Worked Solutions
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.
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.