Σ TMUA Practice
Home›Mock papers›Paper 2 Set 09

Paper 2 · Set 09

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

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
1Trigonometric functions and equations · spec MM4.5 · Paper 1 style · ★☆☆

Given that \(\sin\theta = \dfrac{3}{5}\) and \(\theta\) is obtuse, find \(\tan\theta\).

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

\(\cos\theta = -\frac45\) (negative in the second quadrant), so \(\tan\theta = \frac{3/5}{-4/5} = -\frac34\).

Full method — how and why
From one ratio to another, with a quadrant condition fixing the signs. Use a right-angled triangle plus CAST.
  1. From \(\sin\theta = \frac35\), the reference triangle has opposite \(3\) and hypotenuse \(5\), so the third side is \(4\) (a 3-4-5 triangle).
    Pythagoras gives the missing side; the \(3\)-\(4\)-\(5\) triple appears constantly.
  2. \(\theta\) is obtuse, so it lies in the second quadrant where \(\sin\) is positive but \(\cos\) and \(\tan\) are negative.
    CAST: only sine is positive in the second quadrant.
  3. So \(\cos\theta = -\frac45\) and \(\tan\theta = \dfrac{\sin\theta}{\cos\theta} = \dfrac{3/5}{-4/5} = -\dfrac34\).
    The fives cancel; the sign comes from the quadrant.
Where it goes wrong
  • Ignoring 'obtuse' and answering \(\frac34\).
  • Inverting the ratio and answering \(-\frac43\).
Takeaway Build the reference triangle from the given ratio, then let the quadrant decide the signs.
Answer A · Trigonometric functions and equations (MM4.5) · question TRG-06
2Circles and circle theorems · spec MM3.2 · Paper 1 style · ★☆☆

A circle passes through \((0, 0)\), \((6, 0)\) and \((0, 8)\). Find the coordinates of its centre.

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

The triangle has a right angle at the origin, so \((6, 0)\)-\((0, 8)\) is a diameter; the centre is its midpoint \((3, 4)\).

Full method — how and why
Three points determine a circle, but here two of the chords are along the axes - and the right angle at the origin gives the centre instantly.
  1. Notice the triangle \((0,0)\), \((6,0)\), \((0,8)\) has a right angle at the origin.
    The two legs lie along the axes, so the angle between them is \(90^\circ\).
  2. A right angle inscribed in a circle stands on a diameter, so the hypotenuse from \((6,0)\) to \((0,8)\) is a diameter.
    This is the converse of the angle-in-a-semicircle theorem.
  3. The centre is the midpoint of that diameter: \(\left(\frac{6+0}{2}, \frac{0+8}{2}\right) = (3, 4)\).
    Centre is the midpoint of any diameter.
Where it goes wrong
  • Averaging all three points to get \((2, \frac83)\) - that is the centroid, not the circumcentre.
  • Taking the midpoint of the wrong pair, e.g. \((3,0)\).
Takeaway If three given points form a right angle, the hypotenuse is a diameter and its midpoint is the centre.
Answer E · Circles and circle theorems (MM3.2) · question CIR-15
3Exponentials and logarithms · spec MM5.2 · Paper 1 style · ★☆☆

Solve \(\log_2 (x^2 - 4) - \log_2 (x - 2) = 3\).

  1. A\(x = 2\)
  2. B\(x = 6\)
  3. C\(x = -10\) or \(x = 6\)
  4. D\(x = 8\)
  5. E\(x = 10\)
  6. F\(x = 4\)
Worked solution

For \(x > 2\): \(\log_2\frac{(x-2)(x+2)}{x-2} = \log_2(x + 2) = 3 \Rightarrow x + 2 = 8 \Rightarrow x = 6\).

Full method — how and why
Combine the logs; the algebraic fraction simplifies dramatically because the numerator factorises.
  1. Combine: \(\log_2\dfrac{x^2-4}{x-2} = 3\).
    Difference of logs is the log of the quotient.
  2. Factorise and cancel: \(\dfrac{(x-2)(x+2)}{x-2} = x + 2\) (valid since \(x \neq 2\)).
    The domain requires \(x > 2\) for \(\log_2(x-2)\) to exist, so cancelling is legitimate.
  3. Convert: \(\log_2(x+2) = 3 \Rightarrow x + 2 = 8 \Rightarrow x = 6\).
    \(2^3 = 8\).
  4. Check the domain: \(x = 6\) gives \(x^2 - 4 = 32 > 0\) and \(x - 2 = 4 > 0\) ✓.
    Always verify against the original.
Where it goes wrong
  • Forgetting the domain and including a spurious negative solution.
  • Cancelling incorrectly to get \(x - 2\) or \(x^2 - 2\).
Takeaway Factorise before cancelling inside a log, and confirm every solution keeps all arguments positive.
Answer B · Exponentials and logarithms (MM5.2) · question LOG-15
4Identifying errors in proofs · spec Err2 · Paper 2 style · ★☆☆

A student proves that the equation \(x^2 + 2x + 3 = 0\) has no real solutions.

  1. IComplete the square: \(x^2 + 2x + 3 = (x + 1)^2 + 2\).
  2. II\((x + 1)^2 \geq 0\) for all real \(x\).
  3. IIISo \((x + 1)^2 + 2 \geq 2 > 0\) for all real \(x\).
  4. IVHence the equation has no real solutions.

Which of the following is correct?

  1. AThe proof is completely correct.
  2. BThe first error is on line I.
  3. CThe first error is on line II: \((x + 1)^2\) can be negative.
  4. DThe first error is on line III.
  5. EThe first error is on line IV: a positive expression can still equal zero.
Worked solution

Every step is valid.

Full method — how and why
Another valid proof. Completing the square and bounding below is a standard, correct technique.
  1. Line I: \((x+1)^2 = x^2+2x+1\), so \(x^2+2x+3 = (x+1)^2 + 2\) ✓.
    Half the coefficient of \(x\) is \(1\); correct by \(3 - 1 = 2\).
  2. Line II: any real square is \(\geq 0\) ✓.
    True for every real \(x\).
  3. Line III: adding \(2\) gives \(\geq 2\), and \(2 > 0\) ✓.
    Adding a constant to both sides of an inequality is always valid.
  4. Line IV: an expression that is always at least \(2\) can never equal \(0\), so there are no real roots ✓.
    That is the definition of having no real solutions. (Check: discriminant \(4 - 12 = -8 < 0\) ✓.)
Where it goes wrong
  • Claiming \((x+1)^2\) can be negative - not for real \(x\).
  • Claiming 'a positive expression can still equal zero' - it cannot; that is what positive means.
  • Assuming there must be an error.
Takeaway Completing the square gives a rigorous lower bound; if the minimum is positive, there are no real roots.
Answer A · Identifying errors in proofs (Err2) · question ERR-09
5Simultaneous equations and inequalities · spec MM1.5 · Paper 2 style · ★☆☆

\(x\) is a real number. \(P\): '\(x^2 < 4\)'. \(Q\): '\(x < 2\)'. 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

\(x^2 < 4\) means \(-2 < x < 2\), which implies \(x < 2\) (sufficient); but \(x = -5\) satisfies \(Q\) and not \(P\) (not necessary).

Full method — how and why
Solve the quadratic inequality properly, then compare the two solution sets as intervals.
  1. \(x^2<4\) means \(|x|<2\), i.e. \(-2
    A quadratic inequality of this shape gives an interval between the roots.
  2. Every \(x\) in \((-2,2)\) satisfies \(x<2\), so \(P \Rightarrow Q\): sufficient.
    \((-2,2)\) is contained in \((-\infty,2)\).
  3. Is it necessary? Take \(x=-5\): \(x<2\) holds but \(x^2 = 25 \not<4\). So \(Q \not\Rightarrow P\): not necessary.
    The larger set contains points outside the smaller one.
  4. So \(P\) is sufficient but not necessary.
    Whenever one solution set sits strictly inside another, the inner condition is sufficient but not necessary.
Where it goes wrong
  • Solving \(x^2<4\) as \(x<2\) and concluding the conditions are equivalent - that drops the lower bound.
  • Solving it as \(x<-2\) or \(x>2\), which is the solution of \(x^2>4\).
Takeaway Compare conditions as sets: a subset is sufficient for the superset, the superset necessary for the subset.
Answer C · Simultaneous equations and inequalities (MM1.5) · question X2-14
6Chains of reasoning · spec Prf5 · Paper 2 style · ★☆☆

Seven consecutive integers are written down. Which of the following statements is/are true?

  1. ITheir sum is divisible by \(7\).
  2. IITheir product is divisible by \(7\).
  3. IIITheir sum is odd.
  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

The sum is \(7m\) where \(m\) is the middle integer (I). Seven consecutive integers cover every remainder mod \(7\), so one is a multiple of \(7\) (II). \(7m\) is odd only when \(m\) is odd, so III need not hold.

Full method — how and why
Seven consecutive integers: the sum telescopes about the middle, and the seven values cover all remainders mod \(7\).
  1. Let the middle be \(m\); the seven are \(m-3, \ldots, m+3\), and the offsets cancel, so the sum is \(7m\). I is true.
    Symmetric representation makes the cancellation immediate.
  2. II: seven consecutive integers take all seven remainders on division by \(7\), so exactly one is a multiple of \(7\); hence \(7\) divides the product. True.
    Among any \(k\) consecutive integers there is exactly one multiple of \(k\).
  3. III: the sum is \(7m\), which is odd only when \(m\) is odd. Take \(1,\ldots,7\): sum \(28\), even. Not always true.
    One counterexample settles it.
  4. So I and II.
    Match to the list.
Where it goes wrong
  • Accepting III after testing \(2,\ldots,8\) (sum \(35\), odd) and no other case.
  • Confusing the sum and product claims - I is about \(7m\), II about a multiple of \(7\) among the factors.
Takeaway \(k\) consecutive integers: their sum is \(k \times\) middle (for odd \(k\)), and exactly one of them is a multiple of \(k\).
Answer E · Chains of reasoning (Prf5) · question RSN-21
7Identifying errors in proofs · spec Err2 · Paper 2 style · ★☆☆

A student solves \(x^2 = 4x\).

  1. IDivide both sides by \(x\): \(x = 4\).
  2. IISo the only solution is \(x = 4\).

Which of the following is correct?

  1. AThe first error is on line I: dividing by \(x\) discards the solution \(x = 0\).
  2. BThe solution is completely correct.
  3. CThe first error is on line II.
  4. DThe first error is on line I: \(x^2 \div x = 2x\).
  5. EThe first error is on line I: it should give \(x = 2\).
Worked solution

Factorise instead: \(x(x - 4) = 0\) gives \(x = 0\) or \(x = 4\).

Full method — how and why
Dividing an equation by a variable destroys the root where that variable is zero.
  1. Line I divides \(x^2 = 4x\) by \(x\). That is only legitimate when \(x \neq 0\) - and \(x = 0\) does satisfy the equation. First error.
    \(0^2 = 4 \times 0\) ✓, so \(x=0\) is a genuine solution that the division removes.
  2. The correct method: bring everything to one side and factorise. \(x^2 - 4x = 0\), so \(x(x-4) = 0\).
    Factorising preserves every root, because a product is zero exactly when a factor is.
  3. So \(x = 0\) or \(x = 4\) - two solutions.
    Line II's 'only solution' is wrong as a consequence of line I.
Where it goes wrong
  • Blaming line II - it faithfully reports what line I produced; the first error is the division.
  • Thinking \(x^2 \div x = 2x\); it is \(x\).
  • Trying to 'fix' it by remembering to add \(x = 0\) afterwards - correct in this case, but factorising is the reliable habit.
Takeaway Never divide an equation by a variable expression. Factorise instead: division loses roots, squaring adds them.
Answer A · Identifying errors in proofs (Err2) · question ERR-12
8Ratio, proportion and percentages · spec M3.8 · Paper 2 style · ★☆☆

Which of the following statements is/are true?

  1. IIncreasing a quantity by \(50\%\) and then decreasing the result by \(50\%\) returns it to its original value.
  2. IIA \(10\%\) increase followed by a \(10\%\) increase is a \(21\%\) increase overall.
  3. IIIIf \(A\) is \(25\%\) more than \(B\), then \(B\) is \(20\%\) less than \(A\).
  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: \(1.5 \times 0.5 = 0.75\). II: \(1.1^2 = 1.21\), true. III: \(A = 1.25B \Rightarrow B = 0.8A\), true.

Full method — how and why
Three percentage claims. Each is settled by converting to multipliers.
  1. I: \(+50\%\) then \(-50\%\) gives \(1.5 \times 0.5 = 0.75\), i.e. \(75\%\) of the original. False.
    The decrease acts on the larger amount, so it removes more than the increase added.
  2. II: \(+10\%\) twice gives \(1.1^2 = 1.21\), a \(21\%\) increase. True.
    The extra \(1\%\) is the interest on the first year's interest.
  3. III: \(A = 1.25B\) means \(B = \dfrac{A}{1.25} = 0.8A\), which is \(20\%\) less than \(A\). True.
    Percentage changes are not symmetric because the base changes.
  4. II and III only.
    Match to the options.
Where it goes wrong
  • Accepting I because the percentages 'cancel'.
  • Rejecting III on the grounds that \(25\%\) up should mean \(25\%\) down.
Takeaway Always convert to multipliers: percentage changes are asymmetric because the base changes.
Answer G · Ratio, proportion and percentages (M3.8) · question RAT-15
9The 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
10Mathematical proof · spec Prf1 · Paper 2 style · ★☆☆

Prove by cases that \(n^3 - n\) is divisible by \(3\) for every integer \(n\). Which set of cases is both sufficient and necessary to consider?

  1. A\(n = 3k + 1\), \(n = 3k + 2\)
  2. B\(n\) even, \(n\) odd
  3. C\(n\) positive, \(n\) zero, \(n\) negative
  4. D\(n = 3k\), \(n = 3k + 1\), \(n = 3k + 2\)
  5. E\(n\) prime, \(n\) not prime
Worked solution

\(n^3 - n = (n - 1)n(n + 1)\) is a product of three consecutive integers; splitting by remainder on division by \(3\) shows one factor is a multiple of \(3\) in every case.

Full method — how and why
Proof by exhaustive cases: the cases must cover every integer (sufficient) with none redundant (necessary), and they must be chosen to match the divisor in question.
  1. The claim is about divisibility by \(3\), so split by remainder on division by \(3\): \(n = 3k\), \(3k+1\), \(3k+2\).
    Every integer has exactly one of these three forms, so the cases are exhaustive and non-overlapping.
  2. Check each: \(n^3 - n = (n-1)n(n+1)\). If \(n = 3k\) the middle factor is a multiple of \(3\); if \(n = 3k+1\) then \(n - 1 = 3k\); if \(n = 3k+2\) then \(n+1 = 3(k+1)\).
    Three consecutive integers always include a multiple of \(3\) - the case split makes that explicit.
  3. So the product is divisible by \(3\) in every case.
    The proof is complete because the cases cover all integers.
Where it goes wrong
  • Splitting into even/odd - that is the wrong modulus and tells you nothing about divisibility by \(3\).
  • Using only \(3k+1\) and \(3k+2\), which omits the multiples of \(3\) themselves - not exhaustive.
  • Splitting by sign or by primality: exhaustive, but useless for this claim.
Takeaway For divisibility by \(m\), split into the \(m\) residue classes \(mk, mk+1, \ldots, mk+(m-1)\).
Answer D · Mathematical proof (Prf1) · question PRF-07
11The logic of arguments · spec Arg2 · Paper 2 style · ★☆☆

For real numbers \(a, b\): \(P\): '\(ab = 0\)'; \(Q\): '\(a = 0\)'. 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

\(a = 0 \Rightarrow ab = 0\), so \(P\) is necessary for \(Q\). But \(ab = 0\) could be because \(b = 0\) with \(a \neq 0\), so \(P\) is not sufficient.

Full method — how and why
This one reverses the usual direction - the check must be done carefully rather than by pattern-matching.
  1. Test '\(P\) sufficient for \(Q\)', i.e. \(ab = 0 \Rightarrow a = 0\): take \(a = 5\), \(b = 0\). Then \(ab = 0\) but \(a \neq 0\). Not sufficient.
    The zero-product rule gives \(a = 0\) or \(b = 0\), not \(a = 0\).
  2. Test '\(P\) necessary for \(Q\)', i.e. \(Q \Rightarrow P\): if \(a = 0\) then \(ab = 0 \cdot b = 0\). True, so necessary.
    Anything times zero is zero.
  3. So \(P\) is necessary but not sufficient for \(Q\).
    Note the direction: the weaker condition (\(ab=0\)) is necessary for the stronger one (\(a=0\)).
Where it goes wrong
  • Answering 'sufficient but not necessary' by reflex - in this question the roles are swapped from the common pattern.
  • Thinking \(ab = 0\) requires both to be zero.
Takeaway \(ab = 0 \Rightarrow a = 0\) or \(b = 0\); and always check both directions explicitly rather than assuming the familiar answer.
Answer B · The logic of arguments (Arg2) · question LGC-22
12Pythagoras, mensuration and 3D · spec M5.16 · Paper 2 style · ★☆☆

\(P\): 'a solid cylinder and a solid cube have equal volumes'. \(Q\): 'the cylinder and the cube have equal surface areas'. 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

Equal volumes say nothing about surface areas (a long thin cylinder has a large surface area), and equal surface areas say nothing about volumes. Neither implies the other.

Full method — how and why
Volume and surface area are independent quantities. Neither condition constrains the other, so test both directions.
  1. Not sufficient: take a cube of side \(1\) (volume \(1\), surface \(6\)) and a cylinder of volume \(1\) that is long and thin - say radius \(0.1\), height \(\frac{1}{0.01\pi} \approx 31.8\), whose surface area is far larger than \(6\).
    Equal volumes with wildly different shapes give wildly different surface areas.
  2. Not necessary: two solids can share a surface area while having different volumes, by the same flexibility of shape.
    Stretching a cylinder thin at fixed surface area shrinks its volume.
  3. So \(P\) is neither necessary nor sufficient for \(Q\).
    Match to the list.
Where it goes wrong
  • Assuming 'bigger volume means bigger surface area' - among different shapes, no such rule holds.
  • Testing only one direction and then guessing the other.
Takeaway Volume and surface area are independent across shapes; a sphere minimises surface area for a given volume, and everything else can be made worse without limit.
Answer D · Pythagoras, mensuration and 3D (M5.16) · question X2-29
13Mathematical proof · spec Prf1 · Paper 2 style · ★☆☆

Which of the following is a correct proof that the sum of any three consecutive integers is divisible by \(3\)?

  1. AAny three integers include one that is a multiple of \(3\), so the sum is a multiple of \(3\).
  2. BLet the integers be \(n - 1\), \(n\), \(n + 1\). Their sum is \(3n\), which is divisible by \(3\).
  3. CLet the integers be \(n\), \(2n\), \(3n\). Their sum is \(6n\), which is divisible by \(3\).
  4. DThe middle integer is the mean of the three, so the sum is three times an integer... but this needs the middle integer to be a multiple of \(3\).
  5. E\(1 + 2 + 3 = 6\), \(2 + 3 + 4 = 9\) and \(3 + 4 + 5 = 12\) are all divisible by \(3\).
Worked solution

Writing the three as \(n-1, n, n+1\) gives the sum \(3n\) for every integer \(n\) - general and correct. Checking \(1+2+3\), \(2+3+4\), \(3+4+5\) is only cases; \(n, 2n, 3n\) are not consecutive; and 'any three integers include a multiple of \(3\)' is invalid reasoning (containing a multiple of \(3\) does not make the sum one).

Full method — how and why
Choosing the right algebraic representation is most of the work. Centre the variables on the middle term.
  1. Represent three consecutive integers as \(n-1\), \(n\), \(n+1\).
    Centring on \(n\) makes the \(\pm1\) cancel; \(n, n+1, n+2\) also works but gives \(3n+3\), needing one more line.
  2. Add: \((n-1) + n + (n+1) = 3n\).
    The \(-1\) and \(+1\) cancel.
  3. \(3n\) is \(3\) times an integer, hence divisible by \(3\), for every integer \(n\).
    General, so it is a proof.
Where it goes wrong
  • Using \(n, 2n, 3n\) - those are multiples, not consecutive integers.
  • Checking three numerical cases.
  • Arguing 'any three integers include a multiple of \(3\)' - false for \(1,2,4\), and in any case containing a multiple of \(3\) would not make the sum a multiple of \(3\).
  • The 'mean' argument that ends by demanding the middle integer be a multiple of \(3\) - it confuses '\(3 \times\) an integer' with '\(3 \times\) a multiple of \(3\)'.
Takeaway Consecutive integers: use \(n-1, n, n+1\). Their sum is \(3n\), always divisible by \(3\).
Answer B · Mathematical proof (Prf1) · question PRF-13
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
15Simultaneous equations and inequalities · spec MM1.5 · Paper 2 style · ★★☆

Consider the statement: for all real \(x\), if \(x > 2\) then \(x^2 - 3x + 2 > 0\). Which of the following is true?

  1. AThe statement is true, and so is its converse.
  2. BThe statement is true, but its converse is false.
  3. CThe statement is false, but its converse is true.
  4. DThe statement and its converse are both false.
Worked solution

\(x^2 - 3x + 2 = (x-1)(x-2) > 0\) exactly when \(x < 1\) or \(x > 2\). So \(x > 2\) does imply it (true), but the converse fails, e.g. \(x = 0\) gives \(2 > 0\) without \(x > 2\).

Full method — how and why
A Paper 2-style wrapper on a quadratic inequality: decide the statement and its converse separately.
  1. Factorise the quadratic: \(x^2 - 3x + 2 = (x-1)(x-2)\), which is positive exactly when \(x < 1\) or \(x > 2\).
    Knowing precisely where the expression is positive lets you test both directions quickly.
  2. The statement: does \(x > 2\) guarantee positivity? Yes - \(x > 2\) lies inside the region \(x > 2\). True.
    An implication is true when the hypothesis region sits inside the conclusion region.
  3. The converse: does positivity guarantee \(x > 2\)? No - \(x = 0\) gives \((0-1)(0-2) = 2 > 0\) but \(0 \not> 2\). False.
    The positivity region is strictly larger than \(x > 2\), so a counterexample must come from the extra part \(x < 1\).
  4. So the statement is true but its converse is false.
    Report both verdicts in the order the options use them.
Where it goes wrong
  • Assuming the converse holds because the implication does - converses are logically independent.
  • Only testing values greater than \(2\), which can never refute the converse.
Takeaway An implication is true when one region is contained in the other; its converse needs the reverse containment, so look for a counterexample in the leftover region.
Answer B · Simultaneous equations and inequalities (MM1.5) · question SIM-18
16Chains of reasoning · spec Prf5 · Paper 2 style · ★★☆

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

  1. I\(a \leq 3\)
  2. II\(c \geq 5\)
  3. III\(b = 4\)
  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

Since \(a < b < c\), \(3a < 12\) so \(a \leq 3\) (I); \(3c > 12\) so \(c \geq 5\) (II). III fails: \(1 + 2 + 9 = 12\) has \(b = 2\).

Full method — how and why
Extremal reasoning: squeeze \(a\) and \(c\) using the ordering, then test the remaining claim with an example.
  1. I: since \(aTrue.
    Replacing \(b\) and \(c\) by the smallest value \(a\) gives a strict under-estimate of the sum.
  2. II: similarly \(3c > a+b+c = 12\), so \(c>4\), i.e. \(c \geq 5\). True.
    Replacing \(a\) and \(b\) by the largest value \(c\) over-estimates the sum.
  3. III: must \(b=4\)? Try \(1+2+9 = 12\) with \(1<2<9\) ✓ and \(b=2\). False.
    One valid triple with \(b \neq 4\) settles it.
  4. So I and II.
    Match to the list.
Where it goes wrong
  • Writing \(3a \leq 12\) instead of \(3a<12\); the strict ordering \(a
  • Assuming \(b\) must be the mean of the three - true only for equally spaced values.
Takeaway Bound an ordered sum by replacing every term with the smallest (or largest): \(3a < a+b+c < 3c\).
Answer E · Chains of reasoning (Prf5) · question X2-26
17Graphs of functions · spec MM8.6 · Paper 2 style · ★★☆

\(p(x)\) is a polynomial of degree \(4\) with real coefficients. Which of the following statements must be true?

  1. I\(p(x) = 0\) has at least one real root.
  2. II\(p(x) = 0\) has at most four real roots.
  3. IIIThe graph of \(y = p(x)\) has at least one turning point.
  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 fails: \(x^4 + 1\) has no real roots. II: a degree-4 polynomial has at most 4 roots. III: an even-degree polynomial tends to the same infinity at both ends, so it must turn at least once.

Full method — how and why
Three claims about a quartic. Two follow from degree; the first needs a counterexample.
  1. I: must a degree-\(4\) polynomial have a real root? No: \(x^4+1 \geq 1 > 0\) for all real \(x\). False.
    Even-degree polynomials can stay entirely above the axis - unlike odd-degree ones.
  2. II: a polynomial of degree \(n\) has at most \(n\) real roots, so at most \(4\). True.
    Each root supplies a linear factor; more than \(4\) would exceed the degree.
  3. III: as \(x \to \pm\infty\), a quartic tends to the same infinity at both ends (the sign of the leading coefficient).
    Even powers dominate and are positive at both extremes.
  4. A continuous function going to the same infinity both ways must turn round somewhere, so it has at least one turning point. True.
    It has a global minimum (or maximum) at a finite point.
  5. So II and III.
    Match to the list.
Where it goes wrong
  • Accepting I by analogy with odd-degree polynomials, which must have a real root.
  • Reading II as 'exactly four roots' - repeated roots and non-real roots reduce the count.
Takeaway Odd degree guarantees a real root; even degree guarantees a turning point. Degree \(n\) caps the real roots at \(n\).
Answer G · Graphs of functions (MM8.6) · question X2-21
18Probability · 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
19Polynomials, 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
20The logic of arguments · spec Arg1 · Paper 2 style · ★★★

Consider the statement \(S\): for all real \(x\), if \(x > 0\) then \(x + \dfrac{1}{x} \geq 2\). Which of the following statements is/are true?

  1. I\(S\) is true.
  2. IIThe converse of \(S\) (for all real \(x\), if \(x + \frac1x \geq 2\) then \(x > 0\)) is true.
  3. IIIThe contrapositive of \(S\) is true.
  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

For \(x > 0\), \(x + \frac1x - 2 = \frac{(x-1)^2}{x} \geq 0\), so \(S\) is true, and so is its contrapositive (III). For the converse: if \(x < 0\) then \(x + \frac1x < 0 < 2\), so \(x + \frac1x \geq 2\) forces \(x > 0\) (II true).

Full method — how and why
One inequality, examined three ways. The contrapositive is free once the statement is settled; the converse must be tested separately.
  1. I: for \(x > 0\), write \(x + \dfrac1x - 2 = \dfrac{x^2 - 2x + 1}{x} = \dfrac{(x-1)^2}{x}\).
    Put over a common denominator - the numerator turns out to be a perfect square.
  2. Since \((x-1)^2 \geq 0\) and \(x > 0\), the fraction is \(\geq 0\), so \(x + \frac1x \geq 2\). True.
    A non-negative numerator over a positive denominator is non-negative. Equality at \(x = 1\).
  3. III: the contrapositive of a true statement is automatically true.
    No work needed - contrapositive and original are logically equivalent.
  4. II (the converse): suppose \(x + \frac1x \geq 2\); must \(x > 0\)? If \(x < 0\) then both \(x\) and \(\frac1x\) are negative, so \(x + \frac1x < 0 < 2\) - the hypothesis fails. And \(x = 0\) is not allowed. So \(x > 0\). True.
    Proved by eliminating the alternatives, which is itself a contrapositive argument.
  5. So all three.
    Match to the list.
Where it goes wrong
  • Assuming the converse is false because converses 'usually' are - here it genuinely holds.
  • Multiplying the inequality \(x + \frac1x \geq 2\) by \(x\) without knowing the sign of \(x\); that is precisely what is in question.
Takeaway \(x + \frac1x \geq 2\) for \(x>0\) is worth knowing; and a statement always shares its truth value with its contrapositive, never necessarily with its converse.
Answer H · The logic of arguments (Arg1) · question LGC-09
After submitting, every question shows the correct answer and its worked solution; wrong answers are added to your needs work list and correct ones to secure.