1.1.1

x² + 9x + 14 = 0
(x + 7)(x + 2) = 0
x = -7 or x = -2

✓ factors
✓ x = -7
✓ x = -2

(3)

1.1.2

4x² + 9x - 3 = 0

OR

✓ substitution

✓ simplification

✓ x = 0,29

✓ x = - 2,54

OR

✓ for adding 81 on both sides
                    64

✓ simplification

✓ x = 0,29

✓ x = - 2,54

(4)

1.1.3

√x ² - 5 = 2√x
x ² - 5 = 4x
x ² - 4x - 5 = 0
(x - 5)(x + 1) = 0
x = 5 or x = -1
x = 5

✓ x² - 5 = 4x
✓ standard form
✓ both answers
✓ select x = 5
(4)

1.2

3x - y = 4
y = 3x - 4
x ² + 2xy - y ² = -2
x ² + 2x(3x - 4) - (3x - 4)² = -2
x ² + 6x ² - 8x - (9x ² - 24x + 16)= -2
7x ² - 8x - 9x ² + 24x - 16 = -2
- 2x ² + 16x - 14 = 0
x ² - 8x + 7 = 0
(x - 7)(x - 1) = 0
x = 1   or   x = 7
y = 3(1) - 4      y = 3(7) - 4
y = -1   or    y = 17

OR

3x - y = 4
x = y + 4
         3
x² + 2xy - y² = -2
x ² + 2xy - y ² = -2

y ² + 8 y + 16 + 6 y ² + 24 y - 9 y ² = -18
- 2 y ² + 32 y + 34 = 0
y ² - 16 y - 17 = 0
(y - 17)(y + 1) = 0
y = -1        or        y = 17
x = - 1 + 4               x = 17 + 4
           3                             3
x = 1              or           x = 7

✓ y subject of formula

✓ substitution

✓ correct standard form

✓ factors

✓ x-values

✓ y-values

OR

✓ x subject of formula

✓ substitution

✓ correct standard form

✓ factors

✓ y-values

✓ x-values

(6)

1.3.1

x ² + 8x + 16 > 0
(x + 4)(x + 4) > 0
x ∈ R, x ≠ - 4     or
x ∈ (-∞; - 4) or x ∈ (- 4;∞) or
x < - 4 or x > - 4

OR

✓ (x + 4)(x + 4)
✓✓ any one of the solutions

OR

✓ (x + 4)(x + 4)
✓✓ any one of the solutions

(3)

1.3.2

For two negative unequal roots:

0 < p < 16

OR

x² + 8x + 16 = p
x² + 8x + 16 - p = 0
0 < 16 - p < 16
- 16 < - p < 0
0 < p < 16

OR
x ² + 8x +16 - p = 0

x = - 8 ± √64 - 4(16 - p)
                    2
0 < 64 - 4(16 - p) < 64
0 < 4 p < 64
0 < p < 16

OR
x ² + 8x +16 = p
x² + 8x +16 - p = 0
Roots are real and unequal :
8² - 4(16 - p) > 0
4 p > 0
p > 0
Roots are : - 8 ± √ 4 p
                         2

For both roots to be negative :
√4p < 8
4 p < 64
p < 16
0 < p < 16

✓0

✓ 16

✓✓ 0 < p < 16   (4)

OR

✓ 0

✓ 16

✓✓ 0 < p < 16               (4)

✓ 0

✓ 16

✓✓ 0 < p < 16               (4)

✓ 0

✓ 16

✓✓ 0 < p < 16              (4)

[24]

2.1.1

first differences: –9; –15; –21
second difference = – 6

✓ first differences

✓ – 6

(2)

2.1.2

T_n = an² + bn + c 

a = second difference = -3
                   2

3a + b = -9
3(- 3)+ b = -9
b = 0
a + b + c = 5
- 3 + 0 + c = 5
c = 8
T_n = -3n ² + 8

OR

(n - 1)(n - 2)d

T_n = T₁ + (n - 1)d₁ + (n - 1)(n - 2)d₂
                                            2
= 5 + (n - 1)(- 9) + (n - 1)(n - 2)(- 6)
                                          2
= 5 - 9n + 9 - 3n ² + 9n - 6
T = -3n ² + 8

✓ a = -3

✓ b = 0

✓ c = 8

✓ T_n = -3n² + 8

OR

✓ a = -3

✓ b = 0

✓ c = 8

✓ T_n = -3n² + 8

(4)

2.1.3

- 3n² + 8 = -25 939
- 3n² = -25947
n² = 8649
n = -93 or n = 93
The 93^rd term has a value of –25 939

✓ T_n = -25 939

✓ n² = 8649

✓ answer

(3)

2.2.1

2k - 7 ; k + 8 and 2k -1
k + 8 - (2k - 7) = 2k -1- (k + 8)
- k +15 = k - 9
2k = 24
k = 12
2k - 7; k + 8 and 2k - 1
17 ; 20 ; 23.......
d = 3
T₁₅ = 17 + 14(3)
= 59

✓ k + 8 - (2k - 7) = 2k -1- (k + 8)

✓ k = 12

✓ 17

✓ d = 3

✓ T₁₅ = 59

(5)

2.2.2

Sequence is 17 ; 20 ; 23 ; 26 ; 29 ; 32 …….
Every alternate term of the sequence will be even
20 + 26 + 32 + .......
S₃₀ = 30  [2(20) + (29)(6)]
           2
= 15[40 + 174]
= 3210

OR
T ₃₀ = 20 + 29(6)
= 94
S₃₀ = 30  (20 + 194)
           2
= 3210

✓ 20 + 26 + 32 + .......

✓ a = 20 d = 6

✓ subst into correct formula

✓ answer (4)

✓ a = 20 d = 6

✓ T₃₀ = 94

✓ S₃₀ = 30  (20 + 194)
           2
✓ answer (4)

[18]

a + ar = 2
a(1 + r ) = 2
a =    2   
      1 + r

OR

  a    - 2 = 1 
1- r           4
4a - 8(1- r ) = 1- r
4a - 8 + 8r = 1- r
4a = 9 - 9r
a = 9 - 9r
         4

OR
S_n = a(r ⁿ -1)
           r -1

2 = a(r ² -1)
          r -1

2 = a(r -1)(r +1)
           r -1 2
= a(r +1)
a =  2   
      r +1

OR

  ar   ² = 1 
1 - r        4
a = 1 - r
       4r ²

✓ a + ar = 2

✓ a =   2   
         1+ r
(2)

✓    a   - 2 = 1  
    1- r          4

✓ a = 9 - 9r
             4
(2)

OR

✓ 2 = a(r ² -1)
             r -1

✓ a =   2   
          1+ r
(2)

OR

✓ ar  ²  =  1 
   1- r        4

✓ a = 1- r
          4r ²
(2)

3.2

OR

OR

OR

✓ S_∞ = 2 + 1 
                  4
✓   a    = 9 
    1- r     4

✓ substitution of a into the correct formula

✓ 9r ² = 1
✓ r = 1 
         3
✓ a = 3 
         2

(6)

OR 

✓ S = 2 + 1 
            4
✓   a    = 9 
    1- r     4
✓ r = 9 - 4a
        9

✓ substitution of a into the correct formula
✓ a = 3 
          2
✓ r = 1 
         3
(6)

OR

✓ r = 2 - a
           a
✓   ar  ²    = 1
    1- r          4
✓ substitution of a

✓ (2a - 3)(a - 3) = 0

✓ a = 3
          2
✓ r = 1 
         3
(6)

OR

✓ S_∞ = 2 + 1 
              4
✓  a    = 9
   1- r     4
✓ substitution of a
✓ 9r ² = 1
✓ r = 1 
         3
✓ a = 3 
          2
(6)
[8]

4.1

f (x) = -ax² + bx + 6
f ^/ (x) = -2ax + b
- 2ax + b = 3
at x = -1
2a + b = 3               [1]

f (- 1) = 7 
            2
- a - b + 6 = 7 
                   2
- 2a - 2b + 12 = 7
2a + 2b = 5                [2]
[2]- [1]
b = 2
2a + 2 = 3
a = ½

OR
f ^/(x) = -2ax + b
3 = 2a + b
b = 3 - 2a
 7 = -a(-1)²  + (3 - 2a)(-1)+ 6
 2
a + 3 = 7 
            2
a = ½
b = 2

✓ – 2ax + b

✓✓ 2a + b = 3

✓ - a - b + 6 = 7 
                   2

✓ solve simultaneously 

(5)

✓ – 2ax + b
✓✓ 2a + b = 3
✓  7 = -a(-1)²  + (3 - 2a)(-1)+ 6
     2
✓ solve simultaneously

(5)

4.2

f (x) = - ½ x ² + 2x + 6
x - intercepts :
- ½ x ² + 2x + 6 = 0
-x ² + 4x +12 = 0
x ² - 4x -12 = 0
(x - 6)(x + 2) = 0
(- 2 ; 0)      (6 ; 0)

✓ - ½ x² + 2x + 6 = 0
✓ (- 2 ; 0)
✓ (6 ; 0)
(3)

4.3

OR

OR

4.4

4.6

4.4:
f:
✓ shape
✓ x- intercepts
✓ y- intercept
✓ (2 ; 8)
(4)

4.6:
g:
✓ x- intercept
✓ y- intercept
(2)

4.7

x ≤ -2 or - 1 ≤ x ≤ 6

OR/OF

(-∞ ; - 2] or [-1; 6]

✓ x ≤ -2

✓✓ - 1 ≤ x ≤ 6

(3)

[23]

5.1

y ∈ R ; y ≠ -1
OR
y < -1 or y > -1
OR
y ∈ (- ∞; - 1) or y ∈ (-1; ∞)
OR
R -{-1}

✓✓ answer (2)

5.2

D(2 ; - 1)
g (x) =  2   - 1
          x - 2

✓ D(2 ; - 1)
✓    2     - 1
    x - 2
(2)

f(x) = log₃x
log₃t=1
 t=3

OR
g (x) =  2   - 1
          x - 2
1 =    2   - 1
       t - 2
2 =    2  
       t - 2
2t - 4 = 2
t = 3

5.4

x = log₃ y
y = 3^x

✓ interchange x and y
✓ y = 3^x (2)

5.5

3^x < 3¹
x < 1

OR
3^x < 3¹
x ∈ (- ∞ ; 1)

✓ 3^x < 3¹
✓ x < 1
✓ 3^x < 3¹
✓ x ∈ (- ∞ ; 1)
(2)

5.6

Equation of the axis of symmetry: y = -x + 1
x-intercept of the axis of symmetry is at x = 1
f has an x-intercept at B(1 ; 0) which is the same as
the x-intercept of the axis of symmetry Point of intersection: B (1 ; 0)

OR
Since BE = ED = 1 and D lies on the axis of symmetry and the gradient of
the axis of symmetry is –1, B will also lie on the axis of symmetry. But B also
lies on f. Therefore B(1 ; 0) is the point of intersection between f and the axis
of symmetry with a negative gradient.

✓✓equation of axis of symmetry

✓ B or (1 ; 0)

OR

✓✓ BE = ED = 1
✓ B or (1 ; 0)

6.1

A = P(1 + i)n

✓  r 
   12
✓ n = 36

✓ correct substitution into formula

✓ 1 +r  = ³⁶√36 1,214672
        12
✓ 6,5%
(5)

6.2.1

6.2.2

Amount paid for the year : (5 536,95´12) = R 66 443,40
= 192 296,17

Interest = (5 536,95´12)- (235 000 -192 296,17)
= 66 443,40 - 42 703,83
= 23 739,57

OR

OR

✓ R 66 443,40

OR

✓ R 66 443,40
✓ n = – 42
✓ substitution into correct formula
✓ R192 296,20
✓ R42 703,80
✓ R23 739,60
(6)

OR
✓ R62 648,18
✓ R172 351,82
✓ R192 296,17
✓ R66 443,40

✓ 235 000 – 192 296,17
✓ R23 739,57
(6)

[15]

7.1  

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

OR

f / (x) = lim   f (x + h) - f (x)
           h→0         h
=  lim   2(x + h)² - (x + h) - (2x² - x) 
   h→0                h                                    

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

✓ 2x² + 4xh + 2h ² - x - h
✓ 4xh + 2h ² - h
✓ f^'(x) = lim f (x + h) - f (x)
   h→0               h
✓ subst. into formula
✓ lim (4x + 2h - 1)
   h→0
✓ 4x – 1

OR
✓ f^'(x) = lim f (x + h) - f (x)
   h→0               h
✓ subst. into formula
✓ 2 x ² + 4xh + 2h ² - x - h
✓ 4xh + 2h ² - h
✓ lim(4x + 2h -1)
   h→0
✓ 4x – 1
(6)

7.2.1

D_x[(x + 1)(3x - 7)]
= D_x(3x² - 4x - 7)
= 6x - 4

✓ 3x² - 4x - 7
✓ 6x - 4
(2)

7.2.2

[12]

8.1

Point of inflection at x = 2

✓ x³ - 6x² + 9x

✓ 3x ² -12x + 9

✓ 6x -12

✓ 6x -12 = 0

✓ explanation

(5)

8.2

✓ shape

✓ (0 ; 0)

✓ (3 ; 0) as TP

✓ (1 ; 4)

(4)

8.3

f concave up for x > 2
y = - f (x) will be concave down for x > 2

✓✓ x > 2

(2)

8.4.1

(3;7)

✓ 3

✓ 7                             (2)

8.4.2

Do not agree with Claire as her statement is incorrect. Between x = 1
and x = 3 the graph of f is decreasing. Therefore at x = 2 the gradient
will have a negative value.

OR
f ^/ (2) = 3(2)² - 12(2) + 9
= -3
≠1

✓ no

✓ justification

(2)

[15]

y = x² + 2
P(x ; x² + 2 )
B(0 ; 3)
PB² = (x - 0)²  + (x² + 2 - 3)²
= x² + x⁴ - 2x² + 1
= x⁴ - x² + 1
PB will be a minimum if PB² is a minimum
d (PB² ) = 4x³ - 2x 
    dx
4x³ - 2x = 0
x(2x² -1)= 0
x = 0 or x² = ½
x = 1 
     √2

OR
Gradient of tangent to curve = 2x
Gradient of line joining B and the curve = x² + 2 - 3
                                                                      x - 0
= x²  -1
      2
Shortest distance will be where tangent to curve is perpendicular
to the line joining P and the curve.
x² -1 = - 1 
  x         2x
2x(x² -1)= -x
2x³ - 2x = 0
x(2x² -1)= 0
x = 0   or  x² = ½
x = 1 
     √2

OR

✓ (x - 0)²  + (x² + 2 - 3)²
✓ x ⁴ - x² + 1
✓ 4x³ - 2x
✓ d (PB² ) = 0
        dx
✓ x = 1 
        √2 


answer

OR
✓ = 2x
✓ = x² -1 
          x 
✓ x² -1 = - 1 
      x         2x
✓ 2x³ - 2x = 0
✓ x = 1 
        √2

OR

[7]

10.1

8 values need to be placed in correct position:

2 or 3 correct: 1 mark
4 or 5 correct: 2 marks
6 or 7 correct: 3 marks
8 correct: 4 marks
(4)

10.2

(49 - x) + x + 8 + 4 + 5 + 2 + (60 - x) + 14 = 100
- x + 142 = 100
x = 42

✓ setting up equation
✓ answer

10.3

P (use only one application) = 7 + 2 + 18
                                                     100
=  27    or 27%
   100

✓ 7 + 2 + 18
         100
✓ answer (2)
[8]

11.1

5 x 5 x 10 x 9

= 2250

✓ 5 x 5
✓ 10 x 9
✓ 2250 (3)

11.2

Codes of two letters and five digits will ensure unique numbers for 700 000 clients.

✓ 5 x 5 x 10 x 9 x 8 x 7 x 6
✓✓ five digits(3)
[6]