1. According to the topic, if x is constant in the interval [1, +∞) and f(x)-0, then f(x) must be increasing function in the interval [1, +∞) and F (1). F (x1)-f (x2) = (x1-x2) (x1* x2-a)/(x1x2) 0 holds, that is, x1* x2-a. a≦ 1

discuss

If a>0, then x is in the interval [1, +∞), and f (x) = x+2+a/x > 0 also holds.

Generally speaking, a & gt-3

2. Similarly, if 2 ≦ x2 < x 1, then f(x 1)-f(x2) is substituted for simplification, so that f (x1)-f (x2) = (x1-x2). And f(x) is increasing function in the interval [2, +∞), then f (2) =11￠ a is constant, so a.

3. Let 2 ≦ x2 < x 1 ≦ 5, and you can get f (x1)-f (x2) = (x2-x1)/[(x1-kloc-0). 0, it can be seen that this function is a subtraction function, so f (x)