MATHEMATICS PAPER 2
GRADE 12
MEMORANDUM
NOVEMBER 2021

QUESTION 1

1.1.1
x =396
      18
x = 22
(2)

1.1.2
σ = 10,1707 ≈ 10,17 (1)

1.1.3
x +  σ = 32,17
= 5 days (2)

1.2
22 x 18 = 396 ordered
20 x 18 = 360 sold/verkoop
Total not sold: 36
OR
22–20 = 2
2x18 = 36
(2)

1.3.1
Option B
Any one of the following reasons:

• Median= 18,5
• Q₁ = 14
• IQR = 21
• Mean > Median, therefore the data is skewed to the right
  (2)

1.3.2 Data is positively skewed/skewed to the right
[10]

QUESTION 2

SCATTER PLOT

1 mark: 3 to 5 points plotted correctly
2 marks: 6 to 9 points plotted correctly
3 marks: all points plotted correctly
(3)

2.2
a = 90,478... ≈ 90,48
b = -1,773... ≈ -1,77
yˆ = 90,48 - 1,77x
(3)

2.3 y = 23,069… ≈23,07 units(calculator)
OR
y = 90,48 - 1,77(38)
y = 23,22 units
(2)

2.4 r = – 0,94
The value of r indicates a strong relationship between the cost per 5 litre and the number of units sold therefore there is a good chance of the prediction being accurate.(2)
[10]

QUESTION 3

3.1.1
m_BE = m_CE = 0 -( - 2) 
                       12 -4
= ¼
OR
m_BE = m_CE =-2 - 0
                     4 - 12
= ¼ (2)

3.1.2
m_AB = tan81,87º 
m_AB = 7 (2)

3.2
y = mx + c 
0 = ¼(12) + c
c = -3 
y = ¼x - 3
or
y - y₁ = m(x - x₁)
y - 0 = ¼(x - 12)
y = ¼x - 3
OR
y = mx + c 
-2 = ¼(4) + c
c = -3
y = ¼x - 3
or
y - y₁ = m(x - x₁)
y - (-2) = ¼(x - 4)
y = ¼x - 3
(2)

3.3.1
y = ¼x - 3
k = ¼k - 3 
¾k = -3
k = -4
B = (-4; -4) 
OR
m_BE = ¼ 
0 - k = ¼
12 - k
-4k = 12 - k
k = -4
B = (-4; -4)
OR
m_BE = ¼
k = ¼
k - 12
4k = k - 12
k = -4
OR
m_AB = tan81,87º 
m_AB = 7
m_AB = 10 - k
           -2 - k 
7(-2 - k) = 10 - k
-14 - 7k = 10 - k
-6k = 24
k = -4
B (-4; -4)
OR
EB: y = ¼x - 3 and AB: y = 7x + 24
¼x - 3 = 7x + 24
²⁷/₄x = -27
x = k = -4
B = (-4;-4)
(2)

3.3.2
In ΔAFG:
m_AC = 10 - (-2) = -2
             -2 -4
tanθ = m_AC = -2
θ = 180 - 63.43...º
θ = 116,57º
Aˆ =116,57º - 81.87º[ext∠ of Δ]
Aˆ = 34.70º
OR
In ΔABC:
a = BC = 2√17; b = AC 6√5; c = AB 10√2
a² = b² + c² -2bCosA
(2√17)² = (6√5)²+(10√2)²- 2(6√5)(10√2).cosA
cos A = (6√5)²+(10√2)²-(2√17)²
                     2(6√5)(10√2)
= 0.8222...
A = 34.7º
(4)

3.3.3
M[12 + (-2) ; 10 + (0)]
         2               2 
Diagonals intersect at the point (5 ; 5)

3.4.1
BE = ET
4√17 = √(12 - p)² + (0 - p)²
(4√17)² = (√(12 - p)² + (0 - p)²)
272 = 144 - 24p + p² + p²
p² = - 12p - 64 = 0
(p - 16)(p + 4) = 0
p = 16 or p = – 4 (n.a.)
∴T(16; 16)
(5)

3.4.2

1. (x - 12)² + y² = (4√17)² = 272(2)
2. m_radius = ¼
   m_tangent = -4
   y = -4x + c
   - 4 = -4(-4) + c
   c = – 20
   y = – 4x – 20
   OR
   y - y₁ = -4(x - x₁)
   y - (-4) = -4(x - (-4))
   y = -4x - 20 (3)
   [24]
   

QUESTION 4


4.1 Radius = 4 units (1)

4.2.1
CD ⊥ CN
∴C( -1 ; 7) (2)

4.2.2
CD = 6 units
∴D(5 ; 7)
(2)

4.2.3
⊥h = 5units
DC = 6 units
Area ΔBCD = ½(6)(5)
=15units²
OR
⊥h = 5units
DC = 6 units
Area ΔBCD = ½[Area of ||^m]
=½(5)(6)
=15units²
(3)
OR
Let andgle of inclionation of BC = a
tan a = ⁵/₃
a = 59,036..º
BCD = 180º - a
BCD = 180º - 59,036º
BCD = 120,96º
Area ΔBCD = ½(√34)(6)sin120,96º
= 15 units²
(3)

4.3.1
M(3 ; –1) [reflection of N(–1 ; 3) about the line y = x]
MN = √(3 - (-1))² + (-1 - 3)²
MN = √32 = 4√2 = 5,66units
(3)

4.3.2
M(3 ; –1)
m_MN = 3 - (-1) = -1
             -1 - 3
MN: -1 = -(3) + c
c = 2
y = -x + 2
or
y - 3 = -1(x + 1)
y - 3 = -x - 1 
y = -x + 2

x = -x + 2 
2x = 2
x = 1
y = 1
midpoint (1; 1)
OR

N(-1 ; 3)
y_F = y_N = 3
Reflected about y = x
F(3 ; 3)
midpoint (-1 + 3  ; -1 + 3) = (1 ; 1)
                    2           2 
OR

NAMF is a square (NA=NF=AM=MF and NA ⊥ AM)
Midpoint NM= (1 ; 1)
= Midpoint of AF
(4)
[15]

QUESTION 5
5.1
sin140.sin(360º - x)
    cos 50.tan(-x)
= sin40º(- sin x)
   sin40º(- tan x)
- sin x
- sin x
 cos x
= cos x
(6)

5.2
LHS = -2sin²x + cos x + 1  RHS = 2cos x - 1
             1 - cos(540º - x)
LHS = 2(1 - cos²x) + cos x + 1
                    1 - (- cos )
LHS = -2 + 2cos²x + cos x - 1 
                      1 + cos x 
LHS = (2cosx - 1)(cos x + 1)
                   1 + cos x 
LHS = 2cos x - 1
LHS = RHS
(4)

5.3.1
sin 36º = √1 - p²
tan36º = √1 - p²
                   p

OR
cos²36º = 1 sin²36º
cos36º = √1 - (1 - p²)
= p
tan 36º = sin36º
               cos 36º
= √1 - p²
      p
(3)

5.3.2
cos108º
= -cos72º
= –cos (2×36°)
= -(2cos²36º - 1) 
= -2p² + 1 
OR
cos108º
= -cos72º
= –cos (2×36°)
= (1 - 2sin²36º)
= 1 + 2(√1 - p²)²
= 1 + 2(1 - p²)
= -2p² + 1
OR
cos108º
= -cos72º
= –cos (2×36°)
= -(cos²36ºsin²36º) 
= - (p² - (√1 - p²)²)
= - (p² -(1 - p²))
=  -2p² + 1
OR
cos108º
= cos(2 x 54º)
= 2cos² 54º - 1
= 2(1 - p²) - 1
= 1 - 2p²
OR
cos108º = cos(72º + 36º)
= cos72º cos36º - sin 72º sin 36º
= (2cos²36º 1) cos36º -(2sin 36º cos36º) sin 36º
= 2cos³36º - cos36º - 2cos36ºsin²36º
= 2p³ - p - 2p(√1 - p²)²
= 2p³ - p - 2p + 2p³
= 4p³ - 3p
(4)
[17]

QUESTION 6
6.1.1
cos(a + β)
= cos(a - (- β))
= cos a cos(- β) + sin a sin(- β)
= cos a cosβ +sin a(-sinβ)
= cos a cosβ - sin a sinβ
(3)

6.1.2
2 cos6 cos 4 - cos10x + 2sin²x
= 2 cos 6x cos 4x - cos(6x + 4x) + 2sin²x
= 2 cos 6x cos 4x - (cos 6x cos 4x - sin 6x sin 4x)  + 2sin²x
= cos 6x cos 4x + sin 6x sin 4x + 2sin²x
= cos 2 + 2sin²x
= 1 - 2sin²x + 2sin²x
= 1
(5)

6.2
tan x = 2sin 2x
 sin x  = 2(2sin x cos x)
cos x
sin x = 4sin x cos²x
4sin x cos²x - sin x = 0
sin x (4cos²x - 1) = 0
sin x = 0
or
cos²x = ¼
cos x = - ½
x = 180° + k.360°; k∈Z
or
x = 120° + k.360°; k∈Z
x = 240° + k.360°; k∈Z
OR
tan x = 2sin 2x
 sin x  = 2(2sin x cos x)
cos x
sin x = 4sin x cos²x
4sin x cos²x - sin x = 0
4sin x(1 - sin²x) - sin x = 0
3sin - 4sin³ = 0
sin x (3 - 4sin²x) = 0
sin x = 0
or
sin²x = ¾
sin x = ^√3/₂ or sin x = -^√3/₂
x = 180º + k.360º, k∈Z
or
x= 120º + k.360º,k∈Z
or
x = 240º + k.360º,k∈Z (7)
[15]

QUESTION 7
7.1

both turning points
both x intercepts (–30° & 150°)
shape (3)

7.2
Period = 120° (2)

7.3
x = -30º (1)

7.4
Range of g: y ∈ [-1;1]
Range of ½g: y ∈ [-½ ; ½]
Range of ½g + 1: y ∈ [-½ ; ³/₂] 
OR
Range of ½g + 1: ½ ≤ y ≤ ³/₂ (2)
[8]

QUESTION 8

8.1
In ΔSQR:  
 QS  =       QR      
sin x   sin(90º + x)
 QS  =   5   
sin x   cos x
QS = 5sin x 
          cos x
QS = 5tan x
(3)

8.2
        QT         =   TS  
sin(180º - 2x)    sin x
  QT    = 5tan x
sin 2x     sin x
QT = 5tan x sin 2x
               sin x
QT =  
QT = 5sin x (2sin x)
                sin x
QT = 10sin x (5)
OR
QT² = QS² TS² - 2QS.TScosQST
QT² = (5 tan )² + (5tan )^{2 -} 2(5 tan x).(5 tan x)cos(180º - 2x)
QT² = 50 tan² x - 50 tan² x(cos2x)
QT² = 50 tan² x(1 + cos2x) 
QT² = 50 tan² x(1 + 2cos² x - 1)
QT² = 50 tan² x(2cos²x) 
QT² =100 sin²x (cos2x)
                cos²x
QT² = 100sin²x
QT² = 10sin x
OR
TS² =QS² +TQ² - 2QS.TQ.cos x
(5tan x)² = (5tan x) + TQ² - 2(5tan x).TQ.cos
0 = TQ - 2(5tan x).TQ.cos x
0 = TQ[TQ - 10 tan x .cos x]
TQ = 10 tan x.cos x(TQ ≠ 0)
= 10 sin x  .cos
        cos x
= 10sin x
(5)

8.3
Area of ΔTQR = ½.TQ.QRsinTQR
= ½(10sin 25º)(5)(sin70º)
= 9,93unit
(2)
[10]

QUESTION 9

9.1
tangents from same(common) point (1)

9.2.1
Sˆ = SRT 
S₂ = 51º
[∠s opp equal sides]
[sum of ∠s in Δ]
(2)

9.2.2
S₂ + S₃ = 93º
S₃ = 42º

S₁ = 87º
S₃ = 180º - (87º + 51º)
S₃ = 42º
[∠s of cyclic quad]
[∠s on a str line]
[5]

QUESTION 10

10.1
line from centre ⊥ to chord(1)

10.2
A₁ = 90º[sum of ∠s in Δ]
M₁ = 180º - 2x [∠ at centre=2× at circumf]
(3)

10.3
CAD = 90º
A₂ = 90º - (90º - x)
A₂ = C = x
AD  is a tangent [converse tan-chord theorem]
OR
EMD = 2x [adj suppl ∠s]
A₂ = x [∠ at centre= 2×∠ at circumf]
A₂ = C = x
AD is a tangent [converse tan-chord theorem]
OR
M₃ = 180º - 2x[vert. opp]
A₃ = 90º - x [∠ at centre= 2×∠ at circumf= 2]
BAE = 90º [∠ in semi-circle]
A₂ = C = x
AD is a tangent [converse tan-chord theorem]
OR
CD || AB [midpt. Thm]
BAE = 90º [∠ in semi-circle]
A₃ = D = 90 - x [alt.∠s;CD||AB]
AD is a tangent [converse tan-chord theorem]
OR
CAD = 90º [∠ in semi circle]
AC = diameter [converse ∠ in semi circle]
AD is a tangent [converse radius ⊥ tangent]
(4)

10.4
AF = FE and BM = ME [given & radii]
FM = ½AB = 12 units [Midpt Theorem]
EM = MB = CM = 18 units [radii]
EB = 36 units [diameter = 2 radius]
AE² = (36)² – (24)² [Pythagoras]
AE = 12√5 or 26,83 units
OR
AF = FE and BM = ME [given & radii]
FM = ½AB = 12units
AB = 12 units [Midpt Theorem]
EM = MB = CM = 18 units [radii]
FE² = (18)² – (12)² [Pythagoras]
FE = 6√5
AE = 12 5 or 26,83 units (5)
[13]

QUESTION 11
11.1

Construction: Draw diameter BF and draw BE
BFK = 90º  or DFK = 90º - BFD [radius ⊥ tangent]
BEF = 90º[∠ in semi-circle]
DEF = 90º - BED
= 90° – BFD [∠s same segment]
DFK = DEF
(5)
OR

Construction: Draw radii DO and OF
OFK = 90º or DFK = 90º - OFD radius ⊥ tangent]
ODF = OFD [∠s opp = sides]
DOF = 180º - 2OFD [∠s of Δ]
DEF = 90º- OFD [∠ at centre = 2×∠ circumf]
DFK = DEF
(5)
OR

Construction: Draw diameter BF and join BD.
BFK = 90º or DFK = 90º - BFˆD [radius tangent]
FDB = 90º [∠ in half circle/semi-sirkel]
B = 90º -BFD
DFK = B
but B = E[∠s same segment]
DFK = E
(5)

11.2

K₄ = S₁ [tan chord theorem]
M₂ + M₃ = S1 [corresp ∠s; MN || KS]
K₄ = M₂ + M₃ = NML
(4)

K₄ = M₂ + M₃ = NML
KLMN is a cyclic quad [ext∠ of quad = opp int ∠]
OR
N₁ = K₁ + K₂ = NKS [corresp ∠s;MN || KS]
NKS = KLS [tan chord theorem]
N₁ = KLS
KLMN is a cyclic quad [ext ∠ of quad = opp int ∠]
OR
NKL 180º - K₄ [adj. suppl.]
NKL 180º - NML [proved]
KLMN is a cyclic quad [opp. s supplementary] (1)

11.2.2
In ΔLKN|||ΔKSM:
N₃ = M₃ [∠s in the same seg]
L₁ = M₂ [∠s in the same seg]
= K₂ [alt ∠s; MN||KS]
NKL = MSK [∠s of Δ]
cLKN|||ΔKSM
OR
In ΔLKN|||ΔKSM:
N₃ = M₃ [∠s in the same seg]
NKL=M₁ [ext ∠ of cyclic quad]
= S₂ [corresp ∠s; KS || NM]
∠LKN|||ΔKSM [∠,∠,∠]
OR
In LKN|||ΔKSM:
N₃ = M₃ [∠s in the same seg]
K₄ +NKL=S₁ +S₂ [∠s on straight line]
NKL=S₂ [ K₄ = S₁]
∠LKN|||ΔKSM [∠,∠,∠]
(5)

11.2.3
^LK/_KS = ^KN/_SM
=¹²/_KS = ⁴/₃
SM = 6
^LT/_NL = ^LS/_ML
^LT/₁₆ = ¹³/₁₉
LT = ²⁰⁸/₁₉ = 10,95
[line || one side of Δ]
(4)
[23]
TOTAL: 150