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 

1.1.2

NOTE: Penalise 1 mark if the  rounding to TWO decimal  places is incorrect.
1.1.2 b
x = 0,14 or x = -3,64 

✔subs into correct formula
1.1.2 c
✔x = 0,14 
✔x = −3,64  

OR
✔for adding 49/16 on  both sides 
1.1.2 c
✔x = 0,14 
✔x = −3,64  (4)

1.2

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

1.3
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 1.4 a

1.3 b

✔ 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 1.5.2

✔ √3x −5 = x −3 
✔ 3x − 5 = (x − 3)2
✔ factors 
✔x = 7  (4) 

[26]

QUESTION 2

2.1.1 

2.1

✔ r = 1/3
✔substitution into correct   formula 
2.1.1 B
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

2.2

✔expandingn S
✔reverse writing 
2.2 B(4) 
✔expandingn Sn
✔reverse writing 
2.2 C(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

3.2 3.3 and
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 −Tn−1 = 3n - 4
2a = 3     and b - a = -4 
a = 3/   and    b = − 5/2
121a + 11b + c =125 
121(3/2) + 11(-5/2) + c = 125
✔c = −29 
tn2

✔✔ 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

4.5 a

OR

4.5 b

✔ 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

5.4 actual

 ✔1< x 
✔x < 3 (2) 

OR

✔ 1 < x 
✔ x < 3 (2) 

[8]

QUESTION 6

6.1

q = 1

✔q =1 (1)

6.2

6.2

✔ 0 =      a      +  1
            0 + p 
✔a = −p 
✔ substitution 
✔ a = 2 
✔p = −2  (5)

6.3

6.3

✔y = 1 
✔x = 2  
✔shape 
✔(0 ; 0)  
 (4) 

[10]

QUESTION 7

7.1

7.1

✔ n = 60 and i =  0,06 / 0,005
                             12 
✔ C
orrect 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

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  7.2.3

✔ n = - 083826912
✔ substitute into correct formula
✔ answer
✔ 2652,00 ( 1 + 0,15 / 12 ) 1
✔ answer 

OR

7.2.3 an (5)

[16]

QUESTION 8
8
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

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

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  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

Last modified on Friday, 13 August 2021 10:35