1.1 

Mean = 48 

✔ 48  (1)

1.2 

SD/SA = 22,08
Penalty of 1 mark for incorrect rounding  

✔✔22,08  (2)

1.3 

Girls performed better.  
The girls’ mean percentage is bigger than that of boys and the girls  standard deviation is smaller than that of boys  

✔Girls 
✔Reason (2)

1.4 

51 – 48 = 3  
∴each boy’s percentage must be increased by 3.  

✔3  (1)

1.5 

Boys’ standard deviation will remain the same  

✔remain the same   (1)

[7]

2.1 

✔frequency  
✔cumulative  frequency   (2)

2.2 

✔grounding at (0 ; 18)  
✔upper limits  
✔shape  (3)

2.3 

48 ≤ x < 58 

✔answer  (1)

2.4 

60 – 49 = 11 senior citizens 

✔49  
✔answer(2)

2.5 

Q1 = 39  
Q2 = 50  
Q3 = 58 

✔value of  Q1 
✔value of Q2
✔value of  Q3  (3)

2.6 

✔minimum and  maximum  
✔ box (2)

[13]

3.1.1 

✔substitution 
✔answer  (2)

3.1.2 

✔substitution
✔gradient of KM (2)

3.1.3 

m_LM = y₂ - y₁
            x₂ - x₁
 = -4 - 2
     7 - 10
= 2
tan α =  2 
∴ α =  63,43º

✔gradient of LM 
✔ tan α =  2 
✔value of α  (3)

3.1.4 

tanθ = - 11
              7 
Ref ∠ = 57,53º
∴θ = 122,47º
LMK = 122,47º -  63,43º
 = 59,04º

✔tanθ = - 11
                 7 
✔reference angle 
✔value of θ
✔value of LMK∧ (4)

3.2 

✔ y = 2x + 7
✔ y = -x - 1
           2   2
✔value of x
✔value of y 

OR 

✔m_KL = -5
              10 
✔m_NM = -5
               10
✔value of x
✔value of y 

OR
✔Midpoint of KM  = 
✔ Midpoint of LN = 
✔value of x  
✔value of y (4)

3.3 

m_LM × m_MN = 2 × (-½)
= -1
∴LMN = 90º

OR LMN is a right angle

✔product of gradients
✔conclusion (2)

3.4 

LMN = 90º and LMK = 59,04º
∴KMN = 90º - 56,04º = 30,96º
Area of KMN = ½  × √170  ×  5√5  ×  sin30,96º
=37,50 square units

✔KL  = NM = 5√5
✔KM  = √170  
✔LMN = 90º and LMK = 59,04º
✔KMN = 30,96º
✔Area of KMN Δ  (5)

[22]

4.1 

M (2 ; 4) 

✔value of x 
✔value of y  (2)

4.2 

r2 = (5 - 2)² + (0 - 4)² 
= 25
∴ (x - 2)² + (y - 4)² = 25

✔ (x - 2)²
✔ (y - 4)²
✔25     (3)

4.3 

m_GH = 8 - 0
          -1 - 5
= - 8 = - 4
     6       3
m_tan =  ³/₄
y - 8 = ³/₄(x - 1)
∴y = ³/₄x + ³⁵/₄

✔m_GH 
✔m_tan 
✔substitution 
✔equation  (4)

4.4 

At J, y = 0  
(x - 2)² + (0 - 4)² = =25
(x - 2)² = 9
x - 2 = ±3
x = 5 or x = -1
∴ J (-1;0) 

✔ y = 0  
✔substitution 
✔ x = – 1   (3)

4.5 

HJG is a right angled triangle (8,6  and 10).  So the rotation of J around M will complete a rectangle Hence, j ((-1 + 6; 0 + 8)) = J'(5;8)

✔value of x 
✔value of y  (2)

4.6 

x² + y² - 12x - 2y + 17 = 0
x² - 12x + y² - 2y = -17
x² - 12x + 36 + y² - 2y + 1 = -17 + 36 + 1
(x - 6)² + (y - 1)² = 20
Distance between the centres:  

∴ the centre lies on the original circle

✔completing the square
✔factorisation: x
✔factorisation: y 
✔distance formula 
✔conclusion  (5)

[19]

5.1.1 

  (2)

5.1.2 

sin84º = sin2 × 42º
= 2sin42ºcos42º

✔double angle
✔expansion
✔substitution   (3)

5.1.3 

sin3º = sin(45º - 42º
= sin45ºcos42º - -cos45ºsin42º

✔3° = 45° – 42°  
✔expansion
✔substitution
✔substitution  (4)

5.2 

sin(x - 450º).tan(180º + x).sin(90º - x)
                      cos(-x)
-cos x.tan x.cos x
         cos x
-cos x.sin x
    cos x
-sin x

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

5.3.1 

cos (A + B) = cos A cos B - sin A sin B 

✔expansion  (1)

5.3.2 

✔compound angle identity  
✔cos double angle identity  
✔sin double angle identity  
✔ (1 - cos²α)  (4)

5.4 

✔
✔ cos2 θ = cos²θ -  sin²θ 
✔substitution
✔standard form  
✔values of p

OR

✔ 2cos²θ − 1 
✔✔substitution 
✔standard form
✔values of p

OR

✔
✔ 1 - 2sin²θ 
✔substitution  
✔standard form  
✔values of p  (5)

[25]

6.1 

y∈[-1;1] 
OR
-1 ≤ y ≤ 1

✔answer (1)

6.2 

g:  
✔asymptotes at -90° and 90°  
✔x-intercepts  
✔shape  (3)

6.3 

180° 

✔180°  (1)

6.4 

x = – 45° 

✔– 45°  (1)

6.5 

x∈[45º ; 90º) 
OR
45º ≤ x < 90º

✔critical values  
✔notation (2)

6.6 

h(x) =sin (x – 45°) + 1 

✔h(x) =sin (x – 45°) + 1  

(1)

[9]

7.1 

InΔ KLN : 
tan w =   h   
              LN
LN =    h    
        tan w

✔ LN =    h     (1)
            tan w

7.2 

✔correct sine rule  
✔substitution  
✔isolating LM  
✔answer  (4)

7.3 

LM = h sin(y + z)   and 
         tan w sin z
h = 38m,   w = 21º,  y = 52º   and z = 59
∴LM = 38.sin(52º + 59º)
            tan21º sin59º
= 107,82m

✔substitution  
✔answer (2)

[7]

8.1 

the centre of a circle  

✔answer (1)

8.2.1 

AC = 10 cm (line from centre ⊥ chord)   

✔length of AC  
✔Reason  (2)

8.2.2 

(x + 5)² = 10² + x²
x² + 10x + 25 = 100 + x²
10x = 75
 ∴ x = 7,5cm
∴radius = 7,5cm +5cm 
= 12,5cm 

✔radius = (x + 5)  
✔applying Pythagoras theorem
✔value of x
✔length of radius (4)

[7]

9.1 

interior opposite angle 

✔answer (1) 

9.2.1 

R₂ = Q₂ = 41° (∠s in the same seg)
P₄ = Q₁ = 41° (tan-chord theorem)
T₁ = P₄ = 41° (∠s opp. = sides)/(∠e teenoor gelyke sye)  

OR

T₁ = R₂ = 41° (tan – chord theorem)/(raaklyn-koord stelling) 

✔Statement  
✔Reason   
✔Statement 
✔Reason  
✔Statement 
✔Reason (6)

9.2.2(a) 

T₂ + 34° = 78°   (tan - chord theorem )
∴T₂ = 44°

✔Statement 
✔Reason (2)

9.2.2(b) 

41° + Q₂ + 44° + 34° = 180°   ( opp. ∠ s of a cyclic quad.) 
∴Q₂ = 61°

✔Statement 
✔Reason (2)

9.2.2(c) 

T₄ = 41°  + 61°     ( ext. ∠s of a cyclic quad.) 
∴T₄ = 102°  
OR
T₄ + 44°  + 34°  = 180°   (int.∠s of a Δ)

✔Statement  
✔Reason 

OR

✔Statement 
✔Reason (2)

9.2.2(d) 

W + 41°+ 41° = 180° ( int .∠s of a Δ)
∴ ܹW = 98° 

✔Statement
✔ Reason (2)

9.2.3(a) 

Q₂ =  61º      and   T₂ =  44º 
∴ Q₂  ≠  T₂
∴  QR is not parallel to PT (alt. ∠s are not equal) 

✔Q₂ ≠  T₂
✔alt. ∠s are not equal  (2)

9.2.3(b) 

R₂ + W =  41º +  98º
= 139º ≠ 180º 
∴ PRTW is not a cyclicquad. (Opp. ∠s are not supp.) 

✔R₂ +W ≠ 180º
✔ PRTWis not a cyclicquad. (2)

9.2.3(c) 

R₁ = T₂ = 44°(∠s in same seg) 
R₁ + R₂ = 44° + 41°  
               = 95° ≠ 90°  
∴ TQ is not a diameter (∠subt. by TQ is not a right angle) 

✔R₁ + R₂ ≠  90º
✔TQ is not a diameter (2)

[21]

10.1 

Construction: Draw heights h and l on KR and KS  respectively. Join LS and MR  
Proof:

But: area of ΔLRS =  area of ΔMSR
Area of ΔKRS = Area of ΔKRS    
Area of ΔLRS     Area of ΔMSR 
∴ KR = KS
   RL    SM

✔construction 
✔ratio of areas   
✔ratio of areas  
✔same base and same height  
✔ Area of ΔKRS = Area of ΔKRS    (5)
    Area of ΔLRS     Area of ΔMSR 

10.2.1 

AB = 5 units 

✔5  (1)

10.2.2(a) 

C is common 
CBA = CFB (both =90°) 
CAB = CBF (sum of ∠s of Δ)
∴ ΔCBA|||ΔCFB (∠, ∠, ∠)

✔Statement /  Reason
✔Statement / Reason
✔Reason  (3)

10.2.2(b) 

CB = CA  (|||Δs)
CF    CB 
CB² = CF.CA
∴ CF = CB²
            CA

✔proportion
✔reason
✔ CB² = CF.CA (3)

10.2.3 

 CF = CB²
          CA
 CF = (12)²
           13
≈ 11units

✔substitution g 
✔length of CF   (2)

10.2.4 

AF = 13 – 11 = 2 unit

✔length of AF (1)

10.2.5 

CB  =  CA = (prop. theorem ; DF||BA )
BD      AF 
12 = 13
BD    2 
∴ BD = 24
            13

CF = CB (prop. theorem ; DF ||BA )
FE    BD 
11 = 12
FE   24 
        13 
∴FE = 22 units
          13 

✔proportion  
✔reason 
✔length of BD  
✔proportion 
✔length of FE (5)

[20]

TOTAL: 150