Question

The number of distinct real values of x, satisfying the equation max{x, 2} - min{x, 2} = |x + 2| - |x - 2|, is

Answer 1

We are given the equation $\max\\{x, 2\\} - \min\\{x, 2\\} = |x+2| - |x-2|$, and we need to find the number of distinct real solutions.
Let's analyze both sides of the equation:
The expression $\max\\{x, 2\\} - \min\\{x, 2\\}$ represents the absolute difference between $x$ and 2, i.e., $|x - 2|$.
The right-hand side of the equation $|x + 2| - |x - 2|$ is more complicated; so we need to analyze it case by case based on the value of $x$.

Case 1 : $x > 2$. In this case: $\max\\{x, 2\\} = x$, $\min\\{x, 2\\} = 2$. So the left-hand side becomes $x - 2$.
On the right-hand side: $|x + 2| = x + 2$, $|x - 2| = x - 2$. So the right-hand side becomes $(x + 2) - (x - 2) = 4$.
Equating both sides:
$x - 2 = 4 \implies x = 6$
Thus, $x = 6$ is a solution for $x > 2$.

Case 2 : $x < 2$. In this case: $\max\\{x, 2\\} = 2$, $\min\\{x, 2\\} = x$. So the left-hand side becomes $2 - x$.
On the right-hand side: $|x + 2| = x + 2$, $|x - 2| = 2 - x$. So the right-hand side becomes $(x + 2) - (2 - x) = 2x$.
Equating both sides:
$2 - x = 2x \implies 2 = 3x \implies x = \frac{2}{3}$
Thus, $x = \frac{2}{3}$ is a solution for $x < 2$.

Conclusion: The solutions are $x = 6$ and $x = \frac{2}{3}$, so the number of distinct real solutions is 2.

Answer 2

We are given the equation $\max\\{x, 2\\} - \min\\{x, 2\\} = |x+2| - |x-2|$, and we need to find the number of distinct real solutions.
Let's analyze both sides of the equation:
The expression $\max\\{x, 2\\} - \min\\{x, 2\\}$ represents the absolute difference between $x$ and 2, i.e., $|x - 2|$.
The right-hand side of the equation $|x + 2| - |x - 2|$ is more complicated; so we need to analyze it case by case based on the value of $x$.

Case 1 : $x > 2$. In this case: $\max\\{x, 2\\} = x$, $\min\\{x, 2\\} = 2$. So the left-hand side becomes $x - 2$.
On the right-hand side: $|x + 2| = x + 2$, $|x - 2| = x - 2$. So the right-hand side becomes $(x + 2) - (x - 2) = 4$.
Equating both sides:
$x - 2 = 4 \implies x = 6$
Thus, $x = 6$ is a solution for $x > 2$.

Case 2 : $x < 2$. In this case: $\max\\{x, 2\\} = 2$, $\min\\{x, 2\\} = x$. So the left-hand side becomes $2 - x$.
On the right-hand side: $|x + 2| = x + 2$, $|x - 2| = 2 - x$. So the right-hand side becomes $(x + 2) - (2 - x) = 2x$.
Equating both sides:
$2 - x = 2x \implies 2 = 3x \implies x = \frac{2}{3}$
Thus, $x = \frac{2}{3}$ is a solution for $x < 2$.

Conclusion: The solutions are $x = 6$ and $x = \frac{2}{3}$, so the number of distinct real solutions is 2.