Find W/L for CMOS Circuit: Triode/Cutoff Modes

[jean28]

How to find W/L??

Homework Statement

The exercise already tells me that the value of W/L is 2. However, I can't seem to understand where that result came from. I'd like to know how I can get to that conclusion.

Here is the exercise with the graph of the inverter:
Inverter image:
http://i1226.photobucket.com/albums/ee410/jean28x/image2_zps9e7f8724.png

Exercise, part 1:
http://i1226.photobucket.com/albums/ee410/jean28x/image_zps136b69f7.png

Exercise, part 2:
http://i1226.photobucket.com/albums/ee410/jean28x/image_1_zps32b9a8ee.png

Homework Equations

Here is a picture of relevant formulas of MOSFETS:
http://i1226.photobucket.com/albums/ee410/jean28x/image_2_zpse3d8245e.png

There is also the formula in the hint that is given in the exercise:
http://i1226.photobucket.com/albums/ee410/jean28x/image_1_zps32b9a8ee.png

The Attempt at a Solution

So, using the formula given in the hint, I substitute the values (I already know the 48 KΩ part):

48k = 1/((125μ)(W/L)(2.5 - 0.5))

Solving for W/L, I get:

12 = 1 / (W/L)

W/L = 1 / 12 = 0.08333

However, the answer says that W/L is supposed to be 2. What am I doing wrong here?

Thanks a lot.P.S.

It might be worth noting that, since this is an exercise working with CMOS circuits, then the relevant state that it should be analyzed in is in Triode and Cutoff modes.

 

[jean28]

Here I will upload the pictures, just in case the links don't work.

 

[jean28]

Last one.