Analysis: (i) By proving the necessary and sufficient conditions, it is concluded that {xn} is a descending sequence if and only if c < 0；;

(ii) from (i), c≥0, from ① when c=0, ② when c > 0, 0 < c < 1, c≤ 1/4.

It is proved that xn+1> xn.lim n→∞ xn+1= lim n→∞ (-x2n+xn+c)? Lim n →∞ xn = C. When c > 1/4, the bright sequence {xn} is the contradiction of decreasing sequence. When 0 < c ≤ 1/4, the sequence {xn} is an increasing sequence.

Personally, I think the last question in Hunan is really a bit difficult. Let me explain.

20 12 Hunan

Analysis: (1) First determine a > 0, and then derive the derivative function to determine the monotonicity of the function. When x= 1a ln 1 a, the minimum value of f(x) is f (1aln1a) =1a-.

Construct a new function g(t)=t-tlnt, then G ′ (t) =-lnt, determine the monotonicity of the function, find the maximum value of the function, and then get the value set of A;

(2) According to the meaning of the question, k = eax2-eax1x2-x1,and construct a new function φ (x) = f' (x)-k = aeax-eax2-eax1.

x2-x 1，

Then φ (x1) =-eax1x 2-x1[ea (x2-x1)-a (x2-x1)-1], φ (. The function F(t)=et-t- 1 can be constructed to prove that φ (x 1) < 0 and φ (x2) > 0, and then the existence of x0∈(x 1, x2) can be obtained, so that f ′ (x0) >.

Note that the two letters after the letter are both squares.