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设函数F(x)=23sinxCosx?2sin2x+1(x∈R),则F(x)...

∵f(x)=3sin2x+cos2x=2sin(2x+π6),∴f(x)=2sin(2x+π6),T=π故选B.

(1)因为f(x)=23sinxcosx+1?2sin2x=3sin2x+cos2x=2sin(2x+π6),故 函数f(x)的最小正周期为T=π. 由2kπ?π2≤2x+π6≤2kπ+π2,k∈Z,得f(x)的单调递增区间为[kπ?π3,kπ+π6],k∈Z.(2)根据条件得μ=2sin(4x+5π6),当x∈[0,π8]时,4x+5π6∈[56π,...

(1)∵函数f(x)=23sinxcosx?2sin2x+1=3sin2x+cos2x=2sin(2x+π6),令 2kπ-π2≤2x+π6≤2kπ+π2,k∈z,解得 kπ-π3≤x≤kπ+π6,k∈z.故函数f(x)的单调增区间为[kπ-π3,kπ+π6],k∈z.(2)∵x∈[0,π2],∴2x+π6∈[π6 ,7π6],故当2x+π6=7π6,即x=π2时,...

(1)f(x)=23sinxcosx-2sin2x+1=3sin2x+cos2x=2sin(2x+π6),∴T=2π2=π.∵x∈[0,π2],∴π6≤2x+π6≤7π6,∴当2x+π6=π2时,即x=π3时函数有最大值,最大值为2;当2x+π6=7π6时,即x=π2时,函数有最小值为-1;(2)f(x0)=65,x0∈[π4,π2],则sin(2x...

(1)∵f(x)=sin2x+23sinxcosx+3cos2x,∴f(x)=1+3sin2x+2cos2x=3sin2x+cos2x+2=2sin(2x+π6)+2,故函数的最小正周期为T=2π2=π,由2kπ-π2≤2x+π6≤2kπ+π2,可得kπ-π3≤x≤kπ+π6,故函数单调递增区间为:[kπ-π3,kπ+π6](k∈Z)(2)∵x∈[-π6,π3],...

(I)因为f(x)=2sin2x+23sinxcosx+1=1?cos2x+3sin2x+1=2sin(2x?π6)+2由2kπ?π2≤2x?π6≤2kπ+π2(k∈Z)得kπ?π6≤x≤kπ+π3(k∈Z)所以f(x)的单调增区间是[kπ?π6,kπ+π3](k∈Z);(Ⅱ)因为0≤x≤π2,所以?π6≤2x?π6≤5π6所以?12≤sin(2x?π6)≤1所以f(x)=2sin(2...

(Ⅰ)f(x)=2sinxcosx+23sin2x?3=sin2x?3cos2x=2sin(2x?π3),----(4分)∴周期T=π.----(6分)(Ⅱ)根据f(x)=2sin(2x?π3),可得当2x?π3=2kπ+π2,k∈Z,即x=kπ+5π12时,----(10分)f(x)max=2.----(13分)

(Ⅰ)f(x)=sin2x+3sin2x+12(sin2x-cos2x)=1?cos2x2+3sin2x-12cos2x,=3sin2x-cos2x+12=2sin(2x-π6)+12,∴f(x)的周期为π,由-π2+2kπ≤2x-π6≤π2+2kπ得:-π6+kπ≤x≤π3+kπ,k∈Z.∴f(x)的单调递增区间为[-π6+kπ,π3+kπ]k∈Z.(Ⅱ)由f(x0)=2...

(1)f(x)=3sin2x+cos2x=2sin(2x+π6)∴f(x)的最小正周期为T=2π2=π,令sin(2x+π6)=0,则x=kπ2?π12(k∈Z),∴f(x)的对称中心为(kπ2?π12,0),(k∈Z);(2)∵x∈[?π6,π3]∴?π6≤2x+π6≤5π6∴?12≤sin(2x+π6)≤1∴-1≤f(x)≤2∴当x=?π6时,f(x)的最...

(1)由数f(x)=23sinxcosx+2cos2x-1,得f(x)=3sin2x+cos2x=2sin(2x+π6),所以函数f(x)的最小正周期为π;∵2kπ-π2<2x+π6<2kπ+π2,k∈Z∴x∈(kπ-π3,kπ+π6),k∈Z又x∈[0,π2],f(x)=2sin(2x+π6)在[0,π2]上的单调递增区间为(0,π6);...

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