Question

求解析！谢谢了！！！

Answer 1

18
1)a垂直于b
√5coa2a+√15sin2a=0

tan2a=-√3/3

π/2<a<π
π<2a<2π
2a=11π/6

a=11π/12
2)a*b=√5coa2a+√15sin2a=-8√5/5
coa2a+√3sin2a=-8/5
cos(2a-π/3)=-4/5
5π/12<a<2π/3
π/2<2a-π/3<π
sin(2a-π/3)=3/5
sin2a=sin[(2a-π/3)+π/3]=1/2sin(2a-π/3)+√3/2cos(2a-π/3)
=1/2*3/5-√3/2*4/5
=(3-4√3)/10
19
1)平面ADEF垂直于平面ABCD,且DE垂直于AD,DB在平面ABCD上
所以,DB垂直于DE
因DB垂直于CD
CD与DE交于点D
即DB垂直于平面CDE
2)取DB中点为O,连接OH,OG
OH是中位线,OH//DE
OG是中位线,OH//BF
又四边形BCEF是平行四边形
BF//CE
所以,OH//CE
OH和OG交于点O ,DE和CE交于点E,
即平面HGO//平面CDE
所以,GH//平面CDE
20
1)f(x)=x^2-(m+1)x+m
f(2+cos2a)=(2+cos2a)^2-(m+1)(2+cos2a)+m<=0
整理得:
m>=[(cos2a)^2+3cos2a+2]/(1+cos2a)[(2+cos2a)(1+cos2a)]/(1+cos2a)=2+cos2a
m>=2+cos2a
cos2a最大值=1
即有:m>=3
2)f(x)-4=0
x^2-(m+1)x+m+4=0
tanA+tanB=m+1 ,tanAtanB=m-4
0<A<π/2,0<B<π/2,0<C<π/2
tanC=-tan(A+B)=-(tanA+tanB)/(1-tanAtanB)>0
-(tanA+tanB)/(1-tanAtanB)>0
(1+m)/(m-5)>0
(m+1)(m-5)>0
m>=5
是)f(x)-4=0才能作