Question

f(x)=根号3sinxcosx+cosx的平方+m

（1）求最小正跟周期及单调递增区间（2）若x属于【-π/6,π/3】，f(x)的最小值为2，求此时f(x)的最大值并指出取何值时f（x）去最大值

Answer 1

f(x)=√3sinxcosx+cos²x+m
=√3/2sin2x+1/2(1+cos2x)+m
=√3/2*sin2x+1/2*cos2x+m+1/2
=sin(2x+π/6)+m+1/2
(1)f(x)最小正周期T=2π/2=π
由2kπ-π/2≤2x+π/6≤2kπ+π/2,k∈Z
得 kπ-π/3≤x≤kπ+π/6, k∈Z
∴f(x)单调递增区间为
[kπ-π/3,kπ+π/6] k∈Z
(2)
∵x属于【-π/6,π/3】
∴2x+π/6∈[-π/6,5π/6]
∴2x+π/6=-π/6时，f(x)的最小值-1/2+m+1/2=m
∵f(x)的最小值为2 ∴m=2
∴2x+π/6=π/2,x=π/6时,f(x)取得最大值7/2

Answer 2

f(x)=√3/2*sin2x+1/2/*cos2x+1/2+m
=sin(2x+π/3)+1/2+m
1).T=π
单增【-5/12π+kπ,π/6+kπ】
2).m=3/2
x=π/6时取最大值3