【高一数学】20题
3个回答
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f(x)=ab=cos^wx+√3sinwxcoswx
=1/2(1+cos2wx)+√3/2sin2wx
=√3/2sin2wx+1/2cos2wx+1/2
=sin(2wx+π/6)+1/2
T=2π/2w=π
W=1
f(x)=sin(2x+π/6)+1/2
2x+π/6在[2kπ-π/2,2kπ+π/2]是单调递增
x在[kπ-π/3,kπ+π/6]是单调递增
2)对称点为(kπ,0)
2wx+π/6=kπ
x=(6k-1)π/12w
(6k-1)π/12w=π/6
2w=6k-1
w=(6k-1)/2
k=1时有最小,w=5/2
=1/2(1+cos2wx)+√3/2sin2wx
=√3/2sin2wx+1/2cos2wx+1/2
=sin(2wx+π/6)+1/2
T=2π/2w=π
W=1
f(x)=sin(2x+π/6)+1/2
2x+π/6在[2kπ-π/2,2kπ+π/2]是单调递增
x在[kπ-π/3,kπ+π/6]是单调递增
2)对称点为(kπ,0)
2wx+π/6=kπ
x=(6k-1)π/12w
(6k-1)π/12w=π/6
2w=6k-1
w=(6k-1)/2
k=1时有最小,w=5/2
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(1) f(x)=a*b
=cos^2(ωx)+√3sinωxcosωx
=1/2(cos2ωx+1)+√3/2sin2ωx
=sin(2ωx+π/6)+1/2
T=π
2π/(2ω)=π
ω=1
f(x)=sin(2x+π/6)+1/2
-π/2+2kπ=<2x+π/6<=π/2+2kπ
-2π/3+2kπ=<2x<=π/3+2kπ
-π/3+kπ=<x<=π/6+kπ
单调递增区间:[-π/3+kπ,π/6+kπ] (k∈Z)
(2) f(x)=sin(2ωx+π/6)+1/2
2ω×π/6+π/6=π/2+kπ
2ω+1=3+6k
2ω=6k+2
ω=3k+1
∵ω>0
∴当k=0时,ω取得最小值:ωmin=1
=cos^2(ωx)+√3sinωxcosωx
=1/2(cos2ωx+1)+√3/2sin2ωx
=sin(2ωx+π/6)+1/2
T=π
2π/(2ω)=π
ω=1
f(x)=sin(2x+π/6)+1/2
-π/2+2kπ=<2x+π/6<=π/2+2kπ
-2π/3+2kπ=<2x<=π/3+2kπ
-π/3+kπ=<x<=π/6+kπ
单调递增区间:[-π/3+kπ,π/6+kπ] (k∈Z)
(2) f(x)=sin(2ωx+π/6)+1/2
2ω×π/6+π/6=π/2+kπ
2ω+1=3+6k
2ω=6k+2
ω=3k+1
∵ω>0
∴当k=0时,ω取得最小值:ωmin=1
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