We need to find how many different real values of $x$ satisfy this equation: $$\newline$$ $\max\{x, 2\} - \min\{x, 2\} = |x + 2| - |x - 2|$ $$\newline$$ The equation changes behavior at critical points where the expressions inside the max/min and absolute value functions change sign: $x = 2$ and $x = -2$. $$\newline$$ This gives us three ranges to check: $x \geq 2$, $-2 < x < 2$, and $x < -2$. $$\newline$$ Your basic concepts with modulus should be covered to understand this solution. We will find the RHS & LHS for the 3 ranges to find $x$ and then check if the $x$ found is valid for the range. $$\newline$$

$$\newline$$ When $x \geq 2$: $$\newline$$ Left side: Since $x \geq 2$, we have $x$ is at least as large as $2$ $$\newline$$ $\max\{x, 2\} = x$ and $\min\{x, 2\} = 2$ $$\newline$$ So left side = $x - 2$ $$\newline$$ Right side: Since $x \geq 2$: $$\newline$$ $x + 2 \geq 4$, so $|x + 2| = x + 2$ $$\newline$$ $x - 2 \geq 0$, so $|x - 2| = x - 2$ $$\newline$$ Hence, both mod open as positives. $$\newline$$ So right side = $(x + 2) - (x - 2) = 4$ $$\newline$$ Our equation becomes: $x - 2 = 4$ $$\newline$$ Therefore $x = 6$ $$\newline$$ Since $6 \geq 2$, this solution is valid for this case. $$\newline$$

$$\newline$$ When $-2 < x < 2$: $$\newline$$ Left side: Since $x < 2$: $$\newline$$ $\max\{x, 2\} = 2$ and $\min\{x, 2\} = x$ $$\newline$$ So left side = $2 - x$ $$\newline$$ Right side: Since $-2 < x < 2$: $$\newline$$ $x + 2 > 0$, so $|x + 2| = x + 2$ $$\newline$$ $x - 2 < 0$, so $|x - 2| = 2 - x$ $$\newline$$ So right side = $(x + 2) - (2 - x) = 2x$ $$\newline$$ Our equation becomes: $2 - x = 2x$ $$\newline$$ Therefore $2 = 3x$, so $x = \dfrac{2}{3}$ $$\newline$$ Since $-2 < \dfrac{2}{3} < 2$, this solution is valid for this case. $$\newline$$

$$\newline$$ When $x < -2$: $$\newline$$ Left side: Since $x < -2 < 2$: $$\newline$$ $\max\{x, 2\} = 2$ and $\min\{x, 2\} = x$ $$\newline$$ So left side = $2 - x$ $$\newline$$ Right side: Since $x < -2$: $$\newline$$ $x + 2 < 0$, so $|x + 2| = -x - 2$ $$\newline$$ $x - 2 < 0$, so $|x - 2| = 2 - x$ $$\newline$$ So right side = $(-x - 2) - (2 - x) = -4$ $$\newline$$ Our equation becomes: $2 - x = -4$ $$\newline$$ Therefore $x = 6$ $$\newline$$ We need $x < -2$ for this case, but we got $x = 6$. Since $6 > -2$, this solution is invalid for this case. $$\newline$$

$$\newline$$ The summary table makes it easier to understand what is happening here: $$\newline$$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$ $\newline$

Case	Condition	Left Side	Right Side	Equation	Solution	Valid?
1	x ≥ 2	x - 2	4	x - 2 = 4	x = 6	✓
2	-2 < x < 2	2 - x	2x	2 - x = 2x	x = 2/3	✓
3	x < -2	2 - x	-4	2 - x = -4	x = 6	✗

$$\newline$$ Therefore, the number of distinct real values of $x$ satisfying the equation is 2.

CAT (QA) PYQs

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