（教师版）浙江省宁波市镇海中学2026届高三模拟预测数学试题与解析

1.已知集合A={y|y=-x2{,B={x|x>2}，则下列说法正确的是A.A∩B=BB.AUB=RC.([RA(UB=BD.([RB(∩A=A【解析】A={y|y=-x2{={y|y≤0{,B={x|x>2},:A∩B=g,AUB≠R,CRA=(0,+∞(,(CRA(UB=(0,+∞(≠B,CRB=(-∞,2],(CRB(∩A=(-∞,0[=A.A.1.4mB.2.8mC.4.2mD.5.7m【解析】以拱顶为原点,竖直向下为y轴,设抛物线为y=ax2(a>0(,y=1时水:(1,1(在抛物线上,:a=1,水面下降1米后,y=2,x2=2,水面宽为22≈2.8米.A.B.C.D.(x2-1(ex【解析】由图像知,x→+∞⇒f(x(→0+,且f(x(>0;x→-∞⇒f(x(→+∞,并有两个零点,选项B中f(x,零点为x=±1x→-∞⇒f(x(→+∞,x→+∞⇒f(x(→0+,与图像相符.A.πB.2πC.3πD.4πA.充分不必要条件B.必要不充分条件C.充要条件D.既不充分也不必要条件(1+sinB(cosAcosB-cosA=sinBcosA-sinAcosB,cosB-cosA=sinB-A∴A=B,所以为充要条件.A.3B.C.23D.3+1则PF12+PF22=F1F22,x+c2+9+x-c2+9=4c∴(c-x((c+x(=9,∴c-x=3,∴c=23,x=3.PF=6,PF2=23,2a=PF1-PF2=6-23,a=3-3,8.已知函数f(x(=ex-x+1,g(x(=klnx,h(x(=kx-k,在区间(0,+∞(上恒有f(x(≥h(x(≥g(x(,求kA.[0,e2-1[B.(0,e2-1[C.(0,e-1[D.[0,e-1[【解析】方法一:由h(x(≥g(x(⇒k(x-lnx-1(≥0恒成立,x=1显然成立,x>0且x≠1时,x-lnx-1>0,∴必有k≥0由f(x(≥h(x(⇒ex-x+1≥k(x-1(,∵ex-x+1>x+1-x+1=2,当0<x≤1时,左边>2,右当x>1时,kmin,而e2-1∴0≤k≤e2-1,A正确.方法二:h(x(-g(x(=k(x-1-lnx(,lnx≤x-1,等号在x=1时成立.h(x(≥g(x(恒成立⇔k≥0.k=0时,f(x(≥0=h(x(=g(x(成立.以下设k>0,令F(x(=f(x(-h(x(=ex-(k+1(x+k+1.F,(x(=ex-k-1,F聂(x(=ex>0,F,(x(=0⇒x=ln(k+1(F(x(min=F(ln(k+1((=(k+1((2-ln(k+1((k>0,k+1>0.F(x(≥0恒成立⇔2-ln(k+1(≥0.ln(k+1(≤2⇒k≤e2-1综上,0≤k≤e2-1.方法三:h(x(≥g(x(⇒kx-k≥klnx⇒k(x-1-lnx(≥0.∵x∈(0,+∞(时x-1-lnx≥0恒成立,∴k≥0.x∈(1,2(时,φ,(x(<0,φ(x(单调递减;x∈(2,+∞(时,φ,(x(>0,φ(x(单调递增.∴φ(x(min=φ(2(=e2-1.∴k≤e2-1.当x∈(0,1(时,x-1<0,ex-x+1>0,且k≥0,∴ex-x+1≥k(x-1(恒成立.当x=1时,e≥0恒成立.综上,0≤k≤e2-1.A.χ2独立性检验方法不适用于普查数据C..的最小值为1D..的最大值为32=2+y2=4.,则三个向量的终点构成边长为2的等边三角形,所以.的最小值为-1,最大值为3,C错,D对.11.已知无穷数列{an{前n项和为Sn,若存在i,j∈{1,2,…,n{,当i≠j时,lSil=lSjl,则称{an{为“绝对数A.已知数列an=2n-5(n∈N*(，则数列{an{为“绝对数列”B.若数列{an{和{bn{均为“绝对数列”,则{an+bn{为“绝对数列”C.若等比数列{an{为“绝对数列”,则公比为-1D.存在两个公差均不为0的等差数列{an{和{bn{,使得数列{an{,{bn{,{an+bn{和{anbn{均为“绝对数列”【解析】方法一:对于A,:a1=-3,a2=-1,a3=1,：S1=S3,：lS1l≤lS3l，A正确.对于B,取an=2n-5(受A的启发),bn=5-n,an+bn=n,显然{an+bn{不为“绝对数列”,B错.对于C,取an=(-2(n-1,则a1=1,a2=-2,：S1=1,S2=-1,lS1l=lS2l符合{an{为“绝对数列”,但q=-2≠-1,C错.对于D,取an=bn=n-2,则对{an{,{bn{而言,S1=S2,：lS1l=lS2lan+bn=2(n-2(,anbn=(n-2(2,也都有lS1l=lS2l,符合{an{,{bn{,{an+bn{,{anbn{均为“绝对数列”,D正确.方法二:对A,an=2n-5,Sn=n(n+1(-5n=n2-4n=n(n-4(,S1=-3,S3=-3,lS1l=lS3l,A正确.对B,取an=n-2,bn=3-n,设前n项和分别为An,Bn,Cn,AnA1=A2=-1,所以{an{为绝对数列;BnB2=B3=3,所以{bn{为绝对数列;an+bn=1,Cn=n,i<j时,lCil=i,lCjl=j,所以lCil≠lCjl,B错误.对C,取等比数列an=(-2(n-1,则q=-2,S1=1,S2=-1,lS1l=lS2l,但q≠-1C错误.对D,取an=n-2,bn=2n-4,公差分别为1,2,A1=A2=-1,B1=B2=-2,an+bn=3n-6,C1=C2=-3,anbn=2(n-2(2,P1=2,P2=2,D正确.方法三:对于A,an=2n-5⇒Sn=n2-4nS1=-3,S3=-3⇒lS1l=lS3l,A正确.对于B,取等差数列an=2n-5⇒Sn=n2-4n,有S1=S3=-3取等差数列bn=-2n+7⇒其前n项和Tn=-n2+6n,=T对于C,取a1=1,q=-2⇒S1=1,S2=-1⇒lS1l=lS2l,此时公对于D,取an=n-2,bn=n-3a2=0⇒S1=S2=-1b3=0⇒其前n项和Tn满足T2=T3=-3an+bn=2n-5⇒其前n项和Mn=n2-4n⇒M1=M3=-3anbn=(n-2((n-3(⇒第2项为0⇒其前n项和Hn满足H1=H2=2四者均满足条件,D正确.lzl=l1+i【解析】由ab+5=3a+2b得(a-2((b-3(=1,a>2,所以ba+b=5+(a-2(+方法三:设经过n次交换后,黑球在甲手变形得到pnpn因为p所以pnn化成pnn,n=5→p：、3sinC-cosC=1,2sin(C-1,sin(C-=,而-<C-<,：C-=→Cπ在△ABC中,c2=a2+b2-2aba2+b2-ab:△ABC为锐角三角形解法2::CE平分LACB：====sin(=sinAcosA=+2taA在锐角△ABC中tanA(2)求直线DB1与平面AB1E所成角的正弦值.BBBB1MCEAFDB1CCEMGHBAFD=3,：而CEBC.丫FM≠平面AB1E,AGC平面AB1E,丫AEC平面ABCD平面BB1E丄平面ABCD过B1作B1H丄BE于点H,B1H丄平面ABCD,丫B1E=BE=BB1=2,：△BB1E为等边三角形,：B1H=3,设D到平面AB1E的距离为h,DB1与平面AB1E所成角为θ,：sin17.设数列{an{的前n项和为Sn,a1=-50,当n≥2时满足nSn-(n+1(Sn-1=n3-n.(1)求Sn；(1)nSn-(n+1(Sn-1=n(n+1((n-1(,n≥2，两边同除以n(nn-1,∴n≥2时,Sn=(n+1n2-n-25(,而S1=-50也满足上式,18.在平面直角坐标系xOy中,抛物线y2=4x上点P(t2,2t(处的切线与双曲线x2-y2=1相交于不同的两点A,B.【解析】方法一:(1)y2=4x在P(t2,2t(处的切线方程为即x-ty+t2=0,易知t≠0.设A(x1,y1(,B(x2,y2((2)A(x1,y1(,B(x2,y2(,AB中点R(x0,y0(联立⇒(t2-1(y2-2t3y+t4-1=0,Δ=4t6-4(t2-1((t4-1(=4(t4+t2-1(>0lABl=、1+t2.ly1-y2l=、1+t2.,y0==-=1+t2.(3(3)设直线AB与x轴交于E,∴E(-t20(方法二:(1)抛物线y2=4x在P(t2,2t(处的切线为ty=x+t2由题意,t≠0,所以x=ty-t2设A(x1,y1(,B(x2,y2(,且xi=tyi-t2,i=1,2.联立得(ty-t2(2-y2=1,(t2-1(y2-2t3y+t4-1=01,于是y1+yy1y2=t2+1,(y1-y又x1+x2=t(y1+y2(-2t所以AB中点为M-,-(若P为AB中点,则t∴t2=2,∴t=±2若△ABC为正三角形,且C在x轴上,则Mt2(y1-y2(2此时t4+t符合题意.ΔOABx1y2-x2y1l19.在平面直角坐标系xOy中,曲线Γ1:y=ax(a>1(与Γ2:xy=m(m>0(交于点A.3若直线OA与Γ1交于另一点B,且lOBl≥2lOAl,【解析】方法一:(1)Γ1:y=ex,y,=ex,k=1,Γ1在(0,1(处的切线方程为y=x+1.(2)设A(x0,ax(,Γ1:y=ax,y,=axlna，∴Γ1在A处的切线斜率kOA=ax0lna⇒x0lna=1,ax0=e,又∵Γ1与Γ2交于点A,∴A也在Γ2上,∴x0.ax=m⇒ex0=m∴am=aex=(ax(e=ee.(3)设A(x1,y1(,B(x2,y2(,∵O,A,B三点共线，令g(tg,(t运用了lnt,当且仅当t=1时取“=”)：g(t(在[2,+∞(上单调递减,0<g(t(≤g(2(=ln2→0<lny1≤ln2,1<y1≤2又:A也在Γ2上,：x1y1=m,：am=axy=(ax(y=y=eylny而y1lny1在y1∈(1,2[上单调递增,：0<y1lny1≤ln4,：1<am≤4.(2)设A(x1,ax(,x1>0,k=lna，:OA与Γ1相切于Akax1kx1=1.2,m=x1ax,：ln(am(=mlna=x1axlna=kx1ekx=e(3)设A(x1,ax(,B(x2,ax(，由lOBl≥2lOAl，可设x2≥2x1>0，令r,