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引例:若不等式x2+ax+1≥0对一切x∈(0,41]成立,则a的最小值为()A.0B.-2C.52D.-3这是高中数学中典型的一个恒成立的问题,主要解法有两种:解法一:分离参数法:将参数a分离出来,得到1a(x)x,然后再求函数1a(x)x的最大值。解法二:直接考虑[x2+ax+1]min≥0,此方法则涉及到对a进行分类讨论,较为繁琐。这道题给人的感觉是,解决恒成立问题的时候,解法一比解法二要好的多,而且从高中三年的练习中,确实大部分恒成立的问题用分离参数法更合适。但是否所有的恒成立问题都是如此呢?
Example: If the inequality x2 + ax + 1 ≥ 0 for all x∈ (0,41] holds, then the minimum value of a is () A.0B.-2C.52D.-3 This is a typical constant in high school mathematics There are two main solutions to the problem: solution one: separation parameter method: separate the parameter a, get 1a (x) x, and then find the maximum value of the function 1a (x) x Solution 2: Consider [x2 + ax + 1] min ≥ 0, this method involves the classification of a discussion, more complicated .This question gives the impression that the solution to the problem of permanent establishment, the solution is better than a solution two, and from In the three years of high school, it is true that most of the constitutive problems established by the method of separation of parameters are more suitable, but whether all the problems of constancy are true?