A Level Mathematics: Pure Mathematics 2 — Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — Cambridge International holds copyright in its own papers. Use these alongside the official past papers available free from your board.

Related: Pure Mathematics 2 study guide and revision notes

---

Section A — Algebra and logarithms

1. Solve the equation |2x − 3| = 7. [3]

2. Use the factor theorem to show that (x − 2) is a factor of f(x) = x³ − 3x² − 4x + 12, then fully factorise f(x). [4]

3. Solve the equation 3^(2x) = 5, giving your answer correct to 3 significant figures. [3]

4. Solve log₂(x) + log₂(x − 2) = 3 for x > 2. [4]

Section B — Trigonometry, calculus and numerical methods

5. Simplify sec²θ − tan²θ, and hence solve sec²θ = 2 + tanθ for 0° ≤ θ ≤ 360°. [5]

6. Differentiate y = e^(2x) sin x with respect to x. [3]

7. Find ∫ (1/(2x + 1)) dx. [2]

8. Show by a change of sign that the equation x³ − x − 1 = 0 has a root between x = 1 and x = 2. Use the iteration x_(n+1) = (x_n + 1)^(1/3) starting from x₀ = 1.5 to find this root correct to 3 decimal places, showing each iteration. [5]

9. Find the exact value of ∫₀^(π/4) sec²x dx. [3]

---

Answers

1. |2x − 3| = 7 means 2x − 3 = 7 or 2x − 3 = −7 [1]. Solving each: 2x = 10, so x = 5 [1]; or 2x = −4, so x = 5 or x = −2 [1].

2. f(2) = (2)³ − 3(2)² − 4(2) + 12 = 8 − 12 − 8 + 12 = 0 [1], so by the factor theorem (x − 2) is a factor [1]. Dividing f(x) by (x − 2) gives x² − x − 6 [1], which factorises as (x − 3)(x + 2). So f(x) = (x − 2)(x − 3)(x + 2) [1].

3. Taking logs of both sides: 2x ln 3 = ln 5 [1]. So x = ln 5 / (2 ln 3) = 1.6094 / 2.1972 [1] = 0.732 (3 s.f.) [1] (x = 0.73249 before rounding).

4. log₂(x) + log₂(x − 2) = log₂(x(x − 2)) = 3 [1], so x(x − 2) = 2³ = 8 [1]. This gives x² − 2x − 8 = 0, which factorises as (x − 4)(x + 2) = 0, so x = 4 or x = −2 [1]. Since x > 2 is required, x = 4 [1] (x = −2 is rejected as it does not satisfy the domain of the original logarithms).

5. Using the identity sec²θ = 1 + tan²θ, sec²θ − tan²θ = 1 [1]. Substituting sec²θ = 1 + tan²θ into sec²θ = 2 + tanθ gives 1 + tan²θ = 2 + tanθ, so tan²θ − tanθ − 1 = 0 [1]. Using the quadratic formula, tanθ = (1 ± √5)/2, giving tanθ = 1.618 or tanθ = −0.618 [1]. For tanθ = 1.618: θ = 58.3° or 238.3° [1]. For tanθ = −0.618: θ = 148.3° or 328.3° [1] (all four values in range, accept to 1 d.p.).

6. Using the product rule with u = e^(2x), v = sin x: du/dx = 2e^(2x), dv/dx = cos x [1]. dy/dx = u(dv/dx) + v(du/dx) = e^(2x)cos x + 2e^(2x)sin x [1] = e^(2x)(cos x + 2 sin x) [1].

7. ∫ 1/(2x + 1) dx = (1/2) ln|2x + 1| + c [2] (one mark for recognising the 1/2 scaling factor from the chain rule in reverse, one for the correct ln form with the constant of integration).

8. Let f(x) = x³ − x − 1. f(1) = 1 − 1 − 1 = −1 (negative). f(2) = 8 − 2 − 1 = 5 (positive) [1]. Since f(x) changes sign between x = 1 and x = 2, a root lies in this interval [1]. Iterating x_(n+1) = (x_n + 1)^(1/3) from x₀ = 1.5: x₁ = (2.5)^(1/3) = 1.3572, x₂ = (2.3572)^(1/3) = 1.3309, x₃ = (2.3309)^(1/3) = 1.3259, x₄ = (2.3259)^(1/3) = 1.3249, x₅ = (2.3249)^(1/3) = 1.3248, x₆ = (2.3248)^(1/3) = 1.3247 [2] (one mark for a correct first iterate, one for the rest correct to 4 d.p.). Since x₅ and x₆ both round to 1.325, the iteration has converged to x = 1.325 (3 d.p.) [1].

9. The derivative of tan x is sec²x, so ∫ sec²x dx = tan x + c [1]. Evaluating between the limits: [tan x] from 0 to π/4 = tan(π/4) − tan(0) = 1 − 0 [1] = 1 [1].

A note on the numerical methods question

Question 8 illustrates the two habits the study guide and revision notes both flag as essential for 2.6: carrying more decimal places through each intermediate iteration than the final answer requires (dropping precision too early is a common way marks are lost, since small rounding errors compound across several iterations), and correctly interpreting what the sign change actually demonstrates — that a root exists somewhere within the interval, not what its precise value is. A full-mark answer states the sign-change conclusion explicitly before moving into the iteration itself, rather than treating the two steps as one combined statement.

The same discipline applies across this whole practice set: question 5 rewards recognising which trigonometric identity turns an otherwise unmanageable equation into a standard quadratic in tanθ, and question 7 rewards spotting the chain-rule scaling factor before integrating, rather than reaching for a more complicated substitution than the question actually requires. Checking which representation of an expression is easiest to work with before starting a calculation is the single habit that connects every sub-topic tested here.