已知向量a=(2根号3sinx,cos^2x),b=(cosx,2),函数f(x)=a点乘b1）求函

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1向量a=(2根号3sinx,cos^2x),b=(cosx,2),
f(x)=a●b
=2√3sinxcosx+2cos²x
=√3sin2x+cos2x+1
=2(√3/2sin2x+1/2cos2x)+1
=2sin(2x+π/6)+1
由2kπ+π/2≤2x+π/6≤2kπ+3π/2,k∈Z
得kπ+π/6≤x≤2kπ+2π/3,k∈Z
∴函数f(x)的单调递减区间为
[kπ+π/6,2kπ+2π/3],k∈Z
2 将函数y=f(x)图像向左平移π/12个单位
得到y=2sin[2(x+π/12)+π/6]+1=2sin(2x+π/3)+1
图像,将所得图像上各点的横坐标缩短为
原来的1/2倍,纵坐标不变,得到函数
g(x)=2sin(4x+π/3)图像
∵x∈[0,π/4]
∴4x∈[0,π]
∴4x+π/3∈[π/3,4π/3]
∴4x+π/3=π/2时,g(x)max=3
4x+π/3=4π/3时,g(x)min=1-√3
∴g(x)值域为[1-√3,3]
======以下答案可供参考======
供参考答案1:
f(x)=a.b.
=2√3sinxcosx+2cos^2x.
=√3sin2x+cos2x+1.
=2[(√3/2)sin2x+(1/2)cos2x]+1
=2sin(2x+π/6)+1.
1. f(x)=2sin(2x+π/6)+1.
∵x∈( 2kπ+π/2,2kπ+3π/2),sinx为减函数，
∴函数f(x)的单调递减区间为
：(2K+1)π+π/6,2π+3)π+π/6.).2.f(x)=2sin(2x+π/6)+1 --->左移π12，--->f(x)=2sin2(x+π/12+π/12)+1=2sin(2x+π/3)+1.--->再将图像上的横坐标缩短为原来的1/2,--->f(x)=2sin(2*2x+π/3)+1,
∴g(x)=asin(4x+π/3)+1.
g(x)在[0,π/4]上的值域为：
g(x)=2sin(4*0+π/3)= √3+1 (x=0).
g(x)=2sin(4*π/4+π/3)=2sin(π+π3).
=2sin(π+π/3)=-sinπ/3.
=-√3+1
∴ g(x)在 x∈[0,π/4}区间上的值域为[√3+1,-√3+1].