# Mathematics Grade 12 P1 Pretrial FS Mock 2023 Paper & Memo

> Grade 12 Mathematics Paper 1 question paper and memo from the Pretrial FS Mock Exam 2023, 25 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/grade-12-mathematics-2023-pretrial-fs-mock-exam-question-paper-and-memorandum-paper-1/cfdf5033-dc2b-4b51-b1f6-28b8134246e8
- Updated: 2026-09-28
- Subject: Mathematics
- Grade: Grade 12
- Document type: Question paper and memo
- Year: 2023
- Exam period: Pretrial FS Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 25

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## About this document

Mathematics · Grade 12 · Pretrial FS Mock Exam · 2023. Question paper and memorandum, 25 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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education

Department of
Education

FREE STATE PROVINCE

MARKS: 150
TIME: 3 HOURS

Stanmorephysics

This paper consists of 12 pages.

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Gr. 12 Mock Exam

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the question paper.

1.

This Question Paper consists of 11 questions. Answer all questions.
AnswenALL the questions.

Cleartyshow all the calculations, diagrams, graphs, et cetera, which you have
used in determining the answers.

An approved scientific calculator (non-programmable and non-graphical) may be
used, unless stated otherwise.

If necessary, answers should be rounded off to TWO decimal places, unless stated
otherwise.

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

Diagrams are not necessarily drawn to scale.
Full marks will not necessarily be given to answers only.

It is in your own interest to write legibly and to present the work neatly.

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Gr. 12 Mock Exam

QUESTION 1

11

13

1.5

Given x* +2x=0

Solve for x:

2x* —3x-—7=0 (correct to TWO decimal places)

Given: k+5 -4

1.3.1 Solve for k.

14

1.3.2 Hence, or otherwise, solve for x if Vx+5+5= :
vx+5

Solve for x and y simultaneously if:
x—2y—3=0 and
4x? —Sxy+y? =0

The roots of a quadratic equation is given by x =
of k for which the equation will have real roots.

—2 tv4—-20k
2

. Determine the value(s)

Q)

(3)

(4)

(3)

(3)

(7)

(2)
[24]

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Gr. 12 Mock Exam
QUESTION 2
2.1 The first four terms of a quadratic sequence are: 1; —5;—13; —23 ....
2.1.1 jate the general term of the sequence. (4)
2.1.2 h term has a value of -6437 G3)
2.2. 2k +1; 3k ; 5k—S5 are the first three terms of an arithmetic sequence
2.2.1 Calculate the value of k. (2)
2.2.2. Write down this sequence in sigma notation for the first 20 terms. (3)
2.3 Given the geometric series: 8x7 + 4x’ + 2x" +...
2.3.1 For what value(s) of x will the series converge? (3)
8
23.2 Calculate the values of x if S.= 3. (4)
2.4 ‘The first, third and thirteenth terms of arithmetic sequence are the first 3terms of a

geometric sequence. If the first term of both sequences is 1, determine:

2.4.1 the first three terms of the geometric sequence if r > 1
2.4.2 the sum of 7 terms of the geometric sequence if the sequence
Al 2;

(6)

(2)

[27|

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Gr. 12 Mock Exam

QUESTION 3
Sketched below are the graphs of f(x) = 2% and g(x) = —(x — 1)? + q, where q isa

constant. >
The graphs off, intersect at C and D.
C is the y-intercepfjof both fand g.

D is the point of

3.1. Show that g = 2. (2)
3.2 Write down the coordinates of the turning point of g. (2)
3.3 Determine the value(s) of ¢ for g(x) = t if the roots are equal. ()
3.4 Write down the equation of f~?(x) in the form y = ... (2)

3.5 Sketch the graph of f~* on the same system of axes. Indicate the x-intercept and the (3)

coordinates of one other point on your graph.

3.6 Write down the equation of f if A(x) = g(x +1) -2. (2)
3.7. How can the domain of h be restricted so that h 1 will be a function? qd)
[13]

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Gr. 12 Mock Exam

QUESTION 4

The ske! low shows the graph of f(x)= oe A and B are the x- and the y-intercepts
eee

of f.
eames ae ove Z
3
41 Write the equation of fin the form f(x) = part +4q. (3)
4.2 Determine the equations of the asymptotes of f- (2)
43 Write down the coordinates of A, the x-intercept of f (2)

(7]

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Gr. 12 Mock Exam

QUESTION 5
5.1 Given: f(x) = 3-*
5.1.1 Write the equation of f~1(x), the inverse of f(x), in the form (2)
x) =
5.1.2 the same set of axes, sketch the graphs of | (x) and #"(x) [4]

our ANSWER BOOK. Clearly indicate the intercepts with the
s and one other point on the graphs and label the TWO graphs.

5.2 A linear function satisfies the following conditions: p(—3) = 10 and (4)
p'(x) = —2. Determine the inverse of p in the form p~1(x) = ++
[10
QUESTION 6

6.1 On 30 June 2013 and at the end of each month thereafter, Asif deposited R2 500 into
a bank account that pays interest at 6% per annum, compounded monthly. He wants
to continue to deposit this amount until 31 May 2018.

Calculate how much money Asif will have in this account immediately after
depositing R2 500 on 31 May 2018. (3)

62 Onl February 2018, Genevieve took a loan of R82 000 from the bank to pay for her
studies. She will make her first repayment of R3 200 on 1 February 2019 and
continue to make payments of R3 200 on the first of each month thereafter until she
settles the loan. The bank charges interest at 15% per annum, compounded monthly.

6.2.1 Calculate how much Genevieve will owe the bank on 1 January 2019. (3)
6.2.2 How many instalments of R3 200 must she pay? (5)
6.2.3 Calculate the final payment, to the nearest rand, Genevieve has to pay to (5)

settle the loan.

[16]

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Gr. 12 Mock Exam
QUESTION 7
7.1 If f(x) = —2x?, determine f’(x) from first principles. (5)
7.2 Determine:
7210 @ = 2e?+1 (3)
vx
7.2.2 x — 2)? (3)
. 1
7.3 Given: yaa Q)
Prove that the gradient of the curve is negative at each point on the curve where
x>0.
[14]
QUESTION 8

The following information about a cubic polynomial, y = f(x) , is given:

8.1

8.2
8.3
8.4

f(-1) =0
f2)=0
fQ)=—-4
f(0) =—2

f'(-1) = fd) =0
ifx <—1then f'(x) >0
ifx > 1 then f’(x) >0

Use this information to draw a neat sketch graph of f using the grid on the (5)

DIAGRAM SHEET

For which value(s) of x is f decreasing?
Use your graph to determine the x-coordinate of the point of inflection of f
For which value(s) of x is fconcave up?

(2)
(2)
(2)
(14)

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Gr. 12 Mock Exam

QUESTION 9

A window framéiwith dimensions y by / is illustrated below.
The frame consjsts/of six smaller frames.

+—_y—_>

91 If 12 m of material is used to make the entire frame, show that y = ‘ (12 — 3h).

9.2 Show that the area of the window is given by A = 3h— ch

9.3 dA
Find — and hence the dimensions, h/ and y, of the frame so that the area of the

window is a maximum.

Q)
(3)
(5)

[10]

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Gr. 12 Mock Exam

QUESTION 10

Consider the word)*CALCULATOR”

10.1

10.2

10.3

How many different word arrangements can be made from all the letters of the word (2)
CALCULATOR?
What is the probability of making a word arrangement that will start and end with (2)
the letter L?
In how many ways can all the letters be arranged if no similar letters should be close (3)
to each other?

7]

QUESTION 11

11.2

Given P(A) =0,5; P(B) =x and P(A or B) =0,88.

Calculate the value(s) of x, if
11.1.1 Event A and B are mutually exclusive. (2)

11.1.2. Event A and B are dependent. (3)
South African Women Football Team known as “Banyana Banyana” are to play a
friendly football match against their Nigerian counterparts known as “Super

Falcons”. There should be a winner on the day of the match. Penalties would decide

the winner in case the match is drawn after regulation time. On the day the match

takes place, there is a 30% chance that it could rain (R), 45% chance that it could be
sunny(S) or it could be cloudy(C). Banyana Banyana has a 24% of winning ona

rainy day, a 65% of winning on a sunny day or a 33% chance of wining on a cloudy

day.

11.2.1 Draw a tree diagram to represent all outcomes of the abave scenario. (3)
11.2.2 What is the probability of Banyana Banyana winning the match? (3)
[11]

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Gr. 12 Mock Exam

Name and Surmame: .........00...066:00cccccccceseessceeeeeeeeeneeecseessestenneesrsseness

Class:

DIAGRAM S
QUESTION 8.

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Gr. 12 Mock Exam

INFORMATION SHEET
ye DEVE —4ac
2a
A= P(l+ni) = P(1—ni) A=P(\-i)" A=P(l+i)"
T, =a+(n- S, =52a+(n-Nd]
T, =ar"" gs <M) pel S,=—4;-ler<l
. r-l l-r
a i
 S(x+h)—f(x)
= in ——————
fQ)- oA
d= (x, -%)' +02 -»,) w(S22222)
y=nmerte y-y, =mMx-x,) m=22 m=tané
x, x,
(x-a) +(y-by =y*
b 2_,? 2 1
InAABC: an me a’ =b'+c°-2be.cosA area AABC = = absin
sin(a+ 8) =sin a.cos f+ cosa.sin B sin(@— 8)=sin a.cos#—cosa.sin B
cos(a+f)= cosa@.cos #—sin a.sin § cos(a—2)=cose.cosf+sin asin §
cos’ a—sin* a
cos 2@ =41—2sin* a sin 2a = 2sin a.cos@
2cos?a-1
a 2
_ x ee
yo oo =f
n n
P(A) = 78) P(A or B) = P(A) + P(B) — P(A and B)
p=atbx pone“)

De-x)°

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education

Department of
Education

FREE STATE PROVINCE

MARKS: 150

This marking guideline consists of 14 pages.

1

### Page 14

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QUESTION 1
11.1] x@+2)=0
r=
x=0 OR/OF x=-2 el

Vx=-2 (2)
vx? + 2x> 0
¥ graph/grafiek/ critical
pts
v¥x>0

OR/OF —+|—+—_|+-

If x > 0 only | mark
Indien x > 0 slegs 1
punt

-2 0
ax>0 (3)
1.2. ~(-3)+,/(-37 - a ¥ substitution into
x= Aa correct formula/
(2) substitusie in
34 V65 korrekte formule
*y 4 ¥ simplify
—~ oe ¥2,77
x¥=2,77 OR/OF x=-1,27 v1.27
If left in surd form only
2 marks.
Indien in wortelvorm,
slegs 2 punte. (4)
1.3.1} 4745k-14=0 ¥ multiplying
(k-2)(k +7) =0 Y factorising
k=2 o k=-T7 ¥ both solutions (3)
13.2] Jx+5=2 of Vxt+5=-7 Y substitution
x+5=4 or _ invalid/no solution ¥ identifying invalid
x=-l Y value of x
KH 2H 3H veasases 1
14 4 o Y equation (3)

4x7 —Sxyty?=0 ..... (2)
From eq (1): x = 2y +3 ...(3)
Subst. (3) into (2):
+ A(2y + 3)? - Sy(2y +3) +y? =0
« 4(4y? + 12y + 9) — 5Sy(2y +3) + y* =0
« l6y? + 48y + 36 — 10y2 —15y+y* =0
= Ty? + 33y + 36 =0
+ (Ty + 12)(y + 3) =0
—12

HY= we OLY S -3

bs => orx = -3

substitution

Y simplification
Y standard form
Y factors

Y both y values

Y both x-valul

(7)

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1.5 For real roots : ¥4-20k>0
4—20k >0
“20k > —4 v Answer Q)
w~k< 1
“te
a ‘
—
QUESTION 2
2.1.1 | 13-5; —-13; -23 ¥ Value of a
-6 -8 -10 ¥ Value of b
ea ie Y Value of c
2a=-2 3a+b=-6 at+b+c=1
a=-1 3(-1)+b=5 -1-3+c=1
b=-3 c=5
Tr=—w —3n +5 ¥ General term (3)
24.2 | -7 = 3n + 5= 643 ¥ equating
w +3n — 648=0 fants
(n+ 27)(n — 24)=0
n=—27 orn=24
2.Trq = -643 y answer (3)
3k —(2k + 1)=5k—5—3k ¥" Difference
k-1=2k-5
v
2.2.1 eet answer (2)
2.2.2 | 912,15 v Sequence
¥ General term
T, =3nt+6
20) ¥ Sigma Notation (3)

### Page 16

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2.3.1

¥ value of r
¥ Condition for converging
series
¥ answer (3)
2.3.2
i ¥ — Substitution into the
S.7 a correct formula
x
i>
2
8
= 1-2)- 8x?
3 2
8 8 2
Le
3 6
16 - 8x = 48x? ¥ — Standard form
2 =
48x* + 8x -16=0 Apaches
6x" +x-2=0
(3x +2)2x -1)=0 ¥ Answers
2 1 4)
x=-= or x==
2
241 a+2d=ar at+l2d=ar a+2d=ar
1+2d=1r 1+12d=1/7 ¥ a+i2d=ar
_r-l Folia v values of d
de 2 142 (3 ei Y — standard form
Tore =r v factors
P= 6s 0 Y values of r
(r= Ir — 5) =0 v sequence
r=lorr=5
eeSen2y
OR
r=1+2d ———» 1+12d=7
1+ 12d=(1 +2dP
BEAL ¥ at+2d=ar
4d —8d=0 Y atl2d=ar
4d(d — 2) =0 Y¥ values of r
WE OGRae Y — standard form
r=14+20) r=1+2(2) Y factors
| =5 v values of d
Le 5: 25 ¥ sequence (6)

ies

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2.42 a(r" —1) ¥ substitute in correct
eae ia | formula
aes
S,= 16° -))
5-1
= 19531 Sane
be Q)
L
Tava}
QUESTION 3
3.1 y=2* ¥ substitute x = 0 and y=0
x=0 into correct formula
-. f(0) =2°=1
2. C(0;1)
y - intercept: a(=—(x-l) +4 i oa
C(0;1)
1=-(0-1)' +q
l=-l+q tS ee
er pr
¥ Symmetry E(4 ; 0)
3.2 g(x)=-(x-1)° +2 v answer /antwoord
DQ; 2)
3.3 t=2 ¥ answer/antwoord (1)
3.4 fol ix=2” ¥ ¥ answer/antwoord
“y= log 2x
3.5 Y form of the graph
¥ /vorm van die grafiek
¥ X—intercept/afsnit
(0:1)
phe (231) ¥ any other coordinate on
—- the graph
: Xx
0/150) /enige ander koordinaat
op die grafiek
—t
iF
3.6 g(x) =—(x—1)? +2 v d 2
g(t+1)-2=-(x- 141)? +2-2 VRMr/anwoord — 2)
“ A(x) = -x?
3.7 Domain: x>0 or x <0 v Answer (1)

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QUESTION 4
4
4.1 fart
x=l y 22143
x-1+3 x-1
J (x)=
an v a oo
x-l
f(x)=——+1 v FG)e 2241
x-1 x-1 (3)
4.2 x=l vx=l
yel v y=1(2)
4.3 o-2%* 7 y=0
x-l
0=2+x
x=-2 ear
A(-2:0) aS 252)
[07]
QUESTION 5

5.1.1 ( )
Pope
)
f"(x)=log,x OF f-'(x)=—log, x

: | Slegs Antwoord:Volpunte

¥ ¥antwoord @)

5.1.2 \ }

(13)\f |

¥ vorm
¥ y-afsnit

{7100 = logax:
3

¥ vorm
T Y x-afsnit
sb (4)
f'@)
5.2 p(-3)=10 en p'(x)=-2
p(x)=-2x+e P ipeg
10=-2(-3)+e Cea
-e=16 “3
omruil vai
p(x)=y=-2x+16 (sam apat
x=-2y+16 ¥ antwoord (4)
rs 1
‘(x)=-ax+8
Pi(x)=-5

[10]

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QUESTION 6

6.1

¥ n= 60 and i= 298 0,005

Y correct substitution
into correct formula

v answer GB)
= RI74 425,08
6.2.1 After eleven months, Genevieve will owe/ Vn=l11
Na elf maande skuld Genevieve Y correct substitution
A =82.000 1+ 0,15 Mt into correct formula
12 ¥ answer
=R94006,79 (3)
6.2.2 x[1-(+i)"] ¥ 94006,79
p=_—___.
is
-"] v substitute into correct
2am -(14918) | formula
12
94006,79 = a
12
94 006,79 0,15 0,15)"
xx 2 = 1-| 14+
3200 12 12 ¥ correct use of logs (logs to
(: : a) =1-0,3672147... be defined)
12
0,15 =!
—nlog| 1+ —— | =log0,6327852.. v n=36,84
12 ¥ 36 installments
—n = —36,8382...
(5)
n= 36,84

Genevieve will have to pay 36 installments of R3 200

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6.2.3

xl-(1+i)”
r

P=

vn = —083826912

¥ substitute into correct

eal (4 01 ee] forrniala
:
¥ answer
P=2652
Outstanding balance after 36 installments is R2 652 i
Final payment will be: v 2652.0 z, ae
A= 2652,00114.2!5) .
a on Calin ¥ answer
=R 2685,00
OR/OF OR/OF
015 2 0,15)"
3.200 [1 us)" -1 94006,7! 2
0,15) 12
Balance : 94006,7{ 1+ - ) - 15 015 36
12 ed 3200 (1+) -1
12 v 12
= R2 651,72 0,15
12
v¥ 2 651,72
Final payment will be:
0,15)"
v ,
A= 26s173{1+ 248) 2osi rf |
Y¥ answer
=R2 685,00
(5)

[16]

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QUESTION 7

7A fi) = lim Lorn F@)

¥ Substitution into correct formula

v :
, . —2(x+h)2—(—2x2) Expansion
P= 1 ¥ Factoring
' « —2atpareh—2h? +2x
P@)= a ¥ Applying limits
f'(x) = lim v Answer
f'(x) = lim
f'@) = 4x
721 | y= Bed ¥ Simplified to binomial.
ct ;
oO Y 3x2

vtix? 3)

1 3

e = 3x2+=-x 2
7.2.2 | Dy[9x? — 12x + 4] 9x? — 12x +4

=18x-12 v 18x

v-12 (3)

73 y=x? v —2x73

WY _ oy 3 = 2 v2

dx ax" = x3 *

Therefore <0 forx >0 v oe <0 forx >0 (3)

[14]

QUESTION 8

[|

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8 ‘ v (-1;0)
‘ v (2;0)
; Y (0:-2)
v4)
v shape
: _ : 0) (5)

8.2 | -1<x< 1 (2)

8.3 Pt. of inflection: “halfway” between turning points(x- mate
values) v¥x= 2
ig weno)

2
8.4 f (x)>0 Yf'(x)>0
ax >0 ¥x>0 (2)
[11]
QUESTION 9
109A.
LI] donat
Et

10

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9.1 There are four horizontal bars and ¥ 4y43h
three vertical bars. Y 4y+3h=12 0)
=> 4y+3h=12
4y =12-3h
y= A (12 —3h) =
4 Im
9.2 Area = lb v Aayh
=yh ¥ Subst of y
=1(12-3iyh Y simplification
eae G)
4
9.3 Area =3h asp Y derivative
roaee |
dA 3
= =3-=h
dh 2 :
aA a ¥ equating to 0
ae
3-2 n=0
i as
~ 1 ¥ subst of h
y=7(12-3@)) Y value of y
33 (5)
2)

[10]

11

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QUESTION10
10.1 10! _
222 453600 yo!
21.21.21
v Answer
naa . 2)
10.2 0080 ore
v Answer
(2)
10.3 Same letters are together in ¥ Same letters together

2.2.2.7! ways =

!
Probability At = <
38

So required probability is =
OR

v Probability of same letters
together

v Answer

OR
¥ Same letter together
v Same letter not

Same letters not together is 41380 together
ways v Answer
Probabilit ired i ad 8)
obability required 18 45
[07]
QUESTION 11
11.1.1 | P(A or B)= P(A) + P(B) subst into correct
0.88 = 0.5+x formula
ee eee answer (2)
11.1.2 | P(A or B) = P(A) + P(B)— P(A).P(B) subst into correct
0.88 =0.5+x—-0.5x formula
0.38 =0.5x 0.5x
x=0,76 answer (3)

12

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¥ 25% for C

11.2.1
¥ Correct outcomes
w— RW
242 to represent
30% R 76 ogee BAL win/lose
9 v Correct diagram
65% a seta I (3)
5% S$
% 35% w— sw
3
Wr ote cw
679
ib
acai 3 |
11.2.2 Probability of winning v

P(W) = P(R and W) or P(S and L) or P(C and W)
=30% x 24% + 45% x 65% + 25% x 67%

1889
= 0.9445 OR 2000

30% x 24% + 45% x 65% + 25% x 67%
v

~ answer

(3)

(14)

13

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## Collection

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

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