MATHEMATICS PAPER 2
GRADE 12
NATIONAL SENIOR CERTIFICATE EXAMINATIONS
MEMORANDUM
MAY/JUNE 2021

NOTE:

• If a candidate answers a question TWICE, only mark the FIRST attempt.
• If a candidate has crossed out an attempt of a question and not redone the question, mark the crossed out version.
• Consistent accuracy applies in ALL aspects of the marking memorandum. Stop marking at the second calculation error.
• Assuming answers/values in order to solve a problem is NOT acceptable.

GEOMETRY

• S - A mark for a correct statement
  (A statement mark is independent of a reason)
• R - A mark for the correct reason
  (A reason mark may only be awarded if the statement is correct)
• S/R - Award a mark if statement AND reason are both correct

QUESTION 1
1.1

1.1.1

1. x = 375
          15
   x = 25 MB
   Answer only: Full marks (2)
2. σ = 17,65 MB (1)

1.1.2
25 + 17,65 = 42,65 
2 days 🗸  (2)

1.1.3
Overall x =  80  x 25
                  100
= 20 MB
375 + x = 20
   30
x = 600 - 375
= 225
maximum total amount of data that Sam must use for the remainder of the month: 225 MB
(3)

1.2

1.2.1
a =  29,35
b = -0,46
y = 29,35 - 0,46x 
(3)

1.2.2
y = 25,20 °C (calculator) OR
y = 29,35 - 0,46(9)
y = 25,21 °C 🗸🗸 answer
(2)
🗸 substitution
🗸 answer
(2)

1.2.3
b < 0 , indicating that as the wind speed increases the temperature decreases.
🗸 interpretation (1)
[14]

QUESTION 2

2.1 Modal class: 10 ≤ x < 15
🗸 answer (1)

2.2 177 learners
🗸 answer (1)

2.3 230 learners
🗸 answer (1)

2.4

🗸grounding at (0; 0)
🗸shape
🗸upper limits
🗸All other points correct

2.5 The median is at position 115.
⸫value of median is 12 days (accept 11 – 14) Answer only: Full marks
🗸 reading off at 115
🗸 answer
(2)
[9]

QUESTION 3

3.1
M(4 + 8 ; -8 + 0)
       2          2
M(6; 4)
x_M
y_M
(2)

3.2
m_NS =0 - (-16) or m_NQ =0 - (-8) or m_QS =-8 - (-16)
             8-0                       8-4                       4-0
= 2
🗸subst N and Q or N and Q or Q and S into gradient formula
🗸answer
(2)

3.3
m_LQ x 2 = -1 [LQ ⊥ NS]
m_LQ = -½
-8 = -½(4) + c OR y + 8 = -½(x +4)
c = -6                    y + 8 = -½x + 2
y = -½x - 6
🗸m_LQ
🗸 substitution of Q
🗸 calculation of c or simplification
(3)

3.4 OS is the radius of circle passing through S
(x - 0)² + (y - 0)² = (16)²
x² + y² = 256
Answer only: Full mark
🗸 identifying radius = 16
🗸 Equation of circle
(2)

3.5
m_RM = m_LQ = -½  [RM || LQ]
- 4 = -½ (6) + c OR y + 4 = -½ (x - 6)
c = -1                      y + 4 = -½ x + 3
y = -½x - 1
🗸 m_RM
🗸 substitution of M(6; - 4)
🗸 coordinates of T
(3)

3.6
T(0; -1) , P(0; -6) and S(0; -16)
PS = 10 units and TS = 15 units
LS = PS = 2 [prop theorem; RM || LP] OR [line || one side of Δ/lyn]
RS    TS    3 
Answer only: Full marks
🗸 PS = 10 units
🗸 TS = 15 units
🗸 answer
(3)

OR
M(6 ; - 4), Q(4 ; - 8) and S(0 ; -16)
MS = √180 = 6√5 and QS = 80 = 4√5
LS = QS = 2  [prop theorem; RM || LQ] OR [line || one side of Δ/lyn]
RS    MS   3 
Answer only: Full marks
🗸MS = 6√5
QS = 4√5
🗸 answer
(3)

3.7
area of PTMQ = area of ΔTSM – area of ΔPSQ
= ½ST. ⊥ h_M -½.PS. ⊥ h_Q
= ½(15)(6) - ½(10)(4)
= 45 – 20
= 25 square units
🗸 area of ΔTSM – area of ΔPSQ
🗸 area ΔTSM = 45
🗸 area ΔPSQ = 20
🗸 answer
(4)

OR
TM = √45 =3√5 = 6,71
MQ= √20 = 2√5 = 4,47
PQ =√20 = 2√5 = 4,47
area of trapezium PTMQ =½(3√5 + 2√5)(2√5)
= ½ (5√5)(2√5)
= 25 square units
🗸 TM  =3√5 
MQ = 2√5
PQ = 2√5 
🗸 area of trapezium = ½
(sum of ||sides)(height)
🗸 substitute into formula
🗸 answer
(4)

OR
MQ √20 = 2√5
PQ √20 = 2√5
TP √5
area of PTMQ = area of ΔMTP + area of ΔPQM
area of PTMQ = ½TP x ⊥ h_M + ½.MQ x PQ
area of PTMQ = ½(5) x 6 + ½(2√5 )(2√5 )
area of PTMQ = 10 + 15 = 25
🗸 area of ΔMTP + area of ΔPQM
🗸 area ΔMTP = 10
🗸 area ΔPQM = 15
🗸 answer
(4)
[19]

QUESTION 4

4.1
PV = r = √10
PV = √(k - (-3))² + (1 - 4)² = √10
(PV)² = (k - (-3))² + (1 - 4)² = 10
k² + 6k + 9 + 9 = 10 OR (k + 3)² + 9 = 10
k² + 6k + 8 = 0               (k + 3)² = 1
(k + 4)(k + 2) = 0            k + 3 = 1 or k + 3 = -1
k = -4 or k = -2
k = -2
🗸 PV = r = √10
🗸 substitution into distance formula
🗸 standard form
🗸 factors
🗸 answer
(5)

4.2
x² + 6x + y² - 8y +15 = 0
y-intercepts: (0)² + 6(0) + y² - 8y + 15 = 0
(y - 3)( y - 5) = 0
y_C = 3 or y_B = 5
BC = 2 units
🗸 x = 0
🗸 factors
🗸 both values
🗸 answer
(4)

4.3.1
m_TC =  3 - 1   
           0- (-2)
= 1
tan a = 1
a = 45º
🗸 substitution into gradient formula
🗸 tan a = 1
🗸 answer
(3)

OR
y = mx + 3
1 = m(-2) + 3
m_TC = 1
tan a = 1
a = 45º
🗸 substitution into equation of a line
🗸 tan a = 1
🗸 answer
(3)

4.3.2
BCV = 135º [ext ∠ of Δ]
VWB = 45º [opp ∠s of cyclic quad]
Answer only: Full marks
🗸 BCV = 135º
🗸 answer
(2)

OR
TCO = 45º [∠s of Δ]
VWB = 45º [ext ∠s of cyclic quad]
Answer only: Full marks
🗸 TCO = 45º
🗸 answer
(2)

4.4.1
Q(-3; - 2)
🗸 xQ 🗸 yQ (2)

4.4.2
(x + 3)² + (y + 2)² = 10
🗸 LHS 🗸 RHS (2)

4.4.3
x = -2 or x = -4 (2)
[20]

QUESTION 5
5.1 tan(-x).cos x.sin(x -180º) - 1
= -tan x.cos x.sin(-(180º - x)) -1
= -sin x .cos x.(- sin x) - 1
   cos x
= sin² x -1
= -cos² x
🗸 –tan x
🗸 –sin x
    cos x
🗸 sin 2 x -1
🗸 answer
(5)

5.2.1
cos 215°
= –cos 35°
= –m
🗸 reduction
🗸 answer
(2)

5.2.2 sin 20°
= cos 70°
= cos 2(35º)
= 2 cos² 35º - 1
= 2m² - 1
🗸 co-function
🗸 double angle expansion
🗸 answer in terms of m
(3)

OR
= sin (55° – 35°)
= sin55°cos35° – cos55°sin35°
= m.m -√1 - m². √1 - m²
= m² - (1 - m²)
= 2m² - 1
🗸 compound angle expansion
🗸 cos 55° √1 - m² or
sin 35° √1 - m²
🗸 answer in terms of m
(3)

5.3
cos 4x.cos x + sin 4x.sin x = -0,7
cos(4x - x) = -0,7
ref ∠ = 45,57…°
3x = 180° - 45,57...° + k.360° or     3x = 180° + 45,57...° k.360°
3x = 134,43° + k.360°            or     3x = 225,57° + k.360°
x = 44,81° + k.120° ; k ∈ Z              x = 75,19° + k.120° ; k ∈ Z
🗸 compound angle
🗸 3x = 134,43° or 225,57°
🗸 x = 44,81° or 75,19°
🗸 + k.120°; k ∈ Z
(4)

5.4
RHS = cos² x - sin² x
LHS = sin 4x.cos 2x - 2cos 4x.sin x.cos x
                                tan 2x
= sin 4x.cos 2x - cos 4x.sin 2x
                sin 2x 
                cos 2x
sin(4x - 2x)(cos 2x)
                   sin 2x
= cos 2x
=cos² x - sin² x
LHS = RHS
🗸 sin 2x
🗸 sin 2x
   cos 2x
🗸 sin(4x - 2x)
🗸 cos 2x
(4)
[18]

QUESTION 6
6.1
1 - 2sin² x = -sin x
2sin² x -sin x -1 = 0
(2sin x +1)(sin x -1) = 0
sin x = -½   or         sin x = 1
ref ∠ = 30°               ref ∠ = 90°
x = 210º + k.360º      x = 90º + k.360º 
or x = 330º  + k.360º 
x = -150º  or x = -30º  or x = 90º 
🗸 identity
🗸 factors
🗸 sin x = -½
🗸 sin x = 1
🗸 –150° and –30°
🗸 90° (A)
(6)

OR
cos 2x = -sin x
cos 2x = -cos(90º - x)
2x = 180º - (90º -  x) + k.360º or 2x = 180º + (90º - x) + k.360º
2x = 90º + x + k.360º  or  2x = 270º - x + k.360º
x = 90º + k.360º              x = 90º + k.12º
x = -150º or x = -30º  or  x= 90º
🗸 co-functions
🗸 2x in quadrant 2
🗸 2x in quadrant 3
🗸 both general solutions
🗸 –150° and –30°
🗸 90° (A)
(6)

OR
cos 2x = -sin x
cos 2x = cos(90º + x)
2x = 90º + x + k.360º or 2x = 360º - (90º + x) + k.360º
x = 90º + k.360º or 3x = 270º + k.360º
x = 90º + k.120º
x = -150º or x =- 30º or x =90º
🗸 co-functions
🗸 2x in quadrant 1
🗸 2x in quadrant 4
🗸 both general solutions
🗸 –150° and –30°
🗸 90° (A)
(6)

OR
cos 2x = -sin x
sin(90º - 2x) = -sin x
90º - 2x =180º x + k.360º or 90º - 2x = 360º - x + k.360º
x = -30º + k.120º x = -270º + k.360º
x = -150º or x = -30º or x=90º 
🗸 co-functions
🗸 90°–2x in quadrant 3
🗸 90°–2x in quadrant 4
🗸 both general solutions
🗸 –150° and –30°
🗸 90° (A)
(6)

6.2.

6.2.1
A(–150°; 0,5) B(–30°; 0,5)
AB = –30° – (–150°)
AB = 120° Answer only: Full marks
🗸 AB = –30° – (–150°)
🗸 answer
(2)

6.2.2
x ∈ (0º; 90º) or x ∈ (90º; 180º)
OR
0º < x < 90º or 90º < x < 180º
🗸 x ∈ (0º; 90º)
🗸 x ∈(90º; 180º)
(2) 

6.2.3

cos 2x = k - 3
k - 3 < -1 or k - 3 > 1
k < 2 or k > 4 Answer only: Full marks
OR
🗸 graph of y = cos2x + 3
🗸 k < 2 🗸 k > 4
(3)
[13] 

QUESTION 7

7.1
In ∆BCE:
  CE   =   BC   
sinB     sinBEC
  CE   =  BC  
sin30º   sin2x
CE = BCsin30º
           sin 2x
🗸 correct use of sine rule
🗸 CE = BCsin30º
                sin 2x

In ∆CDE:
DC = tan DEC
CE
DC = BC.sin30º (tan x)
            sin 2x
DC =        BC      ( sin x )
       4 sin x cos x  cos x 
DC =    BC    
         4cos²x
🗸 correct trig ratio
🗸 Subst CE
🗸 2sin x cos x
🗸 sin x
   cos x
(6)

7.2
DC =     BC     
         4cos² 30º
=     BC   
   4(^√3/₂)²
= BC 
    3
BC = 3DC
But AB = DC [opp sides of rectangle]
BC = 3AB
Area of rectangle = (AB)(BC)
= (AB)(3AB)
=3AB²
🗸 DC = BC 
              3
🗸 BC = 3AB
🗸 substitution into area formula
(3)
[9]

QUESTION 8
8.1

8.1.1
MRP = 90º [∠ in semi circl]
R₂ =21º 
(2)

8.1.2
O₁ = 138º [∠ at centre = 2 × ∠ at circumference/midpts.
(2)

8.1.3
M₁ = 21º [∠s in the same segment]

OR
M₁ + N₁ = 180º - 138º [sum of ∠s in ∆ ]
M₁ = 21º [∠s opp equal sides]
(2)

8.1.4
O₂ = 42º [∠s on a str line]
P = 42º [alt ∠s; NO || PR]
M₂ = 48º [sum of ∠s in ∆]
OR
N₂ = R₂ = 21º [alt ∠s; NO || PR]
N₁ = M₁ = 21º [ ∠s opposite equal sides]
M₂ = 48º [sum of ∠s of Δ NMR//] 
(4)

8.2

8.2
D₁ = 4x [ext ∠ of ∆ ]
D₂ = 180º - 4x [∠s on a str line]
B₁ = 5x [ext ∠of ∆]
B₁ = D₂ [ext ∠ of cyclic quad/buite  v kvh]
180º - 4x = 5x
9x = 180º
x = 20º

OR
C₁ = 3x [ext ∠ of cyclic quad]
B₂ = 4x [ext ∠ of ∆]
C₁ = C₃ = 3x [vert opp ∠s]
D₂ = 5x [ext ∠ of ∆]
4x + 5x = 180º [opp∠ of cyclic quad]
x = 20º
(6)

OR
C₃ = 3x [ext ∠of cyclic quad]
D₁ = 4x [ext ∠ of ∆]
2x + 3x + 4x = 180º [sum of ∠s in ∆]
9x = 180º
x = 20º
(6)
[16]

QUESTION 9
9.1

9.1 Constr: Join BE and CD and draw h1 from E ⊥ AD and h2 from D ⊥ AE
Proof:
area ∆ADE = ½AD x h₁ =AD
area ∆BDE    ½BD x h₁   BD

area ∆ADE = ½AE x h₂ = AE
area ∆DEC    ½EC x h₂   EC

area ∆ADE = area∆ADE [common]
But area ∆BDE = area ∆DEC [same base & height ; DE || BC]
area ∆ADE = area ∆ADE
area ∆BDE    area ∆DEC

AD = AE
BD    EC
🗸 constr
🗸 S 🗸R
(6)

9.2

9.2.1
A₁ = x [corresp ∠s; PQ || CA]
B = x [∠s opp equal sides]
A₂ = x [tan-chord theorem]
P = x [alt ∠s; PQ || CA || CA] 
(6)

9.2.2
B = P [proved in 9.2.1]
A, B, P and R are concyclic
ABPR is a cyclic quadrilateral [conv ∠s in the same segment]
(2)

9.2.3
BA = BC [prop th; AC || QP]
BQ   BR
OR
[line || one side /lyn]
But QR = BR [sides opp = ∠s]
BA = BC
BQ   QR
(3)

OR
In ΔABC and ΔBQR:
A₁ = B = x [proved in 9.2.1]
B = Q = x [proved in 9.2.1]
C₁ = BRQ = 180º - 2x [sum of ∠s of Δ ]
ΔABC ||| ΔBQR
BA = BC
BQ   QR

OR
In ΔABC and ΔBQR:
A₁ = B = x [proved in 9.2.1]
B = Q = x [proved in 9.2.1]
C₁ =  BRQ = 180º - 2x [sum of ∠s of Δ ]
Δ ABC ||| Δ BQR [∠∠∠]
BA = BC
BQ   QR

OR
In Δ ABC and Δ QBR:
B is common
A₁ = Q = x [corres ∠s; PQ || CA]
C₁ = BRQ = 180º - 2x [sum of ∠s of Δ ]
Δ ABC ||| Δ QBR [ ∠∠∠ ]
But QR = BR [sides opp = ∠s]
BA = BC
BQ   QR
(3)
[17]

QUESTION 10

10.1.1
Q₁ + Q₂ = 90º [∠ in semi circle]
M₂ = 90º [co-interior ∠, MS || QR]
SQ is a diameter [converse: ∠ in semi circle]

OR
MS || QR
TS = TM = 1 [prop theorem; SM || QR] OR
SR   MQ    1 [line || one side of Δ]
TM = MQ
M₂ = 90º [Line from centre bisects chord]
SQ is a diameter [converse:∠ in semi circle]

OR
SQ ⊥ QP [tan ⊥ rad]
SQ is a diameter [converse: tan ⊥ rad/] 
(3)

10.1.2
In ΔRTQ and ΔRQP
T = Q₃
Q₁ + Q₂ = 90º
Q₁ + Q₂ = P = 90º
R₁ = R₂
ΔRTQ ||| ΔRQP

RT = RQ
RQ   RP
RT = RQ²
         RP
🗸 S 🗸 R
🗸 S
🗸 S
🗸 R
🗸 ratio
(6)

OR
In ΔRTQ and ΔRQP
T = Q₃  [tan-chord theorem]
Q₁ + Q₂ = 90º [co-interior ∠s, MS || QR] or [∠ in semi circle] 
Q₁ + Q₂  = P= 90º
ΔRTQ ||| ΔRQP [∠∠∠]
RT = RQ
RQ    RP
RT = RQ²
         RP
🗸 S 🗸 R
🗸 S
🗸 S
🗸 R
🗸 ratio
(6)

10.2
QR = 28 units [midpoint theorem]
RP² = 28² - (√640)² [Pythagoras]
RP = 12 units
RT= RQ²
        RP
RT = 28²
         12
RT = 196
          3
Radius =⁹⁸/₃ 3 units   
🗸 S
🗸 RP = 12

[15]

TOTAL: 150