1.1

45 children

🗸answer (1)

1.2

= ∑fx = (4 x 2) + (8 x10) + (12 x 9) + (16 x 7) + ( 20 x 8) + (24 x 7) + (28 x 2)
        n                                                     45
= 692 OR  = 15,38 minutes      Answer only: full marks
       45

🗸692
🗸answer
(2)

1.3

🗸first 4 cum freq correct
🗸last 3 cum freq correct
(2)

Time taken (t) (in minutes)

Number of children

Cumulative frequency

2 < t ≤ 6

2

2

6 < t ≤ 10

10

12

10 < t ≤ 14

9

21

14 < t ≤ 18

7

28

18 < t ≤ 22

8

36

22 < t ≤ 26

7

43

26 < t ≤ 30

2

45

1.5

On graph at the y-value of 22,5 or 23
Median = ± 15 minutes.       Answer only: full marks

🗸 graph
🗸 answer
(2)

[10]

2.1

a = 12,44
b = 0,98
y = 12,44 + 0,98x
Answer only: full marks

🗸 value of a
🗸 value of b
🗸 equation

2.2.1

Percentage =15 x 100
                     50
= 30%

🗸answer

(1)

2.2.2

y = 12,44 + 0,98x
y = 12,4 4 + 0,98(30)
yˆ = 41,84
= 42             Answer only: full marks
OR

y = 41,87 (if using calculator)
y = 42 

OR

y = 21 
      50

2.3.1

standard deviation =13,88

🗸 🗸 answer

(2)

2.3.2

x = 50,67 - 45,67
= 5%

Answer only: full marks

🗸 50,67 - 45,67
🗸 answer

(2)

[10]

3.1.1

Midpoint of EC:
= (-2 + 2 ; 0 + (-3)) = (0 ; - 3)
         2          2                   2

🗸 x value 🗸 y value
(2)

3.1.2

m_DC  = - 3 - (-5) OR - 5 - (-3)
                2-(-2)           -2-2
= 2 = 1                            Answer only: full marks
   4    2

🗸 substitution
🗸 answer
(2)

3.1.3

m_AB = ½  [AB || DC]
y = ½ x + c                                  y - y₁= ½(x - x₁)
0 =½ (-2) + c           OR               y - 0 = ½ (x - (-2))
c =1
∴ y = ½ x + 1

🗸m_AB =½
🗸 substitution of ( - 2 ;0)
🗸equation
(3)

3.1.4

tan a = m_AB = ½
a = 26,57°
θ = 90° + 26,57°      [ext ∠ of Δ]
=116,57°

🗸 tan a = ½
🗸 value of a
🗸 value of q
(3)

B(0 ; 1)
m_BC = 1 - (-3)   OR m_BC = (-3) - 1
              0-2                           2-0
= -2                                      = -2
mAB x mBC  = ½ x -2
= - 1
∴ AB ⊥ BC

🗸 coordinates of B
🗸 mBC = - 2
🗸 product of gradients = –1

(3)

ABˆ C = 90°
∴ EC is diameter [converse : ∠ in semi circle]
∴ centre of circle= æ 0 ; - 3
                                        2

🗸 answer
(1)

🗸 substitution of centre
🗸 correct substitution of E(–1 ; 0), B(0 ; 1) or
C(2 ; –3) to calculate
r ² or r
🗸 value of r ² or r
🗸 equation
(4)

4.1

(x - 2)² + ( y -1)² = 25               (x-2)² + (y-1)²= 25
(-2 - 2)2 + (b - 1)2 = 25            (-2-2)² + (b-1)² = 25 
(b -1)² = 9                     OR     16+b² - 2b + 1 = 25
b - 1 = ± 3                                 b² -2b - 8 = 0
∴ b = 4      or b ≠ - 2                   ∴ b=4 or b ≠ -2

🗸 equation of the circle
🗸substitution of point T
🗸simplification
🗸 answer
(4)

4.2.1

K(2 ; 1 – 5)
∴ K(2 ; –4)            Answer only: full marks

🗸 x value 🗸 y value
(2)

4.2.2

m_MT   =4 -1  =- 3 
           -2 -2      4
m_PL= 4         [radius ^ tangent]
          3
y = 4  x + c
      3
4 = 4  (-2) + c
      3
c = 20 
       3
y = 4  x + 20 
      3        3

OR
m_MT   =4 -1  =- 3 
           -2 -2      4
m_PL= 4         [radius ^ tangent]
          3
y - y₁ =  4  (x - x₁)
              3
y = 4 = 4  (x + 2) 
            3
y = 4  x + 20 
      3        3

OR

P(–11 ; –8)

mPL =  4 - (-8)  
          - 2 - (-11)
= 4 
   3
y = 4  x + c
      3 
- 8 = 4 (-11) + c
        3
c = 20 
      3
y = 4  x + 20 
      3        3

🗸 m_MT

🗸 m_PL = 4 
              3
🗸 substitution of m_PL and the point T
🗸 equation (4)

🗸 m_MT

🗸 m_PL = 4 
              3
🗸 substitution of m_PL and the point T
🗸 equation (4)

(4)

🗸 coordinates of P

🗸 m_PL = 4 
              3
🗸 substitution of m_PL and the point T
🗸 equation (4)

(4)

Coordinates of P:

OR
y_L =  4  (2) +  20  = 28 
        3             3      3 
L(2; 28) and K(2; -4): LK = 28  -(-4)= 40 
        3                                  3             3 

Coordinates of P:

tanθ = 4  ∴θ = 17.1027...º
          13
∴ PKL = 90º + 17.1027...º = 107.1º
Area ΔPKL= ½ (PK)(LK). sinPKL
=½(√185)(40)sin107.10º
                 3
= 86.67 square units

🗸 y_L= 28 
            3
🗸 length of LK 
🗸 x_P 🗸 y_P
🗸 length of ⊥ height
🗸 substitution into the area formula
🗸 answer
(7)

🗸 y_L= 28 
            3
🗸 length of LK 
🗸 x_P 🗸 y_P
🗸 PKL
🗸 substitution into the area
🗸 answer
(7)

The centres of the two circles lie on the same vertical line
x = 2. and the sum of the radii = 10
n -1 = 10                                 1 - n = 10
n =11                or                   n = - 9
Answer only: full marks

🗸 correct method
🗸 sum of radii = 10
🗸 n =11 🗸 n = - 9
(4)

5.1.1

sin191°
= - sin11°

🗸 - sin11°
(1)

5.1.2

cos 22°
= cos(2 ´11°)
= 1 - 2sin ²11°

🗸 answer
(1)

5.2

cos(x - 180°) + √2 sin (x + 45°)
= - cos x + √2(sin x cos 45° + cos x sin 45°)
= - cos x + √2(sin x( 1 )) + cos x ( 1 ))
                               √2                 √2
= - cos x + sin x + cos x
= sin x

OR
cos(x - 180°) + √2sin (x + 45°)
= - cos x + √2(sin x cos 45° + cos x sin 45°)
= - cos x + √2(sin x( 2 )) + cos x ( 2 ))
                               √2                 √2
= - cos x + sin x + cos x
= sin x

🗸 - cos x 🗸 expansion
🗸 special angle ratios
🗸simplification of last 2 terms
🗸answer
(5)

🗸 - cos x 🗸 expansion
🗸 special angle ratios
🗸simplification of last 2 terms
🗸answer
(5)

5.3

sin P + sin Q = sin P + cos P
(sin P + cos P)²  = ( 7 )²
                                5
sin ² P + 2 sin P cos P + cos² P = 49 
                                                      25
2 sin P cos P = 49 - 1
                         25
sin 2P =(49 - 25)
              25    25 
= 24 
   25

🗸 sin Q = cos P
🗸 squaring
🗸 expansion
🗸 sin ² P + cos² P =1
🗸answer
(5)

[12]

6.1

cos(x - 30°) = 2 sin x
cos x cos 30° + sin x sin 30° = 2 sin x
√3 cos x + ½ sin x = 2 sin x
 2
√3 cos x = 3  sin x
2               2
tan x =√3
            3
x = 30° + k.180°;  k ∈ Z

OR
x = 30° + k.360° or x = 210° + k.360° ; k ∈ Z

🗸 expansion
🗸 special ∠ s

🗸 simplification
🗸 equation in tan
🗸 30°
🗸 k.180°; k ∈ Z

OR

🗸 30° and 210°
🗸 k.360° ; k ∈ Z
(6)

🗸 x value 🗸 y value
(2)

🗸 endpoints
🗸 correct interval
(2)

🗸endpoints
🗸correct interval
(2)

y = 2^{2 sin x +3}
Range of y = 2 sin x + 3 : y ∈ [1 ; 5] OR 1 ≤ y ≤ 5
Range: y = 2^2sinx+3 : y ∈[2 ; 32] OR 2 ≤ y ≤ 32
Answer only: full marks

🗸 1 🗸 5
🗸 2 🗸 32
🗸 correct interval
(5)

7.1.1

sinq =   x                sinθ = sin90º
            AC   OR       x          AC

AC =   x              AC =   x  
         sinθ                    sinθ

🗸 trig ratio
🗸 simplification
(2)

7.1.2

cos 60° = x + 2            sin30 = sin90º
                 CE     OR     x + 2     CE
CE =  x + 2                   CE =  x + 2 
        cos 60°                           sin30º
= x + 2 = 2(x + 2)             =2(x + 2)
     1 
     2

🗸 trig ratio
🗸 making CE the subject
(2)

7.2

Area ΔACE =  ½ AC.EC.sin ACE
= ½(  x  )(2(x + 2))sin 2q
       sinθ
= x(x + 2) x 2 sinq cosθ
              sinθ
= 2x(x + 2)cosθ

🗸 use area rule correctly
🗸 substitution of
x    (2(x + 2)) sinθ
🗸 substitution of sin 2θ
(3)

EC = 2(12 + 2) = 28
AE² = AC² + EC² - 2(AC)(EC)cosACE
=(  12   )² + 28² - 2(   12   )(28)cos110°
   sin 55°                 sin 55°
AE = 35,77m

🗸 EC
🗸 use cosine rule correctly
🗸 substitution
🗸 answer
(4)

8.1.1(a)

MOS= 62°  [∠ at centre = 2´ ∠ at circumf]

🗸S 🗸 R
(2)

8.1.1(b)

L = 31°        [equal chords; equal  ∠s]

🗸S 🗸 R
(2)

8.1.2

LN = NP and LO = OM
ON = ½ PM            [ midpoint theorem]
ON = ½ MS            [PM = MS]

OR
N₁ = 90°  [line from centre to midpt chord]
P = 90°   [∠ in semi-circle]
L is common
ΔNLO ||| Δ PLM (∠∠∠)
NL = NO = ½
PL    PM  
ON = ½PM
ON = ½MS  [PM = MS]

🗸LO = OM
🗸S 🗸 R
🗸S
(4)

🗸S R
🗸S/R
🗸S
🗸S
(4)

8.2.1

AN = AK [line || one side of D OR prop theorem; KN ||BM
AM   AB
AN = 3y = 3 
AM    5y    5

🗸R 

🗸S
(2)

8.2.2

AM = 10x     [given]
MC    23x
AM = 5 y =10x    \    y = 2x
LC = MC      [line || one side of D OR prop theorem; KN ||LM
KL    NM
= 23x = 23x = 23 
    2y     4x       4

OR
AM = 10x        [given]
MC    23x
AN = 3y = 6x
MN    2y   4x
LC = MC      [line || one side of D OR prop theorem; KN ||LM
KL    NM
= 23x = 23x = 23 
    2y     4x       4

🗸S
🗸R
🗸S
(3)

OR

🗸S
🗸R
🗸S
(3)

[13]

9.1

B₁ = x       [∠ s  opp = sides]
M₂  = 2x           [ext∠ of ∆]    OR   Mˆ 1 =180° - 2x  [∠s of ∆]
BM = MN       [ 2 tans from a common point]
N   = 180° - 2x = 90° - x          [ ∠s  opp = sides]
               2

OR
NM = BM [ 2 tans from a common point]
B₂  = N₁   [ ∠'s  opp = sides] B₁  = x     [ ∠'s  opp = sides] In ∆ KBN
x + x + B₂  + N₁ =180° [sum of  ∠'s  of ∆]
2x + 2N₁ =180°
x + N₁  = 90°
N₁ = 90° - x

🗸S
🗸S 🗸R
🗸S 🗸R
🗸answer
(6)

🗸S 🗸R
🗸S 🗸R
🗸S
🗸answer
(6)

9.2

MBA  = B₂ + B₃ = 90°   [tangent^diameter]
B₃= 90° - B₂
= 90° - (90° - x) = x
B₃ = Kˆ  =  x
AB is a tangent converse tan-chord theorem

OR
B₂  = N₁
B₁ + B₂ = x + (90° - x) = 90°
KN is diameter [converse ∠ in semi-circle
MBA  = B₂  + B₃  = 90°    [tangent ^diameter]
AB is a tangent converse tan-chord theorem

🗸S 🗸R
🗸S
🗸S
🗸R
(5)

10.1

Constr: Let M and N lie on AB and AC respectively such that
AM = DE and AN = DF. Draw MN.

Proof:
In ∆ AMN and ∆ DEF
AM = DE [Constr]
AN = DF [Constr]
A = D       [Given]
Δ AMN º Δ DEF(SAS)
AMˆ N = E = B
MN || BC [corresp ∠s are equal]
AB = AC       [line || one side of D OR prop theorem; MN ||BC]
AM   AN
AB = AC    [AM = DE and AN = DF]
DE    DF

🗸Constr
🗸Δ AMN Ξ Δ DEF
🗸SAS
🗸MN || BC and R
🗸AB = AC 🗸R
   AM   AN
(6)

10.2.1(a)

DOˆ B= 90°
DGˆ F= Gˆ 3  + Gˆ 4  = 90°    [∠ in semi-circle]
DOˆ B+ DGˆ F=180°
DGFO is a cyclic quad.      [converse: opp ∠s of cyclic quad] OR
∠s of quad = 180°/Ðe van koordevh = 180°]

OR
EOB= 90°
DGF= G₃  + G₄  = 90°    [∠ in semi-circle]
EOB = DGF
DGFO is a cyclic quad.   [converse: ext ∠ = opp int ∠
OR
ext∠ of quad = opp int ∠]

🗸S 🗸R
🗸R
(3)

🗸S🗸 R
🗸R
(3)

10.2.1(b)

F₁   = D          [ext ∠ of cyclic quad]
G₁  + G₂  = D    [tan-chord theorem]
F₁  = G₁  + G₂
GC = CF      [ sides opp equal ∠s]

🗸S 🗸R
🗸S🗸 R
🗸R (5)

10.2.2(a)

AB = DE = 14                        [diameters]
OB = 7 units
BC = OC – OB = 11 – 7          Answer only: full marks
= 4 units

🗸S
🗸S
🗸S
(3)

10.2.2(b)

In Δ CGB and Δ CAG
G₁  = A =  x        [tan-chord theorem]
C = C     [common]
∆CGB ||| ∆CAG     [∠∠∠]
CG = CB
CA    CG
CG = 4 
18    CG
CG ² = 72
CG = √72 or 6√2  or 8,49 units

🗸S/R

🗸S

🗸S

🗸CA = 18

🗸answer

(5)

10.2.2(c)

OF = OC – FC
= 11 – √72
tan E = OF 
            OE
= 11 - √72 = 0,36
        7
E = 19,76°

OR
OF = OC – FC
= 11 – √72
FE² = OE ² + OF²
= 7² + (11 - √72)²
FE = 7,437.. = 7,44
cos E = OE         OR        sin E = OF 
             FE                                    FE
=   7   = 0,94                 = 11 - √72 = 0,338
   7,44                                  7,44
E = 19,76°                      E = 19,76°

🗸OF
🗸trig ratio
🗸substitution
🗸answer
(4)

🗸OF
🗸trig ratio
🗸substitution
🗸answer
(4)

[26]