AQA A-level Mathematics: Language, Sets, Functions and Critiquing Arguments (7357) – 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 – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions follow overarching theme OT1.1 to OT1.5 of the AQA 7357 A-level Mathematics specification (version 1.3), with the notation of Appendix A. They cover precise language, set notation applied to inequalities and probability, functions with domain and range, and critiquing arguments. Work through the study guide first if any topic is new.

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Questions

1. For the expression 6x³ − 5x² + x − 8, state the coefficient of x², the constant term, and the name given to 3 in the first term. Then say whether 6x³ − 5x² + x − 8 = 0 is an expression, an equation or an identity. [4]

2. Find the constants p, q and r such that 4x² − 24x + 31 ≡ p(x + q)² + r. [4]

3. In each case write ⇒, ⇐ or ⇔ in the blank, where x is a real number in (a) and n is an integer in (b). Justify each answer briefly.

(a) x > 5 ___ x² > 25 [2] (b) n is even ___ n + 3 is odd [2]

4. The universal set is ℰ = {x ∈ ℤ⁺: x ≤ 15}. A = {x: x is prime} and B = {x: x is odd}.

(a) List the elements of A ∩ B. [1] (b) Find n(A ∪ B). [2] (c) Describe A′ ∩ B′ in words and list its elements. [2]

5. Let P = {x ∈ ℝ: 3x² − 11x − 4 ≥ 0} and T = {x ∈ ℝ: x² < 25}.

(a) Find P in set notation. [3] (b) Write T as an interval. [1] (c) Find P ∩ T. [2]

6. Events A and B satisfy P(A) = 0.55, P(B) = 0.4 and P(A′ ∩ B′) = 0.25.

(a) Find P(A ∪ B). [1] (b) Find P(A ∩ B). [2] (c) Find P(A ∩ B′). [1] (d) Decide whether A and B are independent. [2]

7. (a) Explain why the rule x ↦ ±√x, for x ≥ 0, is not a function. [2] (b) The function f is defined by f(x) = x² − 4x + 9 for x ∈ ℝ. Find the range of f. [3] (c) Is f one-one? Give a reason. [1]

8. The function p is defined by p(x) = 5 − √(x − 1) for x ≥ 1.

(a) State the domain and range of p. [2] (b) Find p⁻¹(x), stating its domain. [4]

9. f(x) = 2x + 3 for x ∈ ℝ, and g(x) = 1/(x − 1) for x ≠ 1.

(a) Show that fg(x) = (3x − 1)/(x − 1) and state its domain. [3] (b) Find gf(x) and state its domain. [2] (c) Solve gf(x) = 1/4. [2]

10. A student solves 5/(x + 2) ≥ 1 as follows. “Multiply by (x + 2): 5 ≥ x + 2, so x ≤ 3.”

(a) Show by substitution that the student’s answer is wrong. [1] (b) Identify the error. [1] (c) Give the correct solution in set notation. [4]

11. A student claims that 6n + 1 is prime for every positive integer n.

(a) The student checks n = 1, 2, 3 and says the claim is proved. Explain why this is not a proof and disprove the claim. [3] (b) Prove that the sum of the squares of two consecutive integers is odd. [3] (c) The student then writes “so the sum of the squares of any two integers is odd”. Show this is false. [2]

Answers

1. Coefficient of x² is −5 [1]. Constant term is −8 [1]. The 3 is the index (power) [1]. The statement with “= 0” is an equation, true only for particular x [1]. [4] Examiner insight: Keep the sign with a coefficient or constant; “5” instead of “−5” is an easy slip, and naming the word precisely (index, not “exponent” or “number”) shows control of the required vocabulary.

2. Expand: p(x + q)² + r = px² + 2pqx + (pq² + r) [1]. Compare x²: p = 4 [1]. Compare x: 2pq = −24, so 8q = −24, q = −3 [1]. Constant: 4 × 9 + r = 31, so r = −5 [1]. [4] Examiner insight: Because this is an identity, matching coefficients is a clean route; completing the square directly also works. A quick expansion at the end to confirm 4x² − 24x + 31 is a worthwhile check.

3. (a) ⇒ [1]. If x > 5 then x² > 25, but x² > 25 also holds for x = −6, so the reverse fails [1]. (b) ⇔ [1]. An even integer plus 3 is odd, and if n + 3 is odd then n = (n + 3) − 3 is even [1]. [4] Examiner insight: Answers tend to gain credit when each direction is tested separately. A bare symbol with no justification gives little to reward, and one counter-example is enough to reject a direction.

4. (a) Primes up to 15 are 2, 3, 5, 7, 11, 13; odd numbers are 1, 3, 5, …, 15. So A ∩ B = {3, 5, 7, 11, 13} [1]. (b) n(A) = 6, n(B) = 8, n(A ∩ B) = 5 [1]. n(A ∪ B) = 6 + 8 − 5 = 9 [1]. (c) A′ ∩ B′ is the set of numbers in ℰ that are not prime and not odd [1], that is {4, 6, 8, 10, 12, 14} [1]. [5] Examiner insight: Listing A and B first makes the later parts easy to check. Remember 1 is odd and not prime, and 2 is prime and not odd; those two numbers are where listings usually go wrong.

5. (a) 3x² − 11x − 4 = (3x + 1)(x − 4) [1]. The parabola opens upwards with roots −1/3 and 4, so the expression is ≥ 0 outside the roots [1]. P = {x: x ≤ −1/3} ∪ {x: x ≥ 4} [1]. (b) x² < 25 gives −5 < x < 5, so T = (−5, 5) [1]. (c) Keep the overlap of the two sets [1]: P ∩ T = (−5, −1/3] ∪ [4, 5) [1]. [6] Examiner insight: A short sketch of the parabola helps you avoid writing a single inequality such as −1/3 ≥ x ≥ 4. Check each end: −1/3 and 4 belong to P because the inequality is non-strict, while −5 and 5 do not belong to T.

6. (a) P(A ∪ B) = 1 − P(A′ ∩ B′) = 1 − 0.25 = 0.75 [1]. (b) P(A ∩ B) = P(A) + P(B) − P(A ∪ B) [1] = 0.55 + 0.4 − 0.75 = 0.2 [1]. (c) P(A ∩ B′) = P(A) − P(A ∩ B) = 0.55 − 0.2 = 0.35 [1]. (d) Independence needs P(A ∩ B) = P(A)P(B). Here P(A)P(B) = 0.55 × 0.4 = 0.22 [1], which is not 0.2, so A and B are not independent [1]. [6] Examiner insight: State the test you are using before substituting. Concluding with a comparison of the two numbers and a clear sentence in context is safer than writing only “no”.

7. (a) The input x = 4 is mapped to both 2 and −2 [1]. A function must give exactly one output for each input in the domain, so it is not a function [1]. (b) f(x) = (x − 2)² + 5 [1]. The square is at least 0, with equality at x = 2, so the minimum is 5 [1]. There is no upper bound, so the range is f(x) ≥ 5 [1]. (c) No: f is many-one, for example f(0) = f(4) = 9 [1]. [6] Examiner insight: A specific pair of inputs with the same output is a convincing reason for many-one. Stating the range as f(x) ≥ 5 (not x ≥ 5) keeps the letter for the output.

8. (a) The domain is given: x ≥ 1 [1]. √(x − 1) takes every value from 0 upwards, so p(x) takes every value up to 5: range p(x) ≤ 5 [1]. (b) Let y = 5 − √(x − 1), so √(x − 1) = 5 − y [1]. Square: x − 1 = (5 − y)², so x = (5 − y)² + 1 [1]. Hence p⁻¹(x) = (5 − x)² + 1 [1]. The domain of p⁻¹ is the range of p, so x ≤ 5 [1]. [6] Examiner insight: An inverse given without its domain is usually an incomplete answer. Because (5 − x)² + 1 is defined for all real x, the restriction x ≤ 5 has to be stated, not inferred.

9. (a) fg(x) = f(g(x)) = 2 × 1/(x − 1) + 3 [1] = (2 + 3(x − 1))/(x − 1) [1] = (3x − 1)/(x − 1). Domain: x ≠ 1, inherited from g [1]. (b) gf(x) = g(2x + 3) = 1/((2x + 3) − 1) = 1/(2x + 2) [1]. The denominator is zero when x = −1, so the domain is x ≠ −1 [1]. (c) 1/(2x + 2) = 1/4 gives 2x + 2 = 4 [1], so x = 1 [1]. [7] Examiner insight: In a “show that” the final expression is given, so every line of algebra should be visible. Write the order of composition explicitly: fg means g acts first.

10. (a) Take x = −5, which satisfies x ≤ 3. Then 5/(−5 + 2) = −5/3, and −5/3 ≥ 1 is false, so the answer is wrong [1]. (b) Multiplying by (x + 2) is only valid when x + 2 > 0; if x < −2 the inequality reverses (and x = −2 is excluded) [1]. (c) 5/(x + 2) − 1 ≥ 0 [1], so (5 − (x + 2))/(x + 2) = (3 − x)/(x + 2) ≥ 0 [1]. The critical values are x = −2 (excluded, as it makes the denominator zero) and x = 3 (included) [1]. The fraction is positive between them: {x: −2 < x ≤ 3} [1]. [6] Examiner insight: A substitution test is a fast way to expose a flawed solution. Moving everything to one side before combining fractions avoids needing to know the sign of the denominator.

11. (a) A few cases cannot cover infinitely many values of n, so checking n = 1 to 3 is not a proof [1]. The values are 7, 13, 19, all prime [1]. For n = 4, 6 × 4 + 1 = 25 = 5 × 5, which is composite, so the claim is false [1]. (b) Take the integers as m and m + 1 [1]. m² + (m + 1)² = 2m² + 2m + 1 = 2(m² + m) + 1 [1]. This is 2 times an integer plus 1, so it is odd for every integer m [1]. (c) Take 3 and 5: 3² + 5² = 9 + 25 = 34 [1], which is even, so the claim fails [1]. The result in (b) holds only for consecutive integers. [8] Examiner insight: A counter-example needs the specific value and the reason it fails (here, 25 = 5 × 5). In a proof, say what m represents and end with a sentence that links your algebra to the statement.

Where marks are usually lost

• Giving “5” for a coefficient of −5, or confusing the constant term with a coefficient.
• Writing ⇔ where only one direction holds, with no test of the reverse.
• Listing primes carelessly: 1 is left out and 2 is included.
• Combining separate pieces of a solution with “and” instead of “or”, or writing a double inequality that cannot be true.
• Mishandling strict and non-strict ends in an interval.
• Stating a range using x rather than f(x).
• Writing an inverse without its domain, or reading the domain off the formula instead of from the outputs of the original.
• Composing in the wrong order, so that fg is treated as f acting first.
• Multiplying an inequality by an expression of unknown sign.
• Presenting a few cases as a proof, or a counter-example with no working.

Next steps

• Revise with the revision notes and the study guide.
• Return to the AQA A-level Mathematics hub and tick off the checklist.
• Use the free 10-minute diagnostics to locate gaps.
• Book a free trial class for teacher-led help.

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).