已知函数f(x)=sinwx+√3coswx(w>0)

Question

已知函数f(x)=sinwx+√3coswx(w>0)若函数f(x)在区间[π/3,π/2]上为增函数，求满足条件的整数W值

Answer 1

f(x)=sinwx+√3coswx=2[sinwxcos(π/3)+sin(π/3)coswx]=2sin(wx+π/3)
可知：f(x)的增区间为：[(-π/2+2kπ-π/3)w，(π/2+2kπ-π/3)/w]，即[(-5π/6+2kπ)/w，(π/6+2kπ)/w]
f(x)在区间[π/3,π/2]上为增函数，所以[π/3,π/2]在区间[(-5π/6+2kπ)/w，(π/6+2kπ)/w]内，
当k=1时，增区间为[7π/(6w)，13π/(6w)]，所以7π/(6w)<π/3，π/2<13π/(6w)，只有w=4符合。
所以整数w值为4

Answer 2

f(x)=2sin（wx+π/3）增区间[2kπ-π/2,2kπ+π/2],k∈Z2kπ-π/2≤wx+π/3≤2kπ+π/2令k=1得7/2≤w≤13/3