Question

一道高一数学题（请进！请详细说明！谢谢！）

已知w是正数。函数f(x)=2sinwx在区间【π/3，π/4】上是增函数，求w的取值范围。

Answer 1

区间反了

增函数
所以2kπ-π/2<=wx<=2kπ+π/2
π/4<=x<=π/3
w>0
wπ/4<=wx<=wπ/3
他要包含于[2kπ-π/2,2kπ+π/2]
所以2kπ-π/2<=wπ/4
wπ/3<=2kπ+π/2

所以w>=8k-4
w<=6k+3/2
所以
8k-4<=w<=6k+3/2，k是正整数

Answer 2

xy=(2y+x)/3>=2√(2y*x)/3
所以3xy>=2√2*√(xy)
3xy-2√2√(xy)>=0
√(xy)[3√(xy)-2√2]>=0
所以√(xy)<=0,√(xy)>=2√2/3
√(xy)<=0不成立
所以√(xy)>=2√2/3
xy>=8/9
最小值8/9