OCR MEI A Level Mathematics B (H640/03)
The H640/03 paper is one of three components for the OCR MEI A Level Mathematics B
qualification. It is a 2-hour written paper worth 75 marks, representing 27.3% of the total A
Level .
Structure of the Paper
| Section | Content | Marks |
|---|---|---|
| Section A | Pure Mathematics (shorter questions) | ~50 marks |
| Section B | Comprehension (a longer, structured problem based on a pre-release article) |
~25 marks |
Note: The paper includes a formulae sheet, but you must know when and how to apply each
formula .
Core Domains Covered
The pure mathematics content in H640/03 includes the following topics, which form the
basis of this practice test bank:
- Proof
- Algebra and Functions
- Coordinate Geometry
- Sequences and Series
- Trigonometry
- Exponentials and Logarithms
- Calculus (Differentiation and Integration)
- Numerical Methods
,- Vectors
The mechanics and statistics content from Papers 1 and 2 is assumed knowledge but will not
be the focus of questions on Paper 3 .
---
Practice Test Bank: 500 Questions with Verified Answers & Detailed Rationales
This test bank is organized by topic. Each question is designed to reflect the style and
difficulty of the actual exam, with detailed rationales to explain the reasoning behind each
answer.
Section 1: Proof and Algebra (Questions 1–50)
1. Prove that the sum of two odd numbers is always even.
Answer: Let the two odd numbers be \(2m + 1\) and \(2n + 1\), where \(m, n \in
\mathbb{Z}\). Their sum is \((2m + 1) + (2n + 1) = 2m + 2n + 2 = 2(m + n + 1)\). Since \(m + n
+ 1\) is an integer, the sum is even.
Rationale: This is a standard proof using the definition of odd numbers.
2. Prove by contradiction that \(\sqrt{2}\) is irrational.
Answer: Assume \(\sqrt{2}\) is rational. Then \(\sqrt{2} = \frac{a}{b}\) in lowest terms, so \(2
= \frac{a^2}{b^2}\), giving \(a^2 = 2b^2\). Hence \(a^2\) is even, so \(a\) is even. Let \(a =
2k\). Then \(4k^2 = 2b^2\), so \(b^2 = 2k^2\), meaning \(b\) is even. This contradicts the
assumption that \(a\) and \(b\) have no common factors.
Rationale: A classic proof by contradiction.
3. Prove that \(n^3 - n\) is divisible by 6 for all integers \(n\).
, Answer: \(n^3 - n = n(n - 1)(n + 1)\). This is the product of three consecutive integers. Among
any three consecutive integers, at least one is even (divisible by 2) and exactly one is
divisible by 3. Hence the product is divisible by \(2 \times 3 = 6\).
Rationale: Factorization and divisibility rules.
4. Solve the inequality \(2x^2 - 5x - 3 < 0\).
Answer: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). The roots are \(x = -\frac{1}{2}\) and \(x = 3\). The
parabola opens upwards, so the inequality holds between the roots: \(-\frac{1}{2} < x < 3\).
Rationale: Factorizing a quadratic and interpreting the sign of the parabola.
5. Find the remainder when \(x^3 + 2x^2 - 5x + 1\) is divided by \(x - 2\).
Answer: Using the Remainder Theorem, substitute \(x = 2\): \(2^3 + 2(2^2) - 5(2) + 1 = 8 + 8 -
10 + 1 = 7\).
Rationale: The Remainder Theorem states that the remainder is \(f(a)\) when dividing by \(x
- a\).
6. Express \(\frac{3x + 5}{(x - 1)(x + 2)}\) in partial fractions.
Answer: \(\frac{3x + 5}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2}\). Multiply through:
\(3x + 5 = A(x + 2) + B(x - 1)\). Setting \(x = 1\): \(8 = 3A \Rightarrow A = \frac{8}{3}\). Setting
\(x = -2\): \(-1 = -3B \Rightarrow B = \frac{1}{3}\). So \(\frac{8}{3(x - 1)} + \frac{1}{3(x + 2)}\).
Rationale: Standard partial fractions decomposition.
7. Simplify \(\frac{2^{n+2} - 2^n}{2^n}\).
Answer: \(\frac{2^n(2^2 - 1)}{2^n} = 4 - 1 = 3\).
Rationale: Factor out \(2^n\).
8. Solve \(3^{2x} - 10(3^x) + 9 = 0\).
Answer: Let \(y = 3^x\). Then \(y^2 - 10y + 9 = 0\), so \((y - 1)(y - 9) = 0\). Thus \(y = 1\) or \(y
= 9\). So \(3^x = 1 \Rightarrow x = 0\), or \(3^x = 9 \Rightarrow x = 2\).
Rationale: Substitution to form a quadratic in \(3^x\).
The H640/03 paper is one of three components for the OCR MEI A Level Mathematics B
qualification. It is a 2-hour written paper worth 75 marks, representing 27.3% of the total A
Level .
Structure of the Paper
| Section | Content | Marks |
|---|---|---|
| Section A | Pure Mathematics (shorter questions) | ~50 marks |
| Section B | Comprehension (a longer, structured problem based on a pre-release article) |
~25 marks |
Note: The paper includes a formulae sheet, but you must know when and how to apply each
formula .
Core Domains Covered
The pure mathematics content in H640/03 includes the following topics, which form the
basis of this practice test bank:
- Proof
- Algebra and Functions
- Coordinate Geometry
- Sequences and Series
- Trigonometry
- Exponentials and Logarithms
- Calculus (Differentiation and Integration)
- Numerical Methods
,- Vectors
The mechanics and statistics content from Papers 1 and 2 is assumed knowledge but will not
be the focus of questions on Paper 3 .
---
Practice Test Bank: 500 Questions with Verified Answers & Detailed Rationales
This test bank is organized by topic. Each question is designed to reflect the style and
difficulty of the actual exam, with detailed rationales to explain the reasoning behind each
answer.
Section 1: Proof and Algebra (Questions 1–50)
1. Prove that the sum of two odd numbers is always even.
Answer: Let the two odd numbers be \(2m + 1\) and \(2n + 1\), where \(m, n \in
\mathbb{Z}\). Their sum is \((2m + 1) + (2n + 1) = 2m + 2n + 2 = 2(m + n + 1)\). Since \(m + n
+ 1\) is an integer, the sum is even.
Rationale: This is a standard proof using the definition of odd numbers.
2. Prove by contradiction that \(\sqrt{2}\) is irrational.
Answer: Assume \(\sqrt{2}\) is rational. Then \(\sqrt{2} = \frac{a}{b}\) in lowest terms, so \(2
= \frac{a^2}{b^2}\), giving \(a^2 = 2b^2\). Hence \(a^2\) is even, so \(a\) is even. Let \(a =
2k\). Then \(4k^2 = 2b^2\), so \(b^2 = 2k^2\), meaning \(b\) is even. This contradicts the
assumption that \(a\) and \(b\) have no common factors.
Rationale: A classic proof by contradiction.
3. Prove that \(n^3 - n\) is divisible by 6 for all integers \(n\).
, Answer: \(n^3 - n = n(n - 1)(n + 1)\). This is the product of three consecutive integers. Among
any three consecutive integers, at least one is even (divisible by 2) and exactly one is
divisible by 3. Hence the product is divisible by \(2 \times 3 = 6\).
Rationale: Factorization and divisibility rules.
4. Solve the inequality \(2x^2 - 5x - 3 < 0\).
Answer: \(2x^2 - 5x - 3 = (2x + 1)(x - 3)\). The roots are \(x = -\frac{1}{2}\) and \(x = 3\). The
parabola opens upwards, so the inequality holds between the roots: \(-\frac{1}{2} < x < 3\).
Rationale: Factorizing a quadratic and interpreting the sign of the parabola.
5. Find the remainder when \(x^3 + 2x^2 - 5x + 1\) is divided by \(x - 2\).
Answer: Using the Remainder Theorem, substitute \(x = 2\): \(2^3 + 2(2^2) - 5(2) + 1 = 8 + 8 -
10 + 1 = 7\).
Rationale: The Remainder Theorem states that the remainder is \(f(a)\) when dividing by \(x
- a\).
6. Express \(\frac{3x + 5}{(x - 1)(x + 2)}\) in partial fractions.
Answer: \(\frac{3x + 5}{(x - 1)(x + 2)} = \frac{A}{x - 1} + \frac{B}{x + 2}\). Multiply through:
\(3x + 5 = A(x + 2) + B(x - 1)\). Setting \(x = 1\): \(8 = 3A \Rightarrow A = \frac{8}{3}\). Setting
\(x = -2\): \(-1 = -3B \Rightarrow B = \frac{1}{3}\). So \(\frac{8}{3(x - 1)} + \frac{1}{3(x + 2)}\).
Rationale: Standard partial fractions decomposition.
7. Simplify \(\frac{2^{n+2} - 2^n}{2^n}\).
Answer: \(\frac{2^n(2^2 - 1)}{2^n} = 4 - 1 = 3\).
Rationale: Factor out \(2^n\).
8. Solve \(3^{2x} - 10(3^x) + 9 = 0\).
Answer: Let \(y = 3^x\). Then \(y^2 - 10y + 9 = 0\), so \((y - 1)(y - 9) = 0\). Thus \(y = 1\) or \(y
= 9\). So \(3^x = 1 \Rightarrow x = 0\), or \(3^x = 9 \Rightarrow x = 2\).
Rationale: Substitution to form a quadratic in \(3^x\).
