A mark for a correct reason  
(A reason mark may only be awarded if the statement is correct.)

Days

1 

2 

3 

4 

5 

6 

7 

8 

9 

10

Units of blood

45 

59 

65 

73 

79 

82 

91 

99 

101 

106

1.1.1

x =  800 
       10
= 80
Answer only: full marks 

✔ 800 (addition of units) 
✔ answer (CA if ÷ 10) (2)

1.1.2

σ =18,83
No penalty for rounding

✔✔ answer (A)  (2) 

1.1.3

(61,17 ; 98,83) 
Days 1, 2, 8, 9 and 10 lie outside 1 standard deviation from the  mean 
∴5 days
Correct answer only: full marks provided  that 1.1.1. & 1.1.2 both correct

✔ mean – 1 SD 
✔ mean + 1 SD 
✔ answer   (3) 

1.2.1 

Skewed to the left or negatively skewed

✔ answer  (1) 

1.2.2 

A = 65 
B = 99
Answers without labelling: 1/2

✔ answer 
✔ answer (2) 

1.3 

New total = 95 × 10 = 950 
∴Units not counted = 950 – 800 = 150 

✔ answer (CA from 1.1.1)  (1) 

 [11]

2.1 

Outlier  (100 ; 100)
accept: 100 as answer

✔ answer (1)

2.2 

a = 94,50273… 
b = 2,913729…  
yˆ= 94,50 + 2,91x 

• Integral values: max 2/3 
• Swopped a and b: 2/3

✔ value of a 
✔ value of b 
✔ equation  (3)

2.3

yˆ= 2,91(240) + 94,50 (CA from 2.1) 
 = 792,90 
Value = R793 000 

OR

yˆ=793,7978142 (calculator) 
Value = R794 000

• Penalise 1 mark if answer  not in thousands of Rands

✔ substitution 
✔ answer in   thousands  of Rands (2) 

✔✔ answer in  thousands  of Rands  (2)

2.4 

b = 2,913729…  
∴ R2 914 OR/OF R2 910 (calculator)

• Answer only: full marks

✔ value of b 
✔ answer  (2) 

[8]

3.1

x = 3

✔ answer  (1)

3.2

m_QP = tan71,57° 
 = 3

• Answer only: full marks

✔ m_QP= tan71,57°   
✔ answer  (2)

3.3

y = mx+c                                    y − y₁ = m (x − x₁) 
− 2 = 3(−7) + c        or                  y + 2 = 3(x + 7) 
y = 3x +19 

(m CA from 3.2 if > 0) 
✔ substitution of m & Q
✔ equation  (2)

3.4 

(wrong R: CA if x > 0) 
✔ substitution 
✔ answer (in surd form)  (2)

tan(90° − θ) = m_QR
= 0 - (-2)
   3 - (-7)
= ¹/₅

• Answer only: full  
• tanθ =  ¹/₅ ; ¹/₃ 

(wrong R: CA if x > 0)
✔ gradient of QR/RN/QN 
✔ substitution of Q & R 
✔ answer    (3) 

OR

✔ RN 
✔ SR 
✔ diagram (5 & √26) 
✔ use of correct area rule 
✔ substitution of sinθ 
✔ answer  (6) 

✔ RN 
✔ SR 
✔ diagram 
✔ use of correct area rule 
✔ substitution of sinθ 
✔ answer  (6)

✔ SR 
✔✔ ⊥ height 
✔ use of correct area   formula 
✔ substitution of sinθ 
✔ answer  (6)  

 [16]

4.1

OK = √180 or 6√5

✔ answer  (1)

4.2

a² + b² = 180
b = ¹/₂a
a² + (¹/₂a)² = 180
a² + ¹/₄a² = 180
a² = 144     ∴a = -12
b = ¹/₂ (-12)
K (–12 ; – 6) (given) 

• No penalty if x and y are not  converted to a and b
• Error in simplification:  max 2/4

OR 

a² + b² = 180
a = 2b 
(2b)² + b² = 180  
5b² = 180
b² = 36       ∴ b = –6 
a = 2(–6) 
K (–12 ; – 6) (given) 

✔ b in terms of a 
✔ substitution 
✔ a²=  144
✔ substitution   (4) 

✔ a in terms of b 
✔ substitution 
✔ b² =36 
✔ substitution  (4)

✔m_PT= −2 
✔ substitution of m & K(–12 ; – 6)
✔ equation  (3)

OR

3MK = OK 
9MK² = OK² = 180
∴MK2 = 20
Let M(x ; y), then y = ½x:
(x + 12)² + (y + 6)² = 20
(x + 12)² + (  ½x + 6)² = 20
4(x + 12)² + (x + 12)² = 80
(x + 12)±4
x = -16     x#-8 [since M is outside the large circle]
y = -8
M(-16; -8)

✔ 3MK = OK 
✔ OM = ⁴/₃ OK 
✔✔ M = ⁴/₃ (-12 ; -6) 
✔ x-coordinate 
✔ y-coordinate (6) 

✔ 3MK = OK 
✔ MK²= 20 
✔ equation 
✔ substitution 
✔ x-coordinate 
✔ y-coordinate (6) 

✔ 3MK = OK 
✔ ✔✔ diagram with  values OR  valid explanation
✔ x-coordinate 
✔ y-coordinate  (6)

✔ 3MK = OK 
✔ MK²= 20 
✔ equation  
✔ substitution 
✔ x-coordinate 
✔ y-coordinate  (6)

✔ LHS (CA from 4.3.2)
✔ RHS (CA from 4.1)  (2)

✔✔ values 
✔ inequality  (3) 

x² + 32x + (16)² + y² + 16y + (8)² = 256 +64 - 240
(x + 16)² + (y + 8)² = 80
New circle:
Centre(−16 ;−8)&  r = 4√5 
Original circle: M(−16 ; −8)& r = 2√5 
This circle will never cut the circle with centre M as they  have the same centre (concentric circles) but unequal  radii

✔ equation in centre,   radius form 
✔ Centre:(−16;−8) 
✔ r = 4 5(new) 
✔ r = 2 5(original) 
✔ conclusion  (“concentric” must be  stated)  (5)  [24]

5.1.1 

[ no calculator in 5.1] 

cos2θ = - ⁵/₆, where 2θ ∈[180°;270°]

y² = 6² − (−5)² [Pythagoras] 
y = ±√11 
(5 ; y) is in 3rd quadrant: 
∴y = −√11 
sin 2θ = − √¹¹/₆ 

• Getting to sin 2θ = − √¹¹/₆  =: 3/4

OR 

sin²2θ = 1 - cos²2θ 
= 1 − (-⁵/₆)²
= 1 - ²⁵ /₃₆
= ¹¹ / ₃₆
sin 2θ = − √¹¹/₆ 

✔ diagram  (3rd quadrant only) 
✔ using Pythagoras 
✔ y – value 
✔ answer   (4) 

✔sin²2θ = 1 - cos²2θ 
✔ substitution 
✔ value of sin²2θ
✔ answer  (4)

5.1.2

cos2θ =1− 2sin²θ 
2sin²θ = 1 - cos 2θ 

✔cos2θ =1− 2sin²θ 
✔ substitution 
✔ answer   (3)

5.2

sin(180° − x).cos(−x) + cos(90° + x).cos(x −180°)
= sin x.cos x − sin x(−cos x) 
= 2sin x.cos x 
= sin 2x 

• Second line written as: sinx cos x + sinx cos x: max 5/6

✔sin x
✔cos x 
✔ − sin x
✔− cos x 
✔ simplification 
✔ answer  (6)

5.3

sin3x.cos y + cos3x.sin y 
sin(3x + y) 
= sin 270° 
= –1

✔ compound angle 
✔ answer  (2)

5.4.1

2cos x = 3tan x 
2cos x = 3sin x
               cos x
2cos²x = 3sin x 
2(1 - sin²x) = 3sin x 
2 - 2sin²x = 3sin x 
2sin²x =  3sinx - 2 = 0 

✔ tan x  = sin x
                cos x
✔ multiplying by cosθ 
✔cos²x  =1− sin²x  (3)

5.4.2

2sin²x +  3sin x - 2 = 0 
(2sin x −1)(sin x + 2) = 0 
sin x = ¹/₂  or     sin x = −2 (no solution) 
x = 30° + k.360°    or       x =150° + k.360° ; k ∈Z

✔ factors 
✔ both values of sin x
✔ no solution 
✔30°+ k.360° 
✔150°+ k.360° ; k ∈Z   (5)

5.4.3

5y = 30º + k.360º        or         5y = 150º + k.360º
y = 6°+ k.72°               or                y = 30°+ k.72°
∴ y = 144° + 6°           or              y = 144° + 30° 
 y = 150°                     or                    y = 174° 

OR

144° ≤ y ≤ 216° 
720° ≤ 5y ≤ 1080° 
5y = 750° or 5y = 870° 
y = 150° or y = 174°

✔y = 6°+ k.72° 
✔y = 30°+ k.72° 
✔ 150°
✔ 174°    (4) 

✔ 5y = 750°
✔ 5y = 870°
✔ 150°
✔ 174°    (4)

5.5.1

g(x) = −4cos(x + 30°) 
maximum value = 4 

✔ answer   (1)

5.5.2 

range of g(x):  
− 4 ≤ y ≤ 4            OR                y ∈[−4 ; 4] 
∴range of  g(x) + 1:  − 3 ≤ y ≤ 5 OR     y ∈[−3 ; 5] 
Answer only: full marks

✔ range of g(x) 
✔ answer   (2)

5.5.3

y = −4cos(x + 30°) 
shifted to the left: 
y = -4cos( x + 30º +  60º ) 
= -4cos(x + 90º)
= 4sinx
∴h(x) = −4sin x 
Answer only: full marks

✔ shift of 60° to the left 
✔ reduction 
✔ equation of h  (3)  [33]

6.1.1

tan θ = PQ  =  PQ 
            QR       x 
∴PQ = xtanθ 
Answer only: full marks

OR   =   PQ  
sinP     sinPRQ 
∴PQ  = x. sinθ
          sin(90º - θ ) 

✔ trig ratio 
✔ answer  (2) 

✔ trig ratio 
✔ answer (2)

6.1.2

    AR        =     QR       
sinAQR         sinQAR 
AR = xsin(90º + θ) 
             sinθ 
Answer only: full marks  

✔ use of sine rule  
✔ substitution into sine  rule correctly (2)

✔ substitution into trig   ratio and AB as subject
✔ substitution of AR 
✔ co-ratio 
✔sin 2θ = 2sinθ cosθ   (4)

✔ substitution  CA from 6.1.1) 
✔ answer (2)   [10]

7.1.1

T₁ = 70° [ext ∠ of cyclic quad] T 70  

✔ S ✔ R   (2)

7.1.2

Q₁ = Q₂ = 35º [equal chords;equal∠s/] 

✔ S ✔ R  (2)

7.2.1

T2 =Q1 = 35º [alt ∠s ; PQ || TR] 

✔ S ✔ R  (2)

7.2.2

PT= QR  [prop theorem ; PQ || TR] 
TS   RS 
∴ TR = QR  [PT = TR]
   TS    RS 

✔ S ✔ R (2)  

[8]

PTR = 90° [∠ in semi-circlel] 
x = 90º +  P  [ext ∠ of Δ] 
∴ R = x − 90°  
STP = x − 90° [tan chord theorem] 
x + x − 90° + y =180° [sum of  ∠s in ∆]
∴ y = 270° – 2x

✔ S/R 
✔ S/R 
✔ S ✔R 
✔ S 
✔ answer   [6]

9.1 

equiangular ∆s OR (∠∠∠) 

✔ answer (1)

9.2

∴ GE  = DE       [||| Δs] 
   GF     GE 
GE² = 45 ×  80 
GE = 60

✔ proportion 
✔ substitution
✔ answer  (3)

9.3 

In ΔDEH and ΔFGH: 
DHE = FHG [vert opp ∠s =]
DEH = FGH    [||| Δs]  
EDH = GFH    [sum of ∠s  in ∆]
∴ΔDEH ||| ΔFGH  

OR

In ΔDEH and ΔFGH: 
DHE = FHG [vert opp ∠s =]
DEH = FGH            [||| Δs]  
∴ΔDEH ||| ΔFGH    [∠∠∠]

✔ S/R 
✔ S/R 
✔ S  (3) 

✔ S/R 
✔ S/R 
✔ R (3)

9.4

GH =  FG    [||| Δs] 
EH     DE 
      GH        =  80 
  60 -  GH        45
45GH = 80(60 – GH) 
45GH = 4800 – 80GH
125GH = 4800 
GH = 38,4

✔S  
✔ substitution 
✔ answer (3) 

[10]

10.1 

Construction: 
AO is drawn and produced to M 
O₁ = A₁ + B [ext ∠ of Δ] 
But A₁ = B [∠s opp = radii]  
∴ O₁ = 2A₁ 
Similarly : O₂ = 2 A₂ 
∴ O₁ + O₂  = 2A₁ +  2A₂ 
 = 2 (A₁ + A₂) 
 BOC = 2BAC

✔ Constr 

✔ S/R 

✔ S/R 

✔ S 

✔ S  (5)

10.2.1(a)

F₁ = 2x     [∠ centre = 2∠ at circum ]  

✔ S ✔ R  (2)

10.2.1(b)

C = x [∠s in the same seg]

OR

C = x [∠ centre = 2∠ at circum]

✔ S ✔ R  (2) 

✔ S ✔ R (2)

10.2.2

D₃ = x      [∠s opp equal sides] 
E₃ = 2x    [ext ∠ of Δ] 
∴ F₁ = E₃ = 2x
∴ AFED is a cyclic quadrilateral [converse ∠s in the same seg] 

✔ S/R  
✔ S/R 
✔ S  
✔ R (4)

10.2.3

A₂ +  A₃  +  D₁ + F₁= 180°      [sum of ∠s in ∆]
A₂ + A₃ = D₁                            [∠s opp = sides]
∴ A₂ + A₃ = 90° − x 
E₁ =  A₂ + A₃                          [ext∠of cyclic quad] 
= 90° − x 
FKE = 90°                              [line from centre bisects chord]
F₃ = x                                    [sum of ∠s in ∆]

✔ S 
✔ S 
✔ R 
✔ S 
✔ S ✔ R  (6)

10.2.4

BAC = D₃          [∠s in the same seg]
AE = BE            [sides opp equal ∠s] 
area ΔAEB =  ¹/₂(BE)(AE).sinAEB 
area ΔDEC      ¹/₂(EC)(ED).sinDEC 
6,25 =  AE²
           ED²  
∴ AE=  2,5 
   ED 

✔ S 
✔ S 
✔ substitution   into area rule 
✔ simplification  of RHS 
✔ answer  (5) 

[24]