MATHEMATICS PAPER 1 GRADE 12 NATIONAL SENIOR CERTIFICATE MEMORANDUM MAY/JUNE 2019
NOTE:
If a candidate answers a question TWICE, only mark the FIRST Consistent Accuracy applies in all aspects of the marking QUESTION 1
1.1.1
x 2 - 5x - 6 = 0 (x - 6)(x +1) = 0x = 6 or x = –1
OR x 2 - 5x - 6 = 0x = 5 ± √(-5)2 - 4(1)(-6) 2(1)x = 5 ± √49 2x = 6 or x = –1
?factors ?both answers (2)OR ?correct subst into correct formula ?both answers (2) 1.1.2
(3x -1)(x - 4) = 16 3x 2 -13x -12 = 0x = 13 ± √(-13)2 - 4(3)(-12) 2(3)x = 13 ± √313 6x = 5,12 or x = -0,78
OR x 2 = 3x 2 -13x -12 = 0 - 13 x = 4 3x 2 = 13x + (13 )2 = 4 + (13 )2 3 6 6 (x - 13 )2 = 313 6 36 x = 13 ± √313 6x = 5,12 or x = -0,78
?standard form ?correct subst into correct formula ??answers (4)
OR ?standard form ?adding (13 )2 both sides 6 ??answers (4)
1.1.3 x (4 - x ) ≥ 0 - x (x - 4) ≥ 0 or - x (x - 4) ≥ 0x (x - 4) ≤ 0
0 ≤ x ≤ 4 or x ∈ [0 ; 4]
factorisation 0 ≤ x ≤ 4 (3)
1.1.4 52x - 1 =4 5x + 1 (5x + 1)(5x - 1) = 4 5x + 1 5x - 1 = 4 5x = 5x = 1
OR 52x - 1 = 4 5x + 1 52x - 1 = 4.5x + 4 52x - 4.5x - 5 = 0 (5x - 5)(5x + 1) = 0 5x = 5 or 5x ≠ -1x = 1
?factors in numerator ? 5x - 1 = 4 ?answerOR
?standard form ?factors ?answer (3)
1.2 x = 2 - 3y................................................ (1)x 2 + 4xy - 5 = 0..................................... (2) Substitute (1) in (2): (2 - 3y )2 + 4 y (2 - 3y ) - 5 = 0 4 -12 y + 9 y 2 + 8 y -12 y 2 - 5 = 0 - 3y 2 - 4 y -1 = 0 3y 2 + 4 y +1 = 0 (3y + 1)( y + 1) = 0y = - 1 or y = -1 3x = 3 or x = 5OR y = 2 - x. ................................................ (1) 3 3x 2 + 4xy - 5 = 0..................................... (2) Substitute (1) in (2):x 2 + 4x ( 2 - x ) - 5 = 0 3 3 3x 2 + 8x - 4x 2 - 15 = 0 - x 2 + 8x -15 = 0x 2 - 8x + 15 = 0 (x - 5)(x - 3) = 0x = 3 or x = 5y = - 1 or y = -1 3
? y = 2 - x 3 3
?correct subst into correct formula ?either standard form
?x – values?y – values (5)
OR
? y = 2 - x 3 3
?correct subst into correct formula ?either standard form
?x – values?y – values (5)
1.3 ab = 2√10bc = 3√2ac = 6√5ab .bc .ac = 2√10.6√5.3 √2 (abc )2 = 36√100abc =360 = 6√10
OR ac = 6√5 a = 6√5 c bc = 3 2 b = 3√2 c ab = 2√10 (6√5) (3√2 ) = 2√10 c c 18√10 = 2√10.c 2 c 2 = 9c = 3 Volume = abc = 2√10.3 = √360 = 6√10
? volume = abc ?? ab .bc .ac = 2√10.6√5.3√2 ? (abc )2 = 36√100 ? answer (5)
OR ? a = 6√5 c ? b = 3√2 c
? value of c ? Volume = abc ? answer (5)[22]
QUESTION 2
2.1.1
59
? answer (1)
2.1.2
2a = -2 a =-1 3(- 1) + b = 14b = 17 (- 1) + (17) + c = 15c = -1T n = -n 2 + 17n - 1
?second difference of – 2?a ?b ? c (4)
2.1.3
T27 = -(27)2 + 17(27) - 1 = -271
? substitution ? answer (2)
2.2.1
r = - 18 = - 1 36 2
answer (1)
2.2.2
T n = 36 (-½)n-1 9 = 36(-½)n-1 4096 1 = (-½)n-1 16384 (-½)14 = (-½)n-1
n = 15
OR 36 ; - 18 ; 9 ; - 9 ; 9 ; - 9 ; ... ; 9 2 4 8 4096 If you look only at the denominator: 2 ; 4 ; 8 ; … ; 4096
2k = 4096 2k = 212 k = 12n = 15 terms
? T n = 36 (-½)n-1 ? 1 = (-½)n-1 16384 ? answer (3)
OR
? 2k = 4096
? k = 12
? answer
(3)
2.2.3 S = a 1 - r = 36 1 - (-½) = 24
?correct subst into correct formula with – 1 < r < 1 ?answer if – 1 < r < 1 (2)
2.2.4 S250even = -18(( ¼) 250 -1) ¼ - 1 = -24 S250sold = -36(( ¼) 250 -1) ¼ - 1 = 48Sodd = 48 Seven -24 = -2T1 + T3 + T5 + T7 + ... + T499 T2 + T4 + T6 + T8 + ... + T500 = a + ar 2 + ar 4 + ... + ar 498 ar + ar 3 + ar 5 + ... + ar 499 = a + ar 2 + ar 4 + ... + ar 498 r (a + ar 2 + ar 4 + ... + ar 498 ) = 1 r = -2
?r = ¼ and n = 250? S 250 even = -24 ?S 250 odd = 48 answer (4)
?a + ar 2 + ar 4 + ... + ar 498 ?ar + ar 3 + ar 5 + ... + ar 499 ?r (a + ar 2 + ar 4 + ... + ar 498 ) ?answer (4)[17]
QUESTION 3
3.1.1
p + 6 – (2p + 3) = p – 2 – (p + 6) – p + 3 = – 8p = 11
?equating i.t.o p ?simplifying (2)
3.1.2
T n = 25 + (n -1)(-8) = 33 - 8n 33 - 8n < -55 - 8n < -88n > 11 Term 12 will be the first term smaller than –55
?subst into Tn formula?n > 11?n = 12 (3)
3.2
S = n [a + l ] = 6 [(x - 3) + (x -18)]6 2 2 = 6x - 63S = n [a + l ] = 9 [(x - 3) + (x - 27)]9 2 2 9x -135 6x - 63 = 9x -135 3x = 72x = 24S15 = n [a + l ] = 15 [(x - 3) + (x - 45)] 2 2 = 15 [2x - 48] 2 = 15 [2(24) - 48]= 0 = RHS 2
OR
= 21+18 +15 +.....+ -21.
Sn = n [a + l )] 2 = 15 [21- 21] 2 = 0 = RHS
OR (x - 3) + (x - 6) + (x - 9) + (x -12) + (x -15) + (x -18) = (x - 3) + (x - 6) + (x - 9) + (x -12) + (x -15) + (x -18) + (x - 21) + (x - 24) + (x - 27) 3x - 72 = 0 3x = 72x = 24
= 21 + 18 + 15 +.....+ -21.S n = n [a + l ] 2 = 15 [21 - 21] 2 = 0 = RHS
?6x - 63 ?9x -135 ?24 ?15 [(x - 3) + (x - 45)] 2 ?substitution of x (5)
OR ?expansion ?3x - 72 = 0 ?24 ?substitution of x ?sum of 15 terms (5)
OR
?expansion ?3x - 72 = 0 ?24 ?substitution of x ?sum of 15 terms (5)
[10]
QUESTION 4
4.1
y > 0OR y ∈ (0 ; ∞)
?answer (1)OR ?answer (1)
4.2
g;y = (½)x g-1 ; x = (½)y y = log½ x or y = - log2 x or y = log2 1 /x
?x = (½)y ?equation (2)
4.3
Yes. The vertical line test cuts g –1 onceOR Yes. For every x -value there is a unique y -valueOR Yes. g is a one-to-one functionOR Yes. The horizontal line cuts g only once
?yes ?valid reason (2) OR ?yes ?valid reason (2) OR ?yes ?valid reason (2) OR ?yes ?valid reason (2)
4.4.1 y = -log2 x 2 = -log2 a a = 2-2 = ¼ or a = (½)2 = ¼
?correct subst into correct formula (a ; 2) ?answer (2)
4.4.2 M/ (2;¼) or M(2;a)
? answer (1)
4.5 M// (-1; 9 /4 )
?-19 /4 ?? (3)[11]
QUESTION 5
5.1.1
x = -2y = 3
?answer ?answer
(2)
5.1.2
(0;7 /2 ) ?answer (1)
5.1.3
1 + 3 = 0x + 2 1 + 3(x + 2) = 0 3x = -7x = - 7 3x -intercept (-7 /3 ; 0)
?y = 0 ?answer (2) 5.1.4
?asymptotes at y = 3 and x = – 2 ?intercepts at y = 3,5 and x = – 2,3 ?shape (reasonable representation in correct quadrants) (3)
5.2.1 - 2x + 4 = 0 2x = 4x = 2 S(2 ; 0)
?y = 0
?x = 2 (2)
5.2.2 Equation of k :y = a (x +1)2 +18 0 = a (2 +1)2 +18 or 0 = a (- 4 + 1)2 + 18 9a = -18a = -2y = -2(x +1)2 +18
? y = a (x +1)2 +18 ?substitute (2 ; 0) or (– 4 ; 0) ?a (3)
5.2.3 - 2x 2 - 4x +16 = -2x + 4 - 2x 2 - 2x +12 = 0x 2 + x - 6 = 0 (x + 3)(x - 2) = 0x = -3 or x = 2y = -2(-3) + 4 = 10 T(–3 ; 10) ?equating ?standard form ?factors ?choosing x = –3 ?answer (5)
5.2.4 x < -3 or x > 2OR (-∞ ; - 3) ∪ (2 ; ∞) ?answerOR ?answer 5.2.5(a) x < -1OR (-∞ ; - 1)
?answerOR ?answer 5.2.5(b) ? shape of cubic with local min tp moving to local max tp ? turning points at x = 2 and x = –4 ? point of inflection atx = –1 (3)[25]
QUESTION 6
6.1.1
A = P (1 - i )n 79866,96 = 180 000(1 - 0,15)n (1 - 0,15)n = 79866,96 180 000
n = 4,999… yearsn » 5 years
?substitution ?use of logs ?answer (3)
6.1.2
A = P (1 + i )n = 49 000(1 + 0.1 )20 4 = R80 292,21 The money will be enough to buy the car.
?values of i and n ?substitution ?conclusion (consistent with answer)(3) 6.2.1
P = R793 749,25OR Balance Outstanding
= 841 885,56 - 48 136,62 = R 793 748,94
?n = 234 ? i = 0,1025 12 ?substitution in present value formula ?answerOR ?n = 6 in both ? i = 0,1025 12 ? A – F R793 748,94
6.2.2 A = P(1 + i)n = 793749,25(1 + 0.1025)3 12 New instalment
?= 793749,25(1 + 0.1025)3 12 ?n = 231 ?substitution of new P ?substitution of n and i into formula ?answer (5)
[15]
QUESTION 7
7.1
f (x) = x2 + 2 f (x + h) = (x + h)2 + 2 = x2 + 2xh + h2 + 2 f (x + h) - f (x) = x2 + 2xh + h2 + 2 - (x2 + 2) = 2xh + h2 f / ( x) = lim f (x + h) - f (x) h→0 h =lim 2xh + h 2 h→0 h
= lim h(2x + h) h→0 h
= lim (2x + h) h→0 = 2x
OR
f / ( x) = lim f (x + h) - f (x) h→0 h = lim x + 2xh + h2 + 2 - (x2 + 2) h→0 h =lim 2xh + h 2 h→0 h
= lim h(2x + h) h→0 h
= lim (2x + h) h→0 = 2x
?x 2 + 2xh + h 2 + 2 2xh + h 2
?=lim 2xh + h 2 h→0 h
?= lim (2x + h) h→0 ?answer (4)
OR ?x 2 + 2xh + h 2 + 2 ?=lim 2xh + h 2 h→0 h ? = lim h(2x + h) h→0 h ?answer (4)
7.2.1 y = 4x 3 + 2x - 1 dy = 12x 2 - 2x - 2 dx ?+ 2x - 1 ?12x 2 ?- 2x - 2 (3) 7.2.2 7.3 Point of contact: (1 ; 5)m = 2y - y 1 = m (x - x 1 ) or y = 2x + c y - 5 = 2(x -1) 5 = 2 + c c = 3y = 2x + 3 y = 2x + 3 ?m = 2 ?substitution of (1 ; 5) ?answer (3)[14]
QUESTION 8
8.1
h (x ) = -2(x + 3 )(x -1)(x + 3) 2h (x ) = -(2x + 3)(x 2 + 2x - 3)h (x ) = -2x 3 - 7x 2 + 9
OR h (x ) = -(2x + 3)(x -1)(x + 3) h (x ) = -(2x + 3)(x 2 + 2x - 3) h (x ) = -2x 3 - 7x 2 + 9
?? – 2 (x + 3 )(x -1)(x + 3) 2 ? correct simplification (3)
OR
?? - (2x + 3)(x -1)(x + 3) ? correct simplification (3)
8.2
h / (x ) = -6x 2 -14x - 6x 2 -14x = 0 - 2x (3x + 7) = 0x = 0 or x = - 7 3
?first derivative ?= 0 ?both answers (3)
8.3
x < - 7 or x > 0 3OR
x ∈ ( - ∞ ; - 7 ) ∪ ( 0 ; ∞) 3
? ?answer (2)
OR ??answer (2)
8.4 y = 4x + 7 - 6x 2 -14x = 4 0 = 6x 2 +14x + 4 0 = 3x 2 + 7x + 2 0 = (3x +1)(x + 2)x = - 1 or x = -2 3
?y = 4x + 7?h / (x ) = 4 ? standard form ? both answers
(4)
[12]
QUESTION 9
9.1
Volume of Sphere = 4 π(8)3 or =2048π or = 2144,66 3 3
? answer (1)
9.2
r 2 + x 2 = 82 (Pythagoras)r 2 = 64 - x 2
? substitution or reason Pythagoras (1)
9.3
V cone = 1 /3 πr2 h =1 /3 π(64 - x 2 )(8 + x ) = π (512 + 64x - 8x 2 - x 3 ) 3dV = 64π - 16π x - 3πx 2 dx 3 3 3 0 = 64 -16x - 3x 2 0 = (8 - 3x )(x + 8)x = 8 x ≠ 8 3
?h = 8 + x ? 1 /3 π(64 - x 2 )(8 + x ) ?expansion ? dV = 64π - 16π x - 3πx 2 dx 3 3 3? x = 8 3 ? volume of the cone ? 8 or 0,3 27 (7)
[9]
QUESTION 10
10.1
P(One Red and One Blue) = P(Red, Blue) + P(Blue, Red)
=
? addition of products ? answer (4)
10.2.1
a = 0,48 x 250a = 120
? answer (1)
10.2.2
b = 150 P(S) × P(F) = 200 x 150 250 250 = 0,48 = P(S and F) These events are independent
? b ? P(S) × P(F) ? 200 and 150 250 250 ? conclusion (with realistic probabilities)
(4)
[9]
QUESTION 11
11.1
10 × 9 = 90
??10 × 9 (2)
11.2.1
10! =3 628 800
? 10! (1)
11.2.2
2! × 2! × 2! × 2! × 2! ×4! = 768
? 2! × 2! × 2! × 2! × 2! ? 4! ? 2! × 2! × 2! × 2! × 2!×4! or 768 (3)
[6]
TOTAL: 150