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Paper 2 · Set 04

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
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1Circles and circle theorems · spec MM3.2 · Paper 1 style · ★☆☆

The points \((1, 7)\) and \((5, -1)\) are the ends of a diameter of a circle. Find the equation of the circle.

  1. A\((x-3)^2 + (y-3)^2 = 40\)
  2. B\((x-2)^2 + (y-4)^2 = 20\)
  3. C\((x-3)^2 + (y-3)^2 = \sqrt{20}\)
  4. D\((x-3)^2 + (y-3)^2 = 20\)
  5. E\((x-3)^2 + (y-3)^2 = 80\)
  6. F\((x+3)^2 + (y+3)^2 = 20\)
Worked solution

Centre is the midpoint \((3, 3)\); radius\(^2\) \(= (3-1)^2 + (3-7)^2 = 4 + 16 = 20\).

Full method — how and why
A diameter's endpoints give the centre (midpoint) and the radius (half the distance) in one move.
  1. Centre \(=\) midpoint \(= \left(\frac{1+5}{2}, \frac{7 + (-1)}{2}\right) = (3, 3)\).
    The centre is the midpoint of any diameter.
  2. Radius\(^2\): distance from the centre to \((1,7)\) squared \(= (3-1)^2 + (3-7)^2 = 4 + 16 = 20\).
    Going from centre to an endpoint is one radius; squaring avoids surds entirely.
  3. Write the equation: \((x-3)^2 + (y-3)^2 = 20\).
    The right-hand side is \(r^2\).
Where it goes wrong
  • Using the full diameter length squared (\(80\)) as \(r^2\).
  • Writing \(r^2 = \sqrt{20}\) - the square root is already undone by squaring.
Takeaway From a diameter: centre is the midpoint, and \(r^2\) is the squared distance from centre to either end.
Answer D · Circles and circle theorems (MM3.2) · question CIR-04
2Exponentials and logarithms · spec MM5.3 · Paper 1 style · ★☆☆

Solve \(3^{2x} = 4\). Give \(x\) in terms of logarithms to base \(3\).

  1. A\(x = \log_3 4 - 2\)
  2. B\(x = \log_3 2\)
  3. C\(x = \dfrac{\log_3 4}{4}\)
  4. D\(x = \log_2 3\)
  5. E\(x = \dfrac{1}{2}\log_3 2\)
  6. F\(x = 2\log_3 2\)
Worked solution

\(2x = \log_3 4 = 2\log_3 2\), so \(x = \log_3 2\).

Full method — how and why
Solving \(a^{kx} = b\). Take logs to the given base and simplify the right-hand side.
  1. Take \(\log_3\) of both sides: \(\log_3 3^{2x} = \log_3 4\), so \(2x = \log_3 4\).
    \(\log_3 3^{u} = u\) - logs and exponentials with the same base undo each other.
  2. Rewrite the right-hand side: \(\log_3 4 = \log_3 2^2 = 2\log_3 2\).
    The power law \(\log a^k = k\log a\) brings the exponent down.
  3. Divide by \(2\): \(x = \log_3 2\).
    The factor \(2\) cancels exactly, which is why the answer is so clean.
Where it goes wrong
  • Stopping at \(x = \frac12\log_3 4\) and then mis-simplifying to \(\frac12 \log_3 2\).
  • Writing \(\log_3 4 = \log_3 2 \times \log_3 2\).
Takeaway \(\log a^k = k \log a\); rewriting numbers as powers often makes fractions cancel.
Answer B · Exponentials and logarithms (MM5.3) · question LOG-02
3Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★☆☆

The polynomial \(x^3 + ax^2 + bx + c\) has roots \(2\), \(-1\) and \(3\). Find the value of \(a + b + c\).

  1. A\(5\)
  2. B\(-5\)
  3. C\(-3\)
  4. D\(3\)
  5. E\(-7\)
  6. F\(7\)
Worked solution

\(p(1) = 1 + a + b + c\), and \(p(1) = (1-2)(1+1)(1-3) = (-1)(2)(-2) = 4\). So \(a + b + c = 3\). (Directly: \((x-2)(x+1)(x-3) = x^3 - 4x^2 + x + 6\).)

Full method — how and why
You could expand, but \(a + b + c\) is the sum of the non-leading coefficients, which \(p(1)\) delivers instantly.
  1. Note \(p(1) = 1 + a + b + c\), so \(a + b + c = p(1) - 1\).
    Substituting \(x = 1\) turns every power into \(1\), so \(p(1)\) is the sum of all coefficients including the leading \(1\).
  2. Write the factorised form from the roots: \(p(x) = (x-2)(x+1)(x-3)\).
    Monic, because the given polynomial starts with \(x^3\).
  3. Evaluate: \(p(1) = (-1)(2)(-2) = 4\).
    Two negatives and one positive give a positive.
  4. So \(a + b + c = 4 - 1 = 3\).
    Remember to subtract the leading coefficient.
Where it goes wrong
  • Answering \(4\) by forgetting the leading \(1\) in \(p(1) = 1 + a + b + c\).
  • Expanding and mis-collecting; the expansion is \(x^3 - 4x^2 + x + 6\), so \(-4 + 1 + 6 = 3\) ✓.
Takeaway \(p(1)\) is the sum of all coefficients - a one-second route to questions like this.
Answer D · Polynomials, factor and remainder theorems (MM1.6) · question POL-15
4Straight lines · spec MM3.1 · Paper 2 style · ★☆☆

Two distinct lines have equations \(y = m_1 x + c_1\) and \(y = m_2 x + c_2\). Which of the following statements is/are true?

  1. IIf \(m_1 m_2 = -1\) then the lines meet at exactly one point.
  2. IIIf \(m_1 = m_2\) then the lines do not meet.
  3. IIIIf the lines meet on the \(y\)-axis then \(c_1 = c_2\).
  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: perpendicular lines are not parallel, so they meet once. II: equal gradients and distinct lines means parallel and different, so no intersection. III: on the \(y\)-axis \(x = 0\), so \(y = c_1 = c_2\).

Full method — how and why
Three claims about two distinct straight lines. Each follows from what gradients and intercepts mean.
  1. I: \(m_1m_2 = -1\) means the lines are perpendicular, so \(m_1 \neq m_2\) and the lines are not parallel - they meet exactly once. True.
    Two non-parallel straight lines always meet at exactly one point.
  2. II: \(m_1 = m_2\) with the lines distinct means \(c_1 \neq c_2\): parallel and separate, so they never meet. True.
    The word 'distinct' in the question is doing real work - without it the lines could coincide.
  3. III: on the \(y\)-axis \(x = 0\), so the two \(y\)-values are \(c_1\) and \(c_2\). Meeting there forces \(c_1 = c_2\). True.
    Substituting \(x = 0\) into both equations.
  4. So all three.
    Match to the list.
Where it goes wrong
  • Rejecting II by imagining the lines could coincide - 'distinct' rules that out.
  • Rejecting III by thinking they might meet elsewhere; the claim conditions on them meeting on the \(y\)-axis.
Takeaway Same gradient and distinct \(\Rightarrow\) parallel, never meet. Different gradients \(\Rightarrow\) exactly one intersection. Read 'distinct' carefully.
Answer H · Straight lines (MM3.1) · question X2-05
5Integration · spec MM7.5 · Paper 2 style · ★☆☆

The trapezium rule with \(n\) strips is used to estimate \(\displaystyle\int_{0}^{1} \sqrt{x}\,dx\). Which of the following is true for every \(n\)?

  1. AThe estimate is an overestimate.
  2. BThe estimate is an underestimate.
  3. CThe estimate is an underestimate only when \(n\) is even.
  4. DThe estimate is exact.
  5. EIt cannot be decided without calculating.
Worked solution

\(y = \sqrt x\) is concave (bends downwards), so every chord lies below the curve and the trapezia underestimate the area.

Full method — how and why
Whether the trapezium rule over- or under-estimates depends on the concavity of the curve.
  1. The trapezium rule replaces the curve on each strip by the straight chord joining the endpoints.
    Its estimate is the area under those chords.
  2. \(y=\sqrt x\) is concave: it rises steeply then flattens, bending downwards.
    Its second derivative \(-\frac14x^{-3/2}\) is negative for \(x>0\) - or just picture the shape.
  3. For a concave curve every chord lies below the arc, so each trapezium sits below the true region.
    That is what concavity means geometrically.
  4. So the estimate is an underestimate, for every \(n\).
    The number of strips affects accuracy, not direction.
Where it goes wrong
  • Assuming the trapezium rule always overestimates - true only for convex (upward-bending) curves such as \(y=x^2\).
  • Thinking the parity of \(n\) matters - that is Simpson's rule, which needs an even number of strips.
  • Answering 'cannot be decided' when the shape settles it.
Takeaway Trapezium rule: convex curve \(\Rightarrow\) overestimate, concave curve \(\Rightarrow\) underestimate. Sketch and look at the bend.
Answer B · Integration (MM7.5) · question X2-17
6Identifying errors in proofs · spec Err2 · Paper 2 style · ★☆☆

A student solves \(x^2 - 5x + 6 > 0\) as follows.

  1. I\(x^2 - 5x + 6 = (x - 2)(x - 3)\).
  2. IISo \((x - 2)(x - 3) > 0\).
  3. IIISo \(x - 2 > 0\) and \(x - 3 > 0\).
  4. IVSo \(x > 3\).

Which of the following is correct?

  1. AThe first error is on line III: a product is positive when both factors are positive or both are negative, so \(x < 2\) is missed.
  2. BThe solution is completely correct.
  3. CThe first error is on line I.
  4. DThe first error is on line IV: it should be \(x > 2\).
  5. EThe first error is on line II.
Worked solution

The full solution set is \(x < 2\) or \(x > 3\).

Full method — how and why
A quadratic inequality solved as if it were an equation-style 'both factors positive'. Find where the missing branch is lost.
  1. Lines I and II are fine: \(x^2-5x+6 = (x-2)(x-3)\), so the inequality is \((x-2)(x-3)>0\).
    Check: \(-2\times-3 = 6\) and \(-2-3=-5\) ✓.
  2. Line III says both factors must be positive. But a product is positive when both are positive or both are negative. First error.
    Two negatives multiply to a positive - the case the student omits.
  3. The missing branch: \(x-2<0\) and \(x-3<0\), i.e. \(x<2\).
    Both factors negative requires the smaller bound.
  4. Full solution: \(x<2\) or \(x>3\). (Sketch check: the parabola is above the axis outside its roots.)
    A quick sketch of an upward parabola confirms the two outer intervals.
Where it goes wrong
  • Blaming line IV - it correctly follows from line III's (incomplete) conditions.
  • Giving the answer as \(20\).
  • Solving factor-by-factor at all; sketching the parabola is faster and safer.
Takeaway \((x-p)(x-q)>0\) holds outside the roots, \(<0\) between them. Sketch rather than case-split.
Answer A · Identifying errors in proofs (Err2) · question X2-25
7Circles and circle theorems · spec MM3.2 · Paper 2 style · ★☆☆

The circle \(C\) has centre \((a, b)\) and radius \(r\). \(P\): '\(r > |b|\)'. \(Q\): '\(C\) crosses the \(x\)-axis at two distinct points'. 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

The distance from the centre to the \(x\)-axis is \(|b|\); the circle cuts the axis twice exactly when this is less than \(r\).

Full method — how and why
Translate the geometry into a distance comparison: circle versus line depends on centre-to-line distance against the radius.
  1. The perpendicular distance from the centre \((a,b)\) to the \(x\)-axis is \(|b|\).
    The \(x\)-axis is \(y = 0\), so the vertical distance is the size of the \(y\)-coordinate.
  2. If \(|b| < r\) the circle reaches across the axis and cuts it twice; if \(|b| = r\) it touches once; if \(|b| > r\) it misses entirely.
    Three cases, decided by comparing distance with radius.
  3. So 'two distinct points' happens exactly when \(r > |b|\) - the two statements are equivalent.
    Necessary and sufficient.
Where it goes wrong
  • Writing the condition as \(r > b\), which fails for a centre below the axis (\(b\) negative). The modulus matters.
  • Including the tangent case \(r = |b|\), which gives one point, not two.
Takeaway Circle versus line: compare the perpendicular distance from the centre with \(r\) - less gives two points, equal one, greater none.
Answer A · Circles and circle theorems (MM3.2) · question X2-10
8Mathematical proof · spec Prf1 · Paper 2 style · ★☆☆

Which of the following claims is FALSE, and therefore can be disproved by a counterexample?

  1. AIf \(a > b\) then \(a + c > b + c\), for all real \(a, b, c\).
  2. BIf \(a > b\) then \(a^2 > b^2\), for all real \(a, b\).
  3. CIf \(a > b > 0\) then \(a^2 > b^2\).
  4. DIf \(a > b\) then \(a^3 > b^3\), for all real \(a, b\).
  5. EIf \(a > b\) then \(-a < -b\), for all real \(a, b\).
Worked solution

\(a = 1\), \(b = -2\): \(1 > -2\) but \(1 < 4\). The others are true.

Full method — how and why
Scan the claims for the one that squaring breaks - negative numbers are where inequalities misbehave.
  1. '\(a>b \Rightarrow a^2 > b^2\) for all real \(a,b\)': try \(a = 1\), \(b = -2\). Then \(1 > -2\) but \(1 < 4\). False.
    Squaring reverses order once negatives with larger magnitude are involved.
  2. '\(a > b > 0 \Rightarrow a^2 > b^2\)' is true - the positivity condition is exactly what the first claim lacked.
    Comparing the two shows precisely which hypothesis does the work.
  3. '\(a>b \Rightarrow a+c > b+c\)' is true: adding the same number to both sides preserves order.
    Valid for every real \(c\), positive or negative.
  4. '\(a>b \Rightarrow a^3>b^3\)' is true: cubing is strictly increasing for every real number, unlike squaring.
    \(x^3\) keeps the sign of \(x\), so no reversal occurs.
  5. '\(a>b \Rightarrow -a < -b\)' is true: multiplying by \(-1\) reverses the inequality.
    The standard sign rule.
Where it goes wrong
  • Thinking the cubing claim fails - it does not; only even powers destroy order.
  • Thinking multiplying by \(-1\) keeps the direction.
Takeaway Squaring preserves order only for non-negative numbers; cubing, adding and negating (with a flip) always behave.
Answer B · Mathematical proof (Prf1) · question PRF-17
9The logic of arguments · spec Arg1 · Paper 2 style · ★☆☆

Let \(x\) be a real number. Which of the following statements is equivalent to '\(x \geq 2\) if and only if \(x^3 \geq 8\)'?

  1. A\(x^3 \geq 8 \Rightarrow x \geq 2\)
  2. B(\(x \geq 2 \Rightarrow x^3 \geq 8\)) and (\(x^3 \geq 8 \Rightarrow x \geq 2\))
  3. C(\(x < 2 \Rightarrow x^3 \geq 8\)) or (\(x^3 < 8 \Rightarrow x \geq 2\))
  4. D\(x \geq 2 \Rightarrow x^3 \geq 8\)
  5. E\(x \geq 2\) or \(x^3 \geq 8\)
Worked solution

'If and only if' is the conjunction of the implication and its converse.

Full method — how and why
Unpacking 'if and only if' into its two implications.
  1. '\(A\) if and only if \(B\)' means both '\(A\) if \(B\)' (\(B \Rightarrow A\)) and '\(A\) only if \(B\)' (\(A \Rightarrow B\)).
    The phrase is literally built from 'if' and 'only if' - each supplies one direction.
  2. With \(A\): \(x \geq 2\) and \(B\): \(x^3 \geq 8\), that is \((x \geq 2 \Rightarrow x^3 \geq 8)\) and \((x^3 \geq 8 \Rightarrow x \geq 2)\).
    The conjunction of the implication and its converse.
  3. (Both happen to be true here, since cubing is increasing - but the question only asks for the equivalent form.)
    Read what is asked: the structure, not the truth value.
Where it goes wrong
  • Giving only one implication - that is half the statement.
  • Joining the two with 'or' instead of 'and'.
Takeaway \(A \Leftrightarrow B\) is exactly \((A \Rightarrow B)\) and \((B \Rightarrow A)\).
Answer B · The logic of arguments (Arg1) · question LGC-17
10The logic of arguments · spec Arg4 · Paper 2 style · ★☆☆

Let \(S\) be the statement: every even number greater than \(2\) can be written as the sum of two primes. A single counterexample to \(S\) would be

  1. Aan odd number that is not the sum of two primes
  2. Ban even number greater than \(2\) that can be written as the sum of two primes in two different ways
  3. Ctwo primes whose sum is odd
  4. Dan even number that is the sum of two primes
  5. Ean even number greater than \(2\) that is not the sum of any two primes
Worked solution

To refute 'for all \(n\), \(P(n)\)' you exhibit one \(n\) with \(P(n)\) false.

Full method — how and why
What it takes to refute a universal statement - and what it takes to refute *this* one in particular.
  1. The statement has the form 'for all even \(n > 2\), \(P(n)\)'.
    Identify the quantifier and the domain: even numbers greater than \(2\).
  2. Its negation is 'there exists an even \(n > 2\) with \(P(n)\) false'.
    One failure suffices; you do not have to show the property fails often.
  3. So a counterexample is a single even number greater than \(2\) that cannot be written as a sum of two primes.
    It must lie in the stated domain and fail the property.
  4. (This is Goldbach's conjecture - no counterexample is known, which is why it remains a conjecture.)
    Verifying billions of cases is not a proof; one counterexample would be a disproof.
Where it goes wrong
  • Offering an odd number that is not a sum of two primes - outside the stated domain, so irrelevant.
  • Offering an even number that is a sum of two primes - that confirms the statement, it does not refute it.
  • Thinking many confirming examples add up to a proof.
Takeaway To disprove 'for all', produce one object in the domain for which the property fails; confirming instances never prove it.
Answer E · The logic of arguments (Arg4) · question LGC-24
11Chains of reasoning · spec Prf5 · Paper 2 style · ★★☆

\(n\) is a positive integer such that \(n^2\) has exactly \(3\) positive divisors. Which of the following must be true?

  1. A\(n = 3\)
  2. B\(n\) is prime
  3. C\(n\) is odd
  4. D\(n\) is a perfect square
  5. E\(n\) is even
Worked solution

A number with exactly \(3\) divisors is the square of a prime; so \(n^2 = p^2\) and \(n = p\) is prime.

Full method — how and why
Use the structure of the divisor count. Exactly three divisors is a very restrictive condition.
  1. A number with exactly \(3\) divisors must be \(p^2\) for a prime \(p\): its divisors are \(1, p, p^2\).
    If a number had two distinct prime factors it would have at least \(4\) divisors; \(p^3\) has \(4\); \(p\) alone has \(2\).
  2. So \(n^2 = p^2\), and since both are positive, \(n = p\).
    Take the positive square root.
  3. Therefore \(n\) is prime.
    That is the only guaranteed property.
  4. Check the others: \(n = 2\) is even, \(n = 3\) is odd - so neither parity is forced; and \(n\) prime is never a perfect square.
    Both \(n=2\) and \(n=3\) satisfy the condition (\(4\) and \(9\) have three divisors each), which kills the parity options.
Where it goes wrong
  • Answering '\(n\) is a perfect square' by confusing \(n\) with \(n^2\).
  • Answering \(n = 3\) from one example; \(n = 2, 5, 7\) all work too.
Takeaway Exactly \(3\) divisors \(\iff\) the square of a prime. Divisor counts come from the prime factorisation exponents.
Answer B · Chains of reasoning (Prf5) · question RSN-20
12Identifying errors in proofs · spec Err1 · Paper 2 style · ★★☆

Claim: the sum of the first \(n\) odd numbers is \(n^2\).

  1. IThe first odd number is \(1 = 1^2\), so the claim holds for \(n = 1\).
  2. IISuppose the claim holds for some \(n\), i.e. \(1 + 3 + \cdots + (2n - 1) = n^2\).
  3. IIIAdding the next odd number, \(2n + 1\), gives \(n^2 + 2n + 1 = (n + 1)^2\).
  4. IVSo the claim holds for \(n + 1\), and hence for all positive integers \(n\).

Which of the following is correct?

  1. AThe proof is completely correct.
  2. BThe first error is on line II: you cannot assume the claim.
  3. CThe first error is on line III: the next odd number is \(2n - 1\).
  4. DThe first error is on line IV: it only shows the claim for \(n + 1\).
  5. EThe first error is on line I.
Worked solution

This is a valid proof by induction: a base case, and a step showing that truth for \(n\) implies truth for \(n + 1\).

Full method — how and why
A correct proof by induction. Check the base case, the inductive hypothesis, the step and the conclusion in turn.
  1. Line I: base case \(n=1\) gives \(1 = 1^2\) ✓.
    An induction needs a case to start from.
  2. Line II: assuming the result for some \(n\) is the inductive hypothesis - not circular, because it is a supposition inside the step, not a global assumption.
    This is the point most often misidentified as an error.
  3. Line III: the next odd number after \(2n-1\) is \(2n+1\) ✓, and \(n^2 + 2n + 1 = (n+1)^2\) ✓.
    Check the term: odd numbers go up in twos, so \((2n-1)+2 = 2n+1\).
  4. Line IV: base case plus step gives the result for every positive integer ✓. The proof is correct.
    \(n=1\) gives \(n=2\), which gives \(n=3\), and so on without end.
Where it goes wrong
  • Calling line II circular - assuming for one \(n\) in order to deduce \(n+1\) is exactly how induction works.
  • Claiming the next odd number is \(2n-1\); that is the \(n\)th one, already included.
  • Claiming line IV proves only the \(n+1\) case - the chain from the base case gives all \(n\).
Takeaway Induction \(=\) base case \(+\) (truth at \(n\) \(\Rightarrow\) truth at \(n+1\)). The inductive hypothesis is not circular reasoning.
Answer A · Identifying errors in proofs (Err1) · question ERR-15
13Differentiation · spec MM6.3 · Paper 2 style · ★★☆

Which of the following statements about \(f(x) = x^3 - 3x\) is/are true?

  1. I\(f\) is increasing for all \(x > 1\).
  2. II\(f\) has a local maximum at \(x = -1\).
  3. III\(f(x) \geq -2\) for all real \(x\).
  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

\(f'(x) = 3x^2 - 3 = 3(x-1)(x+1)\): positive for \(x > 1\) (I true); \(f'' = 6x < 0\) at \(x = -1\) (II true). III false: the local minimum \(f(1) = -2\) is not a global minimum since \(f(x) \to -\infty\) as \(x \to -\infty\) (e.g. \(f(-3) = -18\)).

Full method — how and why
Three claims about a cubic. The derivative settles the first two; the third is about global behaviour, where cubics are unbounded.
  1. Differentiate: \(f'(x) = 3x^2 - 3 = 3(x-1)(x+1)\).
    The roots \(\pm1\) are the stationary points.
  2. I: for \(x > 1\) both factors are positive, so \(f' > 0\) and \(f\) is increasing. True.
    Sign analysis of the factorised derivative.
  3. II: \(f'' = 6x\), so \(f''(-1) = -6 < 0\): local maximum at \(x = -1\). True.
    Negative second derivative means a local maximum.
  4. III: the local minimum is \(f(1) = -2\), but as \(x \to -\infty\), \(f(x) \to -\infty\) - for instance \(f(-3) = -27 + 9 = -18\). False.
    A local minimum is not a global one; odd-degree polynomials are unbounded below.
  5. I and II only.
    Match to the options.
Where it goes wrong
  • Assuming the local minimum value bounds the whole function.
  • Confusing the maximum and minimum positions - for \(+x^3\) the maximum is at the smaller \(x\).
Takeaway Cubics have no global maximum or minimum; 'local' claims need \(f'\), 'global' claims need the end behaviour.
Answer E · Differentiation (MM6.3) · question DIF-11
14Indices and surds · spec MM1.2 · Paper 2 style · ★★☆

Which of the following statements is/are true for all positive real numbers \(a\) and \(b\)?

  1. I\(\sqrt{a} \sqrt{b} = \sqrt{ab}\)
  2. II\(\sqrt{a + b} = \sqrt{a} + \sqrt{b}\)
  3. III\(\sqrt{a^2 + b^2} < a + b\)
  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 is a law of surds. II fails for \(a = b = 1\): \(\sqrt2 \neq 2\). III: \((a+b)^2 = a^2 + b^2 + 2ab > a^2 + b^2\) since \(ab > 0\), and both sides are positive, so III is true.

Full method — how and why
Three surd 'rules', two of which are genuine. The exam skill is knowing which laws distribute and finding a counterexample for the one that does not.
  1. Statement I: \(\sqrt a\sqrt b = \sqrt{ab}\) is a law of surds, valid for all \(a, b \geq 0\). True.
    Roots distribute over multiplication and division, and the condition \(a, b > 0\) in the question is exactly what the law needs.
  2. Statement II: test \(a = b = 1\): \(\sqrt{1+1} = \sqrt2 \approx 1.41\) but \(\sqrt1 + \sqrt1 = 2\). False.
    One counterexample is enough to kill a 'for all' statement - and the simplest numbers usually work.
  3. Statement III: square both sides. \((a+b)^2 = a^2 + b^2 + 2ab > a^2 + b^2\) since \(ab > 0\), and both sides are positive, so \(\sqrt{a^2+b^2} < a + b\). True.
    Squaring is safe here precisely because both sides are known to be positive; that is what makes the comparison valid in both directions.
  4. I and III are true: the answer is 'I and III only'.
    Match your three verdicts to the option list rather than reasoning about the options themselves.
Where it goes wrong
  • Assuming roots distribute over addition because they distribute over multiplication - II is the classic error this question is built around.
  • Rejecting III because the triangle-inequality picture feels like it should be an equality.
Takeaway Roots split over \(\times\) and \(\div\) but never over \(+\) or \(-\); to compare positive quantities, square them.
Answer F · Indices and surds (MM1.2) · question IND-16
15Differentiation · spec MM6.1 · Paper 2 style · ★★☆

Which of the following statements is/are true for every function \(f\) that has a derivative at every real number?

  1. IIf \(f'(a) = 0\) then \(f\) has a maximum or a minimum at \(x = a\).
  2. IIIf \(f\) has a local minimum at \(x = a\) then \(f'(a) = 0\).
  3. IIIIf \(f'(x) > 0\) for all \(x\), then \(f(3) > f(2)\).
  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 is false: \(f(x) = x^3\) has \(f'(0) = 0\) but no turning point. II is true (a stationary point is required at an interior local extremum of a differentiable function). III is true: a strictly increasing function has \(f(3) > f(2)\).

Full method — how and why
Three general claims about differentiable functions. The first is the classic false one; the others are standard facts.
  1. I: test \(f(x) = x^3\) at \(a = 0\): \(f'(0) = 0\) but the function is increasing throughout, so there is no maximum or minimum. False.
    A stationary point can be a point of inflexion.
  2. II: at an interior local minimum of a differentiable function the derivative must vanish (otherwise you could move downhill in one direction). True.
    This is Fermat's stationary-point theorem, and is why we solve \(f' = 0\) in the first place.
  3. III: if \(f' > 0\) everywhere then \(f\) is strictly increasing, so \(f(3) > f(2)\). True.
    Positive gradient throughout means the value rises as \(x\) rises.
  4. II and III only.
    Match to the options.
Where it goes wrong
  • Accepting I because 'we find maxima and minima by setting \(f' = 0\)' - that finds candidates, not guarantees.
  • Doubting II by thinking of endpoints, which the word 'every real number' excludes.
Takeaway \(f'(a) = 0\) is necessary but not sufficient for a turning point; inflexions are the counterexample.
Answer G · Differentiation (MM6.1) · question DIF-19
16Chains of reasoning · spec Prf5 · Paper 2 style · ★★☆

A sequence of \(10\) integers is such that the sum of any three consecutive terms is positive and the sum of any five consecutive terms is negative. Which of the following statements must be true?

  1. IThe sum of all \(10\) terms is negative.
  2. IIThe sequence contains at least one negative term.
  3. IIIThe sequence contains at least one positive term.
  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

Terms \(1\)-\(5\) and \(6\)-\(10\) are two blocks of five, each with negative sum (I). A five-block has negative sum, so some term is negative (II). A three-block has positive sum, so some term is positive (III).

Full method — how and why
Group the terms into blocks that the given conditions control, then read off each claim.
  1. I: split the ten terms into terms \(1\)-\(5\) and terms \(6\)-\(10\) - two blocks of five consecutive terms.
    Each block has negative sum by hypothesis, and the two blocks together are the whole sequence.
  2. Negative \(+\) negative \(=\) negative, so the total is negative. True.
    The grouping is exact - no term is used twice or left out.
  3. II: terms \(1\)-\(5\) sum to a negative number, so at least one of them is negative (five non-negative terms could not). True.
    A sum is negative only if some summand is.
  4. III: terms \(1\)-\(3\) sum to a positive number, so at least one is positive. True.
    Same argument the other way round.
  5. So all three.
    Match to the list. (Such sequences do exist for \(10\) terms - the conditions become contradictory only for longer sequences.)
Where it goes wrong
  • Trying to construct an explicit sequence; the claims follow from the block structure alone.
  • Thinking the sum of all ten is ambiguous because \(10\) is not a multiple of \(3\) - use the blocks of five, which do fit.
Takeaway Partition the sequence into blocks the hypotheses control; a negative sum forces a negative term, a positive sum a positive one.
Answer H · Chains of reasoning (Prf5) · question RSN-15
17Identifying errors in proofs · spec Err1 · Paper 2 style · ★★☆

Claim: if \(n\) is an integer and \(n^2\) is divisible by \(4\), then \(n\) is divisible by \(4\).

  1. ISuppose \(n^2\) is divisible by \(4\).
  2. IIThen \(n^2 = 4k\) for some integer \(k\).
  3. IIISo \(n = 2\sqrt{k}\), and since \(n\) is an integer \(\sqrt k\) is an integer.
  4. IVHence \(n\) is even, so \(n = 2m\), and \(n^2 = 4m^2\) is divisible by \(4\), so \(n\) is divisible by \(4\).

Which of the following is true?

  1. AThe claim is false (\(n = 2\) is a counterexample) and the first invalid step is on line IV.
  2. BThe claim and proof are correct.
  3. CThe claim is true but the first error is on line III.
  4. DThe claim is false and the first error is on line II.
  5. EThe claim is false and the first error is on line I.
Worked solution

\(n = 2\): \(n^2 = 4\) is divisible by \(4\) but \(n\) is not. Lines I-III are correct deductions (\(n\) even). Line IV concludes '\(n\) divisible by \(4\)' from '\(n^2\) divisible by \(4\)', which is exactly the unproved claim - circular and false.

Full method — how and why
A circular argument: the last line assumes the very thing being proved. Also, the claim is false.
  1. Test the claim: \(n = 2\) gives \(n^2 = 4\), divisible by \(4\), but \(2\) is not. Claim false.
    Establish this first; an error must exist.
  2. Lines I-III: from \(n^2 = 4k\), \(n^2\) is even so \(n\) is even. The algebra \(n = 2\sqrt k\) with \(\sqrt k\) an integer is correct given \(n\) is an integer. Valid.
    These lines legitimately establish that \(n\) is even.
  3. Line IV: it writes \(n = 2m\), computes \(n^2 = 4m^2\) (true but irrelevant), and then concludes '\(n\) is divisible by \(4\)'. That conclusion does not follow from anything. First error.
    Showing \(n^2\) is divisible by \(4\) was the hypothesis; restating it is not progress. Divisibility of \(n\) by \(4\) would require \(m\) even, which is never shown.
  4. Compare with the working proof for \(8 \mid n^2 \Rightarrow 4 \mid n\): there the extra factor forces \(m\) even. Here there is no spare factor.
    Seeing where the analogous proof gets its leverage shows exactly what is missing.
Where it goes wrong
  • Accepting line IV because every individual equation in it is true - the inference is what fails.
  • Blaming line III; introducing \(\sqrt k\) is unusual but not invalid here.
  • Accepting the claim because \(4 \mid n \Rightarrow 4 \mid n^2\) (the converse) is true.
Takeaway A step that restates the hypothesis and then asserts the conclusion is circular. \(4\mid n^2\) gives only \(2 \mid n\).
Answer A · Identifying errors in proofs (Err1) · question ERR-08
18Polynomials, factor and remainder theorems · spec MM1.6 · Paper 2 style · ★★☆

Which of the following statements about the polynomial \(p(x) = x^3 - 6x^2 + 11x - 6\) is/are true?

  1. I\(p(x)\) has three distinct real roots.
  2. II\((x - 4)\) is a factor of \(p(x) - 6\).
  3. III\(p(x) > 0\) for all \(x > 3\).
  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

\(p(x) = (x-1)(x-2)(x-3)\), so I is true. \(p(4) - 6 = 3\cdot2\cdot1 - 6 = 0\), so II is true. For \(x > 3\) all three factors are positive, so III is true.

Full method — how and why
Three statements about one cubic. Factorising it first makes all three immediate.
  1. Factorise: \(p(1) = 1 - 6 + 11 - 6 = 0\), so \((x-1)\) is a factor and \(p(x) = (x-1)(x-2)(x-3)\).
    Test the divisors of \(6\); the resulting quadratic \(x^2 - 5x + 6\) factorises easily.
  2. I: the roots \(1\), \(2\), \(3\) are distinct. True.
    Three distinct linear factors means three distinct roots.
  3. II: \((x-4)\) is a factor of \(p(x) - 6\) iff \(p(4) - 6 = 0\). Now \(p(4) = 3 \times 2 \times 1 = 6\), so \(p(4) - 6 = 0\). True.
    Factor Theorem applied to the shifted polynomial - evaluate the whole expression \(p(x) - 6\) at \(x = 4\).
  4. III: for \(x > 3\) each factor \(x-1\), \(x-2\), \(x-3\) is positive, so the product is positive. True.
    Sign analysis from the factorised form; beyond the largest root everything is positive.
  5. All three: 'I, II and III'.
    Match verdicts to options.
Where it goes wrong
  • Expanding \(p(x) - 6\) and re-testing from scratch instead of just evaluating \(p(4) - 6\).
  • Checking III at one value like \(x = 10\) and calling it proved - the factor argument covers every \(x > 3\).
Takeaway Factorise first: roots, sign and shifted-factor questions then all read off the same form.
Answer H · Polynomials, factor and remainder theorems (MM1.6) · question POL-08
19The logic of arguments · spec Arg3 · Paper 2 style · ★★☆

Consider the statement: there exists a real number \(k\) such that for all real \(x\), \(x^2 + kx + 1 > 0\). Which of the following is true?

  1. AThe statement is true; for example \(k = 0\) works.
  2. BThe statement is false because \(x^2 + kx + 1\) can be negative for large \(x\).
  3. CThe statement is true; every value of \(k\) works.
  4. DThe statement is false because for every \(k\) there is an \(x\) with \(x^2 + kx + 1 \leq 0\).
  5. EThe statement is true; for example \(k = 2\) works.
Worked solution

Need discriminant \(k^2 - 4 < 0\), i.e. \(-2 < k < 2\). So some (not all) values of \(k\) work; \(k = 0\) gives \(x^2 + 1 > 0\). (\(k = 2\) gives \((x+1)^2\), which is \(0\) at \(x = -1\).)

Full method — how and why
An \(\exists \forall\) statement: you only need to produce one \(k\) that works for all \(x\). Then check why the offered reasons are right or wrong.
  1. Try \(k = 0\): \(x^2 + 1 > 0\) for every real \(x\), since \(x^2 \geq 0\).
    One successful witness proves an existential statement.
  2. So the statement is true. Now check the reason offered with each claim.
    In a 'true because...' question both the verdict and the justification must be right.
  3. Is it true for every \(k\)? No: the condition is discriminant \(< 0\), i.e. \(k^2 - 4 < 0\), so only \(-2 < k < 2\).
    A positive quadratic with no real roots never crosses the axis.
  4. And \(k = 2\) gives \((x+1)^2\), which equals \(0\) at \(x = -1\) - so that witness fails the strict inequality.
    The boundary case of the discriminant gives \(\geq 0\), not \(> 0\).
  5. So: true, with \(k = 0\) as the witness.
    Match to the list.
Where it goes wrong
  • Claiming it holds for all \(k\) - \(k = 5\) gives \(x^2+5x+1\), which is negative at \(x = -1\).
  • Offering \(k = 2\) as the witness, which fails only at the single point \(x = -1\) - easy to miss.
  • Thinking a quadratic with a positive \(x^2\) coefficient must eventually be negative; it is large positive for large \(|x|\).
Takeaway To prove \(\exists\), exhibit one witness - but check the witness satisfies the condition strictly where required.
Answer A · The logic of arguments (Arg3) · question LGC-12
20Chains of reasoning · spec Prf5 · Paper 2 style · ★★★

The integers \(a, b, c\) satisfy \(a + b + c = 0\) and \(abc = 12\). Which of the following statements must be true?

  1. IExactly one of \(a, b, c\) is positive.
  2. II\(a^2 + b^2 + c^2\) is even.
  3. IIIOne of \(a, b, c\) is divisible by \(3\).
  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

\(abc > 0\) means either none or exactly two of them are negative; they cannot all be positive with sum \(0\), so exactly two are negative and one positive (I). \(a^2 + b^2 + c^2 = (a+b+c)^2 - 2(ab + bc + ca) = -2(ab + bc + ca)\) is even (II). \(3\) is prime and divides \(abc\), so it divides one of them (III). (Example: \(-1, -3, 4\).)

Full method — how and why
Three claims about integers constrained by a sum and a product. Sign analysis, an algebraic identity, and a prime-divisibility fact.
  1. I: \(abc = 12 > 0\), so the number of negatives is even: \(0\) or \(2\). If none were negative, three positives could not sum to \(0\). So exactly two are negative and one positive. True.
    A positive product means an even count of negative factors; note \(0\) is impossible since \(abc \neq 0\).
  2. II: use \((a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)\). With \(a+b+c=0\), \(a^2+b^2+c^2 = -2(ab+bc+ca)\).
    The identity converts a sum of squares into twice an integer.
  3. That is \(2 \times\) an integer, hence even. True.
    \(ab+bc+ca\) is an integer because \(a,b,c\) are.
  4. III: \(3\) is prime and \(3 \mid 12 = abc\), so \(3\) divides one of the factors. True.
    Euclid's lemma: a prime dividing a product divides one of the factors. This would fail for a composite like \(4\).
  5. So all three. (Check with \(-1, -3, 4\): sum \(0\) ✓, product \(12\) ✓.)
    A concrete example confirms the constraints are satisfiable at all.
Where it goes wrong
  • Assuming exactly one negative because the product is positive - that would make the product negative.
  • Trying to settle II by testing examples rather than using the identity.
  • Thinking III needs a specific example; it follows from primality alone.
Takeaway \(a+b+c=0\) makes \(a^2+b^2+c^2 = -2(ab+bc+ca)\); and a prime dividing a product must divide a factor.
Answer H · Chains of reasoning (Prf5) · question RSN-02
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