摘要:11.设ω>0.函数f(x)=2sinωx在上为增函数.那么ω的取值范围是 答案:0<ω≤ 点评:
网址:http://m.1010jiajiao.com/timu3_id_515141[举报]
已知a>0,函数f(x)=-2asin(2x+
)+2a+b,当x∈[0,
]时,-5≤f(x)≤1.
(1)求常数a,b的值;
(2)设g(x)=f(x+
)且lg[g(x)]>0,求g(x)的单调区间.
网址:http://m.1010jiajiao.com/timu3_id_515141[举报]
已知a>0,函数f(x)=-2asin(2x+
)+2a+b,当x∈[0,
]时,-5≤f(x)≤1.
(1)求常数a,b的值;
(2)设g(x)=f(x+
)且lg[g(x)]>0,求g(x)的单调区间.