.24
15353,VMBCDMDS△BDC.32215在△PBC中,CDPB,22
9
f又MDDC,S△MDC
VBMDC
125MDDC3,28
11255312,hS△MDCh3h3382512即点B到平面MDC的距离为.(12分)5
20.(本小题满分12分)
1解:(Ⅰfxl
x1,令fx0,得x.)e1当x0fx0fx单调递减;e
1当xfx0fx单调递增.e
1因为t0t22,e111(1)当0t时,fxmi
f;eee1(2)当t≥时,fxmi
fttl
te
所以fxmi
11e0tetl
tt≥1e
(6分)
(Ⅱ)由2xl
x≥x2mx3得m≤2l
xx设hx2l
xx
3x
令hx0,x1或x3(舍)得,
x31x3则x0,hxxx2
当x01时,hx0,hx单调递减;当x1时,hx0,hx单调递增,所以hxmi
h14所以m≤hxmi
421.(本小题满分12分)解:)e(Ⅰ
2,2c2,a2,c12
(12分)
则ba2c21
椭圆的方程为
x2y12
x2y1,联立2消去y得:x24x0,Ax1,y1Bx2,y2,3设yx1,
10
f1442则AB01AB333
分)(6
(Ⅱ)设Ax1,y1,Bx2,y2
OAOB,OA0,x1x2y1y20,OB即
x2y21,由a2b2消去y得a2b2x22a2xa21b20,yx1,
由2a224a2a2b21b20,整理得a2b21,
又x1x2a21b22a2,,x1x22a2b2ab2
y1y2x11x21x1x2x1x21,
由x1x2y1y20得2x1x2x1x210,
2a21b22a2210,a2b2ab2
整理得:a2b22a2b20,b2a2c2a2a2e2,代入上式得
2a21111,a21,221e21e
1211≤e≤,≤e2≤,22421341≤1e2≤,≤≤2,2431e2
71r