多模塊知識交匯創(chuàng)新命題（解析版）

近幾年的高考數(shù)學在穩(wěn)定的基礎上越來越體現(xiàn)學的工具性作用,體現(xiàn)了對靈活解題能力的考查.塊的知識,更加體現(xiàn)向量、函數(shù)等知識的工具性作用.11數(shù)列的遞推關系或通項公式.*n是曲線y=x2n-1-2在點(1,-1)處的切線與x軸交點A.B.C.D.∴an===(2n-EQ\*jc3\*hps21\o\al(\s\up3(4),1))EQ\*jc3\*hps15\o\al(\s\up7(2),2)n+1)，a1+a2+a3+?+a50=50+-+-+?+-=50+=，故選：A.2.(24-25高三上·山東濟寧·期中) n=g+g+g+?+g(n∈N*(，則數(shù)列{an{的通項公式為.n=4n-2【解析】函數(shù)f(x)=的定義域為R，f(-x)===-f(x)，由g(x)=f(x-1)+2，得g(x+1)=f(x)+2，則g(-x+1)+g(x+1)=f(-x)+f(x)+2+2=4，由g(-x+1)+g(x+1)=4，得g(x)+g(2-x)=4，22an=g+g+g+?+g，an=g+g+g+?+g，n=4n-2所以數(shù)列{an{的通項公式為an=4n-2.n=4n-2.{an+an+1+an+2{是公差為9的等差數(shù)列，且a1=1.【解析】(1)因為{an+an+1{是公差為6的等差數(shù)列，則(an+1+an+2(-(an+an+1(=an+2-an=6，設a2=m，可得a3=7，a4=m+6，又因為{an+an+1+an+2{是公差為9的等差數(shù)列，則(an+1+an+2+an+3(-(an+an+1+an+2(=an+3-an=9，可得an+3-an+2=3，即an+1-an=3,n≥3，且a4-a1=m+5=9，解得m=4，2=43=7n+1-an=3,n=1,2，n+1-an=3，所以{an{是等差數(shù)列.因為f/(x(=6x2+3>0，則f(x(在R上單調遞增，且f(0(=-2<0,f(1(=3>0，可知f(x(有且僅有一個零點b則Sn=<-，4.(24-25高三上·云南大理·模擬預測)已知函數(shù)f(x(=lnx-mx+1.(2)若f(x(≤0恒成立，求實數(shù)m的取值范圍；，(1+1+1+<e.33所以切線斜率k=f/(e(=，所以切線方程為y-2=(x-e(，即x-y+1=0.f(x(≤0恒成立，即m≥恒成立，只需m≥max即可，設g(x(=，(x>0(，則g/(x(==-，所以g(x(max=g(1(=1，≤0恒成立，即lnx≤x-1，則ln1+<1+-1=，故ln(1++ln(1++???+ln(1+<++???+==1-<1，即ln(1+1+1+<1=lne.故(1+1+1+<e.質.A.2n-1B.2n-1C.2n+1-4D.2nEQ\*jc3\*hps21\o\al(\s\up7(—),n)EQ\*jc3\*hps21\o\al(\s\up7(→),d)EQ\*jc3\*hps21\o\al(\s\up7(→),d)EQ\*jc3\*hps21\o\al(\s\up7(→),d)EQ\*jc3\*hps21\o\al(\s\up7(→),d)EQ\*jc3\*hps21\o\al(\s\up7(→),d)EQ\*jc3\*hps21\o\al(\s\up7(→),d)EQ\*jc3\*hps21\o\al(\s\up7(—),n)EQ\*jc3\*hps21\o\al(\s\up8(→),d)EQ\*jc3\*hps21\o\al(\s\up8(→),d)EQ\*jc3\*hps21\o\al(\s\up8(→),d)EQ\*jc3\*hps21\o\al(\s\up7(→),d)則?EQ\*jc3\*hps21\o\al(\s\up7(→),d)+1=2nEQ\*jc3\*hps21\o\al(\s\up7(→),d)44*(滿足aEQ\*jc3\*hps22\o\al(\s\up7(—),i)1-=EQ\*jc3\*hps22\o\al(\s\up8(→),d)(i=1,2,…,n-1(,EQ\*jc3\*hps22\o\al(\s\up8(→),d)EQ\*jc3\*hps21\o\al(\s\up7(—),i)-=EQ\*jc3\*hps21\o\al(\s\up7(→),d)(i=1,2,…,n-1(—=)EQ\*jc3\*hps21\o\al(\s\up7(→),d)→bn=.=2+(n-1.EQ\*jc3\*hps21\o\al(\s\up7(→),d)，EQ\*jc3\*hps21\o\al(\s\up6(→),d)EQ\*jc3\*hps21\o\al(\s\up6(→),d)EQ\*jc3\*hps21\o\al(\s\up6(→),d)n=2+(n-1.EQ\*jc3\*hps21\o\al(\s\up7(→),d)=1+n-1=n，故S20=EQ\*jc3\*hps22\o\al(\s\up9(—→),nD)EQ\*jc3\*hps22\o\al(\s\up9(—→),nA)EQ\*jc3\*hps22\o\al(\s\up9(—→),nE)A.a3=13B.數(shù)列{an+3{是等比數(shù)列C.an=4n-3D.Sn=2n+2-3n-4EQ\*jc3\*hps21\o\al(\s\up9(—→),nE)EQ\*jc3\*hps21\o\al(\s\up9(—→),nA)EQ\*jc3\*hps21\o\al(\s\up9(—→),nB)EQ\*jc3\*hps21\o\al(\s\up9(—→),nB)EQ\*jc3\*hps21\o\al(\s\up9(—→),nA)EQ\*jc3\*hps21\o\al(\s\up9(—→),nE)EQ\*jc3\*hps21\o\al(\s\up9(—→),nD)EQ\*jc3\*hps21\o\al(\s\up9(—→),nB)EQ\*jc3\*hps21\o\al(\s\up9(—→),nA)EQ\*jc3\*hps21\o\al(\s\up9(—→),nE)又GEQ\*jc3\*hps21\o\al(\s\up9(—→),nD)=an+1.GEQ\*jc3\*hps21\o\al(\s\up9(—→),nA)-2(2an+3(GEQ\*jc3\*hps21\o\al(\s\up9(—→),nE)，n+1=2an+3，:an+1+3=2(an+3(，:an+3=4.2n-1=2n+1，:an=2n+1-3，故C錯誤；:Sn=(22+23+...+2n+1(-3n=-3n=2n+2-3n-4，故D正確.故選：ABD.xnn,0)滿足xn+1>xn,PEQ\*jc3\*hps22\o\al(\s\up10(——),nP)EQ\*jc3\*hps22\o\al(\s\up10(→),1)丄AEQ\*jc3\*hps22\o\al(\s\up10(——),nP)EQ\*jc3\*hps22\o\al(\s\up10(→),1)且EQ\*jc3\*hps22\o\al(\s\up7(——),nP)EQ\*jc3\*hps22\o\al(\s\up7(→),1)EQ\*jc3\*hps22\o\al(\s\up7(——),nP)EQ\*jc3\*hps22\o\al(\s\up7(→),1)*55n+1與xn的關系式；2+4n+4≤xEQ\*jc3\*hps13\o\al(\s\up3(2),1)+xEQ\*jc3\*hps13\o\al(\s\up3(2),2)+xEQ\*jc3\*hps13\o\al(\s\up3(2),3)+?+xEQ\*jc3\*hps13\o\al(\s\up3(2),n)+1≤4n2+6n.n+1-xn=；(2)證明見解析EQ\*jc3\*hps21\o\al(\s\up7(——),nP)EQ\*jc3\*hps21\o\al(\s\up7(→),1)xn+1-xn,-AEQ\*jc3\*hps21\o\al(\s\up7(——),nP)EQ\*jc3\*hps21\o\al(\s\up7(→),1)=(xn+1-an,,EQ\*jc3\*hps21\o\al(\s\up8(——),nP)EQ\*jc3\*hps21\o\al(\s\up8(→),1)EQ\*jc3\*hps21\o\al(\s\up8(——),nP)EQ\*jc3\*hps21\o\al(\s\up8(→),1)EQ\*jc3\*hps21\o\al(\s\up8(——),nP)EQ\*jc3\*hps21\o\al(\s\up8(→),1)EQ\*jc3\*hps21\o\al(\s\up8(——),nP)EQ\*jc3\*hps21\o\al(\s\up8(→),1)因xn+1>xn>0,則得xn+1-an=(*)EQ\*jc3\*hps21\o\al(\s\up11(——),nP)EQ\*jc3\*hps21\o\al(\s\up11(→),1)EQ\*jc3\*hps21\o\al(\s\up11(——),nP)EQ\*jc3\*hps21\o\al(\s\up11(→),1)n+1-xn)2+-2=(xn+1-an)2+2，+1-xn)21+=1+，n+1-xn)2=，則xn+1-xn=；由xn+1-xn=可得xn+1-=xn，兩邊取平方，xEQ\*jc3\*hps12\o\al(\s\up5(2),n)+1+-8=xEQ\*jc3\*hps12\o\al(\s\up5(2),n)，即xEQ\*jc3\*hps12\o\al(\s\up5(2),n)+1-xEQ\*jc3\*hps12\o\al(\s\up5(2),n)=8-<8，又xn+1-xn=，故(xn+1-xn)xn+1=4，因xn+1>xn>0,則4=(xn+1-xn)xn+1<xEQ\*jc3\*hps12\o\al(\s\up3(2),n)+1-xEQ\*jc3\*hps12\o\al(\s\up3(2),n)，即4<xEQ\*jc3\*hps12\o\al(\s\up3(2),n)+1-xEQ\*jc3\*hps12\o\al(\s\up3(2),n)<8，故得4<xEQ\*jc3\*hps12\o\al(\s\up3(2),n)+1-xEQ\*jc3\*hps12\o\al(\s\up3(2),n)<8，4<xEQ\*jc3\*hps12\o\al(\s\up3(2),n)-xEQ\*jc3\*hps12\o\al(\s\up3(2),n)-1<8，?,4<xEQ\*jc3\*hps12\o\al(\s\up3(2),2)-xEQ\*jc3\*hps12\o\al(\s\up3(2),1)<8，將以上n個不等式左右分別相加，可得4n<xEQ\*jc3\*hps12\o\al(\s\up3(2),n)+1-xEQ\*jc3\*hps12\o\al(\s\up3(2),1)<8n，即得4n+2<xEQ\*jc3\*hps12\o\al(\s\up3(2),n)+1<8n+2.2+4n+4≤xEQ\*jc3\*hps12\o\al(\s\up3(2),1)+xEQ\*jc3\*hps12\o\al(\s\up3(2),2)+xEQ\*jc3\*hps12\o\al(\s\up3(2),3)+?+xEQ\*jc3\*hps12\o\al(\s\up3(2),n)+1≤4n2+6n恒成立.EQ\*jc3\*hps12\o\al(\s\up3(2),1)EQ\*jc3\*hps12\o\al(\s\up3(2),2)當n≥2時，10+4×+2(n-1)<10+xEQ\*jc3\*hps12\o\al(\s\up0(2),3)+?+xEQ\*jc3\*hps12\o\al(\s\up0(2),n)+1<10+8×+2(n-1)，即得2n2+4n+4<10+xEQ\*jc3\*hps12\o\al(\s\up3(2),3)+?+xEQ\*jc3\*hps12\o\al(\s\up3(2),n)+1<4n2+6n，不等式成立.2+4n+4≤xEQ\*jc3\*hps12\o\al(\s\up3(2),1)+xEQ\*jc3\*hps12\o\al(\s\up3(2),2)+xEQ\*jc3\*hps12\o\al(\s\up3(2),3)+?+xEQ\*jc3\*hps12\o\al(\s\up3(2),n)+1≤4n2+6n得證.66線交雙曲線C右支異于點Pn-1的點Pn.n-yn-1=-k(xn+xn-1(，且0<k2<2.且2(xn+xn-1((xn-xn-1(=(yn+yn-1((yn-yn-1(=-k(yn+yn-1((xn+xn-1(.所以xn-xn-1=-(yn+yn-1(①,而xn+xn-1=-(yn-yn-1(②,n=-yn+yn-1③,xn-1=-yn+yn-1④,n=-yn+1+yn⑤,由③⑤得：yn=(yn+yn-1(⑥.所以==-，77n三點共線.所以PnPn+1nQn+1.，即數(shù)列{Sn{為常數(shù)列.10.(24-25高三上·新疆烏魯木齊·月考)已知點P1(1,-2(在拋物線C:y2=2px(p>0(上，過點P1作斜率4x.,yn(在拋物線上，則xn=，Qn-1(xn,-yn(，過,yn-1，且斜率為1的直線Pn-1Qn-1的方程為y-yn-1=1=x-可得y2-4y+(4yn-1-y-1(=0，解得y=yn-1或y=-yn-1+4，所以-yn=-yn-1+4，可得yn-yn-1=-4，所以數(shù)列{yn{是以首項為-2，公差為-4的等差數(shù)列，88所以yn=-2-4(n-1(=-4n+2，xn===(2n-1(2；、Qn-1(xn,-yn(在拋物線C:y2=4x上，EQ\*jc3\*hps21\o\al(\s\up7(=),4)EQ\*jc3\*hps21\o\al(\s\up7(4),x)-yn((yn-1+yn(=4(xn-1-xn(.Pn-1Qn-1=-==1可得yn-yn-1=-4，所以數(shù)列{yn{是以首項為-2，公差為-4的等差數(shù)列，所以yn=-2-4(n-1(=-4n+2，xn===(2n-1(2.PnPn+2===，則直線PnPn+2的方程為y-yn=(x-xn(，+2|=1+2|yn+2-yn|=44+(yn+1(2，點Pn+1到直線PnPn+2的距離d=n+1-xn(+yn-yn+1=1+2(-(yn+1+yn(((yn+1(2+4==|2(-(yn+1+yn((+4y==(yn+1(2+48(yn+1(2+4+1Pn+2=PnPn+2===則直線PnPn+2的方程為y-yn=(x-xn(，設直線PnPn+2與QnPn+1相交于Mn+1令x=xn+1，可得yMn+1=yn+=yn+，+1Mn+1|=|yMn+1-yn+1|=2-=+1Pn+2==16-1(2,2-4n(、Pn+1((2n+1(2,-2-4n(、Pn+2((2n+3(2,-6-4n(，=[(2n+1(2-(2n-1(2[(4n-2+4n+2(=32n2，即STPPT=32n2，同理可得STPPT=32(n+1(2，=[(2n+3(2-(2n-1(2[(4n-2+4n+6(=16(2n+1(2，99S△PPP=STPPT+STPPT-STPPT=32n2+32(n+1(2-16(2n+1(2=16.造點Pn(n=2,3,4,?(，過點Pn-1作斜率為-EQ\*jc3\*hps22\o\al(\s\up5(1),2)的直線與拋物線EQ\*jc3\*hps21\o\al(\s\up8(3),4)2=2p1=21=1，EQ\*jc3\*hps21\o\al(\s\up6(x),4)EQ\*jc3\*hps12\o\al(\s\up10(2),n)過Pn-1(xn-1，x-1)，n≥2，且斜率為-EQ\*jc3\*hps21\o\al(\s\up5(1),2)的直線Pn-1Qn-1:y-x-1=-EQ\*jc3\*hps21\o\al(\s\up5(1),2)(x-xn-1)，解得x=xn-1或x=-xn-1-2，∴-xn=-xn-1-2，可得xn=xn-1+2，n=2+2(n-1)=2n，又yn=EQ\*jc3\*hps21\o\al(\s\up8(x),4)EQ\*jc3\*hps15\o\al(\s\up12(2),n)=4EQ\*jc3\*hps21\o\al(\s\up7(n),4)2=n2，∴xnyn=n22n=EQ\*jc3\*hps21\o\al(\s\up7(1),2)(EQ\*jc3\*hps21\o\al(\s\up7(1),n)-n2)，∴Tn=EQ\*jc3\*hps21\o\al(\s\up7(1),2)(1-EQ\*jc3\*hps21\o\al(\s\up7(1),3)+EQ\*jc3\*hps21\o\al(\s\up7(1),2)-EQ\*jc3\*hps21\o\al(\s\up7(1),4)+EQ\*jc3\*hps21\o\al(\s\up7(1),3)-EQ\*jc3\*hps21\o\al(\s\up7(1),5)+?+EQ\*jc3\*hps21\o\al(\s\up7(1),n)-n2)=EQ\*jc3\*hps21\o\al(\s\up7(1),2)(1+EQ\*jc3\*hps21\o\al(\s\up7(1),2)-n1-n2)=EQ\*jc3\*hps21\o\al(\s\up7(3),4)-2(n1)-12(n+2)；222EQ\*jc3\*hps21\o\al(\s\up6(n),n)即(n+1)x-y-n2-2n=0，(n+1)2+1(n+1)2+1點Pn+1到直線PnPn+2的距離為d=|2(n+1)2-(n+1(n+1)2+1(n+1)2+1+2|=(2n+4-2n)2+[(n+2)2-n2]2=4(n+1)2+1，*列{an{滿足以下性質：當直線AnPn的斜率kAP與直線BnQn的斜率kBQ恒滿足kAP=3kBQ時，直線PnQn與x軸交于點Dn(-an+1,0(.n(x2設直線PnQn的方程為x=my+t.消去x并整理，得(m2+4(y2+2mty+t2-aEQ\*jc3\*hps12\o\al(\s\up3(2),n)=0，則Δ=4m2t2-4(m2+4((t2-aEQ\*jc3\*hps12\o\al(\s\up3(2),n)(=4(m2aEQ\*jc3\*hps12\o\al(\s\up3(2),n)-4t2+4aEQ\*jc3\*hps12\o\al(\s\up3(2),n)(，y1y2m2+4，y1y2+=-y1y2m2+4，y1y2所以my1y2=(y1+y2(，=my2y1+(t-an(y1=(y1+y2(+(t-an(y1=an-t?(an+t((y1+y2(-2ty1=an-t=3my2y1+(t+an(y2(y1+y2(+(t+an(y2an+t(an-t((y1+y2(+2ty2an+t，所以t=-an.代入Δ=4(m2aEQ\*jc3\*hps12\o\al(\s\up3(2),n)-4t2+4aEQ\*jc3\*hps12\o\al(\s\up3(2),n)(中驗證，得Δ>0，則直線PnQn的方程為x=my-因為直線PnQn與x軸交于點Dn(-an+1,0(，所以-an+1=-an，則.故數(shù)列{an{是以為公比的等比數(shù)列.+y2=，y1y2=-，所以四邊形BnPnOQn的面積Sn=S△BnPnQn-S△OPnQn=(|BnDn|-|ODn|(|y2-y1|設z=m2+3(z≥3(，則函數(shù)f(z(=z+在[3,+∞(上單調遞增，所以f(z(≥n-1n-1，EQ\*jc3\*hps12\o\al(\s\up3(2),n)所以=433，解概率.n-1.故X的分布列為：X234P 9 49 49(2)由題意知Pn+1=Pn+Pn-1(n≥2(，故Pn+1-Pn=-(Pn-Pn-1(，且P2=+×=，P1=，P2-P1=，故{Pn-Pn-1{是以為首項，-為公比的等比數(shù)列，故Pn-Pn-1=×(-n-2(n≥2(，∴當n≥2時，Pn=P1+(P2-P1(+(P=+×1+(-+(-2+???+(-n-2=+×=+1-(-n-1=--n-1，綜上：Pn=--n-1.則P(A)==，P(AB)==，所以P(B|A(=P(AB)==2(或P(B|A(=nn如圖得P(An(=，P(B3(=P(A1(P(A2(P(B3(+P(A1(P(B2(P(B3(+P(B1(P(B2(P(B3(=××+××+ P(Bn(=P(A1(P(A2(·?·P(An-1(P(Bn(+P(A1(P(A2(·?·P(Bn-1(P(Bn(+?+P(B1(P(B2(·?·P(Bn-1(P(Bn(P(Bn(=+×+×+?+×① 則P(Bn(=+×+?×+②②-①得到：P(Bn(=-P(Bn(=2-，X1234nP 1 2 3 4 nE(X(=+×2+×3+?+×n③則E(X(=+×2+×3+?+×(n-1(+④③-④得：E(X(=+++?+-=-，EE(X(=.(參考數(shù)據(jù)：P(μ-σ≤x≤μ+σ)=0.6827,P(μ-2σ≤x≤μ+2σ)=0.9545，Pμ-3σ≤x≤μ+3σ)=0.9973,=1.524×10-1,=5.081×10-5,=1.694×10-5(.+×(-n,1≤n≤9+×(-n,1≤n≤9-3(,n=10則P(35≤x<40)=P(u+σ≤x<u+2σ)==0.1359.獲得n-2分時擲骰子點數(shù)大于4，Pn=Pn-1+Pn-2(n≥3(，則Pn-Pn-1=Pn-1+Pn-2-Pn-1=-(Pn-1-Pn-2(，n≥3故{Pn-Pn-1{(n≥2(為等比數(shù)列.由P1==+?=，故首項為-=.因此Pn-Pn-1=-n-1，Pn-1-Pn-2=-n-2??P2-P1=，將所有等式相加得Pn-P1=?=，所以Pn=+=+×(-n(1≤n≤9(，8-5(+80000題.=P=PnB|=，又ACC平面AOC，所以BD丄AC，(n+1(2n.2n2n，可得anbn=n.(n+1(2((n+1(2n.2n2n，令f(x(=(x≥1(，則f/(x(=，而1n2>ln、e=，當x≥3時，6-(6x-5(ln2<0，可知f(x(=在[3,+∞(上單調遞減，即當n≥3時，數(shù)列{anbn{單調遞減，1=2=， ()A.B.S3=C.數(shù)列{Vn{是公比為的等比數(shù)列D.數(shù)列{Sn{的前n項和為8π(1-設三棱錐A-BCD的棱長均為a，所以OB=×a=a，AO=AB2-OB2=2a(2=O1是三棱錐A-BCD的內(nèi)切球的球心，O1在AO上，設三棱錐A-BCD的外接球半徑為R，球On的半徑為rn，則由O1B2=OOEQ\*jc3\*hps12\o\al(\s\up3(2),1)+OB2，得R2=a-R(2+a(2，得R=a.所以r1=AO-AO1=a-a=a，所以V1=πrEQ\*jc3\*hps12\o\al(\s\up5(3),1)=π3=6π.故A不正確；在AO上取點E，使得EO1=r1=a，則AE=AO-2r1=a-a=a，即E為AO的中點，則球O2與球O1切于E，則球O2是三棱錐A-B1C1D1的內(nèi)切球，因為E為AO的中點，所以三棱錐A-B1C1D1的棱長是三棱錐A-BCD的棱所以球O2的內(nèi)切球的半徑r2=r1，EQ\*jc3\*hps15\o\al(\s\up4(3),n)EQ\*jc3\*hps12\o\al(\s\up4(2),n)S1+S2+?+Sn=6π(1+++?+=6π?=8π(1-，故D正確.的概率為Pn.又因為所以+1=Pn+(1-Pn(=Pn+，n-1=n-n×所以(iai(=EQ\*jc3\*hps22\o\al(\s\up7(—),3n)EQ\*jc3\*hps22\o\al(\s\up7(——),3n)EQ\*jc3\*hps22\o\al(\s\up7(——),3n)n=3n2-n,cn=n3,dn=cos(n∈N+(，求5++的值.EQ\*jc3\*hps22\o\al(\s\up0(l),n)EQ\*jc3\*hps22\o\al(\s\up0(l),n)EQ\*jc3\*hps22\o\al(\s\up0(l),n)EQ\*jc3\*hps22\o\al(\s\up0(l),n)n=n-1,cn=(-n-1,dn=n-1(n∈N+(，設OEQ\*jc3\*hps22\o\al(\s\up11(—→),Bn)=+++…+aEQ\*jc3\*hps22\o\al(\s\up9(——),3n)3，求EQ\*jc3\*hps22\o\al(\s\up0(l),n)OEQ\*jc3\*hps22\o\al(\s\up10(—→),Bn).EQ\*jc3\*hps22\o\al(\s\up7(——),3m)EQ\*jc3\*hps22\o\al(\s\up7(—→),3m)證：++…+|aEQ\*jc3\*hps22\o\al(\s\up4(——),3m)-+EQ\*jc3\*hps22\o\al(\s\up4(—→),3m)≤、3L.EQ\*jc3\*hps21\o\al(\s\up0(l),n)EQ\*jc3\*hps21\o\al(\s\up9(—→),Bn)=10,c1=1=1,d2=cosπ=-1++=|c1.+b2.+d2.=+10-=|(10,1,-1(|=、102=c1=d1,EQ\*jc3\*hps21\o\al(\s\up7(—),3n)1=nEQ\*jc3\*hps21\o\al(\s\up7(——),3n)EQ\*jc3\*hps21\o\al(\s\up7(——),3n)EQ\*jc3\*hps21\o\al(\s\up0(l),n)EQ\*jc3\*hps21\o\al(\s\up9(—→),Bn)則OEQ\*jc3\*hps21\o\al(\s\up9(—→),Bn)=++…+aEQ\*jc3\*hps21\o\al(\s\up7(——),3n)3==(1+b1+b2+…+bn,1+c1+c2+…+cn,1+d1+d2+…+dn(則x=EQ\*jc3\*hps21\o\al(\s\up0(l),n)1+b1+b2+…+bn(，可得：=1+EQ\*jc3\*hps21\o\al(\s\up2147483646(l),n)=1+=，EQ\*jc3\*hps21\o\al(\s\up1(l),n)EQ\*jc3\*hps21\o\al(\s\up10(—→),Bn)++=3，++=(1,1,1(，則++=、3=L，++=、3L，++≤、3L成立；EQ\*jc3\*hps21\o\al(\s\up7(——),3m)EQ\*jc3\*hps21\o\al(\s\up7(—),3)EQ\*jc3\*hps21\o\al(\s\up7(→),m)=(1+b1+b2+…+bm-1(.+(1+c1+c2+…+cm-1(.+(1+d1+d2+…+dm-1(.令Rm-1=b1+b2+…+bm-1,Sm-1=c1+c2+…+cm-1,Tm-1=d1+d2+…+dm-1,+++…aEQ\*jc3\*hps21\o\al(\s\up7(——),3m)-→1+aEQ\*jc3\*hps21\o\al(\s\up7(—→),3m)=(1+Rm-1(.+(1+Sm-1(.+(1+Tm-1(. =(1+Rm-1,1+Sm-1,1+Tm-1(EQ\*jc3\*hps21\o\al(\s\up7(——),3m)EQ\*jc3\*hps21\o\al(\s\up7(—→),3m)1+Rm-1,1+Sm-1,1+Tm-1(|=L，即(1+Rm-1(2+(1+Sm-1(2+(1+Tm-1(2=L，=(1+b1+b2+?+bm-1(+(1+c1+c2+?+cm-1(+(1+d1+d2+?+dm-1(=(1+Rm-1(+(1+Sm-1(+(1+Tm-1(≤3[((1+Rm-1(2+(1+Sm-1(2+(1+Tm-1(2[=3L，EQ\*jc3\*hps21\o\al(\s\up4(——),3m)EQ\*jc3\*hps21\o\al(\s\up4(—→),3m)EQ\*jc3\*hps21\o\al(\s\up4(——),3m)EQ\*jc3\*hps21\o\al(\s\up4(—→),3m)通過論證找到交點位置.EQ\*jc3\*hps22\o\al(\s\up8(→),d)當y=0時,xOz平面截曲面C所得交線上的點M(x,0,z(滿足x2-=1，EQ\*jc3\*hps21\o\al(\s\up7(→),d)EQ\*jc3\*hps21\o\al(\s\up8(—→),P)EQ\*jc3\*hps21\o\al(\s\up8(—→),P)EQ\*jc3\*hps21\o\al(\s\up7(→),d)EQ\*jc3\*hps21\o\al(\s\up9(—→),P)EQ\*jc3\*hps21\o\al(\s\up7(→),d),y0+1,z0+2(=λ(1,0,2(，所以點P的坐標為(λ-1,-1,2λ-2(，設M(x1,y1,z1(是直線l/上任意一點，直線!的方向向量為,b,c)，EQ\*jc3\*hps21\o\al(\s\up8(—→),M)EQ\*jc3\*hps21\o\al(\s\up8(—→),M)-1,y1,z1(=t(a,b,c(，所以點M的坐標為(1+at,bt,ct(，∵M(x1,y1,z1)在曲面C上，2-=0-2，EQ\*jc3\*hps21\o\al(\s\up7(→),d)為A1B1．點E為AB上一點，過點E在下底面內(nèi)作AB的垂線分別交下底面橢圓于點C，D．記m= .(1)若平面A1AD⊥平面A1AC，求m及二面角C-A1B-D的余弦值；<θ<.⊥底面ADBC，AC,AD?底面ADBC，故AA1⊥AD,AA1⊥AC,可設設橢圓方程為+=1,(a>b>0(，由于e==，故=，則+=1,由于AC⊥AD，根據(jù)橢圓的對稱性可知∠CAE=，故CE=DE=AE，設直線CD:x=-d,(0<d<a(，則CE=AE=a-d，故C(-d,a-d(，將C(-d,a-d(代入橢圓+=1,中可得+=1,故E(-a,0(，則m==，EQ\*jc3\*hps21\o\al(\s\up9(—→),B)EQ\*jc3\*hps21\o\al(\s\up9(—→),D)EQ\*jc3\*hps21\o\al(\s\up7(—→),1B)EQ\*jc3\*hps21\o\al(\s\up7(—→),C)EQ\*jc3\*hps21\o\al(\s\up7(—→),D)EQ\*jc3\*hps21\o\al(\s\up9(A),B)EQ\*jc3\*hps21\o\al(\s\up9(x),a)EQ\*jc3\*hps21\o\al(\s\up9(0),y)EQ\*jc3\*hps21\o\al(\s\up9(A),B)EQ\*jc3\*hps21\o\al(\s\up15(—),D)EQ\*jc3\*hps21\o\al(\s\up9(x),a)EQ\*jc3\*hps15\o\al(\s\up6(0-),x0)EQ\*jc3\*hps21\o\al(\s\up8(=),ay)EQ\*jc3\*hps21\o\al(\s\up9(0),0)設二面角C-A1B-D的平面角為θ,======2+42+(43(232+(-4(2+(43(2由于BE=,AA1=b=23a,A1B=、4a2+b2=EQ\*jc3\*hps21\o\al(\s\up5(19),2)a，==?7 EQ\*jc3\*hps16\o\al(\s\up5(6),7)=EQ\*jc3\*hps16\o\al(\s\up3(8),7)π>1=tan,故==?7 EQ\*jc3\*hps16\o\al(\s\up5(6),7)=EQ\*jc3\*hps16\o\al(\s\up3(8),7)π>1=tan,故θ>將x=a代入橢圓+=1,可得yC=a,yD=-a,EQ\*jc3\*hps21\o\al(\s\up6(—→),1B)EQ\*jc3\*hps21\o\al(\s\up6(—→),C)EQ\*jc3\*hps21\o\al(\s\up6(—→),D)EQ\*jc3\*hps21\o\al(\s\up12(A),B)EQ\*jc3\*hps21\o\al(\s\up12(2),m)EQ\*jc3\*hps21\o\al(\s\up12(-),x)EQ\*jc3\*hps21\o\al(\s\up12(=),m)EQ\*jc3\*hps21\o\al(\s\up14(A),B)EQ\*jc3\*hps21\o\al(\s\up20(—),D)EQ\*jc3\*hps21\o\al(\s\up13(2),m)設二面角C-A1B-D的平面角為θ,所以cos=-<-<cosθ≤-<0=cos,故<θ<，得證.θ(0<θ<的直線l與C交于A，B兩點(其中點AFF故折疊前橢圓C的標準方程+=1.+1(，EQ\*jc3\*hps21\o\al(\s\up6(=),x)EQ\*jc3\*hps21\o\al(\s\up7(8),5)EQ\*jc3\*hps21\o\al(\s\up10(—→),2B1)EQ\*jc3\*hps21\o\al(\s\up7(13),5)EQ\*jc3\*hps21\o\al(\s\up10(—),2)EQ\*jc3\*hps21\o\al(\s\up10(→),1)?FA1=0(-y+3z=0EQ\*jc3\*hps21\o\al(\s\up14(—),2)353x-EQ\*jc3\*hps16\o\al(\s\up9(13),5)y?FA1=0(-y+3z=09+3+1=.EQ\*jc3\*hps21\o\al(\s\up6(13),3)EQ\*jc3\*hps21\o\al(\s\up6(13),3)9+3+1=.EQ\*jc3\*hps21\o\al(\s\up9(→),1)EQ\*jc3\*hps21\o\al(\s\up10(—→),2B1)EQ\*jc3\*hps21\o\al(\s\up7(13),5)EQ\*jc3\*hps21\o\al(\s\up10(—),2)EQ\*jc3\*hps21\o\al(\s\up10(→),1)EQ\*jc3\*hps21\o\al(\s\up11(—→),2B1)EQ\*jc3\*hps21\o\al(\s\up11(—),2)EQ\*jc3\*hps21\o\al(\s\up11(→),1)EQ\*jc3\*hps21\o\al(\s\up7(3),5)EQ\*jc3\*hps21\o\al(\s\up7(9),5).EQ\*jc3\*hps21\o\al(\s\up5(1),2)結合圖象可知設直線l的方程為x=my-1，其中m==，+4(y2-6my-9=0，3m2+4，y1y2=-3m2+4,顯然Δ>3m2+4，y1y2=-3m2+4,F+F+++即(x1-x2(2+(y1-y2(2-(x1-x2(2+yEQ\*jc3\*hps12\o\al(\s\up3(2),1)+yEQ\*jc3\*hps12\o\al(\s\up3(2),2)=①, =(x1-x2(2+(y1-y2(2+(x1-x2(2+yEQ\*jc3\*hps12\o\al(\s\up3(2),1)+yEQ\*jc3\*hps12\o\al(\s\up3(2),2)2，(x1-x2(2+(y1-y2(2+(x1-x2(2+yEQ\*jc3\*hps12\o\al(\s\up3(2),1)+yEQ\*jc3\*hps12\o\al(\s\up3(2),2)=-4y1y2②,由①②得：、(x1-x2(2+(y1-y2(2=-2y1y2，即是(1+m2((y1-y2(2=-2y1y2(2，(1+m2(2+=+2，=+2，即是=+，解得m2=，注意到0<θ<，故tanθ==、=，從而存在滿足條件的θ,且tanθ=.體積的最大值.可得A/O/丄平面ABC，由BCC平面ABC，可得A/O/丄BC，O由A/EC平面A/O/E，可得BC丄A/E，在Rt△ACO/x2=2yy=kx+，消去y可得x2-2kx-x2=2y則x1+x2=2k,x1.x2=-1，可得|AB|=、1+k2、4k2+4=2(k2+1(，+1+、2(R，解得R=，則=，可得h=1-=1-(2+2(r，所以圓柱的體積V=πr2h=π[r2-(2+、2(r3[，令f(r(=r2-(2+2(r3,0<r<，則fl(r(=2r-3(2+2(r2=r[2-3(2+2(r[，當0<r<時，fl(r(>0；當<r<時，fl(r(<0；則f(r(≤f=，以轉化為點A(x,y(與點B(a,b(之間的距離的幾何問題．結合上述觀點，對于函數(shù)f(x(=2+4x+5+、x2-4x+5，下列結論正確的是()A.方程f(x(=5無解B.方程f(x(=6有兩個解【解析】f(x(=x2+4x+5+x2-4x+5=(x+2(2+1+(x-2(2+1，+2=9-4=5，即橢圓方程為+=1，此命名為小恐龍曲線．對于小恐龍曲線C:x2+y3-axy=20，下列說法正確的是()A.該曲線與x=6最多存在3個交點x2+y3-axy=20，令f(y)=y3-6ay+16，則f/(y)=3y2-6a，即存在x0>0，關于y的方程y3-ax0y+xEQ\*jc3\*hps12\o\al(\s\up3(2),0)-20=0有三個實根.令f(y)=y3-ax0y+xEQ\*jc3\*hps12\o\al(\s\up3(2),0)-20，則f/(y)=3y2-ax0，EQ\*jc3\*hps12\o\al(\s\up3(2),0)2+y3=20即函數(shù)y=320-x2的圖象，且f(-x)=320-(-x)2=320-x2=f(x)，所以f(x)是偶函數(shù).x2+y3-3xy-20=0.令f(y)=y3-18y+16，則f(0)=16>0，f(1)=-1<0，f(4)=8>0D.C在y軸左邊的部分到坐標原點O的距離均大于2x3-2x+2的曲線，2對于B，由于曲線C在x3-2x+2的曲線，22-2=2-2=03x2-23-2x+2將切線方程代入曲線方程中，得到：x3-2x+2=x2+x+，即4x3-x2-10x+7=0，顯然x=1是方程的根，(x-1((4x2+3x-7(=0，.當-<x<時g/(x(<0，g(x(單調遞減，g(-2(=-8+4+2=-2<0，g(-2(=-22+22+2=2>0，g(0(=2>0，g=-+2=>0，所以-2<x0<-、2，設曲線上的點為P(x,y(，設x3-2x+2=0的解為x0，則-2<x0<-、2，則x0≤x<0，P到原點O的距離為d=x2+y2，由y2=x3-2x+2可得d=、x2+x3-2x+2，令h(x(=x2+x3-2x+2，因為x0≤x<0，所以取x=，0≤x<時，h/(x)>0，h(x(單調遞增，當<x<0時，h/(x)<0，h(x(單調遞減，h(x0(=xEQ\*jc3\*hps12\o\al(\s\up3(2),0)+xEQ\*jc3\*hps12\o\al(\s\up3(3),0)-2x0+2=xEQ\*jc3\*hps12\o\al(\s\up3(2),0)>2，h(0(=2，所以當x0≤x<0時，h(x(>2，d>2,選項D正確.故選軸是其兩條漸近線.(2)已知點A是曲線C的左頂點．圓E：(x-1(2+(y-1(2=r2(r>0)與直線l：x=1交于P、Q兩點，直線AP、AQ分別與雙曲線C交于M、N兩點．試問：點A到直線MN的距離是否存在最大值？若(EQ\*jc3\*hps21\o\al(\s\up7(y),y)EQ\*jc3\*hps21\o\al(\s\up11(x),y)，即雙曲線y=的兩頂點為,-,-，故實軸長為2a=+2++2=2；曲線C的方程為x2-y2=1；(2)方法一：設A(-1,0(，M(x1,y1(，N(x2,y2(，2-y2=1得(1-k2(x2-2kmx-(m2+1(=0，由①②③得，-2k+2m=-m2+2km-k2，所以(m-k((m-k+2(=0，即m=k或m=k-2，若m=k，則MN：y=k(x+1(過點A，不合題意；若m=k-2，則MN：y=k(x+1(-此時MN方程為y=-2，結合x2-y2=1，解得N(5,-2(，yQ=-(5-1(，r=1-yQ=5，1+r，1+r，4-(1+r(2，代入AP方程得，x1=-1，=-1，則kMN==，所以直線MN的方程為y=x-+1+-=(x+1(-2，所以直線MN過定點G(-1,-2(,所以dmax=|AG|=2．當且僅當MN⊥AG，即kMN==0時取得，解得r=、5,則kAP+kAQ=+=1，依題意，直線MN不過點A，可設MN：m(x+1(+ny=1，曲線C的方程x2-y2=1改寫為[(x+1(-1[2-y2=1，即(x+1(2-2(x+1(-y2=0，聯(lián)立直線MN的方程得(x+1(2-2(x+1([m(x+1(+ny[-y2=0，所以(1-2m((x+1(2-2n(x+1(y-y2=0，則Δ=4n2-8m+4>0，+=kAP+kAQ=-2n=1得n=-，在直線MN：m(x+1(-y=1中，令x=-1，則y=-2，max=|AG|=2，且MN方程為y=-2，解得N(、5,-2(，yQ=-(5-1(，r=1-yQ=5，=2，BEQ\*jc3\*hps22\o\al(\s\up5(—→),E)=λBEQ\*jc3\*hps22\o\al(\s\up5(—→),C)，CEQ\*jc3\*hps22\o\al(\s\up5(—→),F)=λDEQ\*jc3\*hps22\o\al(\s\up5(—→),C)，0<λ<1，直線AE與BF交于點P，雙曲線K:—→—→EQ\*jc3\*hps21\o\al(\s\up9(—→),E)EQ\*jc3\*hps21\o\al(\s\up9(—→),C)EQ\*jc3\*hps21\o\al(\s\up9(—→),F)EQ\*jc3\*hps21\o\al(\s\up9(—→),C)、、所以直線AE:y=λ(x+3(，直線BF:y=1(x-3、、2=(x2-3(，即-y2=1，所以點P在K上.-x0,-y0(.由題意，得直線由題意，得直線MN的斜率存在且不為0，設MN:x=my+n，聯(lián)立-y2x=my+n,得(m2-3(y2+2mny+n2-3=0．由題意，知m2-3≠0，Δ>0.又直線PM:x=y+2，代入-y2=1，并注意到-yEQ\*jc3\*hps15\o\al(\s\up2(2),1)=1，-4x1(y2+4(x1-2(y1y+yEQ\*jc3\*hps12\o\al(\s\up3(2),1)=0，0=0=y277-4x2，-4x2(y1+(7-4x1(y2，所以7(y1+y2(=4(my2+n(y1+4(my1+n(y2，EQ\*jc3\*hps22\o\al(\s\up8(→),b)EQ\*jc3\*hps22\o\al(\s\up8(→),b)EQ\*jc3\*hps22\o\al(\s\up8(→),b)(1)+EQ\*jc3\*hps21\o\al(\s\up7(→),b)=(s,λ(，λEQ\*jc3\*hps21\o\al(\s\up7(→),b)+4=(λs,4(.設D(x,y(，則A—EQ\*jc3\*hps21\o\al(\s\up9(—→),D)=(x-s,y(，BEQ\*jc3\*hps21\o\al(\s\up9(—→),D)=(x+s,y(因為直線AD以+EQ\*jc3\*hps21\o\al(\s\up7(→),b)為方向向量，故(x-s(λ=sy，因為直線BD以λEQ\*jc3\*hps21\o\al(\s\up7(→),b)+4+s(4=λsy，整理得到-=1(x≠s(.又AP:y=-(x-3(，BP:y=-(x-3(，故以M、N為直徑的圓的方程為：(x-6(2+(y--y--=0， ×===x1-3x2-3(x1-3((x2-3((ty1+3((ty2+3(t2y1y2 ×===EQ\*jc3\*hps21\o\al(\s\up6(x),4)EQ\*jc3\*hps21\o\al(\s\up6(=),x2)-EQ\*jc3\*hps21\o\al(\s\up6(+),9y)Δ=(48t(2-4×108(4t2-9(=36×108+48×12t2>0，2=9×=-12x1-3x2-3t2×-+3t×(--+9，故以M、N為直徑的圓的方程可化簡為：(x-6(2+y2-μy-12=0，x1-3x2-3，其中μx1-3x2-3，令{r(y(EQ\*jc3\*hps21\o\al(\s\up10(y),x)EQ\*jc3\*hps21\o\al(\s\up10(0),6)EQ\*jc3\*hps21\o\al(\s\up10(y),x)EQ\*jc3\*hps21\o\al(\s\up10(0),6)EQ\*jc3\*hps22\o\al(\s\up9(—→),P)=+yeEQ\*jc3\*hps22\o\al(\s\up9(—→),P)=(x,y(.EQ\*jc3\*hps22\o\al(\s\up8(→),b)EQ\*jc3\*hps22\o\al(\s\up8(→),b)—→EQ\*jc3\*hps22\o\al(\s\up7(—→),C)EQ\*jc3\*hps22\o\al(\s\up7(—→),D)—→—→—→e+e2=2+22?+1==5.EQ\*jc3\*hps21\o\al(\s\up7(→),b)EQ\*jc3\*hps21\o\al(\s\up7(→),b)EQ\*jc3\*hps21\o\al(\s\up5(→),b)EQ\*jc3\*hps21\o\al(\s\up5(→),b)EQ\*jc3\*hps21\o\al(\s\up7(→),b)EQ\*jc3\*hps21\o\al(\s\up11(→),b)(3)依題意設B(m,0(、C(0,n((m>0,n>0(，EQ\*jc3\*hps21\o\al(\s\up7(—→),D)EQ\*jc3\*hps21\o\al(\s\up7(—→),C)EQ\*jc3\*hps21\o\al(\s\up9(—→),F)EQ\*jc3\*hps21\o\al(\s\up9(—→),B)EQ\*jc3\*hps21\o\al(\s\up9(—→),F)EQ\*jc3\*hps21\o\al(\s\up9(—→),B)EQ\*jc3\*hps21\o\al(\s\up5(1),2)EQ\*jc3\*hps21\o\al(\s\up9(—→),C)EQ\*jc3\*hps21\o\al(\s\up9(—→),B)EQ\*jc3\*hps21\o\al(\s\up5(1),2)EQ\*jc3\*hps21\o\al(\s\up9(—→),C)EQ\*jc3\*hps21\o\al(\s\up9(—→),B)EQ\*jc3\*hps21\o\al(\s\up5(1),2)EQ\*jc3\*hps21\o\al(\s\up9(—→),C)EQ\*jc3\*hps21\o\al(\s\up5(1),2)EQ\*jc3\*hps21\o\al(\s\up9(—→),B)EQ\*jc3\*hps21\o\al(\s\up5(1),2)EQ\*jc3\*hps21\o\al(\s\up5(1),2)EQ\*jc3\*hps21\o\al(\s\up8(—→),E)EQ\*jc3\*hps21\o\al(\s\up8(—→),D)EQ\*jc3\*hps21\o\al(\s\up8(—→),B)77447mn+EQ\*jc3\*hps21\o\al(\s\up10(—→),E)77447mn+在△OBC中依據(jù)余弦定理得m2+n2-mn=1，所以mn=m2+n2-1，代入上式得，OEQ\*jc3\*hps21\o\al(\s\up9(—→),E)?OEQ\*jc3\*hps21\o\al(\s\up9(—→),F)=EQ\*jc3\*hps21\o\al(\s\up6(8),19)m2+EQ\*jc3\*hps21\o\al(\s\up6(5),19)n2-EQ\*jc3\*hps21\o\al(\s\up6(13),76)=EQ\*jc3\*hps21\o\al(\s\up6(1),9)EQ\*jc3\*hps21\o\al(\s\up6(13),76)sinEQ\*jc3\*hps16\o\al(\s\up2(π),3)sin∠COBsin∠BCOEQ\*jc3\*hps21\o\al(\s\up7(2),3)EQ\*jc3\*hps21\o\al(\s\up5(2),3)[13+7sin(2θ-φ([，EQ\*jc3\*hps21\o\al(\s\up5(x),1)EQ\*jc3\*hps12\o\al(\s\up9(2),6)EQ\*jc3\*hps21\o\al(\s\up7(y),8)1；(2)①證明見解析-1,-所以Mn(n,2n-1(，即有=(n,2n-1(，設Nn(xn,yn(，則OEQ\*jc3\*hps21\o\al(\s\up9(→),n)=(xn,yn(，消去n得xEQ\*jc3\*hps12\o\al(\s\up3(2),n)+yEQ\*jc3\*hps12\o\al(\s\up3(2),n)-4xn+2yn=0，設Q(x,y(是切線上除P點外的任意一點，則PEQ\*jc3\*hps21\o\al(\s\up9(—→),Q)=(x-x0,y-y0(,OEQ\*jc3\*hps21\o\al(\s\up9(—→),P)=(x0,y0(，因為QP⊥OP，所以PEQ\*jc3\*hps21\o\al(\s\up9(—→),Q)?OEQ\*jc3\*hps21\o\al(\s\up9(—→),P)=(x-x0,y-y0(?(x0,y0(=0，EQ\*jc3\*hps12\o\al(\s\up3(2),0)EQ\*jc3\*hps12\o\al(\s\up3(2),0)聯(lián)立整理得(2xEQ\*jc3\*h