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已知函数F(x)=23sinxCosx%2sin2x+1(x∈R).(1...

(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)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)∵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,...

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

(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);...

(1)∵f(x)=23sinxcosx+2cos2x-1=3sin2x+cos2x=2(32sin2x+12cos2x)=2sin(2x+π6),∴函数f(x)的最小正周期T=2π2=π;(2)由(1)知,当2x+π6=2kπ-π2(k∈Z),即x=kπ-π3(k∈Z)时,f(x)取得最小值-2.∴f(x)min=-2,此时x的取值集合为{x...

(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,...

函数f(x)=2cos2x+23sinxcosx=1+cos2x+3sin2x=2sin(2x+π6)+1+1所以:函数的周期为:T=π(Ⅱ)由于x∈[?π6,π4]所以:2x+π6∈[?π6,2π3]sin(2x+π6)∈[?12,1]所以函数f(x)的值域为:[0,3]

(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)≤...

(Ⅰ)∵f(x)=23sinxcosx-2cos2x=3sin2x-(1+cos2x)=2sin(2x-π6)-1,∴函数f(x)的最小正周期T=π;由2kπ+π2≤2x-π6≤2kπ+3π2得:kπ+π3≤x≤kπ+5π6,k∈Z.∴函数f(x)的单调递减区间为[kπ+π3,kπ+5π6]k∈Z.(Ⅱ)∵x∈[0,π2],∴2x-π6∈[-π6,5π6],∴-...

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