Understand · explore · practise
13 original multi-part pure maths questions connecting algebra, graphs, trigonometry, geometry and calculus. Separate hints and fully worked answers, with no timer.
Before you startPure 1 foundations and all Pure 2 chapters.
01 / Work across topics
This is an independently written practice set, not an official examination paper. Work at your own pace on paper. Each section has a hint and worked solution that you can open separately. Exact answers are preferred unless the question asks for a decimal.
Read the domain → form an equation → solve → check the interpretation.
k = −2; (x − 1)² = 9. Two intersections.
Points: (−2, −2) and (4, −2).
Compare the vertical distance |k + 2| from the centre with radius 3. The algebra and diagram must give the same number of intersections.
02 / Parametric curve and tangent
The parameter interval controls which branch of the Cartesian curve is present.
A curve has x = t² + 1, y = t³ − 3t, with t > 0. (a) Eliminate t and state the x domain. (b) Find dy/dx. (c) Find the tangent at t = 1.
Use the positive square root t = √(x − 1). Divide dy/dt by dx/dt.
(a) y = (x − 4)√(x − 1), x > 1. (b) dy/dx = (3t² − 3)/(2t), valid throughout t > 0. (c) At t = 1 the point is (2,−2), and the derivative is zero. The tangent is y = −2.
03 / Combine two conditions
Solve each condition and take their intersection.
Solve |2x − 1| < 5 and 3x + 2 ≥ −1 simultaneously. State which endpoints are included.
The modulus inequality becomes a double inequality.
−5 < 2x − 1 < 5 gives −2 < x < 3. The linear inequality gives x ≥ −1. Together: −1 ≤ x < 3. The left endpoint satisfies both conditions; the right endpoint fails the strict modulus inequality.
04 / A line and a circle
Pause, replay or seek freely. The notes explain the same idea and stay in view.
The circle is (x − 1)² + (y + 2)² = 9 and the line is y = k. (a) Find all k giving two intersections, one intersection or no intersection. (b) Find the intersection points when k = 0.
Substitution gives (x − 1)² = 9 − (k + 2)².
(a) Two intersections when −5 < k < 1; one when k = −5 or 1; none when k < −5 or k > 1. Equivalently, compare |k + 2| with 3. (b) When k = 0, (x − 1)² = 5, so the points are (1 − √5,0) and (1 + √5,0).
05 / Differentiate from first principles
Angle units matter in trigonometric limits.
Using the addition formula and the standard limits sin h/h → 1 and (cos h − 1)/h → 0, prove from first principles that d(cos x)/dx = −sin x. State the angle unit required.
Expand cos(x + h) before dividing by h.
[cos(x + h) − cos x]/h = cos x[(cos h − 1)/h] − sin x[sin h/h]. As h → 0 this tends to 0 − sin x = −sin x. Angles must be measured in radians; the quoted limits and derivative formula depend on that unit.
06 / Coefficients and signs
A squared parameter can leave two possibilities; an odd-power coefficient can distinguish them.
In (2 + px)⁵, the coefficient of x² is 720 and that of x³ is negative. Find p and the coefficient of x³.
The x² coefficient is 10×2³p².
80p² = 720, so p = ±3. The x³ coefficient is 10×2²p³ = 40p³. Its negative sign selects p = −3, giving coefficient −1080.
07 / A normal meets the curve again
An intersection already used to construct a line will appear again when solving its intersection equation.
Find the normal to y = x² at (1,1), then find its other intersection with the curve.
The tangent slope is 2; the normal slope is −1/2.
Normal: y − 1 = −(x − 1)/2, so y = (3 − x)/2. Substituting y = x² gives 2x² + x − 3 = (2x + 3)(x − 1) = 0. The known x = 1 root is excluded when asking for the other point, which is (−3/2,9/4).
08 / A finite sum near its limit
The inequality is strict and the number of terms must be an integer.
A geometric series has first term 12 and common ratio 3/4. Find its sum to infinity, then the least n for which its first n terms sum to more than 45.
Sₙ = 48[1 − (3/4)ⁿ]. Remember that ln(3/4) is negative.
S∞ = 12/(1 − 3/4) = 48. Sₙ > 45 requires (3/4)ⁿ < 1/16, hence n > ln(1/16)/ln(3/4) ≈ 9.638. The least n is 10; substitution gives S₉ ≈ 44.396 and S₁₀ ≈ 45.297.
09 / A compound transformation
Solve for the new input that produces the original function input.
The graph y = f(x) contains (−2,5) and has its unique minimum at (2,−1). Find the corresponding point and minimum on y = f(2(x − 3)) + 4.
An original input a occurs when 2(x − 3) = a.
A point (a,b) maps to (a/2 + 3,b + 4). Thus (−2,5) becomes (2,9), and the unique minimum becomes (4,3). The positive horizontal scale and vertical translation preserve which point is the minimum.
10 / Complete trigonometric roots
Factor in a trigonometric quantity, then find every angle in the requested interval.
Solve 2sin²θ − sinθ − 1 = 0 for 0° ≤ θ < 360°.
Let u = sinθ.
(2u + 1)(u − 1) = 0, so sinθ = 1 or −1/2. The complete solution is θ = 90°, 210°, 330°. Neither interval endpoint is a solution.
11 / Calibrate exponential decay
Use the observed ratio to determine the constant before extrapolating.
A quantity Q = 150e⁻ᵏᵗ falls to 96 at t = 4 hours. Find k exactly, Q(8), and its half-life in exact form.
e⁻⁴ᵏ = 16/25.
k = ¼ln(25/16) per hour. Q(8) = 150(16/25)² = 61.44. Half-life T satisfies e⁻ᵏᵀ = 1/2, so T = 4ln2/ln(25/16) hours. The model assumes a constant fractional decay rate.
12 / Two triangles make a quadrilateral
Separate the shared internal edge from the external boundary.
Triangle ABC has AB = AC = 5 cm and angle BAC = 120°. Point D lies on the opposite side of BC from A, with perpendicular distance 2 cm from BC. Find BC and the area of quadrilateral ABDC.
Use the cosine rule for BC, then add the two triangle areas.
BC² = 25 + 25 − 50cos120° = 75, so BC = 5√3 cm. Area ABC = ½×25sin120° = 25√3/4 cm². Area BCD = ½×5√3×2 = 5√3 cm². Total area = 45√3/4 cm².
13 / A physical quadratic model
The algebraic curve can extend beyond the interval where the model represents the object.
A trajectory has height h(x) = −0.2x² + 1.6x + 1.8 metres at horizontal distance x metres from launch. Find (a) the first ground point for x ≥ 0, (b) the maximum height, and (c) both positions at height 4 m before landing.
Complete the square: h = 5 − (x − 4)²/5.
(a) h = 0 gives x = −1 or 9; the physical landing distance is 9 m. (b) The maximum is 5 m at x = 4 m. (c) h = 4 gives (x − 4)² = 5, hence x = 4 ± √5 m. Both lie in 0 ≤ x ≤ 9.
14 / Optimise with a constraint
Eliminate one variable using volume and check that the stationary point is a minimum.
An open-topped box has square base side x metres, height h metres and volume 32 m³. Find the dimensions that minimise material area, neglecting thickness and joins.
Area is x² + 4xh. Use h = 32/x², with x > 0.
S(x) = x² + 128/x. Then S′ = 2x − 128/x² = 0 gives x³ = 64, so x = 4 m and h = 2 m. S″ = 2 + 256/x³ > 0 for all x > 0, so this is the minimum. The minimum area is 16 + 32 = 48 m².
Section 1 of 14 · Work across topics