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Paper 2 · Set 06

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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Answer grid

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1Circles and circle theorems · spec MM3.2 · Paper 1 style · ★☆☆

The circle with centre \((2, -1)\) passes through \((5, 3)\). Find the equation of the tangent to the circle at \((5, 3)\).

  1. A\(4x - 3y = 11\)
  2. B\(3x + 4y = 27\)
  3. C\(x + y = 8\)
  4. D\(3x + 4y = 15\)
  5. E\(3x - 4y = 3\)
  6. F\(4x + 3y = 29\)
Worked solution

Radius direction \((3, 4)\) has gradient \(\frac43\), so the tangent has gradient \(-\frac34\): \(y - 3 = -\frac34(x - 5) \Rightarrow 4y - 12 = -3x + 15 \Rightarrow 3x + 4y = 27\).

Full method — how and why
A tangent is perpendicular to the radius at the point of contact. Find the radius gradient, invert, and write the line.
  1. Vector from centre \((2,-1)\) to the point \((5,3)\): \((3, 4)\), so the radius has gradient \(\frac43\).
    Differences of coordinates give the direction; no need for the circle's equation.
  2. The tangent is perpendicular, so its gradient is \(-\dfrac34\).
    Tangent \(\perp\) radius is the key circle fact here.
  3. Line through \((5,3)\): \(y - 3 = -\frac34(x - 5)\).
    The tangent touches at the given point, so use it in the point-gradient form.
  4. Clear fractions: \(4y - 12 = -3x + 15 \Rightarrow 3x + 4y = 27\).
    Multiply by \(4\), then collect.
Where it goes wrong
  • Using the radius gradient \(\frac43\) for the tangent, giving \(4x - 3y = 11\) - an offered option.
  • Arithmetic slip on the constant, giving \(3x + 4y = 15\).
Takeaway Tangent gradient \(= -1 \div\) (radius gradient); the point of contact is on both the circle and the tangent.
Answer B · Circles and circle theorems (MM3.2) · question CIR-03
2Differentiation · spec MM6.3 · Paper 1 style · ★☆☆

Find the minimum value of \(x + \dfrac{4}{x}\) for \(x > 0\).

  1. A\(5\)
  2. B\(2\sqrt2\)
  3. C\(4\)
  4. D\(0\)
  5. E\(2\)
  6. F\(8\)
Worked solution

Derivative \(1 - \frac{4}{x^2} = 0\) at \(x = 2\); value \(2 + 2 = 4\) (second derivative \(\frac{8}{x^3} > 0\), so a minimum).

Full method — how and why
Minimising \(x + \frac4x\). Differentiate, solve, and confirm it is a minimum.
  1. Write as powers: \(y = x + 4x^{-1}\) and differentiate: \(y' = 1 - 4x^{-2}\).
    \(\frac{d}{dx}(4x^{-1}) = -4x^{-2}\).
  2. Solve \(y' = 0\): \(1 = \dfrac{4}{x^2} \Rightarrow x^2 = 4 \Rightarrow x = 2\) (taking the positive root since \(x > 0\)).
    The restriction \(x > 0\) discards \(x = -2\).
  3. Confirm a minimum: \(y'' = 8x^{-3} > 0\) for \(x > 0\).
    Positive second derivative means the curve bends upwards.
  4. Evaluate: \(y = 2 + \dfrac42 = 4\).
    (Also follows from AM-GM: \(x + \frac4x \geq 2\sqrt{4} = 4\).)
Where it goes wrong
  • Giving the minimising \(x = 2\) instead of the minimum value \(4\) - both are offered.
  • Including \(x = -2\), where the expression equals \(-4\) but \(x > 0\) is required.
Takeaway \(x + \frac kx \geq 2\sqrt k\) for \(x > 0\), with equality at \(x = \sqrt k\) - worth recognising instantly.
Answer C · Differentiation (MM6.3) · question DIF-18
3Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

The roots of \(2x^2 + bx + 18 = 0\) are equal. Find the sum of the possible values of \(b\).

  1. A\(36\)
  2. B\(6\)
  3. C\(0\)
  4. D\(12\)
  5. E\(24\)
  6. F\(-12\)
Worked solution

\(b^2 - 144 = 0\), so \(b = \pm 12\); their sum is \(0\).

Full method — how and why
'Equal roots' is the discriminant-zero condition; the twist is that it gives two values of \(b\) and the question wants their sum.
  1. Write the discriminant with \(a = 2\), \(c = 18\): \(b^2 - 4(2)(18) = b^2 - 144\).
    Substitute carefully - \(4ac\) is \(144\), not \(36\).
  2. Set it to zero: \(b^2 = 144\), so \(b = 12\) or \(b = -12\).
    Equal (repeated) roots means the discriminant vanishes; a square equal to a positive number always has two solutions.
  3. Add them: \(12 + (-12) = 0\).
    The two values are symmetric about zero, so the sum is \(0\) without any further work.
Where it goes wrong
  • Giving only the positive root and answering \(12\).
  • Reading 'sum of the possible values of \(b\)' as 'the value of \(b\)'.
Takeaway \(b^2 = k\) always has two solutions \(\pm\sqrt k\); read whether the question wants one of them, both, or a combination.
Answer C · Quadratic functions (MM1.3) · question QUA-12
4Exponentials and logarithms · spec MM5.1 · Paper 2 style · ★☆☆

For which of the following values of \(x\) is \(3^{x} < 2^{x+1}\)?

  1. I\(x = 0\)
  2. II\(x = 1\)
  3. III\(x = 2\)
  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

\(x = 0\): \(1 < 2\) true. \(x = 1\): \(3 < 4\) true. \(x = 2\): \(9 < 8\) false. So I and II.

Full method — how and why
Three numerical tests. Nothing conceptual - just careful arithmetic with small powers.
  1. I: \(x = 0\) gives \(3^0 = 1\) and \(2^{1} = 2\). Is \(1 < 2\)? Yes. True.
    \(a^0 = 1\) for any \(a\).
  2. II: \(x = 1\) gives \(3^1 = 3\) and \(2^2 = 4\). Is \(3 < 4\)? Yes. True.
    Careful with the index \(x + 1 = 2\).
  3. III: \(x = 2\) gives \(3^2 = 9\) and \(2^3 = 8\). Is \(9 < 8\)? No. False.
    \(3^x\) has overtaken \(2^{x+1}\) by here - exponential growth with the larger base always wins eventually.
  4. I and II only.
    Match to the option list.
Where it goes wrong
  • Computing \(2^{x+1}\) as \(2^x + 1\).
  • Assuming the inequality holds for all \(x\) because it holds for the first two.
Takeaway Substitute each value carefully; a larger base always overtakes eventually, whatever the head start.
Answer E · Exponentials and logarithms (MM5.1) · question LOG-16
5The logic of arguments · spec Arg1 · Paper 2 style · ★☆☆

For a positive integer \(n\), consider the statement: if \(n\) is prime, then \(n\) is odd. 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

\(n = 2\) is an even prime, so the statement is false. The converse, 'if \(n\) is odd then \(n\) is prime', fails for \(n = 9\).

Full method — how and why
Statement and converse again, with the classic \(n = 2\) counterexample.
  1. Statement: primes are odd? \(n = 2\) is prime and even. False.
    \(2\) is the only even prime - it is the standard counterexample and worth memorising.
  2. Converse: odd numbers are prime? \(n = 9 = 3 \times 3\) is odd and composite. False.
    Also \(n = 1\), which is odd and not prime by definition.
  3. So both are false.
    Match to the option list.
Where it goes wrong
  • Forgetting that \(2\) is prime and declaring the statement true.
  • Forgetting that \(1\) is not prime when hunting counterexamples.
Takeaway \(2\) is the even prime and \(1\) is not prime - the two facts behind most prime-related counterexamples.
Answer D · The logic of arguments (Arg1) · question LGC-14
6Sequences and series · spec MM2.3 · Paper 2 style · ★☆☆

Which of the following geometric series is/are convergent?

  1. I\(1 + 0.9 + 0.81 + \ldots\)
  2. II\(3 - 3 + 3 - 3 + \ldots\)
  3. III\(100 + 10 + 1 + 0.1 + \ldots\)
  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 geometric series converges if and only if \(|r| < 1\). I has \(r = 0.9\), III has \(r = 0.1\): both converge. II has \(r = -1\) and does not converge.

Full method — how and why
Convergence of a geometric series depends only on \(|r| < 1\). Compute \(r\) for each and apply the test.
  1. I: \(r = \frac{0.9}{1} = 0.9\), and \(|0.9| < 1\): converges.
    A ratio just below \(1\) still converges, though slowly.
  2. II: \(r = \frac{-3}{3} = -1\), and \(|-1| = 1\) is not less than \(1\): does not converge.
    The partial sums oscillate between \(3\) and \(0\) forever - they never settle, so there is no sum to infinity.
  3. III: \(r = \frac{10}{100} = 0.1\): converges.
    Small ratio, rapid convergence.
  4. I and III only.
    Match the verdicts to the option list.
Where it goes wrong
  • Thinking II converges to \(0\) or to \(1.5\) because the terms 'average out' - the partial sums must approach a single limit.
  • Judging by the size of the terms rather than the ratio: III starts at \(100\) but converges fastest.
Takeaway Convergence is decided by \(|r| < 1\) alone - not by the size of the first term or by whether the terms alternate.
Answer F · Sequences and series (MM2.3) · question SEQ-12
7Identifying errors in proofs · spec Err2 · Paper 2 style · ★☆☆

A student solves \(2\cos^2 x = \cos x\) for \(0 \leq x < 2\pi\).

  1. IDivide both sides by \(\cos x\): \(2\cos x = 1\).
  2. IISo \(\cos x = \dfrac12\).
  3. IIISo \(x = \dfrac{\pi}{3}\) or \(x = \dfrac{5\pi}{3}\).

How many solutions has the student missed, and where is the first error?

  1. ATwo solutions missed; the first error is on line I.
  2. BNo solutions missed; the working is correct.
  3. COne solution missed; the first error is on line III.
  4. DTwo solutions missed; the first error is on line III.
  5. EOne solution missed; the first error is on line I.
Worked solution

Dividing by \(\cos x\) loses \(\cos x = 0\), i.e. \(x = \frac\pi2\) and \(x = \frac{3\pi}{2}\): two solutions.

Full method — how and why
Dividing a trigonometric equation by \(\cos x\) loses the solutions where \(\cos x = 0\).
  1. Line I divides by \(\cos x\), valid only if \(\cos x \neq 0\). First error.
    The equation \(2\cos^2x = \cos x\) is satisfied whenever \(\cos x = 0\), so those solutions are discarded.
  2. The correct method: \(2\cos^2x - \cos x = 0\), so \(\cos x(2\cos x - 1) = 0\).
    Factorise rather than divide - it keeps both branches.
  3. Branch \(\cos x = 0\): in \(0 \leq x < 2\pi\) that is \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\) - two missed solutions.
    Cosine vanishes at the top and bottom of the unit circle.
  4. Branch \(\cos x = \frac12\): \(x = \frac{\pi}{3}\) and \(x = \frac{5\pi}{3}\) - the two the student found.
    Cosine is positive in the first and fourth quadrants, so \(x = \frac\pi3\) and \(2\pi - \frac\pi3\). Four solutions in total.
Where it goes wrong
  • Saying one solution was missed; \(\cos x = 0\) has two solutions in a full revolution.
  • Blaming line III - the student's own two values are correct for their (incomplete) equation.
  • Declaring the working correct because the two values found do satisfy the original equation.
Takeaway Never divide a trigonometric equation by a trig function - factorise, and count the solutions of each branch across the whole interval.
Answer A · Identifying errors in proofs (Err2) · question ERR-18
8Mathematical proof · spec Prf3 · Paper 2 style · ★☆☆

\(T_n\) denotes the \(n\)th triangular number \(\dfrac{n(n+1)}{2}\). A student observes \(T_1 + T_2 = 4\), \(T_2 + T_3 = 9\), \(T_3 + T_4 = 16\) and conjectures that \(T_n + T_{n+1}\) is always a perfect square. Which of the following proves the conjecture?

  1. A\(T_n + T_{n+1} = \dfrac{n(n+1) + (n+1)(n+2)}{2} = \dfrac{(n+1)(2n+2)}{2} = (n+1)^2\)
  2. B\(T_n + T_{n+1} = 2T_n + (n + 1)\), which is a square.
  3. CTriangular numbers are always squares.
  4. D\(T_4 + T_5 = 25\), which is a square, so the pattern continues.
  5. E\(T_{n+1} - T_n = n + 1\), so the sum is a square.
Worked solution

Only the algebraic verification \(T_n + T_{n+1} = \frac{(n+1)(2n+2)}{2} = (n+1)^2\) is general; the rest check one more case, assert something false, or state a true identity without showing the result.

Full method — how and why
Proving a conjecture about a sequence: substitute the formula and simplify algebraically to a general result.
  1. Write both terms: \(T_n = \dfrac{n(n+1)}{2}\) and \(T_{n+1} = \dfrac{(n+1)(n+2)}{2}\).
    Replace \(n\) by \(n+1\) throughout the given formula.
  2. Add over the common denominator: \(\dfrac{n(n+1) + (n+1)(n+2)}{2}\).
    Same denominator, so just add numerators.
  3. Factor out \((n+1)\): \(\dfrac{(n+1)\left[n + (n+2)\right]}{2} = \dfrac{(n+1)(2n+2)}{2}\).
    Spotting the common factor avoids expanding and re-factorising.
  4. \(2n+2 = 2(n+1)\), so the expression is \((n+1)^2\) - a perfect square for every \(n\). Proved.
    The \(2\) cancels the denominator.
Where it goes wrong
  • Checking \(T_4 + T_5 = 25\) and stopping - one more case is still not a proof.
  • 'Triangular numbers are always squares' - plainly false (\(T_2 = 3\)).
  • '\(T_n + T_{n+1} = 2T_n + (n+1)\), which is a square' - the identity is right but the conclusion is asserted, not shown.
  • Quoting the difference \(T_{n+1} - T_n = n+1\), which is true but says nothing about the sum.
Takeaway Substitute the general formula and simplify; consecutive triangular numbers sum to \((n+1)^2\).
Answer A · Mathematical proof (Prf3) · question PRF-20
9Angles, polygons, congruence and vectors · spec M5.3 · Paper 2 style · ★☆☆

\(ABCD\) is a quadrilateral. \(P\): 'the diagonals of \(ABCD\) bisect each other'. \(Q\): '\(ABCD\) is a rectangle'. Which one of the following is true?

  1. AP is necessary and sufficient for Q.
  2. BP is necessary but not sufficient for Q.
  3. CP is sufficient but not necessary for Q.
  4. DP is neither necessary nor sufficient for Q.
Worked solution

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

Full method — how and why
A quadrilateral hierarchy question. Bisecting diagonals characterise parallelograms, a larger class than rectangles.
  1. Every rectangle is a parallelogram, and a parallelogram's diagonals bisect each other. So \(Q \Rightarrow P\): \(P\) is necessary.
    Opposite sides equal and parallel forces the diagonals to share a midpoint.
  2. Is \(P\) sufficient? Take a non-rectangular parallelogram, e.g. a rhombus with angles \(60^\circ/120^\circ\): diagonals bisect each other, but it is not a rectangle.
    Bisecting diagonals say nothing about the angles.
  3. So \(P\) is necessary but not sufficient.
    For a rectangle you also need the diagonals to be equal in length.
Where it goes wrong
  • Answering 'necessary and sufficient' by confusing 'bisect each other' with 'bisect each other and are equal'.
  • Getting the direction backwards: the broader property (parallelogram) is necessary for the narrower one (rectangle).
Takeaway Diagonals bisect each other \(\iff\) parallelogram; add equal diagonals for a rectangle, add perpendicular for a rhombus.
Answer B · Angles, polygons, congruence and vectors (M5.3) · question X2-20
10Exponentials and logarithms · spec MM5.2 · Paper 2 style · ★☆☆

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

  1. I\(\log_{10}(x + y) = \log_{10} x + \log_{10} y\)
  2. II\(\log_{10}(xy)^2 = 2\log_{10} x + 2\log_{10} y\)
  3. III\(\log_{10} x - \log_{10} y = \log_{10} x \div \log_{10} y\)
  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

II is correct: \(\log(xy)^2 = 2\log(xy) = 2\log x + 2\log y\). I and III confuse the laws: \(\log\) of a sum has no simple form, and a difference of logs is \(\log\) of a quotient, not a quotient of logs.

Full method — how and why
Three log 'laws', two of which are fabrications. Test each against the genuine rules.
  1. I: \(\log(x+y)\) has no expansion. Test \(x = y = 1\): \(\log 2 \approx 0.30\) but \(\log 1 + \log 1 = 0\). False.
    The product law is \(\log(xy) = \log x + \log y\); the sum inside is a different animal entirely.
  2. II: \(\log(xy)^2 = 2\log(xy) = 2(\log x + \log y) = 2\log x + 2\log y\). True.
    Power law first, then product law - both applied correctly.
  3. III: \(\log x - \log y = \log\frac xy\), which is not the same as \(\frac{\log x}{\log y}\). Test \(x = 100\), \(y = 10\): left \(= 1\), right \(= \frac21 = 2\). False.
    Difference of logs is the log of a quotient, not the quotient of logs.
  4. II only.
    Match to the options.
Where it goes wrong
  • Accepting I by analogy with the product law.
  • Accepting III because 'subtraction corresponds to division' - it does, but inside the log.
Takeaway The laws act on what is inside the log: \(\log(xy)\), \(\log\frac xy\), \(\log x^k\). Nothing simplifies \(\log(x+y)\).
Answer C · Exponentials and logarithms (MM5.2) · question LOG-07
11Identifying errors in proofs · spec Err2 · Paper 2 style · ★☆☆

A student tries to solve \(\sin\theta = \sin 40^\circ\) for \(0^\circ \leq \theta < 360^\circ\).

  1. ISince \(\sin\theta = \sin 40^\circ\), we have \(\theta = 40^\circ\).
  2. IIThe sine function has period \(360^\circ\), so \(\theta = 40^\circ + 360^\circ k\).
  3. IIIIn the given interval the only solution is \(\theta = 40^\circ\).

Which of the following is correct?

  1. AThe first error is on line I: \(\sin\theta = \sin\alpha\) does not imply \(\theta = \alpha\); \(\theta = 140^\circ\) is also a solution.
  2. BThe solution is completely correct.
  3. CThe first error is on line II: the period of sine is \(180^\circ\).
  4. DThe first error is on line III: \(\theta = 400^\circ\) should be included.
  5. EThe first error is on line II: it should be \(\theta = 40^\circ + 180^\circ k\).
Worked solution

'If \(\sin\alpha = \sin\beta\) then \(\alpha = \beta\)' is a classic invalid deduction; \(\sin(180^\circ - 40^\circ) = \sin40^\circ\) too.

Full method — how and why
\(\sin\) is not one-to-one, so \(\sin\theta = \sin\alpha\) has more solutions than \(\theta = \alpha\).
  1. Line I claims \(\sin\theta = \sin 40^\circ \Rightarrow \theta = 40^\circ\). This is the error.
    Sine takes the same value twice per revolution: \(\sin(180^\circ - \alpha) = \sin\alpha\).
  2. So \(\theta = 140^\circ\) also satisfies the equation.
    \(\sin 140^\circ = \sin 40^\circ\) - check on the graph or with the symmetry of the unit circle.
  3. Lines II and III are then correct statements in themselves (sine does have period \(360^\circ\)), but they are built on the incomplete line I.
    The question asks for the first error, which is line I.
  4. The full solution set in \(0^\circ \leq \theta < 360^\circ\) is \(\theta = 40^\circ\) and \(\theta = 140^\circ\).
    Only two, because the sine is positive in the first and second quadrants.
Where it goes wrong
  • Claiming the period of sine is \(180^\circ\) - that is tangent.
  • Claiming \(\theta = 400^\circ\) was missed - it is outside the stated interval.
  • Declaring the work correct because each individual line looks plausible.
Takeaway \(\sin\theta = \sin\alpha\) gives \(\theta = \alpha + 360^\circ k\) or \(\theta = 180^\circ - \alpha + 360^\circ k\). Never cancel a sine.
Answer A · Identifying errors in proofs (Err2) · question ERR-03
12The logic of arguments · spec Arg2 · Paper 2 style · ★☆☆

For real numbers \(a\) and \(b\), let \(P\): '\(a = b\)' and \(Q\): '\(a^2 = b^2\)'. Which one of the following is true?

  1. AP is necessary and sufficient for Q.
  2. BP is necessary but not sufficient for Q.
  3. CP is sufficient but not necessary for Q.
  4. DP is neither necessary nor sufficient for Q.
Worked solution

\(a = b \Rightarrow a^2 = b^2\) (sufficient), but \(a = 1\), \(b = -1\) gives \(a^2 = b^2\) without \(a = b\) (not necessary).

Full method — how and why
The squaring trap again, in its purest form.
  1. Sufficiency: if \(a = b\) then squaring both sides gives \(a^2 = b^2\). True.
    Doing the same thing to equal quantities keeps them equal.
  2. Necessity: does \(a^2 = b^2\) force \(a = b\)? Take \(a = 1\), \(b = -1\): \(1 = 1\) but \(a \neq b\).
    \(a^2 = b^2\) only gives \(a = \pm b\).
  3. So \(P\) is sufficient but not necessary.
    Squaring loses sign information, so it cannot be reversed.
Where it goes wrong
  • Treating squaring as reversible - it is exactly this that makes squaring an equation introduce extra solutions.
  • Only trying positive values.
Takeaway \(a = b \Rightarrow a^2 = b^2\), but \(a^2 = b^2 \Rightarrow a = \pm b\). Squaring is a one-way street.
Answer C · The logic of arguments (Arg2) · question LGC-10
13Ratio, proportion and percentages · spec M3.8 · Paper 2 style · ★☆☆

Which of the following statements is/are true?

  1. IIncreasing a quantity by \(50\%\) and then decreasing the result by \(50\%\) returns it to its original value.
  2. IIA \(10\%\) increase followed by a \(10\%\) increase is a \(21\%\) increase overall.
  3. IIIIf \(A\) is \(25\%\) more than \(B\), then \(B\) is \(20\%\) less than \(A\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

I is false: \(1.5 \times 0.5 = 0.75\). II: \(1.1^2 = 1.21\), true. III: \(A = 1.25B \Rightarrow B = 0.8A\), true.

Full method — how and why
Three percentage claims. Each is settled by converting to multipliers.
  1. I: \(+50\%\) then \(-50\%\) gives \(1.5 \times 0.5 = 0.75\), i.e. \(75\%\) of the original. False.
    The decrease acts on the larger amount, so it removes more than the increase added.
  2. II: \(+10\%\) twice gives \(1.1^2 = 1.21\), a \(21\%\) increase. True.
    The extra \(1\%\) is the interest on the first year's interest.
  3. III: \(A = 1.25B\) means \(B = \dfrac{A}{1.25} = 0.8A\), which is \(20\%\) less than \(A\). True.
    Percentage changes are not symmetric because the base changes.
  4. II and III only.
    Match to the options.
Where it goes wrong
  • Accepting I because the percentages 'cancel'.
  • Rejecting III on the grounds that \(25\%\) up should mean \(25\%\) down.
Takeaway Always convert to multipliers: percentage changes are asymmetric because the base changes.
Answer G · Ratio, proportion and percentages (M3.8) · question RAT-15
14Mathematical proof · spec Prf1 · Paper 2 style · ★☆☆

Which of the following is a counterexample to the claim 'if \(n\) is a positive integer, then \(2^n + 1\) is prime'?

  1. A\(n = 8\)
  2. B\(n = 1\)
  3. C\(n = 4\)
  4. D\(n = 3\)
  5. E\(n = 2\)
Worked solution

\(2^3 + 1 = 9 = 3 \times 3\) is not prime. (\(n = 1, 2, 4, 8\) give \(3, 5, 17, 257\), all prime.)

Full method — how and why
To disprove a universal claim, find one value where the conclusion fails - so test values until one breaks.
  1. Evaluate \(2^n+1\) for each candidate: \(n=1 \to 3\), \(n=2 \to 5\), \(n=3 \to 9\), \(n=4 \to 17\), \(n=8 \to 257\).
    Systematic evaluation is faster than trying to be clever here.
  2. \(9 = 3 \times 3\) is not prime, so \(n = 3\) breaks the claim.
    That is the counterexample: hypothesis satisfied (\(3\) is a positive integer), conclusion false.
  3. The others - \(3, 5, 17, 257\) - are all prime, so none of them refutes anything.
    (\(257\) is prime: it is not divisible by \(2,3,5,7,11,13\), and \(17^2 = 289 > 257\).)
Where it goes wrong
  • Assuming a large value like \(n = 8\) must fail because the numbers get big - size is not the issue.
  • Stopping at the first odd value of \(2^n+1\); every one of them is odd, since \(2^n\) is even.
Takeaway Notice the pattern: \(2^n+1\) can only be prime when \(n\) is a power of \(2\) - and \(3\) is not, which is exactly why it fails.
Answer D · Mathematical proof (Prf1) · question PRF-02
15Chains of reasoning · spec Prf5 · Paper 2 style · ★★☆

Three statements are made about a positive integer \(N\):

  • \(N\) is a perfect square.
  • \(N\) is a multiple of \(6\).
  • \(N\) is less than \(50\).

Exactly two of the statements are true. Which of the following could NOT be \(N\)?

  1. A\(144\)
  2. B\(12\)
  3. C\(36\)
  4. D\(48\)
  5. E\(16\)
  6. F\(25\)
Worked solution

\(36\) makes all three statements true, not exactly two. Each of the others makes exactly two true: \(16\) and \(25\) (square, \(< 50\)), \(48\) and \(12\) (multiple of \(6\), \(< 50\)), \(144\) (square, multiple of \(6\)).

Full method — how and why
Test each candidate against all three statements and count how many are true. You want the one that does not give exactly two.
  1. \(36\): perfect square (\(6^2\)) ✓, multiple of \(6\) ✓, less than \(50\) ✓ - three true.
    The condition demands exactly two, so \(36\) is ruled out.
  2. \(16\): square ✓, multiple of \(6\) ✗, \(<50\) ✓ - two. Allowed.
    Check each of the three statements, do not stop early.
  3. \(25\): square ✓, multiple of \(6\) ✗, \(<50\) ✓ - two. \(48\): square ✗, multiple of \(6\) ✓, \(<50\) ✓ - two. \(12\): the same pattern - two.
    All fine.
  4. \(144\): square (\(12^2\)) ✓, multiple of \(6\) ✓, \(<50\) ✗ - two. Allowed.
    So only \(36\) fails.
Where it goes wrong
  • Reading 'exactly two are true' as 'at least two', which would admit \(36\).
  • Missing that \(144 = 12^2\) is both a square and a multiple of \(6\).
  • Looking for the value making the fewest statements true instead of the most.
Takeaway 'Exactly \(k\)' excludes \(k+1\) as firmly as \(k-1\); tabulate every statement for every candidate.
Answer C · Chains of reasoning (Prf5) · question RSN-06
16Chains of reasoning · spec Prf5 · Paper 2 style · ★★☆

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

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

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

Full method — how and why
Inclusion-exclusion: count what you don't want and subtract. Take care with 'less than \(1000\)'.
  1. The range is \(1\) to \(999\), so \(999\) integers.
    'Less than \(1000\)' excludes \(1000\) itself.
  2. Multiples of \(2\): \(\left\lfloor \frac{999}{2}\right\rfloor = 499\). Multiples of \(5\): \(199\). Multiples of both, i.e. of \(10\): \(99\).
    Divide and round down. \(10\) is the LCM of \(2\) and \(5\).
  3. Divisible by \(2\) or \(5\): \(499 + 199 - 99 = 599\).
    Subtract the overlap, which was counted twice.
  4. Neither: \(999 - 599 = 400\).
    Sanity check: in each block of \(10\) consecutive integers exactly \(4\) (ending in \(1,3,7,9\)) survive, and \(999 = 99\) full blocks plus \(991\)-\(999\), giving \(99 \times 4 + 4 = 400\) ✓.
Where it goes wrong
  • Using \(1000\) as the count and getting \(400\) by luck with wrong intermediate figures - or getting \(399\) by an off-by-one at the top.
  • Forgetting to subtract the multiples of \(10\), giving \(999 - 698 = 301\).
Takeaway \(|A \cup B| = |A| + |B| - |A \cap B|\), with the intersection counted via the LCM; then subtract from the total.
Answer D · Chains of reasoning (Prf5) · question RSN-12
17Quadratic functions · spec MM1.3 · Paper 2 style · ★★☆

The function \(f(x) = x^2 - 6x + 13\) is defined for all real \(x\). Which of the following statements is/are true?

  1. I\(f(x) > 0\) for all real \(x\)
  2. IIThe graph of \(y = f(x)\) is symmetric about the line \(x = 3\)
  3. IIIThe equation \(f(x) = 4\) has exactly one real solution
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Worked solution

\(f(x) = (x-3)^2 + 4 \geq 4 > 0\), so I is true. The axis of symmetry is \(x = 3\), so II is true. \(f(x) = 4\) only when \((x-3)^2 = 0\), i.e. \(x = 3\): exactly one solution, so III is true.

Full method — how and why
Three statements about one quadratic. Completing the square once answers all three at a stroke.
  1. Complete the square: \(x^2 - 6x + 13 = (x - 3)^2 + 4\).
    This single form contains the range, the line of symmetry and the vertex - so do it first, before looking at the statements.
  2. I: \((x-3)^2 \geq 0\) so \(f(x) \geq 4 > 0\) for every \(x\). True.
    The minimum value is \(4\), comfortably positive; equivalently the discriminant \(36 - 52 < 0\), so the curve never reaches the axis.
  3. II: the vertex is at \(x = 3\), and a parabola is symmetric about the vertical line through its vertex. True.
    Symmetry about \(x = h\) is immediate from the \((x - h)^2\) form.
  4. III: \(f(x) = 4\) means \((x-3)^2 = 0\), so \(x = 3\) only - exactly one solution. True.
    A horizontal line through the vertex meets the parabola once; any higher line meets it twice.
  5. All three hold, so the answer is 'I, II and III'.
    Decide each statement independently, then match to the option list.
Where it goes wrong
  • Assuming \(f(x) = 4\) has two solutions because quadratics 'usually' do - \(y = 4\) is the tangent line at the vertex.
  • Reading the symmetry line off as \(x = -3\).
Takeaway Completing the square answers questions about sign, range, symmetry and repeated roots all at once.
Answer H · Quadratic functions (MM1.3) · question QUA-06
18Indices and surds · spec MM1.1 · Paper 2 style · ★★★

For how many integer values of \(n\) with \(1 \leq n \leq 100\) is \(\sqrt{n} + \sqrt{n+1}\) rational?

  1. A\(9\)
  2. B\(1\)
  3. C\(10\)
  4. D\(0\)
  5. E\(2\)
  6. F\(100\)
Worked solution

For the sum to be rational, both \(\sqrt n\) and \(\sqrt{n+1}\) would need to be rational (if one is irrational, \(\sqrt n + \sqrt{n+1}\) squared gives \(2n + 1 + 2\sqrt{n(n+1)}\) and \(n(n+1)\) is never a perfect square for \(n \geq 1\) since \(n^2 < n(n+1) < (n+1)^2\)). No two consecutive positive integers are both squares, so the answer is \(0\).

Full method — how and why
A 'for how many' question that is really about irrationality. The answer is \(0\), and seeing why needs one sharp observation about consecutive integers.
  1. Suppose \(\sqrt n + \sqrt{n+1} = r\) is rational and square it: \(2n + 1 + 2\sqrt{n(n+1)} = r^2\).
    Squaring is the standard way to get at a sum of surds; the rational parts separate from the single remaining surd.
  2. Rearrange: \(\sqrt{n(n+1)} = \dfrac{r^2 - 2n - 1}{2}\), so \(n(n+1)\) would have to be a perfect square.
    A rational square root of an integer is an integer, so \(n(n+1)\) must be a perfect square.
  3. But \(n^2 < n(n+1) < (n+1)^2\) for every \(n \geq 1\), so \(n(n+1)\) lies strictly between consecutive squares.
    There is no integer square strictly between \(n^2\) and \((n+1)^2\) - this is the contradiction.
  4. Hence no \(n\) in \(1 \leq n \leq 100\) works: the answer is \(0\).
    The bound \(100\) is a red herring; the argument rules out every positive integer.
Where it goes wrong
  • Testing \(n = 3\) (\(\sqrt3 + 2\)) and concluding 'sometimes it works' because one of the two roots is nice.
  • Answering \(9\) or \(10\) by counting perfect squares up to \(100\) - that would answer a different question.
Takeaway \(n(n+1)\) is never a perfect square for \(n \geq 1\), because it is squeezed strictly between \(n^2\) and \((n+1)^2\).
Answer D · Indices and surds (MM1.1) · question IND-12
19Identifying errors in proofs · spec Err1 · Paper 2 style · ★★★

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

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

Which of the following is true?

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

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

Full method — how and why
A subtle one: the claim is true but one line asserts more than was proved. Distinguish a false conclusion from an unjustified step.
  1. Check line I: \(f'(x) = 3x^2 - 6x + 3 = 3(x^2-2x+1) = 3(x-1)^2\). Correct.
    Differentiate term by term, then factorise.
  2. Check line II: a square is non-negative, so \(f'(x) \geq 0\), with equality only at \(x = 1\). Correct.
    The word 'only' is right - \((x-1)^2 = 0\) just at \(x=1\).
  3. Line III says '\(f'(x) > 0\) for all \(x\)' - false at \(x = 1\), where \(f'(1) = 0\). So line III overstates line II.
    It asserts a strict inequality that was never established.
  4. Is the claim itself true? Yes: \(f' \geq 0\) everywhere with equality at a single isolated point still gives \(f(a) < f(b)\) for \(a
    A derivative vanishing at one point does not create a flat stretch; there is no turning point, because \(f'\) does not change sign.
Where it goes wrong
  • Concluding the claim is false because \(f'(1) = 0\) - a stationary point of inflexion is not a turning point.
  • Blaming line I or II, both of which are correct.
  • Calling the proof completely correct, ignoring the strictness error in line III.
Takeaway \(f' \geq 0\) with equality only at isolated points still means strictly increasing - but a proof must not claim strict positivity it has not shown.
Answer A · Identifying errors in proofs (Err1) · question ERR-06
20The logic of arguments · spec Arg2 · Paper 2 style · ★★★

Let \(P\): 'the integer \(n\) is divisible by \(4\)' and \(Q\): 'the integer \(n^2\) is divisible by \(8\)'. 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 \(4 \mid n\) then \(16 \mid n^2\), so \(8 \mid n^2\): \(P\) is sufficient. Conversely, if \(8 \mid n^2\) then \(n\) is even, say \(n = 2m\), and \(8 \mid 4m^2\) forces \(m\) to be even, so \(4 \mid n\): \(P\) is also necessary. (Tempting wrong answer: 'sufficient but not necessary' - but no counterexample exists.)

Full method — how and why
A genuine 'necessary and sufficient' case, dressed to look like the usual 'sufficient only'. You must prove both directions rather than assume a counterexample exists.
  1. Sufficiency: if \(4 \mid n\), write \(n = 4k\). Then \(n^2 = 16k^2\), which is divisible by \(8\). True.
    \(16\) is a multiple of \(8\), so divisibility by \(8\) follows immediately.
  2. Necessity: suppose \(8 \mid n^2\). Then \(n^2\) is even, so \(n\) is even - write \(n = 2m\).
    An odd number squared is odd, so \(n\) cannot be odd.
  3. Then \(n^2 = 4m^2\), and \(8 \mid 4m^2\) forces \(2 \mid m^2\), so \(m\) is even, say \(m = 2j\).
    Divide the divisibility through by the common factor \(4\).
  4. So \(n = 4j\), i.e. \(4 \mid n\). Both directions hold: necessary and sufficient.
    No counterexample exists, which is what distinguishes this from the usual pattern.
Where it goes wrong
  • Answering 'sufficient but not necessary' out of habit, without actually hunting for a counterexample. Test \(n = 2\): \(n^2 = 4\), not divisible by \(8\) - so it is not a counterexample; \(n = 6\): \(36\), not divisible by \(8\). The hunt keeps failing, which is the clue.
  • Assuming \(8 \mid n^2 \Rightarrow 8 \mid n\) - false (\(n = 4\) gives \(n^2 = 16\)).
Takeaway Before declaring a condition 'not necessary', actually look for a counterexample; repeated failure to find one is a signal to prove the converse instead.
Answer A · The logic of arguments (Arg2) · question LGC-15
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