四川省资阳市2022-2023学年高三数学（文）上学期第一次诊断考试试卷（PDF版附答案）

2020级注意事项：1．答卷前，考生务必将自己的姓名、准考证号填写在答题卡上。2．回答选择题时，选出每小题答案后，用铅笔把答题卡上对应题目的答案标号涂黑。如需改动，用橡皮擦干净后，再选涂其它答案标号。答非选择题时，将答案写在答题卡上。写在本试卷上无效。3．考试结束后，将答题卡交回。1U=−−{2，10123}，，，，Mxx={|0}Nx=x{|2−2}ZMN()=UA{3}B{23}，C{123}，，D{223}−，，i2zz1iz3i12122121AiBiCiDi555555553αOxαP2π2π(cossin)，sintan333333ABCD22221xx4y()y22AxByCDyx15sincos+=0πcos2=5167716ABC−D−252525256p:a1qfx()=+axcosxpqABCD7CDABECDFBEAB=aAC=bAF5151Aab+Bab+82425151Cab+Dab+8444文科数学第1页（共4页） 8“”堉2aa…12aaaa13158125A2B212732C2D29SA30B70C110D14012110ab+=2ab+ab2A8B16C24D3211fx()Rfx(2)−fx(2)f−x()+0−=x−−[2，1]131fx()ax4=−−xa0a1f(2)−=4|()|fk=ak=1A16B20C24D28π3π12fx()=+sinxcosx0fx()()，241A(04]，B(0]，3515C[3]，D(0，][，3]232文科数学第2页（共4页） yx≥2，13xy，x20≥，2xy_______yx≤2，3π14+=(tan1)(tan−−=1)_______4115abc|||||ab=a=b+=|2||abc+−=||c2_______2216xax−a2x4lna________.6017.12{}anSSSSaaann396511818.12△ABCABCabcb(2cos)Aa−=B3sin1A2DBCADBACAD=2bc=2a1912{}anSaa=+322aSa=+nn31nn11{}an123n2{}bn++++=an+1−2{}bnnTnbbbb123n文科数学第3页（共4页） 20123fx()x1ax=−+1a=1(10)，yfx=()ll2a≤0x0fx()xcos2112x+2fx()ax2=++xe1fx()a2fx()xxxxxxa−−3312122122[44]10G=2cos2π1l=GOPG6Q|PQ|2π2ABG∠AOB△AOBS323[45]10fx()|2xx=−+2+||21|.1fx()x4≤+322fx()Tabcabc++=Tab+ac≤.2文科数学第4页（共4页） 资阳市高中2020级第一次诊断性考试文科数学参考答案及评分意见评分说明：1．本解答给出了一种解法供参考，如果考生的解法与本解答不同，可根据试题的主要考查内容比照评分参考制定相应的评分细则。2．对计算题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应得分数的一半；如果后继部分的解答有较严重的错误，就不再给分。3．解答右端所注分数，表示考生正确做到这一步应得的累加分数。4．只给整数分。选择题和填空题不给中间分。一、选择题：本大题共12小题，每小题5分，共60分。1-5BDACC6-10AACBB11-12CD二、填空题：本大题共4小题，每小题5分，共20分。3511381421516(e−)4，226017.12aaa········································································25118{}aqnSSS396SSS+=2·················································································4369q=13a6a29+a=q1···························6111369aq(1aq)−(1a−q)−(1)111+=211−1−q−qq3693693611−2(1+−qq)q=−qqq+=212+=qq·······················94743aaaq+=+aq=+aqq(1)581114610=aq==qaq2a221111aaa··································································12511818.121bA(2cos)a−=B3sinsin(2cos)B−=A3sinsinAB······························································20BπsinB0π3sincosAA+=22sin()A2+=···············································46ππ7πππ0AπA+A+=66662文科数学答案第1页（共5页） πA=······················································································631π12ABDABAD=BDhsinh△ABCBC2621π1ACDACAD=CDhsin262BDAD1==············································································7CDAC221ADAB=+AC············································································833242124AD=AB+AC+ABAC99941422π12244=2cos+c+ccc=cc=3b=23·················109993922222πabc=bc+−=+−=2cos(3)(23)32393a=3·······················································································12191212aSa=+n≥22aSa=+nn1nn−−11122aaa−=aa=2n≥2··································1nnn−1nn−1{}an2··························································2aa=+3243aa2=+3111a=2····························································································31n{}anan=2···································································4123nn+12++++=−22bbbb123n11n=1=2b1=·································································5b121231n−nn≥2++++=−22bbbb1231n−nn+1nnn1=(2−2)(2−−2)=2b=b=nn1bn22nb=*nnnN··········································································72123nT=++++n23n222211121nn−T=++++········································82n231nn+2222211(1−)11111nn22nn+2T=++++−=−=−12n222232n2n++112n12n+11−2文科数学答案第2页（共5页） ···········································································································11n+2T=−2···········································································12nn22012321a=1fx()x1x=−+fx()3x=−1································1332(1)xx，x−+y−(x−x+1)(3=x−1)(xx−)···········200000003232(10)，−+xx−=xx−−1(31)(1)23xx0−=················30000003323323x0=0x0=f(0)1=f(0)1=−f()=f()=·······42282423yx=−+1yx=−(1)··············································64322gx()x1cosax=−x+−x0gx()3xa=x−+sin················72ux()gx()3==x−+axsinx0ux()6xx=+cos·························8ππ0xux()0x≥ux()0······································922x0ux()0ux()gx()(0，)+ga(0)=−·····10因为a≤0ga(0)0=−≥gx()g(0)0≥gx()(0，)+gx()g(0)0=a≤0x0fx()cosx·································122112x+2x+11fx()ax2=++fx()a=−+xxeex+1fx()fx()0≥a≥············································2xex+1xgx()=gx()=−xxeex0gx()0gx()x0gx()0gx()x=0gx()g(0)1=a≥1a[1，)+································································4x+12fx()fx()0=a=xe1x=0gx()g(0)1=x−1gx()0x→+gx()0→gx()=a01a−10xx12··························61(01)，(20)，yx=−+121x+11x1hx()=gx()(−−x+1)=+x−1x0hx()=−+xx2e2e2x1x−1ux()hx()==−+x0ux()=x0xxe2e01xux()0ux()hx()x1ux()0ux()hx()文科数学答案第3页（共5页） 11x0ux()uh(1)(1)==−+0e21x0ux()hx()=0≥，则hx()h(0)0=，即hx()gx()(=−−x+1)021x0gx()x1−+21−+=xa1xxa=−22··················································9442yx=+1x+11tx()gx()=−−=x1x1(x1)(−−=1)+−−10xxxeex−10x0e1tx()gx()=−−x10−10xgx()x1+xa+=1x3xa3=−1−10xxxx1342x−xx−x=−33a··························································12214322[44]10π1==2cos26ππ==2cos1P(1，)·················································1P36GQ(20)，·····················································222π||1PQ2212cos=+−523=−·············································462π2A()A，0≤≤πB()B，+································532π=2cos2=+2cos2()AB312π3△AOBS==||sin||········································6ABAB23434π13=+4|cos2cos(2)|=|3cos2−(cos2+sin2)|432233233=−|cos2+sin2cos2|=−|(1cos4)++sin4|224433π=|3sin4−cos4−1|=|2sin(4−)1|−···········9446π3π5π334−==S=max62124文科数学答案第4页（共5页） 33△AOB·························································10423[45]101311x−fx()x2x=−22x+−14−+≤−−≤x············125211−≤x≤1fx()x2x=−22x++14++≤−≤x≤1················2225x1fx()2xx=x−+22++14≤1x≤······························3335{|xx}−≤≤···········································5532fx()|2xx=x−x+2+1||2−−||2+2(=1)|≥231(22)(xx−+1)02≤−≤x≤1“”2fx()T3abc++=3············································71121123abac+=ab+++acab+2(=ac++)()()abc2=≤222222213b==ca=“”2432abac+≤.·········································································102文科数学答案第5页（共5页）