3.如图,已知直三棱柱ABC-A1B1C1中,AB=AC,D为

BC中点,F为BB1上一点,BF=BC=2,FB1=1.

(1)  求证:AD⊥平面BB1C1C;

(2)  若E为AD上不同于A、D的任一点,求证:EF⊥FC1;

(3)  若A1B1=3,求FC1与平面AA1B1B所成角的大小.

解:如图,(1)∵AB=AC且D为BC中点,∴AD⊥BC

∵ABC-A1B1C1是直三棱柱,∴平面ABC⊥平面BB1C1C

∴AD⊥平面BB1C1C.

(2)连结DF,DC1,由已知可求得DF=,FC­1=DC1=,

DF2+FC12=DC12,∴∠DFC1=90°,即DF⊥CF1,由三垂线

定理知EF⊥FC1.

(3)作C1G⊥A1B1, 垂足为G,则C1G⊥平面AA1B1B,∴∠C1FG即为所求的角.

在Rt△ABD中,可求得AD=2.由C1G·A1B1=AD·BC

得 C1G = ∴sin∠C1FG=

∴∠C1FG=arcsin.

*选做题:设函数f(x)的定义域为R,对于任意实数x、y,总有f(x+y)=f(x)f(y),且当x>0时,0<f(x)<1.

(1)  证明:f(0)=1,且x<0时,f(x)>1;

(2)  证明:f(x)在R上单调递减;

(3)  设M={(x,y)|f(x2)f(y2)>f(1)},N={(x,y)|f(ax-y+2)=1,a∈R},若M∩N=φ,试确定a的取值范围.

证明:(1)在f(x+y)=f(x)f(y)中,令x=1,y=0,得f(1)=f(1)f(0),因为0<f(1)<1,所以f(0)=1.

取y=-x>0,则f(x-x)=f(x)f(-x)=1,即f(x)=,∵0<f(-x)<1,∴f(x)>1.

(2)设x1<x2,则x2-x1>0,于是,0<f(x2-x1)<1,f(x1)>0,

∴f(x2)-f(x1)=f[(x2-x1)+x1]-f(x1)=f(x2-x1)f(x1)-f(x1)=f(x1)[f(x2-x1)-1]<0,

∴f(x)在R上单调递减.

(3)解:由f(x2)f(y2)>f(1),得f(x2+y2)>f(1),即x2+y2<1;

由f(ax-y+2)=1=f(0),得ax-y+2=0

由若M∩N=φ,得,解得-≤a≤.

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