We have a function f(x) equals e to the x times 2x minus 1, minus ax plus a. To find when this function has exactly two zeros, we rewrite the equation f(x) equals 0 as e to the x times 2x minus 1 equals a times x minus 1. This transforms our problem into finding when these two curves intersect at exactly two points.
因为x等于1不是f(x)等于0的根，所以我们可以用x减1去除两边。这样得到a等于e的x次方乘以2x减1，再除以x减1。设这个函数为g(x)。现在问题变成：对于哪些a值，水平线y等于a与曲线y等于g(x)恰好有两个交点？让我们分析这个函数g(x)。
Now let's analyze the derivative of g(x). After calculation, we get g prime of x equals e to the x times x times 2x minus 3, all divided by x minus 1 squared. Setting g prime equals zero, we find critical points at x equals 0 and x equals 3 halves. By analyzing the sign of the derivative, we determine the monotonicity: g(x) increases for x less than 0, decreases from 0 to 1, continues decreasing from 1 to 3 halves, and increases for x greater than 3 halves.
Now let's identify the key values. At x equals 0, we have a local maximum with g(0) equals 1. At x equals 3 halves, we have a local minimum with g(3/2) equals 4 e to the 3 halves power. The function has a vertical asymptote at x equals 1, where the left limit approaches negative infinity and the right limit approaches positive infinity. As x approaches negative infinity, g(x) approaches 0, and as x approaches positive infinity, g(x) approaches positive infinity.
Now we can determine when the horizontal line y equals a intersects the curve at exactly two points. When a is between 0 and 1, the line intersects the curve twice: once on the left branch and once on the right part before x equals 1. When a is greater than 4 e to the 3 halves power, the line intersects twice on the right branch: once between 1 and 3 halves, and once after 3 halves. Therefore, the range of a is the union of intervals: 0 to 1, and 4 e to the 3 halves power to positive infinity.