Modulus Functions (Cambridge International A-Level Mathematics): Exam Questions

Solve the equation |3x - 2| = 7.

Solve the following.

(i) |4x + 3| < 9

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

(iii) |x + 8| = |2x + 10|

Solve the equation |6 - 2x| = 4.

On the same diagram, sketch the graphs of y = |6 - 2x| 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 = |6 - 2x| 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 $y = \text{f} ( x )$, where $\text{f} ( x ) = 2 x - 3$, is shown below.

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

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

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

Solve the equation |4x + 2| = 5.

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

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

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

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

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

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

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

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

Solve the equation |x - 4| = 2x - 5.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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