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NATIONAL
SENIOR CERTIFICATE

GRADE 12/GRAAD 12

MATHEMATICS/WISKUNDE

MEMORANDUM P2/V2
SEPTEMBER 2023

MARKS/PUNTE: 150

This memorandum consists of 19 pages/Hierdie nasienriglyne bestaan uit 19 bladsye.

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Mathematics P2/Wiskunde V2 2 Limpopo DoE/September 2023 Memo

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QUESTION/VRAAG 1
1.1
all dots
PREDICTION OF FINAL MARKS
correct/alle punte
100 korrek
90
80
9 – 11 dots
70 correct/ punte
FINAL MARK

60 korrek
50
40 6 – 8 dots
30 correct/ punte
20 korrek
10
0 1 – 6 dots correct/
0 10 20 30 40 50 60 70 80 90 100 punte korrek
SBA MARK

(4)
1.2 r = 0,94 answer/antw
(2)
1.3 The strong correlation between the SBA mark and the final reason/rede
mark implies that the points lie close to the least squares conclusion/konklu
regression line. Hence the prediction is reliable. sie
(2)
1.4 a = 14, 49 value/waarde a
b = 0,86 value/waarde b
y = 14, 49 + 0,86 x equation/vgl
(3)
1.5 y = 14, 49 + 0,86(66) subst/vervang
y = 71, 25% answer/antw
(2)
OR/OF

y = 71,18% answer/antw
(2)
[13]

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Mathematics P2/Wiskunde V2 3 Limpopo DoE/September 2023 Memo

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QUESTION/VRAAG 2
2.1.1 3 + 4 + 4 + 6 + 10 + 12 + 12 + 4 + y
=7
9
55 + y 55 + y
=7  =7
9 9
55 + y = 63
y =8
value of/waarde y
(2)
2.1.2 Median = 6 6
(1)
2.2.1 3 + 4 + 4 + 4 + 6 + 8 + 10 + 12 + 12 + 7 − n + 7 + n
x=
11
77 77
x=
11
x =7 7
(2)
OR/OF

63 + 7 − n + 7 + n 77
x=
11
77
x=
11
x =7 7
(2)
2.2.2 x − x = 3 equation/vgl
7 − x = 3
x = 4 answer/antw
(2)
OR/OF

x +  x = 11 equation/vgl
7+ x = 11
x = 4 answer/antw
(2)
[7]

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Mathematics P2/Wiskunde V2 4 Limpopo DoE/September 2023 Memo

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QUESTION/VRAAG 3

y
D

E
63,43°
O C(4 ; 0) x

B( 2 ; 7)

3.1 1 1
y − y1 = m( x − x1 ) m(AC) = −  m(AC) = −
2 2
y − (−7) = 2( x − (−2)) m(BD) = 2  m(BD) = 2
y + 7 = 2( x + 2) subt m and point B
/verv m en punt B
BD : y = 2 x − 3
answer/antw
(4)
OF/OR
1
1  m(AC) = −
y = mx + c m (AC) = − 2
2
 m(BD) = 2
−7 = 2(−2) + c m(BD) = 2
subt m and point B
−7 = −4 + c /verv m en punt B
c = −3
answer/antw
BD : y = 2 x − 3
(4)

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Mathematics P2/Wiskunde V2 5 Limpopo DoE/September 2023 Memo

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3.2 1
− x + 2 = 2x − 3 equating/vgl
2
−x + 4 = 4x − 6
−5 x = −10
x=2 value of/waarde x
subst into BD: y = 2(2) − 3
y =1 value of/waarde y
E (2 ; 1)
(3)
3.3 A ( −4 ; 4 ) through translation value of/waarde x
value of/waarde y
(2)
OR/OF
P(0 ; 2)
Find a point P so that CE = EP
P(0 ; 2)
Then CP = PA A( − 4 ; 4)
A ( −4 ; 4 ) (2)
3.4
AC = ( −4 − 4 ) + ( 4 − 0 ) =4 5
2 2
length/lengte AC

BE = ( −2 − 2 ) + ( −7 − 1) =4 5 length/lengte BE
2 2

Area of kite = 2  area of ABC

1 
= 2  4 5  4 5 
2  method/metode
= 80 80
Enlarged by scale factor 2: 80  4 = 320 sq unit answer/antw
(5)
OR/OF

length/lengte AC
AC = ( −4 − 4 ) + ( 4 − 0 ) =4 5
2 2

 PR = 8 5 length/lengte PR

BE = ( −2 − 2 ) + ( −7 − 1) =4 5
2 2

 QS = 8 5 length/lengte QS
Area of kite = 2  area of PQR
1 
= 2 8 5 8 5  method/metode
2 
= 320 answer/antw
(5)
[14]

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Mathematics P2/Wiskunde V2 6 Limpopo DoE/September 2023 Memo

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QUESTION/VRAAG 4
4.1

4.1.1 ( −13 − 2 ) + ( 5 − m ) = 17 2
2 2
subst R and N/verv
(a)
225 + 25 − 10m + m = 289
2 R en N
m 2 − 10m − 39 = 0 simplify/vereenv

( m − 13)( m + 3) = 0 std form/std vorm

m = 13 or m = −3
NA value/value m

OR/OF (4)

( −13 − 2 ) + ( −11 − m ) = 17 2
2 2

225 + ( −11 − m ) = 289 subst S and N/verv

2

S en N
( −11 − m ) = 64
2

−11 − m = 8 simplify/vereenv
m = −19 or m = −3 std form/std vorm
NA
value/waarde m
(4)

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Mathematics P2/Wiskunde V2 7 Limpopo DoE/September 2023 Memo

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OR/OF

( −13 − 2 ) + ( −11 − m ) = ( −13 − 2 ) + ( 5 − m )

2 2 2 2
equating
225 + m 2 + 22m + 121 = 225 + m 2 − 10m + 25 simplify/vereenv
32m = −96
m = −3 value/waarde m
(4)
OR/OF

RS ǁ y − axis  x = −13 same x − values use line x = −13

Draw line NK ⊥ RS Use NK
 NK bisect RS at ( −13 ; −3 ) line from centre of circle ⊥ midpt of NK
to chord
 N(2 ; −3 ) y = m = −3
(4)
4.1.1 ( x − 2 ) + ( y + 3) = 289 answer/antw
2 2

(b) (1)
4.1.2 −3 − 5 subst in gradient
m ( NR ) =
(a) 2 − (−13) formula/verv in
gradiënt formule
−8
m ( NR ) = gradient of NR
15 (2)
4.1.2 −3 − (−11)
m ( NS) =
(b) 2 − (−13)
8
m ( NS) = gradient of NS/
15 gradiënt van NS
(1)
4.1.3 NR ⊥ PR NS ⊥ PS
tan  def
15 15
 m(PR) =  m(PS) = − value/waarde 
8 8
tan  =
15
tan  = −
15 tan  def
8 8 value/waarde 
 = 61,93  = 180 − 61,93
 = 118, 07
P̂1 = 118, 07 − 61,93 = 56,14 method/metode
 P̂2 = 180 − 56,14 = 123,86 answer/antw
(6)

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Mathematics P2/Wiskunde V2 8 Limpopo DoE/September 2023 Memo

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OF/OR

Inclination of NS = 28,07° 28,07°

Inclination of NR = 151,93° 151,93°
ˆ = 28, 07 + 28, 07
RNS ˆ = 2  28, 07
 RNS
ˆ = 56,14
RNS 56,14°
NRPS is cyclic quadrilateral Opp ' s suppl Opp ' s cyclic
 P̂2 = 123,86 quad/teenoorst ' e
kvh
answer/antw
(6)
4.1.4 Reflection about x – axis: (2 ; − 3) → (2 ; 3)
Shift 2 units up: (2 ; 3) → (2 ; 5)
( x − 2 ) + ( y − 5) = 289
2 2
Circle M: equation/vgl
(2)
4.2

 1   x
C 2 A 4 O

4.2.1 Diameter = 8 + 4 + 2 + 1 ….= sum of diameters/

a som middellyne
S =
1− r
8
S = subst in sum
1
1− formula/verv in som
2 formule
S = 16
 OC = 16
(2)
4.2.2 90°, radius ⊥ tangent answer/antw
reason/rede
(2)

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Mathematics P2/Wiskunde V2 9 Limpopo DoE/September 2023 Memo

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4.2.3 AC = 16 − 4 = 12 AC
AB = 4 AB
BC2 = AC2 − AB2 (Pyth)
BC = 12 − 4
2 2 2

BC2 = 128
BC = 8 2 BC
4 2
 tan C = = tan ratio
8 2 4
(4)
4.2.4 y − y1 = m( x − x1 )
2 subst/verv m
y−0 = ( x − (−16))
4 subst point/verv
2 punt
y= x+4 2 equation/vgl
4
(3)

OR/OF

2
y= x+c
4 subst/verv m
2
0= (−16) + c subst point/verv
4 punt
c=4 2
2
y = x+4 2 equation/vgl
4

(3)
[27]

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Mathematics P2/Wiskunde V2 10 Limpopo DoE/September 2023 Memo

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QUESTION/VRAAG 5
5.1

P(3 ; 4)

R(m ; 12)

5.1.1 4
tan  = answer/antw
3
(1)
5.1.2 sin ( 90 +  )
= cos  reduction/reduksie
3 r = 5
= answer/antw
5
(3)
5.1.3 12 + 13sin  = 0
12
sin  = − std form/std vorm
13
m 2 = 132 − ( −122 ) (Pyth) subst into Pyth
m 2 = 25
simpl/vereenv
m = 5
 m = −5 answer/antw
(4)
5.1.4 cos ( +  )
= cos  cos  − sin  sin  expansion/uitbrei
 3   −5   4   −12 
=    −    subst/vervang
 5   13   5   13 
3 48
=− +
13 65
33
=
65 answer/antw
(3)

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Mathematics P2/Wiskunde V2 11 Limpopo DoE/September 2023 Memo

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5.2.1 4sin150.23tan 225

= (2 )2 sin 30
.23tan 45 sin 30°
tan 45°
1
2  special values/
= 2 2
.23(1) spesiale waardes
= 2.23
simpl/vereenv
= 16
=4
answer/antw
(5)
5.2.2 tan(180 + x) cos x
sin(180 + x) cos x − cos(540 + x) cos(90 + x)
(tan x)(cos x)
= tan x
(− sin x)(cos x) − (− cos x)(− sin x)  − sin x
sin x  − cos x
.cos x  − sin x
= cos x
− sin x cos x − cos x sin x sin x

sin x cos x
=
−2sin x cos x
1 answer/antw
=−
2 cos x (6)
5.3 1 − cos 2 x − sin x
sin 2 x − cos x
cos 2x expansion/
1 − (1 − 2sin 2 x) − sin x
= uitbrei
2sin x cos x − cos x sin 2x expansion/
1 − 1 + 2sin 2 x − sin x uitbrei
=
2sin x cos x − cos x
2sin 2 x − sin x
=
2sin x cos x − cos x simpl/vereenv
sin x(2sin x − 1)
=
cos x(2sin x − 1)
factors/faktore
sin x
=
cos x
= tan x
(4)

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Mathematics P2/Wiskunde V2 12 Limpopo DoE/September 2023 Memo

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5.4 1
sin P sin Q − cos P cos Q =
2
1
 cos P cos Q − sin P sin Q = − rearrange terms/
2 herrangskik terme
1
cos ( P + Q ) = −
2
P + Q = 180 − 60 cos identity/
identiteit
P + Q = 120 .................1 2nd quadrant/
1
sin ( P − Q ) =
kwadrant
2 equation/vgl
P − Q = 30 ..................2 1st quadrant/
 2P = 150 kwadrant
P = 75
75°
Q = 45 45°
(7)
[33]
QUESTION/VRAAG 6
6.1 a = 2 and b = 1 value/waarde a
value/waarde b
(2)
6.2.1 2sin x − cos x = 0
2 sin x = cos x
sin x 1
=
cos x 2
1
tan x =
1  tan x =
2 2
x = 26,57 answer/antw
(2)
6.2.2 x   0 ; 180 interval
notation/notasie
(2)
6.2.3 y = 22 f ( x )−1
 1 
y  − ; 7 
 2  answer/antw
(2)
[8]

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Mathematics P2/Wiskunde V2 13 Limpopo DoE/September 2023 Memo

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QUESTION/VRAAG 7

30°

A 30° E
F
7.1 BF AF
In ABF: =
sin A sin B
BF y
= correct subst in
sin  sin 30
sine rule/korrekte
y sin 
BF = verv in sinreel
1 subst special
2 value/ verv
BF = 2 y sin  spasiale waarde
answer/antw
(3)
7.2 DF DE
In EDF: =
sin E sin E
DF x
= correct subst in
sin  sin 30 sine rule/korrekte
x sin 
DF = verv in sinreel
1
2
DF = 2 x sin  answer/antw
(2)
7.3 BD = BF + FD
2 2 2
(Pyth)
 subst/verv in
BD 2 = ( 2 y sin  ) + ( 2 x sin  )
2 2
Pyth
BD 2 = 4 y 2 sin 2  + 4 x 2 sin 2 
(1)

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Mathematics P2/Wiskunde V2 14 Limpopo DoE/September 2023 Memo

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7.4 In BDC : BD2 = (4 x) 2 − (3 x) 2

BD2 = 16 x 2 − 9 x 2
BD2 = 7 x 2  BD2 = 7x2

7 x 2 = 4 y 2 sin 2  + 4 x 2 sin 2  equating

7 x − 4 x sin  = 4 y sin 
2 2 2 2 2

simpl/vereenv
x 2 (7 − 4sin 2  ) = 4 y 2 sin 2 
4 y 2 sin 2 
x =
2

(7 − 4sin 2  )
4 y 2 sin 2 
x=
7 − 4sin 2  (3)

OF/OR

CD 2 = BD 2 + BC2
16 x 2 = 4 y 2sin 2  + 4 x 2sin 2 + 9 x 2 subst/verv in
Pyth
16 x 2 − 9 x 2 − 4 x 2sin 2  = 4 y 2sin 2 
7 x 2 − 4 x 2sin 2  = 4 y 2sin 2 
x 2 ( 7 − 4sin 2  ) = 4 y 2sin 2 
simpl/vereenv
4 y 2sin 2 
x2 =
7 − 4sin 2 
2
 i.t.o x
4 y 2sin 2 
x=
7 − 4sin 2  (3)
[9]

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Mathematics P2/Wiskunde V2 15 Limpopo DoE/September 2023 Memo

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QUESTION/VRAAG 8

E 3
2
1 O
1

1 2
32°
A B C

8.1 ˆ = Eˆ = 32
ABF tan-chord theorem S R
1
(2)
8.2 Eˆ 1 = B
ˆ = 32
1  ’s opp equal sides S/R
F̂ = 116 sum  ’s of  S/R
(2)
8.3 ˆ + Fˆ = 180
D opp  ’s of cyclic quad S R
D̂ = 64
(2)
8.4 ˆ = 2 D
O1
ˆ midpt  = 2  circumf  S R
Ô1 = 128 (2)
8.5 ˆ +B
Eˆ 2 + O ˆ = 180 sum  ’s of  S/R
1 2

Eˆ 2 = B
ˆ
2  ’s opp equal sides S/R
2Eˆ + 128 = 180
2

2Eˆ 2 = 52
Ê 2 = 26
(2)
OR/OF

ˆ +B
ABF ˆ +B ˆ = 90 rad ⊥ tangent
1 2 S/R
ˆ
32 + 32 + B = 90
2

B̂2 = 26
ˆ = Eˆ = 26
B  ’s opp equal sides
2 2 S/R
(2)
[10]

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Mathematics P2/Wiskunde V2 16 Limpopo DoE/September 2023 Memo

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QUESTION/VRAAG 9
9.1

2
O1
N

Construction: Join LO and ON constr/konstr

ˆ = 2 K
O 1
ˆ  at centre = 2   at circumf S R
Oˆ = 2 M
2
ˆ  at centre = 2   at circumf S/R
Oˆ +Oˆ = 360 revolution R
1 2

2Mˆ + 2Kˆ = 360 equating

ˆ + Kˆ = 180
M
(5)

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Mathematics P2/Wiskunde V2 17 Limpopo DoE/September 2023 Memo

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9.2

9.2.1 Bˆ =M ˆ =x  ’s in the same segment S R

1

ˆB = Rˆ = x alt  ’s, TR BS S R
3

ˆ = Rˆ = x
B tan-chord theorem S R
1
(6)
9.2.2 Equal chords equal circumf  ’s R
(1)
9.2.3(a) Â1 = x + y ext  of  S/R
(1)
9.2.3(b) N̂1 = x + y ext  of  S/R
(1)
9.2.4 In  SAR and  KNR
(i) Rˆ 1 = Rˆ 3 both = x S
(ii) Aˆ =N ˆ both = x + y S
1 1

(iii) K̂ = Sˆ
2 sum  ’s of  S
  SAR ǀǀǀ  KNR    (3)

OR/OF

In  SAR and  KNR

(i) Rˆ 1 = Rˆ 3 both = x S
ˆ =N
(ii) A ˆ both = x + y S
1 1

 SAR ǀǀǀ  KNR    R

(3)
9.2.5 K̂ 2 = Sˆ  SAR ǀǀǀ  KNR S
SAKR is a cyclic quad ext  of quadrilateral OR R
converse ext  of cyclic quad
(2)
[19]

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Mathematics P2/Wiskunde V2 18 Limpopo DoE/September 2023 Memo

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QUESTION/VRAAG 10

10.1 P̂1 = C ext  of cyclic quad S R

C (
ˆ = 180 − Rˆ + Rˆ
1 2 ) opp  ’s of cyclic quad S R

(
 Pˆ1 = 180 − Rˆ 1 + Rˆ 2 ) method/metode

(ˆ +B
Rˆ 1 + Rˆ 2 = 180 − A1
ˆ
1 ) sum  ’s of 

 Pˆ1 = 180 − 180 − A


ˆ +B
1
ˆ 
1
( )
 Pˆ = 180 − 180 + A
1 (
ˆ +B ˆ
1 1 )
ˆ +B
 Pˆ1 = A ˆ (5)
1 1

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Mathematics P2/Wiskunde V2 19 Limpopo DoE/September 2023 Memo

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OR/OF S R
S R
ARB ˆ +B
ˆ = 180 − A
1
ˆ (
1 ) method/metode
ˆ
ˆ = 180 − C
ARB
180 − A1 (
ˆ +B )ˆ
ˆ = 180 − C
1

Cˆ =Aˆ +Bˆ
1 1
ˆC = P
ˆ (5)
1

P ˆ =A ˆ +B ˆ
1 1 1
10.2 In  ARP and  ABR  identify/
(i) A ˆ =A ˆ common  identifiseer  ’s
ˆ ˆ
(ii) P1 = A1 + B1 ˆ proven S
180 − Pˆ1 = 180 − A1(
ˆ +B ˆ
1 ) method/metode

 Pˆ 4 = Rˆ 1 + Rˆ 2
(iii) Rˆ 1 = B ˆ
1 sum  ’s of 
  ARP ǀǀǀ  ABR  R
AR AP
= ǀǀǀ  ’s
AB AR
S
AR 2 = AB.AP
(5)
OR/OF

In  ARP and  ABR

(i) Aˆ =Aˆ common  Identify/
ˆ ˆ
(ii) P1 = A1 + B1ˆ proven identifiseer  ’s

180 − Pˆ1 = 180 − A1 (

ˆ +B ˆ
1 ) S
method/metode
 Pˆ 4 = Rˆ 1 + Rˆ 2
  ARP ǀǀǀ  ABR 
AR AP
= ǀǀǀ  ’s R
AB AR
S
AR 2 = AB.AP
(5)
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