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Paper 2 · Set 02

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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1Differentiation · spec MM6.3 · Paper 1 style · ★☆☆

The curve \(y = x^2 + kx\) (with \(k < 0\)) crosses the \(x\)-axis at the origin and at one other point \(P\). The gradient of the curve at \(P\) is \(6\). Find \(k\).

  1. A\(6\)
  2. B\(-6\)
  3. C\(-12\)
  4. D\(-3\)
  5. E\(3\)
  6. F\(-2\)
Worked solution

\(P = (-k, 0)\). \(y' = 2x + k = -2k + k = -k = 6\), so \(k = -6\).

Full method — how and why
Find the second crossing point in terms of \(k\), then impose the gradient condition there.
  1. Find the \(x\)-intercepts: \(x^2 + kx = x(x + k) = 0\), so \(x = 0\) or \(x = -k\). Thus \(P = (-k, 0)\).
    With \(k < 0\), \(-k\) is positive - the curve crosses to the right of the origin.
  2. Differentiate: \(y' = 2x + k\).
    The gradient at any point.
  3. Evaluate at \(P\): \(y'(-k) = -2k + k = -k\).
    Substituting \(x = -k\).
  4. Set equal to \(6\): \(-k = 6\), so \(k = -6\).
    Consistent with the given condition \(k < 0\) ✓.
Where it goes wrong
  • Using \(x = k\) for the second intercept, giving \(k = 2\) or \(3k = 6\).
  • Answering \(6\) and ignoring the stated sign of \(k\).
Takeaway Find the point in terms of the parameter first, then apply the gradient condition; check any sign restriction at the end.
Answer B · Differentiation (MM6.3) · question DIF-17
2Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★☆☆

Find the quotient when \(2x^3 - 3x^2 + 4x - 5\) is divided by \((x - 1)\).

  1. A\(2x^2 + x - 3\)
  2. B\(2x^2 - 5x + 9\)
  3. C\(2x^2 - x + 3\)
  4. D\(2x^2 - x - 3\)
  5. E\(2x^2 + x + 3\)
  6. F\(2x^2 - x + 5\)
Worked solution

\(2x^3 - 3x^2 + 4x - 5 = (x-1)(2x^2 - x + 3) - 2\): check the expansion \(2x^3 - x^2 + 3x - 2x^2 + x - 3 = 2x^3 - 3x^2 + 4x - 3\), then subtract \(2\) for the remainder.

Full method — how and why
A division question where only the quotient is wanted. Comparing coefficients is quicker and safer than long division.
  1. Write \(2x^3 - 3x^2 + 4x - 5 = (x - 1)(Ax^2 + Bx + C) + R\).
    Dividing a cubic by a linear factor gives a quadratic quotient and a constant remainder.
  2. Compare \(x^3\) terms: \(A = 2\).
    Only one product contributes to the highest power, so the leading coefficient comes free.
  3. Compare \(x^2\) terms: \(B - A = -3\), so \(B = -1\).
    The \(x^2\) term comes from \(x \cdot Bx\) and \(-1 \cdot Ax^2\).
  4. Compare \(x\) terms: \(C - B = 4\), so \(C = 3\). The quotient is \(2x^2 - x + 3\).
    The \(x\) term comes from \(x \cdot C\) and \(-1 \cdot Bx\). (The constant term then gives \(R = -2\), but the question does not need it.)
Where it goes wrong
  • Getting the sign of \(B\) wrong, giving \(2x^2 + x + 3\) - which is offered.
  • Reporting the remainder instead of the quotient.
Takeaway Comparing coefficients from the top down finds a quotient one coefficient at a time, with no long-division layout to get wrong.
Answer C · Polynomials, factor and remainder theorems (MM1.6) · question POL-07
3Graphs of functions · spec MM8.6 · Paper 1 style · ★☆☆

A polynomial of degree \(5\) with real coefficients has how many real roots at least, and at most?

  1. Aat least \(1\), at most \(5\)
  2. Bexactly \(5\)
  3. Cat least \(1\), at most \(4\)
  4. Dat least \(0\), at most \(4\)
  5. Eat least \(0\), at most \(5\)
  6. Fat least \(2\), at most \(5\)
Worked solution

An odd-degree polynomial tends to opposite infinities at each end, so it crosses the axis at least once; a degree-\(5\) polynomial has at most \(5\) roots.

Full method — how and why
Two standard facts about polynomial roots: the maximum is the degree, and odd degree forces at least one real root.
  1. At most: a degree-\(5\) polynomial has at most \(5\) roots, since each root contributes a linear factor.
    More than \(5\) factors would raise the degree above \(5\).
  2. At least: as \(x \to +\infty\) and \(x \to -\infty\), an odd-degree polynomial tends to opposite infinities.
    The leading term \(ax^5\) dominates and changes sign with \(x\).
  3. A continuous function that is positive somewhere and negative somewhere must cross zero, so there is at least one real root.
    Intermediate value reasoning; this is why every odd-degree polynomial has a real root.
  4. At least \(1\), at most \(5\).
    Match to the options.
Where it goes wrong
  • Saying 'at least \(0\)' by analogy with even degrees like \(x^2 + 1\), which genuinely have none.
  • Saying 'exactly \(5\)', which counts complex roots or ignores repeats.
Takeaway Odd degree \(\Rightarrow\) at least one real root; degree \(n\) \(\Rightarrow\) at most \(n\) real roots.
Answer A · Graphs of functions (MM8.6) · question GRA-10
4Sequences and series · spec MM2.2 · Paper 2 style · ★☆☆

An arithmetic sequence has first term \(a\) and common difference \(d\), both non-zero. Which of the following statements is/are true?

  1. IIf \(d > 0\) then every term after some point is positive.
  2. IIThe sum of the first \(n\) terms is a quadratic function of \(n\).
  3. IIIIf \(a\) and \(d\) are integers then \(S_n\) is an integer for every \(n\).
  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: terms \(a + (n-1)d\) eventually exceed any bound. II: \(S_n = \frac n2(2a + (n-1)d)\) is quadratic in \(n\). III: \(S_n\) is a sum of integers.

Full method — how and why
Three claims about an arithmetic sequence, settled by the \(n\)th-term and sum formulas.
  1. I: the terms are \(a+(n-1)d\) with \(d>0\), which increases without bound, so beyond some point every term is positive. True.
    A positive common difference eventually overcomes any negative start.
  2. II: \(S_n = \frac n2\left(2a+(n-1)d\right) = \frac d2 n^2 + \left(a - \frac d2\right)n\) - a quadratic in \(n\) (with \(d \neq 0\), so the \(n^2\) term survives). True.
    Expand and collect powers of \(n\); the hypothesis \(d \neq 0\) is what makes it genuinely quadratic.
  3. III: \(S_n\) is a sum of \(n\) integers \(a, a+d, \ldots\), hence an integer. True.
    Sums of integers are integers - no need to worry about the \(\frac n2\) in the formula, since the bracket is even when \(n\) is odd and vice versa.
  4. So all three.
    Match to the list.
Where it goes wrong
  • Rejecting III because the formula contains \(\frac n2\) - the product still comes out whole.
  • Rejecting II if \(d\) were \(0\) (linear) - but the question states \(d \neq 0\).
  • Reading I as 'every term is positive' rather than 'every term after some point'.
Takeaway \(S_n\) for an arithmetic series is quadratic in \(n\); check the stated conditions (\(a, d\) non-zero) before dismissing a claim.
Answer H · Sequences and series (MM2.2) · question X2-23
5Triangles, radians and sectors · spec MM4.1 · Paper 2 style · ★☆☆

In triangle \(ABC\), \(P\): '\(a^2 = b^2 + c^2\)'. \(Q\): 'angle \(A = 90^\circ\)'. 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

Pythagoras and its converse: the cosine rule gives \(\cos A = \frac{b^2 + c^2 - a^2}{2bc}\), which is \(0\) exactly when \(a^2 = b^2 + c^2\).

Full method — how and why
Pythagoras and its converse both hold, so this is a genuine equivalence - best seen through the cosine rule.
  1. The cosine rule gives \(a^2 = b^2+c^2-2bc\cos A\), so \(\cos A = \dfrac{b^2+c^2-a^2}{2bc}\).
    Rearranging the cosine rule turns both statements into one equation.
  2. \(a^2 = b^2+c^2\) makes the numerator zero, so \(\cos A = 0\) and (in a triangle) \(A = 90^\circ\).
    \(2bc \neq 0\) since side lengths are positive; and \(0^\circ
  3. Conversely \(A = 90^\circ\) gives \(\cos A = 0\), so \(b^2+c^2-a^2 = 0\).
    The same equation, read the other way.
  4. So \(P\) and \(Q\) are equivalent: necessary and sufficient.
    This is Pythagoras together with its converse.
Where it goes wrong
  • Assuming only the forward direction (right angle \(\Rightarrow\) Pythagoras) and answering 'necessary but not sufficient'.
  • Forgetting that \(a\) must be the side opposite \(A\) for the relation to hold.
Takeaway The cosine rule contains Pythagoras and its converse: \(\cos A = 0 \iff a^2 = b^2+c^2\), with \(a\) opposite \(A\).
Answer A · Triangles, radians and sectors (MM4.1) · question X2-15
6Sequences and series · spec MM2.3 · Paper 2 style · ★☆☆

A geometric series has first term \(a \neq 0\) and common ratio \(r\). \(P\): '\(-1 < r < 1\)'. \(Q\): 'the series has a sum to infinity'. 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 geometric series converges if and only if \(|r| < 1\), which is exactly \(-1 < r < 1\).

Full method — how and why
A definition-driven case: the convergence condition for a geometric series is exactly the stated inequality, so both directions hold.
  1. A geometric series with \(a \neq 0\) has a sum to infinity if and only if \(|r| < 1\).
    The partial sums \(\frac{a(1-r^n)}{1-r}\) converge precisely when \(r^n \to 0\).
  2. \(|r| < 1\) is the same statement as \(-1 < r < 1\).
    Modulus inequalities unfold into a double inequality.
  3. So \(P\) and \(Q\) are equivalent: necessary and sufficient.
    An 'if and only if' condition gives both directions at once.
Where it goes wrong
  • Forgetting the case \(r = \pm1\): \(r=1\) gives partial sums \(na\), which diverge; \(r=-1\) oscillates. Both are correctly excluded.
  • Answering 'sufficient but not necessary' from habit - here no counterexample exists.
  • Overlooking why \(a \neq 0\) is stated: with \(a = 0\) every \(r\) would give a (trivial) sum.
Takeaway Geometric series converge \(\iff |r|<1\), and then \(S_\infty = \frac{a}{1-r}\). An 'if and only if' condition is necessary and sufficient.
Answer A · Sequences and series (MM2.3) · question X2-04
7Chains of reasoning · spec Prf5 · Paper 2 style · ★☆☆

A \(4 \times 4\) grid of squares is coloured so that every row and every column contains exactly two black squares. How many black squares are there in the grid?

  1. A\(6\)
  2. B\(16\)
  3. C\(10\)
  4. D\(8\)
  5. E\(4\)
  6. F\(12\)
Worked solution

Each of the \(4\) rows contains exactly \(2\) black squares: \(8\) in total.

Full method — how and why
Count by rows - the column condition is extra information, not needed for the count.
  1. There are \(4\) rows and each contains exactly \(2\) black squares.
    Every black square lies in exactly one row, so rows partition them.
  2. Total \(= 4 \times 2 = 8\).
    No double counting.
  3. Check with the columns: \(4 \times 2 = 8\) as well - consistent, which is why such a colouring can exist.
    If the two counts disagreed, no such grid would exist.
Where it goes wrong
  • Adding rows and columns to get \(16\) - that counts every black square twice.
  • Answering \(4\) by counting rows rather than squares.
Takeaway Count via one partition only; use the other as a consistency check, not as extra squares.
Answer D · Chains of reasoning (Prf5) · question RSN-22
8Chains of reasoning · spec Prf5 · Paper 2 style · ★☆☆

The digits \(1\) to \(9\) are placed in a \(3 \times 3\) grid so that every row and every column has the same sum. What is that sum?

  1. A\(45\)
  2. B\(15\)
  3. C\(20\)
  4. D\(18\)
  5. E\(12\)
  6. F\(10\)
Worked solution

The three rows together contain every digit once, so \(3 \times\) (row sum) \(= 1 + 2 + \cdots + 9 = 45\), giving \(15\).

Full method — how and why
Count the same total two different ways - the standard double-counting move.
  1. The three rows between them contain each digit \(1\) to \(9\) exactly once.
    Every cell lies in exactly one row, so the rows partition the grid.
  2. So the three row sums add to \(1+2+\cdots+9 = 45\).
    Use \(\frac{9 \times 10}{2} = 45\).
  3. All three row sums are equal, so each is \(\frac{45}{3} = 15\).
    And the columns give the same total, so they must each be \(15\) too - consistent.
Where it goes wrong
  • Answering \(45\), the grand total rather than one row's sum.
  • Trying to construct an actual magic square; the total settles it with no construction needed.
Takeaway Double-count a total: rows partition the grid, so (number of rows) \(\times\) (row sum) \(=\) grand total.
Answer B · Chains of reasoning (Prf5) · question RSN-07
9Chains of reasoning · spec Prf5 · Paper 2 style · ★☆☆

The sum of \(5\) consecutive odd numbers is \(S\). Which of the following is always a factor of \(S\)?

  1. A\(4\)
  2. B\(3\)
  3. C\(10\)
  4. D\(25\)
  5. E\(15\)
  6. F\(5\)
Worked solution

\((m - 4) + (m - 2) + m + (m + 2) + (m + 4) = 5m\) where \(m\) is the middle (odd) number; \(5m\) is odd, so \(10\) is never a factor.

Full method — how and why
Represent the numbers symmetrically about the middle so everything cancels, then read off the guaranteed factor.
  1. Let the middle number be \(m\) (odd). The five are \(m-4, m-2, m, m+2, m+4\).
    Consecutive odd numbers differ by \(2\); centring makes the offsets cancel in pairs.
  2. Add: the \(\pm2\) and \(\pm4\) cancel, leaving \(S = 5m\).
    So \(5\) is always a factor.
  3. Is \(10\) always a factor? \(m\) is odd, so \(5m\) is odd and never divisible by \(2\). No.
    For instance \(1+3+5+7+9 = 25\).
  4. Are \(3\), \(15\) or \(25\) always factors? \(25\) needs \(5 \mid m\), \(3\) needs \(3 \mid m\) - neither is guaranteed (\(m = 7\) gives \(S = 35\)).
    Only the factor \(5\) survives every case.
Where it goes wrong
  • Answering \(25\) from the single example \(1+3+5+7+9 = 25\) - that is \(5m\) with \(m=5\), a coincidence.
  • Answering \(10\) by assuming an even total; five odd numbers sum to an odd number.
Takeaway An odd count of equally spaced terms sums to (count) \(\times\) (middle term). Test the extra factors against a second example.
Answer F · Chains of reasoning (Prf5) · question RSN-16
10The logic of arguments · spec Arg1 · Paper 2 style · ★☆☆

Consider the statement about a real number \(x\): if \(x^2 = x\), then \(x = 1\). Which one of the following is true?

  1. AThe statement is true and its converse is true.
  2. BThe statement is true and its converse is false.
  3. CThe statement is false and its converse is true.
  4. DThe statement is false and its converse is false.
Worked solution

\(x = 0\) satisfies \(x^2 = x\) but not \(x = 1\), so the statement is false. The converse, 'if \(x = 1\) then \(x^2 = x\)', is true.

Full method — how and why
Judge a statement and its converse separately. Each needs either a proof or a counterexample.
  1. The statement: if \(x^2 = x\) then \(x = 1\). Solve \(x^2 = x\): \(x(x-1) = 0\), so \(x = 0\) or \(x = 1\).
    Never divide by \(x\) - that is exactly how the root \(x = 0\) gets lost.
  2. \(x = 0\) satisfies the hypothesis but not the conclusion, so the statement is false.
    A single counterexample settles it.
  3. The converse: if \(x = 1\) then \(x^2 = x\). Substituting gives \(1 = 1\). True.
    Verifying one value is enough because the hypothesis pins \(x\) down completely.
  4. So: false statement, true converse.
    Match to the option list.
Where it goes wrong
  • Dividing \(x^2 = x\) by \(x\), getting only \(x = 1\), and wrongly declaring the statement true.
  • Assuming a false statement must have a false converse - the two are independent.
Takeaway A statement and its converse have no logical connection; test each on its own, and never divide an equation by a variable.
Answer C · The logic of arguments (Arg1) · question LGC-04
11Mathematical proof · spec Prf1 · Paper 2 style · ★★☆

A student proves that \(\sqrt{4}\) is irrational by copying the standard proof for \(\sqrt2\): 'Suppose \(\sqrt4 = p/q\) in lowest terms; then \(p^2 = 4q^2\), so \(p^2\) is even, so \(p\) is even, \(p = 2k\); then \(4k^2 = 4q^2\), so \(k^2 = q^2\)...' At which point does the argument break down compared with the \(\sqrt2\) proof?

  1. A\(\sqrt4\) cannot be written as \(p/q\).
  2. B\(p^2 = 4q^2\) does not imply that \(p\) is even.
  3. C\(k^2 = q^2\) does not show that \(q\) is even, so no contradiction arises.
  4. DIt is not valid to assume \(p\) and \(q\) have no common factor.
  5. EThe argument is valid, so \(\sqrt4\) is irrational.
Worked solution

For \(\sqrt2\) the step \(p^2 = 2q^2\) leads to \(2k^2 = q^2\), making \(q\) even. Here \(k^2 = q^2\) carries no parity information, so the contradiction never appears - as expected since \(\sqrt4 = 2\).

Full method — how and why
Diagnosing why a proof technique fails when transplanted. Follow the argument line by line and find the step that no longer bites.
  1. Suppose \(\sqrt4 = \frac pq\) in lowest terms; then \(p^2 = 4q^2\), so \(p^2\) is even, so \(p\) is even - this much is fine.
    \(4q^2\) is certainly even, and an odd number squared is odd, so \(p\) must be even.
  2. Substituting \(p = 2k\): \(4k^2 = 4q^2\), i.e. \(k^2 = q^2\).
    Here the \(4\) cancels entirely, because the constant was \(4\) and not \(2\).
  3. Compare with \(\sqrt2\): there \(p^2 = 2q^2\) gives \(4k^2 = 2q^2\), i.e. \(q^2 = 2k^2\) - a spare factor of \(2\) forcing \(q\) even.
    That leftover factor of \(2\) is the whole engine of the classical proof.
  4. With \(\sqrt4\) no factor is left over, so \(k^2 = q^2\) tells you nothing about the parity of \(q\), and no contradiction ever appears.
    Correctly so: \(\sqrt4 = 2\) is rational, so no proof of irrationality could possibly succeed.
Where it goes wrong
  • Claiming \(p^2 = 4q^2\) fails to make \(p\) even - it does not fail; that step is valid.
  • Claiming the 'lowest terms' assumption is illegitimate - every rational can be written that way.
  • Accepting the argument and concluding \(\sqrt4\) is irrational, when \(\sqrt4 = 2\).
Takeaway The \(\sqrt2\) proof works because \(2\) is prime; the leftover prime factor is what forces the second parity. Check where an adapted proof loses its engine.
Answer C · Mathematical proof (Prf1) · question PRF-19
12Differentiation · spec MM6.3 · Paper 2 style · ★★☆

The function \(f\) has \(f'(x) = (x - 1)(x - 3)^2\). Which of the following statements is/are true?

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

\(f'\) changes from negative to positive at \(x = 1\) (minimum). At \(x = 3\) the squared factor keeps \(f' \geq 0\) on both sides, so no turning point: II is false, and \(f' \geq 0\) for \(x > 1\) (zero only at \(3\)) makes III true.

Full method — how and why
Classify stationary points by whether \(f'\) changes sign - a squared factor does not change sign.
  1. \(f'(x) = (x-1)(x-3)^2\), so \(f'=0\) at \(x=1\) and \(x=3\).
    Stationary points are the roots of \(f'\).
  2. Sign of \(f'\): \((x-3)^2 \geq 0\) always, so the sign is that of \((x-1)\). Negative for \(x<1\), positive for \(x>1\) (zero at \(x=3\)).
    An even-power factor contributes no sign change.
  3. I: \(f'\) goes negative \(\to\) positive at \(x=1\), so a local minimum. True.
    Falling then rising.
  4. II: at \(x=3\), \(f'\) is positive on both sides, so no turning point - a stationary point of inflexion. False.
    The squared factor is precisely why there is no maximum here.
  5. III: for \(x>1\), \(f' \geq 0\) with equality only at the isolated point \(x=3\), so \(f\) is increasing. True.
    An isolated zero of the derivative does not interrupt increase.
  6. So I and III.
    Match to the list.
Where it goes wrong
  • Assuming every root of \(f'\) is a turning point and alternating min/max - the repeated root breaks the alternation.
  • Rejecting III because \(f'(3)=0\); increase needs \(f' \geq 0\) with no interval of equality.
Takeaway Odd-power factors of \(f'\) change sign (turning point); even-power factors do not (stationary inflexion).
Answer F · Differentiation (MM6.3) · question X2-16
13Mathematical proof · spec Prf2 · Paper 2 style · ★★☆

For a positive integer \(n\) it is known that: (i) if \(n\) is a multiple of \(4\) then \(n\) is not a multiple of \(6\); (ii) \(n\) is a multiple of \(12\) or \(n\) is a multiple of \(4\). What can be deduced?

  1. AThe premises are inconsistent unless \(n\) is a multiple of \(4\) but not of \(6\); in particular \(n\) cannot be a multiple of \(12\).
  2. B\(n\) is not a multiple of \(4\).
  3. C\(n\) is a multiple of \(12\).
  4. D\(n\) is a multiple of \(6\).
  5. ENothing can be deduced.
Worked solution

If \(n\) were a multiple of \(12\) it would be a multiple of both \(4\) and \(6\), contradicting (i). So by (ii) \(n\) is a multiple of \(4\), and by (i) not a multiple of \(6\).

Full method — how and why
Deduction from two premises. Test the strongest-looking option against premise (i) first.
  1. Suppose \(n\) were a multiple of \(12\). Then \(12 = 4\times3\) and \(12 = 6 \times 2\), so \(n\) is a multiple of both \(4\) and \(6\).
    Exploring the tempting case first often resolves the whole question.
  2. But (i) says a multiple of \(4\) is not a multiple of \(6\) - contradiction. So \(n\) is not a multiple of \(12\).
    A supposition leading to a contradiction is ruled out.
  3. Premise (ii) offers two alternatives and the first is now impossible, so \(n\) must be a multiple of \(4\).
    Disjunctive syllogism: '\(A\) or \(B\)' plus 'not \(A\)' gives \(B\).
  4. Then (i) applies: \(n\) is a multiple of \(4\) but not of \(6\).
    That is the full deduction.
Where it goes wrong
  • Concluding \(n\) is a multiple of \(12\) by picking the first branch of (ii) without testing it.
  • Declaring the premises inconsistent - they are perfectly consistent (e.g. \(n = 4\) or \(n = 8\)).
  • Concluding nothing, missing that one branch of the 'or' is eliminated.
Takeaway With 'A or B', eliminate one branch and the other must hold; always test the branch that clashes with the other premise.
Answer A · Mathematical proof (Prf2) · question PRF-12
14The logic of arguments · spec Arg1 · Paper 2 style · ★★☆

Which of the following statements is/are logically equivalent to 'if it is raining, then the ground is wet'?

  1. IIt is not raining, or the ground is wet.
  2. IIIf the ground is not wet, then it is not raining.
  3. IIIIf the ground is wet, then it is raining.
  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

'\(A \Rightarrow B\)' is equivalent to 'not \(A\), or \(B\)' (I) and to its contrapositive (II). III is the converse, which is not equivalent.

Full method — how and why
Two standard equivalents of \(A \Rightarrow B\): the contrapositive, and the disjunctive form 'not \(A\), or \(B\)'.
  1. I: 'not \(A\) or \(B\)'. An implication only fails when \(A\) is true and \(B\) false - exactly when '\(\neg A\) or \(B\)' fails. Equivalent.
    Check the one bad case: raining and ground dry makes both statements false; every other case makes both true.
  2. II: 'if the ground is not wet, it is not raining' - the contrapositive. Equivalent.
    Swap and negate.
  3. III: 'if the ground is wet, it is raining' - the converse. Not equivalent; the ground could be wet from a burst pipe.
    A concrete alternative cause is the cleanest refutation.
  4. So I and II.
    Match to the list.
Where it goes wrong
  • Rejecting I because it looks nothing like an implication - test it against the single case that breaks an implication.
  • Accepting III because wet ground is evidence of rain; evidence is not implication.
Takeaway \(A \Rightarrow B\) is equivalent both to 'not \(A\), or \(B\)' and to \(\neg B \Rightarrow \neg A\). The converse is never among the equivalents.
Answer E · The logic of arguments (Arg1) · question LGC-20
15Identifying 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
16Quadratic functions · spec MM1.3 · Paper 2 style · ★★☆

The quadratic equation \(ax^2 + bx + c = 0\) (with \(a \neq 0\)) has real coefficients. Which of the following statements is/are always true?

  1. IIf \(ac < 0\), the equation has two distinct real roots.
  2. IIIf \(b = 0\), the equation has no real roots.
  3. IIIIf the equation has a repeated root, then \(b^2 = 4ac\).
  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: \(b^2 - 4ac > 0\) whenever \(ac < 0\), so true. II: \(x^2 - 4 = 0\) has \(b = 0\) and real roots, so false. III: a repeated root means the discriminant is zero, so true.

Full method — how and why
Three general claims about \(ax^2 + bx + c = 0\). Each is a discriminant statement in disguise; the false one needs a counterexample.
  1. I: if \(ac < 0\) then \(-4ac > 0\), so \(b^2 - 4ac \geq -4ac > 0\): two distinct real roots. True.
    \(b^2 \geq 0\) always, so a positive \(-4ac\) forces the whole discriminant positive. Geometrically, \(a\) and \(c\) of opposite signs means the curve crosses the axis.
  2. II: test \(b = 0\) with \(x^2 - 4 = 0\), which has roots \(\pm 2\). False.
    One counterexample settles a 'for all' claim. With \(b = 0\) the discriminant is \(-4ac\), whose sign depends on \(a\) and \(c\).
  3. III: a repeated root means the discriminant is zero, i.e. \(b^2 - 4ac = 0\), i.e. \(b^2 = 4ac\). True.
    This is the definition of the repeated-root case, just rearranged.
  4. I and III only.
    Match the verdicts to the options.
Where it goes wrong
  • Believing II because \(b = 0\) 'removes' the linear term and the graph 'sits on the axis' - it does not; \(x^2 - 4\) is a counterexample.
  • Mis-stating III as \(b^2 = -4ac\).
Takeaway Statements about roots translate into statements about \(b^2 - 4ac\); to refute one, look for the simplest numbers that fit the hypothesis.
Answer F · Quadratic functions (MM1.3) · question QUA-17
17Triangles, radians and sectors · spec MM4.1 · Paper 2 style · ★★☆

A triangle has sides \(a\), \(b\), \(c\) with \(a^2 + b^2 < c^2\). Which of the following statements is/are true?

  1. IThe angle opposite \(c\) is obtuse.
  2. II\(c\) is the longest side.
  3. IIIThe triangle cannot be isosceles.
  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

Cosine rule: \(\cos C = \frac{a^2 + b^2 - c^2}{2ab} < 0\), so \(C\) is obtuse (I). The largest angle is opposite the longest side, so \(c\) is longest (II). III is false: \(a = b = 1\), \(c = 1.5\) works.

Full method — how and why
A condition on the sides. The cosine rule converts it into a statement about an angle; then test the third claim with a counterexample.
  1. I: \(\cos C = \dfrac{a^2+b^2-c^2}{2ab}\), and \(a^2 + b^2 < c^2\) makes the numerator negative, so \(\cos C < 0\) and \(C\) is obtuse. True.
    The denominator \(2ab\) is positive, so the sign of \(\cos C\) is the sign of \(a^2 + b^2 - c^2\).
  2. II: a triangle can have at most one obtuse angle, and the largest angle is opposite the largest side. Since \(C\) is obtuse it is the largest angle, so \(c\) is the longest side. True.
    Two obtuse angles would already exceed \(180^\circ\).
  3. III: try \(a = b = 1\) and \(c = 1.5\): then \(a^2 + b^2 = 2 < 2.25 = c^2\) ✓ and the triangle is isosceles. False.
    A single counterexample kills the claim. Check the triangle inequality too: \(1 + 1 > 1.5\) ✓.
  4. I and II only.
    Match to the option list.
Where it goes wrong
  • Assuming an obtuse triangle cannot be isosceles.
  • Forgetting to check the triangle inequality when constructing a counterexample.
Takeaway \(a^2+b^2 < c^2\) \(\Leftrightarrow\) the angle opposite \(c\) is obtuse; obtuse triangles can still be isosceles.
Answer E · Triangles, radians and sectors (MM4.1) · question TRI-12
18Graphs of functions · spec MM8.7 · Paper 2 style · ★★☆

The graphs of \(y = f(x)\) and \(y = g(x)\) meet at exactly three points. Which of the following statements must be true?

  1. IThe equation \(f(x) = g(x)\) has exactly three real solutions.
  2. IIThe graph of \(y = f(x) - g(x)\) crosses the \(x\)-axis exactly three times.
  3. III\(f(x) - g(x)\) is a polynomial of degree at least \(3\).
  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

I is the definition of an intersection. II fails because the difference could touch the axis rather than cross it at one of the points. III fails because \(f\) and \(g\) need not be polynomials at all (e.g. \(\sin x\) and a line).

Full method — how and why
Intersections and roots. Distinguish a root from a crossing, and resist assuming the functions are polynomials.
  1. I: the graphs meet where \(f(x)=g(x)\), so three intersection points means exactly three solutions. True - it is the definition.
    Each intersection point has a distinct \(x\)-value, since a graph has one \(y\) per \(x\).
  2. II: at a root of \(f-g\) the graph could touch the \(x\)-axis and turn back rather than cross - a repeated root.
    Meeting and crossing are different: \(y = x^2\) meets the axis but does not cross it.
  3. So the number of crossings could be fewer than three. False.
    For instance \(f-g = x^2(x-1)\) meets the axis at \(0\) and \(1\) but crosses only at \(1\).
  4. III: nothing says \(f\) and \(g\) are polynomials. Take \(f(x)=\sin x\) and a suitable straight line \(g\), meeting three times. False.
    Read the hypotheses: only 'the graphs of \(y=f(x)\) and \(y=g(x)\)' is given.
  5. So I only.
    Match to the list.
Where it goes wrong
  • Accepting II by treating 'crosses' as a synonym for 'meets'.
  • Accepting III by assuming a degree just because three intersections are mentioned.
Takeaway Intersections \(=\) solutions of \(f=g\); but crossing the axis needs a sign change, and never assume a function type that was not stated.
Answer B · Graphs of functions (MM8.7) · question X2-09
19The logic of arguments · spec Arg3 · Paper 2 style · ★★☆

Which of the following statements is/are true?

  1. IFor every real number \(x\), there exists a real number \(y\) such that \(y > x\).
  2. IIThere exists a real number \(y\) such that for every real number \(x\), \(y > x\).
  3. IIIFor every real number \(x\), there exists a real number \(y\) such that \(xy = 1\).
  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 true (take \(y = x + 1\)). II claims a single largest real number exists: false. III fails for \(x = 0\).

Full method — how and why
Quantifier order matters. '\(\forall x \exists y\)' and '\(\exists y \forall x\)' say completely different things.
  1. I: for each \(x\), can I find a bigger \(y\)? Take \(y = x + 1\). True.
    \(y\) is allowed to depend on \(x\), because \(x\) is chosen first.
  2. II: is there a single \(y\) bigger than every real \(x\)? That would be a largest real number. Taking \(x = y\) contradicts \(y > x\). False.
    Here \(y\) is fixed before \(x\), so it must work for all \(x\) at once - including \(x = y\).
  3. III: for each \(x\) is there \(y\) with \(xy = 1\)? For \(x = 0\), \(0 \cdot y = 0 \neq 1\). False.
    'For every \(x\)' includes the awkward cases; zero has no reciprocal.
  4. So I only.
    Match to the list.
Where it goes wrong
  • Reading I and II as the same statement - the difference is the order of the quantifiers, nothing else.
  • Forgetting \(x = 0\) in III and accepting \(y = \frac1x\) uncritically.
Takeaway In \(\forall x \exists y\) the \(y\) may depend on \(x\); in \(\exists y \forall x\) one \(y\) must serve every \(x\). And 'for every' always includes the edge cases.
Answer B · The logic of arguments (Arg3) · question LGC-05
20Mathematical proof · spec Prf2 · Paper 2 style · ★★☆

Every element of a set \(S\) of integers is either a multiple of \(3\) or a multiple of \(5\). Every multiple of \(3\) in \(S\) is negative. \(S\) contains the number \(10\). Which of the following statements must be true?

  1. I\(S\) contains a positive multiple of \(5\).
  2. II\(S\) contains no positive multiple of \(3\).
  3. III\(S\) contains a negative number.
  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

\(10 \in S\) is a positive multiple of \(5\) (I). Multiples of \(3\) in \(S\) are negative, so none is positive (II). III fails: \(S = \{10\}\) satisfies everything.

Full method — how and why
'Must be true' with a concrete object given. Some claims follow from the given element; others need a counterexample set.
  1. I: \(10 \in S\) and \(10 = 2 \times 5\) is a positive multiple of \(5\). Must be true.
    The element is handed to you; nothing more is needed.
  2. II: the second condition says every multiple of \(3\) in \(S\) is negative, so none can be positive. Must be true.
    Note this is vacuously fine even if \(S\) contains no multiples of \(3\) at all.
  3. III: must \(S\) contain a negative number? Try \(S = \{10\}\): every element is a multiple of \(3\) or \(5\) ✓, there are no multiples of \(3\) to be negative ✓, and \(10 \in S\) ✓. No negative number. Not necessarily true.
    A single valid \(S\) with no negative element refutes 'must be true'.
  4. So I and II.
    Match to the list.
Where it goes wrong
  • Reading 'every multiple of \(3\) in \(S\) is negative' as promising that \(S\) has a multiple of \(3\) - a universal statement can be true with nothing to apply it to.
  • Confusing 'could be true' with 'must be true' for III.
Takeaway A conditional statement about members of a set is vacuously true when no member qualifies; to refute 'must be true', build a legal example that fails it.
Answer E · Mathematical proof (Prf2) · question PRF-18
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