxxx−−−−′′′=+−=+=<−−.···········10分所以()g

x在(0,1)上单调递减，所以()(1)0gxg>=，故21()(2)txtx>−，即122xx+>.·························11分由（1）可知，在1(0,)x上，1()(())0fxatxa′=+<，()fx单调递减，在12(,)xx上，()0fx′>，()fx单

调递增，在2(,)x+∞上，()0fx′<，()fx单调递减，又因为(1)0f=，所以1()(1)0fxf<=，因为102a−<<，所以12a<−，所以12ee1a−<<，而11111(e)(e1)lnee12e0aaaaafa=++−=>，所

以()fx在11(e,)ax上存在点3x，使得3()0fx=，·····························13分同理2()(1)0fxf>=，又12a−>，12ee1a−>>，1111(e)(e1)lnee120aaaafa−−−−=++

−=−<，所以()fx在12(,e)ax−上存在点4x，使得4()0fx=，·····························14分故()fx存在3个零点34,1,xx，注意到111111()(1)ln1((1)ln1)()faaxxxfxxxxxxx=++−=−

++−=−，·15分所以341xx=，所以343312xxxx+=+>.···································16分所以123415xxxx++++>，即5mn+>.········

····························17分{#{QQABRYYQoggoAIBAAAhCAQXKCkOQkACACQgOBEAIMAAAARNABAA=}#}