When x≠0, f(x) is an even function.

Then f(-x)=f(x)

That is log2 (1/Cx+1)-kx = log2 (Cx+1)+kx.

That is, log2 (c x) =-2kx.

That is, xlog2(c)=-2kx (note c >；; 0 and c≠ 1)

That is log2(c)=-2k(I).

When k=0, F (x) = log2 (c x+ 1).

And the function y = b (x-1)-1+log2 (5) is obviously above the fixed point (1,log2 (5)).

According to the topic Log2(5)=log2(c+ 1).

That is c=4.

K=- 1 from (i)

(2) Branch 1 f (x) = log2 (4 x+ 1)-x

According to the topic, there are log2 (4x+1)-x = log2 (a * 2x-4a/3).

That is log2 (4x+1)-log2 (2x) = log2 (a * 2x-4a/3).

That is log2 [(4x+1)/2x] = log2 (a * 2x-4a/3).

That is, (4 x+ 1)/2 x = a * 2 x-4a/3.

That is 4 x+ 1 = a * 4 x-4 a * 2 x/3.

That is, a = (4 x+ 1)/(4 x-4 * 2 x/3)

Let t = 2 x (obviously t >；; 0)

Then a = (t 2+ 1)/(t 2-4t/3) (obviously t≠4/3).

Let g (t) = (T2+1)/(T2-4t/3) (t >; 0 and t≠4/3)

It is easy to get g' (t) = (-4t2/3-2t+4/3)/(t2-4t/3) 2.

Let G'(t)=0, t > be, (2t- 1)(t+2)=0, and the solution is t= 1/2 (note t >; 0)

It's easy to know when 0

Explain that t= 1/2 is the maximum point of G(t), and G( 1/2)=-3.

Note that when t & gt4/3 and t→4/3, G(t)→+∞

It shows that t=4/3 is a vertical asymptote.

Note that when t→+∞, g (t) = (1+1/t2)/(1-4/3t) →1.

It shows that G(t)= 1 is a horizontal asymptote.

Therefore, it can be roughly judged that the image of G(t) has the following characteristics:

At 0

At t> in 4/3, G(t) is a decreasing hyperbola (single branch) with two asymptotes, horizontal and vertical.

Because g(x)=f(x) has one and only one solution.

That is, a = (4 x+ 1)/(4 x-4 * 2 x/3) has one and only one solution.

That is, a = (t 2+ 1)/(t 2-4t/3) has one and only one solution.

That is, there is only one intersection point between G(t) and horizontal straight line y = a.

Understanding a > from the characteristics of G(t) image; 1 or a=-3