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上饶市2023届第一次高考模拟考试数学(文科)参考答案一、单选题题号123456789101112答案DAABDCCCBDDB10.【解析】76,76xx1、1431223726,2、147125277256
,3、14111292117296,4、141121572136max=,无解,11.因为,62==RPC,则PC为球O的直径,所以,90=
=PBCPAC又,4,==⊥BCACBCAC所以,24,52===ABPBPA在PAB中,515sin=PAB所以PAB外接圆直径,3310sin==PABPB故截面圆面积为325.12.因为xxxgsin22sin)(+=是
上的奇函数,且最小正周期,2),1)(cos1cos2(1cos22cos2)('+−=−+=xxxxxg当2,0x时,在3,0x上单调递增,在35,3x上单调递减,在2,35x上单调递增,则,1233)3()(max==g
xg又0)2()()0(===ggg,所以当2,0x时,方程1)(=xg有两个不同的解,所以)(xf在2023,0x上共有2024个零点.二、填空题13.914.0.0215.π316.①②③三、解答
题17.【解析】根据题表格中数据知,男患者“痊愈快”的概率估计为503910078=,.....................3分女患者“痊愈快”的概率估计为251610064=.........................................
.....................6分(2)痊愈快慢性别痊愈快痊愈慢总计男性7822100女性6436100总计14258200...................................8分84
1.376.420599800581421001002264367820022=−=)(K.........................................11分所以有95%的把握认为患者性别与痊愈快慢有关。................
...................................12分18.【详解】(1)因为2122nnnSSS++−+=,所以2112nnnnSSSS+++−−+=,即()()2112nnnnSSSS+++−−−=,.................
.............2分则212nnaa++−=....................................................................................
........................3分又12a=,24a=,满足212aa−=,所以na是公差为2的等差数列................................................................................4
分所以数列na的通项公式21)22nann=+−=(...........................................................6分(2)设数列nb的公比为q,因为11b=,所以2230bbqq+=+=,所以()10q=−或舍去,()1
1nnb−=−..................................................................9分所以()112nnnncabn−==−212
22468221)42.nnTcccnnn=+++=−+−++−−=−(...............................12分19.【解析】(1)在ABC中,因为===120,32ABCBCAB,所以6=AC在BCD中,4,3
2==DCBC且=30BCD,所以2=BD,因为222DCBDBC=+,由勾股定理可得PDCBA=90CBD...................................4分PBDCBCBDPBDBDCBCBDPBD平面平面平面平面平面⊥⊥⊥因
为PBCCB平面,所以PCBPBD平面平面⊥................................................6分(2)设点B到PCD平面的距离d由第(1)可知,=60CDB,所以=120ADB,所以3sin21==PDBPBPDSPDB23
2331v==−PDBC..........................................................................................8分在PCD中,,,4,262===CDPDPC415sin,41cos=−=
PDCPDC,所以15sin21==PDCDCPDSPDC,故dPDCB=−1531v又=−PDBCv21531v==−dPDCB,所以5152=d.......................................................
......................................................12分20.【详解】(1)当1a=时,()lnfxxx=+,设切点为()000,lnxxx+,又1()1fxx=+,切线斜率011kx=+......................
...................................................................................2分切线方程为00001(ln)1()yxxxxx−+=+−.Q过点
(0,0),00001ln1()xxxx−−=+−,0xe=...........................................4分所求直线方程为:11yxe=+....................................
...............................................5分(2)由题意,方程(ln)exaxxx+=,显然0x,0a,方程等价于1lnexxxax+=().........................
.........................................................................................6分记ln()0exxxgxxx+=
(),令()()211ln()0exxxxgxx+−−=,得1ln0xx−−,01.x()gx在区间(0,1)上单调递增,在区间()0+,上单调递减.........................
.......8分又0,x→()gx→−;1(1);ge=,()0.xgx→+→结合图形可知,..........................10分方程()exfxx=有两个不相等的实根时有110,.aeae......
..................................12分(另解:()中令lntxx=+,则1ttae=(),同理可得)21.解:(1)由题意可得+=+=+=222322cbacaa,解得
===312cba,所以椭圆方程为1422=+yx................4分(2)由题意知,直线l的斜率不为0,则不妨设直线l的方程为(2)xmytt=+,联立2214xyxmyt+==+消去x得(
)2224240mymtyt+++−=,()()222244440mtmt=−+−,化简整理得224mt+,设),(),,(2211yxQyxP,则212122224,44mttyyyymm−−+==++,.................
.......................................6分因为以线段PQ为直径的圆经过A,所以0=AQAP,得()()1212220xxyy−−+=,将1122,xmytxmyt=+=+代入上式,得()()2212121(2)(2)0
myymtyyt++−++−=,得()22222421(2)(2)044tmtmmttmm−−++−+−=++,解得65t=或2t=(舍去).所以直线l的方程为65xmy=+,则直线l恒过点).0,56(M.............................
...................................................................9分因为过点A做PQ的垂线,垂足为H,所以H在以AM为直径的圆周上,所以点H的轨迹方程为:,2545822=+−yx除去点)
.0,2(A..............................12分22.【详解】(1)由332xtyt=+=−+,消去参数t得3330xy−−=,即直线l的普通方程为3330xy−−=;.................................
................................2分由2sin4cos=,得22sin4cos=,∵cosx=,siny=,∴24yx=,即曲线C的直角坐标方程24yx=.................
...............................................................4分(2)直线l的斜率为33,则DE的斜率为3−,所以DE的倾斜角为120,故设直线DE的参数方程为12232
xtyt=−=(t为参数),.................................6分代入24yx=,得238320tt+−=,设点D对应的参数为1t,点E对应的参数为2t,则121283323tttt+=−=−,且D在x轴上
方,有10t,20t..................................7分故()21212121212124111174ttttttPDPEtttttt+−−+=−===−−,即11PDPE+的值为74.
..............................................................................10分23.【详解】(1)函数()2123fxxx=
−+−,当32x时,()2123fxxxm=−+−,解得:3124mx+;当1322x时,()2fxm=,且1322x;当12x时,()1232fxxxm=−+−,解得:1142mx−.
()fxm的解集为15,22−,5142m+=且1142m−=−,则6m=;.......................5分(2)1116abc++=,证明:()11111366aab
bccabcabcabcbcacab++=++++=++++++()1133222323662abacbcbacacb+++=+=,当12abc===时等号
成立......................................................................................1