1.1.1

x² − 4x + 3 = 0
(x −3)(x −1) = 0 
x = 3 or x =1

✔factors/correct subt in   formula 
✔x = 3 
✔x =1  (3)

1.1.2

✔substitution into the correct   formula 
✔x = 0,72 
✔x = 0,28   (3)

1.1.3

x2 − 3x − 10 >  0
(x − 5)(x + 2) > 0 

OR

x < −2 or x > 5 

✔factors/ critical values 
✔✔x < −2 or x > 5    (3)

1.1.4

3√x = x − 4 
9x = x² − 8x + 16
x² − 17x + 16 =  0
(x −16)(x −1) = 0 
x =16 or x =1  
         NA

✔ squaring both sides 
✔
✔ factors 
✔ answer with  selection  (4)

OR

OR
✔ standard form 
✔ recognize  
✔ factors 
✔ answer with selection (4)

1.2

2y + 9x² = -1......(1) 
3x − y = 2 ...... (2) 
y = 3x − 2 .......(3) 
2(3x - 2) + 9x² = -1
6x - 4 + 9x² = -1
9x² + 6x - 3 = 0 
3x² + 2x - 1 = 0
(3x - 1)(x + 1) = 0
x = ¹/₃      or   x = -1
y = −1 or y = −5 

OR

2y + 9x² = -1 ....... (1)
3x - y = 2 .........(2) 
x = y + 2 
         3
2y + 9 (y + 2)² = -1
           (   3   )
2y + 9 (y² + 4y + 4) = -1
                     9
2y + y² + 4y + 4 + 1 = 0
y2 + 6y + 5 = 0 
(y  5)(y + 1) = 0
y = -1   or   y = -5
x = ¹/₃      or   x = -1

✔y = 3x − 2 
✔substitution 
✔ standard form 
✔factors 
✔both x values 
✔both y values  (6) 

OR

✔x = y + 2
            3
✔substitution 
✔ standard form 
✔factors 
✔both y values  
✔both x values   (6) 

1.3

✔3^3x = 4
✔
✔3^{3x - 3}   or   3^3x.3^-1
✔answer  (4) 

OR

✔3^{3x - 3}   or   3^3x.3^-1
✔3^3x = 4
✔
✔answer   (4)

[23]

2.1.1 

42 

✔answer  (1)

2.1.2 

2a = 6        3a + b = 1             a + b + c = 2
a = 3           3(3) + b = 1         (3) + (–8) + c = 2 
                   b = –8                      c = 7
T_n =  3n² +  8n + 7

OR

2a = 6 
 a = 3 
T_n =  3n² +  bn + c
T₁ :  3 + b + c = 2          b + c = -1 ....(1)
T₂ :  12 + 2b + c = 3      2b + c = -9 ....(2)
T₂ - T₁ : b = -8

Subst. in (1): – 8 + c = – 1  
                              c = 7
T_n =  3n² +  8n + 7

✔ a = 3  
✔ b = –8 
✔ c = 7 
✔T_n =  an² +  bn + c  (4) 

OR

✔ a = 3  
✔ b = –8 
✔ c = 7 
✔T_n =  an² +  bn + c  (4) 

2.1.3

T₂₀ = 3(20)² -  8(20) + 7 
 = 1047

✔substitution 
✔answer (2)

2.2

T_n= −7n + 42 
− 7n + 42 = −140 
 − 7n = −182 
 n = 26

✔T_n= −7n + 42 
✔− 7n + 42 = −140 
✔n = 26   (3)

2.3

✔
✔simplification of S_n 
✔equating 
✔standard form 
✔factors 
✔answer with  
 selection (6)

[16]

3.1

r = ½ and = S_∞ = 6   
S_∞ =   a       
        1 - r 
6 =   a        
     1 - ½
a = 3 

✔substitution 
✔answer   (2)

3.2

T_n = ar^n-1
T₈ = 3(¹/₂)7
T₈ =  3  
       128

✔✔ T₈ = 3(¹/₂)7  (2)

3.3

✔ r = ¹/₂
✔substitution 
✔simplification 
✔answer   (4)

✔expansion 
✔ expansion 
✔ answer (3)

OR

OR

 
✔4p  (3) 

OR

✔ substitution and answer 
✔substitution and answer 
✔4p  (3)

[11]

4.1 

Yes  
For every x-value there is only one corresponding y value OR   One to one mapping (vertical line test)

✔answer 
✔reason   (2)

4.2 

R(–12 ; –6) 

✔answer (1)

4.3

f (x) = ax² substitute (-6 ; -12) 
−12 = a(−6)² 
a = ⁻¹/₃ 

✔substitution 
✔answer  (2) 

4.4

f : y =  −(¹/₃ ) x²
f-1 : x =  −(¹/₃ ) y²   
y² = -3x
y = ± √− 3x 
Only y = −√-3x  and x ≤ 0

✔swapping x and y 
✔y² =  -3x 
✔y = − √-3x   (3)

[8]

5.1 

Domain: x∈R ; x ≠1 

OR

x∈(−∞;1)∪(1;∞)

✔answer  (1)

5.2

x = 1 
y = 0

✔x = 1 
✔y = 0 (2)

5.3

✔ y intercept  
✔vertical asymptote 
✔shape  (3)

5.4

x ≥ 0 ; x ≠1 

OR

0 ≤x < 1 or x > 1 

OR

x∈[0;1) ∪(1;∞)

✔x ≥0 
✔x ≠1 (2)

OR

✔ 0 ≤ x < 1  
✔ x > 1

[8]

6.1

y = mx + c 
m = 5 + 1
       4 - 0
m = 1
c = 1
g(x) = x + 1

OR

y = mx + c 
5 = m(4) +1 
m = 1 
g(x) = x +1

✔substitution into    gradient formula
✔y-intercept (0 ; 1)  (2) 

OR

✔substitute (4 ; 5) 
✔c = 1 (2)

6.2

x² − 2x − 3 = 0
(x +1)(x −3) = 0 
x = −1 or x = 3 
A(–1 ; 0) B(3 ; 0)

✔y = 0 
✔factors 
✔x-values (3)

6.3

x = −1 + 3   or  x = -b = -(-2)   or  f^I(x) = 2x - 2 = 0   
           2                 2a    2(1)
x = 1
f(x) = x² - 2x - 3 
y = (1)² - 2(1) - 3    or      y = (x² - 2x + (-1)²) - 3 - 1 
y = -4                                  = (x - 1)² - 4
y ≥ -4    or [-4 ; ∞]

✔x -value 
✔ substitution/ completing the square 
✔ answer   (3)

6.4.1 

MN: y = (x² - 2x - 3) - (x + 1)
= x² - 3x - 4
6 = x² - 3x - 4
0 = x² - 3x - 10
0 = (x - 5)(x + 2)
x = 5   or   x = -2
OT = 2  or   OT = 5
               NA

✔ x² - 3x - 4
✔substituting y = 6 
✔values of x 
✔ OT = 2   (4)

6.4.2

y = x +1 substitute x = −2 
 = (–2) + 1 
 = –1 
N(–2 ; –1)

✔substituting x = –2 
✔answer (2)

6.5

OR

x² − 2x − 3 = x + p 
x²− 2x − 3 − x − p = 0
This equation will have equal roots, therefore: 
b² − 4ac = 0
(-3)² - 4(1)(-3 - P) = 0 
9 +12 + 4p = 0 
p =  -21
       4
y = x − 21
            4 

✔f ′(x) = 2x − 2
✔2x − 2 =1 
✔ x = ³/₂
✔
✔answer   (5) 

OR

✔equating 
✔equal roots 
✔ substitution 
✔simplification 
✔answer (5)

6.6

 k < - 21
         4 

✔answer  (1)

[20]

7.1.1

✔0,088 and n = 16 
      4 
✔substitution into  correct formula 
✔answer  (3) 

7.1.2

✔ future value – amount   including interest 
✔ 
✔answer  (3) 

OR

✔ R15 000 including interest   – R100 000 
✔  on P and X in F_V
✔method (3) 

7.2.1

✔i = 0,105   
          12 
✔n = 240 
✔ substitution into  correct formula 
✔ answer  (4) 

7.2.2

 = R5 259 229,61– R4 289 302,47 
 = R969 927,14

✔R14 975,70 in P_v-formula
✔✔n = 96 
✔ substitution into correct formula 
✔ answer  (5) 

OR

✔ n = 144 in A-formula
✔n =144in F_v-formula
✔ R14 975,70 
✔ A – F  
✔ answer (5) 

[15]

8.1

✔x² + 2xh + h² − 5
✔simplification 
✔factorisation 
✔ 
✔2x (5) 

OR

✔x² + 2xh + h² − 5
✔simplification 
✔factorisation 
✔ 
✔2x  (5) 

8.2.1

y = 3x³ + 6x² + x −  4
dy = 9x² + 12x + 1
dx 

✔9x²
✔12x 
✔1    (3) 

8.2.2

y(x −1) = 2x(x −1) 
y = 2x(x - 1) if x # 1
        x - 1
y = 2x
dy = 2
dx 

✔y(x −1) 
✔2x(x −1) 
✔y = 2x 
✔answer   (4)

[12]

9.1.1

g(x) = (x + 5)(x − x₁)²
20 = 5(x₁)²
x₁² =  4 
x₁² = 2 
g(x) = (x +5)(x − 2)²
g(x) = (x + 5)( x² − 4x + 4)
g(x) = x³ + x² − 16x + 20

✔(x +5) 
✔repeated root 
✔x₁ = 2 
✔g(x) = (x + 5)( x² − 4x + 4)   (4)

9.1.2

g(x) = x³ + x² − 16x + 20
g′(x) = 3x² + 2x − 16
3x2 + 2x - 16 = 0
(3x + 8)(x - 2) = 0
x = -8  or  x = 2
      3
R(-8 ; 1372)  or  R(-2,67;50,81)
     3     27
P(2;0)

✔derivative 
✔equating to zero 
✔factors 
✔co-ordinates of R 
✔ co-ordinates of P   (5)

9.1.3

g′′(x) = 6x + 2 
g′′(0) = 2 
∴concave up 

OR

g′′(x) = 6x + 2 
6x + 2 = 0 
x = − ¹/₃ is the point of inflection 
∴concave up

✔g′′(x) = 6x + 2 
✔g′′(0) = 2 
✔conclusion (3) 

OR

✔g′′(x) = 6x + 2 
✔x = − ¹/₃
✔conclusion   (3)

9.2

✔ y – intercept of a cubic   graph 
✔ point of inflection and   stationary point, x = 3 
✔ concave up for x < 3and   concave down for x > 3    (3)

[15]

10.1

AH = 3 
HG   2 

✔ answer (1)

10.2 

Area of a parallelogram = base× ⊥height  

✔ ²/₅ t
✔ ³/₅(5 − t) 
✔A(t) ⁶/₂₅ t² +  ⁶/₅t
✔-¹²/₂₅ t  + ⁶/₅
✔answer   (5)

[6]

11.1.1

7⁵ = 16 807 

✔✔ answer (2)

11.1.2

7 × 6 × 5 × 4 × 3 
= 7!  =  2520 
   2! 

✔7 × 6 × 5 × 4 × 3 or 7!
                                   2!
✔answer (2)

11.2

2×7×1= 14 

✔✔✔2 × 7 × 1 (3)  

[7]

12.1 

P(A or B) = P(A) + P(B) 
 0,74 = 0,45 + y 
 y = 0,29

✔ P(A or B) = P(A) + P(B) 
✔substitution 
✔answer (3) 

12.2

Let the number of mystery gift bags = x 
The total number of bags = 4x 

✔ 4x  

✔ equating to     7     
                       1187 
✔answer (6)

OR
P(gift and gift) = P(gift at first draw) × P(gift at second draw)
  7   =   1  × P(gift at second draw) 
118      4
P(gift at second draw) =   7   ÷  1  
                                       118      4
=  14 
    59
Therefore: P(gift at first draw) = ¹⁵/₆₀
And: 15 bags had mystery gifts inside

OR

✔¼
✔ 1   ×   P(gift at 2^nd draw)
    4 
✔   7   ÷  1  P(gift at 2^nd draw)     
   118      4
✔14 
   59
✔ ¹⁵/₆₀
✔answer (6)

[9]