Understand · explore · practise
14 original multi-part pure maths questions on functions, algebra, geometry, integration, series and modelling, with learner-controlled hints, worked solutions and a modulus graph.
Before you startPure 1 foundations and all Pure 2 chapters.
01 / Make each restriction explicit
This independently written cumulative set connects methods from across pure maths. It is not an official examination paper. Work on paper at your own pace; hints and solutions remain available separately. In modelling questions, explain why the chosen solution fits the situation.
Check inputs, signs, units, bounds and physical meaning.
x = 0; f(x) = −1; displayed output = −1.
Original range: [−1, 3].
The faint dashed graph keeps the original visible for comparison. Modulus reflects only negative outputs in the x axis; the x coordinate stays unchanged.
02 / Fit a quadratic
Vertex form uses the known turning point directly.
A quadratic has vertex (3,−2) and passes through (1,6). Find its equation and both x intercepts.
Write y = a(x − 3)² − 2.
6 = 4a − 2 gives a = 2. Hence y = 2(x − 3)² − 2. Setting y = 0 gives (x − 3)² = 1, so the intercepts are (2,0) and (4,0).
03 / Perpendicular line and area
Find the intercept before using it as a triangle vertex.
A = (1,2) and B = (5,4). The line through A perpendicular to AB meets the x axis at C. Find C and the area of ABC.
The slope AB is 1/2; the perpendicular slope is −2.
The perpendicular is y − 2 = −2(x − 1), or y = −2x + 4, so C = (2,0). Vectors AB = (4,2), AC = (1,−2) are perpendicular; their lengths are 2√5 and √5. Area = ½(2√5)(√5) = 5 square units.
04 / Inverse domain and range
Swap input and output only after identifying the original range.
For f(x) = e²ˣ − 3, x ∈ ℝ, find f⁻¹, its domain and range, and f⁻¹(1).
Solve y + 3 = e²ˣ for x.
f⁻¹(x) = ½ln(x + 3), with domain x > −3 and range ℝ. The original function is strictly increasing and has range (−3,∞). At input 1, the inverse gives ½ln4 = ln2.
05 / Correct a logarithm solution
A logarithm requires each original argument to be positive.
Solve log₂(x − 1) + log₂(x + 1) = 3. A student writes log₂(2x) = 3; identify the error and reject any invalid roots of the correct equation.
The sum of logarithms is the logarithm of a product, not a sum.
The original domain is x > 1. Correctly, log₂[(x − 1)(x + 1)] = 3, so x² − 1 = 8 and x = ±3. Only x = 3 is in the domain. The student incorrectly replaced the product by a sum.
06 / Read a piecewise graph
Pause, replay or seek freely. The notes explain the same idea and stay in view.
f(x) = −x − 1 for −4 ≤ x ≤ 0, and f(x) = x/2 − 1 for 0 < x ≤ 4. Find (a) the range, (b) all zeros, (c) all solutions of f(x) = 1, and (d) the range of |f(x)|.
Check values at the joining point and both endpoints; solve on each branch separately.
(a) The minimum is −1 at x = 0 and maximum is 3 at x = −4, so range [−1,3]. (b) The zeros are −1 and 2. (c) The solutions are −2 and 4. (d) The modulus range is [0,3]; the negative portion folds upwards while zero remains attainable.
07 / Factor a cubic completely
One known root reduces the remaining task to a quadratic.
Show x = 1 is a root of p(x) = x³ − 2x² − 5x + 6. Factor p completely and state all real roots.
Use the factor theorem, then divide by x − 1.
p(1) = 1 − 2 − 5 + 6 = 0. Division gives x² − x − 6 = (x − 3)(x + 2). Hence p(x) = (x − 1)(x − 3)(x + 2); roots −2, 1 and 3.
08 / A physical length from area
An algebraic root can fail the positivity required of a side length.
A triangle has adjacent sides x cm and (x + 1) cm, with included angle 30° and area 3 cm². Find x and the length of its third side.
Area = ½x(x + 1)sin30°. Use the cosine rule afterwards.
x(x + 1)/4 = 3, so x² + x − 12 = (x + 4)(x − 3) = 0. The valid length is x = 3, giving adjacent sides 3 and 4 cm. The third side is √(25 − 12√3) cm by the cosine rule.
09 / A restricted parametric circle
The parameter interval specifies an arc, not the whole circle.
x = −1 + 4cos t, y = 2 + 4sin t, for −π/6 ≤ t ≤ π/2. Find the circle equation, both endpoints and the traced arc length.
The angular change is 2π/3 radians.
(x + 1)² + (y − 2)² = 16. The start point is (−1 + 2√3,0), and the end point is (−1,6). The arc runs anticlockwise as t increases, with length 4(2π/3) = 8π/3.
10 / Improper fraction to a series
Divide before partial fractions, then intersect the convergence conditions.
Express (x² + 2x + 4)/[(x + 1)(x + 2)] in partial fractions, expand through x², and give the interval of validity.
The polynomial part is 1. The proper remainder numerator is −x + 2.
The expression is 1 + 3/(x + 1) − 4/(x + 2). Expanding gives 1 + 3(1 − x + x² + …) − 2(1 − x/2 + x²/4 + …) = 2 − 2x + (5/2)x² + … . Both series converge when |x| < 1; the tighter condition comes from 1/(1 + x).
11 / General midpoint proof
A proof must work for arbitrary vertices, rather than one convenient numerical shape.
In a convex quadrilateral ABCD, M, N, P, Q are the midpoints of AB, BC, CD, DA respectively. Prove MNPQ is a parallelogram using position vectors a, b, c, d.
Subtract the midpoint positions to compare opposite directed sides.
m = (a + b)/2, n = (b + c)/2, p = (c + d)/2, q = (d + a)/2. Hence MN = n − m = (c − a)/2 = p − q = QP. Also NP = (d − b)/2 = q − m = MQ. Both pairs of opposite sides are equal and parallel, establishing the parallelogram.
12 / Approximation against an exact integral
The denominator in percentage error is the exact value.
Use the trapezium rule with two equal strips to estimate ∫₀¹xeˣ dx. Find the exact integral by parts and the percentage error of the estimate.
The ordinates are 0, ½√e and e.
With width 1/2, T = ¼(√e + e) ≈ 1.091750775. By parts, ∫xeˣ dx = xeˣ − eˣ + C, giving an exact integral of 1. The estimate is high by approximately 9.175%. Since (xeˣ)″ = (x + 2)eˣ > 0 on [0,1], the chords lie above the curve.
13 / Compare two deposit patterns
These are sums of deposits only; no interest is included.
Plan A deposits £200 in year 1 and increases the annual deposit by £20 each year. Plan B deposits £200 in year 1 and increases each annual deposit by 5%. Find the total deposited over ten years for each plan.
Plan A is arithmetic; Plan B is geometric. Both include the first year as term 1.
A: S₁₀ = 10[2×200 + 9×20]/2 = £2900. B: S₁₀ = 200(1.05¹⁰ − 1)/0.05 ≈ £2515.58. Under these deposit rules, A totals about £384.42 more over ten years. This comparison does not include interest or returns.
14 / Reciprocal trigonometric model
Prove the denominator stays positive before multiplying an inequality by it.
For 0 ≤ t ≤ 2π, let h(t) = 12/[6 + 3cos t + 4sin t]. Put the denominator in R form, find the range of h, and solve h ≥ 4.
Let α = arctan(4/3). Then 3cos t + 4sin t = 5cos(t − α).
The denominator is 6 + 5cos(t − α), with range [1,11], so it is always positive. Thus h has range [12/11,12]. The inequality h ≥ 4 is equivalent to cos(t − α) ≤ −3/5. Since arccos(−3/5) = π − α, the interval is π ≤ t ≤ π + 2α. Both endpoints are included.
15 / Exact and numerical stationary points
Use an exact solution when available; a numerical update can be checked against it.
h(t) = 20 − 2e⁻ᵗ − eᵗ for 0 ≤ t ≤ 2. Find its maximum exactly. Then apply one Newton step from t₀ = 0.3 to F(t) = e²ᵗ − 2 and compare with the exact stationary time.
h′ = 2e⁻ᵗ − eᵗ; h″ is negative.
h′ = 0 gives e²ᵗ = 2, so t = ½ln2 ≈ 0.346573590. Since h″ = −2e⁻ᵗ − eᵗ < 0 throughout, this is the maximum, with value 20 − 2√2. Newton gives t₁ = 0.3 − (e⁰·⁶ − 2)/(2e⁰·⁶) ≈ 0.348811636. It is close but not exact; one step alone does not certify any requested number of decimal places.
Section 1 of 15 · Make each restriction explicit