A.{1,2,3,5}BA.充分不必要條件C.充要條件4.已知直線l:mx+y-2m+1=0,圓CA.-1B.18.已知定義在R上的函數(shù)f(x(的導(dǎo)函數(shù)為f,(x(,f(x+2(=f(-x(,(x-1(f,(x(≥0,且對(duì)任意的x∈R,有f(3-cos2x(≤f(msin2x(,則實(shí)數(shù)m的取值范圍為A.(-∞,-1]B.[-1,1[C.[1,3[D.[3,+∞)9.已知函數(shù)f(x(=2x3-3x,f,(x(為f(x(的導(dǎo)函數(shù)，則A.fB.函數(shù)f(x(在上單調(diào)遞增C.函數(shù)f(x(的極大值為D.函數(shù)y=f(x為奇函數(shù)C.滿足|PMPN|的點(diǎn)P有兩個(gè)C.滿足|PMPN|的點(diǎn)P有兩個(gè)11.如圖,已知點(diǎn)P是棱長(zhǎng)為的正方體ABCD-A1B1C1D1表面上一動(dòng)A1D11P1PAAD.當(dāng)點(diǎn)P在面CDD1C1上時(shí),若PB+PC則點(diǎn)P的軌跡長(zhǎng)度為右兩支分別交于點(diǎn)P,Q,若P為線段F1Q的中點(diǎn),且|PQ|,|QF2|,|PF2|成等差數(shù)列,則雙曲14.若數(shù)列{an{滿足an+2+an≥2an+1(n∈N*,當(dāng)且僅當(dāng)n為奇數(shù)時(shí)取“=”),則稱{an{為“T數(shù)如圖,在四棱錐P-ABCD中,底面ABCD為矩形,PA⊥底面ABCD,PA=AB=2AD=PPA(2)求二面角A-EF-D的余弦值.AECFBDECFBD已知點(diǎn)P1(x1,y1(,P2(x2,y2(,…,Pn(xn,yn(,…均在拋物線x2=4y上，x1=1，0<xn+1<xn,(1)求數(shù)列{xn{的通項(xiàng)公式；M,且△MF1F2的面積為1.ll(ⅱ)設(shè)O為坐標(biāo)原點(diǎn),若l1與Γ相交于點(diǎn)A,B,l2與Γ相交于點(diǎn)C,D,且直線OA,OB,已知函數(shù)f(xa∈R(.(ⅰ)證明:數(shù)列{xn{是遞減數(shù)列；因?yàn)榈酌鍭BCD為矩形,所以BC丄AB.又PAC平面PAB,ABC平面PAB,PA∩AB=A,所以BC丄面PAB,...................................................................................................................3分因?yàn)锳EC面PAB,所以BC丄AE,.......................................................................................4分在△APB中,因?yàn)锳P=AB,E為PB中點(diǎn),所以AE丄PB,所以AE丄平面PBC...............................................................................................................5分又AEC平面AEF,所以平面AEF丄平面PBC.................................................................6分(2)以D為原點(diǎn)，D,D的方向分別作為x，y軸的正方向，建立如圖所示空間直角坐標(biāo)PPED設(shè)m=(x1,y1,z1(為平面AEF的一個(gè)法向量,EDCy1,z1=-1,此時(shí)m=(4,1,-1(.....CyFBAx又D=(1,1,1(,,設(shè)n=(x2,y2,z2(FBAx面DEF的一個(gè)法向量,則有,令x2=4,得y2=-1,z2=-3,所以n=(4,-1,-3(........................11分設(shè)二面角A-EF-D的平面角的大小為θ,則cosθ=|cos<m,n>|,..............................12分所以，二面角A-EF-D的余弦值為.......................................................................13分16.解:(1)在△ABD中,由余弦定理得,AD2=AB2+BD2-2AB.BDcosB,.............................1分2c××≥-=.............................................................3分所以ac≤4，當(dāng)且僅當(dāng)“=c”時(shí)取“=”,.................................................................................4分△ABC面積的最大值為3.................因?yàn)椤螧AD+++C=π,所以∠BAD=-C,即sin(2C+=1,.................................................................................................................13分解得C=..............................................................................................................................15分17.解:(1)因?yàn)辄c(diǎn)Pn(xn,yn(在拋物線x2=4y上,所以yn且yn>0,因?yàn)閳APn(xn,yn(和圓Pn+1(xn+1,yn+1(外切且圓Pn均與x軸相切,所以|PnPn+1|=rn+rn+1=yn+yn+1,..........................................................................................2分整理得(xn-xn+1(2=4ynyn.............................................................................3分因?yàn)?<xn+1<xn,所以xn-xn..........................................................................4分所以所以xn...............................................................7分(2)由(1)知，.............因?yàn)?..............................................................................11分(1-+-+…+-....................13分即,原題得證................................................................................15分所以Γ的方程為y2=1....................................................................................................4分因?yàn)榇胝淼脁0+y0-3=0(x0≠±1,x0≠3(.............................................7分設(shè)與直線x+y-3=0平行且與Γ相切的直線方程為x+y+m=0,8分.....................................................聯(lián)立得3x8分.....................................................所以|PQ|的最小值為........................................................................10分(ⅱ)設(shè)直線l1的方程為y=k1(x+1(,設(shè)A(x1,y1(,B(x2,y2(,因?yàn)閗OA,kOB均存在,所以點(diǎn)所以x1+xx1x同理可得kOC+kODk2≠±1,..................................................................................14分k2-1((k1+k2(=0,..........................................................................................................15分由x0+y0-3=0,解得P,,.................................................................1 ((或P(0 令g(x(=axex-ex+1,則g/(x(=ex(ax+a-1(,.................................................................1分當(dāng)a≤0時(shí),因?yàn)閤>0,所以ax+a-1<0,即g/(x(<0在(0,+∞(上恒成立,所以g(x(1-aa/(x(<0,g(x(在(0,+∞(上單調(diào)遞減,所以g(x(<g(0(=0,不合題意;...........................................................21-aa/(x(<0,g(x(<g(0(=0,不合題意;(,+∞(時(shí),g/(<g(0(=0,不合題意;≤0,此時(shí)g/(x(>0在(0,+∞(上恒成立,所以.............................................................................................................................................................3分≤0,此時(shí)g/(x(>0在(0,+∞(上恒成立,所以g(x(在(0,+∞(上單調(diào)遞增,所以g(x(>g(0(=0恒成立,符合題意.......................................4分綜上，a的取值范圍為[1,+∞).................................................................................................5分f(x(>1.>0,所以ex=f(x>0,所以ex=f(x1(>1,所以x2>0....................................................8分因?yàn)閤2>0,所以ex=f(x2(>1,所以x3>0.以此類推,xn>0(n∈N*(..............................6分因?yàn)閤n+1-xn=lnexn+1-xn=lnlnexn=ln...................................................8分令g(x(=x-(ex-1(,則g/(x(=1-ex,當(dāng)x>0時(shí),g/(x(<0,所以g(x(在(0,+∞(單減,即g(x(<g(0(=0,即x+1-ex<0(x>0(,所以0<x<ex-1,所以0<,.............................................................