1, a=0 f(x)=x monotonically increases.

2. a>0, let f' (x) =1-a-AlN (x+1) = 0 to get x = e (1-a)/a-1e (1-). - 1

So the original function monotonically increases at (-1, e (1-a)/a- 1) and monotonically decreases at (e (1-a)/a- 1, positive infinity).

(2) f (x) = x-(x+1) ln (x+1) makes the derivative f'(x)=-ln(x+ 1)=0 to get x=0.

Therefore, it monotonically increases at [-1/2,0] and monotonically decreases at [0, 1].

Maximum value f(0)=0

f( 1/2)=- 1/2- 1/2ln/ 12 = 1/2(LN2- 1)

f( 1)= 1-2ln2

f( 1/2)>f( 1)

Therefore, the value range of t is [1/2(ln2- 1), and 0) you can draw continuous images by yourself.

(3) the constructor f (x) = ln (x+1)/x.

X>0, monotonically decreasing, that is, f (m)

ln(m+ 1)/m & lt； ln(n+ 1)/n

You can draw a conclusion by sorting it out.