# Mathematics Grade 12 Paper 2 Gauteng Mock 2015 Memo

> Grade 12 Mathematics Paper 2 memorandum from the Gauteng Mock Exam 2015, 18 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/2015-grade-12-math-trial-exam-paper-2-gp-memo/17c4c98c-2cf6-4ee4-9c44-ac3e7d82d259
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2015
- Exam period: Gauteng Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 18

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## About this document

Mathematics · Grade 12 · Gauteng Mock Exam · 2015. Memorandum, 18 pages. Read online or download the PDF.

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### Page 1

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GAUTENG PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

PREPARATORY EXAMINATION
2015
MEMORANDUM

SUBJECT: MATHEMATICS P2/ WISKUNDE V2 (10612)

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10612/15

GAUTENG DEPARTMENT OF EDUCATION
PREPARATORY EXAMINATION ~ 2015

MATHEMATICS/WISKUNDE
(Second Paper/Tweede Vraestel)

MEMORANDUM

NOTE:

If a candidate answered a QUESTION TWICE, mark the FIRST attempt ONLY.
Consistent accuracy applies in ALL aspects of the memorandum.

Penalise for rounding only in QUESTION 3.6

Assuming answers/values in order to solve a problem is NOT ACCEPTABLE.
S/R refers to STATEMENT as well as REASON

NOTA:

As n kandidaat n vraag TWEEKEER beantwoord, merk slegs die EERSTE poging.
Volgehoue akkuraatheid word in ALLE aspekte van die nasienmemorandum toegepas.
Penalisering vir afronding word slegs in vraag 3.6 toegepas.

Aanvaarding van antwoorde/waardes om n probleem op te los, is ONAANVAARBAAR.
S/R verwys na BEWERING sowel as die REDE.

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QUESTION / VRAAG 1
; nie — ia
i a Y (es) (22,6 : 44,5)
70
= v line of best fit
= 60 beste paslyn/lyn van beste
3 passing
50 «
6
i @
| 3
_ & 30
os
i v
| E 20 +
/ 10
i) 10 20 30 40 |
/ Age in years
i
a Q)
1.2.1 25 years /25 jaar v answer / antwoord ()
1.2.2 The reaction time according to the line of best fit for age v answer / antwoord (ld)
25 should be 51,46. The reaction time of this patient is Any other valid explanation.
£65. Enige ander geldige
Die reaksietyd volgens die lyn van beste passing vir die verduideliking
ouderdom van 25 moet 51,46 wees. Die reaksietyd vir
hierdie pasiént is + 65. on
3 A=-21,03 ¥ A=-21,03
B=2,90 ¥ B=2,90
y = 2,90x — 21,03 Y y= 2,90x — 21,03
(3)
1.4 r=0,95 Vv r=0,95
v very strong / strong
very strong correlation / baie sterk korrelasie baie sterk / sterk (2)
15 The 30 year olds have a higher reaction time on average. [
The interquartile range of this group is bigger YIOR
(IQR = 15)than that of the 15 year olds(IQR=5). The v for comments /
reaction time of the 30 year olds varies more. vir kommentaar
Die 30-jariges het gemiddeld n hoér reaksietyd.
Die interkwartielvariasiewydte van hierdie groep is
groter((IQR=15). as dié van die 15-jariges(IQR=5).. Die (2)
reaksietyd van die 30-jariges varieer meer. (11)

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QUESTION / VRAAG 2
Age in years Number of Cumulative frequency
Ouderdom in people Kumulatiewe
jare Aantal mense Srekwensie Yall
weave 20 au answers/
10<A<20 130 150 alle :
205A <30 152 302 antwoorde
30<A<40 92 394
40<A<60 86 480
60S A<80 18 498 (1)
805A <100 2 500
= _| “plotting
: _| — of points
iu / plot van
/ = || punte
g 400 v Joining
2 _| of points
2 verbind
= _| van punte
. _| “grounding
100 | point at
| @50)
3 os eee |] grond
: i) 20 40 60 0 100 190 | punt by
: Age in years (0; 0)
/ ;
we Se ( 3)
2.3.1 | 26 years old (accept 25 ~ 27) v answer
26 jaar oud (aanvaar 25-27) antwoord
dg)
vw
232 mall, 100 90
500 L
18%
= 18%
‘© The percentage of people 16 and above 18 is 82% Y82%
Die persentasie mense 16 jaar en ouer is 82% (G3)
[3]

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QUESTION / VRAAG 3 A(=3 39) y
B
(e
D(-5; 1) x
10
3.1 =3-5 9+]
MC 4:5) a: (2)
3.2 AM’ = (443) + (5-9) OR/OF DM? =(-5 +4)" + (1-5) | “correct substitution
=1+16 =1+16 into distance formula
=17 =17 korrekte vervanging in
srt 417 afstand formule
Yr=Vi7
OR/OF
AD? = (—543)* + (1-9) ¥ correct substitution
=4+64 into distance formula
= 68 korrekte vervanging in
“AD = V68 afstand formule
: V68
radius = =
= Yre= V7
ad Q)
ers) Yes, the circle will pass through point C vyes /ja
Ja, die sirkel gaan deur punt C
B=C=90°, AD is the diameter __ line subtends equal 2°. y 7
B=C=90°, AD die middellyn lynstuk onderspan gelyke ste iced (2)

Ze

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BO; y)

My XMgpy = —1 VY Mg XMgp =~ 1

-3-0/A-5-0 (25) (=)
GC»), » -3-0 eT)
9-10y+y? =-15

¥ standard form

y? -10y+24=0 standaardvorm
(y-O(y-4) =0 Y factors / faktore
“ y=6ory=4 ~. B(O;6) Y BOO; 6)
OR/OF 6)

AB’ + BD? = AD?

(9— y)? + (3-0)? + (5-0)? +(1— y)? = (5 +3)? +19)? Ons +(-3-0)7 4
81-18y+y? +9+25+1-2y+y? =44+64 (-5-0)2 +(1- y)?
2y* ~20y +48 =0 V (-5+3)? + (1-9)?
e -10y+24=0 ¥ standard form/ Standaardvorm
(y~- 6)(y-4) =0
¥ factors / faktore
y=6ory=4
-. B(O;6) ¥ BO; 6)
. (5)
9-6
AB —3-0
=-1 l
my =] my =)
yr y, =m(x-x,) OR/OF y=-x+c
y-1l =-1(x+5) -5=-l+c ¥ substitution of (—1 ; 5)
y-1l =-x-5 4 vervanging van (~1 ; 5)
y =-x-4 yerx-4 Vy
(3)

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[3.6 _ 1-9 | 1-9
Map = ae Y mp = =4
~54+3 -~5+3
tanO=4 A(43:9) an O=4
b= 76° BOO ; 6)
1-6
Mp, = = 7
tana=1 ¥ tana =I
. v45°
= 45 2
BDA= 76° 45°
be [ v 31° (-1 if NOT rounded)
= ext Zof A 31° (-1 as NIE afgerond)
OR/OF
AB’ = (~3—0)’ +(9~6)? v (-3-0)? +(9-6)?
AB = 18 YAB =418
BD? =(-5-0)? +(1-6)? ¥(-5-0)' +(1-6)?
BD = 50 ¥BD = 50
; a0 Yusing a correct trig ratio
tanBDA = 18 =0,6 gebruik korrekte trig verh
50
A : ¥ 31° (-1 if NOT rounded)
BDA=31 31° (-1 as NIE afgerond)
(6)
[20]
QUESTION / VRAAG 4
41.1 x+y? -~2x+6y =0 :
' V(@- 1-1
(x-1)?-1 eo.
(x—1)’ -1+(y+3)’-9=0 V(y+3"-9
(x-1)? +(y +3)? =149
(x-1)? +(y +3)’ =10
“. centre : (1;—3) ¥ (3-3)
ns v
i radius : ¥10 vi0 @
4.1.2 ~34+4
ae
14+2
ot
3 Ym.
m, Xm, =—1 :
m,=-3
equation of tangent / vergelyking van raaklyn ¥m=-3
p+ 4a — - 2, OR y=-3.
ae x +2) 2 a ¥ substitution of (~2; -4)
yt4=-3x-6 ~4=-3(-2) +e vervanging van (—2; -4)
y =-3x—-10 “c=-l0
y=-3x-10 v y=-3x-10
(4)

6

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10612/15

BCI) hy : :
AG 53)
R@s4
Bs
2x+5y+1=0
4.2.1 |RA=RB
RA’ = RB” Vv RA’= RB?
(p-2) +(q-3)* = (p +1)’ +(q-6)" Y (p-2) +(q-3)
p -4p+4+q -6q+9= p’+2ptl+q’ -12q4+36 Y (ptt)? +(q-6)’
—O6p + 6q = 24
a aa v—6p+6q = 24
oe ad (4)
4.2.2 R(p;q) lies onthe line /lé opdielyn

2x+5y+1=0
2p+5q=-1 .....()
Pada
p=q-4......2) OR
sub(1)into(2) / verv (1) in (2)
2p—2q=-8.....(2) 2p—2q=-8....(2)
2(q-4)+5q=-1 2p+5q=-H.....0)
2q-8+5q=-1 (Q-M  7q= 7
Tq=7 g=l1
q=i
subginto equation(2) /vervginvgl(2)
p=l-4
p=-3 R(-3;1)
radius=RA OR/OF _ radius=RB
RA? =(-3-2)? +(1—3)* RB’ =(-3+1)? +(1-6)”
=25+4 =44+25
RA=29 RB= 29

(x+3) +(5-1)? =29

¥ 2pt+5q=-1

uv 2(q—-4)+5q=-1/

2p-2q=-8
Vq=
¥ p=-3
Vv RA?/ RB?
¥ RA/RB=29

¥ (x43) +(y-1F =29

7)

{19}

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QUESTION / VRAAG 5
Ol) oy 271-0
3y =2x
oe 2 eee
a
tan~=m
tno = Y tana=m
; (2)
S12 / y
Pp
aN Cc x
—_
OP =25532 Byn
op= 13 ¥ OP=¥/13
ie or/of a (with rational denominator) Ye a
(met rasionale noemer) (2)
5.2.1
2 y
iN
5
|
0
QOP =B -a v¥ B-a
oO)
5.2.2) sin OOP = sin(B—a) va 2)
ae ee a = sin Bcos a —cos B sin a
=sin Scosa—cos Ssina Q 3 3
(2) 2) (ae 2) a ig
5 VB) (5 V3 2
a a
ou) 5 13
5V13 ai
513
(4)
[9]

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QUESTION / VRAAG 6

10612/15

6.1

cos(40° — x). cos x — sin(40° — x). sin x
sin 205°. cos 25°

_ cos(40° — x + x)

~ —sin 25°. cos 25°

¥ cos(40° — x +x)

2 cose a v —sin 25°
— 5 (2sin 25°, cos.25°)
_ cos 40°
1 sin 50° v—1 sin soe
2 2
cos 40° sin 50°
= ~  OR/OF 1 ¥ sin50°=cos40° or/of
5 cos 40° a sin 50° cos 40° = sin 50° in numerator
in noemer
=-2 v2
| 6)
6.2.1 LHS /LK = £08 X=sim’ x ¥ cos2x = cos? x~sin? x
cos xX +sinx
_ (cos x —sin x)(cos x +sin x)
i. con one ¥ factorising /faktorisering
=cosx—sinx
=RHS/RK (2)
622 | coe [es a
cosx+sinx 2
: 1 Y cosx—sinx
cos x (cos x—sin x)= 6
cos’ x—cosxsinx = a 1
72 ¥ cos’ x—cos xsin x =>
2cos* x—2cosxsinx=1
2cos’ x-1=2cos xsin x ¥ 2cos*x—-1 = cos2x
cos 2x = sin2x ¥ 2cosxsinx = sin2x
(4)
6.2.3 | cosx (=) al
cosx+sinx) 2
cos2x =sin2x
tan2x=1 oo
2x =45°+k.180° ; keZ V¥ 2x = 45° + k.180°
Pyne = ee, ¥ x=22,5°+k.90° 3; keZ
OR/OF GB)

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cos2x =sin2x
cos2x = cos(90° — 2x) ¥ cos 2x = cos(90° — 2.x)
2x = 90° —2x+k.360° or 2x=2x-90°+k.360° ;
keZ
[ 4x = 90° + k.360°
4x=90°+K.360° ; keZ Y x= 225° + 90°
x= 22,5°+Kk.90° ; keZ (3)
6.3.1 i . il
Area AABC = 5m sin4x v Area AABC = —mnsin4x
2
(2)
6.3.2 | Max area if sin4x =1 ¥sin 4x=1
4x =90° Vx = 22,5° (2)
E 6 Pirate
6.3.3 | Right-angled triangle/Reghoekige driehoek ¥ answer
[18]
QUESTION / VRAAG 7
7A sin(60°~ @) =1
60°-@=90° os
@=-30° 7 (1)
ed 0°
period / periode = 36
= 180° ¥ 180° qd)
73 |-2<y<0 of/or [250] ¥-2<y<0 (qd)
TA 45° <x <135°  or/or v 45° and 135°
150° < x <180° ¥ 150° and 180°
¥ ALL inequalities correct
ALLE ongelykhede korrek
Subtract I mark for extra intervals
Trek 1 punt af vir ekstra intervalle (3)
te) O0e = x 1502 ¥ 90° and/en 150°
v inequalities / ongelykhede (2)
[8]

10

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QUESTION / VRAAG 8
8.1
——=tana
ul v ptana (dy
DE= ptana pe
8.2.1 E, =180°-f (opp Z of cyclic quad v E, =180°—B with reason/met rede
teenoorst. Z van kvh)
F = A, (Z opp equal sides; EH = EF ve BE = A, with reason/met rede

Z teenoor gelyke sye)

£ +A, =180°-(80°- 8) (of W/Z van A) | VY F,+8,=8
Soe (3)
: K=H,= oF
8.2.2 In AEFH
po FS
sin 1p sin(180°— 8) ¥ substitution in sine rule
2 vervanging in sinus formule
sal
7 FEL sin; B FLsind B
sin B Vv p= —_—2_
1 sin B
FH. sin 3h
- 2sin£ pcos! p ¥ sin B= 2sin4 B ee B
2 ”) i 7 yo)
- FH
eos; (3)
623 FH? =p + p’ —2p* cos(180°— 8) ¥ substitution in cosine rule
=2p?+2p*cosf ees in kosinus reél
al ¥2p'+2p’*cosf
= 2p (1+ cos @) ¥ 2p2(1+cos 8)
FH = p/2(1+cos f) (3)
[10]

11

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QUESTION / VRAAG 9

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10612/15

S/R

Statement and reason _S/R__Bewering en rede

ol

(a) AB || CD corr. Z equal / ooreenk Z gelyk
(b) line from centre 1 to chord /Midpt. O ; Midpt. koord
(c) Radius

vreason / rede
v reason / rede

¥ reason / rede

(3)

Join OB / Verbind OB
HB = 12 and/en GD = 5 (line from centre .L to chord /
Midpt. O ; Midpt. koord)

In AOHB
(OHY =(OBY -(BHY — 1 =90"(Pyth)
= (13) ~(12)

=25
OH =5

In AOGD
(0G) = (op) -(Gpy

= (13) -(5)
=144

OG =12

GH = 0G —-OH
=12~-5
=7

v S/R

v Pythagoras

YOH=5

¥OG = 12

¥ GH=7

G)

### Page 14

9.2.1 Tangents from same point / Tans from common pt v reason /rede
Raaklyne vanuit dies. punt
()
Q 110°
A T M
9.2.1 | B.+B,=90° (tan _L rad /raaklyn radius) v S/R
A, + A, = 90° (tan L rad / raaklyn |. radius)
«. AOBT is acyclic quadrilateral/is  koordevierhoek Y SIR
(opposite Z's sapplementary/oorstaande Z'e suppl)
YR
@)
9.2.2 | OA=OB (radii) ¥ OA=OB
teat i : (radii)
r=T, (equal chords subtend equal angels/ YR
gelyke hoeke onderspan deur gelyke koorde (2)
923 | A, +B, =110 (ext Z of A/buite Z van A) non
But Ay =B, (As opp equal sides / 2° teenoor gelyke sye) Y G=55°
ee v SIR
AS B, =C=55° (tan chord / Z tussen raaklyn en koord 4
OR/OF
a v SIR
O, +O, =110° (ext Z of cycl quad/buite Z van kvh) vs
ae A v
O, +0, =2C (Aat centre = 2x Zat circ /midptsZ = 2xomtreksZ) ‘
e=55° .
] (4)
[47]

13

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QUESTION / VRAAG 10

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10612/15

10.1

Construction: Join OD and OB
Konstruksie: Verbind OD en OB

6, =2A (Zat centre = 2x Z at circumference /
MidptsZ = 2 x OmtreksZ)

O, = 2¢ (Zat centre = 2x Z at circumference /
MidptsZ = 2 x OmtreksZ)
0, +0, =360° (Z rounda ptOR Z’ ina rev
Z°om 'n punt OF omwenteling
2A + 2C = 360°
2A+C) = 360°
A+ C=180°

¥ Construction / Konstruksie

v SIR

SIR

v SIR

¥ 2A + 2C = 360°

(5)

14

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10612/15
:
10.2.1 A 1800"
i u ¥ all 3 statements
B=180°-x (opp Z of cyclic quad / al 3 bewerings
C=180°-z teenoorst. Z° van kvh ) v reason
rede (2)
10.2.2 (180° = y) * (180° = x) a (180° -z) ¥ substitution of angles
: vervanging van hoeke
=540°-(x+ y +z) ¥ simplification
= 540° — 180° (sum of Z's of A= 180°) vereenvoudiging
= 360° v_ 180°
v answer / antw (4)
(14)
T
QUESTION / VRAAG 11
G E M
to | Ro =M, (altZ's RE||TM / verw. Z°;RE||TM) y (Sv ®)
A ss S R
R,=T corr Z's RE||TM /ooreenk. Z° RE||TM) —- (4)
12 | EM RE : ~(S)
—~—=—— (line |]one side of A OR prop theorem; RE ||TM oF
EG RG
(lyn || een sy van A)
: eo. YS) YR)
But/Maar RT=RM_ (Z‘ oppequalsides ; M, =T/
Z teenoor gelyke sye;M = T)
EM _RM ty
EG RG

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Hales) In AGYE en AGER
G=6 (common /gemeenskaplik) ¥ G=G(SIR)
E,=R, (given /gegee) VE, =R, (S/R)
¥,=6, +8, (rd Zof A/3e Z van A) Aina ee
equiangular
AGYE|JAGER  (2,2,4) Igelykhoekige A
Le or/of Z,2,Z (R) (4)
114] EG RG_ YG
—= = equiang A's / AGYE||| AGER
EY RE EG (@iang Ill ) u
EG RG (R) equiang A's / AGYE||| AGER
an gelykh A’ AGYE ||| AGER
(D)
5.1) (RG) =(GM)’-(RM) — GRM =90° Pythag. .
; 3 Y substitute into Pyth
=(10)° -(6) vervang in Pyth
=64
lea VY RG=8 (2)
Ce eee ‘4 GE _GR
=— (1 de of A OR th ;RE || TM ¥——=—— fp
iG (line || one side o prop theorem; RE || an /reason / rede
(lyn || een sy van A) “Gh 14
GE 8 Y¥ substitution of values
10 Vi vervanging van waardes
GE= 52 or/of pee or/of 5,7 1units/eenhede ¥ answer /antwoord (4)
[19]
ee s

TOTAL/TOTAAL: 150

16

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- [MATHS P2 ANSWER BOOK GR12 JUNE 2024 Afr+English](https://www.hlayiso.com/library/maths-p2-answer-book-gr12-june-2024-afr-english/4a71e975-4ac4-4f55-8e23-0b8155a5b0c6) — Addendum · 2024 · Eng

## Collection

- [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers)

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