试题分析:
(1)对函数求导,求导函数大于0和小于0的解集,该函数的导函数为二次函数,且含有参数,可以通过判断该二次函数的图像的开口零点个数等确定导函数大于0和小于0的解集,进而得到单调区间.
(2)通过(1)可以得
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时,函数在区间[1,2]的单调性得到最大值求出8(并判断是否符合
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),a<0时,继续通过讨论f(x)的导函数,通过对导函数(为二次函数)的开口 根的个数 根的大小与是否在区间[1,3]来确定原函数在区间[1,2]上的最值,进而得到a的值.
试题解析:
(1)
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1分
因为
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,所以
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对任意实数
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恒成立,
所以
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在
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是减函数 4分
(2)当
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时,由(1)可知,
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在区间[1,2]是减函数
由
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得
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,(不符合舍去) 6分
当
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时,
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的两根
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7分
①当
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,即
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时,
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在区间[1,2]恒成立,
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在区间[1,2]是增函数,由
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得
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9分
②当
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,即
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时
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在区间[1,2]恒成立
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在区间[1,2]是减函数
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,
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(不符合舍去) 11分
③当
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,即
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时,
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在区间
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是减函数,
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在区间
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是增函数;所以
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无解 13分
综上,
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14分