# Mathematics Grade 12 Paper 2 Limpopo Mock 2020 Memo

> Grade 12 Mathematics Paper 2 memorandum from the Limpopo Mock Exam 2020, 15 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/2020-grade-12-math-trial-exam-paper-2-limp-memo/07e3b41d-20e7-48cf-b800-edea0e4468e8
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2020
- Exam period: Limpopo Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 15

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## About this document

Mathematics · Grade 12 · Limpopo Mock Exam · 2020. Memorandum, 15 pages. Read online or download the PDF.

Read it free online on the page above; download the PDF from there.

## Document text

Text extracted from the PDF. Layout, diagrams and tables may not survive extraction.

### Page 1

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LIMPOPO

PROVINCIAL GOVERNMENT
REPUBLIC OF SOUTH AFRICA

OEPARTMENT OF

EDUCATION

NATIONAL

SENIOR CERTIFICATE

GRADE/GRAAD 12

MARKS/PUNTE: 150

This memorandum consists of 15 pages / Hierdie memo bestaan uit 15 bladsye.

### Page 2

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Mathematics P2/Wiskunde V2 2 LimpopoDoE/September 2020
NSC/NSS/Memorandum

QUESTION /VRAAG 1

11.1

A | { v¥ box /mond
/¥ whiskers, on
B —— —— scale/

stert, en op skaal
10 20 30 40 50 60 70 80 90

(2)

1.1.2. | School B/skool B: ¥School /skool B
In B 50% of learners got above 60% whilst 50% of learners of
A above 55% /

50% van B het meer as 60%; by A het 50% meer as 55% J Jany two
median of B > median of A/ reasons

mediaan van 8 > mediaan van A :

75% of B above 30% and at A 75% above 23,5% / Ce eds
75% van B bo 30%; By A75% bo 23,5%

OR/OF VS
Skewed to the right/ skeef na Regs
OR/ OF
vv
Q)
1.2
135 24 + 225x 32 +200 x = J numerator/
560 teller
3240 + 7200+ 200x = 560x 560
360x = 10440
x= 29 v 360x = 10440
v
answer/antwoord
(4)
[9]
QUESTION/VRAAG 2
21 a=2319 b=0,768.. Vavb
y =23,19+ 0,77x
v y =23,194+-0,77x
3)
2.2 | y=23,19+,77(30) v
= 46,29

substitution/vervanging
= 46%
¥answer/antwoord

Q)

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Mathematics P2/Wiskunde V2 3 LimpopoDoE/September 2020
NSC/NSS/Memorandum
23.1) 6=10,99 ¥/answer
/antwoord (2)

2.3.2 | 57,4% —49,4%=8% L8%

eee

50 100

x =4 marks ¥4 marks/ punte

(2)
9}
QUESTION /VRAAG 3
P
po
S
x
0
2
IT (0,-5—
(0,-5. >

3.1 | X+ay=a ¥x=0

otay=a

y=l vy=l

- P(Q;1) v¥ P(0;1) (3)

3.2 | OS:OP=3:1 ¥ S(3;0)

8(3;0):x+ay=a

34a(0) =a ¥ substitution/vervang
a=3 Yva=3
3)
3.3 1
Mps =— > V tips =>
3
Mp =3 o iy, =3
RT: y=3x- st Jequation/ ver gelyking
(3)
34 xX+3y=3 > ys--x4t1
ao x+1=3x al
3
~x43=9x-17 ¥ equating /stel gelyk

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Mathematics P2/Wiskunde V2 4

NSC/NSS/Memorandum

LimpopoDoE/September 2020

Substitute: y = -32) +1

v
simplification/vereenvaudig

y=3
R24)
3
v¥x=2
1
Vy=-
an)
4
a areaAPRT = 2x PT xh v PT =67
ab 62x Yh=2
2 3
20 - :
=| rie units/ vierkante eenhede
J¥answer/ antwoord
(3)
3.6 | PT is the diameter(converse: Z in semi ©)/ YSIR
OPT is die middellyn (omgekeerde: 2 in semi ©)
. 10...
radius = 3 units. ZYanswer/antwoord (2)

[18]

### Page 5

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Mathematics P2/Wiskunde V2 5
NSC/NSS/Memorandum

LimpopoDoE/September 2020

QUESTION/VRAAG 4

Vv
4.1.1 | x? ~6x494y? —4y 442124944 Completing the square
(x3)? + (y-2)? =25 vierkantsvoltooing
C@G;2) ¥LHS/ LK
r=5 /RHS/RK
v¥ C(3;2)
vr=5 (4)
4.1.2 | Y-intercept: /Y as:x =0 vx =0
2
y -4y-12=0
(y-6)(y +2) =0 v “entree
y=—2or y=6~. A(0;6) standaardvorm
6-2 4 ¥ factors/faktore
Mac => TZ
0-3 3 ¥ A(0;6)
3 : 4
Map = ai (tan 1 radius) ¥ yc = 5
-3 3
yy* 6 V tay = a
YR
J equation/
vergelyking
(8)
413) pet A=O: tano => J tang = 3
4 4
0 =36,87° v¥ 0 =36,87°
a= 90° + 36,87° > 7
a =12687° ¥ a = 90° + 36,87
¥answer /
antonard fAN

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Mathematics P2/Wiskunde V2 6
NSC/NSS/Memorandum

LimpopoDoE/September 2020

42 | x?42x4¢]4+y?-4y44=444144
(x+1) +(y—2)9 = 49

Centre / middelpunt D(—1;2) ; radius = 7
cD=3--1=4

ro try =5+7=12> 4

«. the two circles intercept internally/
twee sirkels sny inwendig

“Centre D(-1;2)
vt=7

¥CD=4

Vr thy =12
w>4

¥ conclusion /
gevolgtrekking (6)

[22]
QUESTION/V RAAG 5
5. tanA= EI
2
5
R diagram
vr=13
r=12° +5?
= 169
r=13
13sinA —3tanB
5. 412 5
= 13(—) -3(— in A =—
G) ZF) ¥sin A ic
“3 7 12
ul ¥tanB = a
5 Yanswer /antwoord
6)
5.2 —tan60°—sin x J¥ —tan60°
sin x + 2.¢0s30° .
. v-sinx
_ v3 -sin x
a 3 v¥ 2c0s30°
sin x +20 2 ) ¥ special angles/
_ —(sinx + V3) bekende hoeke
(sin x + ¥3)
=-] ~ factorisation/
faktoriseer
v¥ answer/

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Mathematics P2/Wiskunde V2 7

NSC/NSS/Memorandum

LimpopoDoE/September 2020

5.3.1

cos(x + y) =sin(90°-(x+ y)
=sin((90° — x) + (—y))
= sin(90° — x) cos(—y) + cos(90° — x) sin(—y)
=cosxcos y—sin xsin y

“Correct co-fn/

korrekte ko —
funksie

J regrouping/

herrangskik
¥ Correct expansion/
uitbreiding
¥simplification/
vereenvoudig
(4)
5.3.2 | cos(x— y)—cos(x + y) /Expansion/
=cosxcos y +sin xsin y—(cosxcos y~sin xsin y) brei uit
- snces +sin xsin y—cosxcos y+sin xsin y Jsimpilifieation /
=2sin xsin y
vereenvoudig
- - (2)
5.3.3 | 2sin195°sin 45°
=cos(195° — 45°) — cos(195° + 45°) a
=c0s150°—cos240° ¥ Substitution/
=—cos30° —(—cas60°) vervang
31
Roa ta ¥—cos30°
_-v3+1 v¥—cos60°
2 ¥ Special angles/
bekende hoeke
fanswer/antwoord
(5)

[22]

### Page 8

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Mathematics P2/Wiskunde V2 8 LimpopoDoE/September 2020
NSC/NSS/Memorandum

QUESTION/V RAAG 6

ff 4 _ L |
4 i i
|
| LS H — _ +\} |
a k: 1 —
|

6.1 g: ¥x-intercepts/x—asse J y-intercepts/y — as
¥ turningpoints / draaipunte Q)
6.2 | 2sinx =cosxcos30°-sin xsin30° “Expansion of g/
sine cose sine brei uit
3 gin y= YB cose ¥ Special angles/
2 2 bekende hoeke
V3
sinx 2 5. 3
ae v¥ =sin x = —cosx
cosx 5 2 2
2
tanx = 3 i
5 y sinx
cosx
(4)
6.3 V3
tanx = rs 419,12
x=19,1° or x =180°+19,1° = 199,1° /199,1° (2)
6.4.1 | x €[60°180°] W[240%;270°] ¥ 60°and /en180°
¥ 240° and / en270°
“notation / notasie
G3)
6.4.2 | x [19,1°3199,1°] ¥ criticalvalues
/ kritieke waardes
notation /
notasie (2)

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Mathematics P2/Wiskunde V2 9 LimpopoDoE/September 2020
NSC/NSS/Memorandum

QUESTION /VRAAG 7

InAABC :tanx = eee
CB v CB
h
CB= vcp=—"
tanx tanx

InABCD:CB = BD
CD’ = BC’ + BD’ —2BCx BDxcosy

~ Correct application of cos rule/

h h h h korrekte gebruik van cosreél
+ -2 x x COS Ba
tanx’ Canx? tanx tanx i ¥ substitution
__2h* 2h" cosy ¥ simplify/
tan*x tan’ x vereenvoudig
= 2H" (1-cos y)
tan? x uM
“factorising/ faktoriseer
CD = ——./2(1-cos
ae 20 »)
CD= a fH Teosy
an x

[6]

### Page 10

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Mathematics P2/Wiskunde V2 10
NSC/NSS/Memorandum

QUESTION/VRAAG 8

LimpopoDoE/September 2020

8.1.1 on = R, (tangents from same point)/ raaklyne uit dsde punt) | /SVR
2R, +70° =180° (2'sof APQR/ Z'e van APQR) VS
2R, =110°
R, = 55°
/answer/
antwoord
(4)
8.12) 6 +0, =90° (tangent | radius)/(raaklyn 1 op radius) YSYR
O, +55° =90°
0, = 35°
J answer /
antwoord Q)
8.1.3 | 6 =R, =35° (r=) YS/IR
M,=110  (Z'sof AMQR)
M, =M,=110 (=chords subtend =2’s at centre)/ YM, =110
(= koorde = middelpunts z'e) vR G3)

### Page 11

8.2

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Mathematics P2/Wiskunde V2 11
NSC/NSS/Memorandum

LimpopoDoE/September 2020

8.2.1) ¥, =90° (line from centre to midpoint of chord)/ YSYR
(van midpunt van ©, mdpt van koord) vs
U,+U,=90° (2 insemi ©)
+X, =U, +U, =90° VS
=> RU || SV (corresponding 2's =)/ VR
(ooreenkomstige 2’e =) 6)
8.2.2) R, =y — (tan-chord theorem)/ VSVR
(4 tussen raaklyn en koord)
0, =y {alternate 2’s, RU || SV y
verwisselende 2Z’e, RU || SV ) vs
Oo, = ar (4 at centre = 22at circumference)
4 (middelpunts< = 2xomtreksz) SSSR
= 2”
(3)
(20)

### Page 12

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Mathematics P2/Wiskunde V2 12
NSC/NSS/Memorandum
QUESTION/VRAAG 9
9.1
A

LimpopoDoE/September 2020

9.1

Construction: Draw altitudes A and k.
Konstruksie: trek hoogtelyne hen k.

1
areabADE 9% 4%4 ay
areaABDE 1 Bpxh DB

2

VOAE k
areab\ADE 9° *°™" 4g

areaADEC 1 ecxk EC

But the areaABDE = areaADEC (same base and same
height/

dsde basis en hoogte )

_areahADE _ area\ADE

“@reaABDE  area\DEC

_ AD _AE

“DB EC

/¥ Construction/
konstruksie

vS

vS

YSVR

vS

(6)

### Page 13

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Mathematics P2/Wiskunde V2 13 LimpopeDoE/September 2020
NSC/NSS/Memorandum
9.2
A . J D
M
a c
eee) Lees (line IIto one side of AAMDY | SVR
EM FD 3
(lyn || aan een sy van AAMD)
= = : J answer G)
9.2.2 | AM:MC=7:7 (diagonals of parm) YSVR
CM 7
WE = 3 ¥ answer
G)
9.2.3 | areaABDC = areaAABD (diagonals of parm) vS
areaAFDC _ areaAFDC .
areaABDC areahABD UT)
1 vervang
i x FDx height
= 1... ¥ area formula /
2 * AD x height oppervlak formule
FD ;
= a (between same parallel lines)/
(tussen dieselfde I Lyne) YR
_ 3
71 (4)
[16]

### Page 14

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Mathematics P2/Wiskunde V2 14
NSC/NSS/Memorandum

QUESTION /VRAAG 10

LimpopoDoE/September 2020

10.1 | InAXYCand ADYX :
X, =D, — (tan-chord theorem/ VSVR
4 tussen raaklyn en koord)
on =X, (tan-chord theorem/
4 tussen raaklyn en koord VS
¥,+¥,=¥,+¥,(z'sofa)
+ AXYC [| | ADYX (222)
XY YC
>=
DY YX vS
“XY? = YC.DY
vs
(5)
10.2 | 4,=2%, (2 at centre is twice 2 at circumference) | YSYR
(middelpuntsz = 2x omtreksz)
Similarly/netso B, = 2D, y
But ¥, = 5, (tan-chord theorem/
¢ tussen raaklyn en koord)
4, =B, vS
(4)
103 | CACY cacay | | | YBX) vs
YB YX
R_cY
ad ¥Substitute/ vervang
R cy?
rn? yx? .
_CYcY v Square both sides/
CY. DY kwadreer albei kante
of

### Page 15

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Mathematics P2/Wiskunde V2 15 LimpopoDoE/September 2020
NSC/NSS/Memorandum
ae _ DY /Replace/vervang YX"
R’ CY
“answer /antwoord
(3)
[14]

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