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Question 25

Number 25

The solution set of $\left|\frac{x}{3} - \frac{1}{4}\right| = \frac{1}{12}$ is....

$\{-\frac{1}{2}, 2\}$

$\{2, 3\}$

$\{1, 2\}$

$\{-\frac{3}{2}, \frac{1}{3}\}$

$\{\frac{1}{2}, 1\}$

Explanation

First change it so that the fractional form disappears

\left|\frac{x}{3} - \frac{1}{4}\right| = \frac{1}{12}

\left|\frac{4x - 3}{12}\right| = \frac{1}{12}

|4x - 3| = 1

Its definition is

|4x - 3| = \begin{cases} 4x - 3, & \text{for } x \geq \frac{3}{4} \\ -(4x - 3), & \text{for } x < \frac{3}{4} \end{cases}

For $x \geq \frac{3}{4}$ then

|4x - 3| = 1

4x - 3 = 1

4x = 4

x = 1

For $x < \frac{3}{4}$ then

|4x - 3| = 1

-(4x - 3) = 1

-4x + 3 = 1

-4x = -2

x = \frac{1}{2}

So the solution set is $\left\{\frac{1}{2}, 1\right\}$ .

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Question 25Set 3

Exercises

Number 25

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