Solve it Algebraically? I won't dare 2

The surest way to solve this by plotting the graph of $\alpha = |1-|1-|x-1|||$ . From the graph we note that there are exactly $4$ distinct solutions when $\alpha = 1$ .

When $\space x = \begin{cases} -2 & \Rightarrow \alpha = |1-|1-|-2-1||| & = |1-|1-3|| = |1-2| & = 1 \\ 0 & \Rightarrow \alpha = |1-|1-|0-1||| & = |1-|1-1|| = |1-0| & = 1 \\ 2 & \Rightarrow \alpha = |1-|1-|2-1||| & = |1-|1-1|| = |1-0| & = 1 \\ 4 & \Rightarrow \alpha = |1-|1-|4-1||| & = |1-|1-3|| = |1-2| & = 1 \end{cases}$

Therefore, the answer $2\alpha = \boxed{2}$ .

Let´s write the function step by step to get the answer:

1st: $1-∣x-1∣$

Remember that $∣x∣=$ $\begin{cases} -x\quad if\quad x<0 \\ x\quad if\quad x\ge 0 \end{cases}$ , so $1-∣x-1∣=1-$ $\begin{cases} -(x-1)\quad if\quad x<1 \\ x-1\quad if\quad x\ge 1 \end{cases}$ . This equals to $\begin{cases} x\quad if\quad x<1 \\ 2-x\quad if\quad x\ge 1 \end{cases}$

2nd: $1-∣1-∣x-1∣∣$

In this case, you take the absolute value of all the negative parts of the function above, making these parts positive, so we have to calculate when $x$ and $2-x$ are negative, $x$ is negative when is less than $0$ and $2-x$ is negative when $x>2$ .

With the absolute value, the parts of $x$ from $-\infty$ to $0$ and $2-x$ from $2$ to $\infty$ are positive. If we want to know the new functions in the positive parts, just multiply by $-1$ both functions, so the positive parts are $-x$ and $x-2$ . This creates a new function:

$∣1-∣x-1∣∣=$ $\begin{cases} -x\quad if\quad x<0 \\ x\quad if\quad 0\le x<1 \\ 2-x\quad if\quad 1\le x<2 \\ x-2\quad if\quad x\ge 2 \end{cases}$

Now, since the function is negative and then added by $1$ we have to multiply by $-1$ each part of the function and then add the 1 to each part:

$1-∣1-∣x-1∣∣=$ $1-$ $\begin{cases} -x\quad if\quad x<0 \\ x\quad if\quad 0\le x<1 \\ 2-x\quad if\quad 1\le x<2 \\ x-2\quad if\quad x\ge 2 \end{cases}=$ $1+$ $\begin{cases} x\quad if\quad x<0 \\ -x\quad if\quad 0\le x<1 \\ x-2\quad if\quad 1\le x<2 \\ 2-x\quad if\quad x\ge 2 \end{cases}=$ $\begin{cases} x+1\quad if\quad x<0 \\ 1-x\quad if\quad 0\le x<1 \\ x-1\quad if\quad 1\le x<2 \\ 3-x\quad if\quad x\ge 2 \end{cases}$

3rd: $∣1-∣1-∣x-1∣∣$

Do the same thing like the 2nd step and you´ll get that $\begin{cases} x+1<0\quad if\quad x<-1 \\ 1-x<0\quad if\quad x>1 \\ x-1<0\quad if\quad x<1 \\ 3-x<0\quad if\quad x>3 \end{cases}$ . Note that $x+1$ and $3-x$ have negative parts in the function of the second step, so if we take the absolute value, these parts will be positive (With the absolute value, $x+1$ from $-\infty$ to $-1$ and $3-x$ from $3$ to $\infty$ are positive), so, the functions on these parts are $-(x+1)$ and $x-3$ .

The function is $∣1-∣1-∣x-1∣∣=$ $\begin{cases} -(x+1)\quad if\quad x<-1 \\ x+1\quad if\quad -1\le x<0 \\ 1-x\quad if\quad 0\le x<1 \\ x-1\quad if\quad 1\le x<2 \\ 3-x\quad if\quad 2\le x<3 \\ x-3\quad if\quad x\ge 3 \end{cases}$

4th: ¿The solutions?

Well, let´s analyze where we could see 4 different solutions in the equation $∣1-∣1-∣x-1∣∣=α$ or $\begin{cases} -(x+1)\quad if\quad x<-1 \\ x+1\quad if\quad -1\le x<0 \\ 1-x\quad if\quad 0\le x<1 \\ x-1\quad if\quad 1\le x<2 \\ 3-x\quad if\quad 2\le x<3 \\ x-3\quad if\quad x\ge 3 \end{cases}=α$

Note that if we choose integer values $x∈[-1,3]$ , the function is between 0 and 1, therefore, between these numbers we can choose the value of $α$ because if $α>1$ we can find only two solutions for the equation, and we can´t choose $α∈(0,1)$ because we can find more than 4 solutions to the equation. So if $α=0$ , the equation has three solutions:

$\begin{cases} -(x+1)\quad if\quad x<-1 \\ x+1\quad if\quad -1\le x<0 \\ 1-x\quad if\quad 0\le x<1 \\ x-1\quad if\quad 1\le x<2 \\ 3-x\quad if\quad 2\le x<3 \\ x-3\quad if\quad x\ge 3 \end{cases}=0\rightarrow \begin{cases} -(x+1)=0\quad for\quad x<-1 \\ x+1=0\quad for\quad -1\le x<0 \\ 1-x=0\quad for\quad 0\le x<1 \\ x-1=0\quad for\quad 1\le x<2 \\ 3-x=0\quad for\quad 2\le x<3 \\ x-3=0\quad for\quad x\ge 3 \end{cases}\rightarrow x=\begin{cases} N \\ -1 \\ N \\ 1 \\ N \\ 3 \end{cases}$

So, the answer must be $α=1$ :

$\begin{cases} -(x+1)\quad if\quad x<-1 \\ x+1\quad if\quad -1\le x<0 \\ 1-x\quad if\quad 0\le x<1 \\ x-1\quad if\quad 1\le x<2 \\ 3-x\quad if\quad 2\le x<3 \\ x-3\quad if\quad x\ge 3 \end{cases}=1\rightarrow \begin{cases} -(x+1)=1\quad for\quad x<-1 \\ x+1=1\quad for\quad -1\le x<0 \\ 1-x=1\quad for\quad 0\le x<1 \\ x-1=1\quad for\quad 1\le x<2 \\ 3-x=1\quad for\quad 2\le x<3 \\ x-3=1\quad for\quad x\ge 3 \end{cases}\rightarrow x=\begin{cases} -2 \\ N \\ 0 \\ N \\ 2 \\ 4 \end{cases}$

The letter N in the set of solutions means that the equation doesn´t have a solution considering the interval given for each of them.

Now, since $A=α$ , $2A= 2$

There are different sets of conditions that we can apply and evaluate as follows: $\ (1).... \ x \geq\ 2 \Rightarrow\alpha = |3-x|$ $\ (2).... x < 2 \Rightarrow\ 0 \leq\ ( \alpha = | 1-|x| |) \leq\ 1$ Now we will use these conditions to create conditions for alpha based on the number of real solutions that can be taken from equations (1) and (2): $\alpha = 0 \Rightarrow\ x = -1, 1 , 3 \Rightarrow\ 3 \ solutions$ $0 < \alpha \ < 1 \Rightarrow\ 2 \ solutions \ from (1) \ and \ 4 \ from (2)$ $\alpha = 1 \Rightarrow\ x = -2, 0 , 2 , 4$ $\alpha > 1 \Rightarrow\ 2 \ solutions \ from \ (1) \ but \ none \ from (2)$ $\therefore\ 2 \alpha = 2$