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Paper 2 · Set 12

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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1Straight lines · spec MM3.1 · Paper 1 style · ★☆☆

The points \(A(1, 2)\), \(B(5, 4)\) and \(C(k, 8)\) are collinear. Find \(k\).

  1. A\(12\)
  2. B\(11\)
  3. C\(10\)
  4. D\(9\)
  5. E\(7\)
  6. F\(13\)
Worked solution

Gradient \(AB = \frac{2}{4} = \frac12\). Gradient \(AC = \frac{6}{k-1} = \frac12 \Rightarrow k - 1 = 12 \Rightarrow k = 13\).

Full method — how and why
Collinear points have equal gradients between any pair. Set two gradients equal and solve.
  1. Gradient of \(AB\): \(\dfrac{4 - 2}{5 - 1} = \dfrac{2}{4} = \dfrac12\).
    Any two of the three points determine the line, so compute the gradient from the pair with known coordinates.
  2. Gradient of \(AC\): \(\dfrac{8 - 2}{k - 1} = \dfrac{6}{k-1}\).
    Use the same starting point \(A\) for both gradients to avoid sign confusion.
  3. Set them equal: \(\dfrac{6}{k-1} = \dfrac12\).
    Collinear means the gradient between any pair is the same.
  4. Cross-multiply: \(k - 1 = 12\), so \(k = 13\).
    Check: from \(B(5,4)\) to \(C(13,8)\) the gradient is \(\frac48 = \frac12\) ✓.
Where it goes wrong
  • Forgetting the \(-1\) and answering \(12\).
  • Inverting the gradient and solving \(\frac{k-1}{6} = \frac12\), which gives \(k = 4\).
Takeaway Collinearity is an equal-gradient condition; always verify with the third pair of points.
Answer F · Straight lines (MM3.1) · question LIN-02
2Polynomials, 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
3Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

Let \(f(x) = x^2 - 2x - 8\). For how many integers \(n\) is \(f(n) < 0\)?

  1. A\(8\)
  2. B\(6\)
  3. C\(5\)
  4. D\(7\)
  5. E\(3\)
  6. F\(4\)
Worked solution

\((x-4)(x+2) < 0\) gives \(-2 < x < 4\); the integers are \(-1, 0, 1, 2, 3\): five of them.

Full method — how and why
Counting integers in the region where a quadratic is negative. Solve the inequality, then count carefully at the ends.
  1. Factorise: \(n^2 - 2n - 8 = (n - 4)(n + 2)\).
    Two numbers multiplying to \(-8\) and adding to \(-2\): \(-4\) and \(+2\).
  2. Negative between the roots: \(-2 < n < 4\).
    Upward parabola, so \(f < 0\) strictly between the roots.
  3. List the integers strictly inside: \(-1, 0, 1, 2, 3\) - five of them.
    The endpoints \(-2\) and \(4\) give \(f = 0\), which is not less than zero, so they are excluded; and \(0\) is easy to forget.
Where it goes wrong
  • Including \(-2\) and \(4\) and answering \(7\).
  • Forgetting \(0\) is an integer and answering \(4\).
Takeaway After solving an inequality, list the integers explicitly - endpoints and zero are the two things people drop.
Answer C · Quadratic functions (MM1.3) · question QUA-24
4Identifying 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
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
6Mathematical proof · spec Prf1 · Paper 2 style · ★☆☆

To prove 'if \(n^2\) is even then \(n\) is even' by contradiction, which assumption should be made?

  1. A\(n\) is odd.
  2. B\(n^2\) is even and \(n\) is odd.
  3. C\(n^2\) is even.
  4. D\(n^2\) is odd.
  5. E\(n\) is even and \(n^2\) is odd.
Worked solution

Assume the hypothesis holds and the conclusion fails, then derive a contradiction: \(n\) odd gives \(n^2\) odd, contradicting \(n^2\) even.

Full method — how and why
Setting up a proof by contradiction for an implication: assume the hypothesis and the negated conclusion.
  1. The claim is 'if \(n^2\) is even then \(n\) is even': hypothesis \(n^2\) even, conclusion \(n\) even.
    Identify both parts before negating anything.
  2. For contradiction, suppose the whole implication fails. An implication fails only when the hypothesis holds and the conclusion does not.
    This is the one case that would make \(A \Rightarrow B\) false.
  3. So assume: \(n^2\) is even and \(n\) is odd.
    Both, not just one.
  4. (The proof then runs: \(n = 2k+1 \Rightarrow n^2 = 4k^2+4k+1\), odd - contradicting \(n^2\) even.)
    The contradiction is with the retained hypothesis.
Where it goes wrong
  • Assuming only '\(n\) is odd' and dropping the hypothesis - then there is nothing to contradict.
  • Assuming '\(n^2\) is odd', which negates the hypothesis instead of the conclusion.
  • Assuming '\(n\) even and \(n^2\) odd', which negates the wrong part of each.
Takeaway To contradict \(A \Rightarrow B\), assume \(A\) and \(\neg B\) - keep the hypothesis, negate only the conclusion.
Answer B · Mathematical proof (Prf1) · question PRF-14
7Angles, polygons, congruence and vectors · spec M5.3 · Paper 2 style · ★☆☆

\(ABCD\) is a quadrilateral. \(P\): 'the diagonals of \(ABCD\) bisect each other'. \(Q\): '\(ABCD\) is a rectangle'. 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

Every rectangle is a parallelogram, and a parallelogram's diagonals bisect each other: \(P\) is necessary. A non-rectangular parallelogram has bisecting diagonals without being a rectangle: not sufficient.

Full method — how and why
A quadrilateral hierarchy question. Bisecting diagonals characterise parallelograms, a larger class than rectangles.
  1. Every rectangle is a parallelogram, and a parallelogram's diagonals bisect each other. So \(Q \Rightarrow P\): \(P\) is necessary.
    Opposite sides equal and parallel forces the diagonals to share a midpoint.
  2. Is \(P\) sufficient? Take a non-rectangular parallelogram, e.g. a rhombus with angles \(60^\circ/120^\circ\): diagonals bisect each other, but it is not a rectangle.
    Bisecting diagonals say nothing about the angles.
  3. So \(P\) is necessary but not sufficient.
    For a rectangle you also need the diagonals to be equal in length.
Where it goes wrong
  • Answering 'necessary and sufficient' by confusing 'bisect each other' with 'bisect each other and are equal'.
  • Getting the direction backwards: the broader property (parallelogram) is necessary for the narrower one (rectangle).
Takeaway Diagonals bisect each other \(\iff\) parallelogram; add equal diagonals for a rectangle, add perpendicular for a rhombus.
Answer B · Angles, polygons, congruence and vectors (M5.3) · question X2-20
8The logic of arguments · spec Arg4 · Paper 2 style · ★☆☆

Which of the following is the negation of the statement 'every student in the class scored more than \(50\)'?

  1. AAt least one student in the class scored more than \(50\).
  2. BEvery student in the class scored \(50\) or less.
  3. CEvery student in the class scored less than \(50\).
  4. DAt least one student in the class scored \(50\) or less.
  5. ENo student in the class scored more than \(50\).
Worked solution

The negation of 'for all \(x\), \(P(x)\)' is 'there exists \(x\) such that not \(P(x)\)'. 'Not more than \(50\)' means '\(50\) or less'.

Full method — how and why
Negating a universal statement. 'All' flips to 'at least one', and the property inside flips too.
  1. Write it formally: 'for all students \(s\), mark\((s) > 50\)'.
    Formalising stops the English from misleading you.
  2. Negate: 'there exists a student \(s\) with mark\((s) \not> 50\)'.
    \(\neg \forall x\, P(x) \equiv \exists x\, \neg P(x)\): the quantifier flips and the inside is negated.
  3. Translate \(\not> 50\) into '\(\leq 50\)', i.e. \(50\) or less.
    The negation of 'greater than' is 'less than or equal to' - the boundary belongs to the negation.
  4. So: at least one student scored \(50\) or less.
    One failure is all it takes to make 'every' false.
Where it goes wrong
  • Over-negating to 'every student scored \(50\) or less' - that is far stronger than the negation and is not implied.
  • Writing 'less than \(50\)' and losing the case of exactly \(50\).
Takeaway Negating 'all ... are' gives 'at least one ... is not' - and remember \(\neg(x > a)\) is \(x \leq a\).
Answer D · The logic of arguments (Arg4) · question LGC-03
9Polynomials, factor and remainder theorems · spec MM1.6 · Paper 2 style · ★☆☆

\(p(x)\) is a polynomial. \(P\): '\(p(2) = 0\)'. \(Q\): '\((x - 2)^2\) is a factor of \(p(x)\)'. 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

If \((x-2)^2\) divides \(p\) then certainly \(p(2) = 0\), so \(P\) is necessary for \(Q\). But \(p(x) = x - 2\) has \(p(2) = 0\) without a repeated factor, so \(P\) is not sufficient.

Full method — how and why
The factor theorem gives one direction; a repeated factor is strictly more than a root.
  1. If \((x-2)^2\) is a factor, then \((x-2)\) is a factor, so \(p(2) = 0\): \(Q \Rightarrow P\). \(P\) is necessary.
    Factor theorem: \((x-a)\) is a factor exactly when \(p(a)=0\).
  2. Is \(P\) sufficient? Take \(p(x) = x-2\): \(p(2)=0\), but \((x-2)^2\) does not divide a degree-\(1\) polynomial.
    A single root is not a repeated root.
  3. So \(P\) is necessary but not sufficient.
    Note the direction: the weaker condition is necessary for the stronger one.
Where it goes wrong
  • Answering 'sufficient but not necessary' - the common pattern, but reversed here.
  • Thinking \(p(2)=0\) and \(p'(2)=0\) together would be needed; that is exactly the extra condition for a repeated root, and its absence is the point.
Takeaway \(p(a)=0 \iff (x-a)\) is a factor; a repeated factor also needs \(p'(a)=0\).
Answer B · Polynomials, factor and remainder theorems (MM1.6) · question X2-08
10Circles and circle theorems · spec MM3.2 · Paper 2 style · ★★☆

Two circles have equations \(x^2 + y^2 = 25\) and \((x - 8)^2 + y^2 = 9\). Which of the following statements is/are true?

  1. IThe circles touch externally.
  2. IIThe circles intersect at exactly two points.
  3. IIIThe point \((5, 0)\) lies on both circles.
  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

Radii \(5\) and \(3\), centres \(8\) apart \(= 5 + 3\), so the circles touch externally at one point (I true, II false). That point is \((5, 0)\): \(25 = 25\) and \(9 + 0 = 9\) (III true).

Full method — how and why
Two circles: compare the distance between centres with the sum and difference of the radii.
  1. Read off centres and radii: \((0,0)\) with \(r = 5\); \((8,0)\) with \(r = 3\).
    Both are already in centre-radius form.
  2. Distance between centres: \(8\). Compare with \(r_1 + r_2 = 8\).
    Equal to the sum means the circles touch externally - exactly one common point.
  3. I is therefore true, and II (two intersection points) is false.
    Two intersections would need \(|r_1 - r_2| < d < r_1 + r_2\), i.e. \(2 < d < 8\) strictly.
  4. III: check \((5,0)\) in both: \(25 + 0 = 25\) ✓ and \((5-8)^2 + 0 = 9\) ✓. True.
    The touching point lies on the line joining the centres, \(5\) from one and \(3\) from the other.
  5. I and III only.
    Match to the options.
Where it goes wrong
  • Assuming circles that 'meet' must meet twice.
  • Forgetting to test III in both equations.
Takeaway \(d = r_1 + r_2\): touch externally; \(d = |r_1 - r_2|\): touch internally; strictly between: two points.
Answer F · Circles and circle theorems (MM3.2) · question CIR-11
11Chains of reasoning · spec Prf5 · Paper 2 style · ★★☆

How many positive integers less than \(1000\) are divisible by neither \(2\) nor \(5\)?

  1. A\(399\)
  2. B\(600\)
  3. C\(500\)
  4. D\(400\)
  5. E\(450\)
  6. F\(300\)
Worked solution

Of \(1\) to \(999\): \(499\) multiples of \(2\), \(199\) of \(5\), \(99\) of \(10\). Divisible by \(2\) or \(5\): \(499 + 199 - 99 = 599\). Neither: \(999 - 599 = 400\).

Full method — how and why
Inclusion-exclusion: count what you don't want and subtract. Take care with 'less than \(1000\)'.
  1. The range is \(1\) to \(999\), so \(999\) integers.
    'Less than \(1000\)' excludes \(1000\) itself.
  2. Multiples of \(2\): \(\left\lfloor \frac{999}{2}\right\rfloor = 499\). Multiples of \(5\): \(199\). Multiples of both, i.e. of \(10\): \(99\).
    Divide and round down. \(10\) is the LCM of \(2\) and \(5\).
  3. Divisible by \(2\) or \(5\): \(499 + 199 - 99 = 599\).
    Subtract the overlap, which was counted twice.
  4. Neither: \(999 - 599 = 400\).
    Sanity check: in each block of \(10\) consecutive integers exactly \(4\) (ending in \(1,3,7,9\)) survive, and \(999 = 99\) full blocks plus \(991\)-\(999\), giving \(99 \times 4 + 4 = 400\) ✓.
Where it goes wrong
  • Using \(1000\) as the count and getting \(400\) by luck with wrong intermediate figures - or getting \(399\) by an off-by-one at the top.
  • Forgetting to subtract the multiples of \(10\), giving \(999 - 698 = 301\).
Takeaway \(|A \cup B| = |A| + |B| - |A \cap B|\), with the intersection counted via the LCM; then subtract from the total.
Answer D · Chains of reasoning (Prf5) · question RSN-12
12Trigonometric functions and equations · spec MM4.5 · Paper 2 style · ★★☆

Which of the following is/are true for all real \(x\)?

  1. I\(\sin^2 x + \cos^2 x = 1\)
  2. II\(\sin(2x) = 2\sin x\)
  3. III\(\sin(x + \pi) = -\sin x\)
  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

I is the Pythagorean identity. II is false (e.g. \(x = \frac\pi2\): \(0 \neq 2\)). III: translating \(\sin\) by half a period negates it, so III is true.

Full method — how and why
Three identity claims. Two are standard, one is a classic false 'linearity' claim.
  1. I: \(\sin^2 x + \cos^2 x = 1\) is the Pythagorean identity, true for every \(x\). True.
    It follows from Pythagoras applied to the unit circle.
  2. II: test \(x = \frac{\pi}{2}\): \(\sin\pi = 0\) but \(2\sin\frac\pi2 = 2\). False.
    Trig functions are not linear: \(\sin(2x) \neq 2\sin x\) (the correct identity, off-specification here, is \(2\sin x\cos x\)).
  3. III: shifting \(\sin\) by half a period reflects it: \(\sin(x + \pi) = -\sin x\). True.
    Check at \(x = 0\): \(\sin\pi = 0 = -\sin 0\) ✓; at \(x = \frac\pi2\): \(\sin\frac{3\pi}{2} = -1 = -\sin\frac\pi2\) ✓.
  4. I and III only.
    Match to the options.
Where it goes wrong
  • Accepting II because it 'looks like' a factor coming out.
  • Doubting III - test it at two values to be sure.
Takeaway Functions are almost never linear: \(f(2x) \neq 2f(x)\) in general. Test suspicious identities at \(x = \frac\pi2\).
Answer F · Trigonometric functions and equations (MM4.5) · question TRG-17
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
14Chains of reasoning · spec Prf5 · Paper 2 style · ★★☆

\(x\) and \(y\) are positive integers with \(x^2 - y^2 = 45\). How many possible pairs \((x, y)\) are there?

  1. A\(0\)
  2. B\(6\)
  3. C\(2\)
  4. D\(1\)
  5. E\(3\)
  6. F\(4\)
Worked solution

\((x - y)(x + y) = 45\) with \(0 < x - y < x + y\): \((1, 45), (3, 15), (5, 9)\) give \((x, y) = (23, 22), (9, 6), (7, 2)\). Three pairs.

Full method — how and why
A difference of two squares turns a quadratic condition into a factor-pair search.
  1. Factorise: \(x^2 - y^2 = (x-y)(x+y) = 45\).
    Difference of two squares converts the equation into a product of two integers.
  2. Both factors are positive with \(x - y < x + y\), so list factor pairs of \(45\) with the smaller first: \((1,45), (3,15), (5,9)\).
    \(x, y\) positive forces \(x + y > 0\), and then \(x - y > 0\) too since the product is positive.
  3. Solve each: \(x = \frac{\text{sum}}{2}\), \(y = \frac{\text{difference}}{2}\), giving \((23,22), (9,6), (7,2)\).
    Adding the two equations gives \(2x\); subtracting gives \(2y\). Both factors are odd, so both results are integers.
  4. All three have positive integer \(y\), so there are three pairs.
    Check one: \(9^2 - 6^2 = 81-36 = 45\) ✓.
Where it goes wrong
  • Including \((9,5)\) or other reversed pairs - \(x-y\) must be the smaller factor.
  • Forgetting to check that \(x\) and \(y\) come out as integers; if \(45\) had a factor pair of mixed parity it would have to be discarded.
  • Counting \(45 = 45 \times 1\) and \(1 \times 45\) as two different solutions.
Takeaway \(x^2-y^2 = N\): list factor pairs \(N = de\) with \(d
Answer E · Chains of reasoning (Prf5) · question RSN-04
15Mathematical proof · spec Prf3 · Paper 2 style · ★★☆

A student notices that \(1 + 2 = 3\), \(4 + 5 + 6 = 7 + 8\), \(9 + 10 + 11 + 12 = 13 + 14 + 15\) and conjectures that the pattern continues. Which of the following describes the \(n\)th line of the pattern?

  1. AThe \(n\)th line starts at \(n^2\) and has \(n + 1\) numbers on the left and \(n\) on the right.
  2. BThe \(n\)th line starts at \(n(n+1)\) and has \(n + 1\) numbers on the left.
  3. CThe \(n\)th line starts at \(n^2\) and has \(n\) numbers on the left and \(n + 1\) on the right.
  4. DThe \(n\)th line starts at \(n^2\) and has \(n\) numbers on each side.
  5. EThe \(n\)th line starts at \(2n - 1\) and has \(n + 1\) numbers on the left.
Worked solution

Line \(1\) starts at \(1\), line \(2\) at \(4\), line \(3\) at \(9\): start \(n^2\). Left sides have \(2, 3, 4\) numbers (\(n + 1\)), right sides \(1, 2, 3\) (\(n\)).

Full method — how and why
Describing a pattern precisely: check the start value and the two counts separately against every given line.
  1. Starting values: \(1, 4, 9\) for lines \(1, 2, 3\) - these are \(n^2\).
    Test against all three lines, not just one.
  2. Count the left sides: \(\{1,2\}\), \(\{4,5,6\}\), \(\{9,10,11,12\}\) - that is \(2, 3, 4\) terms, i.e. \(n+1\).
    Count carefully; being off by one here is the whole point of the question.
  3. Count the right sides: \(\{3\}\), \(\{7,8\}\), \(\{13,14,15\}\) - that is \(1, 2, 3\) terms, i.e. \(n\).
    Left has one more term than right, which is what makes the equality possible (the right-hand terms are larger).
  4. So: starts at \(n^2\), with \(n+1\) terms on the left and \(n\) on the right.
    Check the total count: \((n+1) + n = 2n+1\), and indeed line \(3\) runs from \(9\) to \(15\), seven numbers ✓.
Where it goes wrong
  • Swapping the counts (\(n\) left, \(n+1\) right) - the sides would not balance, since the right-hand numbers are bigger.
  • Reading the start as \(2n-1\) or \(n(n+1)\) - both fail at line \(2\) (\(3\) and \(6\) rather than \(4\)).
Takeaway Verify a conjectured pattern against every given line, and check each component (start, length) separately.
Answer A · Mathematical proof (Prf3) · question PRF-11
16Chains of reasoning · spec Prf5 · Paper 2 style · ★★☆

A selection of \(n\) distinct integers is taken from \(1, 2, \ldots, 20\). What is the smallest \(n\) that guarantees two of the chosen integers differ by exactly \(10\)?

  1. A\(2\)
  2. B\(15\)
  3. C\(12\)
  4. D\(11\)
  5. E\(20\)
  6. F\(10\)
Worked solution

Pair up \(\{1, 11\}, \{2, 12\}, \ldots, \{10, 20\}\): ten pairs. Choosing \(11\) numbers forces two from the same pair (pigeonhole); choosing one from each pair gives \(10\) numbers with no such difference.

Full method — how and why
A pigeonhole question. Build the 'boxes' so that two items in one box give exactly what you want, then answer boxes \(+\,1\).
  1. Pair the numbers so that each pair differs by \(10\): \(\{1,11\}, \{2,12\}, \ldots, \{10,20\}\).
    Every number from \(1\) to \(20\) lies in exactly one pair, and two numbers differ by \(10\) precisely when they share a pair.
  2. There are \(10\) pairs, so \(10\) numbers can be chosen with no two in the same pair - for example \(1\) to \(10\).
    This shows \(10\) is not enough, which is the half of the argument people skip.
  3. With \(11\) numbers and only \(10\) pairs, two must come from the same pair - and those differ by \(10\).
    Pigeonhole: more items than boxes forces a repeat.
  4. So the smallest guaranteed \(n\) is \(11\).
    Boxes \(+ 1\).
Where it goes wrong
  • Answering \(10\) - the largest selection that can avoid the condition, not the smallest that forces it.
  • Answering \(2\), confusing 'could happen' with 'guaranteed'.
  • Forgetting to show \(10\) is achievable; without it you have not proved \(11\) is smallest.
Takeaway 'Smallest \(n\) that guarantees' \(=\) (largest bad selection) \(+\ 1\). Construct the boxes so a shared box means success.
Answer D · Chains of reasoning (Prf5) · question RSN-01
17Identifying errors in proofs · spec Err1 · Paper 2 style · ★★☆

Claim: if \(f'(a) = 0\) and \(f''(a) = 0\) then \(f\) has a point of inflexion at \(x = a\).

Which of the following shows the claim is false?

  1. A\(f(x) = x^3 - 3x\) at \(a = 1\)
  2. B\(f(x) = x\) at \(a = 0\)
  3. C\(f(x) = x^2\) at \(a = 0\)
  4. D\(f(x) = x^3\) at \(a = 0\)
  5. E\(f(x) = x^4\) at \(a = 0\)
Worked solution

\(x^4\) has \(f'(0) = f''(0) = 0\) but a minimum at \(0\), not an inflexion. (\(x^3\) does have an inflexion; \(x^2\) has \(f''(0) \neq 0\).)

Full method — how and why
To disprove a claim about a test for inflexion, produce a function satisfying the conditions but failing the conclusion.
  1. Need \(f'(a) = f''(a) = 0\) with no inflexion at \(a\).
    Both hypotheses must hold, and the conclusion must fail.
  2. Take \(f(x) = x^4\) at \(a=0\): \(f'(x) = 4x^3\) so \(f'(0)=0\) ✓; \(f''(x) = 12x^2\) so \(f''(0)=0\) ✓.
    Both derivatives vanish - the hypotheses are satisfied.
  3. But \(x^4\) has a minimum at \(0\) (it is positive on both sides), not a point of inflexion.
    An inflexion requires the concavity to change sign; \(f'' = 12x^2 \geq 0\) on both sides, so it does not.
  4. Check the others: \(x^3\) does have an inflexion at \(0\) (so it confirms, not refutes); \(x^2\) has \(f''(0) = 2 \neq 0\); \(x^3-3x\) at \(a=1\) has \(f'(1) = 0\) but \(f''(1) = 6 \neq 0\); \(f(x)=x\) has \(f'(0) = 1 \neq 0\).
    Each fails one of the two hypotheses, or satisfies the conclusion.
Where it goes wrong
  • Choosing \(x^3\), the standard example of the claim rather than a counterexample.
  • Choosing a function where one of the two derivative conditions fails at the stated point.
Takeaway \(f''(a)=0\) does not guarantee an inflexion; the second derivative must change sign. \(x^4\) at \(0\) is the standard counterexample.
Answer E · Identifying errors in proofs (Err1) · question ERR-17
18Mathematical proof · spec Prf1 · Paper 2 style · ★★☆

Here is a proof that there are infinitely many primes, with the steps out of order.

  1. IThen \(N\) leaves remainder \(1\) on division by each of \(p_1, \ldots, p_n\), so no \(p_i\) divides \(N\).
  2. IISuppose there are only finitely many primes \(p_1, p_2, \ldots, p_n\), and let \(N = p_1 p_2 \cdots p_n + 1\).
  3. IIISo \(N\) has a prime factor that is not in the list, which is a contradiction.
  4. IVBut \(N > 1\), so \(N\) has at least one prime factor.

Which is the correct order?

  1. AII, I, IV, III
  2. BII, III, I, IV
  3. CI, II, IV, III
  4. DII, I, III, IV
  5. EIV, II, I, III
Worked solution

Define \(N\) (II), show no listed prime divides it (I), note that \(N\) nevertheless has some prime factor (IV), and conclude that this factor is a new prime (III). The conclusion III cannot come before both I and IV have been established.

Full method — how and why
Ordering Euclid's proof. Again the skeleton is assume-deduce-contradict, and the dependencies fix the order.
  1. II first: assume finitely many primes \(p_1,\ldots,p_n\) and construct \(N = p_1p_2\cdots p_n + 1\).
    Everything else refers to \(N\), so it must be defined first.
  2. Then I: each \(p_i\) divides \(p_1\cdots p_n\), so dividing \(N\) by \(p_i\) leaves remainder \(1\) - no listed prime divides \(N\).
    This is the key property of the \(+1\).
  3. Then IV: \(N > 1\), so it has at least one prime factor.
    Every integer greater than \(1\) has a prime factor - a separate fact, independent of I.
  4. Finally III: that prime factor is not in the list, contradicting the assumption that the list was complete. Order: II, I, IV, III.
    The contradiction needs both I (none of the listed primes works) and IV (some prime does).
Where it goes wrong
  • Putting III immediately after II, before either supporting fact has been established.
  • Thinking the proof claims \(N\) is prime - it need not be; it merely has a prime factor outside the list.
Takeaway In a contradiction proof, the contradiction is the last line, and it must rest on everything before it.
Answer A · Mathematical proof (Prf1) · question PRF-09
19Binomial expansion · spec MM2.4 · Paper 2 style · ★★☆

Which of the following statements is/are true?

  1. I\(\binom{n}{r} = \binom{n}{n-r}\) for all integers \(0 \leq r \leq n\)
  2. II\(\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}\) for all integers \(0 \leq r < n\)
  3. III\(\displaystyle\sum_{r=0}^{n} \binom{n}{r} = 2^{n}\) for all positive integers \(n\)
  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 the symmetry of Pascal's triangle; II is Pascal's rule (each entry is the sum of the two above it); III is \((1+1)^n\) expanded. All true.

Full method — how and why
Three standard properties of binomial coefficients. Each has a one-line reason and a one-line check.
  1. I: \(\binom nr = \binom n{n-r}\) - choosing \(r\) objects is the same as choosing the \(n-r\) to leave. True.
    Visible as the left-right symmetry of each row of Pascal's triangle.
  2. II: \(\binom nr + \binom n{r+1} = \binom{n+1}{r+1}\) - Pascal's rule, i.e. each entry is the sum of the two above it. True.
    Check with numbers: \(\binom32 + \binom33 = 3 + 1 = 4 = \binom43\) ✓.
  3. III: \(\sum_{r=0}^n \binom nr = 2^n\) - put \(x = 1\) in \((1+x)^n\). True.
    Equivalently, a set with \(n\) elements has \(2^n\) subsets, counted by size.
  4. All three: 'I, II and III'.
    Match to the options.
Where it goes wrong
  • Doubting Pascal's rule because of the index shift on the right - test it with small numbers.
  • Thinking the sum in III is \(2^{n+1}\) by miscounting the \(n+1\) terms.
Takeaway Symmetry, Pascal's rule and row sum \(2^n\) are the three facts about \(\binom nr\) worth knowing cold.
Answer H · Binomial expansion (MM2.4) · question BIN-09
20Identifying errors in proofs · spec Err1 · Paper 2 style · ★★★

Claim: the function \(f(x) = x^3 - 3x^2 + 3x\) is increasing for all real \(x\).

  1. I\(f'(x) = 3x^2 - 6x + 3 = 3(x - 1)^2\).
  2. IISo \(f'(x) \geq 0\) for all \(x\), with equality only at \(x = 1\).
  3. IIISince \(f'(x) > 0\) for all \(x\), \(f\) is increasing.

Which of the following is true?

  1. ALine III overstates what was shown: \(f'(1) = 0\); but \(f\) is still increasing for all \(x\), so the claim is true and the proof needs a better justification.
  2. BThe proof is completely correct.
  3. CThe first error is on line I.
  4. DThe first error is on line II: \(f'(x)\) can be negative.
  5. EThe claim is false: \(f\) has a turning point at \(x = 1\).
Worked solution

\(f\) is (strictly) increasing because \(f'(x) \geq 0\) with equality only at an isolated point, so \(f(a) < f(b)\) whenever \(a < b\); but line III's assertion '\(f'(x) > 0\) for all \(x\)' is false at \(x = 1\). There is no turning point at \(x = 1\) (the derivative does not change sign).

Full method — how and why
A subtle one: the claim is true but one line asserts more than was proved. Distinguish a false conclusion from an unjustified step.
  1. Check line I: \(f'(x) = 3x^2 - 6x + 3 = 3(x^2-2x+1) = 3(x-1)^2\). Correct.
    Differentiate term by term, then factorise.
  2. Check line II: a square is non-negative, so \(f'(x) \geq 0\), with equality only at \(x = 1\). Correct.
    The word 'only' is right - \((x-1)^2 = 0\) just at \(x=1\).
  3. Line III says '\(f'(x) > 0\) for all \(x\)' - false at \(x = 1\), where \(f'(1) = 0\). So line III overstates line II.
    It asserts a strict inequality that was never established.
  4. Is the claim itself true? Yes: \(f' \geq 0\) everywhere with equality at a single isolated point still gives \(f(a) < f(b)\) for \(a
    A derivative vanishing at one point does not create a flat stretch; there is no turning point, because \(f'\) does not change sign.
Where it goes wrong
  • Concluding the claim is false because \(f'(1) = 0\) - a stationary point of inflexion is not a turning point.
  • Blaming line I or II, both of which are correct.
  • Calling the proof completely correct, ignoring the strictness error in line III.
Takeaway \(f' \geq 0\) with equality only at isolated points still means strictly increasing - but a proof must not claim strict positivity it has not shown.
Answer A · Identifying errors in proofs (Err1) · question ERR-06
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