论文部分内容阅读
在解答数学问题的过程中,有时巧妙地利用相关的数学思想方法,可以规避分类讨论,从而快速解题.分离参数法例1已知不等式ax2-2x+2>0对任意的x∈[1,4]恒成立,求实数a的取值范围.分析本题若令y=ax2-2x+2,则需要对a=0和a≠0两种情况进行分类讨论,当a≠0时,还要考虑函数所对应方程的实根的分布情况.若将a进行分离,则可以避免对a的分类讨论.解由题意知,当x∈[1,4]时,要使ax2-2x+2>0恒成立,只需a>2(x-1)x2恒成立,即a>[2(x-1)x2]max.令y=2(x-1)x2=-2(1x-12)2+12.由x∈[1,4],得14≤1x≤1.于是可知,当1x=12时,ymax=12.所以,实数a的取值范围是a>1/2.
In the process of solving mathematical problems, sometimes cleverly use the relevant mathematical ideas and methods, you can avoid the type of discussion, in order to quickly solve the problem.Parameter separation law 1 known inequality ax2-2x +2> 0 for any x∈ [1, 4] constant established, the real number of a range of values. If the analysis of the problem y = ax2-2x +2, you need to a = 0 and a ≠ 0 classification discussed two cases, when a ≠ 0, but also Consider the distribution of the real roots of the function corresponding to the function. If a is separated, then the classification of a can be avoided. Solution From the problem we know that when x∈ [1,4], we should make ax2-2x + 2 > 0 holds, so long as a> 2 (x-1) x2 holds, that is, a> [2 (x- 1) x2] max. Let y = 2 (x- 1) x2 = -2 ) 2 + 12. From x∈ [1,4], we have 14≤1x≤1, so we can see that when 1x = 12, ymax is 12. Therefore, the value of real a is in the range of a> 1/2.