2.4
1~12 BDABDDABC@A@B#D#
13. [5,+(infty)]
14. (frac{x^2}{8})+(frac{y^2}{4} = 1)
15. 0
16. 2 1
解:((1).) (f^{'}(x))=(frac{1}{4}-frac{a}{x^2}-frac{1}{x})
令x=1 (f^{'}(1)={-frac{3}{4}}-a=-2)
解得 (a={frac{5}{4}})
(2).
(x>0)
(f^{'}(x))=(frac{1}{4}-frac{5}{4 imes x^2}-frac{1}{x})=(frac{x^2-5-4x}{4 imes x^2})
令 (x^2-5-4x>0) 解得 (x>5/x<-1)
$ herefore f(x)增区间为(5,+infty)quad f(x)减区间为(0,5) (
最小值为)f(5)=-frac{1}{2}-ln^5$
无最大值
解:((1).) (ecausequad AD=DCquad B为AC中点)
( herefore BDot AC quad-> DC=2quad DB=1 quad BC=sqrt{3})
$ ecauseleft{
egin{aligned}
DBEF为矩形
quad平面BDEF ⊥平面ACD
end{aligned}
ight.
$
-> (EDot平面ACD)
( herefore V_{B-DEC}=frac{sqrt{3}}{2} imes2 imesfrac{1}{3}=frac{sqrt{3}}{3})
(2).
如图以B为原点建立如图所示的直角坐标系
$D=(0,1,0) quad C=(sqrt{3},0,0)quad E=(0,1,2) quad M=(0,0,t) quad A(-sqrt{3},0,0)
BM=frac{1}{2} (
)vec{DM}=(0,-1,t)(
)vec{AC}=(2sqrt{3},0,0)(
)vec{AE}=(sqrt{3},1,2)(
)设面ACE的法向量为vec{n}=(x,y,z)(
)ecauseleft{
egin{aligned}
vec{n}cdotvec{AC}=0
vec{n}cdotvec{AE}=0
end{aligned}
ight.
(
->)vec{n}=(0,-4sqrt{3},2sqrt{3})(
又)ecausevec{DM}=(0,-1,t)(
)vec{DM}=vec{n}(
) herefore t=frac{1}{2}(
) herefore BM=frac{1}{2}$
(1).由题可知(quad a=1quad b=sqrt{3})
( herefore 方程为frac{y^2}{1}-frac{x^2}{3}=1)
(2).
(A(0,1) quad B(0,-1)\
Q(x1,y1) quad P(-x1,y1))
(k_AP=frac{y1-1}{-x1}\
k_BQ=frac{y1+1}{x1}\
y_AP = k_AP*x+1 \
y_AP = k_BQ*x-1\
联立得 x=frac{x1}{y1}quad y=frac{1}{y1}\
又ecause frac{y1^2}{1}-frac{x1^2}{3}=1\
herefore x^2+3y^2=3)
(1)
(ecauseleft{
egin{aligned}
PD⊥底面ABCD\
运算可得DBot BC
end{aligned}
ight.\
)
(ecauseleft{
egin{aligned}
BCot DB\
BCot PD \
PDcap DB=D
end{aligned}
ight.
)
->
BC⊥平面 PBD
(ecause BCsubset 面PBC)
->平面 PBD ⊥平面 PBC ;
(2)
(C作CKot PB 于K)
(ecauseleft{
egin{aligned}
平面 PBD ⊥平面 PBC \
CKsubset 面PBC\
CKot PB\
PB为面PBD和面PBC的交棱
end{aligned}
ight.
)面
->(CKot面PBD)
设PD=t
-> (frac{sqrt{6}}{2}=frac{sqrt{t^2+1}}{sqrt{2}})
->(t=sqrt{2})
(H(0,frac{2}{3},0)
P(0,0,sqrt{2})\
B(1,1,0)
C(0,2,0))
(vec{PC}=(0,2,-sqrt{2}))
(vec{PB}=(1,1,-sqrt{2}))
(vec{PH}=(0,frac{2}{3},-sqrt{2}))
(设面PBH的法向量为vec{n_1}=(x,y,z))
(ecauseleft{
egin{aligned}
vec{n}cdotvec{PB}=0\
vec{n}cdotvec{PH}=0
end{aligned}
ight.
)
->(vec{n_1}=(-frac{sqrt{2}}{3},sqrt{2},frac{2}{3}))
(设面PBC的法向量为vec{n_2}=(x,y,z))
(ecauseleft{
egin{aligned}
vec{n}cdotvec{PB}=0\
vec{n}cdotvec{PC}=0
end{aligned}
ight.
)
->(vec{n_2}=(sqrt{2},sqrt{2},2))
(cos<vec{n_1},vec{n_2}>=frac{vec{n_1}cdot{vec{n_2}}}{|n_1| imes|n_2|}=frac{sqrt{3}}{3})
(二面角H−PB−C的余弦值即为cos<vec{n_1},vec{n_2}>=frac{sqrt{3}}{3})
21.
22.