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已知函数F(x)=2sin2x+23sinxCosx?1.(Ⅰ)求F(x...

(Ⅰ)f(x)=2sin2x+23sinxcosx?1=2×1?cos2x2+3sin2x?1=3sin2x?cos2x=2sin(2x?π6)由?π2+2kπ≤2x?π6≤π2+2kπ,(k∈Z)得?π6+kπ≤x≤π3+kπ,(k∈Z)所以f(x)的单调递增区间是[?π6+kπ,π3+kπ],(k∈Z)(Ⅱ) 由0≤x≤π2得?π6≤2x?π6≤5π6,所以?12≤sin(2x?π6)≤...

(1)因为f(x)=2sin2x+23sinxcosx+1=1?cos2x+23sinxcosx+1…(1分)=3sin2x?cos2x+2=2sin(2x?π6)+2,…(3分)所以f(x)的最小正周期T=2π2=π.…..(4分)(2)因为f(x)=2sin(2x?π6)+2,由2kπ?π2≤2x?π6≤2kπ+π2(k∈Z),…(6分)得kπ?π6≤x≤kπ+π3(...

(1)f(x)=2sin2x+23sinxcosx+1=1-cos2x+3sin2x+1=2sin(2x-π6)+2∴f(x)的最小正周期T=2π2=π;(2)令π2+2kπ≤2x-π6≤3π2+2kπ(k∈Z)解得-π3+kπ≤x≤5π6+kπ(k∈Z),因此,f(x)的单调递减区间为[-π3+kπ,5π6+kπ],(k∈Z)(3)当x∈[0,π2]时,...

(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...

(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...

解(1)∵f(x)=2sin2x+23sinxcosx-1=3sin2x-cos2x=2(sin2xcosπ6-cos2xsinπ6),∴f(x)=2sin(2x-π6).∴函数f(x)的图象可由y=sinx的图象按如下方式变换得到:①将函数的y=sinx图象向右平移π6个单位,得到函数y=sin(x-π6)的图象;②将函数y=...

(Ⅰ)函数f(x)=cos2x?sin2x+23sinxcosx+1=cos2x+3sin2x+1=2sin(2x+π6)+1,…(5分)因此,f(x)的最小正周期为π,最小值为-2+1=-1.…..(7分)(2)由f(α)=2 得2sin(2α+π6)+1=2,即sin(2α+π6)=12.…(9分)而由α∈[π4,π2]得2α+π6∈[23π,76...

(1)f(x)=cos2x?sin2x+23sinxcosx+1=3sin2x+cos2x+1=2sin(2x+π6)+1.(4分)因此f(x)的最小正周期为π,最小值为-1.(6分)(2)由F(a)=2得2sin(2α+π6)+1=2,即2sin(2x+π6)=12,而由a∈[π4,π2],得2a+π6∈[23π,76π].(9分)故2a+π6=56...

(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,...

(Ⅰ)f(x)=cos2x+3sin2x=2sin(2x+π6),(6分)∴f(π12)=2sin(π6+π6)=2sinπ3=3.(8分)(Ⅱ)由(Ⅰ)可知f(x)=2sin(2x+π6),∴函数f(x)的最小正周期T=2π2=π.(11分)函数f(x)的最大值为2.(13分)

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