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已知函数F(x)=23sinxCosx+1%2sin2x,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,...

(I)∵f(x)=23sinxcosx+2sin2x-1=3sin2x-cos2x=2sin(2x-π6),∴函数f(x)的最小正周期为T=π;由-π2+2kπ≤2x-π6≤π2+2kπ,k∈Z,解得:-π6+kπ≤≤π3+kπ,k∈Z,则f(x)的单调递增区间为[-π6+kπ,π3+kπ],k∈Z;(II)函数y=f(x)的图象上各点的纵...

(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...

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

(1)f(x)=23sinx?cosx+2cos2x=3sin2x+cos2x+1=2(32sin2x+12cos2x)+1=2sin(2x+π6)+1,x∈R…(4分)∴f(x)的最小正周期为T=2π2=π.…(6分)(2)∵f(α2)=2sin[2(α2)+π6]+1=2sin(α+π6)+1=13,…(7分)∴sin(α+π6)=?13<0,…(8分)∵α∈[...

(Ⅰ) f(x)=23sinxcosx+2cos2x?1=3sin2x+cos2x=2sin(2x+π6)则g(x)=|2sin(2x+π6)|,∵y=|sinx|的单调递减区间为[kπ+π2,kπ+π],(k∈Z). ∴由kπ+π2≤2x+π6≤kπ+π 得:kπ2+π6≤x≤kπ2+5π12,则g(x)的单调递减区间为[kπ2+π6,kπ2+5π12](k∈Z). (...

(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...

(Ⅰ)因为f(x)=23sinxcosx+2sin2x?1=3sin2x?cos2x=2sin(2x?π6),所以函数f(x)的最小正周期T=2π2=π.…(7分)(Ⅱ)因为x∈[?5π12,π6],所以2x?π6∈[?π,π6],所以sin(2x?π6)∈[?1,12],所以2sin(2x?π6)∈[?2,1],所以函数f(x)的取值范围为[...

(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)=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...

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