Further Functions — Worked Solutions (Preliminary Maths Extension 1)
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Worked examples for Preliminary Maths Extension 1 further work with functions. Each shows where the marks are awarded, the key idea, and the full solution explained by your choice of tutor — Stella, Ella or Cassie.
In short: full step-by-step worked solutions for Preliminary Maths Extension 1 — Further Functions. Every question is worked through with the method and reasoning shown, so you can check how to get the answer, not just the final result.
How to use these
Try each question first, then check your working. Use the tutor tabs to read the full solution in the style that suits you: Stella is direct and challenging, Ella is warm and explains the why, and Cassie is concise and analytical.
Example 1 — Rational inequality
Question
Solve $\dfrac{x+1}{x-2} \geq 3$.
Solution
Never just multiply both sides by $x-2$ — its sign is unknown. Bring everything to one side instead.
$\dfrac{x+1}{x-2} - 3 \geq 0 \;\Rightarrow\; \dfrac{x+1 - 3(x-2)}{x-2} \geq 0 \;\Rightarrow\; \dfrac{-2x+7}{x-2} \geq 0$.
The critical values are $x = \tfrac{7}{2}$ (numerator zero) and $x = 2$ (denominator zero, excluded).
Test the sign on each interval: for $x = 2$ to $\tfrac{7}{2}$ the expression is positive, elsewhere negative.
So $2 < x \leq \tfrac{7}{2}$. Note $x = 2$ is open and $x = \tfrac{7}{2}$ is closed.
The trap here is the denominator. If you multiply both sides by $x-2$ you don't know whether to flip the inequality, because $x-2$ could be positive or negative. So we move everything to one side and combine into a single fraction.
$\dfrac{x+1}{x-2} - 3 \geq 0$. Common denominator: $\dfrac{x+1 - 3(x-2)}{x-2} = \dfrac{x + 1 - 3x + 6}{x-2} = \dfrac{-2x+7}{x-2} \geq 0$.
Now find where the fraction can change sign: the numerator is zero at $x = \tfrac{7}{2}$, and the denominator is zero at $x = 2$ — but $x = 2$ can never be a solution since we can't divide by zero.
Checking a value in each region (say $x = 3$ gives $\tfrac{1}{1} > 0$, good) shows the fraction is $\geq 0$ between the critical values, so $2 < x \leq \tfrac{7}{2}$.
The reason $x = 2$ stays excluded is that the original expression isn't even defined there — the why behind the open circle.
Move to one side; never multiply by an unknown sign.
- $\dfrac{x+1}{x-2} - 3 \geq 0 \Rightarrow \dfrac{-2x+7}{x-2} \geq 0$
- Critical values: $x = \tfrac{7}{2}$ (num), $x = 2$ (denom, excluded)
- Sign test: positive on $(2, \tfrac{7}{2}]$
Solution: $2 < x \leq \tfrac{7}{2}$.
Where the marks go
- 1 mark: Combines into a single fraction $\dfrac{-2x+7}{x-2} \geq 0$
- 1 mark: Identifies critical values $x = 2$ and $x = \tfrac{7}{2}$
- 1 mark: Correct solution $2 < x \leq \tfrac{7}{2}$ with $x = 2$ excluded
Key idea
For a rational inequality, move everything to one side and analyse the sign — never multiply by an expression whose sign you don't know.
Example 2 — Absolute value and parametric
Question
(a) Solve $|2x - 1| = x + 4$.
(b) A curve is given parametrically by $x = 2t$, $y = t^2 - 1$. Find its Cartesian equation.
Solution
(a) Split on the sign of the inside. Either $2x - 1 = x + 4$, giving $x = 5$, or $2x - 1 = -(x+4)$, giving $3x = -3$, so $x = -1$.
Check both against $x + 4 \geq 0$ (the right side must be non-negative): $x = 5$ gives $9 \geq 0$ ✓, $x = -1$ gives $3 \geq 0$ ✓. Both valid: $x = -1, 5$.
(b) Eliminate $t$. From $x = 2t$, $t = \tfrac{x}{2}$. Substitute: $y = \left(\tfrac{x}{2}\right)^2 - 1 = \tfrac{x^2}{4} - 1$.
Always verify absolute-value answers — squaring or splitting can introduce solutions that fail the original.
(a) An absolute value equals its inside or the negative of its inside, so we get two cases. Case 1: $2x - 1 = x + 4 \Rightarrow x = 5$. Case 2: $2x - 1 = -(x + 4) = -x - 4 \Rightarrow 3x = -3 \Rightarrow x = -1$.
Because the left side $|2x-1|$ can never be negative, the right side $x+4$ must be $\geq 0$. Both $x = 5$ and $x = -1$ make $x + 4$ positive, so both are genuine solutions: $x = -1, 5$.
(b) Parametric equations describe $x$ and $y$ through a helper variable $t$; to get the Cartesian form we eliminate $t$. From $x = 2t$ we have $t = \tfrac{x}{2}$, and substituting into $y = t^2 - 1$ gives $y = \tfrac{x^2}{4} - 1$ — a parabola.
Eliminating the parameter just means writing one variable's story directly in terms of the other.
(a) Two cases:
- $2x - 1 = x + 4 \Rightarrow x = 5$
- $2x - 1 = -(x+4) \Rightarrow x = -1$
- Both satisfy $x + 4 \geq 0$ → $x = -1, 5$
(b) Eliminate $t$:
- $t = \tfrac{x}{2}$
- $y = \left(\tfrac{x}{2}\right)^2 - 1 = \tfrac{x^2}{4} - 1$
Where the marks go
- 1 mark: Sets up both cases for the absolute value equation
- 1 mark: Correct solutions $x = -1$ and $x = 5$ (both verified)
- 1 mark: Makes $t$ the subject from $x = 2t$
- 1 mark: Correct Cartesian equation $y = \tfrac{x^2}{4} - 1$
Key idea
$|A| = B$ splits into $A = B$ and $A = -B$ (check $B \geq 0$); a parametric curve becomes Cartesian by eliminating the parameter.
Frequently asked questions
Step-by-step solutions to Further Functions questions in Preliminary Maths Extension 1, with the full method shown for each — so you can follow the reasoning, not just the final answer.
Attempt each question yourself first, then compare your method — not just your answer — against the worked solution. The aim is to learn the approach so you can handle unfamiliar questions, rather than memorising answers.
Yes — they follow the NESA syllabus for Preliminary Maths Extension 1, using the methods and notation expected in exams and assessments.