已知ω>0,函数fx=sin(ωx+π1/4)在(π/2,π)上单调递减,则ω的取值范围当x∈(π/2,π)时,wx+π/4∈(πw/2+π/4,πw+π/4)而函数y=sinx的单调递减区间为[π/2,3π/2]那么πw/2+π/4≥π/2,πw+π/4≤3π/2所以1/2≤w≤5/4,即w
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/10 15:04:14
![已知ω>0,函数fx=sin(ωx+π1/4)在(π/2,π)上单调递减,则ω的取值范围当x∈(π/2,π)时,wx+π/4∈(πw/2+π/4,πw+π/4)而函数y=sinx的单调递减区间为[π/2,3π/2]那么πw/2+π/4≥π/2,πw+π/4≤3π/2所以1/2≤w≤5/4,即w](/uploads/image/z/9592989-69-9.jpg?t=%E5%B7%B2%E7%9F%A5%CF%89%3E0%2C%E5%87%BD%E6%95%B0fx%3Dsin%28%CF%89x%2B%CF%801%2F4%29%E5%9C%A8%28%CF%80%2F2%2C%CF%80%29%E4%B8%8A%E5%8D%95%E8%B0%83%E9%80%92%E5%87%8F%2C%E5%88%99%CF%89%E7%9A%84%E5%8F%96%E5%80%BC%E8%8C%83%E5%9B%B4%E5%BD%93x%E2%88%88%28%CF%80%2F2%2C%CF%80%29%E6%97%B6%2Cwx%2B%CF%80%2F4%E2%88%88%28%CF%80w%2F2%2B%CF%80%2F4%2C%CF%80w%2B%CF%80%2F4%29%E8%80%8C%E5%87%BD%E6%95%B0y%3Dsinx%E7%9A%84%E5%8D%95%E8%B0%83%E9%80%92%E5%87%8F%E5%8C%BA%E9%97%B4%E4%B8%BA%5B%CF%80%2F2%2C3%CF%80%2F2%5D%E9%82%A3%E4%B9%88%CF%80w%2F2%2B%CF%80%2F4%E2%89%A5%CF%80%2F2%2C%CF%80w%2B%CF%80%2F4%E2%89%A43%CF%80%2F2%E6%89%80%E4%BB%A51%2F2%E2%89%A4w%E2%89%A45%2F4%2C%E5%8D%B3w)
已知ω>0,函数fx=sin(ωx+π1/4)在(π/2,π)上单调递减,则ω的取值范围当x∈(π/2,π)时,wx+π/4∈(πw/2+π/4,πw+π/4)而函数y=sinx的单调递减区间为[π/2,3π/2]那么πw/2+π/4≥π/2,πw+π/4≤3π/2所以1/2≤w≤5/4,即w
已知ω>0,函数fx=sin(ωx+π1/4)在(π/2,π)上单调递减,则ω的取值范围
当x∈(π/2,π)时,wx+π/4∈(πw/2+π/4,πw+π/4)
而函数y=sinx的单调递减区间为[π/2,3π/2]
那么πw/2+π/4≥π/2,πw+π/4≤3π/2
所以1/2≤w≤5/4,即w的取值范围是[1/2,5/4].
为什么sin(ωx+π1/4)要符合单调递减区间[π/2,3π/2]?
已知ω>0,函数fx=sin(ωx+π1/4)在(π/2,π)上单调递减,则ω的取值范围当x∈(π/2,π)时,wx+π/4∈(πw/2+π/4,πw+π/4)而函数y=sinx的单调递减区间为[π/2,3π/2]那么πw/2+π/4≥π/2,πw+π/4≤3π/2所以1/2≤w≤5/4,即w
这个相当于函数代换和单调性的综合应用,首先你可以令y=wx+pi/4,由于w>0,所以y的单调性和x的单调性一致,对于函数f(x)=sin(y)来说,此时即为sinx的单调性,而siny在(pi/2+2kpi,3pi/2+2kpi)是单调递减的,此时就可以通过y的单调性来确定x的范围
解析:∵函数fx=sin(ωx+π/4)在(π/2, π)上单调递减
又函数fx初相为π/4,即第一象限角,∴在Y轴二侧,最大值点离Y轴之距小于最小值点离Y轴之距
∴最大值点:ωx+π/4=2kπ+π/2==>x=2kπ/ω+π/(4ω)
π/(4ω)<=π/2==>ω>=1/2
∴最小值点:ωx+π/4=2kπ+3π/2==>x=2kπ/ω+5π/(4ω)
全部展开
解析:∵函数fx=sin(ωx+π/4)在(π/2, π)上单调递减
又函数fx初相为π/4,即第一象限角,∴在Y轴二侧,最大值点离Y轴之距小于最小值点离Y轴之距
∴最大值点:ωx+π/4=2kπ+π/2==>x=2kπ/ω+π/(4ω)
π/(4ω)<=π/2==>ω>=1/2
∴最小值点:ωx+π/4=2kπ+3π/2==>x=2kπ/ω+5π/(4ω)
5π/(4ω)>=π==>ω<=5/4
∴ω的取值范围为1/2<=ω<=5/4
至于为什么sin(ωx+π1/4)要符合单调递减区间[π/2,3π/2]
这是因为函数y=sinx在Y轴右侧第一个周期内的单调递减区间为[π/2,3π/2]
做此类题,首先要非常熟习函数y=sinx的图像,单调区间,将给定函数相位与y=sinx的相位作比较,从而得出所需结论。
收起