MATHEMATICS PAPER 1
GRADE 12
NATIONAL SENIOR CERTIFICATEΒ
MEMORANDUM
JUNE 2018
NOTE:
QUESTION 1
1.1.1 | (π₯ β 2)(3π₯ β 1) = 0 | ββ π₯-values |
1.1.2 | 2π₯2Β + 3π₯ β 7 = 0Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Β Penalise 1 mark for incorrect rounding off x =Β -3 Β± β65 | β substitution ββ π₯-valuesΒ (3) |
1.1.3 | -x2 - 2x +15 < 0 x2 + 2x -15 > 0 | βfactorsΒ βcritical values with method ββ answer (accuracy) (4) |
1.1.4 | β simplifying LHS β simplifying RHSΒ β same baseΒ β equating exponents β answer (5) | |
1.2 | x - 3y = 1................... (1) y2 + 2xy - x2 = -7...... (2) fromΒ (1) : x = 3y +1..... (3) (3) in (2) :Β Β Β Β Β Β Β Β Β Β 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 +Β 3)( y -1) =Β 0 y = -3Β or / ofΒ Β Β y = 1 x = -8 or / of x = 4 | β x subject of the formula Β β substitutionΒ β removing brackets β standard form β π¦-values β x-values (6) |
1.3 | For equal roots Β : | βΞΒ = 0 β both n-values β substitution β x = 0 β x = -2 (5) |
Β | Β | [25} |
Β
QUESTION 2
2.1.1 | 15Β ; 10Β ; 7Β Β Β Β ;Β Β Β π₯Β Β ;Β Β Β 7 | β 2nd differencesΒ β equatingΒ β answerΒ (3) |
2.1.2 | 2π = 2Β Β Β 3π + π = β5Β Β Β a + b + c = 1 | βπ = 1 β π = β8 β π = 22 β answer |
2.1.3 | T50Β =Β (50)2Β Β -Β 8(50) +Β 22 | β substitution β answer |
2.2.1 | a +14d = 74 | β setting up 2 equationsΒ β method β answerΒ (3) |
2.2.2 | a + 6(5) = 34 S40Β =Β 40Β [2Β (4) +Β (40 -1)(5)] | β value of a β substitution into correctΒ formula β answer (3) |
2.2.3 | Β | ββ answerΒ (accuracy as one unit) |
2.3.1 | TkΒ =Β 3Β k =Β 1Β Β x 3k -1 | β factors of 15Β β answerΒ (2) |
2.3.2 | Sn =Β a(rn - 1)Β Β Β Β Β Β Β r - 1 241/5 =1/5(3n - 1) Β Β Β Β Β Β Β Β Β 3-1 482/5 =Β 1/5(3n - 1) 242 = 3n -1 243 = 3n 35 = 3n n = 5 | β a and r β substitution into correct formula β exponential equationΒ Β β answerΒ (4) |
2.3.3 | NO | β NOΒ β reason (2) |
2.4 | P = 91/3 x 91/9 x 91/27 x ....to infinity Sβ =Β Β Β aΒ Β Β | β adding exponentsΒ β sum to infinity β answerΒ β P = 9Β½Β / β9 (4) |
Β | Β | Β [29] |
Β
QUESTION 3
3.1.1 | P (2; 4) | β coordinates of PΒ β coordinates of Q (2) | |
3.1.2 | y = a(x - 2)2 + 4 | β substitution β answer β substitution β answerΒ (4) | |
3.1.3 | xΒ β₯ 2 / xΒ β€ 2 | ββ answer / antwoord | |
3.1.4 | h(x) = 2 f(x) is a maximum when f (x) is a maximum | β max. value of f(x)Β βanswer (2) | |
3.2.1 | π¦ β₯ 1, π¦ β π Β ORΒ π¦ β (1; β) | β answer (1) | |
3.2.2 | p(x) = x2 +1Β Β /Β Β r(x) = x2 + 2x Shift 1 unit to the left and 2units downΒ OR Turning Point of p(x) = (0 ; 1) Shift 1 unit to the left and 2 units down | β calculation β 1 unit to the left β 2 units downΒ (3) | |
Β | Β | [14] |
Β
QUESTION 4Β
4.1 | y =Β -3Β Β + 5 | β y-intercept (1) |
4.2 | Β -3Β Β + 5 = 0 Β Β -3Β Β = -5 -5x - 5 = -3 | β simplificationΒ β answerΒ (2) |
4.3 | Β | β asymptotes β x- and y-intercepts x β shape (3) |
4.4 | Β | β substitution β simplification β reflection (3) |
Β | [9] |
Β
QUESTION 5
5.1 | f (x) = log3 x | β interchanging x and y β answer (2) |
5.2 | f -1 is a reflection of f in the line y = x | β answer (2) |
5.3 | y = log3 x =Β 1Β | β substitution β answer (2) |
5.4 | 0 < x < 1Β ORΒ log3 x < -2 and x > 0 | β βanswer (2) |
5.5 | x β₯ 1 | ββ answer (2) |
Β | [10] |
Β
QUESTION 6
6.1 |
| βformula β substitution β answer (3) |
6.2 | Β | βi and/en n β substitution β answer (3) |
6.3.1 | Β | β A and P β substitution β making n subject of the formulaΒ β answer (4) |
6.3.2 | Β | β substitution β answerΒ (2) |
6.3.3 | Amount = R457 551,55 β R50 710.00 | β answer (1) |
Β | Β | [13] |
Β
QUESTION 7
Β | Penalise 1 mark forΒ incorrect notation in the questionΒ | |
7.1 | π(π₯) = 1 β 3π₯2 π(π₯ + β)Β = 1 β 3(π₯ + β)2 = 1 β 3(π₯2 + 2π₯β + β2) π'(π₯) = limΒ Β π(π₯ + β) -Β π(π₯) = limΒ Β β6π₯β β 3β2 = β6π₯Β Β Answer ONLY: 0 marks | β 1 β 3π₯2 β 6π₯β β 3β2 β substitution β common factor β answer (4) |
7.2 | Β | β y = x + 2 + x -1 β1Β β βπ₯β2 (3) |
7.3 | y = 3x2 - 2x +1 6x - 2 = 4 | β y ' = 6x - 2 β 6x - 2 = 4 β x-coordinate β answer (4) |
Β | Β | Β [11] |
Β
QUESTION 8
8.1.1 | f (x) = a(x + 2)(x -Β 2/3 )(x - 3) f (x) = 3(x + 2) (Β x - 2/3 )(x - 3) | β substitution of x-coordinates β substitution of point β value of a β substitution β removing brackets (5) | |
8.1.2 | π(π₯) = 3π₯3 β 5π₯2 β 16π₯ + 12 πβ²(π₯)Β = 9π₯2 β 10π₯ β 16 = 0 (9π₯ + 8)(π₯ β 2) = 0 π¦ = 4900 (20,16)Β Β B(βΒ 8Β ; Β 20,16) | β πβ²(π₯)Β = 0 β factors β x-values β coordinates of P (4) | |
8.1.3 |
βy-intercept βturning pts β shape (4) | ||
8.2.1 | f (x) = ax3 + bx2 + 3x + 3 Β and g(x) = Β f n(x) = 12x + 4 f '(x) = 3ax2 + 2bx + 3 6a = 12Β and 2b = 4 | β a = 2 β b = 2 (2) | |
8.2.2 | 12x + 4 = 0 x =- 1 | β x = - 1 β answer (2) | |
8.2.3 | f nΒ (x) <Β 0Β for x <Β -Β 1Β Β Β Β (concavedown ) | β answer β answer (2) | |
Β | Β | [19] |
Β
QUESTION 9
9.1 | EF = a - 2x DE = x tan 600 Area = l x b No mark for the answer | β EF = (a β 2x) β tan ratio β answer β substitution (4) | |
9.2 | A(x) = β3ax - 2β3x2 A ' (x) = β3a - 4β3x = 0 x = aΒ | β derivative β f ' (x) =Β 0 β answer β substitution β answer (5) | |
Β | Β | [9] |
Β
QUESTION 10
10.1.1 | P(F and S) =Β 67Β / 0, 28 | β answer (1) |
10.1.2 | P(M ) Β΄ P(not S ) P(MΒ β© not S ) =Β 51Β | β P(M) x P(not S) β answer β answer β conclusion (4) |
10.2.1 | Β | β 1st branchΒ β 2nd branchΒ β outcomes (3) |
10.2.2 | Β | β setting up equation β standard form β answer (3) |
Β | Β | [11] |
TOTAL : 150