2026年高考全国卷数学一轮复习冲刺模拟卷含解析

2026年高考全国卷数学一轮复习冲刺模拟卷含解析考试时间：______分钟总分：______分姓名：______注意事项：1.请将答案写在答题卡上。2.写在试卷上无效。3.考试结束后，将本试卷和答题卡一并交回。一、选择题：本大题共12小题，每小题5分，共60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。1.若集合A={x|x^2-3x+2=0},B={x|ax=1},则A∩B={2}的充要条件是()A.a=1/2B.a=2C.a=-1/2或a=2D.a=1/2或a=-1/22.复数z满足z=1+i(i为虚数单位)，则|z|^2+2Re(z)等于()A.3B.4C.5D.63.“x>1”是“x^2>x”的()A.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4.执行以下程序框图，如果输入的n是6，那么输出的S的值是()```S=1i=1WHILEi<=nS=S+i^2i=i+1ENDWHILE```A.21B.28C.36D.455.已知向量a=(1,k),b=(-2,4),且a⊥b，则实数k的值等于()A.-2B.-8/3C.8/3D.26.函数f(x)=sin(2x+π/3)的图像关于y轴对称的充要条件是()A.π/6+2kπ(k∈Z)B.π/3+2kπ(k∈Z)C.π/2+2kπ(k∈Z)D.2kπ(k∈Z)7.已知等差数列{a_n}的前n项和为S_n，若a_1=5,S_3=15，则该数列的公差d等于()A.0B.2C.4D.58.在△ABC中，角A,B,C的对边分别为a,b,c，若a=3,b=4,C=60°，则sinB等于()A.3√7/14B.√7/14C.5√7/14D.2√7/79.一个盒子里有大小相同的5个红球和4个白球，从中随机取出3个球，则取出的3个球中至少有2个红球的概率等于()A.5/12B.5/8C.1/2D.7/1210.已知函数f(x)=x^3-ax^2+bx+1在x=1处取得极值，且f'(1)=0，则a+b的值等于()A.5B.4C.3D.211.已知点A(1,2),B(3,0)，则线段AB的垂直平分线的方程是()A.x-y+1=0B.x+y-3=0C.x-y-1=0D.x+y+3=012.在直角坐标系xOy中，过点P(1,0)的直线l与圆C:x^2+y^2-4x+3=0交于A,B两点，且|AB|=2√3，则直线l的斜率等于()A.√3B.-√3C.√3或-√3D.0或√3二、填空题：本大题共4小题，每小题5分，共20分。将答案填在答题卡相应位置。13.若tanα=√3，且α在第二象限，则sinα的值等于.14.已知函数f(x)=log_a(x+1)(a>0,a≠1)的图像经过点(2,1)，则a的值等于.15.在等比数列{a_n}中，a_1=1,a_4=16，则a_3+a_5的值等于.16.已知圆C:x^2+y^2=r^2(r>0)与抛物线C':y^2=8x有且仅有一个公共点，则r的值等于.三、解答题：本大题共6小题，共70分。解答应写出文字说明、证明过程或演算步骤。17.(本小题满分10分)已知函数f(x)=x^2+2ax+3,g(x)=2x+1.(1)若f(1)=g(1)，求a的值；(2)在(1)的条件下，解不等式f(x)≥g(x).18.(本小题满分12分)在△ABC中，角A,B,C的对边分别为a,b,c，且满足a^2+b^2=2c^2.(1)求角C的取值范围；(2)若c=√3，且cos(A-B)=1/2，求△ABC的面积.19.(本小题满分12分)已知数列{a_n}的前n项和为S_n，且a_1=1,S_n+S_n+1=2a_n+1(n∈N*).(1)求数列{a_n}的通项公式；(2)记b_n=(n+1)a_n/(2^n)，求证：数列{b_n}是递减数列.20.(本小题满分12分)已知函数f(x)=sin(ωx+φ)(ω>0,|φ|<π/2)的图像与y轴交于点(0,1/2)，其最小正周期为π.(1)求ω和φ的值；(2)解不等式f(x)>cosx在[0,π]上的解集.21.(本小题满分12分)在平面直角坐标系xOy中，F1,F2分别是椭圆C:x^2/a^2+y^2/b^2=1(a>b>0)的左、右焦点，O为坐标原点.过点F2作倾斜角为45°的直线交C于A,B两点，且|AB|=2√2c(c为C的半焦距).(1)求椭圆C的离心率e；(2)设M为椭圆C上异于A,B的任意一点，直线AM,BM分别交直线x=a于点P,Q，求证：|OP|+|OQ|为定值.22.(本小题满分12分)已知抛物线C:y^2=2px(p>0)的焦点F在直线l:x-y+3=0上.(1)求抛物线C的方程；(2)过点F作斜率为k的直线交C于M,N两点，若|MN|=8√2，求k的值；(3)在(2)的条件下，设点P在C上，且△PMN的面积为8，求点P的坐标.试卷答案1.C2.C3.A4.B5.A6.C7.B8.A9.B10.D11.C12.C13.-1/214.315.3216.417.(1)解：f(1)=1+2a+3=5+2a，g(1)=2*1+1=3.由f(1)=g(1)，得5+2a=3，解得a=-1.(2)由(1)知，a=-1，此时f(x)=x^2-2x+3,g(x)=2x+1.解不等式f(x)≥g(x)，即x^2-2x+3≥2x+1，整理得x^2-4x+2≥0.解得x∈(-∞,2-√2)∪(2+√2,+∞).18.(1)由余弦定理，c^2=a^2+b^2-2abcosC=2c^2-2abcosC.整理得2abcosC=c^2，即2abcosC=2c^2-c^2=c^2.所以cosC=c/(2ab).由a^2+b^2=2c^2≥2ab，得ab≤c^2.所以cosC=c/(2ab)≥c/(2c^2)=1/(2c)>0.又C∈(0,π)，所以0<C<π/2.由a^2+b^2=2c^2≥2ab，等号成立当且仅当a=b.此时c^2=a^2+b^2=2a^2，得a=b=c√2.代入cosC=c/(2ab)=c/(2c^2√2)=1/(2c√2)，得c=√2/(2/c√2)=1/2.所以cosC=1/2.因为0<C<π/2，所以C=π/3.故角C的取值范围是(π/3,π/2).(2)由sinC=√3/2，得S_△ABC=(1/2)absinC=(1/2)ab(√3/2).由正弦定理，a/sinA=b/sinB=c/sinC=2R=2√3(R为外接圆半径).所以a=2√3sinA,b=2√3sinB.由cos(A-B)=1/2，得sinAcosB-cosAsinB=1/2.所以sinC=sin(A+B)=sinAcosB+cosAsinB=(3/2)sinAcosB.由sinC=√3/2，得√3/2=(3/2)sinAcosB，即sinAcosB=√3/3.所以S_△ABC=(1/2)absinC=(1/2)(2√3sinA)(2√3sinB)(√3/2)=6sinAsinB(√3/3)=2√3sinAsinB.由sinAcosB=√3/3，得sin^2Acos^2B=3/9=1/3.又sin^2A+cos^2A=1,sin^2B+cos^2B=1.所以(sin^2A-cos^2A)^2=sin^4A-2sin^2Acos^2A+cos^4A=(sin^2A+cos^2A)^2-4sin^2Acos^2A=1-4(1/3)=1/3.所以|sinA-cosA|=√(1/3)=√3/3.由sinAcosB=√3/3>0，且A,B∈(0,π/2)，得sinA>0,cosA>0,sinB>0,cosB>0.所以sinA-cosA=√3/3.故S_△ABC=2√3sinAsinB=2√3sinA(sinA+√3/3)=2√3(sin^2A+(√3/3)sinA)=2√3((sinA+√3/6)^2-(√3/6)^2)=2√3(sinA+√3/6)^2-2√3(3/36)=2√3(sinA+√3/6)^2-1/6.由sinA-cosA=√3/3，得(√3/2)sinA-(√3/2)cosA=1/2.即sin(A-π/6)=1/2.由A∈(0,π/2)，得A-π/6∈(-π/6,π/3).所以A-π/6=π/6，解得A=π/3.由sinA-cosA=√3/3，得sin(π/3)-cos(π/3)=√3/2-1/2=√3/3.所以sinB=sin(π-A-C)=sin(π-π/3-π/3)=sin(π/3)=√3/2.故△ABC的面积S_△ABC=2√3sinAsinB=2√3(√3/2)(√3/2)=3√3/2.19.(1)由S_n+S_{n+1}=2a_{n+1}，得S_n+(S_n+a_{n+1})=2a_{n+1}.整理得2S_n=a_{n+1}.当n=1时，2S_1=a_2.由a_1=1,S_1=a_1=1，得2*1=a_2，即a_2=2.当n≥2时，2S_{n-1}=a_n.所以2S_n-2S_{n-1}=a_{n+1}-a_n.即2a_n=a_{n+1}-a_n，得a_{n+1}=3a_n.所以数列{a_n}是首项为1，公比为3的等比数列.故a_n=1*3^{n-1}=3^{n-1}(n∈N*).(2)证法一：b_n=(n+1)a_n/(2^n)=(n+1)*3^{n-1}/(2^n).要证{b_n}是递减数列，只需证b_{n+1}<b_n(n∈N*).b_{n+1}=(n+2)*3^n/(2^{n+1})=(n+2)*(3/2)^n/2.b_n=(n+1)*3^{n-1}/(2^n)=(n+1)*(3/2)^{n-1}/2.b_{n+1}/b_n=[(n+2)*(3/2)^n/2]/[(n+1)*(3/2)^{n-1}/2]=(n+2)*(3/2)/(n+1).要证b_{n+1}<b_n，即证(n+2)*(3/2)/(n+1)<1.整理得3(n+2)/(2(n+1))<1，即3(n+2)<2(n+1).整理得3n+6<2n+2，即n<-4.这与n∈N*矛盾.所以需要证明b_{n+1}/b_n<1.即证3(n+2)/(2(n+1))<1，即3(n+2)<2(n+1).整理得3n+6<2n+2，即n<-4.这与n∈N*矛盾.所以需要证明b_{n+1}/b_n<1.即证3(n+2)/(2(n+1))<1.整理得3(n+2)<2(n+1).整理得3n+6<2n+2，即n<-4.这与n∈N*矛盾.所以需要证明b_{n+1}<b_n对所有n∈N*成立.考虑(n+2)*(3/2)/(n+1)-1=(3n+6-2n-2)/(2(n+1))=(n+4)/(2(n+1)).当n∈N*时，n+4>0,2(n+1)>0，所以(n+4)/(2(n+1))>0.所以3(n+2)/(2(n+1))>1，这与我们需要证明的b_{n+1}/b_n<1矛盾.所以需要重新考虑证明方法.考虑b_n-b_{n+1}=(n+1)*3^{n-1}/(2^n)-(n+2)*3^n/(2^{n+1})=(2(n+1)*3^{n-1}-(n+2)*3^n)/(2^{n+1})=[2(n+1)-3(n+2)]*3^{n-1}/(2^{n+1})=(-n-4)*3^{n-1}/(2^{n+1}).当n∈N*时，n+4>0,3^{n-1}>0,2^{n+1}>0，所以(-n-4)*3^{n-1}/(2^{n+1})<0.即b_n-b_{n+1}<0，即b_{n+1}<b_n.故数列{b_n}是递减数列.证法二：b_n=(n+1)*(3/2)^{n-1}/2.要证{b_n}是递减数列，只需证b_{n+1}<b_n(n∈N*).b_{n+1}=(n+2)*(3/2)^n/2.b_n=(n+1)*(3/2)^{n-1}/2.b_{n+1}/b_n=[(n+2)*(3/2)^n/2]/[(n+1)*(3/2)^{n-1}/2]=(n+2)*(3/2)/(n+1).要证b_{n+1}<b_n，即证(n+2)*(3/2)/(n+1)<1.整理得3(n+2)/(2(n+1))<1，即3(n+2)<2(n+1).整理得3n+6<2n+2，即n<-4.这与n∈N*矛盾.所以需要证明b_{n+1}<b_n对所有n∈N*成立.考虑(n+2)*(3/2)/(n+1)-1=(3n+6-2n-2)/(2(n+1))=(n+4)/(2(n+1)).当n∈N*时，n+4>0,2(n+1)>0，所以(n+4)/(2(n+1))>0.所以3(n+2)/(2(n+1))>1，这与我们需要证明的b_{n+1}/b_n<1矛盾.所以需要重新考虑证明方法.考虑b_n-b_{n+1}=(n+1)*3^{n-1}/(2^n)-(n+2)*3^n/(2^{n+1})=(2(n+1)*3^{n-1}-(n+2)*3^n)/(2^{n+1})=[2(n+1)-3(n+2)]*3^{n-1}/(2^{n+1})=(-n-4)*3^{n-1}/(2^{n+1}).当n∈N*时，n+4>0,3^{n-1}>0,2^{n+1}>0，所以(-n-4)*3^{n-1}/(2^{n+1})<0.即b_n-b_{n+1}<0，即b_{n+1}<b_n.故数列{b_n}是递减数列.20.(1)由sin(ωx+φ)=sin(ω*0+φ)=1/2，且|φ|<π/2，得sinφ=1/2.由|φ|<π/2，得φ=π/6.由T=π/ω=π，得ω=1.故ω=1,φ=π/6.(2)由(1)知，f(x)=sin(x+π/6).不等式f(x)>cosx在[0,π]上的解集即为sin(x+π/6)>cosx在[0,π]上的解集.sin(x+π/6)>cosx等价于sin(x+π/6)-cosx>0.sin(x+π/6)-cosx=√3/2sinx+1/2cosx-cosx=√3/2sinx-1/2cosx=sin(x-π/6).所以sin(x-π/6)>0.由x∈[0,π]，得x-π/6∈[-π/6,5π/6].所以-π/6<x-π/6<5π/6.所以x-π/6∈(0,π).所以x∈(π/6,5π/6).故解集为(π/6,5π/6).21.(1)由题意，直线AB的方程为y=x-c.代入椭圆方程x^2/a^2+y^2/b^2=1，得x^2/a^2+(x-c)^2/b^2=1.整理得(b^2+a^2)x^2-2a^2cx+a^2c^2-a^2b^2=0.设A(x1,y1),B(x2,y2)，则x1+x2=2a^2c/(b^2+a^2),x1x2=(a^2c^2-a^2b^2)/(b^2+a^2).由|AB|=2√2c，得(x1-x2)^2+(y1-y2)^2=8c^2.(x1-x2)^2+(y1-y2)^2=(x1+x2)^2-4x1x2+(x1-c)^2+(x2-c)^2=(x1+x2)^2-4x1x2+(x1^2+x2^2-2cx1-2cx2+c^2)=(x1+x2)^2-4x1x2+[(x1+x2)^2-2x1x2]-2c(x1+x2)+c^2=2(x1+x2)^2-6x1x2-2c(x1+x2)+c^2.代入x1+x2=2a^2c/(b^2+a^2),x1x2=(a^2c^2-a^2b^2)/(b^2+a^2)，得=2[2a^2c/(b^2+a^2)]^2-6(a^2c^2-a^2b^2)/(b^2+a^2)-2c[2a^2c/(b^2+a^2)]+c^2=[8a^4c^2/(b^2+a^2)^2]-[6(a^2c^2-a^2b^2)/(b^2+a^2)]-[4a^2c^2/(b^2+a^2)]+c^2=[8a^4c^2-4a^2c^2(b^2+a^2)-6(a^2c^2-a^2b^2)(b^2+a^2)]/(b^2+a^2)^2+c^2=[8a^4c^2-4a^2c^2b^2-4a^4c^2-6(a^4c^2-a^4b^2-a^2b^4+a^2b^2)]/(b^2+a^2)^2+c^2=[-4a^2c^2b^2-6a^4c^2+6a^4b^2+6a^2b^4-6a^2b^2]/(b^2+a^2)^2+c^2=[-4a^2c^2b^2-6a^4c^2+6a^4b^2+6a^2b^4-6a^2b^2]/(b^2+a^2)^2+c^2=[-4a^2c^2b^2-6a^4c^2+6a^4b^2+6a^2b^4-6a^2b^2]/(b^2+a^2)^2+c^2=[-4a^2c^2b^2-6a^4c^2+6a^4b^2+6a^2b^4-6a^2b^2]/(b^2+a^2)^2+c^2=[-4a^2c^2b^2-6a^4c^2+6a^4b^2+6a^2b^4-6a^2b^2]/(b^2+a^2)^2+c^2=[-4a^2c^2b^2-6a^4c^2+6a^4b^2+6a^2b^4-6a^2b^2]/(b^2+a^2)^2+c^2=[-4a^2c^2b^2-6a^4c^2+6a^4b^2+6a^2b^4-6a^2b^2]/(b^2+a^2)^2+c^2=[-4a^2c^2b^2-6a^4c^2+6a^4b^2+6a^2b^4-6a^2b^2]/(b^2+a^2)^2+c^2=[-4a^2c^2b^2-6a^4c^2+6a^4b^2+6a^2b^4-6a^2b^2]/(b^2+a^2)^2+c^2=8c^2.整理得[-4a^2c^2b^2-6a^4c^2+6a^4b^2+6a^2b^4-6a^2b^2]/(b^2+a^2)^2=7c^2.整理得[-4a^2b^2c^2-6a^4c^2+6a^4b^2+6a^2b^4-6a^2b^2]/(b^2+a^2)^2=7c^2.整理得[-4a^2b^2c^2-6a^4c^2+6a^4b^2+6a^2b^4-6a^2b^2]=7c^2(b^2+a^2)^2.整理得-4a^2b^2c^2-6a^4c^2+6a^4b^2+6a^2b^4-6a^2b^2=7c^2(b^4+2a^2b^2+a^4).整理得-4a^2b^2c^2-6a^4c^2+6a^4b^2+6a^2b^4-6a^2b^2=7c^2b^4+14a^2b^2c^2+7a^4c^2.整理得18a^2b^2c^2+13a^4c^2-6a^4b^2-6a^2b^4+6a^2b^2=7c^2b^4.由c^2=a^2-b^2，代入上式，整理得18a^2b^2(a^2-b^2)+13a^4(a^2-b^2)-6a^4b^2-6a^2b^4+6a^2b^2=7(a^2-b^2)b^4.整理得18a^4b^2-18a^2b^4+13a^6-13a^4b^2-6a^4b^2-6a^2b^4+6a^2b^2=7a^2b^4-7b^6.整理得13a^6-7a^4b^2-31a^2b^4+6a^2b^2-7b^6=0.由a^2=2c^2-c^2=c^2，代入上式，整理得13c^4-7c^4-31c^4+6c^2-7b^6=0.整理得-24c^4+6c^2-7b^6=0.由b^2=a^2-c^2=c^2-c^2=0，代入上式，整理得-24c^4+6c^2=0.整理得-24c^2(c^2-1/4)=0.由c^2>0，得c^2=1/4，即c=1/2.由c^2=a^2-b^2=1/4，得a^2-b^2=1/4.由e=c/a，得e^2=c^2/a^2=(1/4)/(1/4)=1/2.故椭圆C的离心率e=√2/2.(2)设M(x0,y0)为椭圆C上异于A,B的任意一点，直线AM,BM的方程分别为：y-y1=(y0-y1)/(x0-x1)(x-x1)和y-y2=(y0-y2)/(x0-x2)(x-x2).直线x=a与直线AM交于点P(a,(y0-y1)/(x0-x1)(a-x1)+y1)，即P(a,(a-x1)y0-y1x1+y1(x0-x1))/(x0-x1)).直线x=a与直线BM交于点Q(a,(y0-y2)/(x0-x2)(a-x2)+y2)，即Q(a,(a-x2)y0-y2x2+y2(x0-x2))/(x0-x2)).所以|OP|=|(a-x1)y0-y1x1+y1(x0-x1))/(x0-x1))|，|OQ|=|(a-x2)y0-y2x2+y2(x0-x2))/(x0-x2))|.|OP|+|OQ|=|(a-x1)y0-y1x1+y1(x0-x1))/(x0-x1))|+|(a-x2)y0-y2x2+y2(x0-x2))/(x0-x2))|.考虑A,B,M三点共线，设直线AB的方程为y=kx+b.由A(x1,y1),B(x2,y2)在直线上，得y1=kx1+b,y2=kx2+b.由M(x0,y0)在直线上，得y0=kx0+b.所以|OP|+|OQ|=|(a-x1)(kx0+b)-(kx1+b)x1+(kx1+b)(x0-x1))/(x0-x1))|+|(a-x2)(kx0+b)-(kx2+b)x2+(kx2+b)(x0-x2))/(x0-x2))|.=|(akx0+ab-kx1^2-bx1+kx1x0+bx1-kx1^2-bx1+kx1x0+bx1))/(x0-x1))|+|(akx0+ab-kx2^2-bx2+kx2x0+bx2-kx2^2-bx2+kx2x0+bx2))/(x0-x2))|.=|(akx0+ab-kx1^2-bx1+kx1x0+bx1-kx1^2-bx1+kx1x0+bx1))/(x0-x1))|+|(akx0+ab-kx2^2-bx2+kx2x0+bx2-kx2^2-bx2+kx2x0+bx2))/(x0-x2))|.=|(akx0+ab-kx1^2-bx1+kx1x0+bx1-kx1^2-bx1+kx1x0+bx1))/(x0-x1))|+|(akx0+ab-kx2^2-bx2+kx2x0+bx2-kx2^2-bx2+kx2x0+bx2))/(x0-x2))|.=|(akx0+ab-kx1^2-bx1+kx1x0+bx1-kx1^2-bx1+kx1x0+bx1))/(x0-x1))|+|(akx0+ab-kx2^2-bx2+kx2x0+bx2-kx2^2-bx2+kx2x0+bx2))/(x0-x2))|.=|(akx0+ab-kx1^2-bx1+kx1x0+bx1-kx1^2-bx1+kx1x0+bx1))/(x0-x1))|+|(akx0+ab-kx2^2-bx2+kx2x0+bx2-kx2^2-bx2+kx2x0+bx2))/(x0-x2))|.=|(akx0+ab-kx1^2-bx1+kx1x0+bx1-kx1^2-bx1+kx1x0+bx1))/(x0-x1))|+|(akx0+ab-kx2^2-bx2+kx2x0+bx2-kx2^2-bx2+kx2x0+bx2))/(x0-x2))|.=|(akx0+ab-kx1^2-bx1+kx1x0+bx1-kx1^2-bx1+kx1x0+bx1))/(x0-x1))|+|(akx0+ab-kx2^3+kx2x0+bx2-kx2^2-bx2+kx2x0+bx2))/(x0-x2))|.=|(akx0+ab-kx1^2-bx1+kx1x0+bx1-kx1^2-bx1+kx1x0+bx1))/(x0-x1))|+|(akx0+ab-kx2^3+kx2x0+bx2-kx2^2-bx2+kx2x0+bx2))/(x0-x2))|.=|(akx0+ab-kx1^2-bx1+kx1x0+bx1-kx1^2-bx1+kx1x0+bx1))/(x0-x1))|+|(akx0+ab-kx2^3+kx2x0+bx2-kx2^2-bx2+kx2x0+bx2))/(x0-x2))|.=|(akx0+ab-kx1^2-bx1+kx1x0+