1.1.1

x ² - 6x = 0
x(x - 6) = 0
x = 0 or x = 6

common factor
both answers
(2)

1.1.2

x² + 10x + 8 = 0
x = - b ± √b² - 4ac
             2a
= -10 ±√10² - 4(1)(8)
              2(1)
= -10 ± √68
         2
x = -0,88 or x = -9,12

substitution into the correct formula
- 0,88
- 9,12
(3)

1.1.3

(1- x)(x + 2) < 0
Critical values:
x = 1 or  x = - 2

critical values
correct method
answer
(3)

1.1.4

x +18 = x - 2
x +18 = x² - 4x + 4
0 = x² - 5x -14
(x - 7)(x + 2) = 0
x = 7 or  x ≠ -2

squaring both sides (m)
standard form
factors
both answers
rejection of x = - 2
(5)

x + y = 3
y = 3 - x........... (1)
2x² + 4xy - y = 15................ (2)
Substitute (1) into (2):
2x² + 4x(3 - x)- (3 - x) = 15
2x² +12x - 4x² - 3 + x -15 = 0
- 2x² +13x -18 = 0 2x² -13x +18 = 0
(2x - 9)(x - 2) = 0
x = ⁹/₂ or  x = 2
y = - ³/₂   or  y = 1

OR
x + y = 3
x = 3 - y........... (1)
2x² + 4xy - y = 15................ (2)
Substitute (1) into (2):
2(3 - y)² + 4(3 - y) y - y = 15
2y ² -12y + 18 - 4y ² + 12y - y -15 = 0
- 2y ² - y + 3 = 0
2 y ² + y - 3 = 0
(2y + 3)(y - 1) = 0
y = - ³/₂   or  y = 1
x =⁹/₂ or  x = 2

y subject of the formula
substitution
standard form
factors
x-values
y-values
(6)

OR
x subject of the formula
substitution
standard form
factors
y-values
x-values
(6)

n²⁰⁰ < 5³⁰⁰
(n²)¹⁰⁰  < (53 )¹⁰⁰
(n² )¹⁰⁰ < (125)¹⁰⁰
n² < 125
Maximum value of n is 11.

OR
200 log n < 300 log 5
n < 10^3/₂ log5
n < 11,18
n = 11

OR
n²⁰⁰ < 5³⁰⁰
(n²)¹⁰⁰  < (5³)¹⁰⁰
√n² < 5³ 
n < 5^3/₂ 
n < 11,18
n = 11

OR
n²⁰⁰ < 5³⁰⁰ 
n < 5³⁰⁰/₂₀₀
n < 11,18
n = 11

(3)

OR
use of logs
n < 11,18
11
(3)

OR
(n²)¹⁰⁰  < (5³)¹⁰⁰ 
n < 11,18
n = 11

OR
n < 5³⁰⁰/₂₀₀
n < 11,18
n = 11

2.1

7 ; x ; y ; –11 ; …
a = 7
a + 3d = –11 7 + 3d = –11
d = –6
x = a + d = 7 + (–6) = 1
y = a + 2d = 7 + 2(–6) = –5 

OR
a + 3d = –11
3d = -11- 7
3d = -18 d = -6 x = 1
y = -5

OR
x - 7 = y - x  and  y - x = -11 - y
2x - 7 = y ....(1)      2 y = -11 + x ...(2)
(1) into (2)
2(2x - 7) = -11 + x
4x - 14 = -11 + x
3x = 3
x = 1
y = 2(1) - 7 = -5

7 + 3d = –11
d = –6
value of x
value of y
(4)

OR
3d = -11- 7
d = -6
x = 1
y = -5
(4)

OR
2 equations
substitution
value of x
value of y
(4)

2.2.1

-3 ; 6 ; 27 ; 60 ; …

2a = 12
a = 6 3a + b = 9 3(6) + b = 9
b = –9
a + b + c = –3 6 – 9 + c = –3
c = 0
T_n = 6n² - 9n

second difference
a = 6
b = –9
c = 0
(4)

T₅₀    = 6(50)²  - 9(50)
= 14 550
Answer Only: Full Marks

substitute 50
answer
(2)

9 ; 21 ; 33 ; …
a = 9
d = 12
S_n = ⁿ/₂[2a + (n -1)d ]
S_n = ⁿ/₂ [2(9)+ (n -1)(12)]
= ⁿ/₂n [18 + 12n -12] 
= ⁿ/₂[12n + 6] 
= 6n² + 3n

a and d
substitution into the correct formula
ⁿ/₂[12n + 6]  
(3)

- 3 + S_n = 21060
S_n = 21063
6n² + 3n = 21063
6n² + 3n - 21063 = 0
2n² + n - 7021 = 0
n = - b ± √b² - 4ac
              2a
n = -1 ±   (1)² - 4(1)(-7021)
                   2(2)
n = 59 or n ≠ -119
                       2
n = 59

OR
T_n = 21060
6n² - 9n - 21060 = 0
2n ² - 3n - 7020 = 0
n = 60
59 first differences must be added

(4)

equation
standard form
answer
(4)

3.1

r = 4  = 1 
    12    3
-1 < ¹/₃ < 1
series is convergent ( -1 < r < 1)

12 + 4 + ⁴/₃ +... or 36(¹/₃)
value of r
-1 < r < 1
(3)

3.2

a = 4.3^2-p
r = ¹/₃
S_∞ =   a   
        1 - r
 2  = 4.3^2-p 
 9     1 - ¹/₃
4.3^{2 - p} = 4  
              27
3^{2- p} = 3^-3
2 - p = -3
p = 5

expression for a
substitution of a, r and S_∞
simplification 4.3^{2 - p} = 4  
                                    27
3^{2- p} = 3^-3
answer
(5)

[8]

y = mx + c
2 = -1+ c
c = 3
y = -x + 3

or
y - y₁ = m(x - x₁ )
y - 2 = -1(x -1)
y - 2 = -x +1
y = -x + 3

or
y = -(x - p) + q

= -(x -1) + 2
y = -x + 3

method
correct substitution
answer

or
method
correct substitution
answer

or
method
correct substitution
answer

f (x) =½(x + 5)² - 8
f (x) = ½(x² +10x + 25) - 8
f (x) = ½x² + 5x + 4,5
f ^/ (x) = x + 5
h(x) = 2x - 9 + k
x + 5 = 2
x = -3     y = -6
(-3 ; - 6)

OR
f (x) = h(x)
½(x + 5)² - 8 = 2x - 9 + k
½x² + 3x + ²⁷/₂ - k = 0
x = - 3  = -3  b2 - 4ac = 0
     2(½)
y = -6
(-3 ; - 6)

f ^/ (x)
x + 5 = 2
x = -3   y = -6
(4)

OR
equating 
turning point / △ = 0
x = -3   y = -6 (4)

5.1

A(0 ; 1)

answer (1)

5.2

9 = 3^-x
3² = 3^{- x}
x = -2
B(–2 ; 9)

equating
3² = 3^-x
x = -2
(3)

5.3

x ∈(0;∞) or x > 0

answer (2)

5.4

h(x) = 27.3 ^-x
h(x) = 3-(x - 3)
f shifted 3 units to the right

h(x) = 3^-(x-3)
3 units
right
(3)

5.5

27 < 1                       3^-x+3 < 1
3^x
3^x > 27        or         3- x+3 < 30
3^x > 3³                   - x + 3 < 0
x > 3                           x > 3

OR
The graph shifts 3 units to the right
Thus the y-intercept shift 3 units to the right (3 ; 1)
x > 3

3^x > 27 or     3^-x+3 < 3⁰
3^x > 3³or      - x + 3 < 0
x > 3 (3)

OR
translation
y-intercept
answer 
(3)

[12]

6.1.1

n = 145
i = 0,075
       12
substitution into the correct formula
answer (4)

6.1.2

A = P(1+ i)ⁿ
= 234 888,53(1 + 0.075 )¹²
                               12
= R253 123,54

substitution into the correct formula
answer
(2)

6.2

A = P(1- i)ⁿ
92 537,64 = 250 000(1- 0,22)ⁿ
0,37015056 = (0,78)ⁿ
n = log 0,37015056
             log 0,78
n = 4 years

substitution into the correct formula
correct use of logs
answer
(3)

6.3.1

72
substitution into the correct formula
answer
(3)

6.3.2

Amount paid: R1 500 x 60 = R90 000
Interest
= Amount paid – [Loan – Balance]
= R90  000 – [ R78 173,49323 – R16  945,00629]
= R28 771,51

OR
Balance

Balance = R16 945.00
Amount paid: R1 500 ´ 60 = R90 000
Interest
= Amount paid – [Loan – Balance]
= R90  000 – [ R78 173,49323 – R16  945,00629]
= R28 771,51

substitution (A)
R16 945,00629 (A)
R90 000 – [Loan – Balance]
answer
(4)

OR
substitution
R16 945,00629
R90 000 – [Loan – Balance]
answer
(4)

7.1

f (x) = 2x² - 1
f (x + h) = 2(x + h)²  - 1
= 2(x² + 2xh + h² )²  -1
= 2x² + 4xh + 2h² -1
f (x + h)- f (x) = 2x² + 4xh + 2h² -1- (2x² -1)
= 2x² + 4xh + 2h² -1- 2x²+ 1
= 4xh + 2h²
f ^/(x) = lim f (x + h)- f (x)
           h→0       h
= lim =  4xh + 2h²
  h→0          h
= lim h(4x + 2h)
  h→0     h
= lim (4x + 2h)
  h→0
= 4x

2x² + 4xh + 2h² -1
4xh + 2h²
substitution
simplification
answer
(5)

7.2.1

 d  (⁵√x² + x³)
dx 
 d  (x^2/₅ + x³)
dx 
 d  = ²/₅ x^3/₅ + 3x²
dx

x^2/₅
2/₅ x^3/₅ 
3x²
(3)

7.2.2

f (x)=  4x² - 9
           4x + 6
= (2x - 3)(2x + 3)
         2(2x + 3)
= 2x - 3
      2
= x - ³/₂

f ^/ (x) = 1

(2x - 3)(2x + 3)
2(2x + 3)
simplification to two separate terms
answer
(4)

[12]

8.1

-1 < x < 2

answer (2)

8.2

x = -1 + 2
          2  Answer Only:Full Marks
x = ½

method
answer
(2)

8.3

From the graph x >  ½  
Answer Only: Full Marks

answer
(2)

8.4

g (x) = ax³ + bx² + cx
g ^/ (x) = 3ax² + 2bx + c = -6x² + 6x +12
3a = – 6.    2b = 6   c = 12
a = –2         b = 3
g (x) = -2x³ + 3x² +12x

g ^/ (x) = 3ax² + 2bx + c
a = –2
b = 3
g (x) = -2x³ + 3x² +12x
(4)

8.5

g ^/(½) = 6(½)² + 6(½) + 12
m = ²⁷/₂  or 13,5
y = - 2(½)³ + 3(½)² + 12(½)
y = ¹³/₂ or  6,5
y - y₁ = m(x - x₁ )
y - 6,5 = 13,5(x - 0,5)
y = 13,5x - 0,25

max gradient at x =½
answer
y value
substitution
answer
(5)

[15]

9.1

Total surface area  = 2𝑙w + 2wh + 2𝑙h
but: 𝑙 = 3w
Total surface area = 6w² + 2wh + 6wh
C = 15(6w² )+ 6(2wh + 6wh)
= 15(6w² )+ 6(8wh)
= 90w² + 48wh

2𝑙w + 2wh + 2𝑙h
𝑙 = 3w
15(6w² )
6(2wh + 6wh)
(4)

9.2

5 = 3w²h
h =  5  
     3w²
C = 90w² + 48wh
C(w) = 90w² + 48(  5  )
                             3w²
= 90w² + 80w^-1
C ^/(w) = 180w - 80w^-2
180w - 80w^-2 = 0
180w³ - 80 = 0
w³ = 80 
       180
w =³√ 80 
         180
w = 0,76

h =  5  
     3w²
substitution
C(w) = 90w² + 80w^-1
derivative
equating derivative to zero
value of w
(6)

[10]

10.1

10¹⁰ or 10 000 000 000

answer (2)

10.2.1

8 × 10 × 10     x    8 × 8 × 10    x   2 × 10 × 10 × 10
    Area                    exchange            number
No. of valid 10-digit numbers
= (8 × 10 × 10 )x (8 × 8 × 10 ) x (2 × 10 × 10 × 10)
= 1,024 × 10⁹

8 × 10 × 10
or
8 × 8 × 10
2 × 10 × 10 × 10
1,024 × 10⁹ (A) 
(3)

10.2.2

Probability = 1,024 x 10⁹
                          10¹⁰
=  64  = 0,1024 = 10,24%
   625

1,024 x 10⁹
    10¹⁰
answer (2)

[7]

P (Bull’s eye first shot and second shot)
= 0,5 × 0,5
= 0,25 or ¼

two 0,5’s
0,5 × 0,5

11.2

P (Bull’s eye at least twice in 3 shots)
= (0,5 × 0,5 × 0,5) + (0,5 × 0,5× 0,5) + (0,5 × 0,5× 0,5) + (0,5 × 0,5 × 0,5)
= 0,125 + 0,125 + 0,125 + 0,125
= 0,5 or ½

0,5 × 0,5× 0,5
four events
answer (A)

11.3

Person shoots first:
(0,5) + (0,5)³ + (0,5)⁵ + ...
P =  a  
      1- r
P =   0,5   
     1- 0,25
P = ²/₃ = 0,67

(0,5) + (0,5)³
… + (0,5)⁵ + ...
P =   0,5   
     1- 0,25

[8]