The number of distinct real values of x, satisfying the equation

max{x,2} −min{x,2} = |x+2|−|x−2|

is:

The correct answer is 2.

To find the number of distinct real values of $x$ satisfying the equation:
$max\{x,2\}-min\{x,2\}=|x+2|-|x-2|$
we can analyze the behavior of the expressions on both sides of the equation by dividing the real number line into three intervals based on the critical points $x=-2$ and $x=2$.

First, let us simplify the Left-Hand Side (LHS) of the equation, which is:
$\text{LHS}=max\{x,2\}-min\{x,2\}$
Recall that for any two real numbers $a$ and $b$, the difference between their maximum and minimum is simply the absolute difference between them:
$max\{a,b\}-min\{a,b\}=|a-b|$
Thus, the Left-Hand Side simplifies to:
$\text{LHS}=|x-2|$
Now the original equation becomes:
$|x-2|=|x+2|-|x-2|$
Adding $|x-2|$ to both sides, we get:
$2|x-2|=|x+2|$

We solve this absolute value equation by analyzing it in three distinct intervals:

Case 1: $x\ge 2$
In this interval, we have:
$|x-2|=x-2$
and
$|x+2|=x+2$
Substituting these into our equation gives:
$2(x-2)=x+2$
$2x-4=x+2$
$x=6$
Since $6\ge 2$, this is a valid solution.

Case 2: $-2\le x<2$
In this interval, we have:
$|x-2|=-(x-2)=2-x$
and
$|x+2|=x+2$
Substituting these into our equation gives:
$2(2-x)=x+2$
$4-2x=x+2$
$3x=2$
$x=\frac{2}{3}$
Since $-2\le \frac{2}{3}<2$, this is also a valid solution.

Case 3: $x<-2$
In this interval, we have:
$|x-2|=2-x$
and
$|x+2|=-(x+2)=-x-2$
Substituting these into our equation gives:
$2(2-x)=-x-2$
$4-2x=-x-2$
$x=6$
However, the condition for this case is $x<-2$. Since $6$ is not less than $-2$, there is no solution in this interval.

Thus, the distinct real values of $x$ that satisfy the equation are $x=6$ and $x=\frac{2}{3}$. The number of distinct real values is 2.