Question

设向量A=(1,COS2θ),B=(2,1),C=(4SINθ,1),D=(1/2SINθ,1)其中θ∈(0,Л/4)。

（1）求a*d-c*d的取值范围
2）若函数f（X）=|x-1|,比较f(a*d)与f(c*d)的大小

Answer 1

1．∵向量a=(1,COS2θ),b=(2,1),c=(4SINθ,1),d=(1/2SINθ,1),θ∈(0,Л/4)
向量a•向量d-向量c•向量d =1/2sinθ+cos2θ-2(sinθ)^2-1
=1/2sinθ-4(sinθ)^2
设f(θ)=1/2sinθ-4(sinθ)^2,
令f’(θ)=1/2cosθ-8sinθcosθ＝0
Cosθ=0==>θ1=2kπ-π/2, θ2=2kπ+π/2
f(-π/2)=-1/2-4=-5.5, f(π/2)=1/2-4=-3.5
Sinθ=1/16==>θ3=2kπ+arcsin(1/16), θ4=(2k+1)π-arcsin(1/16)
f(arcsin(1/16))=1/32-1/64=1/64, f(π-arcsin(1/16))=1/64
∴a*d-c*d的取值范围为[-5.5,1/64]
2．向量a•向量d =1/2sinθ+cos2θ=1/2sinθ+1-2(sinθ)^2
向量c•向量d =2(sinθ)^2+1
设f(x)=|x-1|
F(向量a•向量d) =|1/2sinθ-2(sinθ)^2|
F(向量c•向量d) =|2(sinθ)^2|
设g(x)= 1/2sinx-2(sinx)^2, g’(x)= 1/2cosx-4sinxcosx=0
Cosx=0==>x1=2kπ-π/2, x2=2kπ+π/2
g(-π/2)=-1/2-2=-2.5, f(π/2)=1/2-2=-1.5
Sinx=1/8==>x3=2kπ+arcsin(1/8), x4=(2k+1)π-arcsin(1/8)
g(arcsin(1/8))=1/16-1/32=1/32, f(π-arcsin(1/16))=1/32
∴g(x)的值域为[-2.5,1/32]
令g(x)= 1/2sinx-2(sinx)^2＝0
Sinx=0==>x1=2kπ, x2=(2k+1)π
Sinx=1/4==> x3=2kπ+arcsin(1/4), x4=(2k+1)π-arcsin(1/4)
∴当x∈(2kπ, 2kπ+arcsin(1/4))∪((2k+1)π-arcsin(1/4), (2k+1)π)时,g(x)>0；
1/2sinx＞2(sinx)^2
F(向量a•向量d)- F(向量c•向量d)= 1/2sinθ-4(sinθ)^2<0
即F(向量a•向量d)＜ F(向量c•向量d)
当x∈(2kπ+arcsin(1/4),(2k+1)π-arcsin(1/4))∪((2k+1)π,2(k+1)π)时,g(x)<0；
1/2sinx＜2(sinx)^2

F(向量a•向量d)- F(向量c•向量d)= 2(sinx)^2-1/2sinx-2(sinx)^2＝-1/2sinx
∴x∈(2kπ+arcsin(1/4),(2k+1)π-arcsin(1/4))时, -1/2sinx<0
即F(向量a•向量d)＜ F(向量c•向量d)
x∈((2k+1)π,2(k+1)π)时, -1/2sinx＞0
即F(向量a•向量d)＞ F(向量c•向量d)
当x1=2kπ, x2=(2k+1)π时，F(向量a•向量d)＝ F(向量c•向量d)
综上：θ∈[2kπ, (2k+1)π]时，F(向量a•向量d)＜＝ F(向量c•向量d)
θ∈((2k+1)π, 2(k+1)π)时，F(向量a•向量d)＞ F(向量c•向量d)

Answer 2

1．∵向量a=(1,COS2θ),b=(2,1),c=(4SINθ,1),d=(1/2SINθ,1),θ∈(0,Л/4)
向量a•向量d-向量c•向量d =1/2sinθ+cos2θ-2(sinθ)^2-1
=1/2sinθ-4(sinθ)^2
设f(θ)=1/2sinθ-4(sinθ)^2,
令f’(θ)=1/2cosθ-8sinθcosθ＝0
Cosθ=0==>θ1=2kπ-π/2, θ2=2kπ+π/2
f(-π/2)=-1/2-4=-5.5, f(π/2)=1/2-4=-3.5
Sinθ=1/16==>θ3=2kπ+arcsin(1/16), θ4=(2k+1)π-arcsin(1/16)
f(arcsin(1/16))=1/32-1/64=1/64, f(π-arcsin(1/16))=1/64
∴a*d-c*d的取值范围为[-5.5,1/64]
2．向量a•向量d =1/2sinθ+cos2θ=1/2sinθ+1-2(sinθ)^2
向量c•向量d =2(sinθ)^2+1
设f(x)=|x-1|
F(向量a•向量d) =|1/2sinθ-2(sinθ)^2|
F(向量c•向量d) =|2(sinθ)^2|
设g(x)= 1/2sinx-2(sinx)^2, g’(x)= 1/2cosx-4sinxcosx=0
Cosx=0==>x1=2kπ-π/2, x2=2kπ+π/2
g(-π/2)=-1/2-2=-2.5, f(π/2)=1/2-2=-1.5
Sinx=1/8==>x3=2kπ+arcsin(1/8), x4=(2k+1)π-arcsin(1/8)
g(arcsin(1/8))=1/16-1/32=1/32, f(π-arcsin(1/16))=1/32
∴g(x)的值域为[-2.5,1/32]
令g(x)= 1/2sinx-2(sinx)^2＝0
Sinx=0==>x1=2kπ, x2=(2k+1)π
Sinx=1/4==> x3=2kπ+arcsin(1/4), x4=(2k+1)π-arcsin(1/4)
∴当x∈(2kπ, 2kπ+arcsin(1/4))∪((2k+1)π-arcsin(1/4), (2k+1)π)时,g(x)>0；
1/2sinx＞2(sinx)^2
F(向量a•向量d)- F(向量c•向量d)= 1/2sinθ-4(sinθ)^2<0
即F(向量a•向量d)＜ F(向量c•向量d)
当x∈(2kπ+arcsin(1/4),(2k+1)π-arcsin(1/4))∪((2k+1)π,2(k+1)π)时,g(x)<0；
1/2sinx＜2(sinx)^2

F(向量a•向量d)- F(向量c•向量d)= 2(sinx)^2-1/2sinx-2(sinx)^2＝-1/2sinx
∴x∈(2kπ+arcsin(1/4),(2k+1)π-arcsin(1/4))时, -1/2sinx<0
即F(向量a•向量d)＜ F(向量c•向量d)
x∈((2k+1)π,2(k+1)π)时, -1/2sinx＞0
即F(向量a•向量d)＞ F(向量c•向量d)

Answer 3

问题里怎么没b呢