1.1.1

3x² + 10x + 6 = 0

OR

x² + 10 x + 100 = -2 + 100
         3         36             36

✓substitution into correct formula

✓x = -2,55

✓x = - 0,78

(3)

✓for adding 100 on both sides
                    36
✓x = -2,55

✓x = - 0,78

(3)

1.1.2

√6x ² - 15 = x + 1

6x ² - 15 = (x + 1)²

6x² - 15 = x² + 2x + 1

5x² - 2x - 16 = 0

(5x + 8)(x - 2) = 0

x = - 8 or x = 2
        5

∴ x = 2

✓concept of squaring both sides

✓standard form (accurate)

✓factors

✓both answers

✓correct selection

(5)

1.1.3

x ² + 2x - 24 ≥ 0

(x + 6)(x - 4) ≥ 0

✓factors

✓✓x ≤ -6 or x ≥ 4

(3)

1.2

y = -5x + 3

3x ² - 2x(- 5x + 3) = (- 5x + 3)²  - 105

3x ² + 10x ² - 6x = 25x ² - 30x + 9 - 105

- 12x ² + 24x + 96 = 0

x ² - 2x - 8 = 0

(x - 4)(x + 2) = 0

x = -2 or x = 4

y = 13 or y = -17

OR

x = 3 - y
        5

✓y subject of formula

✓substitution

✓simplification

✓factors

✓values of x

✓values of y

(6)

✓x subject of formula

✓substitution

✓simplification

✓factors

✓values of y

✓values of x

(6)

1.3.1

p² - 48 p - 49 = 0

(p - 49)(p + 1) = 0

p = -1 or   p = 49

✓factors

✓p = -1

✓p = 49

(3)

1.3.2

7^x= -1           or          7^x= 49

no solution                   x = 2

✓7 ^x= -1 or 7 ^x= 49

✓no solution

✓x = 2

(3)

[23]

2.1.1

3; 2; k; ...

r = 2
     3

✓r = 2 / 0,67
         3
(1)

2.1.2

r = T₃
     T₂

T3 = r x T₂

= 2 x 2
   3

= 4
   3

Thus k = 4
               3

✓2 x 2
  3

✓ 4   / 1,34
   3

(2)

2.1.3

OR

OR

2.2.1

T_n = a + (n -1)d

T₁₈ = 100 + (18 -1)(150)
= R 2 650

✓substitution of n, a and d into AS

✓ 2 650

2.2.2

S_n = n [2a + (n - 1)d ]
        2

30 500 = n [2(100) + (n - 1)(150)]
               2

61 000 = n(150n + 50)

61 000 = 150n² + 50n

3n² + n - 1 220 = 0

(3n + 61)(n - 20) = 0

n = - 61 or n = 20
         3
N/A

x = 100 + (20 -1)(150)

= R 2 950

✓ substitute 30 500, a and d into sum formula for AS

✓ simplification

✓ factors or quad formula

✓ n = 20

✓ substitution Tn of AS

✓ 2 950

(6)

[15]

3.1

First differences:   17; 15

Second difference: –2

T_n = an² + bn + c

a = second difference = - 2 = -1
                    2                   2

3a + b = 17

3(-1)+ b = 17

b = 20

a + b + c = 0

-1+ 20 + c = 0

c = -19

T_n= -n² + 20n -19

OR

First differences:   17; 15

T_n = T₁ + (n - 1)d₁ + (n - 1)(n - 2) d₂
                                           2            

= (0) + (n - 1)(17) + (n - 1)(n - 2) (- 2)
                                        2

= 17n - 17 - n² + 3n - 2

= -n² + 20n - 19

✓17; 15

✓value of a

✓value of b

✓value of c

(4)

✓17; 15

✓value of a

✓value of b

✓value of c

(4)

3.2

56 = -n² + 20n - 19

n² - 20n + 75 = 0

(n - 15)(n - 5) = 0

n = 5 or n = 15

✓T_n = 56

✓factors

✓both answers

(3)

3.3

OR

✓✓symmetry of terms

✓T10

✓81

(4)

✓writing out the symmetry of terms

✓56 + 65 + 72 + 77 + 80 + 81

✓80 + 77 + 72 + 65 + 56

✓81

(4)

[11]

4.1

A (4; 3)

✓(4; 3) (1)

4.2

✓x = 0

✓y = 3
        2
(2)

4.3

0 =     6     +3
         x - 4

- 3 =    6   
        x - 4

- 3(x - 4) = 6

- 3x + 12 = 6

x = 2

C(2 ; 0)

✓y = 0

✓x = 2

(2)

4.4

                                0 -  3
                                      2
Average gradient =  2 - 0

= - 3
     4

0 -  3✓
      2
2 - 0

- 3✓
  4

(2)

4.5

y = -x + 7

OR

m = -1
∴ y - 3 = -(x - 4)
y = -x + 7

✓m = -1

✓y = -x + 7

OR

✓m = -1
✓y = -x + 7

(2)

5.1

f:

✓x-intercepts

✓y-intercept

✓shape

✓TP

g:

✓x-intercept and y-intercept

✓shape

(6)

5.2

y = -20 1
            4

✓✓y = -20 1  / -81
                  4     4
(2)

5.3

- 20 1  < k < -14
       4

✓- 20 1  < k < -14
       4

✓k < -14

(2)

5.4

Reflecting in the x-axis: y = -2x + 14

y = -2(x + 7) + 14

Shifting 7 units to the left: = -2x - 14 + 14

= -2x

✓y = -2x + 14

✓y = -2x

(2)

[12]

6.1

f : y = b^x

f ^-1 : x = b^y

y = log_b x

✓interchange x and y

✓answer

(2)

6.2

y = x

✓answer (1)

6.3

P(0; 1)

✓answer                (1)

6.4

T(1; 0)

y = mx + c
y = -x + 1

✓coordinates of T

✓y = -x +1          
(2)

6.5

✓- x +1 = x

✓x = ½

✓y = ½

✓substitution

✓b = ¼

(5)

✓- x +1 = x

✓x = ½

✓y = ½

✓substitution

✓b = ¼

(5)

[11]

7.1

✓substitution of A, P &

✓n in correct formula

✓answer

(3)

7.2

It will take him 34 months to pay back the loan.

✓i = 0,24 / 0,02 / 1
        12              50

✓substitution of P , x and i in correct formula

✓33,43

✓answer

(4)

7.3

✓ i = 0,075 / 0,01875
            4

✓ n = 4 x 6,5 = 26

✓substitution into correct formula

✓115 902,69

✓substitution into correct formula

✓115 902,69

(6)

[13]

8.1

OR

8.2

y = 12x²+ 2x +1
           6x

= 2x + 1 + 1
          3    6x

= 2x + 1 + 1 x ^-1
           3    6

dy = 2 - 1 x ^-2
dx         6

= 2 - 1  
        6x ²

✓12x ² + 2x + 1         
    6x       6x    6x

✓ 1 x^-1
    6

✓2

✓ - 1 x^-2
      6

(4)

8.3

y = x³ + bx² + cx - 4

y ^/ = 3x² + 2bx + c
y ^// = 6x + 2b

At point of inflection:

y ^// = 6x + 2b = 0

Substitute x = 2:
6(2)+ 2b = 0

2b = -12

b = -6

y = x³ - 6x² + cx - 4
Substitute (2; 4):

4 = 2³ - 6(2)² + c(2)- 4

2c = 24

c = 12

y = x³ - 6x² + 12x - 4

✓y ^/ = 3x² + 2bx + c

✓y^// = 6x + 2b

✓y ^// = 0

✓sub x = 2 into y ^// = 0

✓value of b

✓substitute (2; 4)

✓value of c

(7)

[16]

9.1

(0 ; 1)

✓answer (1)

9.2

f (x) = x³ - x² - x + 1

f (x) = x² (x -1)- (x -1)

f (x) = (x -1)(x² -1)

f (x) = (x -1)(x -1)(x + 1)

f (x) = 0

(x -1)(x -1)(x + 1) = 0

x-intercepts:    (–1; 0); (1; 0)

OR

f (x) = x³ - x² - x + 1

f (x) = (x - 1)(x² - 1)

f (x) = (x - 1)(x - 1)(x + 1)

f (x) = 0

(x - 1)(x - 1)(x + 1) = 0

x-intercepts:    (–1; 0); (1; 0)

OR

✓(x – 1)

✓(x ² -1)

✓(x -1)(x -1)(x +1)

✓(–1; 0)

✓(1; 0)

(5)

✓(x – 1)

✓(x ² -1)

✓(x -1)(x -1)(x +1)

✓(–1; 0)

✓(1; 0)

(5)

f (x) = x³ - x² - x + 1

f (x) = (x + 1)(x² - 2x + 1)

f (x) = (x + 1)(x - 1)(x - 1)

f (x) = 0

(x - 1)(x - 1)(x + 1) = 0

x-intercepts:    (–1; 0); (1; 0)

✓(x + 1)

✓(x² - 2x + 1)

✓(x -1)(x -1)(x +1)

✓(–1; 0)

✓(1; 0)

(5)

9.3

f (x) = x³ - x ² - x + 1

f ^/ (x) = 3x ² - 2x -1

f ^/ (x) = 0 (3x + 1)(x -1) = 0

✓f ^/ (x) = 3x² - 2x -1

✓f ^/ (x) = 0

✓factorisation

✓x value

✓x value

✓y = 32
         27

(6)

9.4

✓y– and x–intercepts

✓shape

✓ turning points

(3)

9.5

f ^'(x) < 0

- 1 < x < 1
  3

OR

(– 1 ; 1)
    3

✓x > - 1
           3
✓ x < 1

(2)

✓(– 1
     3

✓1)

(2)

[17]

10.1

60 = 2b + 2r + ½ (2πr )

2b = 60 - 2r - r

b = 30 - r - ½ πr

✓ 60 = 2b + 2r + ½ (2πr )

✓ b = 30 - r - ½ πr

(2)

10.2

Area = area of rectangle + area of semicircle

(6)
[8]

11.1

8 x 7 x 6 x 5 x 4       or     8!
                                        3!

= 6720

✓8 x 7 x 6 x 5 x 4 / 8!
                               3!
✓ 6720

(2)

11.2

P(A and B) = P(A)´ P(B)

= 0,4 ´ 0,35

= 0,14

P(A or B) = P(A)+ P(B)- P(A and B)

= 0,4 + 0,35 - 0,14

= 0,61

✓ 0,4 x 0,35

✓0,14

✓substitution

✓answer                        (4)

11.3.1

100 % - 20%                      1- 0,2

= 80%               or              = 0,8

OR

30% + 50% = 80% or/of 0,3 + 0,5 = 0,8

✓100 % - 20% or 1 – 0,2

✓80% or 0,8

✓30% + 50% or  0,3 + 0,5

✓80% or 0,8                (2)

11.3.2

0,3 x 0,35 = 0,105

= 10,5%

✓0,3

✓0,35

✓0,105 = 10,5%

(3)

11.3.3

(0,2 x 0,35) + (0,3 x 0,65) + (0,5 x 0,9)

= 0,715

= 71,5%

✓0,2 ´ 0,35

✓0,3´ 0,65

✓0,5´ 0,9

✓answer

(4)

[15]

150

1.1.1

• incorrect rounding 2/3 – only rounding penalization
• use of calculator 2/3 – this is where use of calculator for factors get used
• answer in surd form 2/3 ( at least simplified under square root)

1.1.2

• CA mark only if quadratic equation
• check answers
• if 6x ² -15 = x + 1        breakdown 0/3
• both answer must be seen before selection if no factors are shown
• if in the context of their incorrect sum, both of the answers are NA, both need to be shown as NA

1.1.3

• (x + 6)(x - 4) ≥ 0
• x ³ 4 or / and x ³ -6 , award 1/3 marks (factors)
• x £ 4 or / and x £ -6 , award 1/3 marks (factors)
• - 6 £ x £ 4 , award 1/3 marks (factors)
• x £ -6 and x ³ 4 , award 2/3 marks
• equal is left out: –1

Answer only 3/3

1.2

NB: At the second error related to a mark (two skills) – no further marking.

If incorrect algebra leads to the equation being linear: max 2/6 These marks will be the changing of the formula and the substitution mark.

1.3.2

CA from 1.3.1

• If 7 ^x = p can award 1 mark for the concept
• If answer x = 2 only 2/3

2.1.2

CA from 2.1.1

Answer only 2/2

2.1.3

Answer only 1/4

• If n = 7  2/4
• Incorrect working that leads to use of logs and an not a natural number max 2/4

2.2.1

Answer only 2/2

2.2.2

• Answer only 1/6
• Sn has to equal 30 500 otherwise a BD

3.2

n = 5 only 1/3

3.3

Answer only 1/4

4.1

x = 4 ; y = 3    1/1

4.3

y = 0 can be implied

4.4

CA from 4.2 and 4.3

5.1

Only working out, but no sketch max 4/6 – loose shape mark per graph not sketched

5.2

CA from turning point in 5.1

5.3

CA from sketch (TP to y-intercept)

5.4

Answer only 2/2

6.1

Answer only 2/2

If answer not in terms of b max 1/2

6.3

Coordinate from not needed

7.1

• Interchange A and P – breakdown 0/3
• Wrong formula 0/3
• Early rounding: answer is 12,93% – 2/3

7.3

• i and n incorrect – learner can still get the substitution mark 1/6
• If quarterly is taken as monthly consistently in both parts 5/6
  
  
  

8.1

8.2

• If function and derivative is mixed but splitting of fractions is evident max ¾
• If they start with differentiation – breakdown 0/4

8.3

y ^// = 0 can be implied

9.2

No working is shown(calculator used)

• If the cubic becomes a quadratic 2/5
• If three brackets 5/5

9.3

9.4

If dots only indicated on the graph 1/3 – x and y-intercepts

9.5

Only CA from a cubic graph

Each answer gets evaluated independently

10.2

Derivative equal to zero is an independent mark

A^/ (r ) = 0 can be implied if working is correct

11.1

Answer only 2/2

2 or 0 marks

11.3.2

Do not penalize rounding

11.3.3

Do not penalize rounding