2020年高考真题数学(江苏卷)含答案

2020年普通高等学校招生全国统一考试（江苏卷）注意事项4第1第2345=VSh5．=−1,0,1,2},B={0,2,3}AIB=．1A▲=(1+i)(2−i)2z▲．i34,2a,3−a,5,6▲a．425▲．y−2x5▲．-1-5xy2−=a=ya52▲．()fx=x▲．28sin(+)=▲．43ππ=3sin(2﹢)yyx46▲．{ad{b}q{ab}nnnnnS=n▲．2n+▲．y4x2y2=，=，∠=90D到P3PA=mPB+(−m)PCm▲．23，B：+−(y=)236P(0)x22▲．ABCBFBC11111C；1111Cc=a3,c=45与OO(在DO1到(与D到()=h1401F到()与F到OO()=−b+b.3B到OOOO和为E在3(为OEk2x2y2中，已知椭圆E:FFA在1243EFFE．21△AFF12QPxEOP3ME与△SS=3MSS1122=f(x),y=g(x)h(x)=kx+b(k,bR)与在区间D上恒有已知关于x的函数yf(x)h(x)g(x)．()()=2+2，xD=−2+2，=(−+)(fxxxgxx()=2−+，()=ln，()=fxgxkxhx−,=(0+)kkxkDxxf(x)=x()4−2，()=42−8，()=43−xgxxhxttxttt−34+22(02)，D=m,n−2，．−7nmnN*()a=1nSλ与k，a1nn111S−S=akkkn+n1n1λλan30a~2aan3nn～0aann求λ5分.11．{0,2}5．−323324．932136．7．48．9．123−25．=−x2445184．．或0．1055-5-15分..1又，AB111111BC1.1又⊥I=ABCBCACC,,11111.,ABB11==，Bb=a+c−2acB=+−9223222.5bbc，52得=，5sinC=.54△=−5.CC故==则,433从=−=−=，1cosADCtanADC2==−.554而2=−−C)=−+C−=1−C.A,B,D,F11111111则.3800111由==240=+=+O1设y800221EF=160−y=160+x−6x.38002=1设y2401111=(80−x)=−x+4.2240401和f(x)，131x−6x)+k(−x+4x)322则3x−x+x3233k800令f(xf(x)为CD和1E:+=1，2ab2c43则=△AFF+12E=.x4则==x(x−=(x−2)2−4在=.x24x2y2+=1F,FAEE:4312⊥，AFFF2123FA212yMxySS1M2，55xyxy由由2x2y2+=1432得7x−12x−4=0=2或=−.2xxx2y2+=174312l:3x−4y−6=0y=0或=−.y7212或(−,−).M77f+2+−+2，x2x2x取=000=0．xbbx2x2(2−)0=22−+2k2kxx2x．.1令u(x)=x−1−lnxu'(x)=1−xx()=(1)=0则−1uxuxx(x)g(x)k0f(x)h(x)f−+1−kx2x−+)+1+0kxx2k1+k0=0，kx2+)−+)0−13．k2kkk0k）①当1t2由g(x)h(x)4−8)3+2t2x2ttxt−3−4t−t−842x2−t3−t)x+)4令−)−−2−则=−5+3+8．t3t2t4t2t6t4t2记()=−5+3+2),tt6t4t2t则()=6−+6=2−−0'tt5t3ttt2t2t)在(()(1)2t)7．t)，xxx12−−=7．nmxx2101tf(−h(=t+t−t−t−1．432设()=3+4−2−4−1，(+−4−4=+−vttt3ttt42t3t2ttt23令t)=0=．t33t)0vt)当(0，)v，t3当(31)v，t)0vt)t3v=1，v(1)=001vt)0．t()=(++−1)0vtt2tt-10-则f(1)−h(1)01(n)．，［-22，．n−m2+17mnf(x)，g(x)n−m7−2t0．n−m7201a}−=a=a，Sn1Sann1nn1n1−=(a0n1若aa=−=−=−，1a00aa1n13221a}n=113a}(nN)~2*n333−=−=−．SnSSaSSSn1n3n1n1n3n1n1S3S−1=−1．a0SS0n1nn1nS3SnnS133令=−=−b1(b2−=−．(b1)(b1)bnb1n122S3nnnnnnSS=4+==，bn22n1n1SSnn所以数列S}4nn=1),===Sa1=aS4n14(n2).11nn3n2−a}(nN)~3*n则111−=S3−．SSSS−S=a333n13n3n1n1nn1nSS0a=11=−−1．n1aSS0n133n1n1nSSnn-11-S令=−1=−1)(−=(−1)）ccc3nccccn133333Snnnnnnn0或=1{}cann0）+231*(−+c+=0，cc2−13nnn+231++10=1，cccc2−13nnnn{}10）an+2301+c+1=0c2−13nn11=或=．t3SSSSn1nn1n{}{}SSnn个，则对应的{}an{}~3”数列，01．an数学Ⅱ(附加题)C1a(2,1)M=B(3,．−1ba，bM．M−1ππ(,)26:cos=2C:=4sin上(,)lAB1300，2，12-12-lC设R2|x．x514213nX2np1q．nn11222pq与2pqXX用n)．nnnn]2a−1=3,1=−4=．−12M=−22152-51M=．−155]cos=2=4；=4sin=20，C366112=2．2cos=2,得=2=1．=4sin,00，，所以=2，2．4(22,)．4]22+2+40；xxx3当−102+2−4−10；xxxx当−1−2−2−4−2−1．xxxx2{|−2}．xx322．．以OBOCOA．，CB=CD=5，100020011则-14-AB，|AB|AB||DE|DE||1+0−2|15DE===|cos．531515．151BC=，BC20BF4111BF=BC=(−,,0)42．4又故71．DFDBBF(,,0)42=+=设=，，n(xyz)1111x+y+z=，DEn=，DF111则即711x+y=0,n=0,42111取==–7，==−5)．x2z5yn(21111DC设=，，120n(xyz)22220，n=，++=DEDCx2yz则即取==–1，=–1，x2yz222n=0,x+2y=0,222222=−−1)．(2n2|nn||4+7−5|6故|cos===．12|n||n|12239=sin−=1cos2．13-15-CC111==）p，，32112711q+0(1−p−q)=p+q=，321CCCC39271111111111+232213CCCCCCCC11111216．91n21211q+0(1−p−q)=p+q321CCn1CCn1n1n13n19n113CCCC11111+232213CCn1CCCCn1CCn1n112=−q+9n13241+=+−+399n1n13n1+qnnn1n1113111+()n1n33nn31331−=−−−=，)q59n1