AQA A-level Mathematics: Language, Sets, Functions and Critiquing Arguments (7357) – Revision Notes

These notes condense overarching theme OT1.1 to OT1.5 of the AQA 7357 A-level Mathematics specification (version 1.3), including the set and function notation in Appendix A. The study guide has the full explanations and worked examples. Test yourself with the practice questions.

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1. Language and syntax (OT1.1, OT1.2)

Word	One-line meaning	Example in 5x³ − 2x + 9
Term	Piece added or subtracted	5x³, −2x, 9
Coefficient	Number multiplying a variable	5 and −2
Constant	Fixed number	9
Variable	Letter that can change	x
Index	Power	3
Expression	No equals sign	5x³ − 2x + 9
Equation	True only for some values	5x³ − 2x + 9 = 0
Identity	True for all values, uses ≡	(x + 2)² − 1 ≡ x² + 4x + 3
Function	One output for each input	f(x) = 5x³ − 2x + 9

Symbol	Read as	Check
⇒	implies	Does the left side force the right?
⇐	is implied by	Does the right side force the left?
⇔	if and only if	Both directions must be true
≡	is identical to	Equal for every x
∴ or “so”	therefore	Next line follows from the last

Method for an identity: compare coefficients.

1. Expand the side with brackets.
2. Match the coefficient of each power of x, then the constant.
3. Solve the resulting simple equations.
4. Check by expanding your answer.

Quick reminder: 3(x − 1)² + 5 ≡ 3x² − 6x + 8, so matching 3x² − 6x + 8 to a(x + b)² + c gives a = 3, b = −1, c = 5.

2. Set notation (OT1.3)

Symbol	Meaning
x ∈ A, x ∉ A	x is, or is not, an element of A
A ⊆ B	A is a subset of B (equality allowed)
A ⊂ B	A is a proper subset of B
{x: …}	the set of all x such that …
n(A)	number of elements in A
∅	empty set
A′	complement of A
A ∪ B, A ∩ B	union, intersection
ℕ, ℤ, ℤ⁺, ℤ⁺₀	natural numbers {1, 2, 3, …}, integers, positive integers, non-negative integers
(x, y)	the ordered pair x, y
ℚ, ℝ	rational numbers, real numbers
[a, b], (a, b)	closed interval, open interval of real numbers
[a, b), (a, b]	half-open intervals

Worked reminder. Let the universal set be the integers from 1 to 12, with A = multiples of 3 and B = even numbers.

A = {3, 6, 9, 12}        B = {2, 4, 6, 8, 10, 12}
A ∩ B = {6, 12}          A ∪ B = {2, 3, 4, 6, 8, 9, 10, 12}
n(A ∪ B) = 8             (A ∪ B)′ = {1, 5, 7, 11}

Check: n(A ∪ B) = n(A) + n(B) − n(A ∩ B) = 4 + 6 − 2 = 8.

Inequality solutions. “And” means intersection, “or” means union.

Inequality	Solution	Set notation
x² − 9x + 14 ≤ 0	2 ≤ x ≤ 7	{x: 2 ≤ x ≤ 7} or [2, 7]
|x − 3| > 2	x < 1 or x > 5	{x: x < 1} ∪ {x: x > 5}

Probability notation. Events are sets, so P(A ∪ B) = P(A) + P(B) − P(A ∩ B) and P(A′) = 1 − P(A). With P(A) = 1/2, P(B) = 7/20 and P(A ∩ B) = 3/20:

P(A ∪ B) = 1/2 + 7/20 − 3/20 = 14/20 = 7/10
P(A ∪ B)′ = 3/10
P(A ∩ B′) = 1/2 − 3/20 = 7/20

3. Functions (OT1.4)

• A function gives exactly one output for every input in its domain. The set of outputs is the range.
• Notation: f(x) is the value at x. f: x ↦ y means f maps x to y. f⁻¹ is the inverse. gf(x) means g(f(x)).
• One-one: each output comes from one input. Many-one: two inputs can give the same output. Only one-one functions have inverses.
• Domain rules: no division by zero, no square root of a negative.

Method for the range.

1. Write the domain.
2. Sketch the graph or complete the square.
3. Find the turning point, asymptotes and the values at the ends of the domain.
4. State the range as an inequality in f(x).

Method for the inverse.

1. Set y equal to f(x).
2. Make x the subject.
3. Swap the letter y for x to obtain f⁻¹(x).
4. Give f⁻¹ the domain that matches the outputs of f.

Worked reminder. h(x) = √(x + 4) − 1, x ≥ −4.

h(−4) = −1, h is increasing, so range: h(x) ≥ −1
y = √(x + 4) − 1 ⇒ (y + 1)² = x + 4 ⇒ x = (y + 1)² − 4
h⁻¹(x) = (x + 1)² − 4 = x² + 2x − 3,  domain x ≥ −1

The domain restriction on h⁻¹ matters. Without it, x² + 2x − 3 is not the inverse of h.

Composite functions: order matters. With k(x) = 3x + 4 and m(x) = x²:

km(x) = k(m(x)) = 3x² + 4
mk(x) = m(k(x)) = (3x + 4)² = 9x² + 24x + 16

These are different functions. The inner function acts first and is written nearest x.

4. Critiquing arguments (OT1.5)

Fault	What is wrong	Fix
Proof by example	Cases do not cover all values	Algebraic proof, or one counter-example to disprove
Dividing by an expression	May be zero (lost root) or negative (flips inequality)	Factorise, or split into cases
Squaring both sides	Can add a root that fails the original	Test each answer in the original
Converse used	A ⇒ B does not give B ⇒ A	Check each direction
Gap in logic	Step not justified	State the reason for each step

Reminder 1. Claim: 2ⁿ > n² for every positive integer n. Check n = 1 (2 > 1, true), n = 2 (4 > 4, false). Counter-example n = 2 disproves it. (Also n = 3 gives 8 < 9.)

Reminder 2. “x² = 7x, so x = 7.” Cancelling x loses the solution x = 0. Correct: x(x − 7) = 0, so x = 0 or x = 7.

Reminder 3. To prove the difference between consecutive squares is odd: (n + 1)² − n² = 2n + 1, which is odd for every integer n.

When you critique, name the first faulty line, say why it fails, and show a correct version or a counter-example.

Must-know distinctions

Pair	Difference
Equation and identity	An equation has particular solutions; an identity holds for every value
Range and domain	Domain is the allowed inputs; range is the outputs that result
⊆ and ⊂	⊆ allows the sets to be equal; ⊂ does not
∪ and ∩	Union is “in at least one set” (or); intersection is “in both” (and)
A′ and ∅	A′ is everything outside A; ∅ is the set with no elements
km and mk	Different functions in general: the inner function acts first
Counter-example and proof	One counter-example disproves; no number of examples proves

Method for turning an inequality into set notation.

1. Solve the inequality and sketch if it is quadratic.
2. Decide whether the answer is one block (and) or two pieces (or).
3. Write one block as {x: a < x < b}, or as an interval. Write two pieces joined by ∪.
4. Check strict and non-strict ends against the original sign.

Quick self-test

1. In 4x² − 3x + 7, state the coefficient of x and the constant term.
2. Is x² − 1 = (x − 1)(x + 1) an equation or an identity? Which symbol could replace “=”?
3. Write {x ∈ ℝ: −1 < x ≤ 6} as an interval.
4. Let ℰ = {1, 2, …, 10}, A = {2, 4, 6, 8, 10} and B = {1, 2, 3, 4, 5}. Find A ∩ B, and n(A′ ∩ B′).
5. Is {3, 6} a subset or a proper subset of {3, 6, 9}? Which symbol do you use?
6. State the domain of f(x) = √(x − 8) + 2 and its range.
7. Find the range of g(x) = (x + 1)² − 4 for x ≥ 0.
8. Find f⁻¹ for f(x) = 2x − 7.
9. With k(x) = 3x + 4 and m(x) = x², find km(−1).
10. A student says “x² > 9, so x > 3”. Give a counter-example and the correct solution set.

Answers

1. Coefficient of x is −3, constant term is 7.
2. It is an identity (true for all x), so write ≡.
3. (−1, 6].
4. A ∩ B = {2, 4}. A ∪ B = {1, 2, 3, 4, 5, 6, 8, 10}, which has 8 elements, so (A ∪ B)′ = {7, 9} and n(A′ ∩ B′) = 2.
5. Proper subset, written {3, 6} ⊂ {3, 6, 9}.
6. Domain x ≥ 8, range f(x) ≥ 2.
7. g(0) = −3 and g increases for x ≥ 0, so the range is g(x) ≥ −3.
8. y = 2x − 7 gives x = (y + 7)/2, so f⁻¹(x) = (x + 7)/2.
9. m(−1) = 1, then k(1) = 7, so km(−1) = 7.
10. x = −4 gives 16 > 9 but is not greater than 3. Correct solution: {x: x < −3} ∪ {x: x > 3}.

Where marks are usually lost

• Writing “or” as “and” when giving two separate ranges of x.
• Leaving the domain off an inverse function.
• Stating a range from the formula without testing the ends of a restricted domain.
• Using = in place of ≡ when asked to show an identity, or ⇔ for a one-way step.
• Treating a few checked values as a proof.
• Cancelling a factor that could be zero, and losing a solution.
• Squaring an equation and not testing the answers in the original.
• Composing functions in the wrong order.
• Giving a set answer without the braces or brackets that show what is included.

Official syllabus

Source: the AQA A-level Mathematics 7357 specification, first teaching 2017, version 1.3. Parts used: OT1 (section 3.1.1), content items B5 and B8, and Appendix A (sections 6.1, 6.4 and 6.11).