# Mathematics Grade 12 Paper 2 MP June 2023 Paper & Memo

> Grade 12 Mathematics Paper 2 question paper and memo from the MP June Exam 2023, 34 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/grade-12-mathematics-2023-mp-june-exam-question-paper-and-memorandum-paper-2/6f4e1e08-1d1c-427e-9dfc-cf1db90109f7
- Updated: 2026-09-28
- Subject: Mathematics
- Grade: Grade 12
- Document type: Question paper and memo
- Year: 2023
- Exam period: MP June Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 34

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## About this document

Mathematics · Grade 12 · MP June Exam · 2023. Question paper and memorandum, 34 pages. Read online or download the PDF.

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### Page 1

We Stgumorephysics.com

A, education

} MPUMALANGA PROVINCE
i REPUBLICOF SOUTH AFRICA

MARKS: 150
TIME: 3 hours

This question paper consists of 13 pages and 1 information sheet
and an answer book is provided.

### Page 2

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INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

1. The question pay nsists of 10 questions.

2. Answer ALL tions in the SPECIAL ANSWER BOOK provided.

3. Clearly show ALL calculations, diagrams, graphs, etc. which you have used in determining the answers.
4. Answers only will NOT necessarily be awarded full marks.

5. You may use an approved scientific calculator (non-programmable and non-graphical) unless stated
otherwise.

6. Ifnecessary, round off answers to TWO decimal places, unless stated otherwise.
7. Diagrams are NOT necessarily drawn to scale.

8. Write neatly and legibly.

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QUESTION 1

MDE/June 2023

An estate agent did a survey on the salaries of the people renting homes in a complex. She
selected 12 homes where two salary earners live for her survey. The data collected is recorded

in the table below.
s a
Salary of } 5 oq@ia100 | 5100 | 3560 | 6250 | 7400 | 4210 | 3200 } 2600 | 1000 | 4100 } 8000
person | ly
eel 4500 | 8320 | 6500 | 3500 | 1500 } 4200 | 6420 | 3520 | 10500 | 11000 | 7800 | 19350
1.1 Determine the median salary for person 1 in this data. (2)
1.2 Determine the mean income for person 2 in this data. (2)
1.3 Determine the number of salaries for person 2 that are above ONE standard deviation
from the mean. (3)
1.4 Determine the equation of the least squares regression line for the given data in the
table. (3)
1.5 Draw a scatter plot and the least squares regression line for this data on the grid
provided in the ANSWER BOOK. (4)
1.6 Explain what will happen to the least squares regression line if the income of the
person receiving an income of R19350 decreases by 50%. (2)
{16]

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QUESTION 2

The ogive below shows the money collected by parents during a fundraising event at a school.

MDE/June 2023

fan Ogive
10 + a a
e : I:
g 50
5
a
& 40
ev
: |
£ a
2 30 + —
“| = —-
© 20
10 See SE :
0 oe aon = a (|
0 500 1000 1500 2000 2500 3000 3500
Amount collected
2.1 Determine the modal class of this data. qi)
2.2 Use the ogive to determine the total number of parents who raised R1000 or more. (3)
2.3 Use the ogive to determine the values of a, b and c in the five number summary,
given in the table below.
RO a b ec R3000
(4)
[8]

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SC’

QUESTION 3

In the diagram below ABCD is a parallelogram with A(—3; 0), B(L; 3), C(3; —2) and D(x; y).

3.1

3.2

3.3

3.4

3.5

3.6

A(-3 ; 0)

BCI ;3)

we

D(x; y)

Write down the coordinates of D.

CB +2)

(2)

Calculate the gradients of AB and BC and state if ABCD is a rectangle. Give a

reason for your answer.

Determine the coordinates of M, the midpoint of AB.

(4)

(2)

Find the equation of line MN passing through M, which is perpendicular to AB. (3)

Calculate the size of BCD.

Determine the area of ABCD.

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(5)

(5)
[21]

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QUESTION 4

4.1 A(3; 2) is the centre of the circle through B and C. BC is a diameter of the circle.

x
B(S ;-1)
4.1.1 Write down the equation of the circle in the form (x — a)* + (y — b)? = r? (4)
4.1.2 determine the coordinates of C. (2)
4.1.3 Determine the equation of the tangent to the circle at C. (4)

4.1.4 Determine the equations of the tangents to the circle which gradients are zero.
Give your answers in the simplest surd form. (2)

4.1.5 A new circle is drawn. A point P(x; y) on the circumference of the new circle, is
such that it is always 4 units from the circumference of the original circle, and
outside the original circle. Determine the equation if this new circle. (2)

4.2 — Determine whether the circles with equations A:(x+3) + y*=4 and

B: x? +y?+4y+3=0 will intersect or not. Show ALL calculations. (6)
[20]

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QUESTION 6

The graph of f(x) = —2 cos(x + 15°)is given.

A
B
c

6.1 Give the amplitude of f(x). (i)
6.2 Give the period of f(x). (ld)
63 If g(x)=fx) + 2, give the range of g(x). (2)
6.4 Give the coordinates of:

6.4.1 A, anx- intercept of f (2)

6.4.2 B, the y- intercept of f (2)

64.3 C,aturning-point of f (2)

6.4.4. The x- coordinate of D, if D has coordinates (x; 0,87). Show your
calculations (3)
[13]

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QUESTION 7
AD is a vertical pole and points B and C are in the same horizontal plane as D, the foot of the tower., and

DAC=x, CBD=2x, BDC=90°+x and AC=2

A

7.1 Show that BC=1. 6)
72 Show that BD = 253 @)
cos x

19]

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NSC

Give reasons for your statements and calculations in QUESTIONS 8, 9 and 10.

Question 8
8.1 In the diagr: is the centre of the circle. KL // OM, NLT is a tangent to the circle at L.
KPM =35°

Determine, giving reasons, the sizes of the following angles:

8.1.1 Oe (2)
B12 Oy 3)
8.13 Li 2)
8.14 § (2)

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Ni

8.2 Inthe circle below with centre O, DE = EF = 12, FH= 25, OG 1 FH, OE = 5 and OG =x.

D
F
4
H
Determine, giving reasons, the length of:
8.2.1 OD. (4)
8.2.2 GH. (2)

[15]

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QUESTION 9

9.1 InAKLMand ASTR, R=$, L=T, M=R. Prove that = =5. (6)

K s
/\ T R
L M

9.2 In the sketch EG // FH and GI // HJ, AB = BE and AC = CD. BC = 6, DE=x, EF =x — 3,
DG=y+5, GH=y, DI=2a+2and IJ =2a-3.

Calculate the values of :

9.21 x (2)
9.2.2 y (3)
9.2.3 a (3)

[8]

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QUESTION 10
In the sketch, QS = SR and POT = STR = 90°.

P
T
al 4 |
Q ' Ss R
10.1 Prove that APQR /// ASTR (4)
10.2 Hence, prove that PR. RT = SQ.RQ (2)
[16]
TOTAL: 150

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FORMULA SHEET
__ ~btvb?-4ac
x= 2a
A=P(1+ni) = P(1— ni)

Ty, =a+(n—-1)

T, = ar?

Fs x{(1+ ia 1]
fi@= tim + Nase)
d = J (2-1)? + 02-1)?

yemxtce y— yy = m(x— x4)

(x-a)?+(y—-b)* =r?

In AABC: “W-=—2- =
n ‘ Sind sinB sinC
a® = b* +c? — 2be.cosA
1
area A ABC = zab.sinc

sin(a@ + B) = sina.cosf + cos a@.sinB

cos(a + 8) = cosa.cos 8 — sina sinB

cos 2a = cos*a — sin’a

=1-2sin?a
= 2cos*a—1

_ _Lfx

i=—

n
n(A)
P(A) = RG)
J=atbx

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MDE/June 2023
A=P(1—i)” A=P(1+i)"
n
n = 5 [2a + (n—1)d]
a(r™ — 1)
Sn = aq ire
So=Trpici<r<t
1- ;y-n
paw +0)
i
2A 2)
a( 2 ' 2
m= 22 m = tan@
x2 — 4

sin(a — 8) = sina.cos B — cosa.sinB

cos(a ~ 8) = cosa.cosB +sinasinp

sin 2a = 2sinacos@

yer —x)?

ot= _i=i

n

P(A of B) = P(A) + —P(AenB)

b _r@-HO-/)
Ge = #)?

### Page 14

Mi theTnerias 1 Gtinchedel Vi j MDE/June 2022
° From Stanmerspnysics. com ox

QUESTION/VRAAG 5
5.11

(3)
5.1.2

(2)
5.2

laYaYay
(5)

Copyright reserved/Kopiereg voorbehou HN LL Please turn over/Blaai om asseblief
MPUMALANGA

### Page 15

4, education

MPUMALANGA PROVINCE
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

GRADE 12

MATHEMATICS P2

JUNE 2023

MARKING GUIDELINE

MARKS: 150 marks

This question paper consists of 13 pages and an information-sheet.

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NOTE:

Consistent a
Assuming v;

Pepypr

LET WEL:

If a candidate answered a question TWICE, mark only the FIRST attempt.
If a candidate crossed out an answer and did not redo it, mark the crossed-out answer,

cy applies to ALL aspects of the marking memorandum.
answers in order to solve a problem is unacceptable.

MDE/June 2023

5. As 'n kandidaat 'n vraag TWEE keer beantwoord het, sien slegs die EERSTE poging na.
6. As 'n kandidaat 'n antwoord deurgehaal en nie oorgedoen het nie, sien die deurgehaalde

antwoord na.

7. Volgehoue akkuraatheid is op ALLE aspekte van die memorandum van toepassing.
8. Dit is onaanvaarbaar om waardes/antwoorde te veronderstel om 'n probleem op te los.
9

. Write neatly and legibly.
QUESTION/VRAAG 1
1.1 87110
2 R517 YR 1259,17
(2)
1.2 SD = R4579,26 ¥SD
Above ONE standard deviation = mean+1SD ¥ boundary
= R7259,17 + R4579, 26
= R11838,43
Only ONE household v answer
3)
1.3 | y=atbx ¥ a=6102,1123,47
as6Natt Yb =0,2792,85
b=0,27

y =6102,11+0,27x

Y y=6102,11+0,27x

(3)

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1.4

Scatter plot of income per household

Income of parent 2

0 2000 4000 6000 8000 10000

Income of parent 1

¥ Y points correctly
plotted

¥ V regression line

4)
1.5 | The gradient of the line will become smaller, causing the data to be
more symmetrical about the regression line, and there will be no
P ¥ V answer
outlier,
(2)
[14]

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MDE/June 2023

QUESTION/VRAAG 2
Ogive
” Toi |
60 AapE
B : :
§ 50 "i
3
£ 4
:
E 30 + < =
3 20 = | = + ——
10 + = : —— + 7
0 Sek a |
0 500 1000 1500 2000 2500 3000 3500
Amount collected
2.1 | 1000 <x < 1500 Y answer
(1)
2.2 | 60-15 =45 parents ¥15 and 60
¥ answer
(3)
2.2 | a= lower quartile = 1000 ¥ method
YQl
b = median = 1350 ¥Q2
c= upper quartile = 1850 ¥Q3
(4)
[7]

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QUESTION 3
B(1; 3)

A(-3; 9) a

MDE/June 2023

AA

O(a; y)

3.1 |D(-1; -5) vv (2)
3.2 ~ 023 7
Map = Ra 4 map
v
_ 3-(-2)_ 5 Mec
Mec = 3
3 5 15
Sy22 23+ -1Vv
aX 7re #-1 #-1
:. AB not perpendicular to BC, . ABCD is not a rectangle Not rectangley (4)
—34+1 0+3.
3.3 [Myg= Cs S=(-1: >) vv @)
3.4 Map ==, « m,=-5 mv
3 4
y—7=-3@-CD) Subst ¥
= 4,443
y 3 * 3 2 v @)
=—4y4H
=—5% tz ae
3.5 tana = -: py
a = —68,198 ...° + 180° = 111,80°
tan =* (AB// CD)
B = 36,87° v

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MDE/June 2023
NSC-Marking ¢ uideline
©. =111,80° — 36,87°
= 74,93° (ext Z of A) v
BCD =105,07° _ (angles on straight line) v (5)

3.6

BC? = (1=3) + (3 — (-2))

= 45035 = 29

BC =

cp? =)(34 (-1))* + (-2 = (-5))?
=16+9=25

cD= s

Area ABCD == BC.CD.sin BCD.

= 5.V29. 5 . sin 105,07°
= 13 units?

Subst in formula ¥

BC ¥

CD ¥

Subst in sine form
v

iv (5)

[21]

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“QUESTION 4

4.1

A(3; 2)

5; -1)

MDE/June 2023

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4.1.1 |r? =(5—3)*+(-1-2)? v
=449=13 v
“. (x — 3)? + (y — 2)? = 13 vv (4)
4.1.2 |} C=(1;5) (symmetry) vv (2)
4.1.3 a as v
Mac = 73 <2
2
Mtangent ™ 3 v
y-S= 2a -1) v
y=sx-i+5
=x+4e v @)
4.1.4 |y = V13 and horizontal lines 13 from centre.
y=2+ VI3 and v
v 2)
y=2- vi3
lar
41.5 | (-3)?+(y- 2)? = 44 vi3)? centre
Or (x — 3)? + (y — 2)? = 57,84 radius (2)
4.2 x? +y*+4y+3=0
x? + y? + 4y + 2? = —-3 +2?
x?+(y+2)2=1 v
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\C-Marking Guideline
Centres: A: (3; 0) and B: (0; —2) v
Distance between centres AB = /(3 — 0)? + (2— 0)? = Vi3 y
= 3,61

Radii=1 axeh2 y
Sum of fl

>!

fag y
Sum of radii-< Distance between centres v 6)
.. Circles do not intersect

[20]

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i MDE/June 2023
merephysigs.com
QUESTION 5
5.1 sk
cos 51° = 7
4
39
1
1-k?
51° .
k
5.1.1 y=P-kh Y Pythagoras
yavl-kP
tan 219° = tan (180° + 39°)
= tan39° v¥—tan39°
Vie answer
(3)
5.1.2 | sin(-411°) =sin(-411 ° +360°)
=sin(-51°)
=~-sin51° Y reduction
¥ answer
2
-- ue (2)
5.1.3 | 0089? =cos(60°-51°) y compound angles
= cos60°cos51° +sin 60°sin 51° oa
1 BY fi-R Y substitution
= (5) (k) (2) Y substitution
3(1—K?
1k v8(-#)
2 2
4y0n0

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5.2 | sin(45°+x)-sin(45° - x)
= (sin 45cosx + cos 45sin x)(sin 45cos x —cos45sinx) | expansion
¥ expansion
2 2. 2 2.
= Y2 oosx +2 sinx V2 osx 2 sin x V2
2 2 2 2 ‘>
1 1.) Y simplification
=o sin x answer
; ©)
2
53 sinx+sin2x =tanx
1+cosx+cos2x
LHS = sin x +sin 2x
1+cosx+cos2x
__ sinx+2sinxcosx ¥ 2sin x08 x
1+cosx+2cos? x-1 ¥ 20s’ x1
sinx +2sinxcosx
~  Cosx+2eos* x Y simplification of
denominator
i 142
= sinx(1+ 2006) ¥ common factor
cos x(1+2cos) ¥ common factor
_ _ v sinx =tan
= tanx = RHS cosx *
(6)
54 | sinl5°=sin(45°-30°) ¥ sin(45°-30°)
= sin 45cos30—cos45sin30 ¥ expansion
(213) (ZI | ¥ substitution
=| — |] — |-| > substituti
2 2 2 \2
_ vo-v2 ¥ simplification /6
4 (4)

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sin’ x —2sinxcosx + cos’ x-1=2sin’ x-2
in? x—2sinxcosx+2=0
~sin’ x—-sinxcosx+1=0
sin’ xcosx +sin? x +cos’ x= 0
cos’ x —sinxcosx =0

cosx{cosx—sinx) =0

MDE/June 2023
SC-Marl leline
5.5 (sinx—cosx) -1_,
sin? x—-1
sin? x - sin xcosx + cos’ x1 = 2(sin® x-1) V sin? x—2

Y sin? x +cos? x

¥ standard form
¥ common factor

cosx=0 or cosx —sinx =0 V x=4904360k keZ
x = +90° + 360k cosx =sinx
tanx=1 ¥ tanx =1
x =45°+180k,keEZ V¥ x= 45°+180K
)
[31]

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Mathipasintoa i MDE/June 2023
a ded from Stanmorephysics. com une:
QUESTION 6

The graph of f(x) = —2 cos(x + 15°)is given.

(x}0, 87)
A
c
6.1 |Amplitude = 2 v (1)
6.2 ‘eriod = 360° v (e5)
6.3 |Range of g(x): y € [0; 4] vv (2)
6.4.1 |... A=(—105°; 0) vv (2)
6.4.2 [B= (0, -1,93) vv (2)
6.43 |C= (165°; —2) vv ro)
6.4.4 | —2cos(x + 15°) = 0,87
cos(x + 15°) = —0,435 v
x + 15° = 115,785 ...° + 360k, k EZ

x = 100,79° + 360k ¥v
-. D=(100,79°; 0,87) v G)
{13]

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QUESTION 7

MDE/June 2023

AD is a vertical pole and points B and C are in the same horizontal plane as D, the foot of the
tower. DAC = x, BDC = 2x,BDC = 90° +x and AC =2.

A

cos3x cosx

Icos 3x
=2cos2x ~ 1.

cosx

TA Nein = 2 v
2 v
CD =2sinx
ce v
BC __ op Subst in Sin-rule
sin(90°+x) sin 2x Cosx  ¥
Bo _?sinx 2 sinx cosx ¥
cosx 2sinx cosx
2sinx. cosx Simplification ¥
BC == _
2sinx cosx (6)
=1
7.2 |BCD = 180° — (90° + x) — 2x int Zs of A
= 90° -3x v
In ABCD:
BD _ 1 .
sin(90°3x) _ sin(90°+x) bst in sine-form
BD 1

‘o-functionsY (3)

[9]

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Question 8

\C-Mar! leline

MDE/June 2023

8.1 In the diagram, O is the centre of the circle. KL // OM, NLT is a tangent to the circle at L.

8.1.1 |6, =70° Z at centre circle =2xZ atciroumf |SV RY (2)
8.1.2 |K, =70° alt.Zs, KL // MO S+RY
L, =70° Zs opp = radii S+RY
2
6, = 40° int Zs of A S+RY (3)
8.1.3 |Z, =20° radius L tangent SV RY (2)
8.1.4 |S = 135° opp Z of cyclic quad Sv RY (2)

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ig Guideline

MDE/June 2023
NSC-Mar!
8.2
D
F
mn
8.2.1 |ORD = 90° Centre circle midpoint chord |S“ RY
OD? = 12? + 5? Theorem of Pythagoras v
= 169
OD = 13 v (4)
8.2.2 |FG= 12,5 Centre circle L chord Sv RV (2)
[15]

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MDE/June 2023
NSC-Marking Guideline
QUESTION 9
9.1 InAKLM and ASTR, R = $, £ =f, M = R. Prove that = = =. (6)
K s
A B
T
L M

9.1 |Construction

In AKAB and ASTR:

AK = ST construction

KB=SR construction

R=§ given

. AKAB = ASTR SAS

sAL=f

Bu? =7 given

AL= 7?

“AB // LM corresponding angles =

KA _ KB i si

“la = kM line // one side of A 6

ST SR = =

an = kM AK = ST and KB=SR

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NSC-Marking Guideline

92. Inthe sketch EG // FH and GI// HJ. AB = BE and AC = CD. BC = 6, DE =x, EF =x - 3,

DG=y+5,GH=y, DI=2a+2 and = 2a —3.

9.1 jx=12

midpt theorem SY RY (2)

9.2 = = line // to one side of A SvoRY

(3)
9.3 |2at2 _ y+ line // to one side of A Sv RV
2a-3 y

2

4
2a—3—«S 3
6a+6=8a— 12
2a=18
a=9

i

@)
(14]

WEEE]

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ing Guideline

NSC-Mar!

QUESTION 10

MDE/June 2023

HE
0 ; 2 e
10.1 [In APQR and ASTR
Q=T 90° given S+RVY
R=R common Z S+RY
P=s int Zs of A S+R VY
APQR /// ASTR AAA RV @)
a imilari y
or ar RS similarity
«. PR.RT = RS.RQ
but =9 RS =QS Y @)
©. PR.RT = QS.RQ
[10]
TOTAL: 150

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MATHEMATICS P2 JUNE 2021 GRID

1 2 3 4 TOTAL
1.1 median 2
1.2 mean 2
1.3 standard deviation 3
1.4 regressionline 3
1.5 scatter plot, 4
1.6 interprete——> 2
2.1 modal class 1
2.2 interprete ogive 3
2.3 interprete ogive 4
3.1 coordinates 2
3.2 Rectangle 4
3.3 midpoint 1
3.4 equation of line 2
3.5 size of angle 3
3.6 area triangle 3
4.1.1 equation circle 4
4.1.2 coordinates 2
4.1.3 equation tangent 4
4.1.4 horisontai tangents 2
4.1.5 new circle 2
4.2 circles intersect 6
5.1.1 reduction 3
5.1.2 reduction 2
5.1.3 compound angle 4
5.2 compound expansion 5
5.3 identities 6
5.4 identities 4
5.5 general solution 7
6.1 amplitude 1
6.2 period 1 4
6.3 range 2 I
6.4.1 x-intercept
6.4.2 y-interccept 2
6.4.3 turning point
6.4.4 x-coordinate 3
7.1 show length 5 12
7.2 show that
8.1.1 angle at centre 3
8.1.2 int angles triangle 3

### Page 34

Downloaded from Stanmorephysics.com

8.1.3 tan perp radius 3

8.1.4 opp angles cyclig 2 11

8.2.4 Pyth

8.2.2 centre circle perp chord 5

9.2.1 cyclic quad 4

9.2.2 congruency— 3

9.2.3 angle at Genitre 5

9.2.4 same septne n 5 22
9
pllaTayl

10.1 properties of parm I 1

10.2 ‘one line // to 3 line in triangle

10.3 area rule 2 6

11,1 similarity 3

11.2 ratio 3

11.3 ratio 5 11

TOTAL 29 55 58 14 66

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