1.1.1

x² + 5x - 6 = 0
(x + 6)(x -1) = 0
x = -6 or x = 1

factors
x = -6   
x = 1  (3)

1.1.2

4x ² + 3x - 5 = 0
x = b ± √b ² - 4ac
             2a
x = - 3 ± √(3) ² - 4(4)(-5)
                 2(4)
x = -3 ± √89
          8
x = -1,55    or x = 0,8

substitution into the correct formula

 x = -1,55    
x = 0,8     (3)

1.1.3

4x ² -1 < 0
(2x +1)(2x -1) < 0
- 1 < x < 1 
 2           2

factors
method
answer 
(3)

1.1.4

(√√32 + x)( √√32 - x) = x
√32 - x² = x
32 - x² = x²
- 2x ² = -32
x ² = 16
x = ±4
x = 4

√32 - x²
squaring both sides
x² = 16
x = 4 (selection)   
(4)

y + x = 12
y = -x +12.......... (1)
xy = 14 - 3x........ (2)
Sub (1) into (2)
x(-x +12) = 14 - 3x
- x² +12x -14 + 3x = 0
- x ² +15x -14 = 0
x² -15x +14 = 0
(x -14)(x -1) = 0
x = 14    or    x = 1
y = -2  or     y = 11

OR

y + x = 12
x = - y +12.......... (1)
xy =14 - 3x......... (2)
Sub (1) into (2)
y(- y +12) = 14 - 3(- y +12)
12y - y ² -14 + 36 - 3y = 0
- y ² + 9y + 22 = 0
y ² - 9y - 22 = 0
( y + 2)( y -11) = 0
y = -2  or     y = 11
x = 14    or    x = 1

substitution
simplification

both values of x
both values of y  (5)

OR

x subject of the formula
substitution
simplification 
both values of y
both values of x (5)

identifying multiples of 3
ten multiples of 3
powers of 3
answer
(4)

2.1.1

209 ;   186

209   
186  (2)

2.1.2

2a = 2         3a + b = -31      a + b + c = 321
a = 1          3(1) + b = -31   1 + (-34) + c = 321
                       b = -34                   c = 354
T_n = n² - 34n + 354

2^nd diff = 2
a = 1
b = –34
c = 354
(4)

2.1.3

n² - 34n + 354 = 74
n² - 34n + 280 = 0
(n -14)(n - 20) = 0
n = 14     or      n = 20

equating T_n to 74
standard form
14
20  (4)

2.1.4

f ^/(n) = 0 2n - 34 = 0
2n = 34
n = 17
Term 17 will have the smallest value

OR
n = - b
      2a
n = 34 
       2
n = 17
Term 17 will have the smallest value

OR
n = 14 + 20 = 17
           2
Term 17 will have the smallest value

2n - 34 = 0
answer         (2)

OR

substitution
answer     (2)

OR

substitution
answer              (2)

a = ⁵/₈ ; r = ½;  n = 21
Sn = a(1 - r ⁿ )
            1 - r

= 1,2499...
= 1,25

r
substitution into the correct formula
answer
(3)

T_n >   5   
       8192
ar^n-1 >   5   
          8192

n -1 < 10    or   - n + 1 > -10
n < 11                  n < 11
n = 10                  n = 10
OR

8 ; 16 ; 32 ; … ; 8192
8.2^n-1 < 8192
2^n-1 < 1024
2^{n -1} < 2¹⁰
n -1 < 10
n < 11
n = 10

substitution into the correct formula
method /same base or log

calculating n

answer

(4)

OR

substitution into the correct formula

method

calculating n

answer    (4)

3.1

= 1 - ¹/₉ 
= ⁸/₉

answer                       (3)

3.2

= 52 m²                        = 52 m² 
    3                                  3
for both sides = 2 × 52 = 104 = 34,67m²
                                3       3

OR
 2  x (1+ 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 +10 +11+12) x 2
 9
= 34,67 m²

OR
T1 = 2 x 12  = 8                   l = 2 x 1 = 2 
               9      3                        9          9
2S₁₂ = 
= 34,67 m²

a
d
substitution into the correct formula
answer

answer for both sides

(6)

OR

a
(1 + …. + 12)
x 2
answer                     (6)

OR
a
T₁ = ⁸/₃  l = ²/₉
substitution into correct formula
answer                    (6)

[9]

4.1

p = -1

p = –1                     (1)

4.2

y =   a    
     x  - 1
- 3 =   a   
        0 -1
a = 3
y = x ² + bx - 3
0 = (1)²  + (1)b - 3
b = 2

coordinates D(0 ; –3)
substitute (0 ; –3)
substitute (1 ; 0)

(3)

4.3

y = x ² + 2x - 3
axis of sym: x = - b
                          2a
x = - 2 
     2(1)
x = -1
y = (-1)² + 2(-1) - 3 = -4
C(-1; - 4)

OR

dy = 0
dx
2x + 2 = 0
x = -1
y = (-1)² + 2(-1) - 3 = -4
C(-1; - 4)

substitution

x = -1

substitution

y = –4 
 (4) 

OR

derivative
x = -1
substitution
y = –4   
(4)

4.4

y ∈ [-4;∞)  or  y ≥ -4

- 4
answer                      (2)

4.5

m = tan45° = 1
y = mx + c
- 4 = (1)(-1) + c
c = -3
y = x - 3

gradient
subs m and (-1 ; - 4)
equation         (3)

4.6

No, the line passes through C and D

OR
No, a tangent through turning point C will have a gradient of 0

No
reason                      (2)

OR
No
reason                      (2)

f (m - x) = f [-(x - m)]
f is reflected in the y-axis and translated 1 unit to the left and 4 units upwards.
Therefore: m = -1
q = 4

OR
Substitute x = 0  and     q = 4 for one x- intercept
h(x) = (m - x)² + 2(m - x) - 3 + q
h(0) = (m - 0)² + 2(m - 0) - 3 + 4
0 = m² + 2m +1
0 = (m +1)²
m = -1
q = 4

value of m
value of q
OR
value of m
value of q
(4)

[19]

5.1

f (x) = k ^x
16 = k ⁴
k = 2

substitution (4 ; 16)
answer       (2)

5.2

f : y = 2 ^x
f ^-1 :       x = 2 ^y
y = log ₂ x

x = 2^y
y = log₂ x 
(2)

5.3

asymptote
shape
for any two valid points
eg.(16 ; 4) or
(2 ; 1) or (4 ; 2)
or (1 ; 0)
(4)

5.4.1

x ∈ (1 ; ∞) or   x > 1

1
answer
(2)

5.4.2

0 < x ≤ ½   or   x ∈ [0; ½]

½
answer
(2)

2^x - 2^-x = 15 
                4
2x - 1  = 15 
      2^x     4

2^2x -1 = 15  x 2^x
              4
4.2^2x - 4 = 15´ 2^x
4.2^2x -15.2^x - 4 = 0
(4.2^x +1)(2^x - 4) = 0
4.2^x +1 = 0 or 2^x - 4 = 0
2^x = -1 or 2^x = 2²
        4
N/A       x = 2

OR
2 ^x - 2^{- x} = 15 
                  4
2x - 1  = 15 
      2x     4
Let k = 2^x
k ² - 1 = 15 x k
             4
4.k ² - 4 = 15 x k
4.k ² -15.k - 4 = 0
(4.k +1)(k - 4) = 0
k = -1 or k = 4
      4
2 ^x = -1 or 2 ^x = 2²
         4
N/A     x = 2

2^x - 2^-x = 15 
                4

standard form
factors
answer
(4)

OR

2^x - 2^-x = 15 
                4
standard form
factors

answer
(4)
[16]

6.1

Kuda :     A = P(1+ in)

= 5 000(1+ 0,083´ 4)

= R6 660,00

Final Answer: R6 660,00 + R266,40

= R6 926,40

OR

Kuda :    A = P(1+ in) x 1,04

= 5 000(1+ 0,083´ 4) x 1,04

= R6 926,40

Thabo :     A = P(1 + i)ⁿ

= 5 000 (1 + 0.081 )^{12 x 4}
                       12
= R6 905,71

Kuda will have a better investment

substitution into the correct formula
final answer

OR

substitution into the correct formula

final answer

substitution into the correct formula

answer

conclusion 
 (5)

6.2.1

n = 157,40
n = 158 payments

OR

Number of payments = 13,11686841 x 12=157,40
n = 158 payments

0,1
12

substitution into the correct formula
simplification
use of logs

answer   (5) 

OR

substitution into the correct formula

simplification

0,1
12

substitution into the correct formula
simplification
use of logs
(5)

F = x[(1 + i)ⁿ - 1]
              i

= R162 503,51

OR

F = R881 818,77.....
Amount available for withdrawal
= R1 044 322,28 – R 881 818,77
= R162 503,51 

OR

Outstanding balance with monthly repayment of R5 066,35

Outstanding balance with monthly repayment of R6 000

Amount available for withdrawal
R404 666,23 – R242 162,72 = R162 512,18

R933,64
n = 108
substitution into the correct formula
answer      (4)

OR

n = 108
substitution
into correct formula
substitution into correct formula
final answer

OR

n = 108
substitution into the correct formula
substitution into the correct formula
final answer
(4)

7.1

f (x) = 4 - 7x
f '(x) = lim    f (x + h) - f (x)
           h→0         h

= lim   4 - 7(x + h) - (4 - 7x)
  h→0                h

= lim    h(-7) 
  h→0     h

= –7

4 - 7(x + h)

substitution

simplification

answer
(4)

7.2

y = 4x⁸ +  √x³
= 4x⁸ + x^3/₂

 dy = 32x⁷ +  3 x^½
dx                 2

x^3/₂

32x⁷

 3 x^½
 2

7.3.1

y = ax² + a
dy = 2ax + 0
dx
dy = 2ax
dx

2ax 
(1)

7.3.2

y = ax² + a
dy = x² + 1
da

answer    (2)

Substitute (2 ; b) in y = x + 12
                                            x
b = 2 + 12
             2
b = 8
m_tangent =dy 
                 dx

m_tangent =1 - 12  = -2
                      x²
m_perp = ½

Equation of perpendicular line:
y - y₁ = m(x - x₁)   OR     y = mx + c
y - 8 = ½ (x - 2)          8 = ½ (2)+ c
y = ½ x + 7                         c = 7
y = ½ x + 7 

8.1

36cm

answer     (1)

8.2

\t = 6        (-2t² + 3t - 6) have no real roots
Insect reaches the floor only once.

only once      (3)

8.3

h(t) = -2t³ +15t ² - 24t + 36
h¢(t) = -6t ² + 30t - 24
- 6t² + 30t - 24 = 0
t ² - 5t + 4 = 0 (t - 4)(t -1) = 0
t = 4     or     t = 1
Only   t = 4 because maximum value required
h = -2(4)³ +15(4)² - 24(4) + 36 = 52 cm

expansion
- 6t ² + 30t - 24 = 0 

both values
answer
(4)

[8]

9.1

f ^/(x) = 9x²
3x³ = 9x²
3x³ - 9x² = 0
3x²(x - 3) = 0
x = 0     or    x = 3

f ^/(x) = 9x²
x = 0
x = 3 
(3)

9.2.1

For f and f ^/

answer     (1)

9.2.2

The point (0 ; 0) is :
A point of inflection of  f
A turning point of f ^/

f : inflection point
f ^/ : turning point (2)

9.3

f ^// (x) = 18x
Distance = f ^//(1) - f ^/ (1)
= 18(1) - 9(1)²
= 9

f ^// (x) = 18x
substitution
answer    
(3)

9.4

3x³ - 9x² < 0
3x² (x - 3) < 0
but 3x² > 0

x - 3 < 0
x < 3 , x ≠ 0

3x³ - 9x² < 0
factors

x < 3
x ≠  0                     (4)

[13]

10.1

P(same day) = 4 or 1 
                        16    4
or 0,25 or 25%

4 numerator
16 denominator

(2)

10.2

P(2 consecutive days) = 3 x 2 = 3 
                                         16      8

3 x 2
answer

(3)

[5]

11.1.1

P( A) x P(B)       independent events
= 0,40 x 0,25 = 0,1

0,1

0,15 and 0,3

0,45     (3)

11.1.2

P(A or not B) = P(A) + P(not B)- P(A and not B)
= 0,4+ 0,75 - 0,3
= 0,85

OR

P(A or not B) = 1 - P(only B)
= 1 - 0,15
= 0,85

OR
From Venn diagram:
0,3 + 0,1 + 0,45 = 0,85

substitution
answer
(2)

OR

1 – 0,15
answer
(2)

OR
substitution
answer                 (2)

11.2

(5 × 1 × 5) + (5 × 1 × 6 )+ (5 × 1 × 6) + (5 × 1 × 5) =110
110 × 5 = 550 > 500
Not possible, because not enough space

OR

(5 × 2 × 5) + (5 × 2 × 6) = 110
110 × 5 = 550 > 500
Not possible because not enough space

OR

5 x 4 x 6 = 120
5 x 2 = 10
120 -10 = 110
110 × 5 = 550 > 500
Not possible because not enough space

5 × 1 × 5
5 × 1 × 6
5 × 1 × 6
5 × 1 × 5
110
conclusion         (6)

OR
5 × 2 × 5
5 × 2 × 6
110
conclusion          (6)

OR

5 x 4 x 6 = 120
5 x 2 = 10
120 -10
= 110
conclusion  (6)