电子文档交易市场
安卓APP | ios版本
电子文档交易市场
安卓APP | ios版本

类型自编高考试题(4)

收藏

编号:337166925    类型:共享资源    大小:789.28KB    格式:PDF    上传时间:2022-09-28
  
5
金贝
分享到微信 分享到微博 分享到QQ空间
关 键 词:
试卷
资源描述:
fjddy1g?pK(4)m:120,:150.!JK(40)1.?8A=x|x+1|10,B=x|x=4k+1,k Z.KA B?()A.5B.6C.8D.102.?iJ,K|(1 i)2|?()A.2B.2C.22D.43.?y2=4px(p 0)?O?:m?l2,Kp?()A.14B.12C.1D.24.?an?,a2,a3,3a3?,a2=1.Ka4?()A.1B.3C.5D.75.n?|,?1,XJ?VO0.4,0.5,0.7,KT?V()A.0.14B.0.21C.0.7D.0.916.uf(x)=sinx+lnx,eL?m,?f(x)3mN4?()A.?23,76?B.?,32?C.?53,136?D.?2,52?7.%xO?u#u1893cM?56x?4?L,N?8)?HG.xU?3?mD4*Ly,?6,ez6Ua?.?*?;x?K,?56,x”()Xe.O*?,e?56?,?1?.uy,56 0f(x)=0knx1,x2,x3,um(1,1)S.K33|d|,|c|,|b|,3a,?()A.3|d|B.|c|C.|b|D.3a1fjddy2?!JK(20),?(u19.?x y y2B.2x1 2y1C.ln|x|ln|y|D.cos(x2)2,P8Ba=(x,y)|x2+ya 1,y 0,f(a)8Ba?:?.e?(?k()A.e(x,y)Sa,K0 yx+2 2(x,y)Sa?yx+2=1ln3.C.f(a)uaN4O.D.?a 2,vf(a)1f(x)=lnx,?1 x 0,f(x)=sin?x 6?,|x|2,?K?ocI.K?ocIN,T?ocI?L.ADCB2fjddy3n!)K(70)17.(10)?maxa,b=a,a b,b,a 4,y:(x 1)3 x3 7x 7 x3.(2)ex 2,y:x3 x3 7x 7 (x+1)3.(3)e?x,yvx3+y3=7maxx,y+7,x,y?.18.(10)3ncABC A1B1C1,4ABC?n?/,BAC=90,AB=AA1.?D,E,FOB1A,C1C,BC?:.(1)y:DE kABC.(2)y:B1FAEF.19.(12)56X?:,4X)D.56o?u?u609,u?u70.,?56?100)?o?Xe.36395160616465676768686870707070717172727273737374747475757576767676767777777878787878787979797979798080808080808080808080818182828282838383838484858585858585868686868787878888898990909091919293959599(1)?!(?(J=).(2)?ng,O?2n,2n+1?0,b 0)?:,PV-?:(A,B-),|PA|PB|=2,TV-?l%2.PF(m,0),m 1.AP,BPx=1mO?u:M,N.(1)V-?.(2)y:AM FNP?(=mk?).(3)r1(2)?wum?f(m).y:f(m)74.21.(14)34ABC,?2A=3B,?A,B,C?Oa,b,c.(1)?M=cos(x y),N=cos(x+y).M,N5Lcosxcosysinxsiny.(2)y:(a2 b2)(a2+ac b2)=b2c2.(3)Uk2a=3b?XJk,c?;XJvk,nd.22.(12)?a (0,1)(1,+),3(0,+)?f(x)=xa ax x+a.(1)f(x)L(1,f(1)?l?,yl?u0.(2)y:f(x)?k2:.(3)ebf(x)?:,y:a+b 1+ab.23.(5)NK:y?a 1,3(0,+)?f(x)=xa ax x+ak3:.4fjddy5Y)=1.B.J:nmmmm;E,:nmmmm;O:nmmmm.J:A=11,9,B=1,3,5,7,9,11,13,.A B=1,3,5,7,9,11.2.B.J:nmmmm;E,:nmmmm;O:nmmmm.J:|(1 i)2|=|1 i|2=(2)2=2.3.C.J:nmmmm;E,:nnmmm;O:nmmmm.J:TK?:(p,0),O?:m?l2p=2.Kp=1.4.C.J:nmmmm;E,:nnmmm;O:nnmmm.J:a23=3a2a3,Ka36=0,a3=3a2 a2+a4=2a3=6a2 a4=5a2.da2=1a4=5.5.D.J:nnmmm;E,:nnmmm;O:nmmmm.J:1 0.6 0.5 0.3=0.91.6.A.J:nnnmm;E,:nnmmm;O:nmmmm.J:f0(x)=cosx+1x.,B,f0?32?=23 0,CD,f0(2)=1+12 0.7.B.J:nnnmm;E,:nnnnm;O:nnnmm.J:dBQOBBQ,OQ=pOB2+BQ2=3.Ka=3.?y29+x2b2=1(0 b 3).B?OQ?l1 223?.dq5,B?I 223,83!.“?6481+89b2=1.)?b2=7217.l?c2=a2 b2=8117,l%e1=ca=317.K3e1=17.L17?4.8.D.J:nnnnm;E,:nnnmm;O:nmmmm.J:U?f(x)=a(x x1)(x x2)(x x3),?Xx1x2x3=da,l?da?=|x1x2x3|1.=3|d|3a.duf(x)?4:3(1,1)S,Kf0(x)=3ax2+2bx+c?y1,y23(1,1)S.?c3a?=|y1y2|1=|c|3a.?2b3a?=|y1+y2|y1|+|y2|2=|b|3a,d?3a.9.AC.J:nmmmm;E,:nnmmm;O:nmmmm.10.AD.J:nnmmm;E,:nnmmm;O:nmmmm.J:?4PAC3ABCD?K,?3ABB1A1?K.11.BD.J:nnmmm;E,:nnnmm;O:nnmmm.J:A,e?CL(c/a,0),Kca(a+b+c)=0,l?c=0a+b+c=0.ec=0,dlL:(1,1).B,elL:(1,1),Ka+b+c=0,?CL(c/a,0).C,e?Cxk?:uy,Kca 0,?b(.UlL1o.D,eL1!?!o,Kcb 0.ca 0,7k,x1x2=ca 2Ba?.5yx+2L:(x,y):(2,0)?,=A,JB(1ln3 1).Oa*?/f(a)4O,?C.?DA2 a x (0,2).K?ocI:?.?l2 x2.PO?ocI?.%,P?ocIP EFGH.K?ocI?ph=OE=r?2 x2?2x24=q2 2x (0,2),x=2 h22.ocI?NV=13x2h=h(2 h2)26=h5 4h3+4h6.PV=V(h)(LV uh?),KV0(h)=16(5h4 12h2+4)=16(5h2 2)(h2 2).?h2=25N.P?r,K?n,(h r)2+2x2!2=r2,z?r=h2+x24h=h2+(2 h2)28h=12h+h38=135252.?L4r2=4 169 5252 2=338125.17.:2+2+6;J:nnmmm;E,:nnmmm;O:nnmmm.(1)(2)=,.(3)”?xx,y?.Kx3+y3=7x+7=x3 7x 7=y3.ex 4,Kd(1)y (x 1,x),U?,ex 2,Kd(2)y (x,x+1),U?.dxU1,0,1,2,3,4.dy3O1,7,13,13,1,29.2?x ykx=3v,dy=1.n(x,y)U?(3,1)(1,3).18.:5+5;J:nnmmm;E,:nnmmm;O:nmmmm.(1)PGAB?:,?DG,CG.dD,GOAB1,AB?:?DG k BB1,DG=12BB1.qdnc5CE k BB1CE=12BB1,KDG k CEDG=CE,Ko/DECG1o/,KDE k CG.dCG ABC,DEABC,KDE kABC.6fjddy7(2)”?AF=1.KK8?AB=B1B=2,AB1=2,BF=1,B1F=3,EF=62,B1E=322.?y?AF2+B1F2=AB21,B1F2+EF2=B1E2.AFB1FEFB1F.B1FAEF.?19.:3+4+5;J:nnmmm;E,:nnmmm;O:nnnnm.(1):79.5;:80;:78.36.(2)?LXe:(nATx0?,p?)36,3738,3950,5160,6164,6566,6768,6970,7172,7374,7576,771112223666878,7980,8182,8384,8586,8788,8990,9192,9394,9596,9798,9974.7fjddy821.:3+6+5;J:nnnnm;E,:nnnnn;O:nnnnn.(i)=z?,Lcos(x y)cos(x+y)m=.sinxsiny=12(cos(x y)cos(x+y)cosxcosy=12(cos(x y)+cos(x+y)(ii)()2A=3B,2C=2 5B.d?un,y?du(sin2A sin2B)(sin2A+sinAsinC sin2B)=sin2B sin2C.(1)Iy(1).,k(sin2A sin2B)(sin2A+sinAsinC sin2B)=?1 cos2A21 cos2B2?1 cos2A2+sinAsinC 1 cos2B2?=?=12(cos2B cos2A)?cos2B cos2A2+12(cos(A C)cos(A+C)?=z?=14(cos2B cos2A)(cos2B cos2A+cos(A C)+cosB)=z=14(cos2B cos3B)(cos2B cos3B cos4B+cosB)=14(cos22B+cos23B 2cos2B cos3B cos2B cos4B+cos2B cosB+cos3B cos4B cos3B cosB)=14?cos4B+12+cos6B+12(cosB+cos5B)12(cos2B+cos6B)+12(cosB+cos3B)+12(cosB+cos7B)12(cos2B+cos4B)?=14?1 cos2B cos5B+12(cos3B+cos7B)?.?z?.,sin2B sin2C=14(1 cos2B)(1 cos2C)=14(1 cos2B)(1 cos5B)=14(1 cos2B cos5B+cos2B cos5B)=14?1 cos2B cos5B+12(cos3B+cos7B)?=.=z?d(1)?(.(iii)U.?un,K/2a=3b0?du2sinA=3sinB,=2sin32B 3sinB=0.duC=52B (0,),KB?0,25?.f(x)=2sin32x3sinx,x?0,25?.f0(x)=3cos32x 3cosx=4cos3x2 2cos2x2 3cosx2+1.?t=cosx2?cos5,1?,5?t cos5 cos4=22.Kf0(x),g(t)=4t3 2t2 3t+1,?g0(t)=12t2 4t 3=12(t 16)2103 g0 22!=6 22 3 0,Kg(t)3m?cos5,1?N4O,l?g(t)g(1)=0,=f0(x)0,Kf(x)3m?0,25?4,f(x)1,Kg0(a)=lna 0,?g(a)uaN4,g(a)1+ab,?du(1 a)(1 b)0,=0 a 1a 10 b 1?0 a 1:x 1?f(x).dlna 0ax alna,xa1 1.Kd(1),f0(x)=axa1 axlna 1 a alna 1 0.f(x)3(1,+)N4,?f(x)f(1)0.?,rxb,=f(b)10 b 1:5?ba ab b+a=0,Kawg(x)=bx xb b+x?:,5?g(x)?urf(x)?r?5?ab,u?g?.n,(1 a)(1 b)12?1n21(n+1)2?.(2)y:nPk=11k2kOa
展开阅读全文
提示  金锄头文库所有资源均是用户自行上传分享,仅供网友学习交流,未经上传用户书面授权,请勿作他用。
关于本文
本文标题:自编高考试题(4)
链接地址:https://www.jinchutou.com/shtml/view-337166925.html
关于金锄头网 - 版权申诉 - 免责声明 - 诚邀英才 - 联系我们
手机版 | 川公网安备 51140202000112号 | 经营许可证(蜀ICP备13022795号)
©2008-2016 by Sichuan Goldhoe Inc. All Rights Reserved.