GRADE 12 MATHEMATICS PAPER 1 MEMORANDUM - NSC PAST PAPERS AND MEMOS JUNE 2016
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GRADE 12
JUNE 2016
MEMORANDUM
NATIONAL SENIOR CERTIFICATE
NOTE
- If a candidate answered a question TWICE, mark the FIRST attempt ONLY.
- Consistent accuracy(CA) applies in ALL aspects of the memorandum.
- If a candidate crossed out an attempt of a question and did not redo the question, mark the crossed-out attempt.
- The mark for substitution is awarded for substitution into the correct formula.
QUESTION 1
| 1.1.1 |  | ✓ factorisation ✓✓ 𝑥-values (3) |
| 1.1.2 |
| ✓ standard form ✓ substitution ✓✓ 𝑥-values (4) |
| 1.1.3 |  | ✓ critical values with method ✓✓ answer (3) |
| 1.1.4 | 3𝑥 . 3𝑥+1 = 27𝑥 32𝑥+1 = 33𝑥 ∴ 2𝑥 + 1 = 3𝑥 𝑥 = 1 | ✓ 32𝑥+1 ✓ 33𝑥 ✓ equating ✓ answer(4) |
| 1.2 | 3 + 𝑦 = 2𝑥 If formula is used, award factor’s mark for substitution. | ✓ substitution ✓ removing brackets ✓ standard form ✓ factors ✓ 𝑥-values ✓ 𝑦-values (6) |
| 1.3 | 𝑓(𝑥) = (𝑥 − 2)(𝑥2 − 6𝑥 + 10) Consider the quadratic factor : (𝑥2 − 6𝑥 + 10) Δ = 𝑏2 − 4𝑎𝑐 = (−6)2 − 4(1)(10) = 36 − 40 = −4 => Δ < 0, therefor NO Solutions 𝑥 = 2 is the only solution. | ✓ substitution into delta ✓ answer ✓ conclusion (4) [23] |
QUESTION 2
| 2.1.1 |  | ✓ first differences ✓ second differences (2) |
| 2.1.2 | Next term= 25 | ✓ answer(1) |
| 2.1.3 |  | ✓ 𝑎 = 3 2 ✓ 𝑏 = − 11 2 ✓ 𝑐 = 4 ✓ answer(4) |
| 2.1.4 | 𝑇30 = 3 (30)2 −11(30) + 4 2 2 = 1189 | ✓ substitution (2) |
| 2.2.1 | 𝑇3 − 𝑇2 = 𝑇4 − 𝑇3 𝑏 − 13 = 27 − 𝑏 2𝑏 = 40 𝑏 = 20 13 − 𝑎 = 7 𝑎 = 6 | ✓ method ✓ values of a and b (2) |
| 2.2.2 | 𝑎 + (𝑛 − 1)𝑑 = 𝑇𝑛 6 + (𝑛 − 1)(7) = 230 7𝑛 = 231 𝑛 = 33 | ✓ formula ✓ substitution ✓ answer (3) |
| 2.3 | 𝑟 =1 − 𝑘 5 −1 < 1 − 𝑘 < 1 5 −5 < 1 − 𝑘 < 5 −6 < −𝑘 < 4 −4 < 𝑘 < 6 | ✓𝑟 = 1−𝑘 5 ✓ substitution ✓ answer (3) |
| 2.4.1 |  | ✓method ✓substitution ✓simplification ✓answer (4) |
| 2.4.2 |  |
|
| 2.4.3 | 𝑆∞ = 𝑎 1 − 𝑟  | ✓ substitution ✓ answer (2) [25] |
(2)
QUESTION 3
| 3.1.1 | 𝑦 = −2 (horizontal asymptote ) | ✓ answer(1) |
| 3.1.2 | 𝑥 = 1 (vertical asymptote) | ✓ answer(1) |
| 3.2 | 3 − 2 = 0 | ✓ y = 0 ✓ x – intercept ✓ y – intercept (3) |
| 3.3 |  | ✓ asymptotes ✓ x-and y-intercepts ✓shape (3) |
| 3.4 | (4 ; 5) | (2) |
| [10] |
QUESTION 4
| 4.1 |  | ✓ 𝑥 -coordinate ✓y-coordinate ✓coordinates of C (3) |
| 4.2 | −𝑥2 − 2𝑥 + 3 = 0 𝑥2 + 2𝑥 − 3 = 0 (𝑥 + 3)(𝑥 − 1) = 0 𝑥 + 3 = 0 𝑜𝑟 𝑥 − 1 = 0 𝑥 = −3 𝑜𝑟 𝑥 = 1 𝐴(−3; 0) 𝐵(1; 0) | ✓standard form ✓factors ✓ both values (3) |
| 4.3 | m = 2 & c = 6 y = 2𝑥 + 6 | ✓value of m ✓value of c (2) |
| 4.4 |  | ✓method ✓substitution ✓answer (3) |
| 4.5 | 𝑥> 1 | ✓✓answer (2) |
| [13] |
QUESTION 5
| 5.1.1 |  | ✓substitution ✓answerd (2) |
| 5.1.2 |  | ✓swop 𝑥 and 𝑦 ✓ answer (2) |
| 5.1.3 |  | ✓equating ✓answer (2) |
| 5.1.4 | 𝑥 ∈ R | answer(1) |
| 5.2 |  | ✓asymptote ✓negative y-intercept ✓shape (3) |
| [10] |
Related Items
QUESTION 6
| 6.1.1 |
effective rate = 8,77%p.a | ✓ formula ✓ substitution ✓ answer (3) |
| 6.1.2 |  | ✓formula ✓substitusie ✓answer(3) |
| 6.2 |  | ✓substitution ✓simplification ✓correct use of logs ✓answer (4) |
| 6.3 |  | ✓✓ setting up equation ✓ x the subject of the formula ✓ answer (4) |
[14]
QUESTION 7
| 7.1 |
= 4𝑥 - 3 Answer ONLY: 0 marks | ✓ substitute (𝑥 + ℎ) Penalise 1 mark for incorrect use ✓ simplification
|
| 7.2 |  | Penalise 1 mark for incorrect notation
|
[9]
QUESTION 8
| 8.1.1 | 𝑓(𝑥) = 𝑥3 − 4𝑥2 − 11𝑥 + 20 𝑓′(𝑥) = 3𝑥2 − 8𝑥 − 11 = 0 (3𝑥 − 11)(𝑥 + 1) = 0 3𝑥 − 11 = 0 𝑜𝑟 𝑥 + 1 = 0  | ✓ 𝑓′(𝑥) = 0 ✓ factors ✓ x-values ✓ y-values ✓ coordinates (5) |
| 8.1.2 | 𝑓′′(𝑥) = 6𝑥 − 8 = 0 6𝑥 = 8 𝑥 =8 or/of 1 1 or/of 1, 33 6 3 OR  | ✓ 𝑓′′(𝑥) = 0 ✓ answer (2) |
| 8.1.3 | 𝑚 = 𝑓′(2) = 3(2)2 − 8(2) − 11 = −15 𝑓(2) = (2)3 − 4(2)2 − 11(2) + 30 = 0 𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1) 𝑦 − 0 = −15(𝑥 − 2) 𝑦 = −15𝑥 + 30 | ✓ 𝑚 = −15 ✓ 𝑓(2) = 0 ✓ substitution ✓ answer (4) |
| 8.1.4 | 36 units downwards 14,8 units upwards | ✓ answer ✓ answer (2) |
| 8.2.1 | 𝑚 = −9 | ✓ answer (1) |
| 8.2.2 | 𝑥 = 1 𝑜𝑟 𝑥 = 5 | ✓✓ answers (2) |
| 8.2.3 | 𝑥 <1 or 𝑥 > 5 | ✓ 𝑥 < 1 ✓ 𝑥 > 5 (2) |
[18]
QUESTION 9
| 9.1 | 𝑉 = 𝑥2ℎ 8 = 𝑥2ℎ ∴ ℎ = 8 𝑥2 | ✓substitution ✓answer (2) |
| 9.2 | 𝐴 = 2𝑥2 + 4𝑥ℎ No mark for the answer | ✓ 2𝑥2 ✓ 4𝑥ℎ ✓ substitution (3) |
| 9.3 | 𝐴(𝑥) = 2𝑥2 + 32𝑥−1 𝐴′(𝑥) = 4𝑥 − 32𝑥−2 = 0 4𝑥3 − 32 = 0 4𝑥3 = 32 𝑥3 = 8 = 23 ∴ 𝑥 = 2 | ✓ 𝐴′(𝑥) = 0 ✓ standard form ✓exponential law ✓ value of x ✓ dimensions (5) |
[10]
QUESTION 10
| 10.1.1 | 𝑃(𝐴 and 𝐵) = 𝑃(𝐴) × 𝑃(𝐵) = 0,4 × 0,5 = 0,2 | ✓ rule ✓ answer(2) |
| 10.1.2 | 𝑃(𝐴 or 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 and𝐵) = 0,4 + 0,5 − 0,2 = 0,7 | ✓ rule ✓ answer(2) |
| 10.1.3 | 𝑃(𝑛𝑜𝑡 𝐴 and 𝐵) = 1 − 𝑃(𝐴 or 𝐵) = 1 − 0,7 = 0,3 | ✓ rule ✓ answer (2) |
| 10.2.1 |  | ✓ first branch If probabilities not listed, maximum 1 mark |
| 10.2.2 | 𝑃(𝑌𝑒𝑙𝑙𝑜𝑤 from 𝐵𝑎𝑔 𝐴)= 2 5 | ✓answer (1) |
| 10.2.3 | 𝑃(𝑃𝑖𝑛𝑘) = 3 + 5 10 18 = 26 45 (0,58) | ✓ 3 10 ✓ 5 18 ✓answer (3) |
| 10.3.1 | 𝑎 = 288 − 𝑥 𝑏 = 372 − 𝑥 | ✓ answer ✓ answer (2) |
| 10.3.2 | 288 − 𝑥 + 𝑥 + 372 − 𝑥 + 56 = 600 −𝑥 = −116 𝑥 = 116 | ✓ equation ✓ answer (2) |
| 10.3.3 | No P(A and B) ≠ 0 | ✓ answer (1) [18] |
| TOTAL: 150 |