【文档说明】福建省龙岩市第一中学2023届高三上学期第二次月考数学答案.docx，共(7)页，327.597 KB，由小赞的店铺上传

转载请保留链接：https://www.doc5u.com/view-6851f0875801fc6baf386a33a8fc929e.html

以下为本文档部分文字说明：

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222-5cos216bcaAbc+==.故5cos1

6A=···················································································································

·····················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=+−++222223

2cossincossinsin2cos22222xxxxxx=+−++()2231cos2cossinsin222xxxx+=−++π13sin232x=++·································

··························································································3令π2π,3xkk+=Z，解得ππ

62kx=−+，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,121

2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=+，()log0,0,1byaxabb=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=，10

b=−，120c=∴()221110120107022yxxx=−+=−+，,()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70元．·······································7（3）令()()()1701

010210fxgxxxx==−+−−(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)2f

xxaxaxa=−+−，可得2()21fxaxax=−+−．因为(1)2211faaa=−+−=+，13(1)21122faaa=−+−=−，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()0fx=，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ln210a+−，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11AB

AC=，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1AA

BC⊥；·························································································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117142cos42AHA

D===，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−，12

14214,2,,,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RtRtDEHDOB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()esinxgxax=+，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，()0fx；1,2xx，()0f

x，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()11

2211112esin21e2sincos1xxfxaxaxx=−−=−−，由（1）知()10fx=，所以11ecosxax=，··················································································

············································7则()112112e2esin1xxfxx=−−，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微信公众号www.xiangxue100.com