The Grade 12 Mathematics examination consists of two 3-hour papers worth 150 marks each (300 total). Paper 1 tests Algebra, Equations & Inequalities (25±3 marks), Patterns & Sequences (25±3 marks), Functions & Inverses (35±3 marks), Financial Mathematics (15±3 marks), Differential Calculus (35±3 marks), and Probability (15±3 marks). Paper 2 tests Statistics (20±3 marks), Analytical Geometry (40±3 marks), Trigonometry (40±3 marks), and Euclidean Geometry & Proportions (50±3 marks).
How DBE Marks Mathematics (Understanding CA & Method Marks)
Before studying the worked examples, you must understand how DBE markers grade your script:
- Method Mark (M): Awarded for applying a correct mathematical method, even if an arithmetic slip occurs. You cannot earn accuracy marks if the method mark is lost.
- Accuracy Mark (A): Awarded for correct numerical or algebraic results.
- Consistent Accuracy Mark (CA): If you make an early calculation error in sub-question 1.1, markers follow your incorrect value into 1.2. Provided your working in 1.2 is mathematically valid and of equal difficulty, you earn full CA marks. Never leave a follow-up question blank just because you doubted your first answer!
Worked Example 1: Calculus from First Principles (Paper 1 · 5 Marks)
Write the limit definition clearly. This secures the first method mark.
Substitute \((x + h)\) wherever \(x\) appears in the original function and expand carefully:
f(x + h) = 2(x² + 2xh + h²) - 3x - 3h
f(x + h) = 2x² + 4xh + 2h² - 3x - 3h
Subtract the original function \( (2x^2 - 3x) \). Note how the non-\(h\) terms cancel out completely:
= 4xh + 2h² - 3h
Crucial examiner rule: You must keep writing "lim (h → 0)" until you actually evaluate the limit!
f'(x) = lim (h → 0) (4x + 2h - 3)
• Mark 1 (M): Correct formula stated.
• Mark 2 (M): Correct substitution of \(f(x+h)\).
• Mark 3 (A): Correct simplification of numerator (\(4xh + 2h^2 - 3h\)).
• Mark 4 (M): Factoring out \(h\) and cancelling.
• Mark 5 (A): Final answer \(4x - 3\). (Penalty -1 if "lim h→0" was omitted prematurely).
Worked Example 2: Financial Maths — Loan Outstanding Balance (Paper 1 · 6 Marks)
(a) His monthly repayment \(x\).
(b) The outstanding balance of the loan immediately after making the 36th payment.
1. Deposit = 0.15 × 350,000 = R52,500.
2. Present Value (Loan P) = R350,000 - R52,500 = R297,500.
3. Interest rate \( i = \frac{0.115}{12} \). Total payments \( n = 5 \times 12 = 60 \).
297,500 = x[ 1 - (1 + 0.115/12)^(-60) ] / (0.115/12)
297,500 = x × 45.41407...
x = 297,500 / 45.41407... = R6,550.83
The fastest and most reliable method in the NSC memorandum is to calculate the present value of the remaining unmade payments. Remaining payments = \( 60 - 36 = 24 \) payments left.
Balance = 6550.83[ 1 - (1 + 0.115/12)^(-24) ] / (0.115/12)
Balance = 6550.83 × 21.3653...
Outstanding Balance = R139,960.52
• Deposit deduction & P = R297,500 (1 Mark)
• Correct formula selection and \(i/12\) substitution (1 Mark)
• \(x = R6,550.83\) (1 Mark)
• Recognizing \(n_{remaining} = 24\) (1 Mark)
• Substitution into Present Value formula (1 Mark)
• Final balance \(R139,960.52\) (1 Mark)
Worked Example 3: Euclidean Geometry Circle Theorem Proof (Paper 2 · 6 Marks)
Draw circle with centre O and tangent ABC touching at B. Draw chord BD subtending angle \( \angle BED \) at the circumference. Construction: Draw diameter BF through O and connect FD.
2. ∠B&sub1; + ∠B&sub2; = 90° [where ∠B&sub1; = ∠FBD and ∠B&sub2; = ∠DBC]
3. ∠BDF = 90° [Reason: Angle in semi-circle]
4. In ΔBDF: ∠F + ∠B&sub1; + ∠BDF = 180° [Reason: Sum of angles in Δ]
∴ ∠F + ∠B&sub1; + 90° = 180°
∴ ∠F + ∠B&sub1; = 90°
5. Since both (∠B&sub2; + ∠B&sub1; = 90°) and (∠F + ∠B&sub1; = 90°):
∴ ∠B&sub2; = ∠F
6. But ∠BED = ∠F [Reason: Angles in same segment / subtended by same chord BD]
7. ∴ ∠DBC = ∠BED [Tan-chord theorem proven]
Official Past Papers & Memoranda Links
Download original National Senior Certificate question papers directly from the official Department of Basic Education archive: