设函数f(x)=ax+cosx,x∈[0,π]. (Ⅰ)讨论f(x)的单调性; (Ⅱ)设f(x)≤1+sinx,求a的取值范围.(1)解析:∵函数f(x)=ax+cosx,x∈[0,π].F’(x)=a-sinx当a=0时,f(x)=cosx∴f(x)在[0,π]上单调减; 当a>0时,令f’(x)=a-sinx=0A∈(0,1)x1=arcsina,x2=π-arcsinaf’’(x)=-cosx==>f”(x1)0∴f(x)在x1处取极大值,在x2处取极小值∴x∈[0,x1)或x∈[x2, π]时,单调增;x∈[x1,x2)时,单调减;A∈[1,+∞), f’(x)>=0,f(x)...
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