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Paper 2 · Set 03

Paper 2: Mathematical Reasoning. 20 questions, 75 minutes. Answer on the grid or click the options; nothing is marked until you press Submit.

75 minutes20 marksNo calculator
75:00 PDF

Answer grid

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1Indices and surds · spec MM1.1 · Paper 1 style · ★☆☆

Simplify \(\left(\dfrac{a^{\frac{1}{2}} b^{-2}}{a^{-1} b^{\frac{1}{2}}}\right)^{4}\).

  1. A\(a^{2}b^{-10}\)
  2. B\(a^{2}b^{-6}\)
  3. C\(a^{6}b^{10}\)
  4. D\(a^{6}b^{-6}\)
  5. E\(a^{6}b^{-10}\)
  6. F\(a^{-2}b^{10}\)
Worked solution

Inside the bracket: \(a^{\frac12 + 1} b^{-2 - \frac12} = a^{\frac32} b^{-\frac52}\). Raise to the 4th power: \(a^{6} b^{-10}\).

Full method — how and why
Index laws with two variables. Simplify inside the bracket first, then apply the outside power to each index.
  1. Simplify the \(a\)'s inside: \(\dfrac{a^{1/2}}{a^{-1}} = a^{1/2 - (-1)} = a^{3/2}\).
    Dividing subtracts indices, and subtracting a negative adds - the commonest slip in the question.
  2. Simplify the \(b\)'s: \(\dfrac{b^{-2}}{b^{1/2}} = b^{-2 - 1/2} = b^{-5/2}\).
    Same rule; keep the two variables in separate columns so signs do not get mixed up.
  3. Raise the bracket to the 4th power: \(\left(a^{3/2}b^{-5/2}\right)^4 = a^{6}b^{-10}\).
    \((a^m)^n = a^{mn}\) applies to each base separately: \(\frac32 \times 4 = 6\) and \(-\frac52 \times 4 = -10\).
Where it goes wrong
  • Getting \(a^{-1/2}\) at step 1 by subtracting the wrong way round, leading to \(a^{-2}b^{10}\) - which is offered.
  • Applying the power \(4\) to only the first factor.
Takeaway Simplify inside the bracket to a single power of each base before applying the outer index.
Answer E · Indices and surds (MM1.1) · question IND-13
2Exponentials and logarithms · spec MM5.2 · Paper 1 style · ★☆☆

Find the value of \(\dfrac{\log_5 27}{\log_5 3}\).

  1. A\(9\)
  2. B\(\dfrac{27}{3}\)
  3. C\(\log_5 9\)
  4. D\(3\)
  5. E\(\log_5 24\)
  6. F\(24\)
Worked solution

\(\log_5 27 = \log_5 3^3 = 3\log_5 3\), so the quotient is \(3\).

Full method — how and why
A quotient of two logs with the same base. Rewriting the numerator's argument as a power makes it cancel.
  1. Write \(27 = 3^3\), so \(\log_5 27 = \log_5 3^3 = 3\log_5 3\).
    The power law applies to the argument, whatever the base of the log.
  2. Form the quotient: \(\dfrac{3\log_5 3}{\log_5 3} = 3\).
    The common factor \(\log_5 3\) cancels (it is non-zero).
Where it goes wrong
  • Subtracting to get \(\log_5 24\) - a quotient of logs is not the log of a quotient.
  • Dividing the arguments to get \(\frac{27}{3} = 9\) in log form.
Takeaway \(\frac{\log_c a}{\log_c b} = \log_b a\): quotients of logs change the base, they do not divide the arguments.
Answer D · Exponentials and logarithms (MM5.2) · question LOG-09
3Differentiation · spec MM6.3 · Paper 1 style · ★☆☆

For what range of values of \(x\) is \(f(x) = 2x^3 - 15x^2 + 24x + 1\) decreasing?

  1. A\(-4 < x < -1\)
  2. B\(2 < x < 3\)
  3. C\(x > 1\)
  4. D\(x < 1\) or \(x > 4\)
  5. E\(x < 4\)
  6. F\(1 < x < 4\)
Worked solution

\(f'(x) = 6x^2 - 30x + 24 = 6(x - 1)(x - 4) < 0\) for \(1 < x < 4\).

Full method — how and why
'Decreasing' is the condition \(f'(x) < 0\) - a quadratic inequality once the derivative is found.
  1. Differentiate: \(f'(x) = 6x^2 - 30x + 24\).
    Term by term.
  2. Factorise: \(6(x^2 - 5x + 4) = 6(x-1)(x-4)\).
    Taking out the \(6\) makes the roots visible.
  3. Solve \(f'(x) < 0\): the upward parabola is negative between its roots, so \(1 < x < 4\).
    The gradient is negative exactly where the function falls.
Where it goes wrong
  • Solving \(f'(x) > 0\) and answering \(x < 1\) or \(x > 4\) - which is offered.
  • Forgetting the factor \(6\) when factorising, giving the wrong roots.
Takeaway Decreasing \(\Leftrightarrow f' < 0\); find the roots of \(f'\) and use the shape of its graph.
Answer F · Differentiation (MM6.3) · question DIF-05
4Identifying errors in proofs · spec Err2 · Paper 2 style · ★☆☆

Claim: if \(a > b\) then \(\dfrac{1}{a} < \dfrac{1}{b}\).

  1. ISuppose \(a > b\).
  2. IIDividing both sides by \(ab\): \(\dfrac{1}{b} > \dfrac{1}{a}\).
  3. IIISo \(\dfrac1a < \dfrac1b\).

Which of the following is correct?

  1. AThe claim is false in general and the first error is on line II, since \(ab\) may be negative or zero.
  2. BThe claim and proof are correct.
  3. CThe first error is on line III.
  4. DThe claim is true but the first error is on line I.
  5. EThe claim is false and the first error is on line III.
Worked solution

\(a = 1\), \(b = -1\): \(a > b\) but \(\frac1a = 1 > -1 = \frac1b\). Dividing by \(ab\) without knowing its sign is invalid (and \(b = 0\) would be division by zero).

Full method — how and why
The same unknown-sign trap as ERR-05, now with division, plus a false claim.
  1. Test the claim: \(a = 1\), \(b = -1\). Then \(a > b\) but \(\frac1a = 1\) and \(\frac1b = -1\), so \(\frac1a > \frac1b\). Claim false.
    Opposite signs are the natural test for a reciprocal claim.
  2. Line I is just the supposition - fine.
    No deduction yet.
  3. Line II divides by \(ab\) and keeps the direction. That needs \(ab > 0\), which is not given; \(ab\) could be negative (as in the counterexample) or zero. First error.
    Dividing by a negative reverses the inequality, and dividing by zero is undefined.
  4. The claim is true if \(a\) and \(b\) have the same sign - which is the condition that was silently assumed.
    Naming the missing hypothesis is the cleanest way to describe this kind of error.
Where it goes wrong
  • Accepting the argument because it 'works for positive numbers' - the claim is stated for all reals.
  • Blaming line III, which is a trivial restatement of line II.
Takeaway Reciprocals reverse order only within a fixed sign: for \(a,b>0\) (or both negative), \(a>b \Rightarrow \frac1a<\frac1b\).
Answer A · Identifying errors in proofs (Err2) · question ERR-10
5The logic of arguments · spec Arg2 · Paper 2 style · ★☆☆

Let \(f\) be a function defined for all real numbers. \(P\): '\(f'(a) = 0\)'; \(Q\): '\(f\) has a stationary point at \(x = a\)'. Which one of the following is true?

  1. AP is necessary and sufficient for Q.
  2. BP is necessary but not sufficient for Q.
  3. CP is sufficient but not necessary for Q.
  4. DP is neither necessary nor sufficient for Q.
Worked solution

A stationary point is by definition a point where the derivative is zero, so \(P\) and \(Q\) are equivalent.

Full method — how and why
Sometimes the two statements are the same by definition - recognise that and both directions are immediate.
  1. Recall the definition: \(f\) has a stationary point at \(x = a\) precisely when \(f'(a) = 0\).
    That is what 'stationary' means - the rate of change is zero.
  2. So \(P \Rightarrow Q\) and \(Q \Rightarrow P\) both hold trivially.
    Definitions are biconditional.
  3. Hence \(P\) is necessary and sufficient for \(Q\).
    Match to the option list.
Where it goes wrong
  • Confusing this with 'turning point': \(f'(a) = 0\) does not guarantee a maximum or minimum (\(y = x^3\) at \(x=0\)). But the question says stationary point, not turning point.
  • Worrying about functions that are not differentiable - the question states \(f\) is defined and \(f'(a)\) is referred to, so differentiability is assumed.
Takeaway A definition gives both implications at once; but stationary \(\neq\) turning - the extra condition is a sign change in \(f'\).
Answer A · The logic of arguments (Arg2) · question LGC-19
6Identifying errors in proofs · spec Err2 · Paper 2 style · ★☆☆

A student argues: 'Every prime number is odd, because if it were even it would be divisible by \(2\) and so not prime.' Which of the following is correct?

  1. AThe argument is wrong because \(2\) is even and prime: being divisible by \(2\) does not stop \(2\) itself being prime.
  2. BThe argument is correct.
  3. CThe argument is wrong because some odd numbers are not prime.
  4. DThe argument is wrong because \(1\) is odd and not prime.
  5. EThe argument is wrong because \(0\) is even.
Worked solution

A number divisible by \(2\) is not prime unless it equals \(2\). The conclusion 'every prime is odd' is false.

Full method — how and why
An argument that proves 'almost' the right thing. Find the single case it mishandles.
  1. The reasoning is: even \(\Rightarrow\) divisible by \(2\) \(\Rightarrow\) not prime.
    Trace the chain and test each link.
  2. The second link fails for one number: \(2\) is divisible by \(2\) and is prime, because divisibility by itself is allowed.
    A prime has exactly two divisors, \(1\) and itself; for \(2\) those are \(1\) and \(2\).
  3. So the conclusion 'every prime is odd' is false - \(2\) is an even prime.
    A single counterexample refutes it.
  4. The correct statement is 'every prime greater than \(2\) is odd', and the argument does establish that.
    Identifying the corrected claim shows precisely what the flaw was.
Where it goes wrong
  • Objecting that some odd numbers are not prime - true, but irrelevant; the claim is not the converse.
  • Objecting via \(1\) or \(0\) - neither is a counterexample to 'every prime is odd'.
Takeaway Check whether an argument's chain breaks at a boundary case. \(2\) is the even prime that breaks this one.
Answer A · Identifying errors in proofs (Err2) · question ERR-16
7The logic of arguments · spec Arg2 · Paper 2 style · ★☆☆

For a real number \(x\), \(P\): '\(x > 0\)'. \(Q\): '\(x^3 > 0\)'. Which one of the following is true?

  1. AP is necessary and sufficient for Q.
  2. BP is necessary but not sufficient for Q.
  3. CP is sufficient but not necessary for Q.
  4. DP is neither necessary nor sufficient for Q.
Worked solution

Cubing preserves sign, so \(x > 0 \Leftrightarrow x^3 > 0\).

Full method — how and why
Cubing preserves sign in both directions, so this is an equivalence - unlike squaring.
  1. If \(x>0\) then \(x^3 = x \cdot x \cdot x > 0\): sufficient.
    A product of three positives is positive.
  2. Conversely, if \(x \leq 0\) then \(x^3 \leq 0\) (negative cubed is negative, and \(0^3=0\)). So \(x^3>0\) forces \(x>0\): necessary.
    Proved by contrapositive - check the negative and zero cases rather than hunting for a counterexample.
  3. So \(P\) is necessary and sufficient.
    Cubing is a strictly increasing, sign-preserving operation.
Where it goes wrong
  • Importing the squaring counterexample (\(x=-2\)) - for cubes \((-2)^3 = -8 < 0\), so it does not apply.
  • Answering 'sufficient but not necessary' by pattern-matching to \(x^2\) questions.
Takeaway Odd powers preserve sign and order; even powers do not. \(x>0 \iff x^3>0\).
Answer A · The logic of arguments (Arg2) · question X2-24
8Identifying errors in proofs · spec Err2 · Paper 2 style · ★☆☆

A student attempts to solve the equation \(\sqrt{x + 3} = x - 3\) for real \(x\).

  1. ISquaring both sides: \(x + 3 = (x - 3)^2 = x^2 - 6x + 9\).
  2. IISo \(x^2 - 7x + 6 = 0\).
  3. IIISo \((x - 1)(x - 6) = 0\).
  4. IVHence the solutions are \(x = 1\) and \(x = 6\).

Which of the following best describes the solution?

  1. AThe solution is wrong, and the first error occurs on line IV.
  2. BThe solution is completely correct.
  3. CThe solution is wrong, and the first error occurs on line I.
  4. DThe solution is wrong, and the first error occurs on line II.
  5. EThe solution is wrong, and the first error occurs on line III.
Worked solution

Squaring can introduce extraneous solutions, so line IV should check them: \(x = 1\) gives \(\sqrt4 = 2 \neq -2\). Only \(x = 6\) works. Lines I-III are valid deductions.

Full method — how and why
Squaring an equation is not reversible: it can create solutions that do not satisfy the original. The error is the missing check, not the algebra.
  1. Check line I: squaring both sides of \(\sqrt{x+3} = x-3\) gives \(x + 3 = x^2 - 6x + 9\). Valid.
    \(A = B \Rightarrow A^2 = B^2\) is always a legitimate deduction - it is the converse that fails.
  2. Check lines II and III: \(x^2 - 7x + 6 = 0\) factorises as \((x-1)(x-6) = 0\). Valid.
    \(-1 \times -6 = 6\) and \(-1 - 6 = -7\) ✓.
  3. Line IV asserts both are solutions. Substitute \(x = 1\) into the original: \(\sqrt4 = 2\), but \(x - 3 = -2\). \(2 \neq -2\), so \(x=1\) is extraneous.
    A square root is non-negative, so the right-hand side must be \(\geq 0\), forcing \(x \geq 3\).
  4. Substitute \(x = 6\): \(\sqrt9 = 3\) and \(6-3 = 3\) ✓. So only \(x = 6\) is a solution, and the first error is line IV.
    Lines I-III produce candidates; the check is a required final step.
Where it goes wrong
  • Blaming line I - squaring is a valid implication, just not an equivalence.
  • Declaring the solution correct because the factorising is right; correct algebra can still give a wrong answer set.
  • Checking the candidates in the squared equation rather than the original - both pass there, which proves nothing.
Takeaway After squaring, always substitute back into the original equation; squaring can only add solutions, never lose them.
Answer A · Identifying errors in proofs (Err2) · question ERR-01
9Exponentials and logarithms · spec MM5.3 · Paper 2 style · ★★☆

Consider the statement: for all real \(x\), \(2^{x} > x^2\). Which of the following is a counterexample?

  1. A\(x = 5\)
  2. B\(x = 3\)
  3. C\(x = 0\)
  4. D\(x = 10\)
  5. E\(x = 1\)
Worked solution

\(2^3 = 8 < 9 = 3^2\), so \(x = 3\) breaks the claim. The other options satisfy it: \(2 > 1\), \(32 > 25\), \(1 > 0\), \(1024 > 100\).

Full method — how and why
Exponential versus quadratic growth: the exponential eventually wins, but there is a window where it loses. Test values inside that window.
  1. Check the candidates: \(x=0\) gives \(1>0\) ✓; \(x=1\) gives \(2>1\) ✓; \(x=5\) gives \(32>25\) ✓; \(x=10\) gives \(1024>100\) ✓.
    Evaluate both sides for each - no cleverness needed.
  2. \(x=3\): \(2^3 = 8\) and \(3^2 = 9\). Since \(8 < 9\), the claim fails. Counterexample.
    Hypothesis (\(x\) real) satisfied, conclusion false.
  3. Why here? \(2^x\) and \(x^2\) cross at \(x=2\) (both \(4\)) and again near \(x=4\) (both \(16\)), so the quadratic is ahead between them.
    Knowing where the curves cross tells you exactly where to hunt.
Where it goes wrong
  • Testing only large \(x\), where the exponential dominates and the claim holds.
  • Overlooking the window: the two sides are equal at \(x=2\) and \(x=4\), so the quadratic leads strictly between them - that is where a counterexample must lie.
  • Assuming exponentials always beat polynomials - true only eventually.
Takeaway \(2^x\) and \(x^2\) meet at \(x=2\) and \(x=4\); between them the quadratic is larger. 'Eventually bigger' is not 'always bigger'.
Answer B · Exponentials and logarithms (MM5.3) · question X2-22
10Probability · spec M7.7 · Paper 2 style · ★★☆

Events \(A\) and \(B\) have \(P(A) > 0\) and \(P(B) > 0\). \(P\): '\(A\) and \(B\) are mutually exclusive'. \(Q\): '\(A\) and \(B\) are not independent'. Which one of the following is true?

  1. AP is necessary and sufficient for Q.
  2. BP is necessary but not sufficient for Q.
  3. CP is sufficient but not necessary for Q.
  4. DP is neither necessary nor sufficient for Q.
Worked solution

Mutually exclusive gives \(P(A \text{ and } B) = 0 \neq P(A)P(B)\), so they cannot be independent: \(P\) is sufficient. But two overlapping events can also fail to be independent, so \(P\) is not necessary.

Full method — how and why
Relating mutual exclusivity to independence. One implication holds; the converse has easy counterexamples.
  1. If \(A,B\) are mutually exclusive, \(P(A\cap B)=0\). But \(P(A)P(B)>0\) since both probabilities are positive.
    The positivity hypothesis is what makes the argument work.
  2. So \(P(A\cap B) \neq P(A)P(B)\): they are not independent. \(P \Rightarrow Q\): sufficient.
    Independence is exactly the equality that fails here.
  3. Necessity: find dependent events that are not mutually exclusive. Roll a die with \(A\): 'even' and \(B\): 'at most \(3\)'. Then \(P(A\cap B) = P(\{2\}) = \frac16\), but \(P(A)P(B) = \frac12 \times \frac12 = \frac14\).
    Unequal, so they are dependent - yet they overlap, so they are not mutually exclusive.
  4. So \(P\) is sufficient but not necessary.
    Match to the list.
Where it goes wrong
  • Treating mutually exclusive and independent as opposites in a way that makes them necessary for each other - dependence has many other causes.
  • Forgetting the hypothesis \(P(A),P(B)>0\); with a zero probability, events can be both mutually exclusive and independent.
Takeaway For events of positive probability, mutually exclusive \(\Rightarrow\) dependent, but not conversely.
Answer C · Probability (M7.7) · question X2-19
11Quadratic functions · spec MM1.3 · Paper 2 style · ★★☆

\(p\) and \(q\) are real numbers. Which of the following statements is/are true?

  1. IIf \(p^2 < 4q\) then \(x^2 + px + q > 0\) for all real \(x\).
  2. IIIf \(x^2 + px + q > 0\) for all real \(x\) then \(q > 0\).
  3. IIIIf \(q < 0\) then \(x^2 + px + q = 0\) has two real roots of opposite sign.
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

I: negative discriminant and positive leading coefficient. II: put \(x = 0\). III: the product of the roots is \(q < 0\) (and the discriminant \(p^2 - 4q > 0\)), so the roots are real with opposite signs.

Full method — how and why
Three claims about a quadratic. Each is settled by the discriminant, by a well-chosen substitution, or by the root-product relationship.
  1. I: \(p^2 < 4q\) means the discriminant \(p^2 - 4q < 0\), so \(x^2+px+q\) has no real roots; with a positive \(x^2\) coefficient the graph lies entirely above the axis. True.
    No roots plus an upward parabola means always positive - the graph never reaches the axis.
  2. II: if \(x^2+px+q > 0\) for all \(x\), then in particular at \(x = 0\): \(q > 0\). True.
    Substituting a convenient value is the quickest way to extract information from a universal statement.
  3. III: the product of the roots is \(q\). If \(q<0\) then \(p^2 - 4q > p^2 \geq 0\), so the roots are real and distinct, and their product being negative makes them opposite in sign. True.
    For \(x^2+px+q\) the sum of roots is \(-p\) and the product is \(q\).
  4. So all three.
    Match to the list.
Where it goes wrong
  • Reading I's condition as \(p^2 > 4q\) - that would give two real roots and the opposite conclusion.
  • Forgetting to check in III that the roots are real; a negative product alone would be meaningless for complex roots.
  • Trying to prove II with a general argument when \(x=0\) settles it in one line.
Takeaway Discriminant controls the number of roots; \(x=0\) extracts the constant term; sum and product of roots are \(-b/a\) and \(c/a\).
Answer H · Quadratic functions (MM1.3) · question X2-01
12Functions and transformations · spec MM8.2 · Paper 2 style · ★★☆

Let \(f(x) = x^2\) for all real \(x\), and let \(g\) be the function whose graph is obtained by translating the graph of \(y = f(x)\) by \(2\) units in the positive \(x\)-direction and then reflecting in the \(y\)-axis. Which of the following statements is/are true?

  1. I\(g(x) = (x + 2)^2\)
  2. II\(g(x) = (x - 2)^2\)
  3. IIIThe graph of \(g\) has its vertex at \((-2, 0)\)
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

Translation gives \((x - 2)^2\); reflecting in the \(y\)-axis replaces \(x\) by \(-x\): \((-x - 2)^2 = (x + 2)^2\), with vertex at \((-2, 0)\).

Full method — how and why
Composite transformations, applied in the stated order. Each transformation acts on the function you currently have.
  1. Translate \(y=x^2\) by \(2\) in the positive \(x\)-direction: replace \(x\) by \(x-2\), giving \(y = (x-2)^2\).
    A shift right subtracts inside the bracket - counter-intuitive but standard.
  2. Now reflect in the \(y\)-axis: replace \(x\) by \(-x\), giving \(y = (-x-2)^2\).
    Reflection in the \(y\)-axis is the substitution \(x \to -x\).
  3. Simplify: \((-x-2)^2 = \left(-(x+2)\right)^2 = (x+2)^2\). So \(g(x) = (x+2)^2\) - I is true, II is false.
    Squaring removes the overall minus sign.
  4. III: \((x+2)^2\) has its vertex where \(x+2=0\), i.e. at \((-2,0)\). True.
    Consistent: the vertex moved right to \((2,0)\), then reflected to \((-2,0)\).
  5. So I and III.
    Match to the list.
Where it goes wrong
  • Doing the transformations in the wrong order - reflecting first then translating gives \((x-2)^2\) instead.
  • Shifting right by adding \(2\) inside the bracket.
  • Forgetting to simplify \((-x-2)^2\), and so failing to recognise it as \((x+2)^2\).
Takeaway Right by \(a\): \(x \to x-a\). Reflect in the \(y\)-axis: \(x \to -x\). Apply them in the order given, tracking the vertex as a check.
Answer F · Functions and transformations (MM8.2) · question X2-11
13Chains of reasoning · spec Prf5 · Paper 2 style · ★★☆

\(p\) and \(q\) are prime numbers with \(p + q = 45\). Find \(pq\).

  1. A\(247\)
  2. B\(140\)
  3. C\(102\)
  4. D\(86\)
  5. E\(94\)
  6. F\(50\)
Worked solution

An odd sum of two primes needs one of them to be \(2\) (the only even prime): \(q = 43\), which is prime. \(pq = 86\).

Full method — how and why
Parity applied to primes: the sum of two primes is odd only if one of them is \(2\).
  1. \(45\) is odd, so one prime is even and one odd.
    odd \(+\) odd \(=\) even, even \(+\) even \(=\) even; only odd \(+\) even gives an odd total.
  2. The only even prime is \(2\), so \(p = 2\) and \(q = 43\).
    \(2\) is the unique even prime - every other even number has \(2\) as a proper factor.
  3. Check \(43\) is prime: not divisible by \(2, 3, 5\) and \(7^2 = 49 > 43\), so yes.
    Test primes up to the square root only.
  4. \(pq = 2 \times 43 = 86\).
    The question asks for the product, not the primes.
Where it goes wrong
  • Trying odd pairs like \(2+43\) versus \(5+40\) at random; \(40\) is not prime and no odd pair can work at all.
  • Answering \(43\) or \(45\) instead of the product.
Takeaway An odd sum of two primes forces one of them to be \(2\). Parity settles most prime puzzles instantly.
Answer D · Chains of reasoning (Prf5) · question RSN-13
14Chains of reasoning · spec Prf5 · Paper 2 style · ★★☆

The product of three consecutive positive integers is \(N\), and the middle one of the three integers is \(m\). Which of the following statements must be true?

  1. I\(N\) is divisible by \(6\).
  2. II\(N\) is divisible by \(4\).
  3. III\(N + m\) is a perfect cube.
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

Among three consecutive integers there is a multiple of \(2\) and a multiple of \(3\) (I). II fails for \(1 \cdot 2 \cdot 3 = 6\). III: \(N + m = (m - 1)m(m + 1) + m = m^3 - m + m = m^3\).

Full method — how and why
Three consecutive integers: automatic divisibility by \(2\) and \(3\), but not by \(4\); plus a neat algebraic identity.
  1. Write the three as \(m-1, m, m+1\), so \(N = (m-1)m(m+1)\).
    Centring on the middle term is what makes III come out.
  2. I: among any three consecutive integers there is at least one multiple of \(2\) and one multiple of \(3\), so \(6 \mid N\). True.
    Every second integer is even and every third is a multiple of \(3\).
  3. II: test the smallest case \(1 \times 2 \times 3 = 6\), which is not divisible by \(4\). False.
    Three consecutive integers contain only one or two evens; when only one, it may contribute a single factor of \(2\).
  4. III: \(N + m = (m^2-1)m + m = m^3 - m + m = m^3\) - a perfect cube. True.
    \((m-1)(m+1) = m^2-1\) is the difference of squares; the \(-m\) and \(+m\) cancel.
  5. So I and III.
    Match to the list.
Where it goes wrong
  • Accepting II by testing \(2\times3\times4 = 24\) only - one confirming case is not enough.
  • Expanding \(N\) the long way instead of using \((m-1)(m+1) = m^2-1\).
Takeaway \(k\) consecutive integers are always divisible by \(k!\) - here \(3! = 6\), but not necessarily by \(4\).
Answer F · Chains of reasoning (Prf5) · question RSN-11
15Number, units and estimation · spec M2.3 · Paper 2 style · ★★☆

\(n\) is a positive integer greater than \(1\). Which of the following statements is/are true?

  1. IIf \(n\) is not prime, then \(n\) has a prime factor at most \(\sqrt n\).
  2. IIIf \(n\) has no prime factor at most \(\sqrt n\), then \(n\) is prime.
  3. IIIIf \(n\) is prime, then \(n\) has no factor between \(1\) and \(n\) exclusive.
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

If \(n = ab\) with \(1 < a \leq b\) then \(a \leq \sqrt n\) and \(a\) has a prime factor \(\leq \sqrt n\) (I). II is the contrapositive of I. III is the definition of prime.

Full method — how and why
Three number-theory statements about primality testing, two of which are contrapositives of each other.
  1. I: if \(n\) is composite write \(n = ab\) with \(1
    The smaller factor of a pair cannot exceed the square root, or the product would be too big.
  2. \(a>1\) has a prime factor, which is at most \(a \leq \sqrt n\). So \(n\) has a prime factor \(\leq \sqrt n\). True.
    Every integer greater than \(1\) has a prime factor, and it cannot exceed the number itself.
  3. II: 'no prime factor \(\leq\sqrt n\) \(\Rightarrow\) \(n\) prime' is exactly the contrapositive of I, so it is True automatically.
    Contrapositives share truth values - no separate work needed.
  4. III: 'if \(n\) is prime it has no factor strictly between \(1\) and \(n\)' is the definition of prime. True.
    A prime's only positive factors are \(1\) and itself.
  5. So all three.
    Match to the list.
Where it goes wrong
  • Treating II as the converse of I and rejecting it; it is the contrapositive (negate and swap).
  • Wording III as 'no factors at all', which is wrong: \(1\) and \(n\) are factors.
Takeaway To test \(n\) for primality, trial-divide by primes up to \(\sqrt n\) - and I/II are the two readings of why that works.
Answer H · Number, units and estimation (M2.3) · question X2-18
16Mathematical proof · spec Prf3 · Paper 2 style · ★★☆

The sums \(1\), \(1 + 3\), \(1 + 3 + 5\), \(1 + 3 + 5 + 7\) are \(1, 4, 9, 16\). A student conjectures that the sum of the first \(n\) odd numbers is \(n^2\). Which of the following is a valid justification?

  1. AEach odd number is one more than an even number, and the even numbers sum to \(n(n-1)\)... so the total is \(n^2\) since there are \(n\) of them: \(n(n - 1) + n = n^2\).
  2. BThe conjecture is true because \(n^2\) is always odd when \(n\) is odd.
  3. CThe pattern \(1, 4, 9, 16\) continues with \(25\), so the conjecture is true.
  4. DThe odd numbers form an arithmetic series with first term \(1\) and difference \(2\), so the sum of \(n\) terms is \(\frac n2(2 + 2(n - 1)) = n^2\).
  5. EFour cases have been checked, which is enough for a pattern.
Worked solution

The arithmetic-series argument applies \(S_n = \frac n2(2a + (n-1)d)\) to every \(n\) and simplifies to \(n^2\). (The even-numbers argument also reaches a correct total, but extending the pattern to \(25\), the remark about \(n^2\) being odd, and 'four cases is enough' are not justifications at all.)

Full method — how and why
Distinguishing a justification from a pattern-spot. The valid answer must establish the formula for all \(n\).
  1. The first \(n\) odd numbers are \(1, 3, 5, \ldots\) - an arithmetic series with \(a = 1\), \(d = 2\).
    Recognising the structure is what lets a general formula apply.
  2. Apply \(S_n = \frac n2\left(2a + (n-1)d\right) = \frac n2(2 + 2(n-1))\).
    The standard arithmetic-series sum, valid for every \(n\).
  3. Simplify: \(\frac n2 \times 2n = n^2\).
    \(2 + 2n - 2 = 2n\).
  4. That holds for all \(n\), so it justifies the conjecture.
    By contrast, 'the pattern continues with \(25\)' checks one more case, and 'four cases is enough' explicitly is not a proof.
Where it goes wrong
  • Accepting 'four cases have been checked' - the whole point of this topic is that it is not enough.
  • Accepting '\(n^2\) is odd when \(n\) is odd' - a true but completely irrelevant statement.
  • Being drawn to the even-numbers argument; its calculation happens to work out, but it is presented with an unjustified aside, whereas the series argument is clean and standard.
Takeaway A justification must apply to every \(n\); extending a pattern one more term never does.
Answer D · Mathematical proof (Prf3) · question PRF-05
17GCSE algebra · spec M4.19 · Paper 2 style · ★★☆

The \(n\)th term of a sequence is \(\dfrac{n^2 + 1}{2n}\). Which of the following statements is/are true?

  1. IThe \(5\)th term is \(2.6\).
  2. IIEvery term after the first is greater than the previous term.
  3. IIISome term of the sequence equals \(10\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

I: \(\frac{26}{10} = 2.6\). II: \(u_n = \frac n2 + \frac{1}{2n}\); \(u_1 = 1\), \(u_2 = 1.25\), and for \(n \geq 1\) the increase \(\frac12 - \frac{1}{2n(n+1)} > 0\), so increasing. III: \(n^2 - 20n + 1 = 0\) has no integer solutions (discriminant \(396\) is not a square), so false.

Full method — how and why
Three claims about a sequence. Splitting the formula makes the behaviour transparent.
  1. I: \(u_5 = \dfrac{25+1}{10} = \dfrac{26}{10} = 2.6\). True.
    Direct substitution.
  2. Split the formula: \(u_n = \dfrac{n^2+1}{2n} = \dfrac n2 + \dfrac{1}{2n}\).
    Splitting shows one growing part and one shrinking part - much easier to reason about.
  3. II: \(u_{n+1} - u_n = \frac12 - \frac{1}{2n(n+1)}\), which is positive for all \(n \geq 1\) since \(\frac{1}{2n(n+1)} \leq \frac14\). True.
    The \(\frac n2\) term grows by \(\frac12\) each step, while the \(\frac{1}{2n}\) term shrinks by less than that.
  4. III: solve \(\dfrac{n^2+1}{2n} = 10 \Rightarrow n^2 - 20n + 1 = 0\), whose discriminant \(400 - 4 = 396\) is not a perfect square, so \(n\) is not an integer. False.
    Terms exist only for whole-number \(n\), so a non-integer solution means no term equals \(10\).
  5. I and II only.
    Match to the options.
Where it goes wrong
  • Accepting III because the equation has real solutions - they must be positive integers.
  • Testing II on only the first two terms.
Takeaway Split a rational formula into simpler pieces; and remember sequence indices must be whole numbers.
Answer E · GCSE algebra (M4.19) · question ALG-08
18Graphs of functions · spec MM8.6 · Paper 2 style · ★★☆

The cubic \(y = x^3 + ax + b\) has three distinct real roots. Which of the following statements must be true?

  1. I\(a < 0\)
  2. II\(b \neq 0\)
  3. IIIThe curve has two turning points.
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

Three distinct real roots require two turning points (III), which needs \(y' = 3x^2 + a\) to have two roots, so \(a < 0\) (I). II is false: \(x^3 - x\) has roots \(0, \pm1\) and \(b = 0\).

Full method — how and why
Conditions for a cubic to have three distinct roots. Two claims follow from the shape; the third needs a counterexample.
  1. III: three distinct real roots force the curve to turn twice (up, down, up), so it must have two turning points. True.
    Between consecutive roots the curve must turn around.
  2. I: \(y' = 3x^2 + a\) has two real roots only when \(a < 0\) (so that \(x^2 = -\frac a3\) has solutions). True.
    This follows directly from III: two turning points require \(y'\) to have two roots.
  3. II: test \(y = x^3 - x = x(x-1)(x+1)\): three distinct roots \(0, \pm1\), and here \(b = 0\). False.
    One counterexample is enough; \(b = 0\) simply means the origin is a root.
  4. I and III only.
    Match to the options.
Where it goes wrong
  • Assuming \(b \neq 0\) because 'otherwise the cubic factorises' - it does, and that is fine.
  • Failing to link I to III and guessing the sign of \(a\).
Takeaway Three distinct roots \(\Leftrightarrow\) two turning points with values of opposite sign; that forces \(a < 0\) but says nothing about \(b\).
Answer F · Graphs of functions (MM8.6) · question GRA-15
19Mathematical proof · spec Prf3 · Paper 2 style · ★★☆

Let \(u_n = n^2 - n + 41\). Which of the following statements is/are true?

  1. I\(u_n\) is prime for \(n = 1, 2, 3, 4, 5\).
  2. II\(u_n\) is prime for all positive integers \(n\).
  3. III\(u_{41}\) is divisible by \(41\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

\(u_1 = 41\), \(u_2 = 43\), \(u_3 = 47\), \(u_4 = 53\), \(u_5 = 61\): all prime (I). \(u_{41} = 41^2 - 41 + 41 = 41^2\), divisible by \(41\) (III) and not prime, so II is false - small cases do not prove a general claim.

Full method — how and why
The classic warning about pattern-spotting: a formula can produce primes for many values and still fail.
  1. I: compute \(u_1 = 41\), \(u_2 = 4-2+41 = 43\), \(u_3 = 9-3+41=47\), \(u_4 = 16-4+41 = 53\), \(u_5 = 25-5+41=61\). All prime. True.
    Straight substitution; all five are primes (none divisible by \(2,3,5\) and each is below \(64\) or checkable quickly).
  2. III: \(u_{41} = 41^2 - 41 + 41 = 41^2\). Divisible by \(41\). True.
    The \(-41\) and \(+41\) cancel exactly - which is why \(n = 41\) is the natural place to look for failure.
  3. II: since \(u_{41} = 41^2 = 1681\) has factor \(41\), it is not prime. False.
    One counterexample kills the universal claim, however many earlier cases worked.
  4. So I and III.
    Match to the list.
Where it goes wrong
  • Accepting II because the first several values are prime - this formula is famous precisely because it works for \(n = 1\) to \(40\) and then fails.
  • Trying to check \(u_{41}\) by multiplying out \(1681\) instead of spotting the cancellation.
Takeaway Many confirming cases never prove a universal claim; and when hunting for failure, look for the value that makes terms cancel.
Answer F · Mathematical proof (Prf3) · question PRF-15
20Chains of reasoning · spec Prf5 · Paper 2 style · ★★★

How many ordered pairs of positive integers \((m, n)\) satisfy \(\dfrac{1}{m} + \dfrac{1}{n} = \dfrac{1}{4}\)?

  1. A\(4\)
  2. B\(3\)
  3. C\(2\)
  4. D\(6\)
  5. E\(8\)
  6. F\(5\)
Worked solution

\(4n + 4m = mn \Rightarrow (m - 4)(n - 4) = 16\). Positive factor pairs of \(16\) as ordered pairs: \((1,16), (2,8), (4,4), (8,2), (16,1)\): five. (Negative factor pairs would make \(m\) or \(n\) non-positive.)

Full method — how and why
Clear the fractions, then use the add-a-constant factorising trick and count ordered pairs.
  1. Multiply \(\frac1m + \frac1n = \frac14\) by \(4mn\): \(4n + 4m = mn\).
    Clearing denominators turns it into a polynomial equation in integers.
  2. Rearrange and factorise: \(mn - 4m - 4n + 16 = 16\), so \((m-4)(n-4) = 16\).
    Add \((-4)(-4) = 16\) to both sides to complete the rectangle.
  3. \(m, n\) positive: if \(m - 4\) were negative then \(m \in \{1,2,3\}\) and \(n-4 = \frac{16}{m-4}\) would make \(n\) negative. So both factors are positive.
    Check the negative branch explicitly rather than assuming; e.g. \(m = 3\) gives \(n - 4 = -16\), \(n = -12\) ✗.
  4. Positive factor pairs of \(16\): \((1,16),(2,8),(4,4),(8,2),(16,1)\) - five ordered pairs.
    'Ordered' means \((m,n)\) and \((n,m)\) count separately: \((5,20), (6,12), (8,8), (12,6), (20,5)\).
Where it goes wrong
  • Answering \(3\) by counting unordered pairs when the question says ordered.
  • Forgetting \((4,4)\), the square factor pair.
  • Including negative factor pairs without checking they give positive \(m, n\).
Takeaway \(\frac1m+\frac1n=\frac1k\) becomes \((m-k)(n-k) = k^2\); then count factor pairs, respecting ordered versus unordered.
Answer F · Chains of reasoning (Prf5) · question RSN-18
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