MATHEMATICS
PAPER 1
GRADE 12
NSC PAST PAPERS AND MEMOS
FEBRUARY/MARCH 2018
MEMORANDUM
NOTE:
- If a candidate answers a question TWICE, only mark the FIRST attempt.
- Consistent accuracy applies in ALL aspects of the marking guidelines.
QUESTION 1 |
1.1.1 | x2 - 6x - 16 = 0 (x - 8)(x + 2) = 0 x = -2 or 8 | ✔ factors ✔x = −2 ✔x =8 (3) |
1.1.2 |
NOTE: Penalise 1 mark if the rounding to TWO decimal places is incorrect.
x = 0,14 or x = -3,64 | ✔subs into correct formula ✔ ✔x = 0,14 ✔x = −3,64 OR ✔for adding 49/16 on both sides ✔ ✔x = 0,14 ✔x = −3,64 (4) |
1.2 | | ✔factors ✔✔inequality ✔answer (4) |
1.3 | x = 2y -1 (2y - 1)2 - 7 - y2 = -y 4y2 - 4y + 1 -7 - y2 = -y 3y2 - 3y - 6 = 0 y2 - y - 2 = 0 (y - 2)(y + 1) = 0 y = 2 or y = -1 x = 2(2)-1 or x = 2(-1)-1 x = 3 or x = -3 OR 4x2 - 28 - x2 - 2x -1 = -2x -2 3x2 - 27 = 0 x2 - 9 = 0 (x - 3)(x + 3) = 0 x = 3 or x = -3 y = -3 + 1 or y = 3 + 1 2 2 y = -1 or y = 2
| ✔x = 2y −1 ✔ substitution ✔ correct standard form ✔ factors ✔ y – values ✔ x – values OR ✔ y = x + 1 2 ✔ substitution ✔ correct standard form ✔ factors ✔ x – values ✔ y – values (6) |
1.4 |
| ✔ common factor 32017 ✔ answer OR ✔ common factor 32016 ✔ answer OR ✔ dividing by 32017 ✔ answer (2) |
1.5.1 | 3x - 5 ≥ 0 and x # 3 x ≥ 5/3 and x # 3 | ✔3x −5 ≥ 0 ✔x ≥ 5/3 ✔x ≠ 3 (3) |
1.5.2 | | ✔ √3x −5 = x −3 ✔ 3x − 5 = (x − 3)2 ✔ factors ✔x = 7 (4) [26] |
QUESTION 2 |
2.1.1 | | ✔ r = 1/3 ✔substitution into correct formula
use of logs ✔n = 8 (4) |
2.1.2 | S∞ = a 1 - r = 30 1 - 1/3 = 45 | ✔substitution into correct formula ✔answer (2) |
2.2 | | ✔expandingn Sn ✔reverse writing (4) ✔expandingn Sn ✔reverse writing (4) [10] |
QUESTION 3 |
3.1 | – 1 ; 2 ; 5 Tn = -1 + (n - 1)(3) = 3n - 4 | ✔ 3n ✔ – 4 (2) |
3.2 | T43 = 3(43) - 4 OR T43 = -1 + (43 - 1)(3) = 125 = 125 NOTE: Answer only 2 / 2 | ✔ subs of 43 ✔ answer (2) |
3.3 | | |
3.4 | T11 = (T11 - T10) + (T10 - T9) + (T9 - T8) + ..... + (T3 - T2) + (T2 - T1) + T1 125 = 29 + 26 + 23 + .... 2 + T1 = 10/2 (29 + 2) + T1 = 155 + T1 T1 = -30 NOTE: Answer only 1/6. If they only use 3n - 4 breakdown 0/6 OR Tn = an2 +bn + c ∴ T11 = 121a + 11b + c = 125
Tn −Tn−1 = 3n - 4 2a = 3 and b - a = -4 a = 3/2 and b = − 5/2 121a + 11b + c =125 121(3/2) + 11(-5/2) + c = 125 ✔c = −29
| ✔✔ generating sum ✔ 29+ 26+ 23+ .......2 ✔ 10 (29 + 2) 2 ✔ 155 ✔ – 30 OR ✔121a +11b+c =125 ✔ calculating Tn −Tn−1 in terms of a, b and c ✔ a = 3/2 ✔ b = − 5/2 ✔c = −29 ✔ –30 (6) [13] |
QUESTION 4 |
4.1 | E(4 ; –9) | ✔ x = 4 ✔ y = –9 (2) |
4.2 | f(x) = (x - 4)2 - 9 (x - 4)2 - 9 = 0 (x - 4)2 = 9 x - 4= ±3 x = 7 or x = 1 A(1 ; 0) OR f(x) = (x - 4)2 - 9 0 = x2 - 8x + 16 - 9 0 = x2 - 8x + 16 - 9 0 = x2 - 8x + 7 (x - 7)(x - 1) = 0 x = 7 or x = 1 A(1 ; 0) | ✔ y = 0 ✔x − 4 = ±3 ✔ A(1 ; 0) OR ✔ y = 0 ✔ (x −7)(x −1) ✔ A(1 ; 0) (3) |
4.3 | C(0 ;7) M(8 ;7) NOTE: Answer only 3 / 3 | ✔C(0 ;7) ✔ x = 8 ✔ y = 7 (3) |
4.4 | C(0 ;7) D(4 ; 0) m = 7 - 0 or m = 0 - 7 or 0 = 4m + 7 0 - 4 4 - 0 m = −7/4 m = −7/4 m = −7/4 y - 0 = −7/4(x - 4) y = −7/4x + 7 | ✔D(4 ; 0) ✔ m = − 7/4 ✔ y = −7/4x + 7 (3) |
4.5 | OR | ✔ interchange x and y ✔ simplification ✔ y = −4/7x + 4 OR ✔ straight line through (0 ; 4) and ( 7 ; 0) ✔ substitution ✔ y = −4/7x + 4 (3) |
4.6 | x . f (x) ≤ 0 ∴x ≤ 0 or 1≤ x ≤ 7 | ✔✔x ≤ 0 ✔✔1≤ x ≤ 7 (4) [18] |
QUESTION 5 |
5.1 | aº = 1 T(0 ; 1) | ✔ x = 0 ✔y = 1 (2) |
5.2 | g(x) = ax 9 = a2 a = 3 a >0 | ✔ substitution ✔ a = 3 (2) |
5.3 | y = (1/3)x or y = 3-x | ✔✔ y = (1/3)x (2) |
5.4 | | ✔1< x ✔x < 3 (2) OR ✔ 1 < x ✔ x < 3 (2) [8] |
QUESTION 6 |
6.1 | q = 1 | ✔q =1 (1) |
6.2 | | ✔ 0 = a + 1 0 + p ✔a = −p ✔ substitution ✔ a = 2 ✔p = −2 (5) |
6.3 | | ✔y = 1 ✔x = 2 ✔shape ✔(0 ; 0) (4) [10] |
QUESTION 7 |
7.1 | | ✔ n = 60 and i = 0,06 / 0,005 12 ✔ Correct substitution into correct formula ✔ answer (3) |
7.2.1 | After eleven months, Genevieve will owe: A = 82 000 ( 1 + 0,15 / 12 ) 11 = R 94006,79 | ✔ n = 11 ✔correct substitution into correct formula ✔ answer (3) |
7.2.2 | -n = -36,8382... n = 36,84 Genevieve will have to pay 36 installments of R3 200
| ✔ 94006,79 ✔ substitute into correct formula ✔correct use of logs (logs to be defined) ✔ n =36,84 ✔ 36 installments (5) |
7.2.3 | | ✔ n = - 083826912 ✔ substitute into correct formula ✔ answer ✔ 2652,00 ( 1 + 0,15 / 12 ) 1 ✔ answer OR (5) [16] |
QUESTION 8 |
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QUESTION 9 |
9.1 | f(x) = (x + 2)(x - 1)(x - 4) = (x2 + x - 2)(x - 4) = x3 + x2 - 2x - 4x2 - 4x + 8 = x3 - 3x2 - 6x + 8 b = -3 ; c = -6 ; d = 8 | ✔✔f (x) = (x + 2)(x −1)(x − 4) ✔ expansion ✔ x3 - 3x2 - 6x + 8 (4) |
9.2 | | ✔ f'(x) = 0 ✔ 3x2 - 6x - 6 ✔ substitution into correct formula ✔x = −0,73 (4) |
9.3 | f(x) = x3 - 3x2 - 6x + 8 f(-1) = (-1)3 - 3(-1)2 - 6(-1) + 8 or f(-1) = (1)(-2)(-5) = 10 = 10 f'(-1) = 3 (-1)2 - 6(-1) - 6 = 3 y - 10 = 3(x + 1) y = 3x + 13 | ✔f' (−1) =10 ✔ f' (-1) = 3 ✔ substitution ✔y = 3x +13 (4) |
9.4 | | ✔ f//(x) = 6x − 6 ✔x- intercept ✔y- intercept (3) |
9.5 | f concave upwards f''(x) > 0 6x - 6 > 0 x > 1 NOTE: A nswer only 2/2 | ✔ f''(x) > 0 ✔ x >1 (2) [17] |
QUESTION 10 |
10 | |
QUESTION 11 |
11.1.1 | Let the event Veli arrive late for school be V. Let the event Bongi arrive late for school be B. P(V or B) =1- 0,7 = 0,3 | ✔ answer (1) |
11.1.2 | P(V or B) = P(V) +P(B) - P(V and B) 0,3 = 0,25 + P(B) - 0,15 P(B) = 0,2 | ✔ P(V or B) = P(V) +P(B) –P(V and B) ✔substitution ✔ 0,2 (3) |
11.1.3 | P(V) × P(B) = 0,25 × 0,2 = 0,05 P(V) × P(B) ≠ P(V and B) V and B are NOT independent. | ✔P(V) × P(B) = 0,05 ✔P(V) × P(B) ≠ P(Vand B) ✔NOT independent (3) |
11.2.1 | 6 ! =720 | ✔6! or 720 (2) |
11.2.2 | Number of arrangements = 3! × 3! × 2 = 72 | ✔3!× 3! ✔× 2 ✔ answer (3) |
11.2.3 | P(hearts next to each other) = 3! × 4! 6! = 144 720 = 1/5 or 0,2 or 20% OR P(hearts next to each other) = 4 × 3! × 3! 6! = 144 720 = 1/5 or 0,2 or 20% | ✔ ✔3!× 4! ✔ 1/5 or 0,2 or 20% OR ✔ ✔ ✔ 1/5 or 0,2 or 20% (3) [15] |
TOTAL: 150