Mark Scheme

General instructions to examiners. Method marks (M) are awarded for a correct process regardless of arithmetic slips; accuracy marks (A) require the correct value. Error carried forward (ECF) applies throughout: a candidate whose earlier value is wrong but who uses it correctly in a valid method retains all subsequent method marks and any accuracy marks consistent with their own figure. Penalise premature rounding once only across the whole paper. Answers given with no working score a maximum of half the available marks, rounded down.

Section A — Module 1

Question 1

(a) (8) — Full induction structure is required; a verification for several values of n scores 1 only.

• Base case (2): n = 1: LHS = 1 × 3 = 3. RHS = 1 × 2 × 9 / 6 = 3. LHS = RHS, so the statement is true for n = 1. (M1 for substituting into both sides, A1 for showing equality explicitly.)
• Inductive hypothesis (1): Assume true for n = k, i.e. ∑(r=1 to k) r(r + 2) = k(k + 1)(2k + 7)/6.
• Inductive step (4): ∑(r=1 to k+1) r(r+2) = k(k+1)(2k+7)/6 + (k+1)(k+3) (M1 — adds the (k+1)th term to the hypothesis) = (k+1)[k(2k+7) + 6(k+3)] / 6 (M1 — common factor (k+1)/6 extracted) = (k+1)(2k² + 7k + 6k + 18) / 6 = (k+1)(2k² + 13k + 18)/6 (A1) = (k+1)(k+2)(2k+9)/6 (A1 — factorisation of the quadratic) which is the given formula with n replaced by k + 1, since 2(k+1) + 7 = 2k + 9.
• Conclusion (1): Since the result is true for n = 1, and truth for n = k implies truth for n = k + 1, by the principle of mathematical induction it is true for all positive integers n. The conclusion mark requires this full statement; "hence proved" alone scores zero for it.
• Common errors: stating the hypothesis as "true for all k" rather than "assume true for n = k"; adding (k+1)(k+2) instead of (k+1)(k+3) as the next term; failing to factorise 2k² + 13k + 18 and therefore not demonstrating the required form.

(b) (6) — Two acceptable methods. Method 1 (squaring): Both sides are non-negative, so squaring preserves the inequality. (2x − 1)² < (x + 3)² (M1) → 4x² − 4x + 1 < x² + 6x + 9 (A1) → 3x² − 10x − 8 < 0 (A1) → (3x + 2)(x − 4) < 0 (M1) → critical values x = −2/3 and x = 4 (A1) → since the parabola opens upwards, the expression is negative between the roots: −2/3 < x < 4 (A1). Method 2 (cases): award equivalent marks for a complete and correctly justified case analysis on the three intervals x < −3, −3 ≤ x < 1/2, x ≥ 1/2.

• Common errors: squaring only one side; obtaining the correct critical values but writing x < −2/3 or x > 4 (the complement) — this loses only the final A1; using ≤ instead of <.

(c) (i) (2) — f(3) = 2(27) − 5(9) − 4(3) + 3 (M1) = 54 − 45 − 12 + 3 = 0 (A1). Since f(3) = 0, by the factor theorem (x − 3) is a factor. The concluding statement is required for the A1. Long division reaching a zero remainder is equally acceptable.

(c) (ii) (5) — Dividing f(x) by (x − 3), by long division or by comparing coefficients, gives the quadratic factor 2x² + x − 1 (M1 A1). Factorising: 2x² + x − 1 = (2x − 1)(x + 1) (A1). Hence f(x) = (x − 3)(2x − 1)(x + 1) (A1) and the roots are x = 3, x = 1/2, x = −1 (A1).

• ECF: a candidate with an incorrect quadratic quotient who factorises and solves it correctly retains the final three marks.
• Common errors: sign slips in the synthetic division, most often giving 2x² + x + 1 (which does not factorise, and candidates then abandon the part); giving the factorised form but omitting the roots, which the question explicitly requires.

(d) (4) — log₂[x(x − 2)] = 3 (M1 — correct use of the addition law) → x(x − 2) = 2³ = 8 (M1 — correct removal of logs) → x² − 2x − 8 = 0 → (x − 4)(x + 2) = 0 → x = 4 or x = −2 (A1). x = 4 only, since x = −2 makes log₂ x undefined (A1 — the rejection with a reason is required).

• Common errors: writing log₂ x + log₂(x−2) = log₂ x × log₂(x−2); giving both roots and thereby losing the final mark — this is the single most frequently dropped mark in the question.

Question 2

(a) (i) (2) — α + β = 7/2 (B1); αβ = 4/2 = 2 (B1). Accept 3.5 and 2.

(a) (ii) (3) — α² + β² = (α + β)² − 2αβ (M1 — the identity must be seen or clearly used) = (7/2)² − 2(2) (M1) = 49/4 − 4 = 33/4 (A1). Accept 8.25. ECF from (i).

(a) (iii) (4) — Sum of new roots = α² + β² = 33/4 (B1, ECF). Product of new roots = α²β² = (αβ)² = 4 (M1 A1). New equation: x² − (33/4)x + 4 = 0 (M1), and multiplying through by 4 to obtain integer coefficients as required: 4x² − 33x + 16 = 0 (A1), marks capped at 4.

• Common errors: giving the product as α²β² = αβ = 2; leaving fractional coefficients when the question specifies integers; writing x² − (sum) + (product) with the signs reversed.

(b) (i) (3) — S₄ = a(1 − rⁿ)/(1 − r) with a = 27, r = 2/3, n = 4 (M1) = 27(1 − 16/81)/(1/3) (M1) = 27 × (65/81) × 3 = 65 (A1). Direct addition 27 + 18 + 12 + 8 = 65 earns full marks.

(b) (ii) (3) — Since |r| = 2/3 < 1, the series converges and the sum to infinity exists (B1 — the justification is a separate mark and must reference |r| < 1). S∞ = a/(1 − r) = 27/(1 − 2/3) (M1) = 27/(1/3) = 81 (A1).

(c) (6) —

• Base case (1): n = 1: 1³ + 2(1) = 3 = 3 × 1, which is divisible by 3.
• Hypothesis (1): Assume k³ + 2k = 3m for some integer m.
• Step (3): (k+1)³ + 2(k+1) = k³ + 3k² + 3k + 1 + 2k + 2 (M1 — correct expansion) = (k³ + 2k) + 3k² + 3k + 3 (M1 — the hypothesis expression isolated) = 3m + 3(k² + k + 1) = 3(m + k² + k + 1) (A1), which is a multiple of 3 since m + k² + k + 1 is an integer.
• Conclusion (1): True for n = 1 and truth for n = k implies truth for n = k + 1, so true for all positive integers n by induction.
• Common errors: expanding (k+1)³ as k³ + 1; failing to isolate the hypothesis expression and instead asserting the result; omitting the statement that the bracket is an integer, which is what makes the expression a multiple of 3.

(d) (4) — The critical point is that multiplying both sides by (x − 2) is invalid without a sign analysis, since (x − 2) may be negative. (x + 1)/(x − 2) − 3 ≥ 0 (M1) → [(x + 1) − 3(x − 2)]/(x − 2) ≥ 0 → (7 − 2x)/(x − 2) ≥ 0 (A1). Critical values x = 2 (excluded, denominator zero) and x = 3.5 (included) (M1). Testing intervals gives 2 < x ≤ 3.5 (A1).

• Accept the equivalent method of multiplying by (x − 2)², which is always positive.
• Common errors: multiplying straight through by (x − 2) to get x + 1 ≥ 3x − 6 and hence x ≤ 3.5, losing the lower bound entirely — this is the most common single error in the whole of Module 1 and scores at most M1; including x = 2 in the solution set.

Section B — Module 2

Question 3

(a) (6) — Working from the left-hand side: LHS = (cos 3θ sin θ + sin 3θ cos θ)/(cos θ sin θ) (M1 — single fraction) = sin(3θ + θ)/(cos θ sin θ) (M1 — recognition of the compound-angle form sin A cos B + cos A sin B) = sin 4θ/(cos θ sin θ) (A1) = 2 sin 2θ cos 2θ/(cos θ sin θ) (M1 — double angle applied to sin 4θ) = 2(2 sin θ cos θ)cos 2θ/(cos θ sin θ) (M1 — double angle applied to sin 2θ) = 4 cos 2θ (A1). Restriction: θ ≠ nπ/2 for integer n, since cos θ sin θ must be non-zero. Award the final A1 only if this restriction is stated in some correct form.

• Accept expansion of cos 3θ and sin 3θ into single-angle form as an alternative route, with equivalent marks for the algebra.
• Common errors: working on both sides simultaneously and meeting in the middle without justification (accept, but only if the candidate works down from each side independently); expanding sin 4θ as 4 sin θ cos θ.

(b) (7) — Replace sin²θ using the Pythagorean identity to obtain a quadratic in cos θ. 3(1 − cos²θ) + 4 cos θ = 4 (M1) → 3 − 3cos²θ + 4cos θ − 4 = 0 → 3cos²θ − 4cos θ + 1 = 0 (A1 — signs must be correct) → (3cos θ − 1)(cos θ − 1) = 0 (M1) → cos θ = 1/3 or cos θ = 1 (A1). cos θ = 1 → θ = 0 and θ = 6.28 (2π) (A1 — both required). cos θ = 1/3 → θ = 1.23 rad (A1) and θ = 2π − 1.2310 = 5.05 rad (A1).

• Common errors: substituting sin²θ = 1 + cos²θ; solving for cos θ correctly but working in degrees when radians are demanded (deduct the accuracy marks, retain method); giving only the principal value 1.23 and missing the second-quadrant-equivalent solution 5.05; omitting θ = 2π because the endpoint is thought to be excluded — the range is closed and both endpoints count.

(c) (i) (4) — R cos(θ + α) = R cos θ cos α − R sin θ sin α, so R cos α = 5 and R sin α = 12 (M1). R = √(5² + 12²) = √169 = 13 (A1). tan α = 12/5 (M1) → α = 1.18 rad (1.1760), i.e. 67.4° (A1).

• Common errors: R = √(12² − 5²); tan α = 5/12, giving α = 0.395 — always check which of R cos α and R sin α carries which coefficient.

(c) (ii) (3) — Maximum value 13, occurring when cos(θ + α) = 1 (B1); minimum value −13 (B1). The maximum occurs when θ + α = 0 or 2π, so the smallest positive θ is 2π − 1.1760 = 5.11 rad (B1). ECF from (i) throughout.

• Common error: giving θ = −1.18 without adjusting to the smallest positive value.

(d) (5) — a · b = (2)(1) + (3)(−2) + (−1)(2) (M1) = 2 − 6 − 2 = −6 (A1). |a| = √(4 + 9 + 1) = √14; |b| = √(1 + 4 + 4) = √9 = 3 (A1 — both required). cos θ = −6/(3√14) = −0.5345 (M1) → θ = 122.3°. The acute angle between the lines of the vectors is therefore 180° − 122.3° = 57.7° (A1).

• Accept 122.3° with a clear statement that the acute angle is 57.7°. Award the final mark only if the candidate addresses the word "acute"; a bare 122.3° scores 4 of 5.
• Common errors: computing the cross product instead of the dot product; forgetting that |b| = 3 exactly and carrying √9 through as a surd; sign error on the third component.

Question 4

(a) (i) (4) — Completing the square: (x − 3)² − 9 + (y + 2)² − 4 − 12 = 0 (M1 for both completions) → (x − 3)² + (y + 2)² = 25 (A1). Centre (3, −2) (A1); radius 5 (A1).

• Accept direct use of the standard form x² + y² + 2gx + 2fy + c = 0 with centre (−g, −f) and radius √(g² + f² − c) = √(9 + 4 + 12) = 5.
• Common error: centre given as (−3, 2) — a sign inversion applied twice.

(a) (ii) (2) — Substituting: (7 − 3)² + (1 + 2)² (M1) = 16 + 9 = 25 = r², so P lies on C (A1). Substitution into the original expanded equation to obtain 0 is equally acceptable. The concluding statement is required.

(a) (iii) (5) — Gradient of the radius from centre (3, −2) to P(7, 1) = (1 − (−2))/(7 − 3) = 3/4 (M1 A1). The tangent is perpendicular to the radius at the point of contact, so its gradient is −4/3 (M1 — the perpendicularity statement is where the mark lies). Equation: y − 1 = −4/3 (x − 7) (M1) → 3y − 3 = −4x + 28 → 4x + 3y = 31 (A1).

• ECF from an incorrect centre in (i) is allowed in full.
• Common errors: using the radius gradient as the tangent gradient; taking the negative rather than the negative reciprocal; leaving the answer in point-gradient form when the question specifies ax + by = c.

(b) (4) — From y = 6t, t = y/6 (M1). Substituting into x = 3t²: x = 3(y/6)² = 3y²/36 = y²/12 (M1 A1) → y² = 12x. This is a parabola, with vertex at the origin and axis along the positive x-axis (A1).

• Common errors: eliminating using x rather than y and generating a ± square root unnecessarily; naming the conic an ellipse or hyperbola.

(c) (6) — Let M be the midpoint of the diagonal OB and N the midpoint of the diagonal AC; the proof is complete when M and N are shown to be the same point. OB = OA + AB = a + c, since AB is equal and parallel to OC (M1 A1 — the justification for AB = c is required). Position vector of the midpoint M of OB = ½(a + c) (M1). Position vector of A is a; of C is c. Midpoint N of AC has position vector ½(a + c) (M1 A1). Since M and N have the same position vector, they are the same point; the midpoint of each diagonal is therefore a point lying on both diagonals, so the diagonals bisect each other (A1 — the concluding argument, not merely the observation that the two expressions are equal).

• Common errors: asserting AB = c without justification; computing AC as a + c rather than c − a; stopping at "they are equal" without stating what that proves.

(d) (4) — Expanding the compound angle: sin x cos 30° + cos x sin 30° = 2 cos x (M1) → (√3/2) sin x + (1/2) cos x = 2 cos x → (√3/2) sin x = (3/2) cos x (A1) → tan x = 3/√3 = √3 (M1) → x = 60° and x = 240° (A1 — both required).

• Common errors: dividing by cos x without noting it is non-zero (condone); giving only 60°; expanding sin(x + 30°) as sin x + sin 30°.

Section C — Module 3

Question 5

(a) (4) — Direct substitution gives 0/0, so the factor (x − 2) must be removed (B1 for identifying the indeterminate form or for beginning the factorisation). (x³ − 8)/(x² − 4) = (x − 2)(x² + 2x + 4)/[(x − 2)(x + 2)] (M1 for the difference of cubes, M1 for the difference of squares) = (x² + 2x + 4)/(x + 2) → as x → 2, → (4 + 4 + 4)/4 = 12/4 = 3 (A1).

• Common errors: factorising x³ − 8 as (x − 2)(x² − 2x + 4); substituting x = 2 immediately and writing "0/0 = 0" or "undefined".

(b) (6) — The limit definition must be used explicitly; differentiating by rule scores zero for this part regardless of the answer. f'(x) = lim(h → 0) [f(x + h) − f(x)]/h (M1 — definition stated) f(x + h) = 2(x + h)² + 3(x + h) = 2x² + 4xh + 2h² + 3x + 3h (M1 A1 — correct expansion) f(x + h) − f(x) = 4xh + 2h² + 3h (A1) [f(x + h) − f(x)]/h = 4x + 2h + 3 (M1 — division by h before taking the limit) As h → 0, f'(x) = 4x + 3 (A1).

• Common errors: writing the difference quotient but then setting h = 0 before cancelling, producing 0/0; expanding (x + h)² as x² + h²; omitting the limit notation throughout, which loses the first M1.

(c) (6) — Product rule with u = (2x + 1)³ and v = (x − 4): du/dx = 3(2x + 1)² × 2 = 6(2x + 1)² (M1 for the chain rule, A1 for the factor of 2) dy/dx = 6(2x + 1)²(x − 4) + (2x + 1)³(1) (M1 for correct product rule structure, A1) = (2x + 1)²[6(x − 4) + (2x + 1)] (M1 — common factor extracted) = (2x + 1)²(6x − 24 + 2x + 1) = (2x + 1)²(8x − 23) (A1).

• The final A1 requires the fully factorised form the question demands; an unfactorised but correct expansion scores 5 of 6.
• Common errors: omitting the inner derivative, giving du/dx = 3(2x + 1)²; expanding the whole expression before differentiating (valid, and award full marks if the algebra is correct, but it is far more error-prone).

(d) (5) — Differentiating implicitly term by term with respect to x, using the product rule on xy: 2x + [y + x(dy/dx)] + 2y(dy/dx) = 0 (M1 for the product rule on xy, A1 for the 2y dy/dx term) (dy/dx)(x + 2y) = −(2x + y) (M1 — terms collected) dy/dx = −(2x + y)/(x + 2y) (A1). At (1, 2): dy/dx = −(2 + 2)/(1 + 4) = −4/5 (A1).

• Common errors: differentiating xy as x(dy/dx) only, or as y only — the product rule is where most of the marks are lost; differentiating y² as 2y; substituting the point before differentiating.

(e) (4) — Quotient rule with u = 3x − 2, v = x² + 1: dy/dx = [3(x² + 1) − (3x − 2)(2x)]/(x² + 1)² (M1 for the structure, A1 for both numerator products) = (3x² + 3 − 6x² + 4x)/(x² + 1)² (A1) = (−3x² + 4x + 3)/(x² + 1)² (A1).

• Common errors: reversing the numerator to v'u − u'v, producing the negative of the correct answer; sign slip on −(3x − 2)(2x), giving −6x² − 4x.

Question 6

(a) (i) (5) — dy/dx = 3x² − 6x − 9 (M1 A1). Setting dy/dx = 0: 3(x² − 2x − 3) = 0 → 3(x − 3)(x + 1) = 0 (M1) → x = 3 or x = −1 (A1). When x = 3: y = 27 − 27 − 27 + 5 = −22. When x = −1: y = −1 − 3 + 9 + 5 = 10. Stationary points (3, −22) and (−1, 10) (A1 — both y-values required).

• Common errors: arithmetic slips in the y-values, most often (3, −22) computed as (3, 22); differentiating the constant to 5.

(a) (ii) (4) — d²y/dx² = 6x − 6 (M1 A1). At x = 3: d²y/dx² = 12 > 0, so (3, −22) is a minimum (A1). At x = −1: d²y/dx² = −12 < 0, so (−1, 10) is a maximum (A1).

• A correctly executed first-derivative sign test earns the same marks. The conclusion must name the nature; a sign alone is insufficient.

(a) (iii) (4) — Marks are for the sketch drawn by the candidate; no graph is supplied. Award:

• (1) Correct overall cubic shape with positive leading coefficient — falling from the lower left, rising to a local maximum, falling to a local minimum, then rising to the upper right — with both axes drawn and labelled x and y.
• (1) Local maximum correctly placed in the second quadrant region at approximately (−1, 10) and labelled with its coordinates.
• (1) Local minimum correctly placed to the right of the y-axis and below it at approximately (3, −22) and labelled with its coordinates.
• (1) y-intercept marked at (0, 5), and the behaviour as x → ±∞ indicated, either by arrows on the curve or by an annotation that y → +∞ as x → +∞ and y → −∞ as x → −∞.
• ECF from (i) and (ii) applies in full: a candidate whose stationary points are wrong but whose sketch is consistent with their own values and correctly labelled scores all four marks.
• Common errors: a sketch with the maximum to the right of the minimum, which contradicts the candidate's own working; unlabelled axes; turning points drawn but not annotated with coordinates — the coordinates are explicitly required and their absence costs two of the four marks; drawing the curve as a parabola.

(b) (4) — ∫(3x² − 4x + 1) dx = x³ − 2x² + x (M1 for raising each power correctly, A1 for all three terms). Evaluating between the limits: [x³ − 2x² + x] from 1 to 3 = (27 − 18 + 3) − (1 − 2 + 1) (M1 — upper minus lower, substitution shown) = 12 − 0 = 12 (A1).

• Common errors: including a constant of integration in a definite integral (condone, no penalty, but note it); reversing the limits to give −12; arithmetic slip at the lower limit, where 1 − 2 + 1 = 0 is frequently written as 2.

(c) (i) (3) — Setting x² = x + 2 (M1) → x² − x − 2 = 0 → (x − 2)(x + 1) = 0 (M1) → x = 2, x = −1, giving the points (2, 4) and (−1, 1) (A1 — both coordinate pairs required).

(c) (ii) (5) — Marks are for the candidate's own sketch and for the integration.

• (1) Sketch showing an upward parabola with vertex at the origin and a straight line of positive gradient crossing it at two points, both axes labelled, with the two intersection points marked at (−1, 1) and (2, 4).
• (1) The region between the line and the curve — that is, the region below the line and above the parabola — clearly shaded, and the line correctly drawn above the curve on the interval −1 < x < 2. A candidate who shades the region above the line or under the parabola loses this mark and will almost always then set up the integral with the wrong subtraction.
• (1) Correct integral formed with the upper curve minus the lower curve: Area = ∫(from −1 to 2) [(x + 2) − x²] dx (M1).
• (1) Integration: [x²/2 + 2x − x³/3] (A1).
• (1) Evaluation: at x = 2, (2 + 4 − 8/3) = 10/3; at x = −1, (1/2 − 2 + 1/3) = −7/6. Area = 10/3 − (−7/6) = 20/6 + 7/6 = 27/6 = 4.5 square units (A1).
• Common errors: integrating x² − (x + 2) and giving the answer as −4.5, or as 4.5 without acknowledging the sign — award the marks if the candidate states that area is taken as positive; using limits 0 to 2; forgetting the units.

Level descriptors (extended and multi-step parts)

• Level 4 (top band): Method fully and correctly stated at every stage, with the reasoning visible rather than implied; correct notation throughout, including limit notation, the ∫ symbol with dx, and vector notation; results stated to the required accuracy and with units or set notation where appropriate; proofs carrying the formal conclusion; sketches labelled with axes, intercepts and turning-point coordinates; restrictions and rejected solutions identified with reasons.
• Level 3: Correct method throughout with isolated arithmetic or sign slips; notation mostly sound but occasionally loose; conclusions to proofs abbreviated; sketches correct in shape but incompletely annotated; rejected solutions omitted.
• Level 2: The correct approach is identified but not carried through, or is applied with errors that change the structure of the problem (for example multiplying an inequality by a variable expression without a sign analysis); working fragmentary; answers stated without support.
• Level 1: Isolated correct steps with no coherent method; formulae quoted but not applied.

Sample Answers with Examiner Commentary

Question 2(d) — Sample Answers

Grade I (Distinction) answer

"(x + 1)/(x − 2) ≥ 3.

I cannot simply multiply both sides by (x − 2), because (x − 2) is negative when x < 2 and multiplying an inequality by a negative quantity reverses it. I will instead collect everything on one side.

(x + 1)/(x − 2) − 3 ≥ 0 [(x + 1) − 3(x − 2)]/(x − 2) ≥ 0 (x + 1 − 3x + 6)/(x − 2) ≥ 0 (7 − 2x)/(x − 2) ≥ 0

The critical values are x = 3.5, where the numerator vanishes, and x = 2, where the denominator vanishes. Note that x = 3.5 is included in the solution set because the inequality is not strict, but x = 2 must be excluded because the expression is undefined there.

Testing a value in each interval: x = 0: (7)/(−2) = −3.5, negative — not in the solution set. x = 3: (1)/(1) = 1, positive — in the solution set. x = 4: (−1)/(2) = −0.5, negative — not in the solution set.

Therefore 2 < x ≤ 3.5."

Mark: 4/4. Examiner commentary: A model response. The candidate opens by stating precisely why the obvious first move is illegitimate, which is the entire point of the question — this part exists to discriminate between candidates who understand inequalities and candidates who manipulate them as though they were equations. The algebra is clean and every line is shown. The treatment of the two critical values is exactly right: the candidate justifies the inclusion of 3.5 and the exclusion of 2 separately and for different reasons, rather than applying one rule to both. The interval test is systematic and the final answer uses correct notation.

Grade III (Pass) answer

"(x + 1)/(x − 2) ≥ 3 x + 1 ≥ 3(x − 2) x + 1 ≥ 3x − 6 7 ≥ 2x x ≤ 3.5"

Mark: 1/4. Examiner commentary: This is the error the question is designed to catch, and it is the most common single mistake in Module 1. The candidate has multiplied both sides by (x − 2) without considering its sign. When x < 2 the multiplier is negative and the inequality reverses, so the whole interval x < 2 has been handled incorrectly — and it is precisely that interval which the correct answer excludes. The consequence is an answer that includes every negative number and zero, none of which satisfy the original inequality; substituting x = 0 gives −0.5, which is not greater than or equal to 3, and a candidate who spent ten seconds checking a single value would have caught this. One method mark is available for the correct algebraic manipulation that follows from the invalid first step, but nothing beyond it. Either collect on one side and analyse the sign of the resulting rational expression, or multiply by (x − 2)², which is always positive.

Question 6(c)(ii) — Sample Answers

Grade I (Distinction) answer (extract)

"My sketch shows y = x², a parabola with its vertex at the origin, and the line y = x + 2, which crosses the y-axis at 2 and has gradient 1. From part (i) the two graphs meet at (−1, 1) and (2, 4), and I have marked both. Between those points the line lies above the parabola — I can verify this at x = 0, where the line is at y = 2 and the curve at y = 0 — so the enclosed region I have shaded is bounded above by y = x + 2 and below by y = x².

Area = ∫(from −1 to 2) [(x + 2) − x²] dx = [x²/2 + 2x − x³/3] from −1 to 2 At x = 2: 4/2 + 4 − 8/3 = 2 + 4 − 8/3 = 6 − 8/3 = 10/3 At x = −1: 1/2 − 2 + 1/3 = 3/6 − 12/6 + 2/6 = −7/6 Area = 10/3 − (−7/6) = 20/6 + 7/6 = 27/6 = 4.5 square units"

Mark: 5/5. Examiner commentary: Everything the mark scheme asks for is present. The sketch is properly annotated and the intersection points are carried forward from part (i). The single most valuable sentence is the verification at x = 0 that the line lies above the curve: this is what determines the order of subtraction in the integrand, and candidates who skip it routinely integrate x² − (x + 2) and arrive at −4.5. The fractional arithmetic at the lower limit is set out over a common denominator rather than done mentally, which is exactly the right instinct where a sign error would be invisible. Units are given.

Grade II (Merit) answer (extract)

"Area = ∫(from −1 to 2) [(x + 2) − x²] dx = [x²/2 + 2x − x³/3] = (2 + 4 − 2.67) − (0.5 − 2 + 0.33) = 3.33 − (−1.17) = 4.5"

Mark: 3/5. Examiner commentary: The integration is correct, the limits are correct, the subtraction is the right way round and the final answer is right, so all three of the integration marks are earned. The two marks lost are both for the sketch, which is simply absent — and the question explicitly instructs the candidate to sketch the curve and the line and to shade the enclosed region. This is a recurring and entirely avoidable loss: candidates treat a sketch as scaffolding for their own thinking rather than as an examined component, and discard it once they have the integral. Here it carries 40% of the tariff. Note also the premature rounding to two decimal places mid-calculation; it happens not to affect the answer, but with less forgiving numbers it would, and working in exact fractions costs nothing.

Grade IV (Weak pass) answer (extract)

"Area = ∫ x² dx − ∫ (x + 2) dx from −1 to 2 = [x³/3] − [x²/2 + 2x] = (8/3 − (−1/3)) − ((2 + 4) − (0.5 − 2)) = 3 − 7.5 = −4.5"

Mark: 1/5. Examiner commentary: There is no sketch, and the absence of one is almost certainly the cause of the substantive error: the candidate has subtracted the line from the curve rather than the curve from the line. Had the region been drawn, it would have been immediately visible that the line lies above the parabola throughout the interval. The negative answer should itself have prompted a check, since an area cannot be negative — the sign is a free diagnostic and the candidate has ignored it. One mark is retained for correct integration of the individual terms. The lesson is procedural rather than mathematical: draw the region first, identify which function is on top, and only then set up the integral.

How Unit 1 is assessed

Paper 02 carries 150 marks and contributes 50% of the Unit 1 external assessment mark, alongside Paper 01 (multiple choice, 30%) and the Internal Assessment (20%). All six questions are compulsory, so no Module can be avoided — a candidate who has neglected vectors or the circle has no route past Question 4, and one who cannot differentiate from first principles will lose six marks in Question 5 that no amount of skill with the standard rules will recover.

Four habits account for most of the marks that strong candidates nonetheless drop. First, write the method down. Method marks are awarded for the process, not the answer, and are protected by error carried forward — a candidate who makes an arithmetic slip early but shows correct method throughout can still reach the top band, while a candidate who writes only answers cannot score more than half marks even when every answer is right. Second, finish proofs properly. Both induction proofs in Section A reserve a mark for the concluding statement, and "hence proved" does not earn it. Third, draw the sketch you are asked for, with axes labelled and turning points annotated with their coordinates; the sketches in Question 6 carry eight marks between them and are the most commonly abandoned part of the paper. Fourth, check the domain: reject logarithmic solutions that make an argument non-positive, exclude values that make a denominator zero, state the restriction on an identity, and confirm that an inequality has not been multiplied by an expression of unknown sign. These four checks recover marks in six separate parts of this paper.