Question 22/40: Set 3 - Try Out - Mathematics (TKA)

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All values of $x$ that satisfy $|x| + |x - 2| > 3$ are....

$x < -1$ or $x > \frac{5}{2}$

$x < -\frac{1}{2}$ or $x > 3$

$x < -\frac{1}{2}$ or $x > \frac{5}{2}$

$x < -1$ or $x > 3$

$x < -\frac{3}{2}$ or $x > \frac{5}{2}$

The definition of absolute value $|x - 2|$ is

|x - 2| = \begin{cases} x - 2, & \text{for } x \geq 2 \\ -(x - 2), & \text{for } x < 2 \end{cases}

To facilitate finding the solution, we divide it into 3 regions based on the definition of absolute value above

Region Division

Three regions based on absolute value definition:

x < 0

0 \leq x < 2

, and

x \geq 2

\text{Region I}

\text{Region II}

\text{Region III}

0

0

2

2

\text{Region I:}

(x < 0)

|x| = -x

|x - 2| = -(x - 2) = -x + 2

|x| + |x - 2| > 3

-x + (-x + 2) > 3

-2x + 2 > 3

-2x > 1

x < -\frac{1}{2}

So the solution set is $\{x < 0\} \cap \{x < -\frac{1}{2}\} = \{x < -\frac{1}{2}\}$ .

\text{Region II:}

(0 \leq x < 2)

|x| = x

|x - 2| = -(x - 2) = -x + 2

|x| + |x - 2| > 3

x + (-x + 2) > 3

2 > 3

This means there is no value of $x$ that satisfies region II.

\text{Region III:}

(x \geq 2)

|x| = x

|x - 2| = x - 2

|x| + |x - 2| > 3

x + (x - 2) > 3

2x > 5

x > \frac{5}{2}

So the solution set is $\{x \geq 2\} \cap \{x > \frac{5}{2}\} = \{x > \frac{5}{2}\}$ .

So the total solution is the union of the three regions

\{x < -\frac{1}{2}\} \cup \{x > \frac{5}{2}\}

So the solution is $\{x < -\frac{1}{2} \cup x > \frac{5}{2}\}$ .

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