Displacement Calc: Find x if x'>ab?

[konichiwa2x]

The displacement 'x' of a particle varies with time according to the relation
$$x = \frac{a}{b}(1 - e^-^b^t)$$ Then,

(A) At t=1/b the displacement of the particle is a/b.
(B) The velocity and acceleration of the particle at t=0 are 'a' and 'b' respectively.
(C) The particle cannot reach a point at a distance 'x' from its starting point if x'>ab.
(D) The particle will come back to its starting point as 't' tends to infinity.

I have differentiated to obtain the velocity and also acceleration but I still can't seem to solve this.
My book says the correct answer is (C) but how do you arrive at it? Please help.

 

[Päällikkö]

Have you noticed why a, b and d are not correct?
The particle is at its maximum distance when its velocity is zero.

 

[Saketh]

(A) At t=1/b the displacement of the particle is a/b.

This is just plugging in and checking.

(B) The velocity and acceleration of the particle at t=0 are 'a' and 'b' respectively.

You said you differentiated to get the velocity and acceleration functions, so this is just more plugging in and checking. If you are unsure, you probably differentiated incorrectly.

(D) The particle will come back to its starting point as 't' tends to infinity.

In order for it to come back, its velocity would need to change direction at some point. Does it? And if so, does it reach the starting point again?