# Mathematics Grade 12 P2 Pretrial FS Mock 2023 Paper & Memo

> Grade 12 Mathematics Paper 2 question paper and memo from the Pretrial FS Mock Exam 2023, 29 pages. Read online or get the free PDF download.

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- Updated: 2026-09-28
- Subject: Mathematics
- Grade: Grade 12
- Document type: Question paper and memo
- Year: 2023
- Exam period: Pretrial FS Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 29

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## About this document

Mathematics · Grade 12 · Pretrial FS Mock Exam · 2023. Question paper and memorandum, 29 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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education

Department of
Education

FREE STATE PROVINCE

MARKS: 150

DURATION: 3 HOURS

This question paper consists of 15 pages.

1

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INSTRUCTIONS AND INFORMATION
Read the following instructions carefully before answering the questions.
1. This question paper consists of 9 questions.

2. Answer AEE*the questions.

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

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

5. Answers only will NOT necessarily be awarded full marks.

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

graphical), unless otherwise stated.

7. If necessary, round off answers to TWO decimal places, unless stated otherwise

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QUESTION 1

The speeds, in kilometres per hour, of cyclists that passed a point on the route of the Ironman Race were
recorded and summarised in the table below:

Speed (km/h) EL Frequency i} Cumulative
On Frequency
0<x<10 7 10 10
10<x<20 5 30
20<x<30 © 45
30<x<40 72
40<x<50 170
1.1 Complete the above table in the ANSWER BOOK provided. (2)

1.2. Make use of the axes provided in the ANSWER BOOK to draw a cumulative

frequency curve for the above data. (3)
1.3 Indicate clearly on your graph where the estimates of the lower quartile (QI ) and

median (M) speeds can be read off. Write down these estimates. (2)

1.4. Draw a box and whisker diagram for the data. Use the number line in the

ANSWER BOOK. (2)
1.5 Use your graph to estimate the number of cyclists that passed the point with speeds
greater than 35 km/h. qd)
[10]
QUESTION 2

During the month of June patients visited a number of medical facilities for treatment.

The table shows the number of patients treated on certain dates during the month of June

Dates in the month of June 3 5 8 12 | 15 | 19 | 22 | 26
Number of patients treated 270 | 275 | 376 | 420 | 602 | 684 | 800 | 820
2.1 On DIAGRAM SHEET 2, draw a scatter plot of the given data. (3)

2.2 Determine the equation of the least squares regression line of patients treated (y)

against date (x). (3)
2.3. Estimate how many patients have been treated on the 24th of June. (2)
2.4 Draw the least squares regression line on the grid on DIAGRAM SHEET 2. (3)

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2.5 Calculate the correlation coefficient of the data. Comment on the strength of the

relationship between the variables. (3)

2.6 Given that the mean for patients treated on certain dates is 528,63 calculate how

many patieats were within one deviation of the mean (3)
[17]

QUESTION 3

In the diagram below, A (-3 ; k), B(4 ; 8), C(5 ; 0) and D (-2 ; -4) are vertices of the
parallelogram ABCD. Diagonals AC and BD bisect each other at P. The angles of inclination
of AD and BD are a and B respectively. AD cuts the x-axis at E. F is a point in the fourth quadrant.

e B(4; 8)
A334
Pp
as B
E 0 C(5; 0) x
D(-2;-4)
e F

3.1 Determine the gradient of BC. (2)
3.2 If the distance between points A(-3 ; k) and B(4 ; 8) is 65, calculate the value of k. (4)
3.3 Prove, using analytical geometry methods, that BP | AC. (3)
3.4 Calculate the coordinates of F if it is given that ACFD is a parallelogram. (2)
3.5 Calculate the size of EDO (correct to ONE decimal place). (6)
3.6 Calculate the area of AADC. (4)

[21]

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QUESTION 4

4. A circle with centre at C passes through the origin, O, and also intersects the x-axis at F

and the y-axis at E. The tangent to the circle at B ( 4 ; 6) intersects the x-axis at K and the

F o
c
B(—4; -6) E
L
4.1 Calculate the length of the radius of the circle. (3)
4.2. Determine the equation of the circle in the form (x _ ay + (y - by =r. (4)
4.3. What type of a triangle is AOBL ? Give reason for your answer. (2)
4.4 Determine the equation of the tangent KL. (4)
4.5 Determine the co-ordinates of E. (2)
4.2.6 Determine whether EF is a diameter of the circle. Show all working. (5)
[20]
QUESTION 5
5.1 If tan 58° = m, determine the following in terms of m without usifi
5.1.1 sin 58° (2)
5.1.2 sin 296° (3)
5.1.3  cos2° (3)

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§2 If 5tan@+ 26 =0 and 0°< 6 < 270° , determine with the aid of a sketch and
without using the calculator, the value of :
5.2.1 sinO
5.2.2 cos?
5.13 Afggs0+7V6sind
Bs(E 240°), tan 225°
QUESTION 6
cos(180° + x). tan(360° — x).sin*(90° — 9
5.1 Determine the value of os( x) - ( x).sin ( x) +sim° x
sin(180° — x)
5.2.1 Prove the identity: cos(4 - B)-cos(A + B)= 2sin Asin B
5.2.2 Hence calculate, without using a calculator, the value of
cos15°—cos75°
53 Find the value of tan @, if the distance between A(cos6; sin 2) and BO; 7)
is 86.
QUESTION 7

Consider: I (x)= cos(x — 45°) and J (x)= tan x for xE [- 180° ; 180°]

6.1

6.2

Use the grid provided to draw sketch graphs of f and g on the same set of axes for
xe [-180° 3 180°]. Show clearly all the intercepts on the axes, the coordinates

of the turning points and the asymptotes.

Use your graphs to answer the following questions for x € [- 180° ; 1 80°]

6.2.1 Write the solutions of cos(x - 45°) =0

6.2.2 Write down the equations of asymptote(s) of g.

6.2.3 Write down the range of f.
. . . ° 1
6.2.4 How many solutions exist for the equation cos(x—45 )s tan ?

6.2.5 For what value(s) of x is fi xdg(x)> 0?

(2)
(1)

(4)

[15]

(6)

(3)

(4)

(4)
[17]

(6)

(2)
(2)
(1)

(1)

G3)
[15]

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QUESTION 7

In the figure below Thabo is standing at a point A on top of building AB that is 45 m high. He observes
two cars at C and D respectively. The cars at C and D are in the same horizontal plane as B. The angle

of elevation from{G+to A is 43° and the angle of elevation from D to A is 50° and C AD = 69°

45m

(44
7.1. Calculate the lengths of AC and AD, correct to 2 decimal places. (4)
7.2 Calculate the distance between the two cars, the length of CD. (3)

(7]

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QUESTION 8

PROVIDE REASONS FOR ALL YOUR STATEMENTS AND CALCULATIONS IN
QUESTION 8, 9 AND 10

Tn the diagram, is a cyclic quadrilateral. G is a point on AD such that BG || CD. ECF

is a tangent to tl cle at C. BD is a chord of the circle. GB D =30° and DC F = 60°

8.1 Calculate, with reasons, the size of:
8.1.1 D,
812 «iB,
8.1.3 C,
814 DAB

8.2 Is BD a diameter of the circle? Motivate your answer.

(1)

(2)

(1)

(2)
(2)
18]

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QUESTION 9
9.1. In AABC below, D and E are points on AB and AC respectively such that

AD AE
DE || BC ve the theorem that states that DE =

BEC

(6)

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4 education

Department of
Education

FREE STATE PROVINCE
9.2 In the diagram below, DGFC is a cyclic quadrilateral and AB is a tangent to the circle at B. 2
Chords BD and BC are drawn. DG and CF produced meet at E and DC is produced to A.

EA||GF.

Downloaded from Stanmorephysics.com, ‘

9.2.1 Give areason why B, = D, (1)
9.2.2. Prove that AABCII| AADB. (3)
9.2.3. Prove E, = D, (4)
9.2.4 Prove AE? = ADx AC (4)
9.2.5 Hence, deduce that AE=AB (3)
[21|

10

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SCHOOL NAME:
Name: Grade 12
lnar
4000,
-
Speed (km/h) Frequency (f) Cumulative
Frequency
0<x<10 10 10
10<x<20 30
20<x <30 45
30<x<40 72
40<x<50 170

11

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QUESTION 1.2

Cumulative frequency

100

70

50

30

20

10

Cumulative Frequency Graph

Ee

20

30 40

50 60

70

80

12

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DIAGRAM SHEET 2

QUESTIONS 2.1 AND 2.4

tter plot of dates (in June) versus patients
900 ony
=!
It
800 =)
700
600
=]
2
Fy 500
Fao
&
300
200
100
0 iS 10 15 20
Dates in June

13

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QUESTION 9.1

14

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INFORMATION SHEET: MATHEMATICS

—b+aJb? —4ac

2a
A=P( A=P(l=ni) A=P(l-i)" A=P(l+i)"
n a |
Sige yi-M or =a+(n-Nd 5, ="(2a+(n—1d)
ial =) 2 2
cae" 5-0) pel S, =—%;-I<r<l
r-l l-r
pals -) pla")
i i
(x)= lim F(x+h)- fe)
ho0 h
+
d= V(x, -x,)7 +(¥2 -y,)? m2 272)
a 2
y=mxt+e yy, =m(x—x,) m=227* m=tan@
¥q—%

(x-—af +(y—byY =r?
masse, 2 = 6 2 © qr sh? 40?-2be.cosA area ABC =| absinC

sind sinB sinC 2
sin(a + £8) = sina.cos + cosa.sin # sin(a — 8) = sina.cos # — cos a.sinB
cos(@ + 3)= cos a@.cos #—sina.sin 8 cos(@ — 3)= cos @.cos #+sina.sin 8

2 a]
cos’ a@—sin @

cos 2a =41—2sin’ a sin2a@ = 2sina@.cosa

2cos? a —1
Ie D(z, -3)
y=-—— o =
n n
nat
P = 715) P(A or B) = P(A) + P(B) — P(A and B)
peatbe p= 2G-20-9)
De@-xy

15

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education

Department of
Education

FREE STATE PROVINCE

MARKS: 150

This marking guideline consists of 14 pages.

1

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QUESTION 1
1.1
Speed (km/h) Frequency (/) Cumulative frequency Y column 1 values
O<x 10 10 ¥ column 2 values
fnYaTa
10<x 20 30 (2)
20 < xjS30 45 75
30<x < 40 72 147
40<x<50 23 170
1.2 ¥ Grounding(0; 0)
$98 v¥ Maximum (50; 170)
& a8 - ¥ Curve
G3)
13 | = YQ
# 2
iS
e
Z:
3
Es
1.4 ¥ Q3
1 , 7 Y Box (2)
1.5 164 (Accept 163 — 165) v Answer (1)
[10]

I]

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QUESTION 2

2.1

EE

v¥ v v ALL points plotted
correctly
-1 for every 2 mistakes

-1 if points are joined

1
t
1
t @)
YY t
E soo Ai 175; sa0.a75) |
H A
bade tt
'
i Ea
. +
es i]
a3 H
t
100 +
i
'
ot } 1
o 5 10 41s 20 2 »
Dates in June
2.2 | y=26,88x +161,24 ¥ value of a= 161,24
v value of b = 26,88
¥ equation of the line (3)
2.3. | y=26,88(24) 4161,24 ¥ ssubstitute in formula
y =806,36 y
806 people treated on 24th of June answer as Natural Number (2)
24 | seegrid ¥ straight line
v graph passes through (x ; y)
¥ graph passes through y-
intercept from regression line
formula (3)
2.5 | r=0,98 This is a very strong positive, linear correlation. As the days | YY correct answer
. . . v interpretation (3)
increase the number of patients increase
2.6 | (x-o;x+o)
(528,63 —210,46;528,63 + 210,46)
(318,17 ; 739,09) 17
4 patients. 9,09)
v answer

(3)

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[17]

QUESTION 3
31] 92 ~*daoat
* x x ada
_ 8-0 cal Y correct substitution into the
BC 4-5 correct formula
Y answer
mM, =—-8 Q)
3.2 2 z v substitute A and B into
B=y(-3-4 k-8
A ( ) * ( ) distance formula
65=49+k? -16k +64 Y standard form
2 Y factors
k —-16k +48 =0
v¥ k=4
(k -4\(k -12)= 0
k=4o0r k=12
k=4 4)
3.3 8- (- 4) Y My
Myc
My =2 v Mp XM 4c
-4-9 _ 1
“-3-5 2
1
Myc XM = 2%— =I GB)
AC | BD
3.4 midpoint of AC = midpoint of DC ¥ x-coordinate
Z _ _ ¥ y-coordinate
x+(-3)_=2+5 ) yt4_-4+0
2 2 2 2
x=6o0r y=-8
F(6;-8)
3.5 | m, =m, =-8 Map
tana =-8 vY tana =—
a =180° tan '(8) ¥ a =97,13°
— °
a =97,13° ¥ fRa3,43
veal
t = =2 Ml
an B Mop al er
B=63,43° PS

O, = 63,43°(vert opp Z’)

(6)

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EDO=a-B
EDO =97,13° - 63,43° =33,7°
3.6 = —(—3)/ +(Q—4Y = = Y length of AC
AC = J (5-(-3)) + (0-4) = V80 = 45 Y leneth of DP
DP=, I- 2- ly; == (- 4- 2y = 45 =3V5 Y correct substitution into
i formula
Area of AADG= pac x DP v answer
(4)
Area of AADC = 5 (4y5 avs)
Area of AADC = 30 square units
[08]
QUESTION 4
4
41 = (x, 1) +(v,-y,)
OB & * y +O.-y y v substitution
OB =,|(-4-0) +(-6-0) Y OB=V52
OB = v52 v r=vl13
OB = 213 (3)
r=v13
4.2 ("=F *)
2 2
C(-2;-3)
(x+2) +(y+3) =13
4.3 | Right-angled triangle, tangent perpendicular to the radius. Y right angled A
v reason (2)
44 -
—~ J.
Mog = ox ¥ Mop
—6-0 ¥ Me,
BO yp
3 4-0 Y substitution
Mao = 2 v answer
(4)
m, =-2
‘KL 3
Y-YN= n(x»
yt6= 2 r+ 4)

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__2 2%
rr ae
4.5 F(0;-6) vv answer (2)

46 | (+2) +(063=13

F(-4:0) any
mm evocl a 3
944 2

_-346 3
e240 2

Points F, C and E are collinear
EF is a diameter

v xX,
Y Ve
Y Mac
Y Mey

Y Points F, C and E
are collinear

(5)

[20]
QUESTION 5
5.1.1 tan 58° =m vVl4m =r
x+y =r v answer
P+m=r ;
Vita =r @)
sin 58° = —"
l+m
1+m
5.1.2 | sin 296° =—sin 64° v —sin 64°
sin 296° =—sin 2(32°) Y substitution
sin 296° = —2sin 32° x cos32° ¥ answer
sin 296° =-2x. yp
l+m Vl+m GB)
. 2m
sin 296° =— —,
l+m? lar
52-1 | 5tan@+2V6 =0 lonat

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y ._v49
r=
; 5
——
2vél v sind = 20
sin 9 =———
it v649
; 5
tan 0 = -—— 0°< 6 < 270°
ne 5 (2)
re=yptx
r =(-2V6) +(5)
r =49
r=7
sind= 26
7
5.2.2
cos =— 5 i"
answer
dd)
cos(— 240°), tan 225°
5 26
u- 3 + ra 255) v —sin 30°
cos(90° + 30°) tan(180° + 45°) Y¥ tan45°
—sin 30°. tan 45° u
i v -=—xl
-=xl
Y answer (4)
=-4
[15]
QUESTION 6

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6.1 cos(180°+ x) .tan(360°~x).sin?(90°=2) J sa y —cosx
sin(180°—x) ~ tana
v cos’ x
v sinx
= Y cos x+sin’ x
v answer
(6)
= cos’ x + sin’? x
=]
6.2.1 | cos(A — B)—cos(A + B)
v
=cos A cos B+ sin A sin B — [cos A cos B —sin A sin B] v
v
= cos A cos B + sin A sin B—cos A cos B +sin A sinB
=2sin A sinB G)
6.2.2 | ©os15°—cos 75° = cos(45° — 30°) — cos(45° + 30°) v
v
= 2sin 45°.sin 30° v
v
= axr2 yh or/of 2x-txt
2-2 apy)
eae (4)
2 V2

OR

cos15°— cos 75°
= cos(45° — 30°) — cos(45° + 30°)
= cos 45°cos30°+sin45°sin 30°— [cos 45°cos 30°—

sin45°sin30°]
= 2sin45° sin 30°
=2 2d or/of 2 xxl
2°23 ond)
V2
=— of =
5 of ii

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6.3

AB? = (cos 0-6) Hsin@-7)°

86 = cos’O — 12cos 6+ 36 +sin’0 — 14sin0 +49

86 = 1+ 36; — 12cos 0 —14sin 6
O=-1 14sin 0
14sin6 = -1
sin® _ -12
cos@ 14
tan@ =—°/-0,86
i

LAKK

(4)

### Page 25

\o

cS
= =
a=! R=
228 22
Soh F op
9 foo g
Sead
feces
Esaas
SSssS>S

v shape f

eoSeeS:

Downloaded from Stanmorephysics.com

QUESTION 7

' ' i °
1 1 ' ial
i xr 48
' ' b
H H H har
1" ks
ar Be me Po ase
: . ' by ov
: ' : es
ee eonorer sl psceks =
H ' : eo oii im
: ' ; ES ~ S
H ' ' es Jy - Ss
' ; ' Pee =| So) 8 °
o--4 donenaoee banaes Wyn yy :
, . n x ~~ wy
: e|ig) & *
' i ‘ sss] !
1 ‘ ' ~~ i} Vv
' A |o. Ss se
Eel Reem caren ene nets eer! FSeeRenay eee Cenece eke ee —| &
i ‘ H eioi=| & &®
' 4 ' 1|S\7| BY
' t H ay) fos
i \ "Tllul fy 3%
idea crtinerdetrsindedvaleciel stele ain’ ainid atiladarahatad ben] fe left oF
= lan} iin
= ai ailalaila
i [ey eye ie

Q)
(2)
Q)
(1)
U5]

Y notation
v answer

QB)

v Y answer
v¥_v_answer

v answer
¥ end points

v answer

10

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QUESTION 7

TA
sin 43° = V sin 43° = >
AC AC
AC =66m v AC
. 45 . 45
43° = —— ¥ sin 43° = ——
sin 4c tC
v AD
sin 50° = =
4
AD =58,74m (4)
7.2. | CD? = AC’ + AD’ —2AC.AD cos 69° Y using cosine rule

CD =(66) +(58.74) —2(66)(58.74)cos69°
CD* =5027,72
CD =70.91m

Y substitution

v answer

(3)

11

WEES:

RU

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QUESTION 8

8.1.1] A * ¥ S&R 1
, = B, = 30° (alter angles DC parallel GB) (1)
8.1.2] 4 4 ° vs
B, =C, =60 (tan-chord theorem) YR (2)
8.1.3] 4 ¥ S&R 1
C, =90° (sum of angles in triangle) @
8.1.4 2 n vs
DAB+ C2 =180° YR (2)
DAB =90°
8.2 | Yes is diameter, converse angle in a semi-circle vs
VR (2)

12

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QUESTION 9
9.1
Y constr
vs
vs
YS YR
Cc ¥ conclusion
Construction: Connect DC and BE and draw the altitudes k
and h
i
Area AADE _5*ADxK _ aD (6)
Area ABDE 1xpDxk BD
i
Area AADE _7*%AEXh ag
‘Area ADEC — LxECxh ~ EC
but/maar: Area ABDE = Area ADEC [DE common base and DE | | BC/
DE gemeensk basis en DE | | BC]
Area AADE _ Area AADE
“Area ABDE Area ADEC
AD AE
BD EC
9.2.1 tangent-chord theorem YR (I)
9.2.2 In AABC and AADB: Ag
A =A, [common/gemeenskaplik]
B, =D, [proven/bewys in 10.2.1] vs
:. AABC ||| AADB[Z; Z; 2] YR
OR
In AABC and AADB:
A, =A, [common/gemeenskaplik] “Ss
a" a [proven/bewys in 10.2.1] vs
BCA=B, [Zs of A=180°] mR
:. AABC ||| AADB
9.2.3 E, =F [verwiss Ze/alternate Zs ; EA | | GF] vS YR
F=5 5 [ext Z of cye quad DGFC/buite Z v kdvh YS YR
E,=D,

13

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9.2.4 In AAEC and AADE: vs
[common/gemeenskaplik]
v
[proven/bewys in 10.2.3] s
v
C ||| AADE (4; 2; 2] e
vs
= AD AC
AAEC and AADE:
[common/gemeenskaplik] vs
[proven/bewys in 10.2.3] vs
[4s of A= 180° OR ext 4 of cye quad DGFC/ VR
buite 2 v kdvh DGFC]
.. AAEC ||| AADE
_ AE _ AC
2 AE? = ADx AC
9.2.5 AB LAC [AARC ||| AADR] “S
AD AB
AB? = ADx AC vs
2 “8
= AE~ [from 10.2.4]
-. AB= AE
24)

14

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

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