Why is deflection at the boundary 0 for the given statically indeterminate beam problem? I have been trying really hard to understand the boundary condition applied to the indeterminate beam problems.. although i am citing a particular problem, i have been finding the same approach in many other similar problems and even after referring to various books, and internet sites, all i find is that:-
"IT IS OBVIOUS THAT SLOPE AT x=0 WILL BE ZERO"

and the solution suggests:-

Can someone please explain me how is this boundary condition obtained.
Reference book for the problem:- Timoshenko and Gere 6th edition, Mechanics of Materials
EDIT:- This is how i feel it should look like from intuition.. 

 A: The diagram in your edit would be true if it were pins at the ends, not walls. If we assume that the walls and beam are fused together at the ends, then it means that the walls now resist bending moment. Imagine for the cantilever case (one end to the wall, the other free), if you pull down on the free end, and the fixed end have none zero slope, the cantilever will not be able to resist bending moment.
Also, try to imagine that part of the beam is inserted into a slot in the wall, such that inside the wall is a continuation of the beam itself (so the beam extends from x < 0 to x > L). If there is a non-zero slope, it would imply that there is a discontinuity in the derivative of the deflection y(x), which is not physically possible.  
A: I pulled out my course notes and hopefully the following helps, my conclusion is that it comes down to the boundary conditions:
Beams deflect as cubics, the shape function of a beam element has the form:
$$y=a_1x^3+a_2x^2+a_3x+a_4$$ 
Your question is about the slope, $\frac{dy}{dx}$ which is equal to $\theta$, so if I take the derivative of the shape function I obtain:
$$\frac{dy}{dx}=a_3+2a_2x+3a_1x^2$$
This slope at $x=0$ is simply $a_3$, and this constant is found from the boundary conditions. 
These boundary conditions are imposed in your case by the wall, if the wall is not angled at x=0, neither will the beam. However, if you wanted to, I suppose you could impose a boundary condition to obtain a non-zero slope at $x=0$ by giving the wall a specific angle.
The only other way I can think of to make this intuitive is to consider splitting your beam into multiple sections and analyzing each section individually.
In order for the shape function for the beam as a whole to be a continuous function each section must have the same deflection and slope as the adjacent section at the point of contact. 
