Question

已知f(x)=2sin(wx+派/6),y=f(x)的图像与直线y=1的两个相邻交点的距离最小值等于派/3,则f(x)的

Answer 1

f(x)=2sin(wx+π/6)跟y=1的交点,

先求出其横坐标,

2sin(wx+π/6)=1 即 sin(wx+π/6)=1/2=sinπ/6=sin5π/6

wx+π/6=π/6+2kπ 或者 wx+π/6=5π/6+2kπ 化简得

x1=2kπ/w 或 x2=2π/3w+2kπ/w

依题意,当k取同样的值的时候,x1和x2相差最小,为π/3

所以2π/3w=π/3,解得w=2,

所以f(x)=2sin(2x+π/6)

单调增区间为 2x+π/6∈[-π/2+2mπ,π/2+2mπ],即x∈[-π/3+mπ,π/6+mπ]

单调减区间为 2x+π/6∈[π/2+2mπ,3π/2+2mπ] ,即x∈[π/6+mπ,2π/3+mπ]