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2013年高考新课标全国卷II理科第21题是一道高考压轴题,题目是这样的:已知函数,(1)设x=0是的极值点,求m,并讨论的单调性;(2)当m≤2时,证明>0.分析:这是一道导数的综合应用题,也是一道高考压轴题.本题考查利用导数研究函数的单调性,利用导数求函数在闭区间上的最值,考查不等式的证明、函数与方程思想,分类讨论的数学思想,旨在考查学生分析问题和解决问题的能力.其解决该题的关键在于熟练函数与导数的基础知识,解题的思路如下:(1)求出原函数的导函数,因为x=0是函数的极值点,由极值点处的导数等于求出m的值,代入函数解析式后再由导函数大于和小于求出原函数的单调区间;
2013 New Curriculum National Curriculum II Volume II subjects is a finale question, the question is this: Known functions, (1) Let x = 0 is the extreme point, seeking m, and discuss the monotonic; (2) When m≤2, prove that> 0. Analysis: This is a comprehensive application of derivative problems, but also a college entrance examination pressure axis. The test using the derivative of the monotonicity of the function, the use of derivative function in the closed interval Value, proof of test inequality, function and equation thinking, classification of mathematical thinking, to examine students’ ability to analyze and solve problems.The key to solve the problem lies in the basic knowledge of the function and derivative, the solution is as follows : (1) Derive the derivative function of the original function, because x = 0 is the extreme point of the function, the derivative at the extreme point is equal to the value of m, into the function analytic formula and then by the derivative function is greater than and less than Monotonous interval of the original function;