作业帮设函数f(x)=lnx+ln(2-x)+ax(a>0).(1)当a=1时,求f(x)单调区间；(2)若f(x)在(0,1]上的最大值为1/2,求a
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![设函数f(x)=lnx+ln(2-x)+ax(a>0).(1)当a=1时,求f(x)单调区间；(2)若f(x)在(0,1]上的最大值为1/2,求a](/uploads/image/z/2625657-33-7.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dlnx%2Bln%282-x%29%2Bax%28a%3E0%29.%281%29%E5%BD%93a%3D1%E6%97%B6%2C%E6%B1%82f%28x%29%E5%8D%95%E8%B0%83%E5%8C%BA%E9%97%B4%EF%BC%9B%282%29%E8%8B%A5f%28x%29%E5%9C%A8%280%2C1%5D%E4%B8%8A%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC%E4%B8%BA1%2F2%2C%E6%B1%82a)
设函数f(x)=lnx+ln(2-x)+ax(a>0).(1)当a=1时,求f(x)单调区间；(2)若f(x)在(0,1]上的最大值为1/2,求a
设函数f(x)=lnx+ln(2-x)+ax(a>0).(1)当a=1时,求f(x)单调区间；(2)若f(x)在(0,1]上的最大值为1/2,求a
设函数f(x)=lnx+ln(2-x)+ax(a>0).(1)当a=1时,求f(x)单调区间；(2)若f(x)在(0,1]上的最大值为1/2,求a
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