2020年高考真題數(shù)學(xué)(江蘇卷)含答案

2020年普通高等學(xué)校招生全國統(tǒng)一考試（江蘇卷）注意事項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▲．{adq{ab}nnnnnS=n▲．2n+▲．y4x2y2=，=，∠=90D到P3PA=mPB+(?m)PCm▲．23，B：+?(y=)236P(0)x22▲．ABCBFBC11111C；1111Cc=a3,c=45與OO(在DO1到(與D到()=h1401F到()與F到OO()=?b+b.3B到OOOO和為E在3(為OEk2x2y2中，已知橢圓E:FFA在1243EFFE．21△AFF12QPxEOP3ME與△SS=3MSS1122=f(x),y=g(x)h(x)=kx+b(k,bR)與在區(qū)間D上恒有已知關(guān)于x的函數(shù)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與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==則,433從=?=?=，1cosADCtanADC2==?.554而2=??C)=?+C?=1?C.A,B,D,F11111111則.3800111由==240=+=+O1設(shè)y800221EF=160?y=160+x?6x.38002=1設(shè)y2401111=(80?x)=?x+4.2240401和f(x)，131x?6x)+k(?x+4x)322則3x?x+x3233k800令f(xf(x)為CD和1E:+=1，2ab2c43則=△AFF+12E=.x4則==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則?1uxuxx(x)g(x)k0f(x)h(x)f?+1?kx2x?+)+1+0kxx2k1+k0=0，kx2+)?+)0?13．k2kkk0k）①當(dāng)1t2由g(x)h(x)4?8)3+2t2x2ttxt?3?4t?t?842x2?t3?t)x+)4令?)??2?則=?5+3+8．t3t2t4t2t6t4t2記()=?5+3+2),tt6t4t2t則()=6?+6=2??0'tt5t3ttt2t2t)在(()(1)2t)7．t)，xxx12??=7．nmxx2101tf(?h(=t+t?t?t?1．432設(shè)()=3+4?2?4?1，(+?4?4=+?vttt3ttt42t3t2ttt23令t)=0=．t33t)0vt)當(dāng)(0，)v，t3當(dāng)(31)v，t)0vt)t3v=1，v(1)=001vt)0．t()=(++?1)0vtt2tt-10-則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．a(chǎn)0SS0n1nn1nS3SnnS133令=?=?b1(b2?=?．(b1)(b1)bnb1n122S3nnnnnnSS=4+==，bn22n1n1SSnn所以數(shù)列S}4nn=1),===Sa1=aS4n14(n2).11nn3n2?a}(nN)~3*n則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個，則對應(yīng)的{}an{}~3”數(shù)列，01．a(chǎn)n數(shù)學(xué)Ⅱ(附加題)C1a(2,1)M=B(3,．?1ba，bM．M?1ππ(,)26:cos=2C:=4sin上(,)lAB1300，2，12-12-lC設(shè)R2|x．x514213nX2np1q．nn11222pq與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當(dāng)?102+2?4?10；xxxx當(dāng)?1?2?2?4?2?1．xxxx2{|?2}．xx322．．以O(shè)BOCOA．，CB=CD=5，100020011則-14-AB，|AB|AB||DE|DE||1+0?2|15DE===|cos．531515．151BC=，BC20BF4111BF=BC=(?,,0)42．4又故71．DFDBBF(,,0)42=+=設(shè)=，，n(xyz)1111x+y+z=，DEn=，DF111則即711x+y=0,n=0,42111取==–7，==?5)．x2z5yn(21111DC設(shè)=，，120n(xyz)22220，n=，++=DEDCx2yz則即取==–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