# Mathematics Grade 12 P2 KwaZulu Natal November 2015 Memo

> Grade 12 Mathematics Paper 2 memorandum from the KwaZulu Natal November 2015, 15 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/2015-grade-12-math-trial-exam-paper-2-kzn-memo/3232ccfb-8993-431e-a209-472871bcb9dc
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2015
- Exam period: KwaZulu Natal November
- Paper: 2
- Curriculum: CAPS
- Pages: 15

Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Mathematics](https://www.hlayiso.com/library/grade-12/mathematics) › [Mathematics Grade 12 P2 KwaZulu Natal November 2015 Memo](https://www.hlayiso.com/library/2015-grade-12-math-trial-exam-paper-2-kzn-memo/3232ccfb-8993-431e-a209-472871bcb9dc)

## About this document

Mathematics · Grade 12 · KwaZulu Natal November · 2015. 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

2d Trom hlayiso.com

Basic Education

KwaZulu-Natal Department of Education
REPUBLIC OF SOUTH AFRICA

>

MATHEMATICS P2
PREPARATORY EXAMINATION
SEPTEMBER 2015

MEMORANDUM

NATIONAL
SENIOR CERTIFICATE

| aBed x84 dH WVEEOL 22 PO GLOZ

### Page 2

Mathematics P] D ow n | fe) a d. 2 d. 3 rs nevAdad AVISO°COnT Examination

QUESTION 1
;id .
Scatter plot of Height vs Arm span
200 -
190 |
t
I
E ie
= 180 Al “1-4 points
S correct
S &) v5 —9 points
z 170 correct
< a} ¥ all points
3 correct
160
E (3)
450 160 170 180 190 200
Height (in em)
12 a= -~36, 58 {a= ~36, 57689...) VV a
b= 1,25 (b = 1,25381....) |b
p= 36,38 +1,25x Ca! ¥ equation
Caer ely Yu
Gaya a ie a)
13 f = 36,58 + 1,25(176) eat. |v substitute 176
= 183.42 4. | ¥ answer
@)
OR
yp=184,09 ae) vv answer
_ Q)
1.4 There is strong, positive correlation between height and arm span.@~) v” strong, positive
@)
[10]
Copyright reserved Please turn over

z abed x84 dH WVEEOL 42 PO GLOZ

### Page 3

Mathematics P1 D 0 wn | oad ed f ro m h | AYISPcG QWMNaory Examination

NSC-MEMORANDUM(KZN)
QUESTION 2
2.1
: Cumulative
Daily Sales Frequency Frequency
| ¥ first two
60<5x<70 5 5 ij cumulative
70<x<80 1 16 frequencies
correct
80<x<90 22 387 & | ~ next two
90<x<100 13 s1 cumulative
frequencies
100 <x <110 7 58) correct
110 <x <120 3 61 | v remainder
7 correct (total =
61)
3)
2.2 .
Cumulative frequency graph of Daily Sales
70 -peperrereorpreeerrrarrsr rr rECEIT: perRey
: ¥ grounding at 0
E ¥ plotting
| et cumulative
é Bie frequencies at
= 40 usta upper limits
¢ ae Y points correct
$s E ¥ smooth shape
ki ee of curve
E sth 4
5 es «)
i He 4
ae
arena
40 50 60 70 8&0 90 100 110 130
Daily Sales (in Rands)
2.3 The median for the data is approximately R 87. e&| v reading off
g5-#4 from graph
Cheeest ) a. | RET
2)
2.4 The upper 25% interval is R96 to R120 4.) ¥ 96 to 120
(Range to accept: 94 to 120) a)
[10]
Copyright reserved. Please turn over

€ abed x84 dH WVEEOL 42 PO GLO?

### Page 4

Mathematics P1 D fe) WwW n | fe) ade. df an am, tal AVTSOE Orr?” Examination

QUESTION 3

3.1 tartan Yot9_ 5
oe) W x= 5
Xp =5 x, =4 > Yona
DG 5 4) Q
3.2 _4-©2) 2% | substitution Into
Men = 5-2 gradient formula
_ Mey =
tan a = 2 tan @ = 2op
“= 63,4 as Y answer
(3)
33 May = Mcp = 2 ABIICD, equal gradients RLY in gy = 2
yrlxt+e
0=2%-Dte Oa.) v subst (-1 ; 0)
c=2
yorxt+2 &) ¥ answer
3)
34 _4-©
-2 cay vom =
3
tan (2 of inclination of AD) = :
Z of inclination of AD = 33,7° Be | ¥33,7°
_ A= 63,4 ~ 33,7° BA cmeue
= 29,7° cal v0.7
Copyright reserved Please turn over

y abed x84 dH WVYSE0L 42 PO GLOZ

### Page 5

Mathematics P] D Oo wn | oad ed f ro m h | QYISOxG OD aros Examination

NSC-MEMORANDUM(K2N)
3.5 3AB=DC
“GAB? = DC?
of (x +P +(y- 0] =(3—2) +(442) CO) v substitution
of (x + iy + ] =45
4nifoc, a)
xtl
(241) $Y? Soccer (1)
Y= IN+Qecrerreransen (2)
Swhsiaae (2)in () Sub. 3-8
(x+1+(2x=2) =5 Cal v substitute = y= 2x +2
x? 4241442? +8x+4-5=0
5x7 +10x=0 &.| “ standard form
Sx(x+2)=0
x#0 3; x=-2 |) Vv xy=-2
Substitute x =—2 in (1)
ya} Blv y=
B(-2;-2)
(3)
[16]
Copyright reserved Please turn over

g abed

x84 dH WVSE0L 42 PO GLOZ

### Page 6

Mathematics P1 D own | oaded. £ rQ mn, layiso-corr Examination

QUESTION 4
41 re =(2-47 4G-57 “| ¥ subst into distance formula
=8 a) ¥8
x 2
(x= 2) +(y-3¥ =8 “|v 2)
ALY (y-3Y
z (4)
42 5-3 1m =1
Miyp = rary =] NP
Mpp =—l NP PT, product of gradients = - lea} Vv mp, =-1
yo-xte gt
xe
S=-44e otk ¥ subst (5; 4),
e=9 Boive=9
yYa-xt9 | Md Wy aie + 9
Q=-x+9
x=9
“. TY 50) &.| ¥ coordinates of T
(6)
43 _ fron ase are mp? cas | ¥ substitution into «« &.%
PT = f(9-4)° + (S-0 *
v io ? ( ) distance formula
=v50 fF & | wv f50
=5¥2 ol Q)
44 Area =x PT? “dé | ~ substitution into gyi yu &,
=2x50 area formula
=157 thee | YW 1§7
: be @)
45 BY * ig :
tan wag = 28. PEL ¥ tan Wer = 3S. Met
/50 ¥50 ar
MPP saBir = 21,8° Boo ¥21,8°
e)
46 NP = NM radii Gv SR
PT=T radii @i ¥ SIR
 MNPT isa kite two pairs of adjacent sides.) reason
equal in length
3)
Copyright reserved Please turn over
9 abed x84 dH WYSE0L 42

PO GloZ

### Page 7

Mathematics P1

NSC-MEMORANDUM(K2ZN)

Downloaded from hla@ydsooG Muy examinsio

47 NTP = NTM = 21,8° diagonal of kite Che ” ot on
NPT = NIMT = 90° tangent perpendicular to radius Brew R
NMP = 360° - 90? — 90° —43,6° angles in quadilateral cCaly SR
= 136,4° a@ |v answer
(4)
OR
NP LPT (rad | tan) ¢alv SR
“.TNP = 68,2° (Z sum A) aj“ SR
TNP = TNM (prop of kite) aly sR
MRP = 2(68,2°)
. MNP =136,4° Ou | v answer
@
[23]
Please turn over

Copyright reserved

L abed

x84 dH WVcri0lL 42 PO GLOZ

### Page 8

Mathematics P1

D own | oaded. £ramolal ayisort BEE” Pemination

QUESTION 5
5.1 tan(180° + A). cos (180° — A) . sin (360° - A)
cos (90° — A) .
: a
~ {tan 4) (cos 4) (sin 4) Rv tan A —cosA
sin A an) ¥— sin A“v’ sin A
_ sit A os A S y Sin A
cos A cos A .
= sinA eek. | ¥ answer *
“ ©
5.2.1 cos 52° = cas 2(26°) &.| ¥ writing 52° in
= 2c0s? 26° -1 terms of 26°
; # | “ expansion
=2rY-1
=} -] &) @ answer
: @)
522 | ni sin 7d? & Vv identity
cas71°
__ sin(45° + 26°) &| ~ writing in terms of
cas(45° + 26°) 26°
_ sin 45° cos 26° + cos45°sin 26° & | Vv expansions
00s45* 00s 26°—sin 45° sin 26° ey aR
7
= V2 ; V2 1 v’ substitution
= (2 (\i-*)
2 (2
2 (- +f/b- |
8 ¢ -Jjl-r? )
reafb- 7?
a a) &\ v answer
)
Copyright reserved Please turn over

g abed

x84 dH WY¢er0lL 42 PO GLO?

### Page 9

Mathematics PJ

NSC-MEMORANDUM(KZN)

Downloaded from hl ayYiS.xGONaos Examination

5.3

7 sin 2x
cos2x + sin? x
2sin xcosx

cos* x—sin? x+sin? x

__ 28in xcosx
60s? x
_ 2sinx
"osx

= 2tanx

= RAS

OR

tus =—Si02x_

cos2x+sin“ x
2sin xcosx

2cos? x—1+ sin? x

2sin x cosx

” Beos? x —(1-sin? x)

_ __ 2sinxcosx

~ 2cos? x —cos’ x

_ 2sin xcosx
cos’ x

_ 2sinx

” “cosx

= 2tanx

= RHS

OR
LHS = sin 2x
cos2x+sin? x
_ ___2sinxcosx
~ 1=2sin? x +sin? x
_ 2sinxcosx
~ 1=sin? x
_ 2sinxcosx
cos’ x

_ 2sinx

cosx
= 2tanx

= RHS

PP

as

Ps

Y identity for sin2x
¥ identity for cos 2x

Y simplification

v identity

¥ identity for sin2x
Y identity for cos 2x

¥ simplification

Y identity
(4)

v identity for sin2x
¥ identity for cos 2x

~ simplification

¥ identity
(4)

[19]

Copyright reserved

6 abed

Please tum over

x84 dH WVerOL 22 PO LOZ

### Page 10

Mathematics P1 D own | oa d le af rom, AvESO 206 apppsiory Examination

QUESTION 6

6.1

cos 2x = sin(x - 30°)
= cos[90° - (x — 30°)]
= cos(120° — x)

key angle = 120° — x

2x =120° ~x +3603 Z
3x = 120° +:7.360% xe Z
x= 40° +,120°ne Z

or

2x = 240° +x4n.360% ne Z
xX = 240° + 7.360% € Z

2x = 360° ~ (120° ~x) +2.360°;ne Z

v using co-ratio

¥ 120°-x

¥ 2x =120°-x+ 360°
v x= 40° + 2.120°

¥ -2x =360° - 120° ~ x) + 71.360°
v¥ x= 240° + 1.360°
¥neZ

(7)

SAAS

x-intercepts
turning points
shape = oe

intercepts
tuming points
shape

<< 408

(6)

6.3

—120° < x < -80° or 40° <x < 90°

critical values: —120° ; —80°
critical values: 40° ; 90°
correct notation

SAK

G)

(16

Copyright reserved.

OL abed

Please turn over

x84 dH WVcri0lL 42 PO GLOZ

### Page 11

vatemiser DOWNIoade|d From HlaYiSOs6OM aioe roninaion

NSC-MEMORANDUM(KZN)

QUESTION 7
cml AB
tanx = BD &| ¥ using tan ratio
=.
BD
Bpe—/- ol v pp-
tan x tanx
— (2)
72 BC=BD
CD? = BC? + BD’ -2BC.BD.cos y &! v using cosine
2 1 formula
(2) (4) AA)
© \tanx tanx tanx J tanx J} ” C&.| v substitution 77,).
a ne 2h
Sa ts OS & | “simplification
tan’ x tan? x tan’x
_ 2h cosy
“tan x tan® x”
2H
=——(I-cosy) O& “common factor
tan’ x 4
_ 2h (1-cos.y)
"tan x [6}

Copyright reserved

LL aBed

Please turn over

x84 dH WYPr0L 42 PO GLO?

### Page 12

Mathematics Pi D fe) wn | oaded. fram blayise:cornr Examination

QUESTION 8
8.1
BE
Construction: Draw diameter AD. Join Dio C 3.) ¥ construction
Proof: ;
EAC+DAC =90° tan L radius & VSR
DCA = 90° - Zinsemi circle om ¥ : iR
ADC + DAC = 90° sum of Zs in A
BAC = ADC ale &
But ABC= ADC <s in the same segment ae SYR 6)

Copyright reserved Please tum over

ZL abed x84 dH WVGrOL 42 PO GLOZ

### Page 13

Mathematics P1 D Oo wn | oad ed f ro m h | AVIS OFA Marory Examination

NSC-MEMORANDUM(KZN)

8.2.1 Two tangents drawn from the same external point are equal in &@,| “ answer
length. (¢3)
8.2.2 (a) | § =x angles opposite equal sides &l/“ SYR 2)
8.2.2 (b) | R ,=x tan-chord theorem a iY“ SYR 0)
B22) |v, =\2x ext Z ofA Riv SVR
@)
x = cay v v
8.2.3 R, =0, =x Sy -B, alt Zs; PQ || RS a , : ve
Q,=S,=~x R,=B sin the same segmentebsra FR
oR, =8, 4)
8.2.4 W, =2x ext Z ofA al’ SYR
W, =V,
“.QVSW is acyclic quad converse: ext Z of cyclicquad * “ SVR
(4)
825@ | O,sy tan-chord theorem av SYR By
8250) | P=6,=y corresp Zs; PQ {| RS Ri Y SVR
OR
ext Z of A @)
8.2.6 Jon MtoR and Meo P
é, =x proven in 8.2.3 a
PMR =2x Z at centre =2 x/ at circumference ¥ S/R
PMR = W, = 2x Ry gs
-. PMWR is a cyclic quad converse: angles in same segment iv R (3)
[28]
Copyright reserved Please turn over

€L abed x84 dH WVGHOL 42 PO GLOZ

### Page 14

Mathematics PL

Downloaded. framalalayisoxearmr bens

QUESTION 9
9.1 6, =¢ vert opp angles Siw S wR
eG: ppane wl SYR
Q, =A, tan-chord theorem
A,=R ext angle of cyclic quad aly SVR
56,=R
“PQ=PR sides opp equal angles Rl YR
i)
92 In APBO and APQA
(i) P, is common aly Ss
(ii) Bs Q, tan-chord theorem aly oR
(i) ~=PQB=A, remaining angles intriangle “P & ¥ ware
~. APB APQA equiangular ai i
QI APQ Joi @
a3 PA _PQ from 9.2 @ ¥ deduction
PQ PB
But PQ= PR &)¥ PQ=PR
PA_PR
“PR PB
.. PA, PR and PB form a oy
geometric sequence the ratio is constant me ee in
@)
(14)
Copyright reserved Please turn over

+L aBed

x84 dH WYL70L 42 PO GLO?

### Page 15

GL abed

vatenwiser DOWNloaded From Nlayi$OnCOM ry carina
NSC-MEMORANDUM(KZN)
QUESTION 10
10.1 In AHKG: |
= = = (prop theorem; RD || KG) a/v S/R
DG _? A, ¥ substitution
2 3
“DG = 6 &@) “ answer
@)
10.2 LetFD =»
“EG =6-y) Co| + siatement D&E
aoa
GF GE
G2 K (prop theorem; FE || HK) miv S/R
—
- -yo
DEeMFD |G <> 1 £a.| ¥ substitution
yt+2 2
26-y)=y +2
12 - 2p sp +2 & | v simplification
—3y = -10
10 & | ~ answer
= i. =FD
yrs (5)
[8]
TOTAL: 150
Copyright reserved

x84 dH WYL70L 42 PO GLO?

## Related documents

- [2015 GRADE 12 MATH TRIAL EXAM PAPER 2 KZN](https://www.hlayiso.com/library/2015-grade-12-math-trial-exam-paper-2-kzn/4aa2eb25-577f-4180-861e-a176411663c1) — Mathematics · QP · 2015
- [MATHS P2 GR12 QP AFR SEPT2015](https://www.hlayiso.com/library/maths-p2-gr12-qp-afr-sept2015/3740a254-a75d-4fd7-8875-08e16e21a026) — Mathematics · QP · 2015 · Afr
- [Mathematics P2 June July 2015 Eng](https://www.hlayiso.com/library/mathematics-p2-june-july-2015-eng/f0dc57ae-b1ff-4f0d-8e51-76a1a169619f) — Mathematics · QP · 2015 · Eng
- [Mathematics P2 Feb March 2015 Eng](https://www.hlayiso.com/library/mathematics-p2-feb-march-2015-eng/bbf0bced-85a3-4ad4-bb68-72371428eb70) — Mathematics · QP · 2015 · Eng
- [Mathematics P2 Feb March 2015 Afr](https://www.hlayiso.com/library/mathematics-p2-feb-march-2015-afr/f52dfa4b-dbf2-49fd-b128-54cc089cde28) — Mathematics · QP · 2015 · Afr
- [LP Maths Grade 12 June 2024 P2 and Memo](https://www.hlayiso.com/library/lp-maths-grade-12-june-2024-p2-and-memo/ae975968-03fb-478d-83f9-e424fe245c15) — Mathematics · Memo · 2024

## More from Grade 12 Mathematics

- [MATHS P1 ANSWER BOOK GR12 SEPT 2025 English Afrikaans watermark](https://www.hlayiso.com/library/maths-p1-answer-book-gr12-sept-2025-english-afrikaans-watermark/d78c3fd3-dcd5-4f3f-911b-e5635b4686cc) — Addendum · 2025 · Afr
- [Maths Grade 12 NSC P2 QP September 2025 Gauteng](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-gauteng/1d6bf7fd-d288-4a01-99f7-2c87e247760c) — QP · 2025
- [Maths Grade 12 NSC P2 QP September 2025 Afr Limpopo](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-afr-limpopo/f2f8d481-5d87-4e89-bb21-6077bbcf403d) — QP · 2025 · Afr
- [Maths Grade 12 NSC P1 QP September 2025 Afr Northern Cape](https://www.hlayiso.com/library/maths-grade-12-nsc-p1-qp-september-2025-afr-northern-cape/f5b3a4ce-fa91-4475-96e0-0654a87e1405) — QP · 2025 · Afr
- [Maths Grade 12 NSC P2 QP September 2025 Eng Northern Cape](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-eng-northern-cape/4ede8207-ca82-4419-bae1-fae734498cc7) — QP · 2025 · Eng
- [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)

---

Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. When citing, link the page URL above, not a PDF or /api/ URL.