2020年清华大学强基计划数学试题解答（完整版35道题） PDF版含解析（可编辑）.pdf

1、u 2020 crOyK)PennyC1.x,y v x2+y2 1,K x2+xy y2 ()A.1B.52C.103D.2Y.B.1.d AM GM,x2 y2+(5+2)x5+2 y x2 y2+5+22x2(5+2)2+y2=5(x2+y2)252.x=11045,y=110+45.=52.2.x=r cos,y=r sin,|r|1,R,Kx2+xy y2=r2 cos 2+r22 sin 2 cos 2+12 sin 2 52.r=1,cos =11045,sin =110+45.=52.2.a,b,c,eg ax2+bx+c=0 k,K()A.max a,b,c 12(a+b+c

2、)B.max a,b,c 49(a+b+c)C.min a,b,c 14(a+b+c)D.min a,b,c 13(a+b+c)Y.BCD.K,k b2 4ac.dg5”a+b+c=1.1O ky:min a,b,c 14(a+b+c).d5”a c.K b2 4ac 4c2 b 2c.1=a+b+c c+2c+c=4c c 14.a=c=14,b=12,K.=Ky.q5 14 13,K CD.2O gy:max a,b,c 49(a+b+c).1e b 49,KKy.b=c=49,a=19,K.e b 59.q5 b2 4ac,K1681 4ac 4a 59 aa 19a 49 0 a 0,

3、1949,+.e a (49,+),KKy.e a (0,19),d c 59 a 49,KKy.q5 12 49,K A.3.a,b,c v|a|2,|b|1,|a 2b c|a+2b|,KkU c,|c|()A.42B.26C.0D.2Y.AC.a b,k|a 2b|=|a+2b|.-c=0,|c|=0.dn,|a+2b|a 2b c|c|a 2b|c|a 2b|+|a+2b|.2d Cauchy,|c|2 (|a 2b|+|a+2b|)2 2(|a 2b|2+|a+2b|2)=4|a|2+16|b|2 32|c|42.a b,|a|=2|b|=2,c=a+2b.n,|c|0,42.4.3

4、 ABC,AC=1,BC=3,AB=2,M AB:,y ABC CM,oN B ACM N212,K AB U()A.1B.2C.3D.2Y.BC.:B 3.K:D,KVBACM=13 SACM BD=13 34 BD=212 BD=63.5 BD BC,OD BP,K()A.OP DPB.OP 1+5 cos 1 n u()A.3B.4C.5D.6Y.C.5 3 1,Kn sin 3 1+5 cos 3 n 73 4.q n N+,K n 5.ey n=5,=5 sin 1 1+5 cos 1 sin1 4152.|ysin x 22 x,0 x 4.5 0 1 4 221 4.5,k221

5、 4152 53 arcsin 45,Kf(1)farcsin 45 5 sin 1 5 cos 1 1.=n=5,.11.x,y,z v19x3 13y2 y=119y3 13z2 z=119z3 13x2 x=1,K()A.(x,y,z)k 1|B.(x,y,z)k 4|C.x,y,z knD.x,y,z nY.AD.x,y,z.e x y,K 13x2+x+1 13y2+y+1 z x.z y,K 13z2+z+1 13y2+y+1 y y,ng!K x=y,=x=y=z.5(t+1)3 t3=3t2+3t+1,19x3 13x2 x=1 3 x33=3 x32+3 x3+1=x3+13

6、x33 x=3413 1.w,x,y,z n.12.x1,x2,x21 v 0 xi 1(1,2,21),K 21i=121k=1|xi xk|()A.110B.120C.220D.240Y.C.N.uz(1 i n,xj(j=i),Ku xi 5.553:?,d=I xi 0 1/.6”x1=x2=xt=0,xt+1=xt+2=x21=1,K21i=121k=1|xi xk|=2t (21 t)220.=t=10 11.5.K x1,x2,xn L 1,P P=2 ni=1nj=1|xi xj|.(1)n,k P n2.x1=x2=x n2=0,x n2+1=x n2+2=xn=1.(2)n

7、,k P n2 1.x1=x2=x n+12=0,x n+12+1=x n+12+2=xn=1.13.3IX,IpI:,k:/:/.e:/Sk 8:,.k 10:,K:/()A.10B.11C.12D.13Y.C.d Pick n,S=8+102 1=12.5.|/A/?n.14.,Zn?mK.:“”,:“”,Z:“”.P“wLYf:“k,k 17,K.e A+C+E B D=0,K 2(A+C+E)=27,K.e A+C+E B D=11,K A+C+E=19,B+D=8.A8C0EA0C8E 4 A7C1EA1C7E 12 A6C2EA2C6E 8 A5C3EA3C5E 8 X,Lq 4+

8、12+8+8=32.n,k 64+32=96.d 396 V 96A510=1315.5.What the fuck?|j,k C yi 5,l.19.f(x)=ex+a(x 1)+b 3m 1,3 3:,K a2+b2 ()A.e2B.eC.e22D.e2Y.D.f(x):t,K a(t 1)+b=et.9d Cauchy,e2t a (t 1)+b 1 2 (a2+b2)(t 1)2+1 a2+b2 e2tt2 2t+2.Eg(t)=e2tt2 2t+2,1 t 3,g(t)=2e2t (t2 3t+3)(t2 2t+2)2 0.=g(t)3m 1,3 4O.a2+b2 g(1)=e2.2

9、0.an c n Sn,e an v:?n N+,3 m N+,Sn=am,K an T.eK(k()A.e an=1n=12n2n 2,K an T B.e an=na(a),K an T C.e bn,cn T,K an=bn+cn D.e an,K3 T bn,cn,an=bn+cnY.ABD.u A,m=n+1=.u B,m=n(n+1)2=.u C,bn A,cn 0(n N+),K bn,cn T,an.u D,bn=(2 n)a1,cn=(n 1)(a1+d)=.5.T 3 2015 cu+OyQyL.21.f(x)=2exex+ex+sin x 3m 2,2 M,m,K()A.

10、M+m=2B.M+m=1C.M m=2D.M m=1Y.A.5f(x)+f(x)=2exex+ex+sin x+2exex+ex+sin(x)=2.K M+m=2.1022.A,B O x,y:,e AB C 2x+y 4=0,K C ()A.5B.25C.45D.Y.C.O:,C 2x+y 4=0 u:D.:O 2x+y 4=0 l d,C r.K 2r=CO+CD d r 25.=O,C,P n:.C S=r2 45.23.a,b v a3+b3+3ab=1,a+b kU8 M,K()A.M 8B.M k8,8C.M 8,ke.D.M 8,e.Y.B.5a3+b3+(1)3 3ab (1)

11、=0 (a+b 1)(a b)2+(a+1)2+(b+1)2 =0.K a+b=1 2.=M k8,8.5 1.IXe):a3+b3+c3 3abc=(a+b+c)a2+b2+c2 ab bc ca.5 2.ACcgoK:1O(2019 u+)x,y v x3+27y3+9xy=1,Ke(k()A.x3y 13B.x3y 2764C.x3y 33D.x3y vk2O(2014 nE)a,b,c v a+b+c=0,a3+b3+c3=0,n N+,a2n+1+b2n+1+c2n+1.3O(2016 g)u a,3 b,c v a3 b3 c3=3abc,a2=2(b+c),K a ()A.1B.

12、3C.D.cn4O(2021 b,)m,n v m3+n3+99mn=333,K(m,n)k()|A.32B.33C.34D.cn1124.x,y,en(:1O y2+2x x2+2y U2O y2+4x x2+4y U3O y2+6x x2+6y U()A.0B.1C.2D.3Y.C.u(1O,”x y.K x2 x2+2y x2+2x (x+1)2.=x2+2y.u(2O,”x y.K x2 x2+4y x2+4x (x+2)2.e x2+4y,K x2+4y=(x+1)2 4y=2x+1,)!=x2+4y.u(3O,x=32,y=22,y2+6x=262,x2+6y=342.25.C X

13、 V P(x=k)=12k(k=1,2,3,),Y L X 3,KC Y EY u()A.1B.87C.97D.32Y.B.5EY=12 1+122 2+123 0+n=018n=11 18=87.26.an c n Sn=(1)n an+12n+n 3,t v(t an)(t an+1)a,b,c,e ABC kn,K a u()A.1B.2C.3D.4Y.CD.5 3,4,5 n/=vK,CD(.u A,a=1,K b=c.ABC S,KS=12 1 b2 14=4b2 14.5 4b2 1 3 mod 4.K 4b2 1 U.=S kn.u B,a=2,K c=b c=b+1.e c=b

14、,K S=12 2 b2 1=b2 1 kn.e c=b+1,K p=2b+32.2d,S=2b+322b+32 22b+32 b2b+32 b 1=12b2+12b 94.5 12b2+12b 9 3 mod 4.K 12b2+12b 9 U.=S kn.5.K3 2016 cu+OyQyL.35.f(z)=z10+1z10+12(z5+1z5),K()A.f(z)=0 3)B.f(z)=0 k 20 E)C.f(z)=0 E)1D.f(z)=0 3u 1 E)16Y.BC.P t=z5+1z5,Kf(z)=(t2 2)+t2=0 t=1 334(2,2).e f(z)=0 3)z0.d AM GM,z50+1z50=|z50|+1|z50|2.t (2,2)g!=f(z)=0 3),q5t=z5+1z5 z10 tz5+1=0 z5=t2 1 t22 i,|z|=1.K f(z)=0 k 20 E),1.17