2020年清华大学强基计划数学试题解答（完整版35道题）.pdf

1、u2020crOyK)PennyC1.x,y vx2+y2 1,Kx2+xy y2?()A.1B.52C.103D.2Y.B.1.dAM GM?,?x2 y2+(5+2)x5+2 y x2 y2+5+22?x2(5+2)2+y2?=5(x2+y2)252.?x=11045,y=110+45?.=?52.2.?x=rcos,y=rsin,|r|1,R,Kx2+xy y2=r2cos2+r22sin2 cos2+12sin2 52.?r=1,cos=11045,sin=110+45?.=?52.2.?a,b,c?,e?gax2+bx+c=0k,K()A.maxa,b,c 12(a+b+c)B.ma

2、xa,b,c 49(a+b+c)C.mina,b,c 14(a+b+c)D.mina,b,c 13(a+b+c)Y.BCD.K,kb2 4ac.dg5”?a+b+c=1.1Oky:mina,b,c 14(a+b+c).d5”?a c.Kb2 4ac 4c2 b 2c.?1=a+b+c c+2c+c=4c c 14.?a=c=14,b=12,K.=K?y.q5?1413,KCD.2Ogy:maxa,b,c 49(a+b+c).1eb 49,KK?y.?b=c=49,a=19,K.eb 59.q5?b2 4ac,K1681 4ac 4a?59 a?a 19?a 49?0 a?0,19?49,+?.

3、ea (49,+),KK?y.ea (0,19),dc 59 a 49,KK?y.q5?1249,KA.3.a,b,cv|a|2,|b|1,|a2bc|a+2b|,KkU?c,|c|?()A.42B.26C.?0D.?2Y.AC.?ab,k|a 2b|=|a+2b|.-c=0,?|c|=0.dn?,?|a+2b|a 2b c|c|a 2b|c|a 2b|+|a+2b|.2dCauchy?,?|c|2(|a 2b|+|a+2b|)2 2(|a 2b|2+|a+2b|2)=4|a|2+16|b|2 32|c|42.?ab,|a|=2|b|=2,c=a+2b?.n,|c|?0,42.4.34ABC,

4、AC=1,BC=3,AB=2,?M AB:,y4ABC CM,?oNB ACM?N212,K?AB?U()A.1B.2C.3D.2Y.BC.?:B 3.?K:D,KVBACM=13 S4ACM BD=1334 BD=212 BD=63.5?BD BC,ODBP,K()A.OP DPB.OP 1+5cos1?n?u()A.3B.4C.5D.6Y.C.5?3 1,Knsin3 1+5cos3 n 73 4.qn N+,Kn 5.ey?n=5,?,=5sin1 1+5cos1 sin?1 4?152.|?y?sinx 22 x,0 x 4.5?0 1 422?1 4?.5,k22?1 4?152 5

5、3 arcsin45,Kf(1)f?arcsin45?5sin1 5cos1 1.=?n=5,?.11.x,y,z v19x313y2 y=119y313z2 z=119z313x2 x=1,K()A.(x,y,z)k1|B.(x,y,z)k4|C.x,y,z knD.x,y,z nY.AD.x,y,z?.ex y,K13x2+x+1 13y2+y+1 z x.?z y,K13z2+z+1 13y2+y+1 y y,n?g!Kx=y,=x=y=z.5?(t+1)3 t3=3t2+3t+1,?19x313x2 x=1 3?x3?3=3?x3?2+3 x3+1=?x3+1?3?x3?3 x=341

6、3 1.w,x,y,z?n.12.?x1,x2,x21v0 xi 1(1,2,21),KP21i=1P21k=1|xi xk|?()A.110B.120C.220D.240Y.C.N?.uz(?1 i n,xj(j 6=i)?,K?uxi5.5?53:?,d=Ixi?01?/.6”?x1=x2=xt=0,xt+1=xt+2=x21=1,K21Xi=121Xk=1|xi xk|=2t (21 t)220.?=?t=1011?.5.Kx1,x2,xnL1,PP=2 Pni=1Pnj=1|xi xj|.(1)?n,kP?n2.?x1=x2=xn2=0,xn2+1=xn2+2=xn=1?.(2)?n,

7、kP?n2 1.?x1=x2=xn+12=0,xn+12+1=xn+12+2=xn=1?.13.3?IX,IpI?:,k:?/:/.e:/?Sk8:,.k10:,K:/?()A.10B.11C.12D.13Y.C.dPickn,?S=8+102 1=12.5.|?/A/?n.14.,Zn?mK.:“?”,:“?”,Z:“?”.P“wL?Yf?:“k?,k 17,K.eA+C+E B D=0,K2(A+C+E)=27,K.eA+C+E B D=11,KA+C+E=19,B+D=8.A8C0EA0C8E?4A7C1EA1C7E?12A6C2EA2C6E?8A5C3EA3C5E?8X,Lq?4+12

8、+8+8=32.n,?k64+32=96.?d?396?V96A510=1315.5.What the fuck?|?j,kCy?i5,?,l?.19.f(x)=ex+a(x 1)+b3m1,33:,Ka2+b2?()A.e2B.eC.e22D.e2Y.D.?f(x)?:t,Ka(t 1)+b=et.9dCauchy?,?e2t a (t 1)+b 12(a2+b2)(t 1)2+1 a2+b2e2tt2 2t+2.?Eg(t)=e2tt2 2t+2,1 t 3?,?g0(t)=2e2t(t2 3t+3)(t2 2t+2)2 0.=g(t)3m1,34O.?a2+b2?g(1)=e2.20.?

9、an?cnSn,e?anv:?n N+,3m N+,?Sn=am,KanT?.e?K?(?k()A.ean=(1n=12n2n 2,KanT?B.ean=na(a),KanT?C.ebn,cnT?,Kan=bn+cn?D.ean?,K3T?bn,cn,?an=bn+cnY.ABD.uA,?m=n+1=.uB,?m=n(n+1)2=.uC,?bnA?,cn 0(n N+),Kbn,cnT?,?an?.uD,?bn=(2 n)a1,cn=(n 1)(a1+d)=.5.T?32015cu+?OyQyL.21.f(x)=2exex+ex+sinx3m2,2?M,?m,K()A.M+m=2B.M+m=1

10、C.M m=2D.M m=1Y.A.5?f(x)+f(x)=?2exex+ex+sinx?+?2exex+ex+sin(x)?=2.KM+m=2.1022.?A,B Ox,y?:,eAB?C 2x+y 4=0,K?C?()A.5B.25C.45D.Y.C.?O?:,?C 2x+y 4=0u:D.?:O?2x+y 4=0?ld,?C r.K2r=CO+CD d r 25.?=?O,C,P n:?.?C?S=r245.23.a,bva3+b3+3ab=1,?a+b?kU?8M,K()A.M?8B.M k8,?8C.M 8,ke.D.M 8,e.Y.B.5?a3+b3+(1)3 3ab (1)=0

11、(a+b 1)(a b)2+(a+1)2+(b+1)2=0.Ka+b=12.=M k8,?8.51.IXe):a3+b3+c3 3abc=(a+b+c)?a2+b2+c2 ab bc ca?.52.AC?c?go?K:1O(2019u+?)x,y vx3+27y3+9xy=1,Ke?(?k()A.x3y?13B.x3y?2764C.x3y?33D.x3y vk?2O(2014?nE)?a,b,cva+b+c=0,a3+b3+c3=0,n N+,a2n+1+b2n+1+c2n+1?.3O(2016?g)ua,3b,cva3 b3 c3=3abc,a2=2(b+c),K?a?()A.1B.3C.D

12、.cn4O(2021?b,)?m,nvm3+n3+99mn=333,K(m,n)k()|A.32B.33C.34D.cn1124.?x,y?,en(:1Oy2+2xx2+2y U?2Oy2+4xx2+4y U?3Oy2+6xx2+6y U?(?()A.0B.1C.2D.3Y.C.u(1O,”?x y.Kx2 x2+2y x2+2x (x+1)2.=x2+2y?.u(2O,”?x y.Kx2 x2+4y x2+4x (x+2)2.ex2+4y?,Kx2+4y=(x+1)2 4y=2x+1,)!=x2+4y?.u(3O,?x=32,y=22,?y2+6x=262,x2+6y=342.25.?CX?

13、V?P(x=k)=12k(k=1,2,3,),Y LX?3?,KCY?EY?u()A.1B.87C.97D.32Y.B.5?EY=?12 1+122 2+123 0?+Xn=018n=11 18=87.26.?an?cnSn=(1)nan+12n+n3,tv(tan)(tan+1)a,b,c,?,e4ABC?kn,Ka?u()A.1B.2C.3D.4Y.CD.5?3,4,5?n?/=vK,?CD?(.uA,?a=1,Kb=c.?4ABC?S,KS=12 1 rb214=4b2 14.5?4b2 1 3 mod 4.K4b2 1U?.=S kn.uB,?a=2,Kc=bc=b+1.ec=b,KS

14、=12 2 b2 1=b2 1kn.ec=b+1,Kp=2b+32.2d,?S=s2b+32?2b+32 2?2b+32 b?2b+32 b 1?=12b2+12b 94.5?12b2+12b 9 3 mod 4.K12b2+12b 9U?.=S kn.5.?K32016cu+?OyQyL.35.f(z)=z10+1z10+12(z5+1z5),K()A.f(z)=03)B.f(z)=0?k20?E)C.f(z)=0E)?1D.f(z)=03?u1?E)16Y.BC.Pt=z5+1z5,Kf(z)=(t2 2)+t2=0 t=1 334(2,2).ef(z)=03)z0.dAM GM?,?z50+1z50?=|z50|+1|z50|2.?t (2,2)g!=f(z)=03),q5?t=z5+1z5 z10 tz5+1=0 z5=t2s1?t2?2 i,|z|=1.Kf(z)=0k20?E),?1.17