MATHEMATICS PAPER 1
GRADE 12
NATIONAL SENIOR CERTIFICATE EXAMINATIONS
MEMORANDUM
MAY/JUNE 2021

NOTE:

• If a candidate answers a question TWICE, only mark the FIRST attempt.
• Consistent Accuracy applies in all aspects of the marking memorandum.

QUESTION 1
1.1.1 x² - x - 20 = 0
(x - 5)(x + 4) = 0
x = 5 or x = – 4
factors
x = 5
x = –4 (3)

1.1.2 3x² - 2x - 6 = 0
x =-(-2) ± √(-2)² -4(3)(-6) 
                  2(3)
x = 1 ± √19
           3
x = –1,12 or x = 1,79
substitution
simplification
x = –1,12
x = 1,79 (4)

1.1.3 (x -1)² > 9
x² - 2x - 8 > 0
(x - 4)(x + 2) > 0
Critical values: x = 4 or x = - 2

x <-2 or x > 4

OR
(-∞;-2) or (4;∞)

OR
x - 1 > 3 or x -1 < -3
x > 4 or x < -2
standard form
critical values
x > 4 or x < -2 (4)

OR
(-∞;-2) or (4;∞)

OR
x - 1 > 3
x -1 < -3
x > 4 or x < -2 (4)

1.1.4 2√x + 6 + 2 = x
2√x + 6 = x - 2
4(x + 6) = (x - 2)²
4x + 24 = x² -  4x + 4
x² - 8x - 20 = 0
(x -10)(x + 2) = 0
x = 10 or x ≠ -2
isolating the surd
4x + 24 = x² - 4x + 4
x = 10
x ≠ -2
(4)

1.2 4 x = 2 - y . . . (1)
4x + y² = 8 . . . (2)
∴2 - y + y² = 8
y² - y - 6 = 0
( y - 3)( y + 2) = 0
y = 3 or y = –2
x = -¼ or x = 1

OR
y = -4x + 2 . . . (1)
4x + y 2 - 8 . . . (2)
4x + (- 4x + 2)² = 8
4x +16x² -16x + 4 - 8 = 0
16x² -12x - 4 = 0
4x² - 3x -1 = 0
(4x +1)(x -1) = 0
x = -¼ or x = 1
y = 3 or y = –2 
substitution
standard form
y-values
x-values
(5)

OR
y = -4x + 2
substitution
standard form
x-values
y-values
(5)

1.3 2^x x 3y = (2³ x 3)⁶
2^x x 3y = 2¹⁸ x 3⁶
2^x = 2¹⁸ and 3^y = 3⁶
x = 18 and y = 6
∴ x - y = 18 - 6
∴ x - y = 12

2¹⁸ or 3⁶
x = 18 and y = 6
answer (A)
(4)
[24]

QUESTION 2
2.1.1 

2a = – 8
a = – 4
3a + b = 28
b = 40
a + b + c = 72
c = 36
T_n = 4n² + 40n + 36
second differences = – 8
a = – 4
b = 40
c = 36
(4)

2.1.2
T₁₂ = 36 - 8n = 36 - 8(12) = -60
-4n² + 40n + 36 = -60
n² -10n - 24 = 0
(n - 12)(n + 2) = 0
n = 12
36 – 8n
– 60
standard form
factors
n = 12
(5)

2.1.3
T_n^/ = -8n + 40 = 0
n = 5
T_n = -4(5)² + 40(5) + 36 = 136 

OR
- b = 40 = 5
 2a   8
T_n = -4(5)² + 40(5) = 36
= 136
8n - 40
n = 5
136 (3)

OR
substitution
n = 5
136
(3)

2.1.4 Maximum value = 41
value (1)

2.2
2sin 3x -(-11) =15 - 2sin 3x
4sin 3x = 4
sin 3x = 1
3x = 90º
x = 30∘
equating
4sin 3x = 4
sin 3x = 1
answer
(4)
[17]

QUESTION 3
3.1.1
T_n = ar^n-1 = 2000(¹/₅)^n-1
(1)

3.1.2
T₇ = 2000(¹/₅)^7-1 =  16  
                               125
3.1.3
   16    = 2000(¹/₅)^n-1
15625 
      1       = (¹/₅)^n-1
1953125 
(¹/₅)⁹ = (¹/₅)^n-1
OR
n = 1 = log_¹/5      1      
                     1953125
n – 1 = 9
n = 10
equating
same base / use of log
answer (3)

3.2
S_∞ = 27 =   a   
                1 - r
S₃ = a(1 - r³) = 26
            1 - r
27(1 - r³) = 26 
                  27
1 - r³ = ²⁶/₂₇
r³ = 1 
      27
r = ¹/₃

OR
S_∞ = 27 =   a   
                1 - r
a = 27(1 - r)
But a + ar + ar² = 26
a(1+ r + r² ) = 26
27(1- r)(1+ r + r²) = 26
(1 - r)(1 + r + r²) = ²⁶/₂₇
r² + r + 1 - r³ - r² - r = ²⁶/₂₇
- r³ + 1 = ²⁶/₂₇
r³ = ¹/₂₇
r = ¹/₃

S_∞ = 27 =   a   
                1 - r
S₃ = a(1 - r³) = 26
            1 - r
substitution
r = ¹/₃
(4)

OR
a = 27(1 - r)
a + ar + ar² = 26
substitution
r = ¹/₃
[9]

QUESTION 4
4.1
x + 1 = -x - 7 
2x = - 8
x = -4
y = -3 
f (x) =    -2     - 3
           x + 4
p = 4 and q = -3

OR
p + q = 1 ……(1)
–p + q = –7
q = p – 7 …….(2)
subs. (2) into (1)
p + p – 7 = 1
2p = 8
p = 4
q = –3

p + q = 1
-q = p – 7
substitution
simplification

4.2
y =  -2    - 3
     x + 4
0 =   -2    - 3
     x + 4
-2 - 3(x + 4) = 0
-3x -14 = 0
x = - 14
        3

y = 0
x = - 14
        3
(2)

4.3

horizontal asymptote
vertical asymptote
y intercept
shape
(4)
[10]

QUESTION 5
5.1
-2x² + 4x + 16 = 0
x² - 2x - 8 = 0
(x - 4)(x + 2) = 0 
x = 4 or x = - 2 
A(-2 ;0) and B(4;0) (3)

factors
x = – 2 
x = 4

5.2
f (x) = -2x² + 4x + 16
-  b   = -   -4    = 1
  2a       -2(2)
f (1) = -2(1)² + 4(1) + 16 = 18
C(1;18) 
OR
f (x) = -2x² + 4x +16
f^/(x) = -4x + 4
-4x + 4 = 0
x = 1 
f (1)= -2(1)² + 4(1) + 16 = 18
C(1;18) (2)

5.3 y ≤ 18 
(1)
OR
y∈(-∞;18] 
(1)

5.4 TP (1 ; 18) for f
TP (2 ; 15) for h
p = -1
q = -3
TP for h at (2 ; 15)
p = -1
q = -3
(3)

Related Items

5.5
y = 2x + 4
x = 2y + 4
y = ½x - 2

swop x and y
y = ½x - 2
(2)

5.6
g(x) = 0 or g^-1(x) = 0
x = 4 or x = –2 (product 0 at x-intercepts)
x = 4
x = –2
(2)

5.7
-2x² + 4x + 16 + k = 2x + 4
-2x² + 2x + 12 + k = 0
b² - 4ac < 0
(2)² - 4(-2)(12 + k) < 0
4 + 8(12 + k) < 0
100 + 8k < 0
k < -12,5

OR
g^/ (x) =  2
f^/(x) = -4x + 4 = 2
x = ½
f(½) = 17,5
g(½) = 5
k < -12,5

equating
standard form
b² - 4ac < 0
substitution
answer
(5)

OR
g^/ (x) =  2
f^/(x) = -4x + 4
f(½) = 17,5
g(½) = 5
(5)
[18]

QUESTION 6
6.1.1
y = 3^x
x = 3^y
y = log₃ x
swop x and
equation
(2)

6.1.2
h(x) = 3^x-4 + 2
Transformation: 4 units left, 2 units down
P^/ (2;9)

x = 2 (A)
y = 9 (A)
(2)

6.2
f (x) = 2^{x + p} + q
q = -16
16 = 2^{p + 3} - 16
2^{p + 3}  = 32
2^{p + 3}  = 2⁵
p + 3 = 5
p = 2

q = -16
substitute (3 ; 16)
2^{p + 3}  = 2⁵ or p + 3 = log₂32
p = 2
(4)
[8]

QUESTION 7

7.1

substitution into correct formula
n = 16
simplification
answer (A)
(4)

7.2.1

0,075
   12
substitution into correct Formula
answer
(3)

7.2.2
60 x 9000 = R540 000
A = P(1 + i)ⁿ 
652743,95(1 +0.075)ⁿ = 190 214,14 + 540000
                          12
730 214,14 = 652 743, 95(1 +0.075)ⁿ
                                                 12
1,1186. = (1, 00625)ⁿ
n = log_1,00625 (1,1186)
n = 18 months

60 x 9000 = R540 000
equation
simplification
use of logs18 months (6)

OR
Interest over 5 years = 652 743,95 – 9 000 × 60
= 112 743,95
interest on n years
= 190 214,14 – 112 743, 95 = 77 470,19
652 743,95 + 77 470,19 = 652743,95(1 +0.075)ⁿ
                                                                   12
1,1186.... = (1, 00625)ⁿ
n = log_1,00625 (1,1186)
n = 18 months

OR
60 × 9 000
answer
equating
simplification
use of logs
18 months
(6)
[13]

QUESTION 8
8.1
f(x) = 3x²
f ^/ (x) =  lim f (x + h) - f (x)
            h→0      h
f ^/ (x) =  lim f 3(x + h)² - 3x²
            h→0           h
f ^/ (x) =  lim f 3x² + 6xh + 3h² - 3x²
            h→0               h

  lim  6xh + 3h²
h→0       h
lim  h(6x + 3h)
h→0       h
= 6x

substitution
expansion
simplification
 lim  h(6x + 3h)
h→0       h
= 6x
(5)

8.2.1
f (x) = x² - 3 + 9x^-2
f^/ (x) = 2x - 18x^-3 
(3)

8.2.2
g(x) = (√x + 3)(√x -1)
g(x) = x + 2x^½ -3
g^/ (x) = 1 + x^-½
(4)
[12]

QUESTION 9
9.1
f^/ (x) = 6x² + 6x -12 
6x² + 6x -12 = 0 
x² + x - 2 = 0
(x + 2)(x - 1) = 0
x = -2 or x = 1
y = 20 or y = – 7
A(–2 ; 20) and B(1 ; –7) x
(5)

9.2 
f^// (x)= 12x + 6 
12x + 6 > 0
12x > -6 
x > -½

OR
x =-2 + 1  = -½
        2
(3)

9.3
f^/ (2) = 24
Equation of the tangent:
y - 4 = 24(x - 2)
y = 24x - 44
(3)
[11]

QUESTION 10
10.1 

x = –1 and x = 2
TP at x = –1
TP at x = 1
shape
(4)

10.2.1

Area of segment = ¼ Area of big circle
= ¼ π (x - x² )²

Area triangle ABO counted
= Area Δ = ½ (x - x²)²
Area of shaded region
= ¼ π(x - x² )² – ½ (x - x² )²
=  π - 2 (x - x² )²
      4
=(π - 2)(x² - 2x³ + x⁴)
      4
(5)

10.2.2
Area of shaded region
= (π - 2)(x⁴ - 2x³ + x²)
      4
dA =(π - 2)(4x³ - 6x² + 2x)
dx       4
4x³ - 6x² + 2x = 0
x(2x² - 3x + 1) = 0
x(2x - 1)(x - 1) = 0
x ≠ 0 or x = ½ or x ≠ 1
[13]

QUESTION 11
11.1 P(A) = 1 – P(not A) = 0,6 
P(A and B) = P(A) × P(B)
= 0,6 × 0,3 
= 9 
  50
= 0,18
(3)

11.2.1
a =  15  = 0,1 
      150

11.2.2 m = 1 - 0,7 = 0,3  (1)

11.2.3 0,24 + 0,14 + 0,02 + 0,12 + 0,1 + 2b = 0,7 
2b = 0,08
b = 0,04
0,04 x 150 = 6

11.3.1 9 × 9 × 8 = 648 (2)

11.3.2 2 × 8 × 4 = 64
2 × 8 × 5 = 80
Total number = 64 + 80 = 144  (4)
[15]
TOTAL: 150