12 What actually comes up
Three years of real questions, part by part, with the setup and values from the paper and the grade band each part sat at according to that year's marking schedule. Start with the pattern table — it is the fastest way to decide what to revise.
Across 3 years
9 questions · 45 parts 18 Achieved 27 Merit 9 ExcellenceThe pattern — what to prepare
| Topic | Appears in | Grade level | What to practise |
|---|---|---|---|
| Standard integrals: powers, ekx, trig, reciprocal trig | Every year | Achieved | Rewrite roots and fractions as powers, divide by the number in front of x, and check by differentiating back. The (a) part of every question is one of these. |
| A differential equation dy/dx = f(x) with a starting value | Every year | Achieved–Merit | Integrate, then use the given point to find c. Never assume c = 0. |
| Separating the variables | Every year | Merit–Excellence | Factorise first if you need to, put y with dy and x with dx, integrate both sides with one constant, then use the given point. |
| Writing a differential equation from words | 2023, 2024 | Merit–Excellence | “Proportional to” is “= k ×”. Two readings give two equations for k and c; subtract one from the other to find k. |
| Area under a curve, or between two or three graphs | Every year | Achieved–Excellence | Find every intersection first, decide which graph is on top, split wherever an edge changes, and count a piece below the axis as positive area. |
| Finding a constant or a limit from a definite integral | Every year | Achieved–Merit | Integrate with the letter as a limit, put both limits in (the lower one rarely gives zero), then solve. |
| Fractions: a log, a top-heavy fraction, or a bracket to a power | Every year | Achieved–Excellence | A log comes only when the top is a constant times the derivative of the bottom. Otherwise divide first, or write the bracket as a power. |
| Rewriting trig first: products to sums, squares, odd powers | Every year | Merit–Excellence | Two different angles multiplied: a product-to-sum identity. A square: a double-angle formula, or sec²x − 1 for tan²x. An odd power: keep one factor back to be du. |
| Substitution (u = the inner function) | Every year | Achieved–Excellence | Look for a factor that is a constant times the derivative of the inside, and change the limits when the integral is definite. |
| Kinematics: acceleration to velocity to displacement | Every year | Achieved–Merit | v = ∫ a dt, then s = ∫ v dt, with a constant each time from the starting values. Starting at P does not make the constant zero. |
| Numerical estimate: trapezium or Simpson’s rule | 2023, 2025 | Achieved | h is the gap between measurements. Trapezium: ends once, middles twice, times h/2. Simpson’s: ends once, odd-numbered ×4, even-numbered ×2, times h/3. |
The 2025 paper ↗ Marking schedule ↗
Question 1 — standard integrals, separation of variables
Five independent parts, no shared scenario: a standard integral, a differential equation with a boundary condition, a pair of definite integrals that pin down two constants, a trig area, and a differential equation you have to rearrange before you can separate it.
- Not tried yet: a Find int 6 sec(3x) tan(3x) dx.Worked solution → Achieved
- Not tried yet: b Solve the differential equation dy/dx = 3√x + 2/√x, given that y = 10 when x = 4.Worked solution → Achieved
- Not tried yet: c It is given that int12 (a − 6k/x2)dx = 3 and int1k 6x dx = a, where a and k are constants. Determine the possible value(s) of k.Worked solution → Merit
- Not tried yet: d The graph shows part of y = 4 sin(5x) cos(3x), shaded between the curve and the x-axis from the origin to x = π/6, where the curve comes back down to the axis. Find the shaded area under the curve between x = 0 and x = π/6.Worked solution → Merit
- Not tried yet: e Consider the differential equation y − xy − (1 + x)dy/dx = 0. Given that y = 3 when x = 0, find the value(s) of y when x = 2.Worked solution → Merit to Excellence
Question 2 — substitution-type, constants, kinematics
A power-of-a-linear integral, a rate of change integrated back to its function, a constant found from a definite integral, a kinematics chain (acceleration → velocity → displacement), and an odd-power trig area.
- Not tried yet: a Find int (10/(2x + 1)6) dx.Worked solution → Achieved
- Not tried yet: b The rate of change of a particular function, p, is given by dp/dt = 5 cos(4t). Find the function, given that p = 8 when t = π/24.Worked solution → Achieved
- Not tried yet: c Find the value of the constant k, given that int0k (1/(√(4x + 1))) dx = 1.Worked solution → Merit
- Not tried yet: d A particle’s acceleration can be modelled by a(t) = 2.4e−0.3t − 1, where a(t) is in m s⁻² and t is the time in seconds from the start of timing. Initially, at a fixed point P, the particle had a velocity of 6 m s⁻¹. How far from the point P is the particle 3 seconds after timing started?
- a(t) = 2.4e−0.3t − 1 m s⁻²
- v(0) = 6 m s⁻¹
- t = 3 s
- Not tried yet: e The graph shows y = sin3 x cos3 x, shaded between the curve and the x-axis from the origin to (π/2, 0). Find the shaded area under the curve between x = 0 and x = π/2.Worked solution → Merit to Excellence
Question 3 — trapezium rule, areas, centroid
Opens with a numerical estimate from measured depths, then a log integral left in terms of a limit, a separable differential equation, an area that must be turned into a cubic, and a balance point found from a ratio of two integrals.
- Not tried yet: a The diagram shows the cross-section of a hole dug in the ground, 20 m across the top, with straight sides. The depth of the hole is measured at both edges and every 5 metres in between. Using the trapezium rule, find an estimate for the area of the cross-section of the hole.
- strip width 5 m
- depths, left to right: 2.12, 2.32, 2.65, 2.54, 1.88 m
- Not tried yet: b Find int1k (10/(2x − 1)) dx, giving your answer in terms of k, where k is a constant and k > 1.Worked solution → Achieved
- Not tried yet: c Consider the differential equation dy/dx = (√(4y + 1))/x2. Given that y = 2 when x = 2/3, find the value(s) of y when x = 4/5.Worked solution → Merit
- Not tried yet: d The graph shows part of the curve y = (x2 + 6)/x4, where x > 0, shaded between the curve and the x-axis from x = p to x = 2p. The area of the shaded region is 9/4 units². Prove that 9p3 − 2p2 − 7 = 0.Worked solution → Merit
- Not tried yet: e Murray is planning to hang a piece of his art on a wall. Its shape is the region between the x-axis and the curve f(x) = (1/3)(x2 + 3)2, from x = 0 to x = 3. The x-value of the point to hang it from, so that it balances, is given by (int03 x f(x) dx)/(int03 f(x) dx). Use this formula to find the x-value of the hanging point.
- f(x) = (1/3)(x2 + 3)2
- 0 ≤ x ≤ 3
The 2024 paper ↗ Marking schedule ↗
Question 1 — integrals & areas
A sec–tan integral, an area under a cubic-in-a-bracket, a kinematics chain that starts 360 m from the reference point, a trig differential equation, and an area between two trig curves.
- Not tried yet: a Find int 6 sec(2x) tan(2x) dx.Worked solution → Achieved
- Not tried yet: b The graph shows the function y = 40x(5x2 − 3)3, shaded between the curve and the x-axis from x = −0.4 to the origin. The curve is above the axis there. Find the shaded area under y = 40x(5x2 − 3)3.Worked solution → Achieved
- Not tried yet: c An object’s velocity can be modelled by v(t) = 26.4∛t, where v is in m s⁻¹ and t is the time in seconds since the start of timing. Initially, the object was 360 metres from a point P. Calculate the distance that this object has travelled from the point P when it has reached a velocity of 264 m s⁻¹.
- v(t) = 26.4∛t m s⁻¹
- initial displacement 360 m from P
- v = 264 m s⁻¹
- Not tried yet: d Consider the differential equation dy/dx = 24 cos(3x) sin(x). Given that y = 6 when x = π/3, find the value(s) of y when x = π/2.Worked solution → Merit
- Not tried yet: e The graph shows the curves y = 3 sec2 x and y = 2 tan2 x. The shaded region lies between them, with y = 3 sec2 x on top, from the y-axis to the line x = 1. Find the area of the shaded region enclosed by the two curves, x = 1, and the y-axis.Worked solution → Merit to Excellence
Question 2 — expansion, constants, DE, mass
Expand-then-integrate, a lower limit recovered from a definite integral, a separable equation in y², an area between a curve and a horizontal line, and a physical integral left entirely in symbols.
- Not tried yet: a Find int (3x4 + 4)2 dx.Worked solution → Achieved
- Not tried yet: b Find the value of k, given that intk16 3√x dx = 112.Worked solution → Achieved
- Not tried yet: c Consider the differential equation dy/dx = 12y2e3x. Given that y = 0.5 when x = 0, find the value of y when x = 1/3.Worked solution → Merit
- Not tried yet: d The graph shows part of y = sin2 x together with the line y = 1, which touches the curve at x = −π/2 and x = π/2. The region between them is shaded. Find the shaded area enclosed between y = sin2 x, y = 1, x = −π/2 and x = π/2.Worked solution → Merit
- Not tried yet: e The mass M of a spherical object of radius p can be approximated by M = int0p 4π r2 (a/(1 + br3)) dr, where a, b and p are all positive constants. Using this formula, find an expression for the mass M of a spherical object, giving your answer in terms of a, b, p and π.
- M = int0p 4π r2 (a/(1 + br3)) dr
- a, b, p > 0
Question 3 — exponentials, a coffee, equal areas
Two exponential terms to integrate, a log-form differential equation, an upper limit that appears on both sides, a bisected cosine area, and Newton’s law of cooling built from scratch.
- Not tried yet: a Find int (e2x + 3/e4x)dx.Worked solution → Achieved
- Not tried yet: b Solve the differential equation dy/dx = 5/(4x − 3), given that y = 10 when x = 1.Worked solution → Achieved
- Not tried yet: c Find the value of m, given that int−1m ((4x + 5)/(2x + 3)) dx = 2m.Worked solution → Merit
- Not tried yet: d The graph shows y = 2 cos(x/2) from x = 0 to x = π, where it meets the x-axis. The region under it is split by the line x = k into Area A, from 0 to k, and Area B, from k to π. Find the value of k so that the shaded Area A will be equal to the shaded Area B.Worked solution → Merit
- Not tried yet: e A teacher makes a cup of coffee at the start of interval and leaves it in the staff room, where the temperature is 18 °C. After 30 minutes the coffee is 50 °C. After a further one hour it has cooled to 30 °C. The rate at which the temperature changes at any instant is proportional to the difference between the temperature of the coffee, N, and the temperature of the room. Write a differential equation that models this situation, and then solve it to calculate the temperature of the cup of coffee when it was made.
- room 18 °C
- 50 °C after 30 min
- 30 °C after a further 60 min
The 2023 paper ↗ Marking schedule ↗
Question 1 — a rational integral, kinematics, two bounded areas
A linear-plus-reciprocal integral, a velocity in km hr⁻¹ integrated to a distance from a point 3 km away, an area between a root and a parabola, a separable equation, and an area bounded by three different curves at once.
- Not tried yet: a Find int (3x + 2 + 1/(3x + 2))dx.Worked solution → Achieved
- Not tried yet: b An object’s velocity can be modelled by v(t) = sec2 t, where v is in km hr⁻¹ and t is the time in hours from the start of timing. Initially the object was 3 km from a point P. Find the distance of this object from the point P after π/4 hours.
- v(t) = sec2 t km hr⁻¹
- initial displacement 3 km from P
- t = π/4 hours
- Not tried yet: c The graph shows y = √x and 8y = x2, with the region between them shaded, from the origin to the point where they cross again. Find the shaded area enclosed between y = √x and 8y = x2.Worked solution → Merit
- Not tried yet: d Consider the differential equation dy/dx = y(2x − 3x2). Given that y = 1 when x = 2, find the value(s) of y when x = 1.Worked solution → Merit
- Not tried yet: e The shaded region is bounded by three graphs: y2 = 10 − x, y = 3x and y = −sin((π x)/10). Its left edge is y = 3x, from the origin up to where it meets the top half of the parabola. Its top edge is that half of the parabola, curving down to (10, 0). Its bottom edge is the sine curve, which dips below the x-axis between the origin and (10, 0). Find the area of the shaded region.
- y2 = 10 − x
- y = 3x
- y = −sin((π x)/10)
Question 2 — an exponential integral, a rate constant, a chocolate factory
An exponential integral, a negative-power differential equation, an upper limit that appears on both sides, a trig quotient whose top is a multiple of the derivative of its bottom, and a proportional-to-the-square model built from scratch.
- Not tried yet: a Find int 4e2x − 1 dx.Worked solution → Achieved
- Not tried yet: b Solve the differential equation dy/dx = (4x + 1)−1/2, where x ≥ 0, given that when x = 6, y = 7.5.Worked solution → Achieved
- Not tried yet: c Find the value of k, given that int2k ((6x − 3)/(2x − 3)) dx = 3k.Worked solution → Merit
- Not tried yet: d Find int ((cos 2x + sin 2x)/(cos 2x − sin 2x)) dx.Worked solution → Merit
- Not tried yet: e A cake factory pumps liquid chocolate out of a container so that the rate of change of the volume remaining is proportional to the square of the volume remaining. After one hour of use, p litres remain, where p is a positive constant. After a further one hour, only (4/5) p litres remain. Write a differential equation modelling this, and solve it to calculate how much liquid chocolate was in the container at the start of the day, giving your answer in terms of p.
- rate ∝ (volume remaining)²
- p litres after 1 hour
- (4/5) p litres after 2 hours
Question 3 — Simpson’s rule, a trig area, a factorised differential equation
Opens with Simpson’s Rule on five measured offsets from a garden path, then a root quotient, a product-of-sines area, an exponential-quotient acceleration integrated to a velocity, and a differential equation that has to be factorised before it will separate.
- Not tried yet: a A garden designer measures a section of garden from a straight path: offsets of 10, 13, 16, 15 and 14 metres, taken at 3-metre intervals along the path. Using these measurements, and Simpson’s Rule, find an approximation of the area of the garden section.
- offsets 10, 13, 16, 15, 14 m
- 3 m intervals
- Not tried yet: b Find int ((√x − 3)/√x) dx.Worked solution → Achieved
- Not tried yet: c The graph shows y = 5 sin(3x) sin(x), shaded between the curve and the x-axis from the origin to x = π/3, where it meets the axis. Find the shaded area under y = 5 sin(3x) sin(x).Worked solution → Merit
- Not tried yet: d An object’s acceleration can be modelled by a = e2t/(4e2t − 3), where t ≥ 0, a is in m s⁻² and t is the time in seconds from the start of timing. At t = 0 the object had a velocity of 5 m s⁻¹. Find the object’s velocity when t = 4 seconds.
- a = e2t/(4e2t − 3) m s⁻²
- v(0) = 5 m s⁻¹
- t = 4 s
- Not tried yet: e Consider the differential equation (1 − x2)(1 + y)dy/dx + (1 − x)(1 − y2) = 0. Given that y = 0 when x = 2, find the value(s) of y when x = 6.Worked solution → Merit to Excellence
Read the last column as a to-do list. In every question of 2023–2025, (a) and (b) are Achieved, (c) and (d) Merit, and (e) Merit–Excellence, the same ladder as 91578. Two (e) parts, 2025 and 2024 Question Two, give no Achieved credit at all: their first credit is Merit.
Practice for this standard covers the whole topic — flashcards and quiz.