高校问答东方栀子原本:f(x)=sinax+√3cosax(0<a<1)g(x)=tan(mx+30')(0<m<1)已知f(x),g(x)最小正周期相同,f(1)=g(1)
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1.f(x)=sinax+√3cosax=2sin(ax+π/3)
f(x)最小正周期=2π/a,g(x)最小正周期=π/m.因为f(x),g(x)最小正周期相同，所以2π/a=π/m,即a=2m.f(1)=2sin(a+π/3),g(1)=tan(m+π/6).因为f(1)=g(1),所以2sin2(m+π/6)=tan(m+π/6)→ 4sin(m+π/6)cos(m+π/6)=tan(m+π/6)→ [cos(m+π/6)]^2=1/4.因为0<m<1,所以m=π/6,a=π/3
f(x)=2sin(x+1)π/3,g(x)=tan(x+1)π/6
2.对于f(x),当(x+1)π/3∈(-π/2+2kπ,π/2+2kπ)(k∈z)时f(x)单调递增,解得x∈(-5/2+6k,1/2+6k)(k∈z)时f(x)单调递增,即f(x)单调递增区间为(-5/2+6k,1/2+6k)(k∈z)
1.f(x)=sinax+√3cosax=2sin(ax+π/3)
f(x)最小正周期=2π/a,g(x)最小正周期=π/m.因为f(x),g(x)最小正周期相同，所以2π/a=π/m,即a=2m.f(1)=2sin(a+π/3),g(1)=tan(m+π/6).因为f(1)=g(1),所以2sin2(m+π/6)=tan(m+π/6)→ 4sin(m+π/6)cos(m+π/6)=tan(m+π/6)→ [cos(m+π/6)]^2=1/4.因为0<m<1,所以m=π/6,a=π/3
f(x)=2sin(x+1)π/3,g(x)=tan(x+1)π/6
2.对于f(x),当(x+1)π/3∈(-π/2+2kπ,π/2+2kπ)(k∈z)时f(x)单调递增,解得x∈(-5/2+6k,1/2+6k)(k∈z)时f(x)单调递增,即f(x)单调递增区间为(-5/2+6k,1/2+6k)(k∈z)