2019年高考数学大二轮复习高考阅卷评分指导课(一)函数与导数类解答题课件理.ppt

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1、高考阅卷评分指导课(一) 函数与导数类解答题,【规范解答】,第(2)问得分说明: 正确求出f(x)的最大值得2分; 转化为关于a的不等式得1分; 构造函数并正确求出函数的最大值得2分; 正确写出结论得1分,名师告秘 (1)阅卷速度以秒计,规范答题少丢分; (2)不求巧妙用通法,通性通法要强化; (3)干净整洁保得分,简明扼要是关键.,审题方法 审结论善转换 问题解决的最终目标就是求出结论或说明已给结论正确或错误因而解决问题时的思维过程大多都是围绕着结论这个目标进行定向思考的审视结论,就是在结论的启发下,探索已知条件和结论之间的内在联系和转化规律善于从结论中捕捉解题信息,善于对结论进行转化,使之

2、逐步靠近条件,从而发现和确定解题方向.,满分指导 (1)写全得分步骤 解题过程中得分点的步骤,有则给分,无则没分,所以对于得分点的步骤一定要写全,如第(1)问,求出导函数就得2分,讨论出a0和a0每一种情况的函数单调性都得2分. 第(2)问,求出函数f(x)的最大值就得分;将其转化为关于a的不等式也得分;构造函数g(x)并求导也得分.,(2)写明得分关键 解题过程中的关键点,有则给分,无则没分,所以在答题时一定要写清得分关键点,如第(1)问中,在讨论a0,a0时的单调性时,一定要指明函数f(x)在各个区间的符号,才能指明单调性,否则不给分;第(2)问中必须求出g(x)的最大值为0,否则不给分.,

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