最新高中數學學業(yè)水平考試經典118題(含答案)

2015年學業(yè)水平考試經典118題UABACB)1,U435AAx2x2Bxx22x0AB2,()0,20,20,20,2C3P{x|2x30},a1（））））aP）{a}PaPaPB1f(x)）)4(,-1)(1,+)(1,1)（1+）(,+)C5y(x)2與yxy(x)與yx33x2yx2與y(x)yxy3與32xB0(x1(x6則2x3(xD(0,)()7ylogxy3xy131yx2xD2x8設R3答案D11(x)bb(0)2200fRf0f(1)f11211)3xD0()221當xfxxy164x9[0,)[0,4][0,4)(0,4)Clog210、2112222D,yax,)yxyxaayyyy111O1O1OO1xxxxABCDD(x)logx(a1)[,2]3aa倍,那么a的12、fa232313ex20)x11012123xxe345x2((2)C(x)2xlnx6xm)14f(3,4)C40515,74694sin20cos016)D17、sin24011233222C18、631132222B19sin43cos13cos43sin13)12323322Asin75cos30cos75sin3020、A.1)122322C21ysin2xcos2x224Asin(2x)22y524B(x)x)23f()47(,)(,)(,)(,)44442244Ay3sin2x4cosx)24Btan2x25、y|xk,xR,kZx|x+2k,xR,kZx22k|x,xR,kZx|xk,xR,kZx424C26fx是Cx)yx)27y()364422BABC28設B)ABCab2,45BA()29A.150906030DABCsinA:sinB:sinC5:11:13ABC30sinA:sinB:sinC5:11:135222c0CacosAbcosB2531()A(ac)(ac)b2bcAa、b、cABC32A.150B.1206030B33a20(12)b=(3,4)(ab)(a+b)54,)(10,30)(8,24)Ba2,bma//b|23|ab34()70453525B(|1|2ba,aab,b35、723272C36a(1,2)b(3,2)baba與k,119abk11222D38x31x39(1,1)B(2,y)a=(1,2)//ABa()yC=2)b=x)且a6241)122242aab(與1222{}a2a143a743n1244aa則n2144a}na3aa144,,45Sn則Snn12452n8aa0an255）8）11D8aaq0q，q325220abababa2248a,bab30(,)ab1xaa49fx214b2則y0,b0a,ab51(1x)(2x)2602zx3y則53x)x2246xy20zx02x3y1得z55942361899223Dlll56、,,123,//llll,//lllll1l3l1l122132233////,,llllll2,,,,llllll33123123123,與llllll122333b、57設a、//ba,//abaab,,b//aba,////a////a,b,D58()B59設l,m,n,,//,,l則lll//n//則mnm,mn,l則m,n//,l且()A3460lA(2),Bm)m)A3y20()61x6336Da(a3)x(2a1)y7062、.原直線方程可變形為令.633x4y1204x3y1203x4y1204x3y120464）3）4x3y0）4x3y0D）4x3y20）4x3y5065Al:m2xm2y20l:3xmy10ll,66則m11212()6或166或1B673a3232B68l:2x(my4與直線l:mx3y20m122C(2,(1)2569若Px2y22xy30xy30xy102xy50C70圓xy4x6y022((x2y271、xyC有相同的圓心,且經過點(-2,2)的圓的方程是22((x1)(y2)5(x1)(y2)252222(x1)(y2)5(x1)(y2)252222byab0y2272axx2:)222212745)112;975、m、n,ab1))751212A76a-b()12799977、nPPx22237989,ylog78、,x,則滿足2x（）11ABCD7980某校高三一班有學生人人人人C8118,,,6D82、D83xy235????ybxb為6為(yy,x4aa429.43.59.1y69.49.165.5.y84xy0134y與xy0.95xaa0(x,y)85x,B4y0.5x|xC2xkx3k86fx2在8、若f(x)xa)x2在(,4]上是減函數,則實數的值的集合是a2a3(x)f(x)22=xx,88設fx0R則f(1)3892;2(0)xxf(x)(x3)(x0)fyax32(a且a90lg2alg3blog1291,則a,b2b2aba2)a()()yfxfx.92x(4,3)2sincos93P253,2tanxxx94,543)29542;1sin296342997ABCa:b:c3:2:4cosC1425322ffx98.q3xfx0,1fx在f(x)xf(x)f(x),222q3x2即q3x53q3xq3xq0f2,4p265f2p2.32x22fx.3x2x2212fxx.3x3x0xx1設1212121xxf(x)f(x)xx…xx.123x2x3xx1212211210xx10xx1xx01xx0,,,1f2121212(x)f(x)0f(x)f(x).1212f(x)在0,1-1xf9921x1)1x1x1xf)()f()11)1x1x1x2222又12xx(1x)(1x)f(x)f(x)logloglog1212xx(1x)(1x)212222xx1121221(1x)(1x)(1x)(1x)1log02121(1x)(1x)(1x)(1x)212121xf(x)21xf(x)x2x10021x,x,x(0,1x1212fxfx2xxx22x211112212211xxxxx0=x2221xx121xx121220x2xfx21在mx,101fxxmfx[2,)fx在m4；102k與與b(k3,2k，3abk3102k2(4)0babk19ka1b//abk3(4)2k2100kka31bab與a3bba103a與aab,ab),).ab與a,))101.1ab與aABCDABCD,O是AC與2104BD為BBE11111BDAEC;1DAC1BD;1DOCD1DC11A1B1EDCOABBBDOE1∵E為BB為BDO1∴OE∥BD1BDAEC1BDAEC1ABCDABCD1111BBABCD,1ACABCDAC∴BB1.BDACBDB且BB1∴BD1∴ACACBD1ADBD11ADA∵AC1∴BDDAC111123VS21(Ⅲ)V33DDD111SABCDABCDSA底面ABCD,為SAM105N為CDSBDSAC;SBC‖.SMADNCB底面ABCD∵SA與∵BDSBDSBDSAC1為且2ABCD為CD11且22又SBC‖a}a4n3nS106adn1nnd4,s2nna21naa219,107aaad；，an135418S8解an(nda=an，1(a2d)(a4d)a得111ad9.1ad1annn(nSnd，2187283245680.得S82a}S24,aa16S108an。n425nn434ad24a和d211(ad)(a4d)16112ad121d2a22nn1nS22a5d16nn1a2,a54anS.109a中n14nnaq3q34a113na23,snn12na1，a8S.a1410n1aqqq82a，aq331411(12)S2110231012a162q，前nS3242a，求首項an．5n1n3162,①②a511(13)an242,113a162，1a2．123n將a2242，1313243即n．a2，an1．=r222r222222(B(xy0A、yDxEyF0x2210303DEFD37D6EF03x2y2xy33160EF16DE()()022542A25441∴P()，919252B為CC1C2BCC．C3123,,,,CCCCCC12312312361∴P(B)PC)PC)PC)61231