- 谢谢1.设函数f(x)=2cos^2x+2√3sinxcosx
- 1.设f(x)=2cos^2x+2√3sinxcosx-1(x属于R）的最大值为M，最小正周期为T（1）求M和T的值（2）若有10个互不相等的正数Xi满足f(Xi)=M, 且Xi<10兀(i=1,2,...10)，求的X1+X2+...+X10值
２.已知三角形ＡＢＣ的面积Ｓ满足√３＜＝Ｓ＜＝３，且向量ＡＢ*ＢＣ＝６，ＡＢ与ＢＣ的夹角为Ｏ（１）求Ｏ的取值范围（２）求函数f(O)=sin^2O+2sinO*cosO+3cos^2O的最小值
- 1.设f(x)=2cos^x+2√3sinxcosx-1(x∈R）的最大值为M，最小正周期为T（1）求M和T的值（2）若有10个互不相等的正数Xi满足f(Xi)=M, 且Xi<10π(i=1,2,...10)，求的X1+X2+...+X10值
f(x)= 2cos^x+2√3sinxcosx-1 = (2cos^x-1)+√3(2sinxcosx)
= cos2x + √3sin2x = 2[sin(π/6)cosx+cos(π/6)sin2x]
= 2sin(2x+π/6) ≤ 2
--->M=2, T=2π/2=π
f(Xi)=2sin(2Xi+π/6)=2--->2Xi+π/6=2kπ+π/2--->Xi=kπ+π/6,k∈Z
0＜Xi=kπ+π/6＜10π--->0≤k≤9,正好10个值
--->X1+X2+...+X10 = (π/6)+(π+π/6)+(2π+π/6)+...+(9π+π/6)
= (1+2+3+...+9)π + 10(π/6) = 140π/3
２.已知△ABC的面积S满足√3≤S≤3，且向量AB*BC=6，AB与BC的夹角为θ（１）求θ的取值范围（２）求函数f(θ)=sin^θ+2sinθcosθ+3cos^θ的最小值
AB*BC=6=|AB||BC|cosθ--->|AB||BC|=6/cosθ
S=(1/2)|AB||BC|sinθ=3sinθ/cosθ=3tanθ∈[√3,3]
--->tanθ∈[√3/3,1]--->θ∈[π/6,π/4]
--->2θ∈[π/3,π/2]--->2θ+π/4∈[7π/12,3π/4]
f(θ)=sin^θ+2sinθcosθ+3cos^θ=(1+2cos^θ)+sin2θ
=1+1+cos2θ+sin2θ
=2+√2sin(2θ+π/4)∈[2+√2sin(3π/4),2+√2sin(7π/12)]
--->minf(θ)=2+√2sin(3π/4)=2+1=3