Question

高一数学第十五题

Answer 1

f(x)=sin^2x+2√3sinxcosx+3cos^2x

=1+2cos^2x+√3sin2x
=2cos^2x-1+√3sin2x+2
=cos2x+√3sin2x+2
=2sin(2x+π/6)+2
(1)最大值=2+2=4
此时2x+π/6=π/2+2kπ,k∈Z
x=π/6+kπ,k∈Z
(2)
令π/2+2kπ<=2x+π/6<=3π/2+2kπ,k∈Z
∴π/6+kπ<=x<=2π/3+kπ,k∈Z
减区间是[π/6+kπ,2π/3+kπ],k∈Z
(3)
x∈[0,π/3]
2x∈[0,2π/3]
2x+π/6∈[π/6,5π/6]
∴sin(2x+π/6)∈[1/2,1]
2sin(2x+π/6)+2∈[3,4]
f(X)值域是[3,4]
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