2024届河南省周口市项城市上学期高三期末数学试卷及答案

{#{QQABJQIEggCgABBAAQgCQwWYCkGQkBAAAKoOhAAMoAAAyRFABAA=}#}{#{QQABJQIEggCgABBAAQgCQwWYCkGQkBAAAKoOhAAMoAAAyRFABAA=}#}{#{QQABJQIEggCgABBAAQgCQwWYCkGQkBAAAKoOhAAMoAAAyRFABAA=}#}{#{QQABJQIEggCgABBAAQgCQwWYCkGQkBAAAKoOhAAMoAAAyRFABAA=}#}2024年高考备考精英联赛调研卷高三数学参考答案123456789101112CBDABCDB12CRA={x|xB2−5x−6≤=[B=(2,2+e2)(R)∩B=(2,6].z=a+i(a,b∈R)z+2=−i)⋅za+2=a−ba=4(a+2)+bi=−i)(a−bi)=(a−b)−(a+b)i|z=16+4=25.b=−(a+b)b=23DAααllABBCl⊂αl∩α=AlmllmA∉mlmCDααlmlmD.4AAC=4A20B40C20x=2(x−D(2,22)2+y=y≥2x=2.y=22(x−2+y2=y≥0)22−0k==−2.2−456B2πππππ295α+)=α+)+]=α+=1−2sin2α+)=1−2×=.32129C904000P(2000≤X<3000)=P(X≥4000)=1−0.3×2=0.3300[2000,3000)7=0.2[2000,3000)2300×0.2=60.Dπ122πωπωπf(x)=2sin(ωx+)|=×=f=2sin(ω+)331πωππωππωπ12××|2sin(ω+)=×|ω+)=ω+)=±.0<ω<22333π2πωω=f(x)T==4.28B19{#{QQABJQIEggCgABBAAQgCQwWYCkGQkBAAAKoOhAAMoAAAyRFABAA=}#}f(x)=(x−2+|x−1|1f(x)x=1810f(x)+∞)b=f−=f+c=f()=f().991g(x)=ex1−x−1g'(x)=ex1x∈g>g=0h(x)−g'(x)+∞)g(x)g'=0xx∈+∞)g'(x)>0g'(x)<01e0.1>1+ln1.1>1.h(x)=−x)ex−1(0<x<10h'(x)=−ex<001h=0.9e−1<0)b<a<c.e0.1<910109>e>1+ln1.1>1f+<f(e0.1)<f(99n∑x=00=(−2)n=32×(2)n=5=Ax=1a4=C5i=−1i=0n∑i=−1−(−32)=B3=C522C1×(−2)=−10D.i110aa=2aa23=233=2aa=a23=4A3=26=2431512×−2)6631263n}q==2n3q=n−3=2n−21=S6==B31−223+a411=q2=4CSn=×(2n−a2n−2Sn=22n−2−(2n−=×(2n−2)≥021+a2a2n≥2Sn24D.11f(x)Rf(0)=0f(−=−f=−a−2+3f(x)+∞)f(0)+f(=−a−2=−4a=2Aa=3x>0f(x)=2xxf(x)(−∞,0)Bf(2)4f(−2)f(2)=−f(2)=−(4+a2)=−(a+)≤−4a=2aaa1214CD.=x>0f(x)2x=+f'(x)=(2x−2−x)ln2a2x+∞)f'(x)>0f(x)+∞)f(x)(−∞,0)D12xx1112y221Ay=By=−==−2221y21k⋅k==−AM(x,y)=λ+µ(x,y)=λ(x,y)+µ(x,y)00001122xx412x=λx+µxλ−µ0λ1+µ2()1012ABx2=x=−===y21kOMy=λy+µy1λ1+µx2(λ+µ)1012029{#{QQABJQIEggCgABBAAQgCQwWYCkGQkBAAAKoOhAAMoAAAyRFABAA=}#}x=λx+µx2xB012−y=12y=λy+µy40122(λx+µx)212−(λy+µy)2=1λ(x2−41)+µ22(2221212−42)+2λµxx−λµyy=41214111=1y2=−24λµxx=4λµxx=1λµ1x2121222CD.13.1014.315.2lge)16.3341310b34ub=(−,)(,−)|b|5555a⋅bba⋅b|b|u=⋅=(4)=5a⋅b=.|b||b|143MABMA,B,N1111p723|=(||+||)=|AB||=2+=p=3F(,0).112222=d=|PF|xy2+63y+27=0∆=02x+23y+9=0|3+9|dd2Fl=3.4152lge)f(x)=0y=−lg(2e)x=mx−lg(2e)f(x)g(x)=xg(x)=x.y=−lg(2e)1(x,lgx)g(x)=xg'(x)=00xln101m=1x10ln10y−0=(x−0)y=+0−0ln100ln10ln102lge).10−=−lg(2e)ln1012x=0mm=2lge163hP1S=3S+S=3×(×h2+1×23)+33=33(h2+1+PPABABCABC211111S⋅h=S⋅r=×33(h2+1+⋅r×33⋅h=×33(h2+1+⋅r33333hr=.h2+1+139{#{QQABJQIEggCgABBAAQgCQwWYCkGQkBAAAKoOhAAMoAAAyRFABAA=}#}h2+4h2+4Rr(h2+h2h2+1+2hh(h−R)2+22=R2R===2h2h2+1+1h2+1+1=t(t>2)h2=t2−tRr(t2−t+tt2−t+4(t−2)2+t−2)+414t−21====[(t−2)++2]≥×(24+2)=3t2−t)t−2(t−2)224t−2Rrt−2=t=4h=223.1713020.25021C130C21202449P==≈.0.4905C5042600×=652×23004807802001203205006007H0.1100×(300×120−480×200)780×320×600×50021375χ2==≈52.885>10.8289260.001χ2H0.101812b+23acosAsinB=bcos2Ab+2acosAsinB=b−2sin2)bsin2A+3acosAsinB=02sinBsin2A+3sinAcosAsinB=0.4AB>0A∈(0,π)sinA+3cosA=02πtanA=−3A=.63B=2sinCb=c7123228310=b+cbcsinA=c=839c=4b=8a222−bccosA=1121149{#{QQABJQIEggCgABBAAQgCQwWYCkGQkBAAAKoOhAAMoAAAyRFABAA=}#}a=47.12PAB∩1912ABC⊂⊂3⊂OC∩OD=Ox⊂5CD.6OOyOzxyz=θ.7C(2,D−P(0,2cosθ,2sinθ)=(2,0)OP=2cosθ,2sinθ)CD=(−−8n=(x,y,z)⋅n=2x=002ycosθ+2zsinθ=0n⋅=0z=1n=(0,−tanθ|3tanθ+1|9⋅CDn2sin45°=<,>===|n|⋅||2tan2θ+1×22tanθ=3.10πθ∈(0,π)θ=P3)CP=(−3).3CD=(−262CD=(−−,−,)11|CD|2442PP−(⋅)=62||6.12201n=1n≥2a=S=111n(n+n(n−n2[(n+2−(n]−2a=S−S=[]2−[]2==n3nnn12241=1{an}an=n3459{#{QQABJQIEggCgABBAAQgCQwWYCkGQkBAAAKoOhAAMoAAAyRFABAA=}#}Snn(n+4n21411==(n++2)≥×(2+2)=1n4Snnn=162n+anan13+(n+3n11+1n=(n⋅n1=(−n⋅=(n[+]7n(n3+3n3(n3n11111111n=−+)+(+)−(+)+⋯+[+]=3−12323333343n3(n+(n+35121511512n>−−1>−3(n+n<7(n+<3nnn69n−111+1n311+1n=n1−[+]=−1−[+]=−−1.n3(n3n3(n3(n3+1n>−−−1>−512(n+311−>.11(n+3n>−n6.12512n+anan1n3+(n+311+1n=(n⋅n1=(−n⋅=(n[+]7n(n3+3n3(n3111111111n=−+)+(+)−(+)+⋯+(−n1[+]+(−n[+]32323333343(n−3n3n3(n+(−n=−1.9(n+3512nn<1n>−n>−105121511512nn−1>−3(n+3<n<7(n+n6.n>−n6.12512211OxyxOy1b=2b=1269{#{QQABJQIEggCgABBAAQgCQwWYCkGQkBAAAKoOhAAMoAAAyRFABAA=}#}bcos45°2a==22a=23x2C+y2=1.422C2c1c=151(−0)20)ly=k(x−y=k(x−2y(2k2+x2−k2xk2−2=0+7x+y2=124k222k2k2−23x+x==2−xx==1−8122+2+122+2+12k12k112k3xx=(x+x)−21212211+1y22−2Ay=(x+y=(x−FAMx11+1y22−2y(x−2)121×(3+=×(3−2)=10y(x+4213k[(x+x)−2]−2−kx+2k14(−k)(x−2)x−2−+2k12122=12=1212=(−k)(x+x−+−k12123k[(x+x)−2]−+kx−k2121212−1+x21k≠0=5x+x−6=012x+5x−64125x+x=6.1212221f(x)=x−2f'(x)=−sinx−2f'(x)≤0x∈(,)ππf(x)(,)44πx∈(,π)−x−2≤0479{#{QQABJQIEggCgABBAAQgCQwWYCkGQkBAAAKoOhAAMoAAAyRFABAA=}#}sinx2xm≥−.2sinxπxcosx−sinxg(x)=−x∈(,)g'(x)=−2x42x2πh(x)=xcosx−sinxx∈(,)h'(x)=−xsinx<04πh(x)(,)4πππ2πh(x)<−sin=×(−<044424πg'(x)>0g(x)(,)4sinπ2πg(x)<−=0m≥0m+∞).f(x)=x+41π1π2xπ2m=−x2f'(x)=−sinx52xπ2πϕ(x)=−sinx'(x)=−xπ2ππ2π'(x))'(0)=−1<0'()=>022π'(x))x0.x∈(x,)2πx∈0)'(x)<0'(x)>002πϕ(x)(0,0)(x,)062ππx∈(0,)ϕ(0)=ϕ()=0ϕ(x)<0f'(x)<022πf(x)).2π2xππx∈(,+∞)>1≥sinxf'(x)>0f(x)(,+∞).22ππx≥0f(x)≥f()=724f(x)πx<0f(x)≥.4πf(x)≥.843f(0)=1f(x)=t(x>0)π42<t<189{#{QQABJQIEggCgABBAAQgCQwWYCkGQkBAAAKoOhAAMoAAAyRFABAA=}#}x+x<π12π1<20<1<<2<π2π2x+x<π1−20<π−2<12π2f(x))f(x)>f(π−x)122f(x)=f(x)f(x)>f(π−x)912221π(π−x)2πF(x)=f(x)−f(π−x)=x+F'(x