An Ito integral question

We know:X(t) = 1 + I[sigma*(X(s)^0.5)]dW(s) + bt, and tao = inf{t: X(t)=0}
Sorry, I can't type integral notation, so I use the "I[f(s)]dW(s)" in the 
formula to denote Ito integral for function f(s) from 0 to t.

Question is:
Give necessary and sufficient conditions on sigma and b>0, such that P{tao<
infinite}>0

Can anyone help to calculate it, thank you!
My answer is 0<2b/(sigma^2)<1, is it rite?

We know dXt=bdt+sigma*sqrtXt*dWt X0=1
Given 00,Xt=x1 or Xt=x2}
It can be shown that P[Xtau=x1]=[x2^(1-2*b/sigma^2)-1]/[x2^(1-2*b/sigma^2)-x1^(1-2*b/sigma^2)],when 
2*b/sigma^2!=0.5; 
Now let x1-->0, x2-->infinit, we can get the desired result.

That is a good idea. 
I have several questions about this method.

1. Is the ODE just comes form Ito's lemma and set the drift term zero? 
f'b + f'' x\sigma^2 x / 2 = 0 ? 

2. Solve this ODE, I got f = x^(1 - 2b/sigma^2) 
for (1-2b/sigma^2) > 0, b < sigma^2/2
for (1-2b/sigma^2) < 0, when x < 0 , f is not well defined. What is that 
mean?

3. solve for p, we get,
p = (1 - x2^(1-2b/sigma^2))/ (x1^(1-2b/sigma^2) - x2^(1-2b/sigma^2))

if (1-2b/sigma^2) > 0, when x2->inf, p->1, is that mean the probability is 1?
if (1-2b/sigma^2) < 0, p->0 as x1->0, is that mean the probability is 0?

4. Such a sharp change of probability?