论文部分内容阅读
不等式恒成立问题贯穿在整个高中数学学习中,涉及到函数、方程、不等式、导数、数列等相关知识,是高考中的热点问题,综合性比较强,较难找到解题的切入点和突破口,归纳总结出不等式恒成立的题型和解决方法至关重要,下面笔者将从函数法、分离参数法、数形结合法、最值法四方面来阐述.(一)函数法1.转换主元解决恒成立问题例1设p=(log_2x)~2+(t-2)log_2x+1-t,当t∈[-2,2]时恒有p>0,则x的取值范围是_____分析:令g(t)=(log_2x-1)t+(log_2x)~2-2log_2x
Inequality is the establishment of inequality throughout the high school mathematics learning, involving functions, equations, inequalities, derivatives, series and other related knowledge, is a hot topic in the college entrance examination, relatively comprehensive, difficult to find a solution to the entry point and breakthrough, Summarizes the inequality Heng established the type and solution is essential, the following author from the function method, the separation of parameter method, the number of combination method, the most value method to explain four aspects. (A) Function Method 1. Conversion of the main element To solve the constant problem 1, let p = (log_2x) ~ 2 + (t-2) log_2x + 1-t, when p> 0 for t∈ [-2,2] ____ Analysis: Let g (t) = (log_2x-1) t + (log_2x) ~ 2-2log_2x