Necessary speed for being at rest 

In the figure above mass $m_1$(1.5kg) is rotating with uniform speed in a circular path of radius $r$ and mass $m_2$(2kg) is tied to $m_1$ with a thread. The coefficient of friction on the table is stated to be $0.2$. The problem asks us to find the necessary speed for which mass $m_2$ remains at equilibrium.

The solution given was divided into two cases:
Case-1 The frictional force acting along the centrifugal force which gives the equation $T=\frac{mv^2}{r}+f_k$ which gives us an answer of $3.87\sqrt{r}$.
Case-2 The frictional force acting along the tension which gives us the equation $T+f_k=\frac{mv^2}{r}$ which in turn gives the result $3.33\sqrt{r}$ for speed.
Then they asserted that $3.87\sqrt{r}$ is the maximum speed and $3.33\sqrt{r}$ is the minimum speed and hence showed that this speed will be in the range $[3.33\sqrt{r},3.87\sqrt{r}]$.
My confusion arises in this step. First of all, $f_k=0.2\times m_1\times g$ is not the exact frictional force,rather the maximum frictional force acting. So the $f_k$ which they calculated is the maximum. Now the answer for case-2 turns out to be less than that of case-1 but how does it make it the minimum? I mean how are they so sure that any other velocity in case-1 can't be less than that of case-2?Also i couldn't decipher why they took the velocity of case-2 to be the minimum. Why isn't it possible for any other velocity in case-2 to be less than that?
I would like to draw the kind attention of the physics lovers in this issue.
 A: From what i understood, you must be asking why direction of friction is changing and why speed have to remain in that particular range.
Lets start the question freshly. Firstly, $T=m_2g$
Now, we dont know about direction of friction which is acting, that's why we take 2 cases (extreme values) and we get a particular range. Friction has property to always oppose the change in motion. Tension however will always act inwards in our situation.
CASE-$1$
Lets talk about minimum velocity, at which the $m_1$ has tendency to slip towards the hole in table. That's why friction will act outwards to oppose slipping of $m_1$  body. Note, since $m_1$ is trying to slip (change in motion), so static friction will not act but limiting friction will act (highest value of friction) .So we will get:
$$ T-f_k=\frac{mv_{min}^2}{r}$$
Case-$2$
Now for maximum velocity, the $m_1$ will have tendency to slip outwards, so friction will oppose this by acting inwards (together with tension). Hence we get:
$$ T+f_k=\frac{mv_{max}^2}{r}$$
Now for centrifugal or centripetal, these are frame dependant quanitities, i.e. they change on frame you are working. Its misleading to say that if $f_k$ is acting outwards, then its centrifugal force or vice-versa. By default we always work in ground frame and use centripetal force (we used it here). Centrifugal force is used if we work in the rotating frame. Also working with respect to different frames, must not change our calculations, (unless they are non-inertial i.e. accelerating).
