率为[0.5(×1–0.4)+(1–0.5)×0.4]×0.5×0.4=0.1.119.解:(1)由题设得4(an1bn1)2(anbn),即a1b1(ab).nnnn2又因为a1+b1=l,所以anbn是首项为1,公比为12的等比数列.由题设得4(an1bn1)4(anbn)8,即an1bn1anbn2.又因为a1–b1=l,所以anbn是首项为1,公差为2的等差数列.(2)由(1)知,1ab,anbn2n1.nnn12所以111a[(ab)(ab)]n,nnnnnn222111b[(ab)(ab)]n.nnnnnn22220.解:(1)f(x)的定义域为(0,1),(1,+∞)单调递增.因为f(e)=e110e1,22e1e32f(e)20,22e1e1所以f(x)在(1,+∞)有唯一零点x1,即f(x1)=0.又101,1x11,x1f()lnxf(x)011xx111故f(x)在(0,1)有唯一零点1x1.综上,f(x)有且仅有两个零点.(2)因为1x0elnx0,故点B(–lnx0,1x0x)在曲线y=e上.由题设知f(x0)0,即lnx0x0x011,故直线AB的斜率k11x1110lnx0xxx000lnx1xxxx00000x10.x曲线y=e1B(lnx,)在点0x0处切线的斜率是1x0,曲线ylnx在点A(x,lnx)处切线的斜率也是001x0,所以曲线ylnx在点xA(x,lnx)处的切线也是曲线y=e00的切线.21.解:(1)由题设得yy1x2x22,化简得22xy421(|x|2),所以C为中心在坐标原点,焦点在x轴上的椭圆,不含左右顶点.(2)(i)设直线PQ的斜率为k,则其方程为ykx(k0).ykx由x2y2得142x212k.2记u212k2,则P(u,uk),Q(u,uk),E(u,0).于是直线QG的斜率为k2k,方程为()yxu.2
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