新课标高中数学人教A版集合、函数、导数综合测试题

A5分的）yx23xPxP3939(－；2439－－；2439(－,；(,2424UMN2,3,4,5},MCN{2,4},N则（）UA．．Ｃ．Ｄ．1a2b()c2log21.20.8c）25A．cba．cabbacbcalogx,x01f(x)f(f())3（）2,x0x9114441f(x)x)）1xA．(,．()．(,)yf(x)+）f(x)y=f(x)1的f(x)=2+x2(0,1)）x3f(x)f(x4)f(x),x(0,2)f(x)2x2f(7)在R当=（）298f(x)xx，xxf(x)f(x)A．，当<>）1212121f(x)f(x)(x1)f(x)exf(x)ln(x1)==C.=D2x1f(x)ex(sinxcosx)[0,]2211111[,e](,e)e]e)A．2；．2；．；．.22222245xx1y1,3设UA{xU|xmxCAm2URxfxx()bxc在[1,)a,b,c_3ax2.f(x)a(a0,a1)mg(x)(14m)x在)xa.(x)mxf'(x)8x3,则mfn.mn6{xN|0xA{xN|1x，集合．（本小题满分分）已知全集U，集合B{xN|2x6}.()CAB；(CA)(CB).UUU(x)axbxcxx5f320,b,c求xa02(x)lg1x1xf.(x)f(x).f(x)，f(xy)f(x)f(y)0x19fxyf(x)0f(2.(x)f(x)[2在f在R320．(12分)(2010·)设函数fx)sin－cosxx10<<2π，求函数f()的单调区间与极值．1221．＝x．2求1223当x<x23．423456789BADDBABCBBAA145.)．(1,)．.4U6}，A｝，CAU又B(CA)A.UCA，CB(CA)(CB)UUUU1x1xx1x1001x1.，(x)的定义域为{x|1xf(x)f.1xf(x)f(x)lg1xlg1x1x1x1x1xf(x)lglg()lg10.1x1x1x(x)f(x).f(x).f,x(xxx是12121x1xxx)f(x)f(x)122121x1xxx)121125xx)xx)xx)2(xx)112121221xx)xx)xx)1212121xx1xx1x01x0，.，，122112xx)xx)101.1212xx)xx)1212xx)f(x)f(x)lg10()()即fx＞fx.12xx)121212(x)(在.fxR0，xxxx且x1212210()0()0.fxfxxx21(x)f[(xx)x]f(xx)f(x)f(x)又f即f22112111(x)f(x).()fx12xf(0)f(x)f(x).yy0(0)0.()()()fxfxfxxf(1)得ff(1)2,f(2)2f(1)4,f(x)[2上的值域为[4，2].10ax0.f(x)A{x|xa得.xx(x)fa12x(axxf(x)x,a(aa(axxx(a1)1)x(aa(a1)a1)ax(1a)x(a1)xxxxxxxf(x)f(x).a(aa(1a)a(a1)xxx(x)f.x(axx0aa10.()fxxa(axx(axx0aa10.()fx当xa(ax6x(axA()fxxa(ax(x)mA，fmm0t0.1ykt(0.1，从而10.1ky10t.11y()(0.11()0.1a，a0.1.又taa16161y().tt16t(0ty1.(),(tt111()0.25()得2，ttyt2t0.21644.(xy)f(y)x(x2yf(1)0，.f，y1f(0)f(12f(0)2.得令x0f(x)f(0)x(xf(x)xx2.(x)f(x)xx2.y得f22120xf(x)32xa得x2x2321.xax2xa11334(x)xx1xh(0)，h().h(x)1.记h2224Aa|a.a1(x)xx2xa)x2x②g，222a1a1，或x2a5.xa22Ba|a，或aa|3aCBRa|1a故ACBR7xx)xx)xx)2(xx)112121221xx)xx)xx)1212121xx1xx1x01x0，.，，122112xx)xx)101.1212xx)xx)1212xx)f(x)f(x)lg10()()即fx＞fx.12xx)121212(x)(在.fxR0，xxxx且x1212210()0()0.fxfxxx21(x)f[(xx)x]f(xx)f(x)f(x)又f即f22112111(x)f(x).()fx12xf(0)f(x)f(x).yy0(0)0.()()()fxfxfxxf(1)得ff(1)2,f(2)2f(1)4,f(x)[2上的值域为[4，2].10ax0.f(x)A{x|xa得.xx(x)fa12x(axxf(x)x,a(aa(axxx(a1)1)x(aa(a1)a1)ax(1a)x(a1)xxxxxxxf(x)f(x).a(aa(1a)a(a1)xxx(x)f.x(axx0aa10.()fxxa(axx(axx0aa10.()fx当xa(ax6x(axA()fxxa(ax(x)mA，fmm0t0.1ykt(0.1，从而10.1ky10t.11y()(0.11()0.1a，a0.1.又taa16161y().tt16t(0ty1.(),(tt111()0.25()得2，ttyt2t0.21644.(xy)f(y)x(x2yf(1)0，.f，y1f(0)f(12f(0)2.得令x0f(x)f(0)x(xf(x)xx2.(x)f(x)xx2.y得f22120x