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Paper 2 · Set 05

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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1Triangles, radians and sectors · spec MM4.2 · Paper 1 style · ★☆☆

Convert \(135^\circ\) to radians.

  1. A\(\dfrac{4\pi}{3}\)
  2. B\(\dfrac{3\pi}{8}\)
  3. C\(\dfrac{3\pi}{4}\)
  4. D\(\dfrac{5\pi}{6}\)
  5. E\(\dfrac{2\pi}{3}\)
  6. F\(\dfrac{5\pi}{4}\)
Worked solution

\(135 = \frac34 \times 180\), so \(\frac{3\pi}{4}\).

Full method — how and why
Degree-to-radian conversion: multiply by \(\frac{\pi}{180}\), or spot the fraction of \(180^\circ\).
  1. Express \(135\) as a fraction of \(180\): \(\dfrac{135}{180} = \dfrac34\).
    Since \(180^\circ = \pi\) radians, the same fraction of \(\pi\) gives the answer.
  2. So \(135^\circ = \dfrac34 \pi = \dfrac{3\pi}{4}\).
    Equivalently \(135 \times \frac{\pi}{180} = \frac{3\pi}{4}\).
Where it goes wrong
  • Multiplying by \(\frac{180}{\pi}\) (the reverse conversion).
  • Simplifying \(\frac{135}{180}\) incorrectly to \(\frac23\) or \(\frac56\), both of which are offered.
Takeaway \(180^\circ = \pi\): convert by multiplying by \(\frac{\pi}{180}\), and cancel the fraction fully.
Answer C · Triangles, radians and sectors (MM4.2) · question TRI-05
2Circles and circle theorems · spec MM3.3 · Paper 1 style · ★☆☆

\(TA\) is a tangent to a circle at \(A\), and \(AB\) is a chord. \(C\) is a point on the circle on the opposite side of \(AB\) from \(T\), with angle \(ACB = 65^\circ\). Find angle \(TAB\).

  1. A\(25^\circ\)
  2. B\(65^\circ\)
  3. C\(115^\circ\)
  4. D\(130^\circ\)
  5. E\(50^\circ\)
  6. F\(32.5^\circ\)
Worked solution

Alternate segment theorem: the angle between tangent and chord equals the angle in the alternate segment, so \(TAB = 65^\circ\).

Full method — how and why
The alternate segment theorem, stated almost word for word. The work is in identifying which angle is 'in the alternate segment'.
  1. \(TA\) is tangent at \(A\) and \(AB\) is a chord, so the angle \(TAB\) between them is the tangent-chord angle.
    The theorem applies exactly to this configuration.
  2. The alternate segment is the one on the other side of \(AB\) from \(T\) - which is where \(C\) lies.
    The question places \(C\) deliberately; the angle used must be in the segment away from the tangent's side.
  3. Therefore angle \(TAB\) = angle \(ACB = 65^\circ\).
    Alternate segment theorem: the tangent-chord angle equals the inscribed angle in the alternate segment.
Where it goes wrong
  • Using \(180^\circ - 65^\circ = 115^\circ\), which is the angle on the other side of the tangent.
  • Halving to \(32.5^\circ\) by confusing this with the centre-circumference theorem.
Takeaway Tangent-chord angle \(=\) angle in the alternate segment - the segment on the far side of the chord.
Answer B · Circles and circle theorems (MM3.3) · question CIR-10
3Quadratic functions · spec MM1.3 · Paper 1 style · ★☆☆

Let \(f(x) = x^2 - 2x - 8\). For how many integers \(n\) is \(f(n) < 0\)?

  1. A\(8\)
  2. B\(6\)
  3. C\(5\)
  4. D\(7\)
  5. E\(3\)
  6. F\(4\)
Worked solution

\((x-4)(x+2) < 0\) gives \(-2 < x < 4\); the integers are \(-1, 0, 1, 2, 3\): five of them.

Full method — how and why
Counting integers in the region where a quadratic is negative. Solve the inequality, then count carefully at the ends.
  1. Factorise: \(n^2 - 2n - 8 = (n - 4)(n + 2)\).
    Two numbers multiplying to \(-8\) and adding to \(-2\): \(-4\) and \(+2\).
  2. Negative between the roots: \(-2 < n < 4\).
    Upward parabola, so \(f < 0\) strictly between the roots.
  3. List the integers strictly inside: \(-1, 0, 1, 2, 3\) - five of them.
    The endpoints \(-2\) and \(4\) give \(f = 0\), which is not less than zero, so they are excluded; and \(0\) is easy to forget.
Where it goes wrong
  • Including \(-2\) and \(4\) and answering \(7\).
  • Forgetting \(0\) is an integer and answering \(4\).
Takeaway After solving an inequality, list the integers explicitly - endpoints and zero are the two things people drop.
Answer C · Quadratic functions (MM1.3) · question QUA-24
4Chains of reasoning · spec Prf5 · Paper 2 style · ★☆☆

Each of \(100\) cards has a different integer from \(1\) to \(100\) written on it. What is the smallest number of cards that must be drawn (without looking) to be certain of having two cards whose numbers sum to \(101\)?

  1. A\(51\)
  2. B\(2\)
  3. C\(100\)
  4. D\(99\)
  5. E\(52\)
  6. F\(50\)
Worked solution

The \(50\) pairs \(\{1, 100\}, \{2, 99\}, \ldots, \{50, 51\}\) each sum to \(101\). Fifty cards (one from each pair) may avoid it; \(51\) cannot.

Full method — how and why
Pigeonhole again, with the pairs built to sum to the target.
  1. Pair the numbers to sum to \(101\): \(\{1,100\}, \{2,99\}, \ldots, \{50,51\}\).
    Each of \(1\) to \(100\) lies in exactly one such pair, so there are \(50\) pairs.
  2. Drawing one card from each pair gives \(50\) cards with no pair summing to \(101\) - for example \(1\) to \(50\).
    So \(50\) is not enough; this constructs the worst case.
  3. Any \(51\) cards must include two from one pair, and those sum to \(101\).
    \(51\) cards into \(50\) pairs forces a repeat.
Where it goes wrong
  • Answering \(50\) - the largest number that can fail.
  • Answering \(2\) by thinking about luck rather than certainty.
Takeaway Same recipe as any pigeonhole guarantee: build the pairs, show the worst case, then add one.
Answer A · Chains of reasoning (Prf5) · question RSN-10
5Chains of reasoning · spec Prf5 · Paper 2 style · ★☆☆

In a knockout tournament with \(37\) players, each match eliminates exactly one player and the tournament ends when one player remains. How many matches are played?

  1. A\(36\)
  2. B\(35\)
  3. C\(19\)
  4. D\(37\)
  5. E\(64\)
  6. F\(18\)
Worked solution

Each match eliminates one player and \(36\) must be eliminated, so \(36\) matches - whatever the draw looks like.

Full method — how and why
An invariant argument: count what each match achieves rather than simulating the draw.
  1. At the end one player remains, so \(37 - 1 = 36\) players have been eliminated.
    Count the outcome, not the structure.
  2. Each match eliminates exactly one player.
    Given in the question - this is the invariant.
  3. So the number of matches equals the number of eliminations: \(36\).
    And this holds whatever byes or bracket shape are used, which is why the answer is unique.
Where it goes wrong
  • Trying to build the bracket and getting lost in byes because \(37\) is not a power of \(2\).
  • Halving repeatedly (\(18, 19\)-type answers), which only works for tidy brackets.
Takeaway Count the invariant: a knockout with \(n\) players always takes \(n-1\) matches.
Answer A · Chains of reasoning (Prf5) · question RSN-05
6The logic of arguments · spec Arg1 · Paper 2 style · ★☆☆

The statement 'if a triangle is equilateral then it is isosceles' is true. Which of the following must therefore also be true?

  1. IIf a triangle is isosceles then it is equilateral.
  2. IIIf a triangle is not isosceles then it is not equilateral.
  3. IIIIf a triangle is not equilateral then it is not 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

Only the contrapositive (II) is logically equivalent to the original. I is the converse and III is the inverse; both are in fact false here.

Full method — how and why
Given a true implication, only one of the three relatives is guaranteed. Identify which of I, II, III is the contrapositive.
  1. I is 'isosceles \(\Rightarrow\) equilateral' - the converse. Not guaranteed, and here false (a \(3,3,5\) triangle).
    Swapped but not negated.
  2. II is 'not isosceles \(\Rightarrow\) not equilateral' - the contrapositive. Always guaranteed. True.
    Swapped and negated; logically equivalent to the original.
  3. III is 'not equilateral \(\Rightarrow\) not isosceles' - the inverse. Not guaranteed, and here false (the same \(3,3,5\) triangle).
    Negated but not swapped - and the inverse is just the contrapositive of the converse, so it carries the converse's truth value, not the original's.
  4. So II only.
    Match to the list.
Where it goes wrong
  • Accepting III because both its parts are negated - negation alone is not enough; the parts must also swap.
  • Accepting I because equilateral triangles are also isosceles - that supports the original, not the converse.
Takeaway From \(A \Rightarrow B\) you get exactly one free statement: \(\neg B \Rightarrow \neg A\).
Answer C · The logic of arguments (Arg1) · question LGC-11
7The logic of arguments · spec Arg4 · Paper 2 style · ★☆☆

Which of the following is the negation of '\(x > 3\) and \(y \leq 5\)'?

  1. A\(x < 3\) and \(y > 5\)
  2. B\(x < 3\) or \(y \geq 5\)
  3. C\(x \leq 3\) or \(y \geq 5\)
  4. D\(x \leq 3\) and \(y > 5\)
  5. E\(x \leq 3\) or \(y > 5\)
Worked solution

Not (\(A\) and \(B\)) is (not \(A\)) or (not \(B\)). Not \(x > 3\) is \(x \leq 3\); not \(y \leq 5\) is \(y > 5\).

Full method — how and why
De Morgan's law: negating an 'and' turns it into an 'or', with both parts negated.
  1. \(\neg(A \text{ and } B) \equiv (\neg A) \text{ or } (\neg B)\).
    For the conjunction to fail, it is enough that one part fails.
  2. Negate the first part: \(\neg(x > 3)\) is \(x \leq 3\).
    The boundary \(x = 3\) satisfies the negation, since \(3 > 3\) is false.
  3. Negate the second: \(\neg(y \leq 5)\) is \(y > 5\).
    Strict inequality, because \(y = 5\) satisfied the original.
  4. Combine with 'or': \(x \leq 3\) or \(y > 5\).
    The joining word flips too - this is the half most often forgotten.
Where it goes wrong
  • Keeping 'and': '\(x \leq 3\) and \(y > 5\)' is far too strong - it demands both failures.
  • Getting the boundaries wrong: writing \(x < 3\) or \(y \geq 5\), which mishandles \(x = 3\) and \(y = 5\).
Takeaway De Morgan: not(A and B) \(=\) notA or notB; not(A or B) \(=\) notA and notB. Negate the connective as well as the parts.
Answer E · The logic of arguments (Arg4) · question LGC-13
8Mathematical proof · spec Prf1 · Paper 2 style · ★☆☆

Consider the claim: for every positive integer \(n\), \(n^2 + n\) is even. Which of the following is a complete and valid proof?

  1. A\(n^2 + n\) is even because both \(n^2\) and \(n\) are even.
  2. BIf \(n\) is even then \(n^2\) is even and so \(n^2 + n\) is even.
  3. CSuppose \(n^2 + n\) is odd. Then \(n\) is odd, so \(n^2\) is odd, so \(n^2 + n\) is even, which is a contradiction... unless \(n\) is even.
  4. D\(1^2 + 1 = 2\), \(2^2 + 2 = 6\), \(3^2 + 3 = 12\), which are all even, so the claim holds.
  5. E\(n^2 + n = n(n + 1)\) is the product of two consecutive integers, one of which must be even, so the product is even.
Worked solution

The factorised argument covers every \(n\) at once: \(n(n+1)\) is a product of consecutive integers, so one factor is even. Checking \(n = 1, 2, 3\) is only three cases; 'if \(n\) is even ...' leaves odd \(n\) untreated; 'both \(n^2\) and \(n\) are even' is false for odd \(n\); and the contradiction attempt ends with an unresolved exception.

Full method — how and why
What makes a proof complete: it must cover every case in the claim with a general argument, not a sample of values.
  1. The claim is about every positive integer, so any valid proof must handle all \(n\) at once.
    Checking particular values can only ever disprove a universal claim, never prove one.
  2. Factorise: \(n^2 + n = n(n+1)\) - the product of two consecutive integers.
    Factorising turns a statement about a sum into one about a product, where parity is easy to control.
  3. Of any two consecutive integers, exactly one is even; an even factor makes the product even.
    This single sentence covers both the case \(n\) even and the case \(n\) odd - no case-splitting needed.
  4. That argument is general and complete, so it is the proof.
    Now examine why the others fail, which is the real content of the question.
Where it goes wrong
  • Accepting the check of \(n = 1, 2, 3\): three cases out of infinitely many prove nothing.
  • Accepting 'if \(n\) is even then \(n^2+n\) is even' - true, but it leaves odd \(n\) untreated, so the proof is incomplete.
  • Accepting 'both \(n^2\) and \(n\) are even' - simply false when \(n\) is odd.
  • Accepting the contradiction attempt that ends 'unless \(n\) is even' - a proof cannot end with an unresolved exception.
Takeaway A proof of 'for all \(n\)' must be general. Examples illustrate; only argument proves.
Answer E · Mathematical proof (Prf1) · question PRF-01
9Trigonometric functions and equations · spec MM4.6 · Paper 2 style · ★☆☆

Consider the equation \(\sin x = k\) where \(k\) is a real constant. Which of the following statements is/are true?

  1. IFor every \(k\) with \(-1 < k < 1\) the equation has exactly two solutions in \(0 \leq x < 2\pi\).
  2. IIFor \(k = 1\) the equation has exactly one solution in \(0 \leq x < 2\pi\).
  3. IIIFor every real \(k\) the equation has at least one 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

I and II are read from the graph of \(\sin\) over one period. III is false: \(\sin x = 2\) has no solutions.

Full method — how and why
Read the claims off the graph of \(y = \sin x\) over one period, then remember the range.
  1. I: for \(-1True.
    Sine rises from \(0\) to \(1\), falls to \(-1\) and returns - a strictly-between value is hit once on the way down and once elsewhere in \([0, 2\pi)\).
  2. II: \(k=1\) is reached only at the single peak \(x = \frac{\pi}{2}\). True.
    The maximum is attained once per period.
  3. III: for \(|k| > 1\) there is no solution at all - \(\sin x = 2\) is impossible. False.
    The range of sine is \([-1,1]\).
  4. So I and II.
    Match to the list.
Where it goes wrong
  • Accepting III by forgetting the range restriction.
  • Rejecting II by counting \(x = \frac{\pi}{2}\) and \(x = \frac{5\pi}{2}\) - the second is outside \(0 \leq x < 2\pi\).
  • Rejecting I because \(k=0\) feels special: \(\sin x = 0\) gives \(x = 0\) and \(x=\pi\) in the interval - still exactly two.
Takeaway Per period, \(\sin x = k\) has two solutions for \(|k|<1\), one for \(|k|=1\) and none for \(|k|>1\). Sketch the line across the curve.
Answer E · Trigonometric functions and equations (MM4.6) · question X2-07
10Simultaneous equations and inequalities · spec MM1.5 · Paper 2 style · ★☆☆

\(x\) is a real number. \(P\): '\(x^2 < 4\)'. \(Q\): '\(x < 2\)'. Which one of the following is true?

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

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

Full method — how and why
Solve the quadratic inequality properly, then compare the two solution sets as intervals.
  1. \(x^2<4\) means \(|x|<2\), i.e. \(-2
    A quadratic inequality of this shape gives an interval between the roots.
  2. Every \(x\) in \((-2,2)\) satisfies \(x<2\), so \(P \Rightarrow Q\): sufficient.
    \((-2,2)\) is contained in \((-\infty,2)\).
  3. Is it necessary? Take \(x=-5\): \(x<2\) holds but \(x^2 = 25 \not<4\). So \(Q \not\Rightarrow P\): not necessary.
    The larger set contains points outside the smaller one.
  4. So \(P\) is sufficient but not necessary.
    Whenever one solution set sits strictly inside another, the inner condition is sufficient but not necessary.
Where it goes wrong
  • Solving \(x^2<4\) as \(x<2\) and concluding the conditions are equivalent - that drops the lower bound.
  • Solving it as \(x<-2\) or \(x>2\), which is the solution of \(x^2>4\).
Takeaway Compare conditions as sets: a subset is sufficient for the superset, the superset necessary for the subset.
Answer C · Simultaneous equations and inequalities (MM1.5) · question X2-14
11Angles, polygons, congruence and vectors · spec M5.3 · Paper 2 style · ★☆☆

Which of the following statements is/are true?

  1. IEvery rhombus is a parallelogram.
  2. IIEvery rectangle is a rhombus.
  3. IIIThe diagonals of a kite are perpendicular.
  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 rhombus has two pairs of parallel sides (I). A rectangle need not have equal sides (II false). The diagonals of a kite cross at right angles (III).

Full method — how and why
Quadrilateral classification. Each claim is about whether one family is contained in another.
  1. I: a rhombus has two pairs of parallel sides, which is the definition of a parallelogram. True.
    Rhombus \(=\) parallelogram with all sides equal, so it is a special parallelogram.
  2. II: a rectangle has equal angles but not necessarily equal sides - a \(2 \times 5\) rectangle is not a rhombus. False.
    Only the square lies in both families.
  3. III: a kite's diagonals meet at right angles (the axis of symmetry bisects the other diagonal perpendicularly). True.
    This is a standard kite property.
  4. I and III only.
    Match to the option list.
Where it goes wrong
  • Reversing II: every rhombus is not a rectangle either, but the claim as stated is about rectangles.
  • Thinking a kite's diagonals bisect each other - only one bisects the other.
Takeaway Square \(=\) rectangle \(\cap\) rhombus; kites have perpendicular diagonals but only one is bisected.
Answer F · Angles, polygons, congruence and vectors (M5.3) · question GEO-10
12Identifying errors in proofs · spec Err2 · Paper 2 style · ★☆☆

Claim: if \(x\) and \(y\) are real with \(x^2 = y^2\), then \(x = y\).

Which of the following is a counterexample to the claim?

  1. A\(x = -1\), \(y = -1\)
  2. B\(x = 1\), \(y = 2\)
  3. C\(x = 2\), \(y = -2\)
  4. D\(x = 0\), \(y = 0\)
  5. E\(x = 2\), \(y = 2\)
Worked solution

A counterexample must satisfy the hypothesis (\(x^2 = y^2\)) and fail the conclusion (\(x \neq y\)).

Full method — how and why
A counterexample must satisfy the hypothesis and break the conclusion. Test both halves on each candidate.
  1. Hypothesis: \(x^2 = y^2\). Conclusion to break: need \(x \neq y\).
    Both conditions, or it is not a counterexample.
  2. \(x=2, y=-2\): \(4 = 4\) ✓ hypothesis, and \(2 \neq -2\) ✓ conclusion fails. Counterexample.
    Equal squares with opposite signs is exactly the gap in the claim.
  3. \(x=2,y=2\) and \(x=0,y=0\) and \(x=-1,y=-1\): hypothesis holds but so does the conclusion - these are examples of the claim.
    Confirming instances never refute.
  4. \(x=1,y=2\): \(1 \neq 4\), so the hypothesis fails and the case is irrelevant.
    A case outside the hypothesis says nothing.
Where it goes wrong
  • Picking a pair where the hypothesis fails and thinking the different values refute the claim.
  • Picking a pair where both hold, which supports rather than refutes.
Takeaway \(x^2=y^2 \Rightarrow x = \pm y\). A counterexample satisfies the 'if' and fails the 'then'.
Answer C · Identifying errors in proofs (Err2) · question ERR-14
13Integration · spec MM7.1 · Paper 2 style · ★☆☆

Which of the following statements is/are true?

  1. I\(\displaystyle\int_{-2}^{2} x^3\,dx = 0\)
  2. IIThe area between \(y = x^3\) and the \(x\)-axis from \(x = -2\) to \(x = 2\) is \(0\).
  3. III\(\displaystyle\int_{-2}^{2} x^2\,dx = 2\int_{0}^{2} x^2\,dx\)
  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: odd function over a symmetric interval, so \(0\). II is false: the area is \(2\int_0^2 x^3\,dx = 8\); the integral is \(0\) because the parts cancel. III: even function, true.

Full method — how and why
Three statements distinguishing signed integrals from areas, using odd and even symmetry.
  1. I: \(x^3\) is odd, so the contributions from \([-2,0]\) and \([0,2]\) cancel exactly: the integral is \(0\). True.
    For odd \(f\), \(\int_{-a}^{a} f = 0\) always.
  2. II: the area counts both pieces positively: \(2\int_0^2 x^3 dx = 2 \times 4 = 8 \neq 0\). False.
    Area and integral differ precisely when the curve crosses the axis - this is the distinction the question is testing.
  3. III: \(x^2\) is even, so the two halves are equal: \(\int_{-2}^{2} x^2 dx = 2\int_0^2 x^2 dx\). True.
    For even \(f\), \(\int_{-a}^{a} f = 2\int_0^a f\).
  4. I and III only.
    Match to the option list.
Where it goes wrong
  • Treating 'integral' and 'area' as synonyms and marking II true.
  • Thinking odd symmetry also halves the area.
Takeaway Odd function over a symmetric interval: integral \(0\) but area positive. Even function: integral is twice the half.
Answer F · Integration (MM7.1) · question INT-09
14Indices and surds · spec MM1.1 · Paper 2 style · ★★☆

\(a\) and \(b\) are positive real numbers. Which of the following statements is/are true?

  1. IIf \(a^2 = b^2\) then \(a = b\).
  2. IIIf \(2^a = 2^b\) then \(a = b\).
  3. IIIIf \(a^b = 1\) then \(a = 1\).
  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 holds because \(a, b > 0\) rules out \(a = -b\). II holds because \(2^x\) is one-to-one. III: for positive \(a\) and \(b\), \(a^b = 1\) gives \(b\log a = 0\), so \(\log a = 0\) and \(a = 1\). All three are true.

Full method — how and why
Three cancellation claims, all made safe by the stated positivity. Check what each would need.
  1. I: \(a^2=b^2\) gives \(a = \pm b\) in general - but \(a,b>0\) rules out \(a=-b\), so \(a=b\). True.
    The positivity hypothesis is exactly what repairs the usual counterexample.
  2. II: \(2^x\) is a strictly increasing function, so it takes each value once: \(2^a=2^b \Rightarrow a=b\). True.
    One-to-one functions can be cancelled; equivalently take \(\log_2\) of both sides.
  3. III: \(a^b = 1\) with \(a,b>0\). Take logs: \(b\log a = 0\). Since \(b>0\), \(\log a = 0\), so \(a=1\). True.
    Taking logs turns the exponent into a factor, and a product is zero only if a factor is.
  4. So all three.
    Match to the list.
Where it goes wrong
  • Rejecting I by recalling the counterexample \(2\) and \(-2\) - forbidden here, since both are positive.
  • Rejecting III by thinking of \(a^0=1\) - that needs \(b=0\), which positivity forbids. (And \((-1)^2 = 1\) needs a negative \(a\).)
  • Overlooking how much work the phrase 'positive real numbers' is doing in all three.
Takeaway Read the domain restrictions first - positivity makes squaring and exponentiation reversible, when in general they are not.
Answer H · Indices and surds (MM1.1) · question X2-12
15Mathematical proof · spec Prf3 · Paper 2 style · ★★☆

A student notices that \(1 + 2 = 3\), \(4 + 5 + 6 = 7 + 8\), \(9 + 10 + 11 + 12 = 13 + 14 + 15\) and conjectures that the pattern continues. Which of the following describes the \(n\)th line of the pattern?

  1. AThe \(n\)th line starts at \(n^2\) and has \(n + 1\) numbers on the left and \(n\) on the right.
  2. BThe \(n\)th line starts at \(n(n+1)\) and has \(n + 1\) numbers on the left.
  3. CThe \(n\)th line starts at \(n^2\) and has \(n\) numbers on the left and \(n + 1\) on the right.
  4. DThe \(n\)th line starts at \(n^2\) and has \(n\) numbers on each side.
  5. EThe \(n\)th line starts at \(2n - 1\) and has \(n + 1\) numbers on the left.
Worked solution

Line \(1\) starts at \(1\), line \(2\) at \(4\), line \(3\) at \(9\): start \(n^2\). Left sides have \(2, 3, 4\) numbers (\(n + 1\)), right sides \(1, 2, 3\) (\(n\)).

Full method — how and why
Describing a pattern precisely: check the start value and the two counts separately against every given line.
  1. Starting values: \(1, 4, 9\) for lines \(1, 2, 3\) - these are \(n^2\).
    Test against all three lines, not just one.
  2. Count the left sides: \(\{1,2\}\), \(\{4,5,6\}\), \(\{9,10,11,12\}\) - that is \(2, 3, 4\) terms, i.e. \(n+1\).
    Count carefully; being off by one here is the whole point of the question.
  3. Count the right sides: \(\{3\}\), \(\{7,8\}\), \(\{13,14,15\}\) - that is \(1, 2, 3\) terms, i.e. \(n\).
    Left has one more term than right, which is what makes the equality possible (the right-hand terms are larger).
  4. So: starts at \(n^2\), with \(n+1\) terms on the left and \(n\) on the right.
    Check the total count: \((n+1) + n = 2n+1\), and indeed line \(3\) runs from \(9\) to \(15\), seven numbers ✓.
Where it goes wrong
  • Swapping the counts (\(n\) left, \(n+1\) right) - the sides would not balance, since the right-hand numbers are bigger.
  • Reading the start as \(2n-1\) or \(n(n+1)\) - both fail at line \(2\) (\(3\) and \(6\) rather than \(4\)).
Takeaway Verify a conjectured pattern against every given line, and check each component (start, length) separately.
Answer A · Mathematical proof (Prf3) · question PRF-11
16Mathematical proof · spec Prf1 · Paper 2 style · ★★☆

Which of the following statements can be disproved by a single counterexample?

  1. IEvery prime number greater than \(2\) is odd.
  2. IIEvery integer of the form \(4k + 3\) is prime.
  3. IIIThere exists an even prime 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

Only a 'for all' statement can be disproved by one counterexample. I is true (no counterexample exists). II is false and \(k = 3\) (\(15\)) disproves it. III is an existence statement - a counterexample is meaningless; it is proved by the example \(2\).

Full method — how and why
Only a universal statement that is false can be disproved by a counterexample. Two separate tests.
  1. I: 'every prime \(> 2\) is odd' is universal - but it is true (any even number \(>2\) has factor \(2\)), so no counterexample exists.
    You cannot disprove a true statement.
  2. II: 'every integer \(4k+3\) is prime' is universal and false: \(k = 3\) gives \(15 = 3 \times 5\). Disprovable. Yes.
    One failing case is a counterexample.
  3. III: 'there exists an even prime' is existential, not universal - and it is true (\(2\)). A counterexample is not even the right tool.
    To disprove an existential statement you would have to rule out every case, which is not a 'single counterexample'.
  4. So II only.
    Match to the list.
Where it goes wrong
  • Including I because it is a 'for all' statement - the statement also has to be false.
  • Including III by confusing what disproves an existential claim.
Takeaway Counterexamples disprove false universal statements only. Existential statements are proved by example and disproved only by a general argument.
Answer C · Mathematical proof (Prf1) · question PRF-06
17Chains of reasoning · spec Prf5 · Paper 2 style · ★★☆

A bag contains red and blue balls only. If one red ball is removed, one quarter of the remaining balls are red. If instead two blue balls are removed, one third of the remaining balls are red. How many balls are in the bag?

  1. A\(16\)
  2. B\(20\)
  3. C\(14\)
  4. D\(8\)
  5. E\(17\)
  6. F\(12\)
Worked solution

With \(r\) red and \(b\) blue: \(r - 1 = \frac14(r + b - 1)\) and \(r = \frac13(r + b - 2)\). The second gives \(b = 2r + 2\); substituting into the first, \(4r - 4 = 3r + 1\), so \(r = 5\), \(b = 12\): \(17\) balls.

Full method — how and why
Two conditions, two unknowns: translate each sentence into an equation, then solve simultaneously.
  1. Let \(r\) be red and \(b\) blue. Removing one red: \(r - 1 = \frac14(r + b - 1)\).
    The total drops by \(1\) as well as the red count - both sides of the fraction change.
  2. Removing two blues: \(r = \frac13(r + b - 2)\).
    Here the red count is unchanged; only the total falls, by \(2\).
  3. From the second: \(3r = r + b - 2\), so \(b = 2r + 2\).
    Solve the simpler equation first and substitute.
  4. Substitute into the first: \(4(r-1) = r + (2r+2) - 1 = 3r + 1\), so \(4r - 4 = 3r + 1\) and \(r = 5\).
    Multiply through by \(4\) to clear the fraction before substituting.
  5. Then \(b = 12\) and the total is \(17\).
    Check: remove one red \(\to 4\) red of \(16\) \(= \frac14\) ✓; remove two blue \(\to 5\) red of \(15 = \frac13\) ✓.
Where it goes wrong
  • Forgetting that removing a ball also reduces the total, giving \(r - 1 = \frac14(r+b)\).
  • Answering \(r\) or \(b\) rather than the total.
  • Not checking the answer against both original statements.
Takeaway When items are removed, adjust the total as well as the category; then verify against both conditions.
Answer E · Chains of reasoning (Prf5) · question RSN-19
18Identifying errors in proofs · spec Err2 · Paper 2 style · ★★☆

A student solves the inequality \(\dfrac{x + 1}{x - 2} < 3\).

  1. IMultiply both sides by \((x - 2)\): \(x + 1 < 3x - 6\).
  2. IISo \(7 < 2x\).
  3. IIISo \(x > \dfrac{7}{2}\).

Which of the following is correct?

  1. AThe first error is on line I: multiplying by \(x - 2\) is only valid as written when \(x - 2 > 0\).
  2. BThe solution is completely correct.
  3. CThe first error is on line II: it should be \(7 < 4x\).
  4. DThe first error is on line III: it should be \(x < \frac72\).
  5. EThe first error is on line I: you cannot multiply an inequality by an expression.
Worked solution

For \(x < 2\) the multiplier is negative and the inequality reverses; indeed \(x = 0\) gives \(-\frac12 < 3\), a solution missed by the student. The full solution set is \(x < 2\) or \(x > \frac72\).

Full method — how and why
Multiplying an inequality by an expression whose sign is unknown. The error is the unstated assumption.
  1. Line I multiplies by \((x-2)\) and keeps the direction of the inequality.
    That is only valid if \(x - 2 > 0\); if \(x < 2\) the inequality must reverse.
  2. So line I is the first error: it silently assumes \(x > 2\).
    The line is not wrong for all \(x\), but it is wrong as an unconditional step - which is what makes it an error.
  3. Consequence: the student misses a whole branch. Test \(x = 0\): \(\frac{0+1}{0-2} = -\frac12 < 3\) ✓ - a genuine solution not in \(x > \frac72\).
    A single test value exposes the omission.
  4. The correct solution set is \(x < 2\) or \(x > \frac72\).
    For \(2 < x < \frac72\) the fraction exceeds \(3\); at \(x = 2\) it is undefined.
Where it goes wrong
  • Saying you can never multiply an inequality by an expression - you can, if you case-split on its sign, or multiply by \((x-2)^2\), which is always positive.
  • Blaming line II or III; the arithmetic there is right, given line I.
  • Forgetting that \(x = 2\) must be excluded from any solution set.
Takeaway Never multiply an inequality by an expression of unknown sign: split into cases, or multiply by its square.
Answer A · Identifying errors in proofs (Err2) · question ERR-05
19Sequences and series · spec MM2.1 · Paper 2 style · ★★★

A sequence has \(u_1 = 3\) and \(u_{n+1} = \dfrac{u_n + 5}{2}\). Which of the following statements is/are true?

  1. IThe sequence is increasing.
  2. II\(u_n < 5\) for all \(n\).
  3. III\(u_n = 5 - 2^{2-n}\) for all \(n\).
  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

\(u_2 = 4\), \(u_3 = 4.5\), \(u_4 = 4.75\). If \(u_n < 5\) then \(u_{n+1} = \frac{u_n + 5}{2} < 5\) and \(u_{n+1} - u_n = \frac{5 - u_n}{2} > 0\), so I and II hold. III: \(5 - 2^{2-n}\) gives \(3, 4, 4.5, \ldots\) and satisfies the recurrence: \(\frac{5 - 2^{2-n} + 5}{2} = 5 - 2^{1-n}\). All three are true.

Full method — how and why
A recurrence converging to a limit. Compute terms for intuition, then prove the two general claims and verify the closed form.
  1. Generate terms: \(u_1 = 3\), \(u_2 = \frac{3+5}{2} = 4\), \(u_3 = 4.5\), \(u_4 = 4.75\) - rising towards \(5\).
    The rule averages \(u_n\) with \(5\), so each term sits halfway between the previous one and \(5\).
  2. II: show \(u_n < 5\) always. If \(u_n < 5\) then \(u_{n+1} = \frac{u_n + 5}{2} < \frac{5+5}{2} = 5\); and \(u_1 = 3 < 5\). True.
    This is an induction argument in miniature: true at the start, and preserved by the rule.
  3. I: \(u_{n+1} - u_n = \frac{u_n + 5}{2} - u_n = \frac{5 - u_n}{2} > 0\) because \(u_n < 5\). So the sequence increases. True.
    Compute the difference algebraically rather than relying on the first few terms; it also shows *why* it increases - the gap to \(5\) is always positive.
  4. III: test \(5 - 2^{2-n}\): at \(n=1\) it gives \(5 - 2 = 3\) ✓, and \(\frac{(5 - 2^{2-n}) + 5}{2} = 5 - 2^{1-n}\), which is the formula at \(n+1\) ✓. True.
    Check the starting value and that the formula satisfies the recurrence - both are needed.
  5. All three are true.
    Match to the options.
Where it goes wrong
  • Concluding II from the first four terms alone - the general argument is what makes it certain.
  • Checking III only at \(n = 1\); many formulas share a first term.
Takeaway To prove a claim about every term: check the first case, then show the recurrence preserves the property.
Answer H · Sequences and series (MM2.1) · question SEQ-21
20Integration · spec MM7.1 · Paper 2 style · ★★★

\(f\) is a function defined for all real \(x\) and \(\displaystyle\int_{0}^{4} f(x)\,dx = 0\). Which of the following statements must be true?

  1. I\(f(x) = 0\) for some \(x\) with \(0 \leq x \leq 4\).
  2. IIThe area between the curve and the \(x\)-axis from \(0\) to \(4\) is \(0\).
  3. III\(\displaystyle\int_{0}^{2} f(x)\,dx = -\int_{2}^{4} f(x)\,dx\)
  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

III is just the rule for combining integrals over adjacent ranges. I fails for a function that jumps from \(-1\) to \(1\) without passing through \(0\) (nothing says \(f\) is continuous). II fails: \(f(x) = x - 2\) has integral \(0\) but area \(4\).

Full method — how and why
A hard 'must be true' about integrals. Two claims import assumptions the question never makes.
  1. III: \(\int_0^4 = \int_0^2 + \int_2^4\), and the left side is \(0\), so \(\int_0^2 = -\int_2^4\). True - pure algebra of integrals.
    Splitting an integral at an interior point is always valid where the integral exists.
  2. I: 'zero integral \(\Rightarrow\) \(f\) hits zero' is the intermediate value theorem in disguise - but that needs continuity, which is not given.
    Read the hypotheses: it only says \(f\) is defined for all real \(x\).
  3. Counterexample for I: a step function equal to \(-1\) on \([0,2)\) and \(1\) on \([2,4]\) has integral \(0\) but never equals \(0\). False.
    It jumps straight past zero.
  4. II: 'integral \(0\)' means signed areas cancel, not that the area is \(0\). Take \(f(x) = x-2\): \(\int_0^4 (x-2)dx = 0\), but the area is \(2 + 2 = 4\). False.
    Area counts the region below the axis as positive; the integral counts it as negative.
  5. So III only.
    Match to the list.
Where it goes wrong
  • Accepting I by silently assuming \(f\) is continuous - an assumption the question deliberately withholds.
  • Treating 'integral' and 'area' as the same thing; they agree only when \(f \geq 0\) throughout.
Takeaway Integral \(\neq\) area when the function changes sign; and the intermediate value theorem requires continuity, which must be given.
Answer D · Integration (MM7.1) · question X2-03
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