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Paper 2 · Set 07

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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1Graphs of functions · spec MM8.1 · Paper 1 style · ★☆☆

Which of the following functions is undefined at exactly two real values of \(x\)?

  1. A\(f(x) = \dfrac{1}{(x - 2)^2}\)
  2. B\(f(x) = \dfrac{x}{x^2}\)
  3. C\(f(x) = \sqrt{x - 2}\)
  4. D\(f(x) = \dfrac{1}{x^2 + 4}\)
  5. E\(f(x) = \dfrac{1}{x^2 - 4}\)
Worked solution

\(x^2 - 4 = 0\) at \(x = \pm 2\): two values. \(x^2 + 4\) is never zero; \(\frac{x}{x^2}\) and \(\frac{1}{(x-2)^2}\) are undefined at one value; \(\sqrt{x - 2}\) is undefined for all \(x < 2\).

Full method — how and why
'Undefined' means division by zero (or a negative under a square root). Count the offending values for each option.
  1. \(\dfrac{1}{x^2-4}\): the denominator vanishes at \(x = 2\) and \(x = -2\) - two values.
    Difference of two squares gives two distinct roots.
  2. \(\dfrac{1}{x^2+4}\): the denominator is always at least \(4\), so it is never zero - defined everywhere.
    A sum of a square and a positive constant is never zero.
  3. \(\dfrac{x}{x^2}\) and \(\dfrac{1}{(x-2)^2}\): each has a single problem value (\(0\) and \(2\) respectively).
    A repeated factor still vanishes at only one point.
  4. \(\sqrt{x-2}\): undefined for all \(x < 2\) - infinitely many values, not two.
    Square roots need a non-negative argument.
Where it goes wrong
  • Counting \((x-2)^2\) as two values because of the power.
  • Thinking \(x^2 + 4\) has roots \(\pm 2\).
Takeaway Count the distinct values that break the expression: repeated factors give one, sums of squares give none.
Answer E · Graphs of functions (MM8.1) · question GRA-12
2Integration · spec MM7.2 · Paper 1 style · ★☆☆

Find \(\displaystyle\int \left(x + \dfrac{1}{x}\right)^2 dx\).

  1. A\(\dfrac{x^3}{3} + 2x - \dfrac{1}{x} + c\)
  2. B\(\dfrac{x^3}{3} - \dfrac{1}{x} + c\)
  3. C\(\dfrac{1}{3}\left(x + \dfrac1x\right)^3 + c\)
  4. D\(\dfrac{x^3}{3} + 2x + \dfrac{1}{x} + c\)
  5. E\(\dfrac{x^3}{3} + 2x + c\)
  6. F\(x^3 + 2x - \dfrac1x + c\)
Worked solution

\(\left(x + \frac1x\right)^2 = x^2 + 2 + x^{-2}\), which integrates to \(\frac{x^3}{3} + 2x - x^{-1} + c\). (\(\frac13\left(x + \frac1x\right)^3 + c\) is a common error: you cannot integrate a power of a bracket that way.)

Full method — how and why
Expand before integrating - there is no rule for integrating a power of a bracket on this specification.
  1. Expand: \(\left(x + \frac1x\right)^2 = x^2 + 2\left(x\cdot\frac1x\right) + \frac{1}{x^2} = x^2 + 2 + x^{-2}\).
    The cross term is \(2 \times x \times \frac1x = 2\), a constant - easy to lose.
  2. Integrate term by term: \(\frac{x^3}{3} + 2x + \frac{x^{-1}}{-1} = \frac{x^3}{3} + 2x - \frac1x\).
    \(\int x^{-2}dx = \frac{x^{-1}}{-1} = -x^{-1}\); the index \(-2\) is allowed because it is not \(-1\).
  3. Add the constant: \(\dfrac{x^3}{3} + 2x - \dfrac1x + c\).
    Indefinite integrals always need \(+c\).
Where it goes wrong
  • Choosing \(\frac13\left(x+\frac1x\right)^3 + c\) - you cannot integrate a bracket by raising its power.
  • Sign error giving \(+\frac1x\), or dropping the \(2x\) from the cross term.
Takeaway Expand brackets fully first; \(\int x^{-2}dx = -x^{-1}\), and the cross term of a square is easy to forget.
Answer A · Integration (MM7.2) · question INT-10
3Sequences and series · spec MM2.2 · Paper 1 style · ★☆☆

An arithmetic series has first term \(5\) and common difference \(3\). Find the sum of the first \(40\) terms.

  1. A\(2440\)
  2. B\(2540\)
  3. C\(1270\)
  4. D\(2600\)
  5. E\(2500\)
  6. F\(2580\)
Worked solution

\(S_{40} = \frac{40}{2}\left(2\cdot5 + 39\cdot3\right) = 20(10 + 117) = 2540\).

Full method — how and why
A direct application of the arithmetic sum formula. The only decision is which version of the formula to use.
  1. Identify \(a = 5\), \(d = 3\), \(n = 40\).
    'First term' and 'common difference' name \(a\) and \(d\) exactly; \(n\) is how many terms you are adding.
  2. Use \(S_n = \frac n2\left[2a + (n-1)d\right]\): \(S_{40} = 20\left[10 + 39 \times 3\right]\).
    Use this version when you know \(a\) and \(d\); the other version \(\frac n2(a + l)\) needs the last term first.
  3. Evaluate the bracket: \(10 + 117 = 127\), so \(S_{40} = 20 \times 127 = 2540\).
    Note \((n-1)d = 39 \times 3\), not \(40 \times 3\) - there are \(39\) steps between \(40\) terms.
Where it goes wrong
  • Using \(40d\) instead of \(39d\), giving \(2600\).
  • Forgetting to halve \(n\), or halving the bracket instead.
Takeaway \(S_n = \frac n2[2a + (n-1)d]\) - the number of gaps is \(n - 1\), one fewer than the number of terms.
Answer B · Sequences and series (MM2.2) · question SEQ-01
4Identifying 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
5Polynomials, 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
6Chains of reasoning · spec Prf5 · Paper 2 style · ★☆☆

In a class every student studies at least one of Maths and Physics. \(70\%\) study Maths and \(60\%\) study Physics. A student is picked at random. What is the probability they study both?

  1. A\(0.42\)
  2. B\(0.2\)
  3. C\(0.1\)
  4. D\(0.3\)
  5. E\(0.5\)
  6. Fit cannot be determined
Worked solution

\(P(M \text{ or } P) = 1 = 0.7 + 0.6 - P(\text{both})\), so \(P(\text{both}) = 0.3\).

Full method — how and why
The addition rule, with 'every student studies at least one' supplying the missing total.
  1. 'At least one of Maths and Physics' means \(P(M \text{ or } P) = 1\).
    Nobody is outside the union, so the union has probability \(1\).
  2. Apply the addition rule: \(P(M \cup P) = P(M) + P(P) - P(M \cap P)\).
    Subtract the overlap, which the two percentages each count.
  3. Substitute: \(1 = 0.7 + 0.6 - P(\text{both})\), so \(P(\text{both}) = 1.3 - 1 = 0.3\).
    The excess of \(1.3\) over \(1\) is exactly the double-counted overlap.
Where it goes wrong
  • Multiplying \(0.7 \times 0.6 = 0.42\) by assuming independence - nothing here says the subjects are independent, and the union condition rules it out.
  • Answering 'cannot be determined'; the 'at least one' condition is precisely what makes it determinable.
Takeaway When percentages sum to more than \(100\%\) and everyone is covered, the excess is the overlap.
Answer D · Chains of reasoning (Prf5) · question RSN-14
7The logic of arguments · spec Arg1 · Paper 2 style · ★☆☆

Consider the statement: if \(n\) is a multiple of \(6\), then \(n\) is a multiple of \(3\). Which of the following is its contrapositive?

  1. AIf \(n\) is not a multiple of \(3\), then \(n\) is not a multiple of \(6\).
  2. BIf \(n\) is a multiple of \(3\), then \(n\) is a multiple of \(6\).
  3. CIf \(n\) is not a multiple of \(6\), then \(n\) is not a multiple of \(3\).
  4. D\(n\) is a multiple of \(6\) if and only if \(n\) is a multiple of \(3\).
  5. EIf \(n\) is a multiple of \(3\), then \(n\) is not a multiple of \(6\).
Worked solution

The contrapositive of 'if \(A\) then \(B\)' is 'if not \(B\) then not \(A\)'. It is logically equivalent to the original; B is the converse and C is the inverse.

Full method — how and why
The four relatives of an implication. Given \(A \Rightarrow B\), you must be able to write the converse, inverse and contrapositive without hesitating.
  1. Name the parts: \(A\) is '\(n\) is a multiple of \(6\)', \(B\) is '\(n\) is a multiple of \(3\)'.
    Always label before manipulating - most errors here are bookkeeping errors.
  2. The contrapositive is 'if not \(B\) then not \(A\)': if \(n\) is not a multiple of \(3\), then \(n\) is not a multiple of \(6\).
    Negate both parts and swap them. Doing only one of the two gives the wrong relative.
  3. Sanity-check the meaning: if \(3\) does not divide \(n\), then \(6\) certainly cannot, since every multiple of \(6\) is a multiple of \(3\).
    The contrapositive is logically equivalent to the original, so it must also be true - a free check.
Where it goes wrong
  • Choosing 'if \(n\) is a multiple of \(3\) then \(n\) is a multiple of \(6\)' - that is the converse (swap only), and it is false (\(n=9\)).
  • Choosing 'if \(n\) is not a multiple of \(6\) then \(n\) is not a multiple of \(3\)' - the inverse (negate only), also false (\(n=9\)).
Takeaway Converse: swap. Inverse: negate. Contrapositive: swap and negate - and only the contrapositive is equivalent to the original.
Answer A · The logic of arguments (Arg1) · question LGC-01
8Mathematical 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
9The logic of arguments · spec Arg4 · Paper 2 style · ★★☆

A statement reads: for every positive integer \(n\), there is a prime \(p\) with \(n < p < 2n\). Which of the following is its negation?

  1. AFor every positive integer \(n\), there is a prime \(p\) with \(p \leq n\) or \(p \geq 2n\).
  2. BThere is a positive integer \(n\) such that no prime \(p\) satisfies \(n < p < 2n\).
  3. CThere is a positive integer \(n\) and a prime \(p\) with \(p \leq n\) or \(p \geq 2n\).
  4. DThere is a prime \(p\) such that for every positive integer \(n\), \(n < p < 2n\) fails.
  5. EFor every positive integer \(n\), no prime \(p\) satisfies \(n < p < 2n\).
Worked solution

Negate the quantifiers in turn: 'for every \(n\), there exists \(p\) with ...' becomes 'there exists \(n\) such that for every prime \(p\), not ...', i.e. no prime lies strictly between \(n\) and \(2n\).

Full method — how and why
Negating a nested \(\forall \exists\) statement: work from the outside in, flipping one quantifier at a time.
  1. Formalise: \(\forall n\ \exists p\) (prime) with \(n < p < 2n\).
    Two quantifiers, so two flips.
  2. Flip the outer: \(\exists n\) such that \(\neg[\exists p\) with \(n
    'For all' fails exactly when one case fails.
  3. Flip the inner: \(\exists n\) such that \(\forall\) primes \(p\), \(\neg(nno prime lies strictly between \(n\) and \(2n\).
    'There exists' fails exactly when every candidate fails.
  4. So: there is a positive integer \(n\) for which no prime \(p\) satisfies \(n < p < 2n\).
    (The original is Bertrand's postulate and is in fact true, so this negation is false - but the question is about form.)
Where it goes wrong
  • Keeping 'for every \(n\)' and only negating the inside - that is a much stronger claim, not the negation.
  • Turning it into '\(\exists n\) and \(\exists p\) with \(p \leq n\) or \(p \geq 2n\)', which is trivially true and says nothing: the inner quantifier must become for all primes.
Takeaway Negate nested quantifiers outside-in: each \(\forall\) becomes \(\exists\), each \(\exists\) becomes \(\forall\), and the innermost statement is negated once.
Answer B · The logic of arguments (Arg4) · question LGC-18
10Statistics · spec M6.3 · Paper 2 style · ★★☆

A set of five distinct positive integers has median \(10\) and mean \(10\). Which of the following statements must be true?

  1. IThe largest value is at least \(12\).
  2. IIThe smallest value is at most \(8\).
  3. IIIThe range is at least \(4\).
  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

The two values above the median are distinct integers greater than \(10\), so at least \(11\) and \(12\) (I); similarly the two below are at most \(9\) and \(8\) (II); hence the range is at least \(12 - 8 = 4\) (III).

Full method — how and why
Extremal reasoning with distinct integers: the distinctness forces gaps of at least \(1\) on each side of the median.
  1. Order the five values. The median (third value) is \(10\).
    Five values means the middle one is the third.
  2. The two above are distinct integers greater than \(10\), so at least \(11\) and \(12\). Hence the largest is \(\geq 12\). I true.
    Distinct integers above \(10\) must be \(11\) or more, and two of them cannot both be \(11\).
  3. Symmetrically, the two below are at most \(9\) and \(8\), so the smallest is \(\leq 8\). II true.
    Same argument downwards.
  4. Range \(\geq 12-8 = 4\). III true.
    Combining the two bounds. (The mean being \(10\) is consistent - e.g. \(8,9,10,11,12\) - but is not needed for any of the three claims.)
  5. So all three.
    Match to the list.
Where it goes wrong
  • Forgetting 'distinct' and allowing two values equal to \(11\), which would break I.
  • Assuming the mean condition must be used; here it is extra information.
Takeaway Distinct integers force minimum gaps: \(k\) values above a median \(m\) reach at least \(m+k\).
Answer H · Statistics (M6.3) · question X2-28
11Chains of reasoning · spec Prf5 · Paper 2 style · ★★☆

A rectangle has integer side lengths and its area (in cm\(^2\)) is numerically equal to its perimeter (in cm). How many such rectangles are there (counting \(a \times b\) and \(b \times a\) once)?

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

\(ab = 2a + 2b \Rightarrow (a - 2)(b - 2) = 4\): \((a - 2, b - 2) = (1, 4)\) or \((2, 2)\), giving \(3 \times 6\) and \(4 \times 4\).

Full method — how and why
Turn a two-variable Diophantine equation into a product by the 'add a constant and factorise' trick.
  1. Set up: \(ab = 2a + 2b\) with \(a,b\) positive integers.
    Area equals perimeter, numerically.
  2. Rearrange: \(ab - 2a - 2b = 0\), then add \(4\) to both sides: \(ab - 2a - 2b + 4 = 4\).
    The added constant is the product of the two subtracted coefficients, chosen to make the left side factorise.
  3. Factorise: \((a-2)(b-2) = 4\).
    Check by expanding: \(ab - 2a - 2b + 4\) ✓.
  4. Positive factor pairs of \(4\) with \(a-2 \leq b-2\): \((1,4)\) and \((2,2)\), giving \(3\times 6\) and \(4 \times 4\).
    Negative factors would need \(a < 2\); \(a = 1\) gives \(b - 2 = -4\), i.e. \(b = -2\), not a length.
  5. So two rectangles.
    Check: \(3\times6\) has area \(18\) and perimeter \(18\) ✓; \(4\times4\) has \(16\) and \(16\) ✓.
Where it goes wrong
  • Counting \(3\times 6\) and \(6\times 3\) separately - the question says to count each rectangle once.
  • Forgetting to add \(4\) and being unable to factorise.
  • Including the factor pair \((4,1)\) as a third solution.
Takeaway \(ab + pa + qb = r\) factorises after adding \(pq\): \((a+q)(b+p) = r + pq\). Then list factor pairs.
Answer E · Chains of reasoning (Prf5) · question RSN-09
12Differentiation · spec MM6.3 · Paper 2 style · ★★☆

Let \(f\) be a polynomial. \(P\): '\(f'(x) > 0\) for all real \(x\)'. \(Q\): '\(f(x) = 0\) has exactly one real root'. 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 \(f' > 0\) everywhere, \(f\) is strictly increasing and (being an odd-degree polynomial) crosses the axis exactly once: \(P\) is sufficient. But \(f(x) = x^3\) has exactly one root and \(f'(0) = 0\), so \(P\) is not necessary.

Full method — how and why
A necessary/sufficient question about polynomials. Test each direction, and look for a boundary case to break the converse.
  1. Sufficiency: if \(f'(x)>0\) for all \(x\), then \(f\) is strictly increasing, so it takes each value at most once - at most one root.
    A strictly increasing function cannot return to a value it has already passed.
  2. Moreover such an \(f\) must have odd degree (an even-degree polynomial turns), so it goes from \(-\infty\) to \(+\infty\) and crosses the axis. Exactly one root: \(P \Rightarrow Q\). Sufficient.
    Both halves are needed: 'at most one' plus 'at least one'.
  3. Necessity: find an \(f\) with exactly one root but \(f'\) not always positive. Take \(f(x) = x^3\): one root at \(0\), but \(f'(0) = 0\).
    A stationary point of inflexion satisfies \(Q\) without satisfying \(P\).
  4. So \(P\) is sufficient but not necessary.
    Match to the list.
Where it goes wrong
  • Claiming \(P\) is necessary because 'one root suggests increasing' - \(x^3\) and also \(x^2\)-type shifts break that.
  • Forgetting that \(f'>0\) forces odd degree, so 'at least one root' still needs justifying.
Takeaway Strictly increasing \(\Rightarrow\) at most one root; but a single root does not force a strictly positive derivative (\(x^3\)).
Answer C · Differentiation (MM6.3) · question X2-02
13Binomial expansion · spec MM2.4 · Paper 2 style · ★★☆

The expansion of \(\left(2x - \dfrac{1}{x^2}\right)^{9}\) contains a term in \(x^{k}\) for each of which of the following values of \(k\)?

  1. I\(k = 0\)
  2. II\(k = 3\)
  3. III\(k = -6\)
  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

General term: \(\binom9r (2x)^{9-r}(-x^{-2})^r\) has power \(9 - r - 2r = 9 - 3r\), which takes values \(9, 6, 3, 0, -3, \ldots, -18\). \(k = 0\) needs \(r = 3\): yes. \(k = 3\) needs \(r = 2\): yes. \(k = -6\) needs \(r = 5\): yes. All three.

Full method — how and why
Which powers of \(x\) actually occur. Find the general power as an expression in \(r\) and see which target values it can hit.
  1. General term: \(\binom9r (2x)^{9-r}\left(-x^{-2}\right)^r\), whose power of \(x\) is \((9 - r) + (-2r) = 9 - 3r\).
    Add the powers contributed by each bracket; the \(-\frac{1}{x^2}\) contributes \(-2\) per copy.
  2. As \(r\) runs from \(0\) to \(9\), the power takes the values \(9, 6, 3, 0, -3, -6, -9, -12, -15, -18\).
    They step down by \(3\) - so only multiples of \(3\) appear, all the way down to \(-18\).
  3. \(k = 0\) needs \(r = 3\) ✓; \(k = 3\) needs \(r = 2\) ✓; \(k = -6\) needs \(r = 5\) ✓.
    Each value of \(r\) must be a whole number between \(0\) and \(9\) - that is the only restriction.
  4. All three occur: 'I, II and III'.
    Solve \(9 - 3r = k\) for each target and check that \(r\) is a valid index.
Where it goes wrong
  • Forgetting that the second bracket contributes \(x^{-2r}\), not \(x^{-r}\).
  • Assuming negative powers cannot appear because it is a 'binomial expansion'.
Takeaway Express the power of \(x\) as a linear function of \(r\); a term exists exactly when the required \(r\) is an integer in range.
Answer H · Binomial expansion (MM2.4) · question BIN-12
14Chains of reasoning · spec Prf5 · Paper 2 style · ★★☆

Five friends each shake hands with some of the others (each pair shakes hands at most once). Which of the following is impossible?

  1. AEvery friend shakes hands with exactly two others.
  2. BEvery friend shakes hands with exactly four others.
  3. CThe numbers of handshakes made are \(1, 1, 2, 2, 2\).
  4. DThe numbers of handshakes made by the five friends are \(4, 4, 4, 4, 1\).
  5. EExactly one friend shakes no hands at all.
Worked solution

If four people each shook hands with all four others, everyone shook hands with everyone, so the fifth person also made \(4\) handshakes, not \(1\). (The total \(4+4+4+4+1 = 17\) is odd, which is a second reason.) The others can happen: 'exactly two each' is a pentagon of handshakes, 'exactly four each' is everyone with everyone, \(1,1,2,2,2\) is a chain, and one person shaking no hands just leaves them out.

Full method — how and why
A handshake (graph) consistency question. Test each list for an internal contradiction.
  1. Take \(4,4,4,4,1\): four people each shook hands with all four others.
    With five people, the maximum possible is \(4\) handshakes each.
  2. If person \(A\) shook hands with everyone, that includes the fifth person. The same for the other three.
    Each of the four 'maximum' people contributes one handshake to the fifth person.
  3. So the fifth person made at least \(4\) handshakes, not \(1\). Impossible.
    The list contradicts itself.
  4. The others are all realisable: \(2,2,2,2,2\) is a pentagon; \(4,4,4,4,4\) is everyone with everyone; \(1,1,2,2,2\) is a chain \(A\)-\(B\)-\(C\)-\(D\)-\(E\); and \(0\) for one person just leaves them out.
    Constructing one arrangement is enough to show a list is possible.
Where it goes wrong
  • Rejecting 'everyone shakes four hands' as too many - it is exactly the complete case and is fine.
  • Rejecting the list \(1,1,2,2,2\) without trying to build it; a chain works.
  • Forgetting the parity check: the degree total must be even. \(4+4+4+4+1 = 17\) is odd, which is a second, instant reason it is impossible.
Takeaway Two fast checks on a handshake list: the total must be even, and someone shaking everyone's hand forces everyone else's count up.
Answer D · Chains of reasoning (Prf5) · question RSN-03
15The logic of arguments · spec Arg1 · Paper 2 style · ★★☆

A teacher says: 'You will pass only if you revise.' Which of the following statements means the same thing?

  1. AIf you pass, then you revised.
  2. BIf you revise, then you will pass.
  3. CYou will pass if you revise.
  4. DIf you do not pass, then you did not revise.
  5. EYou will pass if and only if you revise.
Worked solution

'\(A\) only if \(B\)' means '\(A \Rightarrow B\)': passing implies revising. It does not promise that revising guarantees a pass.

Full method — how and why
The phrase 'only if' is the one piece of logical English almost everyone mistranslates. '\(A\) only if \(B\)' means \(A \Rightarrow B\).
  1. Let \(A\) = 'you pass', \(B\) = 'you revise'. The sentence is '\(A\) only if \(B\)'.
    Label first.
  2. '\(A\) only if \(B\)' says that \(A\) can happen only in the case that \(B\) holds - so \(A\) forces \(B\): \(A \Rightarrow B\).
    Compare with '\(A\) if \(B\)', which is the other direction \(B \Rightarrow A\). 'Only if' and 'if' point opposite ways.
  3. Translate back: if you pass, then you revised.
    Revising is necessary for passing, but the teacher promises no guarantee that it is enough.
Where it goes wrong
  • Reading it as 'if you revise, you will pass' - the converse, and a much stronger promise than the teacher made.
  • Reading it as 'if and only if'; 'only if' gives one direction only.
  • Choosing 'if you do not pass then you did not revise' - the inverse, not equivalent (the correct contrapositive is 'if you did not revise, you did not pass').
Takeaway '\(A\) if \(B\)' means \(B \Rightarrow A\); '\(A\) only if \(B\)' means \(A \Rightarrow B\). 'Only if' marks the necessary condition.
Answer A · The logic of arguments (Arg1) · question LGC-06
16Identifying errors in proofs · spec Err1 · Paper 2 style · ★★☆

Claim: for all positive real \(x\), \(x + \dfrac{1}{x} \geq 2\).

  1. ISuppose \(x + \dfrac1x \geq 2\).
  2. IIMultiply by \(x > 0\): \(x^2 + 1 \geq 2x\).
  3. IIISo \(x^2 - 2x + 1 \geq 0\), i.e. \((x - 1)^2 \geq 0\), which is true.
  4. IVHence the claim is proved.

Which of the following best describes the argument?

  1. AThe argument assumes what it is trying to prove (line I); it would be valid if the steps were reversed, starting from \((x - 1)^2 \geq 0\).
  2. BThe argument is completely correct as written.
  3. CThe first error is on line II: multiplying by \(x\) reverses the inequality.
  4. DThe first error is on line III.
  5. EThe claim is false.
Worked solution

Starting from the conclusion and reaching a truth does not prove the conclusion unless every step is reversible and the argument is presented in the forward direction. Here the steps are reversible, so the repair is to write it backwards.

Full method — how and why
Assuming the conclusion. The chain of steps is valid but it runs the wrong way.
  1. Line I supposes \(x + \frac1x \geq 2\) - which is exactly the claim to be proved. First error.
    A proof may suppose the hypothesis (\(x>0\)), never the conclusion.
  2. Lines II and III are then correct algebra, ending at \((x-1)^2 \geq 0\), a truth.
    Deriving a true statement from an assumption proves nothing: false statements can also imply true ones.
  3. The repair: run it backwards. Start from \((x-1)^2 \geq 0\) (known true), expand to \(x^2 - 2x + 1 \geq 0\), so \(x^2 + 1 \geq 2x\), and divide by \(x > 0\) to get \(x + \frac1x \geq 2\).
    Every step is reversible here - the division by \(x\) is safe because \(x>0\) - so the reversal is valid.
  4. So the best description is: it assumes what it is proving, and would be valid reversed.
    Both halves matter: the flaw, and the fact that it is repairable.
Where it goes wrong
  • Calling the argument correct because each line follows from the previous one - direction is what matters.
  • Calling the claim false; it is true, as the reversed proof shows.
  • Thinking multiplying by \(x\) reverses the inequality - \(x>0\) is given.
Takeaway Working backwards is how you find a proof; to present one, start from a known truth and check every step reverses.
Answer A · Identifying errors in proofs (Err1) · question ERR-13
17Mathematical 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
18Graphs 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
19Binomial expansion · spec MM2.4 · Paper 2 style · ★★☆

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

  1. IThe coefficient of \(x^n\) in \((1 + x)^{2n}\) is \(\binom{2n}{n}\).
  2. IIThe sum of the coefficients in the expansion of \((1 - x)^n\) is \(0\).
  3. III\((1 + x)^n \geq 1 + nx\) for all \(x \geq 0\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

I is the general term with \(r = n\). II: put \(x = 1\). III: expanding \((1 + x)^n\) gives \(1 + nx\) plus non-negative terms when \(x \geq 0\).

Full method — how and why
Three binomial facts: the general term, substituting a value, and a term-by-term inequality.
  1. I: the coefficient of \(x^r\) in \((1+x)^m\) is \(\binom mr\). With \(m = 2n\) and \(r = n\) that is \(\binom{2n}{n}\). True.
    Straight from the binomial theorem.
  2. II: the sum of the coefficients of any polynomial is its value at \(x=1\). Here \((1-1)^n = 0^n = 0\) for \(n \geq 1\). True.
    Substituting \(x=1\) makes every power equal \(1\), so the value is the coefficient sum.
  3. III: for \(x \geq 0\), \((1+x)^n = 1 + nx + \binom n2 x^2 + \cdots\), and every omitted term is non-negative. So \((1+x)^n \geq 1+nx\). True.
    All binomial coefficients and all powers of a non-negative \(x\) are non-negative.
  4. So all three.
    Match to the list.
Where it goes wrong
  • Thinking II fails for \(n=0\) - the question says \(n\) is a positive integer.
  • Doubting III for large \(n\); the extra terms only help, since \(x \geq 0\).
  • Confusing \(\binom{2n}{n}\) with \(\binom{2n}{2}\) or \(2\binom nn\).
Takeaway Sum of coefficients \(=\) value at \(x=1\); and \((1+x)^n \geq 1+nx\) for \(x \geq 0\) (Bernoulli) follows from discarding non-negative terms.
Answer H · Binomial expansion (MM2.4) · question X2-13
20Straight lines · spec MM3.1 · Paper 2 style · ★★★

The line through \((a, 0)\) and \((0, b)\), with \(a, b > 0\), passes through \((2, 3)\). Which of the following statements is/are true?

  1. I\(\dfrac{2}{a} + \dfrac{3}{b} = 1\)
  2. II\(a > 2\) and \(b > 3\)
  3. IIIThe smallest possible value of \(ab\) is \(24\)
  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 line is \(\frac xa + \frac yb = 1\), so I holds. Since \(\frac2a < 1\) and \(\frac3b < 1\), II holds. \(b = \frac{3a}{a-2}\), so \(ab = \frac{3a^2}{a-2}\); setting \(a = 2 + t\) gives \(\frac{3(2+t)^2}{t} = 3\left(\frac4t + 4 + t\right) \geq 3(4 + 4) = 24\) by AM-GM (equality at \(t = 2\), \(a = 4\), \(b = 6\)). III holds.

Full method — how and why
The intercept form of a line, plus an optimisation. Statements I and II are quick; III needs a minimisation argument.
  1. I: a line with intercepts \(a\) and \(b\) has equation \(\frac xa + \frac yb = 1\); substituting \((2,3)\) gives \(\frac2a + \frac3b = 1\). True.
    The intercept form is worth knowing precisely for questions like this.
  2. II: since \(\frac2a\) and \(\frac3b\) are both positive and sum to \(1\), each is less than \(1\), so \(a > 2\) and \(b > 3\). True.
    Two positive numbers summing to \(1\) are each below \(1\) - a clean argument with no algebra.
  3. III: from I, \(b = \dfrac{3a}{a - 2}\), so \(ab = \dfrac{3a^2}{a-2}\). Put \(a = 2 + t\) with \(t > 0\): \(ab = \dfrac{3(2+t)^2}{t} = 3\left(\dfrac4t + 4 + t\right)\).
    The substitution \(a = 2 + t\) clears the awkward denominator and exposes the structure \(\frac4t + t\).
  4. By AM-GM, \(\dfrac4t + t \geq 2\sqrt{4} = 4\), with equality when \(t = 2\). So \(ab \geq 3(4 + 4) = 24\), attained at \(a = 4\), \(b = 6\). True.
    \(u + \frac ku \geq 2\sqrt k\) for positive \(u\) - here it gives both the bound and the case of equality, which is what 'smallest possible value' demands.
  5. All three are true.
    Match to the options.
Where it goes wrong
  • Assuming II fails because \(a\) and \(b\) 'could be anything' - the positivity constraint forces both bounds.
  • Claiming III without checking equality is attainable; a lower bound is only the minimum if some case achieves it.
Takeaway \(\frac xa + \frac yb = 1\) is the intercept form; and \(t + \frac4t \geq 4\) (AM-GM) is the standard tool for such minimisations.
Answer H · Straight lines (MM3.1) · question LIN-10
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