Technique Masterclasses

Pure Mathematics 1

Flashcard Bank

Provenance (2026-09-29). This is a reformatting document, not new research. Every card below is a front/back flashcard rebuild of content already verified and written up in this folder's other codex-*.md files for WMA11 (Pearson Edexcel IAL, Unit 1: Pure Mathematics P1), read in full for this pass: codex-formula-and-method-bank.md (Part 1's formula/identity masterlist and Part 2's worked-method reference), codex-complete-worked-paper.md (the full January 2023 worked paper), and the seventeen topic deep-dives — codex-deep-dive-cubic-graph-sketching-from-factored-form.md, codex-deep-dive-differentiating-negative-and-fractional-indices.md, codex-deep-dive-discriminant-with-a-parameter.md, codex-deep-dive-graph-transformations-in-words.md, codex-deep-dive-hidden-higher-degree-equations-from-differentiation.md, codex-deep-dive-hidden-quadratic-substitution.md, codex-deep-dive-integrate-twice-two-constants.md, codex-deep-dive-perpendicularity-proofs-and-coordinate-geometry.md, codex-deep-dive-quadratic-functions.md, codex-deep-dive-quadratic-inequalities-graphically.md, codex-deep-dive-sector-area-and-arc-length.md, codex-deep-dive-simultaneous-equations-by-substitution.md, codex-deep-dive-sine-and-cosine-rules-ambiguous-case.md, codex-deep-dive-splitting-fractions-before-integrating.md, codex-deep-dive-surds-manipulation-and-exact-answers.md, codex-deep-dive-the-derivative-as-a-limit.md, and codex-deep-dive-trig-graphs-period-and-domain.md. No new fact, formula, spec claim, series citation or worked number is introduced here — every card's Back either quotes/paraphrases a formula, definition or condition already stated in one of those files, or re-presents a worked micro-example (real exam numbers or a labelled [VERIDIAN, illustrative] construction) that one of those files already carried through and checked. Tags carried forward unchanged from the source files: [MEMORISE] = on the spec's own "will not appear in the booklet" list; [BOOKLET] = printed in the real exam formula booklet's P1 section; [ASSUMED] = genuine spec content treated as prior knowledge, given in neither place; [real, Series Qn] = a verified real Pearson exam question, worked on its own actual numbers; [VERIDIAN, illustrative] = a fresh, short, invented example built to demonstrate a mechanism, with no claim to resemble a specific real Pearson question — the same disclosure convention the source files use throughout. Where a source file states a genuine, honest evidentiary gap (no real per-question anchor found for some sub-case), that gap is carried forward, not silently filled.

WMA11 CODEX — Flashcard Bank (Definitions, Formulas & Worked Micro-Examples)

Maths Pure 1 (Unit 1). One flashcard per term or formula, front then back, organised by the spec's own five numbered content strands (P1.3: Algebra and functions; Coordinate geometry; Trigonometry; Differentiation; Integration) — the same organisation codex-formula-and-method-bank.md uses. A "worked micro-example" card gives a formula applied to real numbers (a genuine past-paper question, cited by series and question number) or, where no real anchor exists for that exact step, to a short [VERIDIAN, illustrative] construction already built and checked in the source deep-dive — never to a number invented fresh for this deck.

How to use this deck. Read the Front, answer it from memory, then check the Back. Every Back carries its own source citation so a card can be checked against the file it was rebuilt from at any time. A card tagged [ASSUMED] is genuine, examined P1 content that must simply be recalled as prior knowledge — it is not in the exam formula booklet and not on the spec's short memorise-list, but it is still fair game on the paper.

Part 1 — Algebra and Functions (spec 1.1–1.12) — 50 cards

1. Law of indices — multiplication Front: State the law of indices for aᵘ × aᵛ. Back: aᵘ × aᵛ = aᵘ⁺ᵛ. [ASSUMED — codex-formula-and-method-bank.md §1.1.1]

2. Law of indices — division Front: State the law of indices for aᵘ ÷ aᵛ. Back: aᵘ ÷ aᵛ = aᵘ⁻ᵛ. [ASSUMED — §1.1.1]

3. Law of indices — power of a power Front: State the law of indices for (aᵘ)ᵛ. Back: (aᵘ)ᵛ = aᵘᵛ. [ASSUMED — §1.1.1]

4. Zero and negative indices Front: State the value of a⁰, and rewrite a⁻ⁿ without a negative index. Back: a⁰ = 1; a⁻ⁿ = 1/aⁿ. [ASSUMED — §1.1.1]

5. Fractional-index equivalence Front: State the equivalence spec 1.1 requires between a^(m/n) and a root. Back: a^(m/n) ≡ (ⁿ√a)ᵘ ≡ ⁿ√(aᵘ) — i.e. a^(m/n) is equally "the nth root of a, raised to the mth power" or "a raised to the mth power, then the nth root taken." Spec's own wording: "the equivalence of a^(m/n) and (ⁿ√a)ᵘ should be known." [ASSUMED, but the equivalence itself is spec-named — §1.1.1]

6. Domain condition on a^(m/n) Front: What condition on a is required for a^(m/n) to be real-valued when n is even? Back: a ≥ 0 is required — the spec does not spell this out explicitly, but it is a genuine mathematical condition attached to the m/n-index equivalence. [§1.1.1]

7. Surd multiplication law Front: State the surd law for √a × √b. Back: √a × √b = √(ab), for a, b ≥ 0. [ASSUMED — §1.1.2]

8. Surd division law Front: State the surd law for √a ÷ √b. Back: √a ÷ √b = √(a/b), for a ≥ 0, b > 0. [ASSUMED — §1.1.2]

9. Rationalising a monomial surd denominator Front: Rationalise 1/√a. Back: Multiply top and bottom by √a: 1/√a × √a/√a = √a/a. [ASSUMED — §1.1.2]

10. Rationalising a binomial surd denominator Front: Rationalise 1/(a + √b). Back: Multiply top and bottom by the conjugate (a − √b): (a−√b)/((a+√b)(a−√b)) = (a−√b)/(a² − b), using the difference-of-squares identity (a+√b)(a−√b) = a² − b to clear the surd from the denominator. [ASSUMED — §1.1.2]

11. Exact-answer convention Front: What does the general marking convention say happens when a candidate resorts to a rounded decimal where exact/surd working is clearly required? Back: "Marks will normally be lost if the candidate resorts to using rounded decimals." [verbatim general marking principle — §1.1.2 / codex-deep-dive-surds-manipulation-and-exact-answers.md Part 7]

12. Simplifying a surd — worked example Front: Simplify √200. Back: The largest perfect-square factor of 200 is 100 (200 = 100 × 2), so √200 = √100 × √2 = 10√2. Check: 10√2 ≈ 14.142 ≈ √200. [VERIDIAN, illustrative — codex-deep-dive-surds-manipulation-and-exact-answers.md Part 2]

13. Combining surds — worked example Front: Simplify √48 − √27. Back: √48 = √(16×3) = 4√3; √27 = √(9×3) = 3√3. So √48 − √27 = 4√3 − 3√3 = √3. [VERIDIAN, illustrative — codex-deep-dive-surds-manipulation-and-exact-answers.md Part 3]

14. Multiplying surd binomials — worked example Front: Expand (2 + √3)(5 − √3). Back: 2×5 + 2×(−√3) + √3×5 + √3×(−√3) = 10 − 2√3 + 5√3 − 3 = 7 + 3√3. [VERIDIAN, illustrative — codex-deep-dive-surds-manipulation-and-exact-answers.md Part 4]

15. Squaring a binomial surd — the cross-term trap Front: Expand (3 + √2)², and name the error of writing 9 + 2 = 11 instead. Back: (3+√2)² = 9 + 3√2 + 3√2 + 2 = 11 + 6√2. Writing 9+2=11 (or 9+(√2)²=11) drops the 6√2 cross term that (a+b)² = a² + 2ab + b² requires. [VERIDIAN, illustrative — codex-deep-dive-surds-manipulation-and-exact-answers.md Part 4]

16. General quadratic form Front: Write the general form of a quadratic function, and state the one condition on its leading coefficient. Back: f(x) = ax² + bx + c, with a ≠ 0. [ASSUMED — codex-formula-and-method-bank.md §1.1.3]

17. The discriminant Front: Define the discriminant of a quadratic ax² + bx + c. Back: Δ = b² − 4ac. [ASSUMED, spec 1.4 — §1.1.3]

18. The three discriminant cases Front: State what Δ > 0, Δ = 0 and Δ < 0 each tell you about a quadratic's roots. Back: Δ > 0 ⟺ two distinct real roots; Δ = 0 ⟺ one repeated real root (the curve touches the x-axis); Δ < 0 ⟺ no real roots. [spec 1.4 — §1.1.3]

19. The quadratic formula Front: State the quadratic formula for solving ax² + bx + c = 0 (a ≠ 0). Back: x = (−b ± √(b² − 4ac)) / 2a. [MEMORISE — spec P1.2's own "will not appear in the booklet" box, §1.1.3]

20. Completing the square Front: State the completed-square identity for ax² + bx + c. Back: ax² + bx + c = a(x + b/2a)² + (c − b²/4a). [ASSUMED as a technique — §1.1.3]

21. Turning point from vertex form Front: Given f(x) = a(x − h)² + k, state the coordinates and type of the turning point. Back: Turning point (h, k) — a minimum if a > 0, a maximum if a < 0; the line of symmetry is x = h. [codex-deep-dive-quadratic-functions.md §1]

22. Three-route method-mark convention Front: Name the three routes that each earn the method mark for solving a three-term quadratic. Back: Factorisation (with |pq| = |c|, |mn| = |a|), the formula (attempted with values substituted), or completing the square (in the exact form (x ± b/2)² ± q ± c = 0, q ≠ 0). A correct answer with no visible route from this list is treated as an answer, not evidence of method. [verbatim mark-scheme convention — codex-formula-and-method-bank.md §1.1.3 / §2.1.1]

23. The factorisation condition Front: For (mx + p)(nx + q) to be a valid factorisation of ax² + bx + c, what must m, n, p, q satisfy? Back: mn = a and pq = c (given by the mark scheme as |mn| = |a|, |pq| = |c| — magnitude-matched, since the method mark credits finding a factor pair of the right size before the signs are sorted). [codex-deep-dive-quadratic-functions.md §2]

24. Discriminant-factorisation theorem Front: State the connection between a quadratic's discriminant being a perfect square and whether it factorises with rational coefficients. Back: b² − 4ac is a perfect square ⟺ both roots are rational ⟺ a factorisation a(x−α)(x−β) with rational α, β exists. A non-square discriminant makes a rational factorisation provably impossible, not merely hard to find. [codex-deep-dive-quadratic-functions.md §2]

25. Parameter-in-leading-coefficient exclusion Front: When a discriminant condition is solved for a parameter that sits in the quadratic's leading coefficient (e.g. kx² + 6x + 2 = 0), what value must be separately excluded and why? Back: The value making the leading coefficient zero must be excluded, because at that value the equation is linear, not quadratic, and the discriminant test does not apply to it at all. [codex-formula-and-method-bank.md §1.1.3]

26. Worked example — discriminant with a parameter Front: kx² + 6kx + 5 = 0 (k a non-zero constant) has no real roots. Find the range of k. Back: No real roots ⟺ Δ < 0: (6k)² − 4(k)(5) < 0 ⟹ 36k² − 20k < 0 ⟹ 4k(9k−5) < 0. Critical values k = 0 and k = 5/9; negative between them (upward parabola in k). Range: 0 < k < 5/9. [real, Jan 2023 Q4 — codex-complete-worked-paper.md / codex-deep-dive-discriminant-with-a-parameter.md §3.1]

27. Simultaneous equations — the four-stage method Front: State the four stages of solving simultaneous equations by substitution. Back: (1) Isolate one variable in whichever equation makes that easiest; (2) substitute that expression into the other equation; (3) solve the resulting single-variable equation for all valid values; (4) substitute each value back into an original equation to recover the other variable. [codex-deep-dive-simultaneous-equations-by-substitution.md §2]

28. Line meets curve — discriminant of the resultant Front: A line is substituted into a quadratic curve's equation. What does the resultant equation's discriminant tell you about the number of intersections? Back: Δ > 0 gives two distinct real x-values (two crossing points); Δ = 0 gives one repeated value (the line touches the curve, a tangency); Δ < 0 gives no real value (the line and curve never meet). [codex-deep-dive-simultaneous-equations-by-substitution.md §2]

29. Worked example — simultaneous equations word problem Front: A rectangular pitch has perimeter 350 m and area 7350 m². Given x > y, find its length x and width y. Back: 2x + 2y = 350 and xy = 7350. Substituting y = 175 − x into xy = 7350 gives x² − 175x + 7350 = 0, factorising as (x−70)(x−105) = 0. Both x = 70 and x = 105 satisfy the equations; the stated constraint x > y selects x = 105, y = 70. [real, June 2023 Q2 — codex-deep-dive-simultaneous-equations-by-substitution.md §3.5]

30. Quadratic inequality method Front: State the standard method for solving a quadratic inequality px² + qx + r [inequality] 0. Back: Solve px² + qx + r = 0 for its critical values, then use the shape of the parabola (upward if p>0, downward if p<0) to read off which region (outside the roots, or between the roots) satisfies the inequality. [spec 1.7's own worked interpretation — codex-formula-and-method-bank.md §1.1.6]

31. Reciprocal-inequality identity Front: State the spec's own named method for reducing a/x < b to a solvable inequality. Back: Multiply through by x² (always ≥ 0, so the inequality direction never flips): a/x < b becomes ax < bx², x ≠ 0. [spec 1.7's own worked example, verbatim — §1.1.6]

32. Never answer an inequality in terms of "R" Front: When a question asks for the solution set of an inequality (e.g. describing a region R), in what variable must the final answer be stated? Back: In the actual variable (e.g. x) — never as a region label such as "R." This scores M0, a method-mark failure, not merely a lost accuracy mark. [verbatim, real, Oct 2021 Q3(ii) — codex-formula-and-method-bank.md §2.1.3]

33. Worked example — a quadratic inequality not pre-arranged Front: Solve the inequality 4x² − 3x + 7 ≥ 4x + 9. Back: Gather terms: 4x² − 7x − 2 ≥ 0 ⟹ (4x+1)(x−2) ≥ 0, critical values x = −1/4, 2. Since the leading coefficient is positive and the inequality is ≥ 0, this is the outside-the-roots case: x ≤ −1/4 or x ≥ 2. Writing the answer as one chained inequality (e.g. −1/4 ≥ x ≥ 2) is refused, since it asserts something no real x satisfies. [real, June 2023 Q1 — codex-deep-dive-quadratic-inequalities-graphically.md §3.2]

34. Polynomial factorisation, degree n ≥ 3 Front: Factorise x³ + 4x² + 3x (the spec's own worked example for degree-n≥3 factorisation). Back: Extract the common factor x, then factorise the remaining quadratic by inspection: x³+4x²+3x = x(x²+4x+3) = x(x+1)(x+3). The factor theorem is not needed and is not P1 content. [spec 1.10's own worked example, verbatim — codex-formula-and-method-bank.md §1.1.7]

35. Cubic sketching — the four facts factored form hands you Front: List the four facts a factorised cubic's sketch needs, and no more. Back: (1) Every root and its multiplicity; (2) the y-intercept, f(0); (3) the end-behaviour direction, the sign of the leading coefficient; (4) the full sign pattern along the number line. [codex-deep-dive-cubic-graph-sketching-from-factored-form.md Part 1]

36. Simple root vs repeated root Front: State how a cubic's graph behaves at a simple (non-repeated) root versus a repeated (squared) root. Back: A single, non-repeated root: the curve crosses the x-axis there. A repeated (squared) factor: the curve touches the axis there — smoothly, as a proper turning point, never a sharp cusp. [codex-formula-and-method-bank.md §2.1.5]

37. Why a cubic has 0 or 2 turning points, never 1 Front: State the rule for how many turning points a cubic can have, and the one-line reason. Back: A cubic has exactly 0 or 2 turning points, never 1. f′(x) = 3ax² + 2bx + c is itself a genuine quadratic; its discriminant positive gives two genuine sign changes in f′ (two turning points); its discriminant zero or negative gives no sign change in f′ at all (zero turning points) — a quadratic's own discriminant has no case producing "one crossing and nothing else." [codex-deep-dive-cubic-graph-sketching-from-factored-form.md Part 3]

38. Reciprocal graph y = k/x Front: State the asymptotes and symmetry of y = k/x (x ≠ 0). Back: Asymptotes at x = 0 and y = 0; two-fold rotational symmetry about the origin (opposite-sign branches in opposite quadrants for k > 0). [ASSUMED — codex-formula-and-method-bank.md §1.1.8]

39. Reciprocal graph y = k/x² Front: State the asymptotes and branch positions of y = k/x² (x ≠ 0). Back: Asymptotes at x = 0 and y = 0; both branches lie in the upper half-plane (or both lower, if k < 0), since squaring removes the sign of x. [ASSUMED — §1.1.8]

40. Worked example — cubic sketch, three distinct simple roots Front: Sketch f(x) = (x+5)(x−2)(x−4). What is f(0), and how many times does the curve cross the x-axis? Back: f(0) = (5)(−2)(−4) = 40 (matching direct expansion). Three distinct simple roots (x = −5, 2, 4) ⟹ the curve crosses the x-axis three times. [VERIDIAN, illustrative — codex-deep-dive-cubic-graph-sketching-from-factored-form.md Part 2]

41. Worked example — cubic with one simple and one repeated root Front: Sketch f(x) = (x−3)(x+1)². Describe the curve's behaviour at each root. Back: Simple root at x = −1: the curve crosses. Repeated (double) root at x = 3: the curve touches. f(0) = (−3)(1)² = −3. [VERIDIAN, illustrative, mirroring the real Oct 2021 Q6(a) shape f(x)=2(x+1)(x−3)² — codex-deep-dive-cubic-graph-sketching-from-factored-form.md Part 2]

42. Graph transformation — y = af(x) Front: Describe the transformation y = af(x) applies to y = f(x). Back: Vertical stretch, scale factor a, from the x-axis. [spec 1.12, ASSUMED — codex-formula-and-method-bank.md §1.1.9]

43. Graph transformation — y = f(x) + a Front: Describe the transformation y = f(x) + a applies to y = f(x). Back: Vertical translation by (0, a). [spec 1.12, ASSUMED — §1.1.9]

44. Graph transformation — y = f(x + a) Front: Describe the transformation y = f(x + a) applies to y = f(x). Back: Horizontal translation by (−a, 0) — the sign flips relative to how it appears inside the bracket. [spec 1.12, ASSUMED — §1.1.9]

45. Graph transformation — y = f(ax) Front: Describe the transformation y = f(ax) applies to y = f(x). Back: Horizontal stretch, scale factor 1/a, from the y-axis. [spec 1.12, ASSUMED — §1.1.9]

46. The required vocabulary: "translation" Front: What single word does a real mark scheme gate the transformation-description mark on, for a translation? Back: "Translation" — not "shift" or "move." A real examiner report confirms this: "it was extremely rare to see a fully correct description of the transformation... most using the words 'shift' or 'move' instead." [real, Jan 2023 Q7(b) — codex-formula-and-method-bank.md §1.1.9]

47. Invariant line for a stretch/reflection Front: State the invariant line for y = af(x), and for y = f(ax). Back: For y = af(x) (any a): every root of f is unmoved, since a·0 = 0 regardless of a — the x-axis is invariant. For y = f(ax): the point (0, f(0)) is unmoved, since f(a·0) = f(0) regardless of a — the y-axis is invariant. [codex-deep-dive-graph-transformations-in-words.md §3]

48. Hidden quadratic — the three disguises Front: Name the three real, recurring disguises a "hidden quadratic" can take before the discriminant or factorisation can be applied. Back: (1) Line meets curve — set the two y-expressions equal; (2) substitution (e.g. p = 3ˣ turns an exponential equation into a quadratic in p) — solve for the substituted variable, then convert back to the original variable; (3) unknown in a denominator, cleared by multiplying through. [codex-formula-and-method-bank.md §2.1.2]

49. Worked example — exponential substitution Front: Solve 7^(2x) − 6×7^x − 7 = 0. Back: Let u = 7^x: u² − 6u − 7 = (u−7)(u+1) = 0, so u = 7 or u = −1. Reject u = −1 since 7^x > 0 always. 7^x = 7 = 7¹ ⟹ x = 1. [VERIDIAN, illustrative — codex-deep-dive-hidden-quadratic-substitution.md §2]

50. Worked example — root substitution Front: Solve x − √x − 2 = 0. Back: Let u = √x: u² − u − 2 = (u−2)(u+1) = 0, so u = 2 or u = −1. Reject u = −1 since √x ≥ 0. √x = 2 ⟹ x = 4 (check: 4 − 2 − 2 = 0). [VERIDIAN, illustrative — codex-deep-dive-hidden-quadratic-substitution.md §3.3]

Part 2 — Coordinate Geometry (spec 2.1–2.2) — 10 cards

51. Point-gradient form Front: State the point-gradient equation of a straight line through (x₁, y₁) with gradient m. Back: y − y₁ = m(x − x₁). [ASSUMED — codex-formula-and-method-bank.md §1.2.1]

52. General form of a line Front: State the general form of a straight-line equation. Back: ax + by + c = 0. [ASSUMED — §1.2.1]

53. Gradient between two points Front: State the formula for the gradient of the line through (x₁,y₁) and (x₂,y₂). Back: m = (y₂ − y₁) / (x₂ − x₁). [ASSUMED, universal prior knowledge — §1.2.1]

54. Parallel condition Front: State the condition for two lines with gradients m₁ and m₂ to be parallel. Back: m₁ = m₂. [ASSUMED — §1.2.2]

55. Perpendicular condition Front: State the condition for two lines with gradients m₁ and m₂ to be perpendicular. Back: m₁ × m₂ = −1, equivalently m₂ = −1/m₁. [ASSUMED — §1.2.2]

56. Vector dot-product perpendicularity test Front: State the direction-vector dot-product test for two lines being perpendicular. Back: A line with gradient m has direction vector (1, m). Two lines are perpendicular exactly when (1,m₁)·(1,m₂) = 1 + m₁m₂ = 0 ⟺ m₁m₂ = −1 — the same fact as the gradient test, reached algebraically instead of geometrically. [codex-deep-dive-perpendicularity-proofs-and-coordinate-geometry.md §2.2]

57. Vertical-line edge case Front: Why does m₁m₂ = −1 fail to identify the line perpendicular to a horizontal line, and what should be used instead? Back: m₁ = 0 makes m₂ = −1/m₁ undefined — correctly, since the perpendicular to a horizontal line is vertical, and a vertical line's gradient is not a number. Use direction vectors instead: a vertical line has direction (0,1), a horizontal line (1,0), and their dot product is 0 with no division involved. [codex-deep-dive-perpendicularity-proofs-and-coordinate-geometry.md §2.3]

58. Proof requires an explicit closing statement Front: For a "prove PQR = 90°" question, is showing m₁ × m₂ = −1 alone sufficient for full marks? Back: No — the mark scheme requires an explicit closing statement connecting the arithmetic to the geometric conclusion, e.g. "therefore PQ ⊥ QR, so angle PQR = 90°." The calculation alone, with no such sentence, is a real, confirmed way to lose the final mark. [real, Jan 2023 Q2 — codex-formula-and-method-bank.md §1.2.2]

59. Worked example — perpendicularity proof by vectors Front: P(−3,7), Q(9,11), R(12,2). Use vectors to show angle PQR = 90°. Back: PQ→ = (12,4), QR→ = (3,−9). PQ→ · QR→ = (12)(3) + (4)(−9) = 36 − 36 = 0 ⟹ PQ ⊥ QR ⟹ angle PQR = 90°. [real, Jan 2023 Q2 — codex-deep-dive-perpendicularity-proofs-and-coordinate-geometry.md §3.4]

60. Worked example — perpendicular bisector construction Front: A(1,1), B(5,5). C lies on the perpendicular bisector of AB with AC = 6. Find C. Back: Midpoint M = (3,3); AM = 2√2; by Pythagoras, MC² = 36 − 8 = 28, MC = 2√7. The perpendicular bisector's unit direction vector is (1/√2)(1,−1). C = (3,3) ± √14(1,−1), giving C = (3+√14, 3−√14) or (3−√14, 3+√14). [VERIDIAN, illustrative — codex-deep-dive-perpendicularity-proofs-and-coordinate-geometry.md §5]

Part 3 — Trigonometry (spec 3.1–3.3) — 20 cards

61. Sine rule Front: State the sine rule. Back: a/sin A = b/sin B = c/sin C. [MEMORISE — spec P1.2's own "will not appear in the booklet" box, codex-formula-and-method-bank.md §1.3.1]

62. Cosine rule Front: State the cosine rule. Back: a² = b² + c² − 2bc cos A. [BOOKLET — one of only two entries in the real formula booklet's P1 section, §1.3.1]

63. Area of a triangle Front: State the area-of-a-triangle formula using two sides and the included angle. Back: Area = ½ab sin C. [MEMORISE — §1.3.1]

64. The ambiguous case, defined Front: State the mathematical fact behind "the ambiguous case of the sine rule." Back: sin θ = sin(180° − θ) for every θ in (0°,180°) — so solving sin B = k by calculator returns one angle when the data may genuinely admit a second, geometrically distinct triangle with the supplementary angle in its place. [spec 3.1's own named inclusion — codex-deep-dive-sine-and-cosine-rules-ambiguous-case.md §1]

65. SSA triangle-count classification (A acute) Front: Given angle A (acute), opposite side a, and adjacent side b, with h = b sin A, state how many triangles result for a < h, a = h, h < a < b, a = b, and a > b. Back: a < h: 0 triangles (sin B > 1). a = h: exactly 1 (B = 90°). h < a < b: exactly 2 (both candidate angles are valid). a = b: exactly 1 (isosceles; the supplementary candidate is degenerate). a > b: exactly 1 (only the acute candidate keeps the angle sum under 180°). [codex-deep-dive-sine-and-cosine-rules-ambiguous-case.md §2]

66. Worked example — a genuine two-triangle SSA case Front: A = 40°, b = 12, a = 9.86. Show this SSA data gives two valid triangles. Back: h = 12 sin40° ≈ 7.71, and h < 9.86 < 12 = b. Solving sin B = (12 sin40°)/9.86 gives B ≈ 51.47° or B ≈ 128.53°, with third angles ≈ 88.53° or ≈ 11.47° — both valid, two triangles. [VERIDIAN, illustrative — codex-deep-dive-sine-and-cosine-rules-ambiguous-case.md §2]

67. The radian, defined Front: Define one radian. Back: One radian is the angle subtended at a circle's centre by an arc exactly one radius long. [codex-deep-dive-sector-area-and-arc-length.md §1]

68. Full turn in radians Front: State a full turn in radians, and the one-line reason. Back: 2π radians — because the whole circumference is 2πr, and one radian of turning corresponds to an arc of length r, so a full turn corresponds to an arc of 2πr, i.e. 2π radians. [§1.3.2 — codex-deep-dive-sector-area-and-arc-length.md §1]

69. Degree–radian conversion Front: State the conversion factors between degrees and radians. Back: Degrees → radians: multiply by π/180. Radians → degrees: multiply by 180/π. [codex-deep-dive-sector-area-and-arc-length.md §1]

70. Arc length Front: State the formula for arc length, given radius r and angle θ in radians. Back: s = rθ. [MEMORISE — spec P1.2's own box, codex-formula-and-method-bank.md §1.3.2]

71. Sector area Front: State the formula for the area of a sector, given radius r and angle θ in radians. Back: A = ½r²θ. [MEMORISE — §1.3.2]

72. Minor + major angle Front: For any two points A, B on a circle, what does the minor angle plus the major angle at the centre equal? Back: θ_minor + θ_major = 2π (a full turn) — neither is "half" of the other. [codex-deep-dive-sector-area-and-arc-length.md §1]

73. Worked example — sector area rearranged for the radius Front: A sector has area 40 m² and angle 2.4 rad. Find the radius. Back: A = ½r²θ ⟹ r² = 2A/θ = 80/2.4 = 33.3̇ ⟹ r ≈ 5.77 m (2 d.p.). [real, Jan 2023 Q6(a) — codex-deep-dive-sector-area-and-arc-length.md §3.2]

74. Major vs minor angle — the confirmed trap Front: What is the single most common real error on a major-arc/sector question, and how is it avoided? Back: Using the minor angle where the major angle was required (e.g. using 1.64 instead of 2π − 1.64 ≈ 4.64). Check: minor + major sector areas must reconstruct the whole circle, πr² — a wrong split never does. [real, Oct 2021 Q7(b) — codex-deep-dive-sector-area-and-arc-length.md §4]

75. Period of sin x and cos x Front: State the period of sin x and cos x. Back: 2π (360°). [ASSUMED — codex-formula-and-method-bank.md §1.3.3]

76. Period of tan x Front: State the period of tan x, and the conceptual trap examiners confirm around it. Back: π (180°) — the single hardest-documented fact on this paper: many candidates confuse period (the length of one repeating cycle) with domain/interval, wrongly answering with something like −π/2 < x < π/2. [real, Jan 2023 Q9(a) — §1.3.3]

77. General period rule for f(kx) Front: If f has minimal period P, state the minimal period of g(x) = f(kx), k ≠ 0. Back: P / |k|. [codex-deep-dive-trig-graphs-period-and-domain.md §2]

78. tan x domain restriction Front: State the values excluded from the domain of tan x. Back: x ≠ π/2 + nπ, for integer n (where cos x = 0). [codex-deep-dive-trig-graphs-period-and-domain.md §1]

79. Local symmetry of sin and cos Front: State the local (within-period) symmetry identities for sin and cos. Back: sin(π − x) = sin x (a vertical mirror line through each peak/trough); cos(π − x) = −cos x (point symmetry about each zero, e.g. (π/2, 0)). [codex-deep-dive-trig-graphs-period-and-domain.md §5]

80. Odd/even classification Front: Classify sin x, cos x and tan x as odd or even functions. Back: sin x and tan x are odd (f(−x) = −f(x)); cos x is even (f(−x) = f(x)). [codex-deep-dive-trig-graphs-period-and-domain.md §1]

Part 4 — Differentiation (spec 4.1–4.3) — 14 cards

81. The derivative as a limit Front: State the first-principles limit definition of f′(x). Back: f′(x) = lim_{h→0} [f(x+h) − f(x)] / h. [spec 4.1 — codex-formula-and-method-bank.md §1.4.1]

82. The chain rule exclusion Front: What does spec 4.1 explicitly say about the chain rule at P1, and what does that rule out? Back: "Knowledge of the chain rule is not required." A composite like (2x+5)ⁿ or 1/(3x−1) cannot be differentiated by chain-rule shortcut — it must be expanded or rewritten first. [verbatim spec exclusion — §1.4.1]

83. Power rule for differentiation Front: State the power rule for differentiating xⁿ. Back: f(x) = xⁿ ⟹ f′(x) = nxⁿ⁻¹, for all rational n. [MEMORISE — spec P1.2's own box, §1.4.2]

84. Differentiation method-mark convention Front: What is the specific criterion for earning the differentiation method mark? Back: "Power of at least one term decreased by 1 (xⁿ → xⁿ⁻¹)" — earned by any one correctly power-shifted term, not by the whole expression being correct. [verbatim mark-scheme convention — §1.4.2]

85. Tangent equation Front: State the equation of the tangent to y = f(x) at (a, f(a)). Back: y − f(a) = f′(a)(x − a). [ASSUMED, built from 1.4.2 and coordinate geometry — §1.4.3]

86. Normal equation Front: State the equation of the normal to y = f(x) at (a, f(a)), and its one condition. Back: y − f(a) = −1/f′(a) · (x − a), provided f′(a) ≠ 0. [§1.4.3]

87. The zero-gradient edge case Front: If f′(a) = 0, what are the tangent and normal equations at (a, f(a))? Back: The tangent is the horizontal line y = f(a); the normal is the vertical line x = a. A real, confirmed trap: candidates reaching f′(a) = 0 are "thrown by the zero gradient" and fail to write the resulting (now-trivial) tangent equation at all. [real, Oct 2021 Q6(c) — §1.4.3]

88. Rewriting before differentiating Front: Before applying the power rule to √x or 1/x², what must be done first? Back: Rewrite each as a power of x: √x = x^(1/2); 1/x² = x⁻² — nothing can be differentiated by the power rule until it is in xⁿ form. [codex-deep-dive-differentiating-negative-and-fractional-indices.md §1]

89. dy/dx as a rate of change Front: State spec 4.1's own guidance-text description of what dy/dx represents. Back: "dy/dx is the rate of change of y with respect to x." [verbatim spec guidance — codex-deep-dive-the-derivative-as-a-limit.md §1]

90. Worked example — the second derivative Front: For f(x) = x³, find f′(x) and f″(x) directly (not just by the power rule — state what f″ actually is). Back: f′(x) = 3x²; f″(x) is the derivative OF f′(x) — the same limiting-chord process run a second time — giving f″(x) = 6x, matching the ordinary power-rule result. [codex-deep-dive-the-derivative-as-a-limit.md §3]

91. Worked example — differentiating a fraction over a root denominator Front: Differentiate (x² + 5x − 3)/(3√x), x > 0. Back: Rewrite 3√x = 3x^(1/2); split to (1/3)x^(3/2) + (5/3)x^(1/2) − x^(−1/2). Differentiating term by term: dy/dx = (1/2)x^(1/2) + (5/6)x^(−1/2) + (1/2)x^(−3/2). [spec 4.2's own named worked-example expression — codex-deep-dive-differentiating-negative-and-fractional-indices.md §2.3]

92. Worked example — a hidden higher-degree equation from differentiation Front: f(x) = x⁵ − 2x³ + 4x. Find all x for which f′(x) = 3. Back: f′(x) = 5x⁴ − 6x² + 4. Setting equal to 3: 5x⁴ − 6x² + 1 = 0. As a quadratic in u = x²: (5u−1)(u−1) = 0, giving u = 1/5 or u = 1 — both positive, so both survive: x = ±1/√5 or x = ±1. [VERIDIAN, illustrative — codex-deep-dive-hidden-higher-degree-equations-from-differentiation.md §5.1]

93. The parity theorem for hidden quadratics under differentiation Front: State the condition on f(x)'s powers of x that guarantees f′(x) = k hides a quadratic in x². Back: If every power of x in f(x) is an odd integer (plus possibly a constant), every term of f′(x) has an even integer power of x, and multiplying f′(x) − k = 0 through by any even power of x preserves that — so the resulting equation is expressible as a polynomial in u = x². [codex-deep-dive-hidden-higher-degree-equations-from-differentiation.md §2]

94. Domain restriction and the ± step Front: When solving u = x² for x, why must both signs, +√u and −√u, normally be kept? Back: The map x ↦ x² is two-to-one on ℝ{0}: every positive u has exactly two real square roots, and nothing about solving u = x² for x can recover which sign the original x had. Both signs are the default; a domain restriction stated in the original question (e.g. x > 0) is what removes marks for keeping both, never the reverse. [codex-deep-dive-hidden-higher-degree-equations-from-differentiation.md §6]

Part 5 — Integration (spec 5.1–5.2) — 15 cards

95. Constant of integration Front: State spec 5.1's own requirement about a constant every time indefinite integration reverses a differentiation. Back: "Students should know that a constant of integration is required" — every single application of indefinite integration introduces its own new constant. [verbatim spec 5.1 — codex-formula-and-method-bank.md §1.5.1]

96. Power rule for integration Front: State the power rule for integrating xⁿ, and its one exclusion. Back: ∫xⁿ dx = xⁿ⁺¹/(n+1) + c, n ≠ −1. [MEMORISE — spec P1.2's own box, §1.5.2]

97. Why n = −1 is excluded Front: Why can no valid P1 integration question require integrating a term with index exactly −1? Back: ∫x⁻¹ dx = ln|x| + c requires natural logarithms, and logarithms are P2 content — so a term with index exactly −1 is outside what P1 integration can legally be asked to handle. [§1.5.2]

98. Splitting a fraction before integrating — the general rule Front: State how to rewrite Ax^m/(Cx^n) + Bx^k/(Cx^n) as separate powers of x, ready to integrate. Back: Divide each numerator term by the denominator separately, then apply the index law: Ax^m/(Cx^n) + Bx^k/(Cx^n) = (A/C)x^(m−n) + (B/C)x^(k−n). [codex-deep-dive-splitting-fractions-before-integrating.md §2.1]

99. Why "integrate top, integrate bottom, divide" fails Front: Explain, in general, why integrating a fraction's numerator and denominator separately and then dividing cannot give the correct antiderivative. Back: ∫Ax^m dx / ∫Cx^n dx gives a result proportional to x^(m−n) — one power of x lower than the genuine antiderivative, x^(m−n+1), because the method has already "used up" the +1 integration should add before the division that should have happened first ever takes place. [codex-deep-dive-splitting-fractions-before-integrating.md §2.2]

100. Shifted denominator cannot be split additively Front: Why can't 6/(x − 2) be rewritten as 6/x − 3? Back: 6/(x−2) ≠ 6/x − 3 for any x — a shifted denominator (x − a) cannot be pulled apart additively the way a shifted numerator over a fixed denominator can; it can only be handled by cross-multiplying or genuine algebraic manipulation of the whole expression. [real, confirmed WMA11 error, Jan 2023 Q7(c)–(d) — codex-formula-and-method-bank.md §1.5.2]

101. Worked example — splitting a fraction before integrating Front: Find ∫(4x⁵ + 3)/(2x²) dx. Back: Split: 4x⁵/(2x²) + 3/(2x²) = 2x³ + (3/2)x⁻². Integrate term by term: ∫2x³ dx = ½x⁴; ∫(3/2)x⁻² dx = −(3/2)x⁻¹. Answer: ½x⁴ − (3/2)x⁻¹ + c. [real, Jan 2023 Q3 — codex-complete-worked-paper.md / codex-deep-dive-splitting-fractions-before-integrating.md §3.1]

102. Two-stage integration — sort the conditions first Front: Before integrating f″(x) at all, how should a gradient condition f′(a) = m and a point (p,q) on y = f(x) be sorted? Back: A gradient condition f′(a) = m belongs to f′(x) and can only be used once f′(x) exists; a point (p, q) belongs to f(x) and can only be used once f(x) exists. The gradient condition must be applied immediately after the first integration, before integrating a second time. [codex-formula-and-method-bank.md §1.5.2 / §2.5.2]

103. Why c₁ always resurfaces as a linear term Front: Explain, in general, why the first constant of integration (c₁) always appears as a genuine term of f(x), rather than being absorbed at the end. Back: If f′(x) = F(x) + c₁, then integrating again gives f(x) = ∫F(x)dx + c₁x + c₂ — the term c₁x falls straight out of ∫c₁ dx = c₁x regardless of what F(x) is. A single constant added only after both integrations are complete has no linear term to contribute, so it is missing a term the correct f(x) genuinely has. [codex-deep-dive-integrate-twice-two-constants.md §2]

104. Worked example — integrate twice, two constants Front: f″(x) = 4x + x^(−1/2), x > 0. f′(4) = 3 and f(4) = 16. Find f(x). Back: Integrate once: f′(x) = 2x² + 2x^(1/2) + c₁. Using f′(4)=3: 32 + 4 + c₁ = 3 ⟹ c₁ = −33, so f′(x) = 2x² + 2x^(1/2) − 33. Integrate again: f(x) = (2/3)x³ + (4/3)x^(3/2) − 33x + c₂. Using f(4)=16: 128/3 + 32/3 − 132 + c₂ = 16 ⟹ c₂ = 284/3. f(x) = (2/3)x³ + (4/3)x^(3/2) − 33x + 284/3. [real, Jan 2023 Q11(b) — codex-deep-dive-integrate-twice-two-constants.md §3.2]

105. A tangent line can encode both integration conditions at once Front: A tangent to a curve at x = a is given as y = mx + k. What two conditions does this single fact supply for a two-stage integration problem? Back: Its gradient, m, is directly f′(a) (a tangent shares the curve's gradient at the point of contact). Its value at x = a, ma + k, is f(a) (a tangent touches the curve, so they share a height there even though not a gradient anywhere else). [real, Jan 2023 Q11 — codex-deep-dive-integrate-twice-two-constants.md §3.3]

106. Quoting a formula before substituting protects the method mark Front: Why does writing out the quadratic formula in full, in letters, before substituting numbers matter for the method mark? Back: Where a memorised formula is quoted in letters first, the method mark can still be scored even if a later substitution goes wrong; where it is not quoted, the method mark has to be inferred from the working and "may be lost if there is any mistake in the working." [verbatim general marking guidance — codex-deep-dive-quadratic-functions.md §4]

107. Exact-values-only convention for a length-based perpendicularity proof Front: In a "prove perpendicular via Pythagoras" proof, must the squared side lengths be kept exact, or can rounded decimals be compared? Back: Values must appear as exact for the accuracy mark — comparing rounded decimal lengths (e.g. 12.65² + 9.49² ≈ 15.81²) scores one mark fewer than keeping the squared lengths as exact integers (160, 90, 250) throughout. [real, verbatim mark-scheme condition, Jan 2023 Q2 — codex-deep-dive-perpendicularity-proofs-and-coordinate-geometry.md §3.5]

108. Sketch labelling convention on a trig graph Front: What must any hand-drawn sketch of a trig graph carry on both axes, and what is the consequence of omitting it? Back: A labelled scale on both axes — a real, confirmed, examiner-documented mark loss applies to an otherwise-correct sketch with no scale shown. [real, Jan 2019 Q5 — codex-formula-and-method-bank.md §1.3.3]

109. The cosine rule as an alternative route through SSA data Front: Besides the sine rule, what alternative, equally-credited method solves SSA triangle data? Back: The cosine rule, rearranged into a quadratic in the unknown third side c: c² − (2b cos A)c + (b² − a²) = 0 — its two roots (where both positive) correspond to the two triangles the sine rule's two candidate angles would also find. [real, Jan 2019 Q7(a) Alt1(a) — codex-deep-dive-sine-and-cosine-rules-ambiguous-case.md §4]

Genuine gaps and exclusions — no flashcard forced

Consistent with this course's standing discipline against padding a document type that doesn't apply: several sub-cases named in the source deep-dives carry no real per-question WMA11 anchor found in the archive those files searched, and this deck states that plainly on the relevant card rather than upgrading a [VERIDIAN, illustrative] example into a false citation. Named examples carried forward as honest gaps: the single-surd-simplification step in isolation (card 12); combining two surds in isolation (card 13); multiplying/squaring a surd binomial (cards 14–15); a numeric coefficient on a monomial surd denominator, and both conjugate generalisations for a binomial surd denominator; a hidden-cubic-in-x² variant of card 92's mechanism (the spec's own curriculum boundary — the Factor Theorem needed to solve a general cubic in u sits at WMA12, not WMA11 — predicts this shape cannot legally appear, but no real anchor confirms or denies it either way); a root (fractional-index) denominator for the fraction-splitting method of card 98; a bare c = 0 quadratic-factorisation trap tested in its own right (structurally parallel to card 26's k = 0 exclusion, but not independently confirmed); a parameter reaching only the leading coefficient or only the constant term of a discriminant condition (linear-in-parameter, no anchor found); and a symmetric quadratic-in-parameter case with a non-empty excluded region (no anchor beyond a degenerate case already in the lesson file). No diagram-only flashcard was built for any topic, because every real trap this deck could anchor to a diagram (cusp vs smooth turn, unlabelled y-intercept, unscaled trig axes) is already carried on an existing formula or worked-example card instead of a separate card that would just restate the same finding a second time.

Card count

109 flashcards across five parts: Part 1 (Algebra and functions, spec 1.1–1.12, 50 cards), Part 2 (Coordinate geometry, spec 2.1–2.2, 10 cards), Part 3 (Trigonometry, spec 3.1–3.3, 20 cards), Part 4 (Differentiation, spec 4.1–4.3, 14 cards), Part 5 (Integration, spec 5.1–5.2, 15 cards). Every card's Back is a direct reformatting of a formula, definition, condition or worked calculation already present in one of the nineteen codex-*.md source files read in full for this pass — no new fact, formula, series citation or worked number is introduced anywhere in this deck.