压轴解答题(二)1.设椭圆x2a2+y2b2=1(ab0)的右顶点为A,上顶点为B.已知椭圆的离心率为53,|AB|=13.(1)求椭圆的方程;(2)设直线l:y=kx(kx10,点Q的坐标为(-x1,-y1).由BPM的面积是BPQ面积的2倍,可得|PM|=2|PQ|,从而x2-x1=2x1-(-x1),即x2=5x1.易知直线AB的方程为2x+3y=6,由方程组2x+3y=6,y=kx,消去y,可得x2=63k+2.由方程组x29+y24=1,y=kx,消去y,可得x1=69k2+4.由x2=5x1,可得9k2+4=5(3k+2),两边平方,整理得18k2+25k+8=0,解得k=-89,或k=-12.当k=-89时,x2=-90时,g(x)0,所以g(x)在(0,+)上单调递增,且g(1)=0.所以当x(0,1)时,g(x)0,所以当x(0,1)时,f (x)0.故函数f(x)在(0,1)上单调递减,在(1,+)上单调递增.(2)由f(x)=ex-aln x-e,得f (x)=ex-ax.当a0时,f (x)=ex-ax0,所以f(x)在1,+)上单调递增,f(x)min=f(1)=0.(符合题意)当a0时,f (x)=ex-ax,当x1,+)时,exe.(i)当a(0,e时,因为x1,+),所以axe,f (x)=ex-ax0,所以f(x)在1,+)上单调递增,f(x)min=f(1)=0.(符合题意)(ii)当a(e,+)时,存在x01,+),满足f (x0)=ex0-ax0=0,所以f(x)在1,x0)上单调递减,在(x0,+)上单调递增,故f(x0)0,x1+x2=-8kb4k2+1,x1x2=4b2-44k2+1.x1x24+y1y2=0,x1x24+(kx1+b)(kx2+b)=0,得2b2-4k2=1,满足0.SPOQ=12|b|1+k2|PQ|=12|b|(x1+x2)2-4x1x2=2|b|4k2+1-b24k2+1=1.POQ的面积S为定值.4.解析(1)f (x)=ax-1ax2(x0),当a0恒成立,函数f(x)在(0,+)上单调递增;当a0时,由f (x)=ax-1ax20,得x1a,由f (x)=ax-1ax20,得0x1a,函数f(x)在1a,+上单调递增,在0,1a上单调递减.综上所述,当a0时,函数f(x)在1a,+上单调递增,在0,1a上单调递减.(2)当x1e,e时,函数g(x)=(ln x-1)ex+x-m的零点可转化为当x1e,e时,方程(ln x-1)ex+x=m的根.令h(x)=(ln x-1)ex+x,则h(x)=1x+lnx-1ex+1.由(1)知当a=1时,f(x)=ln x+1x-1在1e,1上单调递减,在(1,e)上单调递增,当x1e,e时,f(x)f(1)=0.1x+ln x-10在x1e,e上恒成立.h(x)=1x+lnx-1ex+10+10,h(x)=(ln x-1)ex+x在x1e,e上单调递增.h(x)min=h1e=-2e1e+1e,h(x)max=e.当me时,函数g(x)在1e,e上没有零点;当-2e1e+1eme时,函数g(x)在1e,e上有一个零点.