Question

设w>0，若函数f(x)=asinwx在[-∏/3，∏/4]上单调递增，则w的取值范围是?.

请写出计算过程和结果好吗?

Answer 1

w>0
sinx在(-π/2,π/2)递增
则这里-wπ/2<=wx<=wπ/2
所以-wπ/2<=-π/3<=wx<=π/4<=wπ/2

-wπ/2<=-π/3
w/2<=1/3
w<=2/3

π/4<=wπ/2
w>=1/2

所以1/2<=w<=2/3

Answer 2

(a+b+c)³-a³-b³-c³
=(a+b+c-a)[(a+b+c)²+a(a+b+c)+a²]-(b+c)(b²-bc+c²)
=(b+c)(a²+b²+c²+2ab+2ac+2bc+a²+ab+ac+a²)-(b+c)(b²-bc+c²)
=(b+c)(3a²+3ab+3ac+3bc)
=3(b+c)[a(a+b)+c(a+b)]
=3(a+b)(a+c)(b+c)