w是正实数,如果f(x)=2sinwx在〔-π/3,π/4〕上是增函数,那么W的取值范围
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w是正实数,如果f(x)=2sinwx在〔-π/3,π/4〕上是增函数,那么W的取值范围
w是正实数,如果f(x)=2sinwx在〔-π/3,π/4〕上是增函数,那么W的取值范围
w是正实数,如果f(x)=2sinwx在〔-π/3,π/4〕上是增函数,那么W的取值范围
由f(x)=2sinwx的单调性可知2kπ-π/2≤wx≤2kπ+π/2,所以
(2kπ-π/2)/w≤x≤(2kπ+π/2)/w上是增函数
又已知f(x)=2sinwx在〔-π/3,π/4〕上是增函数,所以〔-π/3,π/4〕落在k=0时的区间上
即-π/(2w)≤x≤π/(2w)
-π/(2w)≤-π/3且π/4≤π/(2w)
2w≤3且2w≤4
所以0
(1/w) *π/2 > π/3,
(1/w) *π/2 > π/4.
0
因为sinx的单调增区间是〔2kπ-π/2,2kπ+π/2〕所以2kπ-π/2<=wx<=2kπ+π/2,解得2kπ/w-π/(2w)<=x<=2kπ/w+π/(2w)
因为f(x)在〔-π/3,π/4〕上是增函数,所以这个区间包含于[2kπ/w-π/(2w),2kπ/w+π/(2w)].这样可以列出不等式组:
(1):2kπ/w-π/(2w)<=-π/3
(2):2kπ/...
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因为sinx的单调增区间是〔2kπ-π/2,2kπ+π/2〕所以2kπ-π/2<=wx<=2kπ+π/2,解得2kπ/w-π/(2w)<=x<=2kπ/w+π/(2w)
因为f(x)在〔-π/3,π/4〕上是增函数,所以这个区间包含于[2kπ/w-π/(2w),2kπ/w+π/(2w)].这样可以列出不等式组:
(1):2kπ/w-π/(2w)<=-π/3
(2):2kπ/w+π/(2w)>=π/4
解这个不等式组就可以了,我试着解了一下,不知对不对,希望对你有帮助吧
8k+2<=w<=3/2-6k,k=-1,-2,-3...
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