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龙岩一中2023届高三上学期第二次月考数学答案1-8：DDCCBABD9.ABD10.AB11.AC12.BC13.4314.()0,−和()1,015.2−16.1317．（1）因为22258bcabc+−=，由余弦定理可得2

22-5cos216bcaAbc+==.故5cos16A=···············································································

·························································4（2）因为sin2sinCB=，所以2cb=.由余弦定理得2222152cos4abcbcAb=+−=，则152ab=.··························

··············································································································6因为ABC的周长为615+，所以1536152bb+=+，解得2b

=.··························8所以ABC的面积为215231212164bb−=··························································

···············1018.（1）因为函数()2ππ2sinsin3sincoscos44fxxxxxx=+−++2222232cossincossinsin2cos22222xxxxxx=+−++

()2231cos2cossinsin222xxxx+=−++π13sin232x=++···································································

························································3令π2π,3xkk+=Z，解得ππ62kx=−+，即对称中心π1π,622kk−+Z，······························

····5当π0,2x时，则ππ4π2,333x+，再结合三角函数图像可得()11,32fx−+所以，函数对称中心：π1π,622k−+，kZ，值域：11,32−+.··················

·························7（2）因为函数()gx的图像与函数()fx的图像关于y轴对称，则()()π13sin232gxfxx=−=−++，··························································

·······························9令ππ3π2π22π232kxk+−++，kZ，解得7ππππ,1212kxkk−+−+Z·······························11当1k=时

，即为5π11π,1212所以当()0,πx时，()gx的单调递增区间：5π11π,1212.·················1219.（1）由题表知，随着时间x的增大，y的值随x的增大，先减小后增大，而所给的函数(0)yaxba=+，()log0,0,1

byaxabb=和(0)aybax=+在(0,)+上显然都是单调函数，不满足题意，故选择()20yaxbxca=++．·················································································

······3（2）把()2,102，()6,78，()20,120分别代入2yaxbxc=++，得42102,36678,40020120,abcabcabc++=++=++=解得12a=，10b=−，120c=∴()221110120107

022yxxx=−+=−+，,()0x+．···································································5∴当10x=时，y有最小值，且min70y=．故当该纪念章上市10天时，市场价最低，最低市场价为每枚7

0元．·······································7（3）令()()()1701010210fxgxxxx==−+−−(10,)x+，······································

··························8因为存在()10,x+，使得不等式()0gxk−成立，则()minkgx．······························································

·····································································9又()()()17017010210235210210gxxxxx=−+−=−−∴当10235x=+

时，()gx取得最小值，且最小值为235，∴235k．··············································································

························································1220.解：（1）由21()2ln(21)(0)2fxxaxaxa=−+−，可得2()21fxaxax=−+−．因为(1)2211faaa=−+−=+，13(1)21

122faaa=−+−=−，所以切点坐标为3(1,1)2a−，切线方程为：()311(1)2ayax−−=+−，因为切线经过(0,0)，所以3112aa−=+，解得4a=．·············

······················································4（2）解：由题可知()fx的定义域为(0,)+，21()[(21)2]fxaxaxx=−−−−，令()0

fx=，则2(21)20axax−−−=，解得1xa=−或2x=，························································6因为0,a所以10a−，所以12a−，令

()0fx，即2(21)20axax−−−，解得：12xa−，令()0fx，即2(21)20axax−−−，解得：1xa−或2x，··················································

·8又()fx的定义域为(0,)+，所以，()fx增区间为(0,2)，减区间为(2,)+．因为()22()211gtttt=−=−−，所以函数()gt在区间(0,2]的最大值为0，····················

··············9函数()fs在(0,2)上单调递增，故在区间(0,2]上max()(2)2ln222fsfa==+−，·······················10所以2ln2220a+−，即l

n210a+−，故1ln2a−，所以a的取值范围是(0,1ln2)−．·····1221.（1）取BC中点O，连接AO，1AO,1AC,因为ABAC=,所以AOBC⊥,························

···········································2因为11AABAAC=，11,ABACAAAA==，所以11AABAAC，所以11ABAC=，所以1AOBC⊥，······

······················································4因为1AOAOO=，1,AOAO平面1AAO，所以BC⊥平面1AAO，因为1AA平面1AAO，所以1AABC⊥；····

·····················································································6（2）连接OD，则平面1AAO即为平面1AADO，由（1）知BC⊥平面1AADO，因为BC平面ABC，

且BC平面11BCCB，故平面1AADO⊥平面ABC，平面1AADO⊥平面11BCCB，过O作1OMAD⊥于M，则OM⊥平面ABC，过1A作1AHOD⊥于H，则1AH⊥平面11BCCB，因为11DOBBAA∥∥知DOBC

⊥，在ABC中：2,22ABACBC===，所以11122BDBSDBDO==△，所以11111111214336BABDABDBBDBAAVVShh−−====△，所以1172AAHh==，···································

·····················································································8法一：设MOD=，则1DAH=，在1RtAHD△中117142cos42AHAD===，

所以214sin,cos22DMDOOMOD====，又12AD=，所以点M为线段1AD的中点，以O为原点，分别以,,OAOBOM分别为x，y，z轴正方向建立空间直角坐标系，1214(2,0,0),(0,2,0),(0,2,0),,0,22ABCA−，1214214,

2,,,0,2222BD−−，设面1ABD的法向量为()1111,,xnyz=，则有11111111121420222142022nBAxyznBDxyz=−+==−−+=，两式相减得：10x=，所以1114202yz−

+=，令12z=，可得：17y=，所以1(0,7,2)n=，设面11CBBC的法向量为()2222,,nxyz=，则有221122220214022nCBynCBxz===−+=，解得：20y=，令21z=，解得：27x=所以2(7,0,1

)n=，设锐二面角为，则有121222cos2122477nnnn===++.················································12法二：过H做HEBD⊥，连接1AE，1AH

⊥面11BCCB，1AHDB⊥，则DB⊥面1AHE，1AEBD⊥，则1AEH即为所求二面角.在1RtADH△中，117,22AHAD==，则12DH=，在RtDOB中，2,2,6DOOBDB===，由RtRtDEHD

OB可得：HEDHOBDB=，36HE=，则1116AE=，11122cos2222HEAEHAE===.···································································

··························1222.解：(1)()ecosxfxax=−，·························································································

·················1①当01a时，因为()0,x，所以cos1ax，1eex，()0fx，()fx在()0,上单调递增，没有极值点，不合题意，舍去；②当1a时，令()=()gxfx，则()esi

nxgxax=+，因为()0,x，所以()0gx，所以()fx在()0,上递增，又因为(0)10fa=−，2e02f=，所以()fx在()0,上有唯一零点1x，且10,2x，所以()10,xx，()0f

x；1,2xx，()0fx，所以()fx在()0,上有唯一极值点，符合题意.综上，(1,)+a.················································

·············································································4(2)由（1）知1a，所以,2x时，()ecos0xfxax=−，所以()10,xx，(

)0fx，()fx单调递减；()1,xx，()0fx，()fx单调递增，所以()10,xx时，()(0)0fxf=，则()10fx，又因为()e10f=−，所以()fx在()1,x上有唯一零点2x，即()fx在(0,)上有唯一零点2x.···············

························6因为()112211112esin21e2sincos1xxfxaxaxx=−−=−−，由（1）知()10fx=，所以11ecosxax=，···························

···································································································7则()112112e2esin1xx

fxx=−−，构造2()e2esin1,0,2ttpttt=−−，·······································8所以()2()2e2e(sincos)2eesincosttttpttttt=−+=−−，记(

)esincos,0,2ttttt=−−，则()ecossintttt=−+，显然()t在0,2上单调递增，所以()(0)0t=，···················

············································································································

·9所以()t在0,2上单调递增，所以()(0)0t=，所以()0pt，所以()pt在0,2上单调递增，所以()(0)0ptp=，·····················································

····························································10所以()()1220fxfx=，由前面讨论可知：112xx，12xx，且()fx在()1,xx单调递增，所以122xx

.································································································································12获得更多资源请扫码加入享学资源网

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