Algebra and Functions (AS Unit 1: Pure Mathematics A) — WJEC A-Level Mathematics
Test yourself on Algebra and Functions (AS Unit 1: Pure Mathematics A) with WJEC A-Level practice questions.
7 days Premium · Then free forever · No card, no charge
Algebra and Functions (AS Unit 1: Pure Mathematics A) explained
Rigorous mathematical communication requires distinguishing between conjunction ('and') and disjunction ('or') when writing inequality solution sets.
Read the full explanation
A single continuous interval where x satisfies two simultaneous constraints requires 'and', written as a double inequality a < x < b or in set notation as {x ∈ ℝ : a < x < b} = {x ∈ ℝ : x > a} ∩ {x ∈ ℝ : x < b}. Conversely, disjoint solution regions—such as the outer tails of a quadratic inequality x² > 4—cannot be satisfied simultaneously by any real number. They must be connected by the word 'or' (x < −2 or x > 2) or expressed as a set union {x ∈ ℝ : x < −2} ∪ {x ∈ ℝ : x > 2}. Writing a combined expression like −2 > x > 2 is mathematically invalid because it asserts −2 > 2.
Your focus
- Distinguish between conjunction ('and') and disjunction ('or') in algebraic solutions.
- Express continuous and disjoint inequality solutions using standard set-builder notation.
- Use set union (∪) and intersection (∩) symbols correctly to represent solution sets.
Algebra and Functions (AS Unit 1: Pure Mathematics A) exam tips
Marking Points
- using 'and' or intersection notation for bounded single-interval inequalities
- using 'or' or union notation for disjoint inequality intervals
- writing complete solution sets using correct set-builder braces and membership symbols
Examiner Tips
- 💡If the solution consists of two separate tails on the number line, write them separated by 'or' or ∪.
- 💡Use set-builder notation {x : ...} carefully, ensuring the condition inside is fully specified.
Common Mistakes
- combining two disjoint inequalities into a nonsensical statement like 2 < x < −2
- using 'and' to connect disjoint outer intervals where no single value of x can satisfy both