一、试题呈现题目(2015年新课标(Ⅰ)理12)设函数f(x)=e~x(2x-1)-ax+a,其中a<1,若存在唯一的整数x_0,使得f(x_0)<0,则实数a的取值范围是().(A)[-3/2e,1)(B)[-3/2e,3/4)(C)[3/2e,3/4)(D)[3/2e,1)二、思路分析本题是有关不等式的存在性恒成立问题,难点是不等式的解是整数,但处理的方法与一般性的恒成立问题并无本质区别,解决这类问题通常有三种思路: First, the questions presented questions (2015 New Curriculum (Ⅰ) Li 12) Let f (x) = e ~ x (2x-1) -ax + a, where a <1, if there is a unique integer x_0, f (x_0) <0, then the real value of a is in the range of () (A) [- 3/2e, 1) B - [3/3 e, 3/4) 2) Analysis of Ideas This question is about the existence of inequalities, the difficulty is inequality solution is an integer, but the method of dealing with the general establishment of the problem is not constant The essential difference, there are usually three kinds of ideas to solve such problems: