辽宁省新民市高级中学2023-2024学年高三10月月考 数学答案和解析

2023-11-13 · U1 上传 · 10页 · 330.2 K

参考答案1.【答案】C24【解答】解:集合A{x|x„4x}{x|0„x„4},B{x|3x40}{x|x},344AB{x|x„4}(,4].故选:C.332.【答案】A34i(34i)(12i)112i112【解答】解:由已知可得复数zi,12i(12i)(12i)55511所以复数z的实部为,故A正确,5112i2112复数z的共轭复数z,虚部为,其在复平面内对应的点(,)在第一象限,故B错误,C错误,55555112|z|()2()25,故D错误,故选:A.553.【答案】D4【解答】解:数列{a}是等差数列,数列{b}是等比数列,且aa,bbb8,nn79326104aa2aaa,可得aaa2,b38,则b2,可得bb1413,313879338136648aaa23813.故选:D.b4b8134.【答案】C3sin2x【解答】解:函数y,可知函数是奇函数,排除选项B,当x时,f()23,排除A,11cosx3312x时,f()0,排除D.故选:C.5.【答案】A1【解答】解:由题意,可知:,1.alog521blog0.50.2log1log215log25log2422511111c0.50.21,b最大,a、c都小于1.alog2,c0.50.2()55.55log2522211而log5log4252,.ac,acb.故选:A.225log2526.【答案】D【解答】解:过E作EF//BC,交AB于F,设A到BC的距离为h,111则SEFh,SCDh,由已知得EFCDBD,ABE2ADC221{#{QQABRQ4QggCAAAAAAQgCAwHyCgIQkAACCIoOBBAEMAAAwBFABAA=}#}11211故E为AD的中点,BE(BABD)(BABC)BABC,ACBCBA,22323111212111113故BEAC(BABC)(BCBA)BABCBABC323232.故选:D.2323623624111117.【解答】解:由题意得acsin120asin60csin60,即acac,得1,222ac11c4ac4ac4a得4ac(4ac)()5…25459,当且仅当,即c2a时,取等号,故选:B.acacacac8.【答案】B【解答】解:作出函数f(x)的图象如图所示,当y0时,f(x)y只有一个根,当0y„1时,f(x)y有二个根,当1y„2时,f(x)y有三个根,当y2时,f(x)y有二个根,令tf(x),11F(t)f(t)3t的零点即为f(t)3t的根,221作出两函数yf(t)与y3t可知两函数有两交点t,t,且2120t11,1t22,1所以函数F(x)f[f(x)]3f(x)的零点个数是5个.故选:B.2二.多选题(共4小题)9.【答案】AB【解答】解:对于A,a2aa2a0,a1或a0,a1是a2a的充分不必要条件,故A正确,对于B,m0,x2,x20,mmm函数yxx22…2m2,当且仅当x2,即x2m时,等号成立,x2x2x22m26,解得m4,故B正确,11对于C,当a2,b1时,则,故C错误,ab对于D,当b0时,满足a//b,b//c,但a//c不一定成立,故D错误,故选:AB.2{#{QQABRQ4QggCAAAAAAQgCAwHyCgIQkAACCIoOBBAEMAAAwBFABAA=}#}10.【答案】BCD【解答】解:2212对于选项A:AEADDEADDCAD(ACAD)ADAC,故选项3333A不正确;DFDE222对于选项B:易证明DEF∽BFA,所以,所以DFFBDB,BFAB335故选项B正确;22122对于选项C:AEBD2,即(ADAB)(ADAB)2,所以ADABADAB2,33312ABAD11所以1ABAD42,解得ABAD1,cosAB,AD,33|AB||AD|212因为AB,AD[0,],所以AB,AD,故选项C正确.333232对于选项D:FBFCDB(FDDC)(ABAD)(BDAB)(ABAD)[(ADAB)AB]55555332926293627(ABAD)(ABAD),AB3ABADAD4,故选项D正确.555252525252525故选:BCD.11.【答案】ABn(14n3)【解答】解:由题设可得:a14(n1)4n3,Sn(2n1),nn2SS10(119)n2n1,数列n的前10项和为1319100,故选项A正确;nn2又若,,成等比数列,则2,即2,解得,故选项正确;a1a3ama3a1am94m3m21B11111(),anan1(4n3)(4n1)44n34n1n1111111111()(1),i1aiai1415594n34n144n1n16由可得:n6,故选项C错误;i1aiai125又,由等差数列的性质知:,,*,amana2a10mn12mnN12m11611161n16m1255()(nm)(17)…(21617),当且仅当时取““,等号取不mn12mn12mn121248n511625到,即,故选项D错误,故选:AB.mn123{#{QQABRQ4QggCAAAAAAQgCAwHyCgIQkAACCIoOBBAEMAAAwBFABAA=}#}12.【答案】BCDlnx【解答】解:f(x),故f(x)在(0,1)递增,(1,)递减,其图像如下:x21易得若f(x)2m有两个不同解,则02m1,则0m,故A错2误,当k„0时,ykx与yf(x)显然有且仅有1个交点,当k0时,则yf(x)与ykx相切时,有且仅有1个交点,lnx1lnx设切点为,,切线方程为00,(x0y0)y2(xx0)x0x01lnx1lnx1e将原点代入:则002,,2(x0)lnx0x0ekx0x022e故k„0或k,则B正确;21xex恒成立,f(x)在(1,)上单调递减,lnx1x1lnex1f(x)f(ex),故C正确;xexex31ln21ln813422lne1ln221lne1ln2122,2e22e22e22e即比较f(22)与f(e)大小,又因为22e,f(x)在(1,)递减,故f(22)f(e),D正确,故选:BCD.三.填空题(共4小题)313.【答案】.515【解答】解:终边过点P(1,2),cos,12225533sin(2)cos22cos212()21.故答案为:.25552514.【答案】.5【解答】解:向量m(1,2),n(2,),若mn,则mn220,1,2mn(0,5),(2mn)m25252mn与m的夹角余弦值为,故答案为:.|2mn||m|554{#{QQABRQ4QggCAAAAAAQgCAwHyCgIQkAACCIoOBBAEMAAAwBFABAA=}#}215.【答案】1.2【解答】解:f(x)Asin(x),12由图像知A1,T4,,284由五点作图法可知,10,,f(x)sin(x),4444222f(1)0,f(2),f(3)1,f(4),f(5)0,f(6),f(7)1,f(8)2222,2f(1)f(2)f(3)f(8)0,又222025286,2f(1)f(2)f(3)f(2022)2520f(1)f(2)f(3)f(6)1,22故答案为:1.216.【答案】2n34n8.【解答】解:,2(1an)EnFnEnC(an12)EnB,又,,三点共线,EnC2(1an)EnFn(2an1)EnBBFnC2(1an)(2an1)1,an132(an3),又a134,是以首项为,公比为的等比数列,n1n1,n1,{an3}42an3422an23又,,,BCnBFnEnCEnBn(EnFnEnB)EnCnEnFn(1n)EnB2(1a)又,nn,又n1,EnC2(1an)EnFn(2an1)EnBan231n2an18(12n)2(12n13)2n24,T4n2n34n8.nn12故答案为:2n34n8.四.解答题(共6小题)abc17.【解答】解:(1)2bcosBacosCccosA,由正弦定理,sinAsinBsinC得2sinBcosBsinAcosCsinCcosA,即2sinBcosBsinAcosCsinCcosA,即2sinBcosBsin(AC)sinB.20B,sinB0.2cosB1,即cosB,25{#{QQABRQ4QggCAAAAAAQgCAwHyCgIQkAACCIoOBBAEMAAAwBFABAA=}#}又0B,B.(2)ACD中,AD5,AC7,DC3,4AD2DC2AC252327212cosADC.0ADC,ADC.2ADDC25323在ABD中,AD5,B,ADBADC,43ADAB5AB56由正弦定理,得,AB.sinBsinADB232225118.【答案】(1){x|xk,kZ}.(2).1273【解答】解:(1)因为a(2cosx,),b(sin(x),1),f(x)ab,233133313f(x)2cosxsin(x)2cosx(sinxcosx)sinxcosx3cos2xsin2xcos2xsin(2x)322222223当sin(2x)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