Westminster International School
Form 3 Mathematics — Homework Answers
Term 2, 2026 · Cambridge IGCSE® Mathematics 0580 — Extended
Worked solutions:
- Each solution follows the formula → values → substitute → arithmetic → answer pattern.
- Supplement marks Extended-syllabus-only content (E2.9 / E2.13).
- All answers verified independently before writing.
Section A — Speed-Time Graphs
33.
[4 marks]
The speed-time graph shows information about a car journey.
(i) Complete these statements about the journey.
Section cannot be correct.
Section shows constant speed.
Section shows deceleration.
(ii) Supplement
Section A shows an acceleration of m/s2.
(iii) Supplement
The distance travelled in the first 30 seconds of the journey is m.
ANSWER
(i) Completing the statements
Section D cannot be correct.
Section B shows constant speed.
Section C shows deceleration.
Why D cannot be correct: Section D has a positive gradient — speed is
increasing. Once a vehicle is braking (decelerating in Section C), its speed can only
decrease. A section showing speed increasing during the braking phase cannot be physically correct
for this journey.
B shows constant speed because the gradient is zero (horizontal line).
C shows deceleration because speed is decreasing (negative gradient).
B shows constant speed because the gradient is zero (horizontal line).
C shows deceleration because speed is decreasing (negative gradient).
(ii) Acceleration of Section A Supplement
-
Formula: $$a = \frac{\Delta v}{\Delta t}$$
-
Read values from graph: speed increases from $0$ to $10$ m/s; time taken $= 30 – 0 = 30$ s.
-
Substitute: $$a = \frac{10 – 0}{30 – 0} = \frac{10}{30}$$
-
Simplify (divide numerator and denominator by 10): $$a = \frac{1}{3} \text{ m/s}^2$$
Acceleration of Section A $= \dfrac{1}{3}$ m/s2 $\approx 0.333$ m/s2
Key idea: On a speed-time graph, acceleration = gradient of the line.
For a straight section, use any two clear data points: $a = \dfrac{v_2 – v_1}{t_2 – t_1}$.
A positive gradient means acceleration; a negative gradient means deceleration.
(iii) Distance in first 30 seconds Supplement
-
Principle: Distance = area under the speed-time graph.
-
Identify the shape: Section A (0 to 30 s) forms a triangle with base $= 30$ s and height $= 10$ m/s.
-
Formula for area of a triangle: $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$
-
Substitute: $$d = \frac{1}{2} \times 30 \times 10$$
-
Arithmetic: $\frac{1}{2} \times 30 = 15$, then $15 \times 10 = 150$ m.
Distance in first 30 s $= \mathbf{150}$ m
Key idea: Area under a speed-time graph = distance travelled.
The shape here is a triangle (the car accelerates from rest), so use
$d = \frac{1}{2} \times \text{base} \times \text{height}$.
Do not confuse with a rectangle (which applies to a constant-speed section).
35.
[3 marks]
Supplement
The speed-time graph shows information about a different journey.
Calculate the total distance travelled.
ANSWER
-
Principle: Total distance = area under the speed-time graph. Split the graph into two simple shapes.
-
Annotated graph — shaded regions show the two areas:
-
Triangle area (0 to 30 s): $$\text{area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 30 \times 12$$ $\frac{1}{2} \times 30 = 15$, then $15 \times 12 = 180$ m.
-
Rectangle area (30 s to 90 s): $$\text{area} = \text{width} \times \text{height} = (90-30) \times 12 = 60 \times 12 = 720 \text{ m}$$
-
Total distance: $$d = 180 + 720 = 900 \text{ m}$$
Total distance $= \mathbf{900}$ m
Method: Area under a speed-time graph = distance travelled (E2.9.4).
Always split the graph into triangles and/or rectangles and sum their areas.
Triangle: $\frac{1}{2} bh$. Rectangle: $lw$.
Section B — Functions
36.
[2 marks]
Supplement
$f(x) = 7x – 8$
Find $f^{-1}(x)$.
ANSWER
-
Write $f(x)$ as $y$: $$y = 7x – 8$$
-
Make $x$ the subject — add 8 to both sides: $$y + 8 = 7x$$
-
Divide both sides by 7: $$x = \frac{y + 8}{7}$$
-
Swap $x$ and $y$ (replace $y$ with $x$): $$f^{-1}(x) = \frac{x + 8}{7}$$
$f^{-1}(x) = \dfrac{x + 8}{7}$
Method for inverse functions (E2.13):
Write $y = f(x)$, rearrange to make $x$ the subject, then swap $x$ and $y$.
Check: $f(f^{-1}(x)) = x$. Here: $f\!\left(\dfrac{x+8}{7}\right) = 7 \cdot \dfrac{x+8}{7} – 8 = x + 8 – 8 = x$ ✓
37.
[4 marks]
Supplement
$f(x) = x^2 – 25$ and $g(x) = x + 4$
Solve $fg(x + 1) = gf(x)$.
ANSWER
-
Find $g(x+1)$ by substituting $(x+1)$ into $g$: $$g(x+1) = (x+1) + 4 = x + 5$$
-
Find $fg(x+1) = f(x+5)$ by substituting $(x+5)$ into $f$: $$f(x+5) = (x+5)^2 – 25$$ Expand using $(a+b)^2 = a^2 + 2ab + b^2$: $$(x+5)^2 = x^2 + 2(x)(5) + 5^2 = x^2 + 10x + 25$$ $$\therefore\; fg(x+1) = x^2 + 10x + 25 – 25 = x^2 + 10x$$
-
Find $f(x)$: $$f(x) = x^2 – 25$$
-
Find $gf(x) = g(x^2 – 25)$ by substituting $f(x)$ into $g$: $$g(x^2 – 25) = (x^2 – 25) + 4 = x^2 – 21$$
-
Set $fg(x+1) = gf(x)$: $$x^2 + 10x = x^2 – 21$$
-
Subtract $x^2$ from both sides: $$10x = -21$$
-
Divide by 10: $$x = -\frac{21}{10} = -2.1$$
$x = -2.1$
Check:
$fg(-2.1+1) = f(g(-1.1)) = f(2.9) = 2.9^2 – 25 = 8.41 – 25 = -16.59$.
$gf(-2.1) = g(-2.1^2 – 25) = g(4.41-25) = g(-20.59) = -20.59 + 4 = -16.59$ ✓
Key idea (E2.13): $fg$ means “first apply $g$, then apply $f$” : $fg(x) = f(g(x))$. Do not confuse with $gf(x) = g(f(x))$.
Key idea (E2.13): $fg$ means “first apply $g$, then apply $f$” : $fg(x) = f(g(x))$. Do not confuse with $gf(x) = g(f(x))$.
38.
[2 marks]
Supplement
$f(x) = 4x + 3$ and $g(x) = 5x – 4$
$fg(x) = 20x + p$
Find the value of $p$.
ANSWER
-
Write $fg(x) = f(g(x))$ — substitute $g(x) = 5x – 4$ into $f$: $$fg(x) = f(5x – 4)$$
-
Apply $f$ — substitute $(5x-4)$ in place of $x$ in $f(x) = 4x + 3$: $$f(5x-4) = 4(5x – 4) + 3$$
-
Expand and simplify: $$= 20x – 16 + 3 = 20x – 13$$
-
Compare with $fg(x) = 20x + p$: $$20x – 13 = 20x + p \implies p = -13$$
$p = -13$
Method (E2.13): For $fg(x)$, replace every $x$ in $f(x)$ with the
expression $g(x)$, then expand fully before comparing coefficients.
Section C — Applied Problem
39.
Supplement
The speed-time graph shows information about a car journey.
(a)
Find the deceleration of the car between 60 and 70 seconds.
ANSWER
(a) Deceleration between 60 s and 70 s
-
Formula: deceleration = magnitude of the gradient of the speed-time graph $$\text{deceleration} = \frac{\Delta v}{\Delta t}$$
-
Read values from graph: speed decreases from $20$ m/s to $0$ m/s; time interval $= 70 – 60 = 10$ s.
-
Substitute: $$\text{deceleration} = \frac{20 – 0}{70 – 60} = \frac{20}{10}$$
-
Calculate: $$\text{deceleration} = 2 \text{ m/s}^2$$
Deceleration $= \mathbf{2}$ m/s2
Key idea (E2.9): On a speed-time graph, deceleration = magnitude of the
(negative) gradient. Use $\dfrac{\Delta v}{\Delta t}$ where $\Delta v$ is the decrease in speed
and $\Delta t$ is the time taken. Quote deceleration as a positive value.
