1.1.1

𝑥² − 4𝑥−12=0
(𝑥 − 6)(𝑥 + 2)=0
𝑥 = 6 or 𝑥 = −2

✔ standard form
✔ 𝑥=6 (CA applies)
✔ 𝑥=−2 (CA applies)

(3)

1.1.2

3𝑥²+2𝑥−6=0
𝑥=−(2) ± √(2)² − 4(3)(−6)
                2(3)
𝑥=−2 ± √766
𝑥=−1,79 or 𝑥=1,12

✔ substitution

✔ x = −1,79 

✔ x = 1,12 

(3)

1.1.3 

 3𝑥² − 1 = 27^-x
                 3
3𝑥² − 1 = 3^−3𝑥 −1
∴ 𝑥² − 1=−3𝑥 − 1
𝑥² + 3𝑥 = 0
𝑥 (𝑥+3) = 0
𝑥 = 0 or  𝑥 = −3

OR

3𝑥² − 1 = 27 − 𝑥3
3𝑥² − 1.3 = 27−𝑥
3𝑥² −1 + 1 = 3^−3𝑥
∴ 𝑥²=−3𝑥
𝑥² + 3𝑥 = 0
𝑥(𝑥+3) = 0
𝑥=0 or 𝑥=−3

✔ 3^−3𝑥−1 
✔ equating exponents
✔factors/faktore 
✔ both x-values

✔𝑥² −1 + 1 = 3^−3𝑥
✔equating exponents 
✔ factors/faktore 
✔both x-values

(4)

✔ 𝑥 =−1
✔ 𝑥=0

(2)

✔ manipulation of equation 
 ✔ standard form
✔ factors
✔ both x-values
✔ choosing x = 2 

(5) 
[17]

𝑥 − 𝑦 =3 ; 𝑥𝑦 = 28
𝑥 − 𝑦 = 3…………………..(1)
𝑥𝑦 = 28………………… (2)
From (1) 𝑥 = 𝑦 + 3
Substitute in (2)
𝑦(𝑦+3) = 28
𝑦² + 3𝑦 − 28 = 0
(𝑦−4)(𝑦+7)  =0
𝑦=4 or  𝑦=−7
𝑥=7 or  𝑥=−4

OR

From  (1) 𝑦 = 𝑥−3
Substitute in (2)
𝑥(𝑥−3) = 28
𝑥² − 3𝑥 − 28 = 0
(𝑥+4)(𝑥−7) = 0
𝑥=−4 or  𝑥=7
𝑦=−7 or  𝑦=4

✔ 𝑥 = 𝑦 + 3
✔ substitute in (2)
✔ standard form
✔ factors
✔ y-value
✔ x-values

✔ 𝑦 = 𝑥 − 3✔ substitute in (2)
✔ standard form
✔factors
✔ x-values
✔ y-values

(6)

✔ standard form/standaardvorm
✔ factors/faktore 
✔ solution/oplossing  
−1 ≤ 𝑥 ≤4
✔ final answer/finale antwoord 
0 < 𝑥 ≤ 4

(4) 
[10]

✔ a = 1 
✔ b = −1 
✔ c = 1 
✔ 𝑇_𝑛 = 𝑛² − 𝑛 + 1
✔ 6321 

(5)

Row 80
6321 6323 6325 6327 ...
𝑆_𝑛=^𝑛/₂[2(𝑎) + (𝑛 − 1)𝑑]
𝑆₈₀ = ⁸⁰/₂ [2(6321)+(80−1)(2)] Row 80
𝑆₈₀ = 512000

OR

Row 80 Term 80
𝑇₈₀ = 6321 + (79×2)
𝑇₈₀ = 6479
𝑆_𝑛 = 𝑛²[𝑎 + 𝑙]
𝑆₈₀ = 8⁰/₂[6321 + 6479] Row 80
𝑆₈₀ = 512000

OR

2𝑎 = 2
𝑎 = 1
3𝑎 + 𝑏 = 4
3 + 𝑏 = 4
𝑏 = 1
𝑎  + 𝑏 + 𝑐 = 1
1 + 1 + 𝑐 = 1
𝑐 = −1

𝑇_𝑛 = 𝑛² + 𝑛 − 1
𝑇𝑛 = 𝑛²+ 𝑛 − 1
𝑇₈₀ = 80² + 80 −1
𝑇₈₀ = 6479
𝑆_𝑛= 𝑛/₂[𝑎+𝑙]
𝑆₈₀=80/₂[6321+6479] Row 80
𝑆₈₀=512000

✔ n= 80 
✔ d = 2 
✔ sub into correct formula   
✔ answer(4) 

✔ calculating term 80 of row  
✔ 6479 
✔ sub into correct formula 
✔ answer (4)

✔ 𝑇_𝑛 = 𝑛² + 𝑛 − 1
✔ 𝑇₈₀ = 6479
✔ sub into formula
✔ answer 

(4) 
[9]

✔ substitution
✔ answer

(3)

𝑝 ;3𝑝 ;5𝑝 ;……………
𝑑 = 2𝑝
𝑆_𝑛=^𝑛/₂[2𝑎 + (𝑛 − 1)𝑑]
𝑆_𝑝=^𝑝/₂[2𝑝 + (𝑝 − 1)2𝑝]
𝑆_𝑝=^𝑝/₂(2𝑝 + 2𝑝²−2𝑝)
𝑆_𝑝=𝑝³

OR

𝑎 = 𝑝
𝑙 = 2𝑝² − 𝑝
𝑆_𝑛 = ^𝑛/₂ [𝑎 + 𝑙]
𝑆𝑝=^𝑝/₂[𝑝+2𝑝² − 𝑝]
𝑆_𝑝=𝑝³

✔ first three terms
✔ 𝑑 = 2𝑝
✔ substitution
✔ answer (4) 

✔ 𝑎 = 𝑝
✔ 𝑙 = 2𝑝² − 𝑝
✔ substitution
✔ answer 

(4) 
[7]

𝑟 = 𝑥+2
       𝑥
𝑇3 = (𝑥+2)²
           𝑥

✔ ratio
✔ answer

(2)

𝑆∞=  𝑎  
     1−𝑟
−8=       𝑥       
        1 − ^(𝑥+2)/_𝑥
−8=     𝑥²           
        𝑥−𝑥−2
𝑥²=16
𝑥  = ±4

✔ both answers 

(4) 
[6]

✔ 𝑖=¹⁵/₁₀₀ and 𝑛 =5
✔ sub into correct formula
✔ answer

(3)

✔ 𝑖=^16.75/₁₂₀₀
✔ 𝑛=−48
✔ sub into correct formula
✔ answer

(4)

𝑃_𝑣 = 𝑥[1 − (1+𝑖)^−𝑛]
                𝑖
𝑃_𝑣 = 1436.29 [1 − (1+^16.75/₁₂₀₀)⁻¹⁸]
                          ^16.75/₁₂₀₀
𝑃_𝑣 = 𝑅22 721,97704
𝑃_𝑣 = 𝑅22 722

OR

Outstanding balance (OB)
OB = 50000(1+ ^16.75/₁₂₀₀)³⁰−[1436.29 [(1 + ^16.75/₁₂₀₀)³⁰−1]]
                                                                     ^16.75/₁₂₀₀
OB =  𝑅 22722,14
OB = 𝑅22722

✔ 𝑛=−18
✔ 𝑖=^16.75/₁₂₀₀
✔ substitution
✔ answer
✔ rounding

(5)

✔ 𝑛=30
✔ 𝑖=^16.75/₁₂₀₀
✔ sub into both formulae
✔ answer
✔ rounding

(5)

✔ 𝐴 = 2𝑥 and 𝑃 = 𝑥
✔ sub into correct formula
✔ using of logs
✔ answer

(4)
[16]

✔answer 

(1)

✔x > -2
✔x # 0

(2)

✔answer 

(1)

✔ sub of −4
✔ sub of point (2;5)
✔ 𝑏 = ±3
✔ answer with correct 𝑏 value

(4)

✔x = -2
✔y = -1

(2)

✔ sub of asymptotes
✔ sub of point (0;-3)
✔ 𝑎=−4

(3)

✔𝒚 = 𝒙 + 𝟏
✔𝑦 = −(𝑥 + 2) −1
✔𝒚 = −𝒙 − 𝟑

(3)

✔swopping of x and y
✔answer

(2)

✔Shape 
✔y-intercept
✔any other correct point

(3)

g(x) = (½)^-x

OR

g(x) = 2x

✔✔Answer

✔✔Answer

(2)

✔✔x > 1

(2)
[25]

✔x = -3

(1)

✔ sub of turning point
(−3;5)
✔ sub of (0;4)
✔ simplification

(3)

✔ Δ=20
✔ irrational
✔ unequal

(3)

𝑔(𝑥) = 2𝑥
𝑥² + 6𝑥 + 4 = 2𝑥
𝑥² + 4𝑥 + 4 = 0
(𝑥+2)² = 0
𝑥 = −2
𝑔(−2)=−4

Point (−2;−4)

OR

𝑓(𝑥) = 𝑥² + 6𝑥 + 4
𝑓′(𝑥) = 2𝑥 + 6 and 𝑚 = 2
2𝑥 + 6 = 2
2𝑥 = −4
𝑥 = −2
𝑦 = −4
Point (−2;−4)

✔ 𝑔(𝑥) = 2𝑥
✔ equating equations
✔ 𝑥 = −2
✔ 𝑦 = −4

✔ derivative
𝑓′(𝑥) = 2𝑥 +6
✔ equating to gradient of g.
✔ x-value
✔ y-value

(4)
[11]

 𝑓(𝑥) = 3𝑥² − 1
𝑓′(𝑥) = lim   𝑓(𝑥+ℎ)−𝑓(𝑥)
          ^ℎ→0          ℎ
=    lim  3(𝑥 + ℎ)² − 1 −(3𝑥² − 1)
     ^ℎ→0                      h
=   lim  3(𝑥² + 2xh + h²) − 1 - 3𝑥² − 1
     ^ℎ→0                      h
=   lim  3𝑥² + 6xh + 3h²− 1 - 3𝑥² − 1
     ^ℎ→0                      h
=       lim     6𝑥ℎ
        ^ℎ→0      h
= 6𝑥

✔ formula
✔substitution of (𝑥+ℎ)
✔ simplification
3𝑥² + 6𝑥ℎ + 3ℎ² − 1 − 3𝑥² + 1
✔ =  lim     6𝑥ℎ
        ^ℎ→0     h
✔ answer

(5)

✔𝑥^½
✔10x
✔½𝑥^−½

(3)

✔ 2
✔ −⁴/₃𝑥⁻¹
✔ answer

(3)

✔ derivative
✔ 3𝑡² ≥ 0

(2)
[13]

𝑓(𝑥) = 𝑥³ − 𝑥² − 8𝑥 + 12
(𝑥−2)(𝑥² + 𝑥 − 6) = 0
(𝑥−2)(𝑥−2)(𝑥+3) = 0
𝑥=2 or 𝑥=2 or 𝑥 = −3
𝐴(−3;0)

OR

                         𝑥+3               
𝑥² − 4𝑥 + 4 √𝑥³ − 𝑥² − 8𝑥 + 12
𝑥³ − 4𝑥² +4𝑥

     3𝑥² − 12𝑥 + 12
    3𝑥² − 12𝑥 + 12  
𝑓(𝑥) = (𝑥² − 4𝑥 + 4)(𝑥 + 3)
𝐴(−3;0)

✔(𝑥−2)
✔(𝑥² + 𝑥 − 6)
✔(𝑥−2)(𝑥−2)(𝑥+3) 
✔coordinates of 𝐴 (−3;0)

✔𝑥² − 4𝑥 + 4
✔✔𝑥 + 3
✔coordinates of 𝐴 (−3;0)

(4)

✔𝑓′(𝑥) 
✔𝑓′(𝑥)  = 0
✔correct 𝑥 value 𝑥 = ⁻⁴/₃
✔y = ⁵⁰⁰/₂₇ 

(5)

 𝑓′′(𝑥) = 6𝑥 − 2
6𝑥 − 2 = 0
𝑥 =¹/₃

OR

𝑥=⁻⁴/₃+2
        2
𝑥=13

✔𝑓′′(𝑥) = 6𝑥 − 2
✔𝑥 =¹/₃
✔finding x value of  midpoint
✔𝑥 =¹/₃

(2)

✔𝑥 <⁻⁴/₃
✔𝑥 > 2

(2)

 ✔answer

(2)
[15]

✔𝐷(0)=3 𝑚

(1)

✔𝐷′(𝑡)
✔𝐷′(3)
✔=−¹⁵/₄ or -3.75

(3)

✔Decreasing

(1)

✔𝐷′(𝑡 )= 0
✔factors
✔t-values
✔answer

(4)
[9]

12.1 

12.1.1 

P(A′) = 1 − P(A) 
= 1 − 0,35 
= 0,65

✔ P(A′) = 1 − P(A) 
✔ answer

(2)

12.1.2 

P(AandB) = 0 
P(AenB) = 0

✔ answer

(1)

12.1.3 

P(A or B) = 0,35 + 0,52 
= 0,87

✔  P(A or B) = P(A) + P(B)
✔ answer

(2)

12.2 

12.2.1 

6! = 720 

✔ 6! or/of 720 

 (1)

12.2.2

4! 
= 24

✔4! 
✔ 24 

(2)

12.2.3

2!.5! =  240   = 1
  6!       720      3

OR 

0,333

✔ 2! 
✔ 5! 
✔ 6! 
✔ answer

(4) 
[12]

TOTAL: 150