You have journeyed through rates of change, limits, derivatives, optimisation, integration, and applications. In this final lesson, we pull everything together: the essential formulae, the most common exam traps, and a set of mixed practice questions designed to test your readiness for the Module 3 assessment.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Before you begin this review, take a moment to reflect. What topic in Module 3 do you feel most confident about? What topic still makes you hesitate? Being honest about your strengths and gaps is the first step toward effective revision.
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Wrong: √(a + b) = √a + √b.
Right: The square root of a sum is not the sum of square roots; √(a+b) cannot be simplified this way.
📚 Core Content
Make sure you work through each example step-by-step before checking the solution. Understanding the reasoning is just as important as getting the right answer.
Even if your final answer is wrong, method marks are awarded for correct setup, correct derivatives/integrals, and correct substitution. Always show full working.
Use this checklist to identify any last-minute gaps:
Earlier you reflected on your strongest and weakest topics in Module 3.
Take a moment now to update that reflection. After working through the mixed practice questions, has your confidence changed? What will you focus on in your final revision before the exam?
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Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Mixed Practice
1. (a) Find the derivative of $f(x) = (2x + 1)^3$ using the chain rule. 1 MARK
(b) Find the gradient of the tangent to $y = x^3 - 3x^2 + 2$ at $x = 1$. 2 MARKS
(c) Find and classify the stationary points of $y = x^3 - 3x^2 + 2$. 3 MARKS
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2. A rectangular box with a square base and open top is to have a volume of 32 cm$^3$. Let the side length of the square base be $x$ cm and the height be $h$ cm. 4 MARKS
(a) Show that the surface area is $S = x^2 + \frac{128}{x}$.
(b) Find the dimensions that minimise the surface area.
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3. Evaluate the following integrals. 4 MARKS
(a) $\int (4x^3 - 2x + 5) \, dx$
(b) $\int_1^3 \frac{2}{x^3} \, dx$
(c) $\int_0^2 (x^2 - 2x) \, dx$
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4. The curves $y = x^2$ and $y = 4 - x^2$ intersect at $x = -\sqrt{2}$ and $x = \sqrt{2}$. 5 MARKS
(a) Find the area enclosed by the two curves.
(b) Find the volume of the solid formed when this region is rotated about the $x$-axis.
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5. A particle moves in a straight line with velocity $v(t) = 3t^2 - 12t + 9$, where $t$ is in seconds and $v$ is in m/s. 5 MARKS
(a) Find the acceleration of the particle at $t = 2$.
(b) Find when the particle is instantaneously at rest.
(c) Find the total distance travelled by the particle in the first 3 seconds.
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Answer in your workbook.
1. B — Chain rule: $4(3x+2)^3 \\cdot 3 = 12(3x+2)^3$.
2. C — Factor: $\frac{(x-2)(x+2)}{x-2} = x+2$, which approaches $4$ as $x o 2$.
3. A — $\\int 2x^3 , dx = \frac{x^4}{2}$ and $\\int (-3x) , dx = -\frac{3x^2}{2}$.
4. B — $y'' = 6x - 6$, so $y''(2) = 6 > 0$, indicating a local minimum.
5. C — Intersections at $x = 0$ and $x = 1$. Area = $int_0^1 (x - x^2) , dx = \frac{1}{6}$.
6. B — $h = 1$, ordinates $0, 1, 4$. Estimate = $\frac{1}{2}(0 + 2 + 4) = 3$.
7. B — $V = \\pi int_0^4 x , dx = \\pi \bigl[\frac{x^2}{2}\bigr]_0^4 = 8pi$.
Q1 (4 marks): (a) $f'(x) = 3(2x + 1)^2 \cdot 2 = 6(2x + 1)^2$ [1]. (b) $y' = 3x^2 - 6x$, so at $x = 1$, $m = 3 - 6 = -3$ [2]. (c) $3x^2 - 6x = 3x(x - 2) = 0 \Rightarrow x = 0$ or $x = 2$ [1]. $y(0) = 2$, $y(2) = -2$ [1]. $y'' = 6x - 6$, so $(0, 2)$ is a local maximum and $(2, -2)$ is a local minimum [1].
Q2 (4 marks): (a) Volume $= x^2h = 32 \Rightarrow h = \frac{32}{x^2}$. Surface area $S = x^2 + 4xh = x^2 + 4x \cdot \frac{32}{x^2} = x^2 + \frac{128}{x}$ [2]. (b) $\frac{dS}{dx} = 2x - \frac{128}{x^2} = 0 \Rightarrow 2x^3 = 128 \Rightarrow x = 4$ [1]. $h = \frac{32}{16} = 2$ [1]. Dimensions: $4$ cm by $4$ cm by $2$ cm [1].
Q3 (4 marks): (a) $x^4 - x^2 + 5x + C$ [1]. (b) $\int_1^3 2x^{-3} \, dx = \bigl[-x^{-2}\bigr]_1^3 = -\frac{1}{9} + 1 = \frac{8}{9}$ [2]. (c) $\bigl[\frac{x^3}{3} - x^2\bigr]_0^2 = \frac{8}{3} - 4 = -\frac{4}{3}$ [1].
Q4 (4 marks): (a) Sketch shows upward parabola $y = 3x^2$ intersecting horizontal line $y = 12$ at $x = -2$ and $x = 2$ [1]. (b) Area $= \int_{-\sqrt{2}}^{\sqrt{2}} (4 - 2x^2) \, dx = \bigl[4x - \frac{2x^3}{3}\bigr]_{-\sqrt{2}}^{\sqrt{2}} = \frac{16\sqrt{2}}{3}$ [3].
Module Review and Exam Preparation
Tick when you've finished all activities and checked your answers.