2021高考数学模拟试卷十五

一、选择题:本大题共10小题,每小题4分,共40分.在每小题给出的四个选项中,只有一项是符合题目要求的.题号12345678910答案CABBCDABCD 二、填空题:本大题共7小题,多空题每题6分,单空题每题4分,共36分.11.-1;　12.;0或π　13.1;λ<-2　14.;515.　16.-1　17.三、解答题:本大题共5小题,共74分.解答应写出文字说明、证明过程或演算步骤.18.解:(Ⅰ)f(x)=2sin xcos x+2cos2x=sin 2x+cos 2x+1=2sin(2x+)+1..........................................................................3分∵x∈[0,],∴2x+∈[,],sin(2x+)∈[-,1],..........................................................6分∴f(x)∈[0,3]..........................................................................7分(Ⅱ)∵f(x+θ)-1=2sin(2x+2θ+),若函数f(x+θ)-1为奇函数,即g(x)=sin(2x+2θ+)为奇函数,.........................................10分由2θ+=kπ(k∈Z),得θ=-(k∈Z)............................................................13分又θ∈[-,],∴θ=-或....................................................................14分19.(Ⅰ)证明:取B1D1的中点E,连接C1E,OA,AE,易知C1E=OA且C1E∥OA,.....................................3分所以C1EAO为平行四边形,所以C1O∥EA,........................................................6分所以C1O∥平面AB1D1....................................................................7分(Ⅱ)解法一:过点C作平面AB1D1的垂线,垂足为G,连接B1G(图略),则∠CB1G就是直线B1C与平面AB1D1所成角的平面角...8分又CG是点O到平面AB1D1的距离的2倍,连接EO,由B1D1⊥EC1,B1D1⊥EO,知B1D1⊥平面AEO,所以平面AEO⊥平面AB1D1,在△AEO中,作OH⊥AE,垂足为H,即OH⊥平面AB1D1.......................11分由题可得AO=,B1C=,AE=2,在Rt△AEO中,OH==,所以点C到平面AB1D1的距离为,............................................................13分所以sin∠CB1G=.......................................................................15分解法二:以O为坐标原点,OA,OB,OE所在的直线分别为x,y,z轴,如图所示,建立空间直角坐标系O-xyz,得A(,0,0),B1(0,1,1),D1(0,-1,1),C(-,0,0),......................9分所以=(-,1,1),=(0,2,0),=(-,-1,-1).............................................................10分设平面AB1D的一个法向量为n=(x,y,z),则......................................................12分得令x=1,有y=0,z=,所以n=(1,0,).............................................................13分记α为直线B1C与平面AB1D1所成角的平面角,则sin α==..........................................15分 20.解:(Ⅰ)设等差数列的公差为d,等比数列的公比为q(q>0),由题得a5=b3-a3=b1-a1,..................................................................3分解得d=3,q=2,所以an=3n-8,bn=2n............................................................7分(Ⅱ)cn=an·bn=(3n-8)·2n,Sn=c1+c2+…+cn=(-5)·2+(-2)·22+…+(3n-8)·2n,①2Sn=(-5)·22+(-2)·23+…+(3n-11)·2n+(3n-8)·2n+1,②由①-②,得-Sn=-10+3(22+23+…+2n)-(3n-8)·2n+1=-22-(3n-11)·2n+1,即Sn=22+(3n-11)·2n+1..................................................................12分易知当1≤n≤3时,(3n-11)·2n+1<0;当n≥4时,(3n-11)·2n+1>0.又S1=-10,S2=-18,S3=-10,所以当n=2时,Sn取到最小值.............................................15分21.(Ⅰ)证明:设lAB:x=my+1(m≠0),代入y2=4x,消x得y2-4my-4=0,设A(x1,y1),B(x2,y2),有y1+y2=4m,y1y2=-4,...........2分所以M的纵坐标yM=2m....................................................................3分lCD:x=-y+1,解得N(-1,2m),.................................................................5分所以yM=yN,所以MN垂直于y轴...............................................................6分(Ⅱ)解:可得lAN:x+1=(y-2m),令y=0,得xQ=-1=.....................................................7分由y1+y2=4m,y1y2=-4,得m=-,又x1=,所以xQ=====-x1.......................................................................10分所以S△AQB=|QF||y1-y2|=(x1+1)|y1-y2|=(x1+1)=2(x1+1)=2(+1)(+)=(+8y1+)................................................................12分记f(y1)=+8y1+,则f'(y1)=3+8-==,令f'(y1)>0,解得>,即y1>,所以f(y1)=+8y1+在(0,)上递减,在(,+∞)上递增,所以(S△AQB)min=f()=..........................15分22.(Ⅰ)解:当a=1时,f(x)=ln(x+2)-2x+1,所以f'(x)=-2,..............................................2分且f(1)=ln 3-1,函数y=f(x)在x=1处的切线斜率k=f'(1)=-,...........................................4分所以函数y=f(x)在x=1处的切线方程为y-(ln 3-1)=-(x-1),即y=-x+ln 3+.........................................................................6分(Ⅱ)证明:令f'(x)=-2a=0,解得x=-2a,所以函数f(x)在区间(-2a,-2a]上单调递增,在区间[-2a,+∞)上单调递减,所以f(x)max=f(-2a)=4a2+a-ln(2a)-1.令h(a)=4a2+a-ln(2a)-1(a>),则h'(a)=8a+1->h'()>0,所以h(a)在区间(,+∞)上单调递增,h(a)>h()=>0,...................................................8分而当x→-2a时,f(x)→-∞,由题意,可以得到x0∈(-2a,-2a).所以当x∈(-2a,x0)时,f(x)<0,则g(x)=-f(x)+2x-1=(1+a)(2x-1)-ln(x+2a),当-2a<x<x0时,g'(x)=2+2a-<2+2a-..........................................................10分要想证明函数g(x)=|f(x)|+2x-1在区间(-2a,x0)上单调递减,只需g'(x)≤0,故只要证明x0≤-2a.记G(a)=f(-2a)=4a2+a--ln(2+2a),G'(a)=8a+1--在区间(,+∞)上单调递增,所以G'(a)>G'()>0,..........................................12分所以G(a)在区间(,+∞)上单调递增,G(a)>G()=1+--ln 3=-ln 3>0,所以f(x0)<f(-2a),x0∈(-2a,-2a),-2a∈(-2a,-2a),且f(x)在区间(-2a,-2a]上单调递增,所以x0<-2a,所以函数g(x)=|f(x)|+2x-1在区间(-2a,x0)上单调递减............................................15分  欢迎访问“高中试卷网”——http://sj.fjjy.org