MATHEMATICS PAPER 1
GRADE 12
NATIONAL SENIOR CERTIFICATE
MEMORANDUM
JUNE 2018
NOTE:
QUESTION 1
1.1.1 | (? − 2)(3? − 1) = 0 | ✓✓ ?-values |
1.1.2 | 2?2 + 3? − 7 = 0 Penalise 1 mark for incorrect rounding off x = -3 ± √65 | ✓ substitution ✓✓ ?-values (3) |
1.1.3 | -x2 - 2x +15 < 0 x2 + 2x -15 > 0 | ✓factors ✓critical values with method ✓✓ answer (accuracy) (4) |
1.1.4 | ✓ simplifying LHS ✓ simplifying RHS ✓ same base ✓ equating exponents ✓ answer (5) | |
1.2 | x - 3y = 1................... (1) y2 + 2xy - x2 = -7...... (2) from (1) : x = 3y +1..... (3) (3) in (2) : y2 + 2 y(3y +1) - (3y +1)2 = -7 y2 + 6 y2 + 2 y - 9 y2 - 6 y -1+ 7 = 0 - 2 y2 - 4 y + 6 = 0 y2 + 2 y - 3 = 0 ( y + 3)( y -1) = 0 y = -3 or / of y = 1 x = -8 or / of x = 4 | ✓ x subject of the formula ✓ substitution ✓ removing brackets ✓ standard form ✓ ?-values ✓ x-values (6) |
1.3 | For equal roots : | ✓Δ = 0 ✓ both n-values ✓ substitution ✓ x = 0 ✓ x = -2 (5) |
[25} |
QUESTION 2
2.1.1 | 15 ; 10 ; 7 ; ? ; 7 | ✓ 2nd differences ✓ equating ✓ answer (3) |
2.1.2 | 2? = 2 3? + ? = −5 a + b + c = 1 | ✓? = 1 ✓ ? = −8 ✓ ? = 22 ✓ answer |
2.1.3 | T50 = (50)2 - 8(50) + 22 | ✓ substitution ✓ answer |
2.2.1 | a +14d = 74 | ✓ setting up 2 equations ✓ method ✓ answer (3) |
2.2.2 | a + 6(5) = 34 S40 = 40 [2 (4) + (40 -1)(5)] | ✓ value of a ✓ substitution into correct formula ✓ answer (3) |
2.2.3 | ✓✓ answer (accuracy as one unit) | |
2.3.1 | Tk = 3 k = 1 x 3k -1 | ✓ factors of 15 ✓ answer (2) |
2.3.2 | Sn = a(rn - 1) r - 1 241/5 =1/5(3n - 1) 3-1 482/5 = 1/5(3n - 1) 242 = 3n -1 243 = 3n 35 = 3n n = 5 | ✓ a and r ✓ substitution into correct formula ✓ exponential equation ✓ answer (4) |
2.3.3 | NO | ✓ NO ✓ reason (2) |
2.4 | P = 91/3 x 91/9 x 91/27 x ....to infinity S∞ = a | ✓ adding exponents ✓ sum to infinity ✓ answer ✓ P = 9½ / √9 (4) |
[29] |
QUESTION 3
3.1.1 | P (2; 4) | ✓ coordinates of P ✓ coordinates of Q (2) | |
3.1.2 | y = a(x - 2)2 + 4 | ✓ substitution ✓ answer ✓ substitution ✓ answer (4) | |
3.1.3 | x ≥ 2 / x ≤ 2 | ✓✓ answer / antwoord | |
3.1.4 | h(x) = 2 f (x) is a maximum when f (x) is a maximum | ✓ max. value of f(x) ✓answer (2) | |
3.2.1 | ? ≥ 1, ? ∈ ? OR ? ∈ (1; ∞) | ✓ answer (1) | |
3.2.2 | p(x) = x2 +1 / r(x) = x2 + 2x Shift 1 unit to the left and 2units down OR Turning Point of p(x) = (0 ; 1) Shift 1 unit to the left and 2 units down | ✓ calculation ✓ 1 unit to the left ✓ 2 units down (3) | |
[14] |
QUESTION 4
4.1 | y = -3 + 5 | ✓ y-intercept (1) |
4.2 | -3 + 5 = 0 -3 = -5 -5x - 5 = -3 | ✓ simplification ✓ answer (2) |
4.3 | ✓ asymptotes ✓ x- and y-intercepts x ✓ shape (3) | |
4.4 |
| ✓ substitution ✓ simplification ✓ reflection (3) |
[9] |
QUESTION 5
5.1 | f (x) = log3 x | ✓ interchanging x and y ✓ answer (2) |
5.2 | f -1 is a reflection of f in the line y = x | ✓ answer (2) |
5.3 | y = log3 x = 1 | ✓ substitution ✓ answer (2) |
5.4 | 0 < x < 1 OR log3 x < -2 and x > 0 | ✓ ✓answer (2) |
5.5 | x ≥ 1 | ✓✓ answer (2) |
[10] |
QUESTION 6
6.1 |
| ✓formula ✓ substitution ✓ answer (3) |
6.2 | ✓i and/en n ✓ substitution ✓ answer (3) | |
6.3.1 | ✓ A and P ✓ substitution ✓ making n subject of the formula ✓ answer (4) | |
6.3.2 | ✓ substitution ✓ answer (2) | |
6.3.3 | Amount = R457 551,55 – R50 710.00 | ✓ answer (1) |
[13] |
QUESTION 7
Penalise 1 mark for incorrect notation in the question | ||
7.1 | ?(?) = 1 − 3?2 ?(? + ℎ) = 1 − 3(? + ℎ)2 = 1 − 3(?2 + 2?ℎ + ℎ2) ?'(?) = lim ?(? + ℎ) - ?(?) = lim −6?ℎ − 3ℎ2 = −6? Answer ONLY: 0 marks | ✓ 1 − 3?2 − 6?ℎ − 3ℎ2 ✓ substitution ✓ common factor ✓ answer (4) |
7.2 | ✓ y = x + 2 + x -1 ✓1 ✓ −?−2 (3) | |
7.3 | y = 3x2 - 2x +1 6x - 2 = 4 | ✓ y ' = 6x - 2 ✓ 6x - 2 = 4 ✓ x-coordinate ✓ answer (4) |
[11] |
QUESTION 8
8.1.1 | f (x) = a(x + 2)(x - 2/3 )(x - 3) f (x) = 3(x + 2) ( x - 2/3 )(x - 3) | ✓ substitution of x-coordinates ✓ substitution of point ✓ value of a ✓ substitution ✓ removing brackets (5) | |
8.1.2 | ?(?) = 3?3 − 5?2 − 16? + 12 ?′(?) = 9?2 − 10? − 16 = 0 (9? + 8)(? − 2) = 0 ? = 4900 (20,16) B(− 8 ; 20,16) | ✓ ?′(?) = 0 ✓ factors ✓ x-values ✓ coordinates of P (4) | |
8.1.3 |
✓y-intercept ✓turning pts ✓ shape (4) | ||
8.2.1 | f (x) = ax3 + bx2 + 3x + 3 and g(x) = f n(x) = 12x + 4 f '(x) = 3ax2 + 2bx + 3 6a = 12 and 2b = 4 | ✓ a = 2 ✓ b = 2 (2) | |
8.2.2 | 12x + 4 = 0 x =- 1 | ✓ x = - 1 ✓ answer (2) | |
8.2.3 | f n (x) < 0 for x < - 1 (concavedown ) | ✓ answer ✓ answer (2) | |
[19] |
QUESTION 9
9.1 | EF = a - 2x DE = x tan 600 Area = l x b No mark for the answer | ✓ EF = (a – 2x) ✓ tan ratio ✓ answer ✓ substitution (4) | |
9.2 | A(x) = √3ax - 2√3x2 A ' (x) = √3a - 4√3x = 0 x = a | ✓ derivative ✓ f ' (x) = 0 ✓ answer ✓ substitution ✓ answer (5) | |
| [9] |
QUESTION 10
10.1.1 | P(F and S) = 67 / 0, 28 | ✓ answer (1) |
10.1.2 | P(M ) ´ P(not S ) P(M ∩ not S ) = 51 | ✓ P(M) x P(not S) ✓ answer ✓ answer ✓ conclusion (4) |
10.2.1 |
| ✓ 1st branch ✓ 2nd branch ✓ outcomes (3) |
10.2.2 | ✓ setting up equation ✓ standard form ✓ answer (3) | |
[11] |
TOTAL : 150