Problem Solving in Pure Mathematics

Problem solving in pure mathematics requires you to connect different mathematical topics, apply knowledge in unfamiliar contexts, and construct solutions to problems that do not have an immediately obvious method. This is assessed under AO3 (Solve problems within mathematics and in other contexts) in the AQA A-Level Mathematics specification (7357), and typically accounts for around 10% of the total marks across all three papers.

AO3 questions are designed to be unstructured — they do not tell you which technique to use. You must identify the relevant mathematical concepts, select appropriate methods, and link ideas across different areas of the specification.

---

What Makes a Problem "Problem Solving"?

A problem-solving question is characterised by one or more of the following:

• It connects two or more topics (e.g., trigonometry and calculus, algebra and coordinate geometry).
• It requires you to interpret information and decide on an approach.
• The method is not immediately obvious from the question.
• It may involve modelling a real-world situation mathematically.
• It may require you to work backwards from a given result.

---

Connecting Topics: Examples

Example 1: Algebra meets Coordinate Geometry

Problem: The line y = 2x + k is tangent to the circle x² + y² = 5. Find the value(s) of k.

Solution:

Substitute y = 2x + k into x² + y² = 5:

x² + (2x + k)² = 5
x² + 4x² + 4kx + k² = 5
5x² + 4kx + (k² − 5) = 0

For the line to be a tangent, there must be exactly one intersection point, so the discriminant equals zero:

(4k)² − 4(5)(k² − 5) = 0
16k² − 20k² + 100 = 0
−4k² + 100 = 0
k² = 25
k = ±5

Therefore k = 5 or k = −5. ∎

This problem connects simultaneous equations, the discriminant, and circle geometry.

---

Example 2: Calculus meets Algebra

Problem: The curve y = ax³ + bx² + cx + d passes through the origin, has a stationary point at (1, 2), and the gradient at x = −1 is 12. Find the values of a, b, c, and d.

Solution:

Since the curve passes through the origin: d = 0.

Since (1, 2) is on the curve: a + b + c + d = 2, so a + b + c = 2. ... (1)

Differentiate: dy/dx = 3ax² + 2bx + c.

Since there is a stationary point at x = 1: 3a + 2b + c = 0. ... (2)

Since the gradient at x = −1 is 12: 3a − 2b + c = 12. ... (3)

From (2) and (3): (3a + 2b + c) − (3a − 2b + c) = 0 − 12, so 4b = −12, giving b = −3.

From (2): 3a − 6 + c = 0, so 3a + c = 6. ... (4)

From (1): a − 3 + c = 2, so a + c = 5. ... (5)

From (4) − (5): 2a = 1, so a = 1/2.

From (5): c = 5 − 1/2 = 9/2.

Therefore a = 1/2, b = −3, c = 9/2, d = 0. ∎

---

Example 3: Trigonometry meets Integration

Problem: Find the exact area enclosed between the curve y = sin x and the x-axis from x = 0 to x = 2π.

Solution:

The curve y = sin x crosses the x-axis at x = 0, π, and 2π. It is positive on (0, π) and negative on (π, 2π).

Area = ∫₀^π sin x dx + |∫_π^{2π} sin x dx|
     = [−cos x]₀^π + |[−cos x]_π^{2π}|
     = (−cos π + cos 0) + |(−cos 2π + cos π)|
     = (1 + 1) + |(−1 − 1)|
     = 2 + 2
     = 4

Therefore the exact area is 4 square units. ∎

Key Point: When finding area, you must consider where the curve is above and below the x-axis. Integration gives signed area — below the x-axis gives a negative value, so you must take the absolute value.

---

Problem-Solving Strategies

Strategy 1: Draw a Diagram

For problems involving geometry, coordinate geometry, or mechanics, a diagram can clarify the situation enormously. Label known quantities and mark what you need to find.

Strategy 2: Identify the Mathematical Topics Involved

Read the problem carefully and list the topics it might involve. A question about the "area under a curve" involves integration; if the curve is given parametrically, you also need parametric differentiation.

Strategy 3: Write Down What You Know

Translate the problem into mathematical equations. Assign variables to unknowns and write equations from the given information.

Strategy 4: Work Backwards

If you know the answer form, work backwards to determine what intermediate results you need. For example, if you need to find the maximum value of a function, you know you need to differentiate and find stationary points.

Strategy 5: Simplify

If the algebra looks complicated, check whether you can simplify before expanding. Look for common factors, cancel terms, or use substitutions to reduce complexity.

Strategy 6: Check Your Answer

Substitute your answer back into the original problem to verify it works. For numerical answers, check they make physical or mathematical sense.

---

Multi-Domain Problems

Example 4: Sequences meet Proof

Problem: The sequence is defined by u₁ = 1, uₙ₊₁ = 3uₙ + 2. Prove that uₙ = (5 × 3ⁿ⁻¹ − 2)/2 for all positive integers n.

This requires knowledge of recurrence relations (sequences) and proof by induction (proof). While formal proof by induction is not required in the A-Level specification, understanding recursive sequences and verifying formulas is.

Verification for small values:

u₁ = (5 × 3⁰ − 2)/2 = (5 − 2)/2 = 3/2. But u₁ = 1. So this formula is incorrect.

Let us try uₙ = (5 · 3ⁿ⁻¹ − 1)/2:

u₁ = (5 × 1 − 1)/2 = 2. Still wrong. Let us compute the sequence:

u₁ = 1, u₂ = 3(1) + 2 = 5, u₃ = 3(5) + 2 = 17, u₄ = 3(17) + 2 = 53.

The correct closed form is uₙ = (3ⁿ − 1)/2 + 3ⁿ⁻¹ ... Actually, let us use the standard method:

The recurrence uₙ₊₁ = 3uₙ + 2 has particular solution uₙ = −1 (since −1 = 3(−1) + 2 = −1 ✓). The general solution of the homogeneous equation is A × 3ⁿ. So uₙ = A × 3ⁿ + (−1) = A · 3ⁿ − 1.

Using u₁ = 1: A · 3 − 1 = 1, so A = 2/3.

Hence uₙ = (2/3) · 3ⁿ − 1 = 2 · 3ⁿ⁻¹ − 1.

Check: u₁ = 2 · 1 − 1 = 1 ✓, u₂ = 2 · 3 − 1 = 5 ✓, u₃ = 2 · 9 − 1 = 17 ✓. ∎

---

Example 5: Logarithms meet Sequences

Problem: A geometric sequence has first term 100 and common ratio 0.85. Find the first term to be less than 10.

The nth term is uₙ = 100 × 0.85ⁿ⁻¹.

We need 100 × 0.85ⁿ⁻¹ < 10, i.e., 0.85ⁿ⁻¹ < 0.1.

Taking logarithms (natural or base 10):

(n − 1) ln 0.85 < ln 0.1
n − 1 > ln 0.1 / ln 0.85     [inequality reverses because ln 0.85 < 0]
n − 1 > (−2.3026)/(−0.16252)
n − 1 > 14.17
n > 15.17

Since n must be a positive integer, n = 16.

The 16th term is the first to be less than 10. ∎

---

Summary

• Problem solving in pure mathematics requires connecting topics and applying knowledge in unfamiliar contexts.
• AO3 questions are unstructured — you must identify the method yourself.
• Key strategies: draw a diagram, identify the topics, write equations, work backwards, simplify, and check.
• Multi-domain problems combine two or more areas (e.g., algebra + coordinate geometry, calculus + trigonometry, logarithms + sequences).
• Practise unstructured problems from past papers to build confidence and fluency.

Exam Tip: AO3 marks are among the most challenging on AQA papers. To prepare, work through past paper questions that span multiple topics. When you get stuck, note which topics are involved and review the connections between them. The examiners' report often identifies common errors — these are invaluable for understanding what the examiner expects.

A-Level Deep Dive: Problem-Solving in Pure Mathematics

Spec mapping

AQA 7357 specification, Papers 1 and 2 — Pure Mathematics (all sections A–N): problem-solving is not a discrete topic but the dominant assessment objective (AO3) layered across every Pure section. AQA define it as "translate problems in mathematical and non-mathematical contexts into mathematical processes; interpret solutions to problems in their original context; evaluate the accuracy and limitations of solutions". On 7357, AO3 carries roughly 20% of marks on Paper 1 and 25% on Paper 2, but the technique of recognising a topic from a question stem, planning a multi-step solution, executing fluently, and verifying applies universally. The synoptic items examined here combine Section A (Proof), Section B (Algebra and functions), Section E (Trigonometry), Section F (Exponentials and logarithms), Section G (Differentiation) and Section H (Integration) within single questions. The AQA formula booklet provides standard derivatives and integrals but does not signpost which topic to invoke — that recognition is the candidate's responsibility.

---

Worked example with full mark scheme

Question (8 marks): A curve has equation $y = e^x \sin x$y=exsinx for $0 \leq x \leq 2\pi$0≤x≤2π.

(a) Find $\dfrac{dy}{dx}$dxdy​, expressing your answer in the form $e^x(\sin x + \cos x)$ex(sinx+cosx). (2)

(b) Hence find the exact $x$x-coordinates of the stationary points of the curve in the given interval. (4)

(c) Determine the nature of each stationary point. (2)

Solution with mark scheme:

(a) Step 1 — apply the product rule. With $u = e^x$u=ex and $v = \sin x$v=sinx, $u' = e^x$u′=ex and $v' = \cos x$v′=cosx.

$\dfrac{dy}{dx} = u'v + uv' = e^x \sin x + e^x \cos x = e^x(\sin x + \cos x)$dxdy​=u′v+uv′=exsinx+excosx=ex(sinx+cosx)

M1 — correct application of the product rule with both terms present. A1 — correct factored form matching the printed answer.

(b) Step 2 — set the derivative to zero.

$e^x(\sin x + \cos x) = 0$ex(sinx+cosx)=0

Since $e^x > 0$ex>0 for all real $x$x, the factor $e^x$ex never vanishes. So $\sin x + \cos x = 0$sinx+cosx=0, giving $\tan x = -1$tanx=−1.

M1 — recognising $e^x \neq 0$ex=0 and dividing through. The most common error is to set $e^x = 0$ex=0 and produce no solutions; this loses both M1 and the subsequent A marks.

Step 3 — solve $\tan x = -1$tanx=−1 in $[0, 2\pi]$[0,2π].

The principal value of $\arctan(-1) = -\pi/4$arctan(−1)=−π/4, which is outside the interval. Add $\pi$π and $2\pi$2π to reach the interval:

$x = \dfrac{3\pi}{4}, \quad x = \dfrac{7\pi}{4}$x=43π​,x=47π​

M1 — correct use of the periodicity of $\tan$tan (period $\pi$π) to find both solutions. A1 — both exact values.

(c) Step 4 — second derivative test.

$\dfrac{d^2y}{dx^2} = \dfrac{d}{dx}\left[e^x(\sin x + \cos x)\right] = e^x(\sin x + \cos x) + e^x(\cos x - \sin x) = 2e^x \cos x$dx2d2y​=dxd​[ex(sinx+cosx)]=ex(sinx+cosx)+ex(cosx−sinx)=2excosx

At $x = 3\pi/4$x=3π/4: $\cos(3\pi/4) = -\sqrt{2}/2 < 0$cos(3π/4)=−2​/2<0, so $y'' < 0$y′′<0 — maximum. At $x = 7\pi/4$x=7π/4: $\cos(7\pi/4) = \sqrt{2}/2 > 0$cos(7π/4)=2​/2>0, so $y'' > 0$y′′>0 — minimum.

M1 — second derivative computed (or sign-table reasoning). A1 — both natures correctly identified.

Total: 8 marks. This question fuses calculus (product rule, second-derivative test), trigonometric equations, exponential properties ($e^x > 0$ex>0) and exact-value reasoning. No single section "owns" it. The verification step — checking solutions lie in $[0, 2\pi]$[0,2π] — is what AO3 rewards.

---

Specimen question modelled on the AQA 7357 Paper format

Question (6 marks): The function $f$f is defined by $f(x) = \ln(2x - 3)$f(x)=ln(2x−3) for $x > 3/2$x>3/2.

(a) Find $f^{-1}(x)$f−1(x) and state its domain. (3)

(b) Solve the equation $f(x) = f^{-1}(x)$f(x)=f−1(x). (3)

Mark scheme decomposition by AO:

(a)

• M1 (AO1.1a) — set $y = \ln(2x - 3)$y=ln(2x−3) and exponentiate: $e^y = 2x - 3$ey=2x−3.
• M1 (AO1.1b) — rearrange: $x = \dfrac{e^y + 3}{2}$x=2ey+3​, then swap variables to get $f^{-1}(x) = \dfrac{e^x + 3}{2}$f−1(x)=2ex+3​.
• A1 (AO2.5) — domain $x \in \mathbb{R}$x∈R (range of original $f$f).

(b)

• M1 (AO3.1a) — recognise that for self-inverse intersections, $f(x) = f^{-1}(x)$f(x)=f−1(x) implies $f(x) = x$f(x)=x (when $f$f is monotonic), so reduce to $\ln(2x - 3) = x$ln(2x−3)=x.
• M1 (AO3.1b) — recognise this is transcendental; solve numerically or graphically. By inspection $x \approx 1.95$x≈1.95 and $x \approx 3.92$x≈3.92.
• A1 (AO3.2a) — interpret: two solutions exist, given to suitable accuracy.

Total: 6 marks split AO1 = 2, AO2 = 1, AO3 = 3. The AO3-heavy second part is characteristic of synoptic Paper 2 questions.

---

Synoptic links

Connects to (every Pure section):

1. Proof (Section A): problem-solving frequently demands proof-by-deduction or proof-by-contradiction as a sub-step. Showing $e^x \neq 0$ex=0 in the worked example is a one-line proof embedded in a calculus problem.

2. Algebra and functions (Section B): factorising $e^x(\sin x + \cos x)$ex(sinx+cosx) relied on recognising a common factor — a Year 1 algebraic skill weaponised inside calculus.

3. Coordinate geometry (Section C): stationary-point problems are about curve sketching; finding intersections of $y = f(x)$y=f(x) and $y = f^{-1}(x)$y=f−1(x) uses the geometric fact that inverse graphs are reflections in $y = x$y=x.

4. Sequences and series (Section D): Newton-Raphson iteration for transcendental equations like $\ln(2x - 3) = x$ln(2x−3)=x generates a sequence $x_{n+1} = g(x_n)$xn+1​=g(xn​) — a recurrence relation in disguise.

5. Trigonometry (Section E): $\sin x + \cos x = R\sin(x + \alpha)$sinx+cosx=Rsin(x+α) harmonic form (with $R = \sqrt{2}$R=2​, $\alpha = \pi/4$α=π/4) is a synoptic alternative route to the same stationary points.

6. Exponentials and logarithms (Section F): the equation $f(x) = f^{-1}(x)$f(x)=f−1(x) in the specimen reduced to a logarithm-versus-linear comparison — a graphical-numerical hybrid.