已知函数f(x)=ln(x+a)-x2-x在x=0处取得极值.(1)求函数f(x)的单调区间;(2)若关于x

(1)
f'(x) = 1/(x + a) - 2x - 1
f'(0) = 1/a - 1 = 0, a = 1

(2)
g(x) = f(x) - (-5x/2 + b) = ln(x + 1) - x² - x + 5x/2 - b = ln(x + 1) - x² + 3x/2 - b
g'(x) = 1/(x + 1) - 2x + 3/2 = (-4x² - x + 5)/(2x + 2) = -(4x + 5)(x - 1)/悔基(2x + 2)
g'(x) = 0, x = -5/4或x = 1
x = 1在区间(0, 2)上
-1 < x < 1时: -(4x + 5) < 0, x - 1 < 0, 2x + 2 > 0, g'(x) > 0, 增函数
1 < x < 2时: -(4x + 5) < 0, x - 1 > 0, 2x + 2 > 0, g'(x) < 0, 减函数
g(1) = ln2 + 1/2 - b为极大值
若关于x的方程f(x= -5x/2 + b在区间(0，2)上有两上不等的实根, 只需g(1) > 0且g(0), g(2)同时<碧举谨0
g(1) = ln2 + 1/2 - b > 0
b < ln2 + 1/2 (i)
g(0) = -b < 0, b > 0 (ii)
g(2) = ln3 -4 + 3*2/2 - b = ln3 - 1 - b < 0
b > ln3 - 1 (iii)
3 > e, ln3 - 1 > 0
结合(ii)(iii)同时成答世立: b > ln3 - 1
结合(i): ln3 - 1 < b < ln2 + 1/2