Question

设函数f(x)=√3cos²x+sinxcosx-√3/2（1）求函数f(x)的最小正周期T，并求出函数f(x)的单调递增区间；

Answer 1

解：
(1)
f(x)=√3cos²x+sinxcosx-√3/2
=(√3/2)[cos(2x)+1]+(1/2)sin(2x)-√3/2
=(√3/2)cos(2x)+(1/2)sin(2x)
=cos(π/6)cos(2x)+sin(π/6)sin(2x)
=cos(2x-π/6)
最小正周期T=2π/2=π
2kπ-π≤2x-π/6≤2kπ时，f(x)单调递增，此时kπ-5π/12≤x≤kπ+π/12
2kπ≤2x-π/6≤2kπ+π时，f(x)单调递减，此时kπ+5π/12≤x≤kπ+7π/12
函数的单调递增区间为[kπ-5π/12，kπ+π/12]，单调递减区间为[kπ+5π/12，kπ+7π/12]，(k∈Z)
(2)
x∈[0，3π]
-π/6≤2x-π/6≤35π/6
当2x-π/6=0，2π，4π时，f(x)取得最大值。
对应的x分别为π/12，13π/12，25π/12，
π/12+13π/12+25π/12=(1+13+25)π/12=39π/12

Answer 2

f(x)=sinxcosx+√3cos²x-√3/2
=1/2sin2x+√3/2[cos2x+1]-√3/2
=1/2sin2x+√3/2cos2x
=sin(2x+π/3)
（1）函数f(x)的最小正周期T=2π/2=π

单调递增区间
2kπ-π/2≤2x+π/3≤2kπ+π/2
即2kπ-5π/6≤2x≤2kπ+π/6
单调递增区间
为
[kπ-5π/12,kπ+π/12],，(k∈Z)
(2)
∵x∈[0,3π）
∴2x+π/3∈[π/3,6π+π/3),
f(x)取到最大值的所有x的和=π/2+(π/2+2π)+(π/2+4π)=15π/2=7.5π