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Paper 2 · Set 01

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

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

The displacement \(s\) metres of a particle at time \(t\) seconds is \(s = t^3 - 6t^2 + 9t\). At what times is the particle momentarily at rest?

  1. A\(t = 2\) only
  2. B\(t = 1\) and \(t = 3\)
  3. C\(t = 1\) only
  4. D\(t = 3\) only
  5. E\(t = 0\) and \(t = 3\)
  6. F\(t = 0\), \(t = 1\) and \(t = 3\)
Worked solution

Velocity \(= \frac{ds}{dt} = 3t^2 - 12t + 9 = 3(t - 1)(t - 3) = 0\) at \(t = 1\) and \(t = 3\).

Full method — how and why
'At rest' means velocity zero, and velocity is the derivative of displacement.
  1. Differentiate the displacement: \(v = \dfrac{ds}{dt} = 3t^2 - 12t + 9\).
    Velocity is the rate of change of displacement.
  2. Factorise: \(3(t^2 - 4t + 3) = 3(t-1)(t-3)\).
    Take the \(3\) out first.
  3. Set \(v = 0\): \(t = 1\) and \(t = 3\).
    'Momentarily at rest' is exactly \(v = 0\).
Where it goes wrong
  • Setting \(s = 0\) instead of \(v = 0\), which gives \(t = 0\) and \(t = 3\) - an offered option.
  • Forgetting one of the two roots.
Takeaway Displacement \(\to\) velocity \(\to\) acceleration by differentiating; 'at rest' means the velocity is zero, not the displacement.
Answer B · Differentiation (MM6.1) · question DIF-06
2Triangles, radians and sectors · spec MM4.2 · Paper 1 style · ★☆☆

A sector of a circle of radius \(6\) cm has area \(15\pi\) cm\(^2\). Find the arc length of the sector.

  1. A\(\dfrac{5\pi}{6}\) cm
  2. B\(10\pi\) cm
  3. C\(5\pi\) cm
  4. D\(\dfrac{5\pi}{2}\) cm
  5. E\(3\pi\) cm
  6. F\(15\pi\) cm
Worked solution

\(\frac12 r^2\theta = 18\theta = 15\pi \Rightarrow \theta = \frac{5\pi}{6}\). Arc \(= r\theta = 6 \cdot \frac{5\pi}{6} = 5\pi\).

Full method — how and why
Sector area gives the angle; the angle gives the arc length. Both formulas need radians.
  1. Use the sector area formula \(A = \frac12 r^2\theta\): \(\frac12(36)\theta = 18\theta\).
    \(r = 6\), so \(\frac12 r^2 = 18\). The formula is only valid with \(\theta\) in radians.
  2. Set equal to the given area: \(18\theta = 15\pi\), so \(\theta = \dfrac{5\pi}{6}\).
    Divide: \(\frac{15\pi}{18} = \frac{5\pi}{6}\).
  3. Use the arc-length formula \(s = r\theta = 6 \times \dfrac{5\pi}{6} = 5\pi\) cm.
    The \(6\) cancels neatly - a sign the angle is right.
Where it goes wrong
  • Quoting the angle \(\frac{5\pi}{6}\) as the arc length - it is offered as a distractor.
  • Using \(A = r^2\theta\) without the \(\frac12\), which halves the angle and gives \(\frac{5\pi}{2}\).
Takeaway \(A = \frac12 r^2\theta\) and \(s = r\theta\) (radians): find \(\theta\) from one, then substitute into the other.
Answer C · Triangles, radians and sectors (MM4.2) · question TRI-03
3Polynomials, factor and remainder theorems · spec MM1.6 · Paper 1 style · ★☆☆

When \(x^3 - 4x^2 + kx + 6\) is divided by \((x - 3)\) the remainder is \(-6\). Find \(k\).

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

Remainder Theorem: \(27 - 36 + 3k + 6 = -6 \Rightarrow 3k - 3 = -6 \Rightarrow k = -1\).

Full method — how and why
A remainder is given, so use the Remainder Theorem: substitute the root of the divisor and set the result equal to the remainder.
  1. The divisor \((x - 3)\) is zero at \(x = 3\).
    The Remainder Theorem says the remainder on dividing by \((x - a)\) is \(f(a)\); you need the value of \(x\) that makes the divisor vanish.
  2. Substitute: \(f(3) = 27 - 36 + 3k + 6 = 3k - 3\).
    Work the numerical terms out first (\(27 - 36 + 6 = -3\)) so only the \(k\) term is left to carry.
  3. Set equal to the given remainder: \(3k - 3 = -6\).
    This is what the Remainder Theorem gives you - an equation, not an identity.
  4. Solve: \(3k = -3\), so \(k = -1\).
    Simple linear equation.
Where it goes wrong
  • Substituting \(x = -3\) (the sign of the root of \(x - 3\) is \(+3\)), which gives a different value of \(k\).
  • Setting \(f(3) = 0\) out of habit - that is the Factor Theorem, and only applies when the remainder is zero.
Takeaway Divisor \((x - a)\) \(\Rightarrow\) substitute \(x = a\); the result is the remainder, which is zero only when \((x-a)\) is a factor.
Answer F · Polynomials, factor and remainder theorems (MM1.6) · question POL-01
4The logic of arguments · spec Arg2 · Paper 2 style · ★☆☆

Let \(P\): 'the quadrilateral \(ABCD\) is a square' and \(Q\): 'the quadrilateral \(ABCD\) has four equal sides'. 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 square has four equal sides (\(P \Rightarrow Q\), sufficient), but a rhombus that is not a square shows \(Q\) does not imply \(P\) (not necessary).

Full method — how and why
The same two-direction test, now in geometry. Look for a shape satisfying one condition but not the other.
  1. Sufficiency: every square has four equal sides, so \(P \Rightarrow Q\). True.
    Straight from the definition of a square.
  2. Necessity: does four equal sides force a square? A rhombus with angles \(60^\circ\) and \(120^\circ\) has four equal sides and is not a square.
    A square also needs right angles; equal sides alone do not fix the angles.
  3. So \(P\) is sufficient but not necessary for \(Q\).
    \(P\) is the stronger condition, so it implies the weaker one but not conversely.
Where it goes wrong
  • Confusing which is the stronger condition: the more demanding property (square) is sufficient for the less demanding one (four equal sides).
  • Failing to produce a concrete counterexample and guessing.
Takeaway A stronger condition is sufficient for a weaker one; the weaker is necessary for the stronger. Name a counterexample to confirm.
Answer C · The logic of arguments (Arg2) · question LGC-07
5Pythagoras, mensuration and 3D · spec M5.16 · Paper 2 style · ★☆☆

Which of the following statements is/are true?

  1. IDoubling the radius of a circle doubles its circumference.
  2. IIDoubling the radius of a sphere multiplies its volume by \(8\).
  3. IIIDoubling the radius of a cylinder while halving its height leaves its volume unchanged.
  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

\(C = 2\pi r\) is linear in \(r\) (I). \(V \propto r^3\) (II). \(\pi(2r)^2\cdot\frac h2 = 2\pi r^2 h\): doubled, not unchanged (III false).

Full method — how and why
Three scaling claims. Each is checked by substituting the new dimensions into the formula.
  1. I: \(C = 2\pi r\) is linear in \(r\), so doubling \(r\) doubles \(C\). True.
    Circumference scales with the first power of the radius.
  2. II: \(V = \frac43\pi r^3\), so doubling \(r\) multiplies the volume by \(2^3 = 8\). True.
    Volume scales with the cube.
  3. III: new volume \(= \pi(2r)^2\left(\frac h2\right) = \pi(4r^2)\left(\frac h2\right) = 2\pi r^2 h\) - double, not unchanged. False.
    The radius is squared, so doubling it quadruples that factor; halving the height only removes a factor of \(2\).
  4. I and II only.
    Match to the option list.
Where it goes wrong
  • Assuming the doubling and halving in III cancel - the radius is squared, so they do not.
  • Confusing area and volume scaling in II.
Takeaway Substitute the new dimensions into the formula; squared quantities change faster than linear ones.
Answer E · Pythagoras, mensuration and 3D (M5.16) · question MEN-11
6The logic of arguments · spec Arg3 · Paper 2 style · ★☆☆

Consider the statement \(S\): for all real \(x\), \(x^2 - 6x + 10 > 0\). Which of the following is true?

  1. A\(S\) is false, because \(x = 3\) gives \(x^2 - 6x + 10 = 0\).
  2. B\(S\) cannot be decided without knowing \(x\).
  3. C\(S\) is false, because a quadratic can be negative.
  4. D\(S\) is true, because \(x^2 - 6x + 10 = (x - 3)^2 + 1\).
  5. E\(S\) is true, because \(x^2 - 6x + 10\) has two real roots.
Worked solution

Completing the square shows the expression is at least \(1\) for every \(x\).

Full method — how and why
A universal claim about a quadratic. Completing the square settles it, and the question also tests whether the reason given is sound.
  1. Complete the square: \(x^2 - 6x + 10 = (x-3)^2 - 9 + 10 = (x-3)^2 + 1\).
    Halve the coefficient of \(x\) to get \(-3\), then correct by \(-(-3)^2 = -9\).
  2. \((x-3)^2 \geq 0\), so the expression is \(\geq 1 > 0\) for every real \(x\). True.
    A square is never negative; the minimum value is \(1\) at \(x = 3\).
  3. Check the false reasons: \(x = 3\) gives \(9 - 18 + 10 = 1\), not \(0\); and the discriminant \(36 - 40 = -4 < 0\), so there are no real roots.
    The claim that it has two real roots is simply wrong, and would in any case not prove positivity.
Where it goes wrong
  • Assuming any quadratic takes negative values - one with a positive leading coefficient and negative discriminant never does.
  • Mis-completing the square as \((x-3)^2 + 10\) or \((x-6)^2+1\).
Takeaway Completing the square proves positivity outright: \(a(x-h)^2 + k\) with \(a>0, k>0\) is always positive.
Answer D · The logic of arguments (Arg3) · question LGC-21
7Mathematical proof · spec Prf1 · Paper 2 style · ★☆☆

Consider the claim: if \(n\) is a positive integer then \(n^2 + n + 1\) is odd. Which of the following is a valid proof?

  1. A\(n^2 + n = n(n + 1)\) is even, being the product of consecutive integers, so \(n^2 + n + 1\) is even plus one, which is odd.
  2. B\(n^2 + n + 1\) cannot be factorised, so it is odd.
  3. CFor \(n = 1, 2, 3\) the values are \(3, 7, 13\), all odd, so the claim holds.
  4. DSuppose \(n^2 + n + 1\) is even. Then \(n\) is even, which is a contradiction.
  5. EIf \(n\) is odd then \(n^2\) is odd and \(n\) is odd, so \(n^2 + n + 1\) is odd.
Worked solution

The argument from \(n^2 + n = n(n+1)\) being even is direct and general. Checking \(n = 1,2,3\) is only cases; 'cannot be factorised' is irrelevant to parity; 'if \(n\) is odd ...' says nothing about even \(n\); and the contradiction attempt never reaches a contradiction, since \(n\) may well be even.

Full method — how and why
Same 'which is a valid proof' pattern as PRF-01: general argument versus cases, irrelevance and incompleteness.
  1. The valid proof: \(n^2+n = n(n+1)\) is a product of consecutive integers, hence even; adding \(1\) gives an odd number.
    General, covering every positive integer with no case split.
  2. Checking \(n=1,2,3\) gives \(3,7,13\) - evidence, not proof.
    Three cases out of infinitely many.
  3. 'It cannot be factorised, so it is odd' - factorisability has nothing to do with parity (\(6 = 2\times3\) and \(9=3\times3\) are respectively even and odd).
    An irrelevant premise cannot support the conclusion.
  4. 'If \(n\) is odd then ... so \(n^2+n+1\) is odd' - correct but only for odd \(n\); even \(n\) is untreated. Incomplete.
    A case-based proof must cover every case.
  5. 'Suppose \(n^2+n+1\) is even. Then \(n\) is even, contradiction' - \(n\) being even is not a contradiction; \(n\) can be even.
    A contradiction proof must actually reach an impossibility.
Where it goes wrong
  • Accepting the odd-\(n\) argument because every line in it is true - truth is not completeness.
  • Accepting the contradiction attempt because it uses the right vocabulary.
Takeaway \(n(n+1)\) is always even, so \(n^2+n+1\) is always odd. A proof must be general, relevant and complete.
Answer A · Mathematical proof (Prf1) · question X2-27
8Statistics · spec M6.4 · Paper 2 style · ★☆☆

Which of the following statements about a scatter graph showing strong positive correlation between hours of revision and exam mark is/are valid?

  1. IStudents who revised more tended to score higher.
  2. IIRevising more causes a higher mark.
  3. IIIA line of best fit could be used to estimate the mark of a student who revised for a number of hours within the range of the data.
  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

Correlation describes a tendency (I) and supports interpolation within the data range (III), but does not by itself establish causation (II).

Full method — how and why
A correlation-and-causation statement test. Decide what correlation does and does not licence.
  1. I: strong positive correlation means higher revision went with higher marks. True - that is exactly what correlation describes.
    'Tended to' is a statement about the trend, not about every student.
  2. II: correlation does not establish causation. False.
    A hidden variable (motivation, prior ability) could drive both, so the data alone cannot prove cause.
  3. III: using a line of best fit within the data range is interpolation. True.
    Interpolation is legitimate; extrapolation beyond the data is what is unreliable.
  4. So I and III only.
    Match to the list.
Where it goes wrong
  • Accepting II because the causal story sounds plausible - plausibility is not evidence.
  • Rejecting III by confusing interpolation with extrapolation.
Takeaway Correlation licenses description and interpolation; it never licenses a causal claim on its own.
Answer F · Statistics (M6.4) · question STA-05
9Mathematical proof · spec Prf1 · Paper 2 style · ★☆☆

A student wants to prove that the square of any odd integer leaves remainder \(1\) when divided by \(8\). Which of the following is the most appropriate first step?

  1. ACheck the claim for \(1, 3, 5, 7\) and \(9\).
  2. BAssume the square leaves a remainder other than \(1\).
  3. CAssume the claim is true for all odd integers.
  4. DWrite the odd integer as \(2k\) for some integer \(k\).
  5. EWrite the odd integer as \(2k + 1\) for some integer \(k\).
Worked solution

A direct proof: \((2k+1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1\), and \(k(k+1)\) is even, so the square is \(8m + 1\). Checking cases is not a proof; assuming the claim is circular.

Full method — how and why
Choosing a proof strategy. For a 'for all odd integers' claim, the standard opening is an algebraic parametrisation.
  1. Represent an arbitrary odd integer as \(2k+1\), \(k\) an integer.
    This one expression covers every odd integer, which is what a universal claim needs.
  2. (For context, the proof then runs: \((2k+1)^2 = 4k^2+4k+1 = 4k(k+1)+1\).)
    Factorising \(4k^2+4k\) as \(4k(k+1)\) exposes the consecutive pair.
  3. \(k(k+1)\) is a product of consecutive integers, hence even, say \(2m\), so the square is \(8m+1\).
    That is remainder \(1\) on division by \(8\), as claimed.
Where it goes wrong
  • Checking \(1,3,5,7,9\) - evidence, not proof.
  • Writing the odd integer as \(2k\), which is the form of an even integer.
  • Assuming the claim is true - circular.
  • Reaching for contradiction ('assume the remainder is not \(1\)'): legal, but here a direct proof is simpler, and the question asks for the most appropriate step.
Takeaway Universal claims about odd/even numbers start by writing the general form: \(2k\) or \(2k+1\).
Answer E · Mathematical proof (Prf1) · question PRF-04
10Identifying 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
11Probability · spec M7.7 · Paper 2 style · ★☆☆

\(P(A) = 0.6\), \(P(B) = 0.5\) and \(P(A \text{ and } B) = 0.3\). Which of the following statements is/are true?

  1. I\(A\) and \(B\) are independent.
  2. II\(A\) and \(B\) are mutually exclusive.
  3. III\(P(A \text{ or } B) = 0.8\)
  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

\(P(A)P(B) = 0.3 = P(A \text{ and } B)\), so independent (I). Not mutually exclusive since they can both happen (II false). \(P(A \text{ or } B) = 0.6 + 0.5 - 0.3 = 0.8\) (III).

Full method — how and why
Three definitions to test: independence, mutual exclusivity and the addition rule.
  1. I: independence means \(P(A \cap B) = P(A)P(B)\). Here \(0.6 \times 0.5 = 0.3\), which matches. True.
    Independence is a numerical condition, not something you can eyeball.
  2. II: mutually exclusive means \(P(A \cap B) = 0\). It is \(0.3 \neq 0\). False.
    They overlap, so they can happen together.
  3. III: \(P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.6 + 0.5 - 0.3 = 0.8\). True.
    Subtract the overlap or you count it twice.
  4. So I and III only.
    Match to the list.
Where it goes wrong
  • Treating independent and mutually exclusive as the same thing - they are near-opposites: two events with non-zero probabilities cannot be both.
  • Forgetting the \(-P(A\cap B)\) and answering \(1.1\), which is impossible for a probability.
Takeaway Independent: \(P(A\cap B) = P(A)P(B)\). Mutually exclusive: \(P(A\cap B) = 0\). Check the numbers, not the words.
Answer F · Probability (M7.7) · question PRO-03
12Functions and transformations · spec MM1.7 · Paper 2 style · ★★☆

Let \(f(x) = \sqrt{x - 3}\) for \(x \geq 3\) and \(g(x) = x^2 + 3\). Which of the following statements is/are true?

  1. I\(g(f(x)) = x\) for all \(x \geq 3\)
  2. II\(f(g(x)) = x\) for all real \(x\)
  3. III\(f(g(x)) = |x|\) for all real \(x\)
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

I: \(g(f(x)) = (\sqrt{x-3})^2 + 3 = x\) for \(x \geq 3\): true. \(f(g(x)) = \sqrt{x^2} = |x|\), so II fails (e.g. \(x = -1\)) and III is true.

Full method — how and why
Two functions that almost invert each other. The asymmetry comes from \(\sqrt{x^2} = |x|\) and from the restricted domain.
  1. I: \(g(f(x)) = \left(\sqrt{x-3}\right)^2 + 3 = (x - 3) + 3 = x\) for \(x \geq 3\). True.
    Squaring a square root recovers the inside, provided the inside is non-negative - guaranteed by the stated domain \(x \geq 3\).
  2. Compute the other order: \(f(g(x)) = \sqrt{(x^2 + 3) - 3} = \sqrt{x^2} = |x|\).
    Do the algebra once and read both remaining statements off it.
  3. II claims this equals \(x\) for all real \(x\): false at \(x = -1\), where \(|x| = 1 \neq -1\). False.
    The square root returns the non-negative value, so it cannot reproduce a negative input.
  4. III says it equals \(|x|\), which is exactly what we found. True.
    Same computation, correct conclusion.
  5. I and III only.
    Match to the options.
Where it goes wrong
  • Assuming \(f(g(x)) = x\) because \(f\) and \(g\) 'look like' inverses - inverses require matching domains and ranges.
  • Writing \(\sqrt{x^2} = x\).
Takeaway \(f(g(x)) = x\) and \(g(f(x)) = x\) are different statements; \(\sqrt{x^2} = |x|\) is what breaks the symmetry here.
Answer F · Functions and transformations (MM1.7) · question FUN-12
13Identifying errors in proofs · spec Err1 · Paper 2 style · ★★☆

Claim: \(\displaystyle\int_{-1}^{1} \frac{1}{x^2}\,dx = -2\).

  1. IAn antiderivative of \(x^{-2}\) is \(-x^{-1}\).
  2. IISo the integral equals \(\left[-\frac1x\right]_{-1}^{1} = -1 - 1 = -2\).
  3. IIIHence the value is \(-2\).

Which of the following is correct?

  1. AThe first error is on line II: the integrand is undefined at \(x = 0\), which lies inside the interval, so the fundamental theorem cannot be applied.
  2. BThe proof is completely correct.
  3. CThe first error is on line I.
  4. DThe first error is on line II: \(-1 - 1 = 0\).
  5. EThe first error is on line III: the value should be \(+2\).
Worked solution

The integrand is positive everywhere it is defined, so a negative answer is a warning sign; the function is not defined on the whole interval and the integral does not exist in the ordinary sense.

Full method — how and why
The fundamental theorem of calculus needs the integrand to be defined (and continuous) throughout the interval.
  1. Spot the warning sign: \(\frac{1}{x^2} > 0\) wherever it is defined, yet the 'answer' is negative.
    An integral of a positive function over an interval cannot be negative - the sign alone signals an error.
  2. Line I is fine: \(\frac{d}{dx}\left(-x^{-1}\right) = x^{-2}\) ✓.
    Differentiate: \(-(-1)x^{-2} = x^{-2}\).
  3. Line II applies the fundamental theorem across \([-1,1]\). But \(\frac{1}{x^2}\) is undefined at \(x = 0\), which lies inside the interval. First error.
    The antiderivative is not continuous on the whole interval, so the theorem does not apply.
  4. The arithmetic in line II is itself right (\(-\frac11 - \left(-\frac{1}{-1}\right) = -1 - 1 = -2\)); the fault is applying the theorem at all.
    Correct arithmetic on an invalid application still gives nonsense.
Where it goes wrong
  • Trying to 'fix' the sign to \(+2\) - the integral does not exist in the ordinary sense, so no finite value is right.
  • Blaming line I; the antiderivative is correct.
  • Ignoring the discontinuity because the limits themselves are perfectly ordinary numbers.
Takeaway Before applying \(\int_a^b f = F(b)-F(a)\), check \(f\) is defined on all of \([a,b]\); and a positive integrand can never give a negative integral.
Answer A · Identifying errors in proofs (Err1) · question ERR-11
14Identifying errors in proofs · spec Err2 · Paper 2 style · ★★☆

A student writes the following argument.

  1. I\(0.1 > 0.01\).
  2. IITaking logarithms to base \(10\) of both sides: \(-1 > -2\).
  3. IIIMultiplying both sides by \(\log_{10}\frac12\), which is negative, reverses the inequality: \(-\log_{10}\frac12 < -2\log_{10}\frac12\).

Which of the following is correct?

  1. AAll three lines are correct.
  2. BLine II is wrong: taking logarithms reverses an inequality.
  3. CLine III is wrong: \(\log_{10}\frac12\) is positive.
  4. DLine II is wrong: \(\log_{10} 0.1 = 1\).
  5. ELine I is wrong.
Worked solution

Logarithms are increasing functions, so line II is valid; \(\log_{10}\frac12 < 0\), so reversing the inequality in line III is correct.

Full method — how and why
Sometimes the argument is simply valid. Check each step against the rules rather than assuming a trap.
  1. Line I: \(0.1 > 0.01\) ✓.
    \(\frac{1}{10} > \frac{1}{100}\).
  2. Line II: \(\log_{10}\) is an increasing function, so it preserves the inequality: \(\log_{10}0.1 = -1\) and \(\log_{10}0.01 = -2\), and indeed \(-1 > -2\) ✓.
    Applying an increasing function to both sides keeps the direction; \(\log_{10} 10^{-1} = -1\).
  3. Line III: \(\log_{10}\frac12 < 0\) because \(\frac12 < 1\). Multiplying an inequality by a negative number reverses it, which is exactly what is written ✓.
    \(-1 > -2\) becomes \(-1 \times c < -2 \times c\) for \(c<0\).
  4. So all three lines are correct.
    The question is testing whether you can certify a valid argument, not only spot faults.
Where it goes wrong
  • Assuming logarithms reverse inequalities - they do not; \(\log\) is increasing on its domain.
  • Thinking \(\log_{10}\frac12\) is positive; logs of numbers below \(1\) are negative.
  • Misreading \(\log_{10}0.1\) as \(1\) rather than \(-1\).
  • Hunting for an error simply because the question implies one might exist.
Takeaway Increasing functions preserve inequalities; multiplying by a negative reverses them. \(\log_b x < 0\) exactly when \(01\)).
Answer A · Identifying errors in proofs (Err2) · question ERR-07
15Binomial expansion · spec MM2.4 · Paper 2 style · ★★☆

Which of the following statements is/are true?

  1. I\(\binom{n}{r} = \binom{n}{n-r}\) for all integers \(0 \leq r \leq n\)
  2. II\(\binom{n}{r} + \binom{n}{r+1} = \binom{n+1}{r+1}\) for all integers \(0 \leq r < n\)
  3. III\(\displaystyle\sum_{r=0}^{n} \binom{n}{r} = 2^{n}\) for all positive integers \(n\)
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

I is the symmetry of Pascal's triangle; II is Pascal's rule (each entry is the sum of the two above it); III is \((1+1)^n\) expanded. All true.

Full method — how and why
Three standard properties of binomial coefficients. Each has a one-line reason and a one-line check.
  1. I: \(\binom nr = \binom n{n-r}\) - choosing \(r\) objects is the same as choosing the \(n-r\) to leave. True.
    Visible as the left-right symmetry of each row of Pascal's triangle.
  2. II: \(\binom nr + \binom n{r+1} = \binom{n+1}{r+1}\) - Pascal's rule, i.e. each entry is the sum of the two above it. True.
    Check with numbers: \(\binom32 + \binom33 = 3 + 1 = 4 = \binom43\) ✓.
  3. III: \(\sum_{r=0}^n \binom nr = 2^n\) - put \(x = 1\) in \((1+x)^n\). True.
    Equivalently, a set with \(n\) elements has \(2^n\) subsets, counted by size.
  4. All three: 'I, II and III'.
    Match to the options.
Where it goes wrong
  • Doubting Pascal's rule because of the index shift on the right - test it with small numbers.
  • Thinking the sum in III is \(2^{n+1}\) by miscounting the \(n+1\) terms.
Takeaway Symmetry, Pascal's rule and row sum \(2^n\) are the three facts about \(\binom nr\) worth knowing cold.
Answer H · Binomial expansion (MM2.4) · question BIN-09
16The logic of arguments · spec Arg3 · Paper 2 style · ★★☆

Which of the following statements is/are true about real numbers?

  1. IFor all \(x\) and all \(y\), if \(x < y\) then \(x^2 < y^2\).
  2. IIFor all \(x\) there exists \(y\) such that \(x < y\) and \(x^2 < y^2\).
  3. IIIThere exist \(x\) and \(y\) with \(x < y\) and \(x^2 > y^2\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

I fails for \(x = -2\), \(y = 1\). II: take \(y = |x| + 1\). III: \(x = -2\), \(y = 1\) works.

Full method — how and why
Three quantified statements about squaring and ordering. Negative numbers decide all three.
  1. I: \(x < y \Rightarrow x^2 < y^2\) for all reals? Take \(x = -2\), \(y = 1\): \(-2 < 1\) but \(4 > 1\). False.
    Squaring is only order-preserving for non-negative numbers.
  2. II: for each \(x\), is there \(y\) with \(x < y\) and \(x^2 < y^2\)? Take \(y = |x| + 1\): then \(y > x\) and \(y > |x|\) so \(y^2 > x^2\). True.
    Choosing \(y\) larger than \(|x|\) guarantees both conditions at once - and \(y\) may depend on \(x\).
  3. III: do some \(x < y\) have \(x^2 > y^2\)? The counterexample from I works: \(x = -2\), \(y = 1\). True.
    Reuse the work - the same pair settles both.
  4. So II and III.
    Match to the list.
Where it goes wrong
  • Accepting I by testing only positive numbers.
  • Rejecting II by picking a bad \(y\); you get to choose \(y\), so choose a helpful one.
Takeaway \(x
Answer G · The logic of arguments (Arg3) · question LGC-16
17Trigonometric functions and equations · spec MM4.4 · Paper 2 style · ★★☆

Which of the following statements about the graph of \(y = \cos x\) is/are true?

  1. IIt is symmetric about the \(y\)-axis.
  2. IIIt is a translation of the graph of \(y = \sin x\) by \(\dfrac{\pi}{2}\) in the negative \(x\)-direction.
  3. IIIIt has period \(\pi\).
  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

\(\cos(-x) = \cos x\) so I is true. \(\sin\left(x + \frac\pi2\right) = \cos x\), so II is true. The period of \(\cos\) is \(2\pi\), so III is false.

Full method — how and why
Three properties of the cosine graph. Each is a standard fact; the false one is about the period.
  1. I: \(\cos(-x) = \cos x\), so the graph is unchanged by reflection in the \(y\)-axis. True.
    Cosine is an even function - its graph is symmetric about the \(y\)-axis, unlike sine.
  2. II: \(\sin\left(x + \frac\pi2\right) = \cos x\), and replacing \(x\) by \(x + \frac\pi2\) shifts the graph \(\frac\pi2\) to the left. True.
    Check a point: \(\sin\) at \(x = \frac\pi2\) equals \(1\), and \(\cos\) reaches \(1\) at \(x = 0\) - shifted left by \(\frac\pi2\) ✓.
  3. III: the period of \(\cos\) is \(2\pi\), not \(\pi\). False.
    \(\cos(x + \pi) = -\cos x\), so a shift of \(\pi\) flips the graph rather than repeating it.
  4. I and II only.
    Match to the option list.
Where it goes wrong
  • Thinking the period is \(\pi\) because \(\cos\) returns to zero every \(\pi\) - the whole shape must repeat.
  • Getting the direction of the shift in II backwards.
Takeaway \(\cos\) is even with period \(2\pi\); \(\cos x = \sin(x + \frac\pi2)\) is a shift to the left.
Answer E · Trigonometric functions and equations (MM4.4) · question TRG-05
18Exponentials and logarithms · spec MM5.2 · Paper 2 style · ★★☆

Which of the following is a counterexample to the claim 'for all positive \(x\) and \(y\), \(\log_{10}(x + y) \leq \log_{10} x + \log_{10} y\)'?

  1. A\(x = 10\), \(y = 100\)
  2. B\(x = y = 2\)
  3. C\(x = 5\), \(y = 5\)
  4. D\(x = 4\), \(y = 4\)
  5. E\(x = y = 1\)
Worked solution

\(x = y = 1\): \(\log 2 > 0 = \log 1 + \log 1\), so the inequality fails. For the others \(x + y \leq xy\) and the claim holds (\(\log\) is increasing).

Full method — how and why
To break an inequality claim, look for the case where the two sides are closest - here small values, where logs go to zero or negative.
  1. Test \(x = y = 1\): left side \(\log_{10}2 \approx 0.30\); right side \(\log_{10}1 + \log_{10}1 = 0\).
    \(\log 1 = 0\) for any base, so the right side collapses.
  2. \(0.30 > 0\), so the claimed inequality \(\leq\) fails. Counterexample.
    Hypothesis satisfied (\(x,y\) positive), conclusion false.
  3. Why the others work: \(\log x + \log y = \log(xy)\), and \(\log\) is increasing, so the claim is just \(x+y \leq xy\). For \(x=y=2\): \(4 \leq 4\) ✓; \(x=y=4\): \(8 \leq 16\) ✓; \(x=y=5\): \(10\leq25\) ✓; \(x=10,y=100\): \(110 \leq 1000\) ✓.
    Converting the log inequality into an algebraic one makes all five cases quick to check.
Where it goes wrong
  • Trying large values, where \(xy\) dwarfs \(x+y\) and the claim comfortably holds.
  • Forgetting \(\log_{10}1 = 0\) and miscomputing the right side.
  • Thinking \(\log(x+y) = \log x + \log y\) - that is the error the whole question is built on.
Takeaway \(\log x + \log y = \log(xy)\), never \(\log(x+y)\). Test inequality claims near the boundary, where the sides are closest.
Answer E · Exponentials and logarithms (MM5.2) · question X2-06
19Mathematical proof · spec Prf1 · Paper 2 style · ★★☆

Consider the claim: if \(x\) and \(y\) are real numbers with \(x + y > 10\), then \(x > 5\) or \(y > 5\). Which of the following proves it?

  1. ASuppose \(x \leq 5\) and \(y \leq 5\). Then \(x + y \leq 10\), contradicting \(x + y > 10\).
  2. BSuppose \(x > 5\). Then \(y > 10 - x\), so \(y > 5\).
  3. CIf \(x + y > 10\) then the average of \(x\) and \(y\) is more than \(5\), so both exceed \(5\).
  4. DTake \(x = 6\), \(y = 6\): then \(x + y = 12 > 10\) and \(x > 5\).
  5. EIf \(x > 5\) then the conclusion holds; if \(y > 5\) then the conclusion holds.
Worked solution

Supposing \(x \leq 5\) and \(y \leq 5\) and deriving \(x + y \leq 10\) is a valid proof by contradiction. 'If \(x>5\) then the conclusion holds' assumes the conclusion; \(x = y = 6\) is a single example; 'suppose \(x>5\), then \(y>5\)' proves something false; and the averaging argument ends with 'both exceed \(5\)', which is wrong.

Full method — how and why
Proving 'if \(P\) then (\(A\) or \(B\))'. Assuming both \(A\) and \(B\) fail and deriving a contradiction is the natural route.
  1. Negate the conclusion: 'not (\(x>5\) or \(y>5\))' means \(x \leq 5\) and \(y \leq 5\) (De Morgan).
    Negating an 'or' gives an 'and' - and it gives you two usable facts instead of a disjunction.
  2. Add the inequalities: \(x + y \leq 10\).
    Inequalities of the same direction may be added.
  3. That contradicts the hypothesis \(x+y > 10\), so the conclusion must hold.
    Hypothesis true and conclusion false is impossible, which is exactly what proves the implication.
Where it goes wrong
  • 'If \(x>5\) then the conclusion holds' - this assumes part of what is to be proved.
  • Testing \(x = y = 6\) - a single example proves nothing about all \(x, y\).
  • 'Suppose \(x > 5\), then \(y > 10 - x\), so \(y > 5\)' - the algebra is wrong and it proves a false statement (both exceeding \(5\)).
  • 'The average exceeds \(5\) so both exceed \(5\)' - the average exceeding \(5\) only forces at least one to.
Takeaway To prove an 'or', negate it into an 'and' and look for a contradiction - De Morgan turns a weak conclusion into strong assumptions.
Answer A · Mathematical proof (Prf1) · question PRF-10
20Chains of reasoning · spec Prf5 · Paper 2 style · ★★☆

\(a\), \(b\) and \(c\) are real numbers with \(a < b < c\). Which of the following statements must be true?

  1. I\(a + b < 2c\)
  2. II\(ab < bc\)
  3. III\(\dfrac{a + c}{2} < c\)
  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: \(a + b < c + c\). III: \(\frac{a + c}{2} < \frac{c + c}{2} = c\). II fails when \(b\) is negative: \(a = -3\), \(b = -2\), \(c = -1\) gives \(ab = 6 > bc = 2\).

Full method — how and why
Inequality manipulation with 'must be true'. Two follow from replacing a term by a larger one; the third involves multiplying, which needs sign information.
  1. I: from \(a < c\) and \(b < c\), add: \(a + b < 2c\). True.
    Adding two inequalities of the same direction is always valid.
  2. III: \(a < c\) so \(a + c < 2c\), and dividing by \(2\) gives \(\frac{a+c}{2} < c\). True.
    Dividing by a positive constant preserves the direction. (The midpoint of two numbers lies below the larger.)
  3. II: \(ab < bc\) would follow from \(a 0\).
    Multiplying an inequality by a negative number reverses it.
  4. Counterexample: \(a=-3, b=-2, c=-1\) satisfies \(a bc\). False.
    All-negative values are the natural place to test a multiplication claim.
  5. So I and III.
    Match to the list.
Where it goes wrong
  • Accepting II by 'multiplying \(a
  • Testing only positive numbers, where II does hold.
Takeaway Adding inequalities is always safe; multiplying by a variable is not - you need its sign.
Answer F · Chains of reasoning (Prf5) · question RSN-08
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.