Modulus Functions (Cambridge (CIE) A Level Maths: Pure 3): Exam Questions

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

(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 $\text{f}\left(x\right)=(x+2{)}^2$ and $\text{g}\left(x\right)=2x+4$.

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

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

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

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

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

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

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

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

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

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

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 $\text{p}\left(x\right)=\left|2x\right|$ for $x\le 0$.

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 $|2x-a|>|x+a|$, 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 $x\ge 0$ and $\text{g}\left(x\right)=2x-3$ for $x\in \mathrm{ℝ}$.

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

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

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