A Level Mathematics: Pure Mathematics 2 — Revision Notes

Condensed for the final weeks. For the full explanation, use the Pure Mathematics 2 study guide.

2.1 Algebra — modulus first, then fluency with division

The modulus function |x| gives the non-negative value of x, so |x| = x for x ≥ 0 and |x| = -x for x < 0. The graph of y = |ax + b| is a “V”: the part of the line y = ax + b that lies below the x-axis is reflected in the x-axis, with the vertex at x = -b/a. Solve equations and inequalities involving a modulus by considering both the positive and negative case, or by squaring both sides, using |a| = |b| ⇔ a² = b². For inequalities use |x − a| < b ⇔ a − b < x < a + b. For example, |3x − 2| = |2x + 7| gives 3x − 2 = 2x + 7 or 3x − 2 = −(2x + 7), so x = 9 or x = −1. (Graphs of y = |f(x)| and y = f(|x|) for non-linear f are not required.)

Polynomial division, the factor theorem and the remainder theorem must also be automatic. Divide a polynomial of degree up to 4 by a linear or quadratic polynomial, identifying the quotient and the remainder. The remainder when f(x) is divided by (ax + b) is f(−b/a), and (ax + b) is a factor exactly when f(−b/a) = 0. Use these to find factors and remainders, solve polynomial equations and find unknown coefficients.

2.2 Logarithms and exponentials — practise as a pair

Convert confidently between index form and logarithmic form — exam questions frequently mix both representations in a single problem rather than testing either alone.

Form	Example
Index	aˣ = b
Logarithmic	logₐ b = x

2.3 Trigonometry — the extra three functions

The secant, cosecant and cotangent functions are the least familiar part of this topic moving on from Pure Mathematics 1. Deliberately practise identities involving all six trig functions, not just sin/cos/tan — this closes the most common gap early.

2.4–2.5 Differentiation and integration — the combination trap

These are the most heavily tested in combination: expect differentiating a product/quotient involving an exponential and a trig function in the same expression, or integrating a function that first needs simplifying via a log law.

Habit: check whether an expression can be simplified algebraically before attempting to differentiate or integrate it directly — saves time, avoids errors.

2.6 Numerical solution of equations — approximating, not solving

Conceptually different from the rest of the topic. A sign change across an interval shows a root lies within it — not what the root’s value is.

Iteration formula: x_(n+1) = g(x_n)

Carry more decimal places than the final answer needs through each step until the last stated iteration — under-carrying decimals is a frequent source of lost marks.

Choosing the easiest representation — a cross-cutting habit

Before calculating, ask which representation is easiest: logarithmic form often simplifies an equation unmanageable in index form; a trig identity can turn an intractable integral into a standard form from 2.5. Examiners reward recognising the most efficient method, not just applying the first technique that comes to mind.

Worked example: combining logs and differentiation

Differentiate y = ln(3x² + 1).

Let u = 3x^2 + 1, so y = ln(u)
dy/du = 1/u
du/dx = 6x
Chain rule: dy/dx = (1/u) x du/dx = 6x / (3x^2 + 1)

This exact chain-rule-with-logarithm pattern recurs throughout 2.4 — practising it until automatic saves significant time under exam pressure.

Worked example: locating a root by sign change

Show that x^3 - x - 1 = 0 has a root between x = 1 and x = 2.

f(x) = x^3 - x - 1
f(1) = 1 - 1 - 1 = -1   (negative)
f(2) = 8 - 2 - 1 = 5    (positive)

Since f(1) is negative and f(2) is positive, and f(x) is continuous between these values, a root lies somewhere in the interval (1, 2) – this sign-change argument is what the syllabus means by “locating roots by sign changes.” It says nothing yet about the root’s precise value; finding that requires the iterative method that follows in the same sub-topic.

Worked example: differentiating a product involving trig and exponential terms

Differentiate y = e^(2x) sin(3x), using the product rule.

Let u = e^(2x),  du/dx = 2e^(2x)
Let v = sin(3x), dv/dx = 3cos(3x)

Product rule: dy/dx = u(dv/dx) + v(du/dx)
            = e^(2x) x 3cos(3x) + sin(3x) x 2e^(2x)
            = e^(2x) [3cos(3x) + 2sin(3x)]

Factoring out the common e^(2x) at the end, rather than leaving the two terms unsimplified, is what a fully worked answer at this level looks like — an unsimplified but algebraically correct answer can still lose a final accuracy mark for presentation.

Worked example: integrating using a logarithm law first

Find the integral of 1/(2x + 1) dx.

d/dx [ln|2x + 1|] = 2 / (2x + 1)        (chain rule, derivative of
                                          the linear inside function
                                          is 2)

So integral of 1/(2x+1) dx = (1/2) ln|2x + 1| + c

Recognising the pattern “integral of f’(x)/f(x) is ln|f(x)|”, adjusted by a constant factor for a linear inner function, is one of the most frequently reused techniques across 2.4-2.5 – it is worth practising until it is recognised on sight rather than derived from scratch each time.

Exam traps

• Treating factor theorem/polynomial division as a fresh challenge each time rather than an automatic first step.
• Testing logarithm/exponential conversion in isolation instead of practising mixed-representation problems.
• Neglecting secant, cosecant and cotangent identities in favour of only sine/cosine/tangent.
• Attempting to differentiate or integrate directly without first checking for algebraic simplification.
• Under-carrying decimal places through iteration steps in numerical methods questions.

Where this sits in the 9709 structure

9709 is modular: Pure Mathematics (Papers 1-3), Mechanics (Paper 4), Probability & Statistics (Papers 5-6). Pure Mathematics 2 (Paper 2) is offered only in the Pure Mathematics only AS Level route (Paper 1 plus Paper 2), which cannot count towards the full A Level. AS Level candidates take two components: Paper 1 with Paper 2, Paper 4 or Paper 5. A Level candidates take four components: Papers 1 and 3 with either Papers 4 and 5 or Papers 5 and 6. Students moving straight from Paper 1 into Paper 2 should expect single questions that combine techniques across several sub-topics at once, rather than the more isolated sub-topic testing typical of Paper 1.

Self-test

1. How do you solve an equation involving a modulus, and how does the remainder theorem give the remainder when f(x) is divided by (ax + b) in 2.1?
2. Why should logarithms and exponentials be practised as a pair rather than separately?
3. What does a sign change across an interval demonstrate in 2.6 — and what does it NOT tell you?
4. What habit should you build before differentiating or integrating any expression in 2.4–2.5?
5. Why is Pure Mathematics 2 a step up from Pure Mathematics 1 in how questions are structured?

Answers: 1. Consider the positive and negative case separately (or square both sides, using |a| = |b| ⇔ a² = b²); the remainder is f(−b/a), and (ax + b) is a factor exactly when f(−b/a) = 0. 2. Because exam questions frequently mix index and logarithmic representations within a single problem, rather than testing either form in isolation. 3. It demonstrates that a root lies within that interval; it does not tell you the root’s actual value. 4. Checking whether the expression can be simplified algebraically first — this saves time and avoids errors. 5. Because Paper 2 combines techniques from across multiple sub-topics within a single question, rather than testing each sub-topic separately as Paper 1 more often does.