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

> Grade 12 Mathematics Paper 2 question paper and memo from the KZN June Exam 2023, 27 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/grade-12-mathematics-2023-kzn-june-exam-question-paper-and-memorandum-paper-2/cb23a845-1414-4cee-8636-af87c69bef35
- Updated: 2026-09-28
- Subject: Mathematics
- Grade: Grade 12
- Document type: Question paper and memo
- Year: 2023
- Exam period: KZN June Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 27

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## About this document

Mathematics · Grade 12 · KZN June Exam · 2023. Question paper and memorandum, 27 pages. Read online or download the PDF.

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

## Document text

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### Page 1

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KWAZULU-NATAL PROVINCE
EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

a "
re "
s "
a n
" =
" .
Seesesseesseees Seeessseeeeest
MARKS: 150

TIME: 3 hours

per consists ooh information sheet,
‘and 1 diagram she

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wae owNloaded from Stanmorephysics.com June 2023 Common Test

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

1. This question paper consists of 9 questions.

2; Answer ALB the questions.

3: Number the answers correctly according to the numbering system used in this question
paper.

4. Clearly show ALL calculations, diagrams, graphs, etc. that you have used in
determining your answers.

3: Answers only will NOT necessarily be awarded full marks.

6. You may use an approved scientific calculator (non-programmable and

non-graphical), unless stated otherwise.

7. If necessary, round off answers to TWO decimal places, unless stated otherwise.
8. Diagrams are NOT necessarily drawn to scale.

2 An information sheet with formulae is included at the end of the question paper.
10. Write neatly and legibly.

Copyright Reserved Please Turn Over

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QUESTION 1

Seven scores for a test out of 50 marks were recorded. However, when recording the data
the statistician had some difficulty with his computer. He listed the following conditions
that he had remembered,

[Scores [ a fot | ¢ | d@ | e | f J g |
> Maximum Valueis 42
> Range is 35
> Median is 23
> The difference between the median and the upper quartile is 14
> The Inter — Quartile Range is 22
> e=2¢
>» The mean is 25.
Determine the values of the scores. (Show all working) (10)
[10]
QUESTION 2
The class teacher of a grade 10 class collected the following data for her learners. She
recorded the time taken by the learners to complete different tasks assigned.
No. of learners
Time in minutes
o0<t<10 5
10<t<20 8
20<t<30 18
30<t< 40
40<t<50
2.1 Calculate the estimated mean time. (3)
2.2 Draw a frequency polygon (Ogive) curve on the system of axes provided, (5)
2.3. Calculate how many learners took more than 60 % of the time to complete the tasks
assigned to them. (3)
(1

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QUESTION 3

In the sketch below, A(-8;6), B(8 ; 10), C and D(—2;0) are the vertices of a trapezium having
AD || BC. T is the midpoint of DB. From the vertex B, BM is drawn perpendicular to the

x—axis and intersects DC in F (8 ;3).

»4
B(8;10)
A(-8;6)
T Cc
i
(8:33)
2: fe)
D(-2;0) M
3.1 Calculate the gradient of AD. (2)
32) Determine the equation of BC in the form y = mx +c. (3)
3.3 Prove that BD 1 AD. GB)
3.4 Calculate the size of BDM. (2)

3.5 If TC || DM and points T and C are symmetrical about the line BM. Calculate the (3)
coordinates of C.

3.6 Calculate the area of ABDF. (5)
[18]

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QUESTION 4

In the diagram below, the circle centred at E (3 ; 1) passes through the point, P (5 ; —5).

4.1

4.2

43

e E(3;1)

P(5;-5)

Determine the equation of :

4.1.1 The circle in the form x? + y? + Ax + By +C=0.

4.1.2 The tangent to the circle at P (5 ; -5) in the form y= mx + c.

A smaller circle is drawn inside the circle. Line EP is the diameter of the small
circle.

Determine the:

4.2.1 Coordinates of the centre of the smaller circle.

4.2.2 Length of the radius.

Hence, or otherwise, determine whether the point C(9 ; 3) lies inside, o} or
on the circle centre, E.

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June 2023 Common Test

(4)
(5)

(4)
(2)

(4)

[19]

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QUESTION 5

5.1

June 2023 Common Test

If tan 12° = q , then determine the value of the following, without using a calculator:

5.1.1 cos1
5.1.2 cos

5.1.3 1—2sin? 6°

(3)
(3)
(2)

(7)

(3)

(2)

(4)
[24]

(6)

np 3 Evaluate the following trigonometric expression without using a calculator:
2 sin? (x — 180°) cos(180° — x)
cos(90° + x) sin x — cos (x — 90) sin (720° — x)
53 cos A 1
Given: (1 —tan A) Fas
mpom 6 {= 2A} cosA +Sin A
5.3.1 Prove the identity.
5.3.2 Write down the maximum value of ———
cosA +Sin A
5.4 Calculate the value of
52°
>dcos* A
38°
QUESTION 6
6.1 Sketch the graphs of f(x) = sin>x and g(x) = cos(x + 60°) for x €
[—180°; 180°] on the grid given at the back.
6.2 Use your graphs to answer the following questions:

6.2.1 Write down the range of the graph of f-
6.2.2 State the period of g.
6.2.3 Determine the values of x for which f(x) = g(x).

6.2.4 If h(x) = g(x + 30’), write down the equation of A.

(2)
Q)
(4)

(3)
[16]

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June 2023 Common Test

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QUESTION 7

In the diagram below, MP is a vertical tower. MQ and MR are wire ropes used to stabilize
MP and are each 19 metres in length. R, P and Q are in the same horizontal plane.
The angle of elevatioméf M from R is y.

RPQ = x and MQ
M
19 19
P
x
: Q
71 Prove that the Area of A PQR = 24si"e0s'y (6)
12 Show that, RQ = 38 cos B (6)
{11]
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QUESTION 8

8.1

8.2

O is the centre of the circle, radius r, and chord AB = ¥128 cm. OCD 1 AB and
OC: CD=3;2. ABD = 27°

A al

D

8.1.1 Calculate the length of the radius of the circle, without using a calculator. (5)

8.1.2 Calculate, with reason and AOD. (2)

Complete the following statement:

(2)
8.2.1 The exterior angle of a cyclic quadrilateral is ..........:.:.cceseeeeeeeeeee seen
8.2.2 TQisachord of the circle PQRST. QAT 1 PAS. Q, = 30° and P = §,.
P
Q R
a) Name 3 angles each equal to 60°. (3)
b) Calculate the size of QRS. (2)
c) Prove that PS || QR. (2)
d) Prove that TR is a diameter of the circle . (2)
[18]

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Math ics P2 ° 23 Ce
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QUESTION 9
9.1 InA ABC, DE || BC. Prove :

AD _ AE

DB EC

re
D E
B c (6)

9.2 In the diagram, O is the centre of the circle ACDE. EC is produced to B. TAB isa
tangent to the circle at A. EC and AD intersect at H. F is a point on BA.
ED || FH || TAB. EOHCB is a straight line. AH : HD = 5: 7.

ieee 4

T >> rn B
9.2.1 Prove AABC ||| AEBA (4)
9.2.2 Calculate r, the radius of the circle, if AB = 5 metres and BC etres. (4)
9.2.3 Write down, with reason the value of AF : FE. (2)
9.2.4 . Area of AAFH 4
Determine Area of AAED © ”
9.2.5 Calculate the length of OH. (3)

[23]
TOTAL MARKS: 150

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INFORMATION SHEET: MATHEMATICS
P(1=ni) A=P(\-i)" A=P(l+i)"
S,= 322+ (Dd)
al alr" -1 a
T, =ar' Be A r#1 S,=——3 -l<r<l
ee FI l-r
Pe (eo) pe -1] poil=d+)")
i ry
. f(xth)- f(x)
' = lm Pe ete SE me
re) 40 h
d= (x, -x,)? +(, -y,)? (222% 2
y=mr+e y-y, =m(x-x,) = 22221 m= tan 0
X2 7x

(x-a) +(y-sf =r?

a b c

sinB sinC

In AABC:
sin(a + #) = sin a.cos # + cosa.sin 8
cos(a + 8)=cosa.cos # -sin a.sin 8

cos? a@—sin* a
cos2a@ =41-2sin? a

2cos* a1

Dix

n

p=atbx

x=

Copyright Reserved

a2 =b +c? -2be.cosA area AABC =~ ab.sin Gg

sin(@ — £) = sin a.cos f — cosa.sin B

cos(a — 8) = cosa@.cos # +sin a.sin 8

sin 2@ = 2sin @.cosa@

SG)

oo =.

P(A or B) = P(A) + PB) — B)
p= X09)

d@-x)

Please Tum Over

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Mathematjes P2 1 .
Jownloaded from Stanmorephysics.com

June 2023 Common Test

DIAGRAM SHEET
QUESTION 2
—
Ti minutes No. of learners
< 10 5
10<t<20 8
20<t<30 18
30<t<40
40<t<50 2
QUESTION 6.1
Y
LS
HH
i n°) + r +o re) > & * fe it ,

0s : pas t

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FINAL

KWAZULU-NATAL PROVINCE

S=EDUCATION
ApAREPUBLIC OF SOUTH AFRICA
i,

NATION
SENIOR CERTIFICATE

MARKS: 150

N.B. This marking guideline consists of 16 pages.

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QUESTION 1

Ll | g=42 AY Value of g
Minimum : .@= 42 — 35 =7 AY Value of a
d=23 AY Value of f
f-23 = 14.Fherefore f= 37 AY Value of d
f—b = 22 jay Therefore b = 15 CAY Value of b
7 +15 +)C#23 + 2c+374+42 _ CAY Setting up equation

~ CAY Simplifying
3c +124 25 numerator
7 ~ CAY equation inc
3c +124 =175
3c =51
c=17 CAY Value of c
Therefore e = 34 CAY Value of e (10)
[10]
QUESTION 2
Time (in minutes) Number of Learners Midpoint of fx Cf
Interval(x)
0<t<10 5 5 25 5
10<t<20 8 15 120 13
20<t<30 18 25 450 31
30<t<40 7 35 245 38
40<t<50 2 45 90 40
Total 40 930

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2.1 Estimated Mean = 22 = 23,25 Av930 Av40
40 CAV Answer (only if (3)
denominator is 40)
(Answer only full marks)
2.2 DO NO RK THIS QUESTION
2.3 60 % of 50 minutes = 30 minutes AY Calculation

9 learners

Av 30 minutes
AY Answer (AO full marks)

(3)

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QUESTION 3
yf
B(8;10)
A(-8:6)
F (8 3 ;)
"3
« A 3 >
D(-2;0) M x
v
3.1 6-0 6 AY subst. of points A and D
mAD = —8+2 —-6 ol CAY Answer
(2)
3.2. | mAD = mBC = --1 .....(AD || BC) CAY gradient of BC
y=mxt+c
10 = —-1(8) +c CAY subst. of point C and
18=c gradient
y=-x +18 CAY Answer (3)
3.3 BD = 10-0 10 1 AY subst. of points A and D
ee B42 10 Av Gradient of BD
Since the product of the gradients of lines AY Justification
AD and BD =-1
Therefore BD 1 AD (3)
3.4 |tan@=1 CAvtan@ =1
BDM = 45° CAV Answer (2)
3.5 | T(3;5) ... midpoint A Y coor es
C(13 ; 5) CA CA ordinates of C (3)

i

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3.6

2

.. Area of A BDF = 10V2 x

= 33,3 units?

3

1 2 20
= —8)2 —3-) =6-=—
BF= |(8-8) +(10 35) 65=5
Area of ABDF = 5 (10) (F) == 33: square units.
OR
Area of A = Area of A DBM ~ area of A DFM
1 1 10
= — (10) (10) — — (10)| —
+o) 00) 4 «0)(2)
= 100 333 units ?
33
OR
1
mpr = —
A 1
tan FDM = —
3
FDM = 18.43°
BD = 10V2
pr — Lovio
3
10v10

sin 26.57°

AY Substitution
AY BF value

CAY Subst. into formula
CAY Simplification
CAY Answer

(5)

[18]

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QUESTION 4
4.1
y
* E(3;1)
fe) x
P(5;-5)

4.1.1] gpa re) —5)?+(1+5)2 AY Substitution

EP = /(—2)? + (6)?

CAY EP value

EP = V40

(x - 3)? + (y— 1)? = 40 CAY Substitution

x —6x4+9+4+y? —2y +1=40

x? +y? - 6x —2y -30=0 CAY Answer (4)
4.1.2 . 1+5 AY Gradient of radius

Radius: m = >—= = -3

3-5
r _ 1 CAY Gradient of tangent
angent: m = 3 (provided it is positive)
y=mxt+c
1

-5=3(5) +e

-15 =5 + 3c CAY Subst. Of point and

-20 = 3c gradient.

o- 62

3
4 2
y= 3° - 65 CAM c, jue
CAY er (5)

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4.2.1 | Coordinates of Centre of smaller circle: AY midpoint formula
(= =) = (4;-2) AY Substitution
2’ 2 , CA CAY V Answer
(AO full marks) (4)
i—)
4.2.2 | Radius of | circle: 2V10 Avi (2vi0)
Radius of er circle: r = 5 (2V 10) = V10 units 2
CA V Answer
OR
Radius = / (4-3) +(-2-1)
= Vi0
OR
Raduis = /(4—5) + (-2-5)
- Jo @)
43 Jae = (9-3) +1) AY Substitution
= {6 +2?
= /36+4
= V40 CAV EC = V40
_ + EC= radius CA Y equating to radius
..C lies on the circle, centre E CA VY Conclusion
OR
xr -6x+9+y—2vt+1=40
x+y’ 6x—2y- 30=0
OR
(x —3) +(y-1) =40
LHS = (9-3) + (3-1)? =40
RHS = 40
-. LHS = RHS
«. It lies on the circle
(4)
aaa [19]

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QUESTION 5

5.1

Cos 24° = | —2sin?

12°

2
= 1 -2| 4
l+q°

2
core =| u }-[
l+¢
-J! ¢
l+q 1+¢
-le@¢
l+q?

1
5.1.1 | cos 192 . AY Ji+@
=-—cos 12 .
1 Av —cos 12
te
CAY Answer
(G3)
5.1.2 | cos 24°
= 2cos?12°-1 AV Double angle
2
-2 ( 1 ) 4 Expansion
y 1+q? CAV Substitution
"ire"
CAY Answer
OR

(3)

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5.1.3 | 1 — 2sin? 6°
=cos12° AY Double angle
1
= CAV Answer
Vit @
4 : : (2)
5.2 2 sin?(x — 180°) cos(180° — x)
cos(90° + x — cos (x — 90) sin (720° — x)
2( )?.(— cos x) AY (-sinx)?
~ =sinx]sin — (sin x. —sin x) Av’— cosx
. AvY—sinx
—2 sin? x cos x Av'sinx
= 7 Av-sinx
—2sin? x cos x
; CAY——_—
= undefined CAY Answer (7)
53.1 cos A
LHS: (1 — tanA i
( an [ss ‘) avs” A
. cos A
_ ;- sn A cos A AV cos? A—sin2 A
cos A }\ cos? A-sin? A
= (cos Asin A __£0s A . A Y simplification
cos A (cos A —sin A) (cos A + sin A)
_ 1
cos A+sin A
=RHS (3)
5.3.2 | DO NOT MARK THIS QUESTION
5.4__| DO NOT MARK THIS QUESTION

[18]

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QUESTION 6

6.1

a 50 -120 -90 -60 -30
-0.5:

AS

30 60 90 120 150

Graph of f: A 1 mark for x — intercepts
A 1| mark for shape

Graph of g: A 1 mark for end points
A | mark for x — intercepts

A 1 marks for minimum and maximum points

A | mark for y — intercept (6)
62.1] ye€[-1;1] CAY [-1; 1] (2)
A V notation
6.2.2 | 360° AY Answer (dl)
1 °
6.2.3 sin5x = cos(x + 60°)
1 ° °. YCo - i
sin5x = sin [90° — (x + 60°)] Av Co - Ratio
sin5x = sin [30° — x] AY Equation in sine only
1 1 °
5x = 30 —x AYSXx = 30—— x
3 °
3% = 30
° Av Ans
=2
x= 20 CA( AG-full: marks) (4)
6.2.4 | h(x) = g(x + 30)
h(x) = cos(x + 30 + 60) A ¥ substitution
h(x) = cos(x + 90) AY cos(x + 90)
h(x) = —sinx AY Answer (AO full marks) (3)

[14]

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QUESTION 7
M
19 19
P
x
Q
R
TA PR
—=cosy
19 AY Trig ratio
PR= 19 cosy AV Length of PR
Now PR = PQ .....(AMPR = AMPQ ...90deg HS) Av Congruent triangles
1 .
Area of A PQR = 5 (19 cos y)(19 cos y) sinx AY Reason SAS

_ 361 sinx cos? y

AY Substitution into Area Formula

; )
7.2. | RQ? = 19? + 19? — 2(19)(19) cos(180° — 28) AY Size of angle RMQ
RQ? = 361 + 361 — 722(—cos 28) Av Subst. Cosine rule
RQ? = 722 +722 cos 2B AY Reduction
RQ? = 722 + 722(2cos? B — 1) AY Simplifying
RQ? = 1444 cos? B AY Double angle expansion a

RQ = 38cosB
OR

MR? = RQ? + MQ? —2RQ. MQ cos
192 = RQ + 192 — 2(RQ) (19) cosB
0=RQ-38RQ. cosB
38 RQcos B _ RQ?
RQ RQ

RQ = 38 cosf

AY 1444 cos? B

[1]

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QUESTION 8
8.1
8.1.1 oc 3 3r AY OC and CD in

CD 2 2r terms of r.

AC = CB = 4y2 cm... (line from centre | chord) AV SIR

In AAOC:

2_ (3,)7 2

re= ( r) + (4v2) ... Pythagoras AY Pythagoras (S/R)

16 , =32

25’ CAY Simplifying

r? =2xS5 50 cm? CAY Answer

r=5V2 cm

x (5)

8.1.2 | AOD = 54° (angle at the centre 2 x angle at the circum) AVS AYR (2)

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8.2.1 | (Exterior angle of cyclic quad = int. opp. A) Svv (2)
8.2.2
a) |P Sv
QTs Sv
S, sv (3)
b) | QRS = 120’...[Opposite angles of cyclic quad. ] Sv RV (2)
c) | QRS +$, = 120° + 60° = 180° SY
PS || QR (Co-Interior angles are supplementary) RY
(2)
d) | Q, = 90’ .....(Corresponding Angles ; PS || QR) S/RY
TR is a diameter of the circle. (Conv. Angle in the semi-circle) | RY (2)
OR
§, = 60° ... given
Q, = 30° ... given
Q, + Q, + S, =180° opp 2's of a cyclic quad
“.Q, =90°
-. TR is a diameter (Z's subt by 90°)
Conv Z's on semi circle)
nn
! [18]

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QUESTION 9

9.1

Proof:

Area A ADE

Area A DBE 5pB x hy

1

Area A AED _ a AE xh,
Area A DBE 5 EC h

Area A DBE = Area A ECD
Area A ADE _ Area A AED

Area ADBE Area A ECD
AD _ AE
“DB EC

= — (Common vertex E, same
DB height h2)

AE (Common vertex D, same

~ EC

height /)

(Common base DE,
same height, DE //
BC)

Area of A ADE is common
and Area A DBE = Area A
ECD

SY Construction

YSYR

S/RY

YSVYR

(No Marks if no
construction
indicated in
drawing or
words)

(6)

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9.2

T

9.2.1 | In A’s ABC and EBA
1) B=B.....[Common] S/RY

2) A, =E£,... [Tangent — Chord Theorem] SYRY
3) C, = EAB.... [Remaining Angles of A’s]

AABC ||| AEBA...[AAA] RY (4)

9.2.2 | AABC ||| AEBA
AB BC LAC tar tri SRY
AB EA similar triangles

2r
5 3
5 SY Substitution

N
ol

rt+

= (2 +2) =25
3 T3)>
4rt ios
3° 9
16r? _ os
7 =

7? = 25x 2
16

S iswer
= {25x a 15 t
r= 16 = 4 metres (4)

SY Multiplication

9.2.3 | Let AH = 5a and HD = 7a

AF AH _Sa_5 a oremPFH II ED YSYR (2)
FE AD 7 Ja 7 7 [Prop eorem; FH || ED]

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9.2.4 5
AAFH = J SAHE .... (Common Vertex; Equal Heights]

57/5
AAFH = 1 AAED) .... [Common Vertex; Equal Heights]

AAFH _
AAED —

OR

Area A AFH _ } AF. AHSin A,
AreaA AED 1 AE. AD Sin A,

(AE) si AF AH
“.| —= | since — =—
AE FE AD

12a
25

144

YSYR

vS/IR

vs

(4)

9.2.5 | Let OH =x
Then HC = r- x = = — x metres

AF _ BH
FE HE
15
5 4a7*
7 15...

atx
s($+)=7(B-)
a °*)= 4 x

(F+5x)=(+ 7x)
aX)" \qa

... +. [Prop. Theorem; FH || AB

¥S/R

vs

(3)

[22]

THIS PAPER SHOULD BE MARKED OUT OF 139

TOTAL MARKS: 150

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

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