You've covered inequalities, inverse functions, composite functions, parametric equations, and a suite of graphical relationships. Now it's time to tie it all together. This final lesson shows you the question types examiners actually use, the marks that are on the table, the traps students fall into, and the techniques that separate top-band answers from average ones.
What is the most common mistake students make when solving rational inequalities, and how do you think it can be avoided? Write your gut answer before reading on.
Knowing what examiners typically ask lets you study smarter. Here are the recurring question types for Module 1, the usual format, and the marks on offer:
| Topic | Typical question | Marks |
|---|---|---|
| Inequalities | Solve and express in interval notation | 2–3 |
| Rational inequalities | Use sign table, exclude denominator zeros | 3–4 |
| Absolute value | Solve using cases or squaring | 2–3 |
| Inverse functions | Find algebraically, state domain | 3–4 |
| Domain restriction | Restrict quadratic, find inverse | 3–4 |
| Graphical relationships | Sketch $y = 1/f(x)$, $y = |f(x)|$ | 3–4 |
| Parametric equations | Eliminate parameter, sketch curve | 3–4 |
Exam checklist: domain stated? Denominator zeros excluded? Interval notation used? All steps shown?; Rational inequality: move to one side → sign table → mark critical values → test intervals
Pause, copy the exam checklist into your book: state the domain; exclude denominator zeros from solution sets; use interval notation; show every step of the sign table; for rational inequalities always move all terms to one side first.
Quick check: When solving a rational inequality such as $\dfrac{x-1}{x+2} \ge 0$, what is the correct first move?
Worked examples · 3 in a row, reveal as you go
Consider $f(x) = x^2 - 4x + 3$. (a) Solve $f(x) \le 0$. (b) Find the largest domain containing $x = 4$ for which $f$ has an inverse, and find $f^{-1}(x)$. (c) Hence sketch $y = \sqrt{f(x)}$.
Solve $\dfrac{2x - 1}{x + 3} \ge 1$ and express your answer in interval notation.
Find the inverse of $f(x) = \dfrac{3x - 1}{x + 2}$ for $x \ne -2$. State the domain of $f^{-1}$.
Did you get this? True or false: when solving $\dfrac{x-2}{x-5} \ge 0$, the value $x = 5$ must be excluded from the solution.
Misconceptions to fix · the 4 traps that cost marks
Fill the gap: For $f(x) = x^2 - 6x + 5$, the vertex is at $x = $ , so to find an inverse containing $x = 10$ we restrict to $x \ge $ .
Activities · practice under exam conditions
Solve $\dfrac{x+1}{x-2} \le 3$ and express your answer in interval notation.
Find the inverse of $g(x) = \sqrt{2x + 5}$ and state the domain and range of $g^{-1}$.
Given $f(x) = x^2 - 2x - 3$, describe the key features of the graphs of $y = |f(x)|$ and $y = 1/f(x)$. You do not need to sketch, but identify all key points.
A curve is given parametrically by $x = 2\cos\theta$, $y = 3\sin\theta$. Eliminate the parameter and identify the curve, stating any restrictions.
Identify which Module 1 topic you find most difficult, explain why, and state one specific thing you will do to address it before the exam.
Did you get this? True or false: the domain of $f^{-1}$ is always the same as the domain of $f$.
Odd one out: Three of these are valid techniques for Module 1 exam questions. Which one is NOT?
Earlier you were asked about the most common mistake in rational inequality questions. The answer: multiplying by an expression containing $x$ without knowing its sign. If the expression is negative, the inequality flips, and students who multiply directly get the wrong solution without realising it.
The fix is simple: use a sign table every time. It takes about the same time as multiplying and is always correct.
Which topic area do you feel most confident about, and which needs more practice before the exam?
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Solve $\dfrac{2x - 1}{x + 3} \ge 1$ and express in interval notation. (3 marks)
Q2. Find the inverse of $f(x) = \dfrac{3x - 1}{x + 2}$ for $x \ne -2$. State the domain of $f^{-1}$. (3 marks)
Q3. Given $f(x) = x^2 - 2x - 3$, sketch $y = |f(x)|$ and $y = 1/f(x)$ on separate axes, showing all key features including asymptotes, intercepts, and invariant points. (4 marks)
Activity 1: 1. Move to one side: $\dfrac{x+1}{x-2}-3=\dfrac{x+1-3(x-2)}{x-2}=\dfrac{-2x+7}{x-2}\le 0$. Sign table: critical values $x=2$ (excluded), $x=3.5$. Solution: $x\in(-\infty,2)\cup[3.5,+\infty)$, wait, check: $\dfrac{-2x+7}{x-2}\le 0$ means expression $\le 0$. Numerator zero at $x=\tfrac{7}{2}$. For $x<2$: test $x=0$: $7/(-2)<0$ ✓. For $2<x<\tfrac{7}{2}$: test $x=3$: $1/1>0$ ✗. For $x>\tfrac{7}{2}$: $-ve/+ve<0$ ✓. Solution: $x\in(-\infty,2)\cup[\tfrac{7}{2},+\infty)$.
2. $g(x)=\sqrt{2x+5}$, $x\ge-\tfrac{5}{2}$. Let $y=\sqrt{2x+5}$. Swap: $x=\sqrt{2y+5}$. Square: $x^2=2y+5$, so $g^{-1}(x)=\dfrac{x^2-5}{2}$. Domain of $g^{-1}$: $x\ge0$ (range of $g$). Range of $g^{-1}$: $y\ge-\tfrac{5}{2}$.
3. $f(x)=x^2-2x-3=(x-3)(x+1)$. Zeros: $x=3,\,-1$. Vertex: $x=1$, $f(1)=-4$. For $|f(x)|$: same as $f$ outside $[-1,3]$; reflected upward inside $[-1,3]$; touches $x$-axis at $x=-1$ and $x=3$. For $1/f(x)$: vertical asymptotes at $x=-1$ and $x=3$; horizontal asymptote $y=0$; invariant points where $f(x)=\pm1$.
Q1 (3 marks): $\dfrac{2x-1}{x+3}-1=\dfrac{x-4}{x+3}\ge0$ [1]. Sign table: critical values $x=-3$ (excl.) and $x=4$ [1]. Solution: $x\in(-\infty,-3)\cup[4,+\infty)$ [1].
Q2 (3 marks): Swap, cross-multiply: $x(y+2)=3y-1$, $xy-3y=-2x-1$, $y=\dfrac{-2x-1}{x-3}$ [2]. Domain of $f^{-1}$: $x\ne3$ [1].
Q3 (4 marks): Zeros of $f$ identified as $x=-1,3$ [1]. $|f|$ sketch: correct reflection with touch-points [1]. $1/f$ asymptotes at $x=-1,3$, $y=0$ [1]. Invariant points and correct end behaviour indicated [1].
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Start the module quiz →Congratulations on completing Module 1: Further Work with Functions!
Key skills you have developed across these 15 lessons:
Ready to tackle Module 2? Head to the first lesson to begin Trigonometric Functions.
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