Modulus Functions (Cambridge International AS Mathematics): Exam Questions

Solve the equation \left|3 x - 2\right| = 7.

Solve the following.

(i) \left|4 x + 3\right| < 9

(ii) \left|2 x\right| + 5 \geq 15

(iii) \left|x + 8\right| = \left|2 x + 10\right|

Solve the equation \left|6 - 2 x\right| = 4.

On the same diagram, sketch the graphs of y = \left|6 - 2 x\right| and $y = 4$. Label the coordinates of the points where the two graphs intersect each other and the coordinate axes.

Consider the graphs of y = \left|6 - 2 x\right| and $y = k$, where $k$ is a constant. For which values of $k$ will the two graphs have

(i) no points of intersection,

(ii) one point of intersection,

(iii) two points of intersection?

The graph of y = \text{f} \left(x\right), where \text{f} \left(x\right) = 2 x - 3, is shown below.

Determine the coordinates of the points marked $A$ and $B$.

(i) On the diagram above, sketch the graph of y = \left|\text{f} \left(x - 1\right)\right|.

(ii) Determine the coordinates of the image of the points $A$ and $B$ under the transformation in part (i).

Solve the equation \left|4 x + 2\right| = 5.

The functions $\text{f}$ and $\text{g}$ are defined for all real $x$ by \text{f} \left(x\right) = \left( x + 2 \right)^{2} and \text{g} \left(x\right) = 2 x + 4.

On the same axes, sketch the graphs of y = \text{f} \left(x\right) and y = \left|\text{g} \left(x\right)\right|, labelling the points at which the graphs intersect the coordinate axes.

Solve the equation \text{f} \left(x\right) = \left|\text{g} \left(x\right)\right|.

Solve the equation \left|2 x - 3\right| = \left|x + 1\right|.

Solve the inequality \left|3 x + 2\right| \leq \left|x - 4\right|.

Solve the equation \left|x - 4\right| = 2 x - 5.

The graph of y = \text{f} \left(x\right), where \text{f} \left(x\right) = \left|3 x + 1\right|, is shown below.

(i) On the diagram above, sketch the graph of y = \left|x - 5\right| and state the coordinates of any points of intersection with y = \text{f} \left(x\right).

(ii) Hence, or otherwise, solve the inequality \left|3 x + 1\right| < \left|x - 5\right|.

The functions $\text{f}$ and $\text{g}$ are defined for all real $x$ by \text{f} \left(x\right) = 3 x - 1 and \text{g} \left(x\right) = 2 x + 2.

On the same axes, sketch the graphs of y = \left|\text{f} \left(x\right)\right| and y = \left|\text{g} \left(x\right)\right|, labelling the points at which the graphs intersect the coordinate axes.

Solve the equation \left|\text{f} \left(x\right)\right| = \left|\text{g} \left(x\right)\right|.

Which of the solutions to \left|\text{f} \left(x\right)\right| = \left|\text{g} \left(x\right)\right| is also a solution to \text{f} \left(x\right) = \text{g} \left(x\right)?

The function $\text{f}$ is defined by \text{f} : x \mapsto \left|3 x - 2\right| for $x \in ℝ$.

Explain why the inverse of \text{f} \left(x\right) does not exist.

Suggest an adaptation to the domain of \text{f} \left(x\right) so that its inverse exists, while also producing the maximum possible range for \text{f} \left(x\right).

Using your adaptation from part (b), find an expression for \text{f}^{- 1} \left(x\right) and state its domain and range.

The function $\text{p}$ is defined by \text{p} \left(x\right) = \left|2 x\right| for $x \leq 0$.

On the same axes, sketch the graphs of y = \text{p} \left(x\right) and y = \text{p}^{- 1} \left(x\right).

Find an expression for \text{p}^{- 1} \left(x\right) and state its domain.

Show that \text{p}^{- 1} \left(x\right) = - \frac{1}{2} \text{p} \left(- \frac{1}{2} x\right).

The constant $a$ is positive. Solve the inequality \left|2 x - a\right| > \left|x + a\right|, giving your answer in terms of $a$.

The functions $\text{f}$ and $\text{g}$ are defined by \text{f} \left(x\right) = \sqrt{x} for $x \geq 0$ and \text{g} \left(x\right) = 2 x - 3 for $x \in ℝ$.

On the same axes, sketch the graphs of y = \text{f} \left(x\right) and y = \left|\text{g} \left(x\right)\right|, labelling the points at which the graphs intersect the coordinate axes.

Solve the equation \text{f} \left(x\right) = \left|\text{g} \left(x\right)\right|.

Which of the solutions to \text{f} \left(x\right) = \left|\text{g} \left(x\right)\right| is not a solution to \text{f} \left(x\right) = \text{g} \left(x\right)?