2016年天津市高考(理科)数学试卷及答案解析

（3）当0?a?33a3a?1??2，由（Ⅰ）和（Ⅱ）知， 时，0?1?334f(0)?f(1?23a3a23a3a)?f(1?)，f(2)?f(1?)?f(1?)， 3333学.科网所以f(x)在区间[0,2]上的取值范围为[f(0),f(2)]，因此

M?max{|f(0)|,|f(2)|}?max{|?1?b|,|1?2a?b|} ?max{|1?a?(a?b)|,|1?a?(a?b)|} ?1?a?|a?b|?1. 41. 4综上所述，当a?0时，g(x)在区间[0,2]上的最大值不小于考点：导数的运算，利用导数研究函数的性质、证明不等式 【结束】

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