# Mathematics Grade 12 Paper 1 KZN Mock 2023 Paper & Memo

> Grade 12 Mathematics Paper 1 question paper and memo from the KZN Mock Exam 2023, 39 pages. Read online or get the free PDF download.

- Page: https://www.hlayiso.com/library/grade-12-mathematics-2023-kzn-mock-exam-question-paper-and-memorandum-paper-1/23e5a387-b156-40df-ba79-15802f0dc067
- Updated: 2026-09-28
- Subject: Mathematics
- Grade: Grade 12
- Document type: Question paper and memo
- Year: 2023
- Exam period: KZN Mock Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 39

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## About this document

Mathematics · Grade 12 · KZN Mock Exam · 2023. Question paper and memorandum, 39 pages. Read online or download the PDF.

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### Page 1

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MARKS: 150

TIME: 3 hours

KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

N.B. This question pap a COHSBES METZ pag s and an information sheet.

ight Reserved

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INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.
1. This question/paper consists of 13 questions.

2. Answer ALL questions.

3. Clearly show ALL calculations, diagrams, graphs, et cetera that you have used in
determining your answers.

4. Answers only will not necessarily be awarded full marks.

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

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

Ts Diagrams are NOT necessarily drawn to scale.

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

question paper. Write neatly and legibly.

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QUESTION 1

1.1 Solve for x:

| (2)
1120 te3y=7 (write your answer to two decimal places). (4)
L130 x—5r44>0 3)
114 3*-10.3'+21=0 ©)
1.2 Solve simultaneously for xand y :
(6)
xt+y=2;  w4+y+6x=4y-4
1.3 +95 _72
The roots of a quadratic equation are x= =a :
For which values of n will the roots be equal? (3)
[23]
QUESTION 2
2.1 Consider the arithmetic sequence:
See 5 see 99.
2.1.1 | Determine the twentieth term of the sequence. (2)
2.1.2 How many terms are in this sequence? (2)
2.2 The first term of an arithmetic sequence is a and the thirteenth term @3=24.
2.2.1 Determine the common difference of the sequence. (2)
2.2.2 Hence determine the sum of the first 200 terms in terms of a. (2)
[8]

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QUESTION 3

The values below are the consecutive terms of a quadratic sequence.
The fourth term is 49.

—3 7-5-5 495/77 lll; USL
3.1 Determinethellhird term of the quadratic sequence. (1)

3.2 Determine the general term, T,, of the quadratic sequence. (4)

tn

3.3 Between which two consecutive terms of the quadratic sequence is the first

difference 418? (3)
[3]

QUESTION 4

4.1 The first two terms of a geometric sequence are x and x+1.

4.1.1 | Write down the common ratio. (1)
4.1.2. Write down the third term. (2)
4.1.3 If x=2, will the sequence converge? Motivate your answer. (2)

4.2 The given sequence below has four terms only, such that the first three terms,
T, ; T, and T, form an arithmetic sequence and the last three terms 7, ; T, and 7,

form a geometric sequence:

6; a; b; 16.
Calculate the values of a and b. (5)

[10]

QUESTION 5

. -3
Given: g(x) =——-1
3-x

5.1 Write down the equations of the asymptotes of g. (2)

5.2 Calculate the x and y intercepts of g. (3)

5.3 Draw the graph of g showing all asymptotes and intercepts with the axes (3)

5.4 Determine the equation of the axis of symmetry of g with a negative gradient. (2)
[10]

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QUESTION 6

The graphs of a parabola f and a straight line g intersect at P and S as shown below.

Points R (-5 0| fee —12) are on the parabola, and Q(0;2) is a point on g.

V is the turning rear the parabola. The straight line makes an angle of 45° with the x- axis.

ae

om
S
TQ ;-12)

ot

Vv
6.1 Show that the equation of g is g(x)=—x+2. (2)
6.2. Write down the coordinates of S. (2)
6.3 Show that the equation of f is f(x) =8x° -12x-8. (4)
6.4 Determine the coordinates of V the turning point of f. (3)

6.5 Use the graph to determine the values of k, if f(x) =k has two distng equal roots. (1)

6.6 For which values of x is f’(x).g'(x) <0? ka (1)

6.7 Anew graph A is obtained by first reflecting the graph of f about the y - axis and then

shifting it in the upward direction so that the x- axis is the tangent of the new graph A.
Write down the coordinates of point T’ (the image of T) which lies on the graph of h. (2)

[15]

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QUESTION 7

The diagram showsjthe,graphs of f(x)=log,x and f-. Point A(—1;4) lies on the graph
of the inverse of

7
rf
AGC 14
f
x
Oo B
7.1. Write down the coordinates of A’, the image of A after reflection in the y = x. dd)

7.2. Show that the value of a is : :

(2)
7.3. Determine the equation of f' in the form y =...... ()
7.4 Calculate the length of AB. (3)
7.5. Write down the value(s) of x for which f(x) <-—-l. (1)

[8]

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QUESTION 8

8.1 Minenhile bought a laptop for R15 800. The laptop depreciates at 12% p.a. ona
reducing balance. After how many years will the value of this laptop be R10 767,26? (3)

8.2 Calculate the/éffective yearly interest rate if an investment offers a nominal interest
rate of 7,64%p.a compounded half yearly. (2)

8.3. Mandla is planning to buy a car in 2 years’ time. He invests R500 in a savings
account at the end of each month for 2 years, starting in one month’s time so that
he can use the money as a deposit for his car. Interest is calculated at 5,8% p.a.
compounded monthly.

8.3.1 | How much money will be in his savings account at the end of 2 years? (3)

8.3.2 The cost of the car is R368 400. Mandla uses all the money from his savings
account as the deposit for his new car. He takes out a loan to pay off the
balance after paying the deposit. Interest is calculated at 10,4% p.a.
compounded monthly. Calculate Mandla’s monthly instalments if the car
must be paid for in 6 years. (4)

8.3.3. Calculate the outstanding balance immediately after paying the 56"
instalment. (3)

[15]

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QUESTION 9
9.1 From first principles, determine the derivative of f(x) = 2x" +9. (5)
9.2 Determine llowing:
9.2.1 :
x if y=x(2x+l). (2)
9.2.2 dy. _
zit yrx=xt3 @)
92300 g |: +3x
= :
bi (3)
9.3 The derivative ofa function f at point P is given below:
2+h) +1|-[2°41
ag PN ol Bel
AO h
9.3.1 | Write down the equation of f. (2)
9.3.2. A tangent is drawn at P. Determine the equation of the tangent to f
at point P. (3)
[18]

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QUESTION 10

The diagram below shows the graph of f(x)=ax’ +bx*. The x- coordinates of L and M, are
1 and 2 respectivelys-Phe average gradient of f between L and M is 5,5. The equation of the

tangent to the cui f at x=6 is y=-18x+e.

N
L
e
oO
10.1 Show that a--1 and b=3
2 (5)
10.2 . 1; dee . .
The equation {(x)= “3% +3x° is given. Determine the coordinates of N,
the turning point of f. (4)
10.3. The graph of f is concave up for x<k. Calculate the value of k. (3)
[12]

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QUESTION 11

A match box consists of an outer cover, open at both ends, into which slides a rectangular box
without a top.

e The length of x is one and a half times its breadth.
e The thickness e material is negligible.
e The volume box is 25cm?.

If the breadth of the box 1s x cm:

11.1 Write down the height of the box in terms of x. (2)
11.2 Show that the amount of the material used for the box is given by: (2)
A= 100 +4,5x7 + 20
x 1,5x
11.3 Hence determine the breadth of the box such that the material used is a minimum. (3)
[7]

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QUESTION 12

A group of 40 people was asked which bus company they liked to travel by: Elto, Greybound or
Translucky bus company.

© 3 people likedravelling by all 3 bus companies.
e 7 people liked-toKravel by Translucky and Elto.
e 3 people did fotitke using any of the bus companies.

e The probability of a randomly selected person liking to travel by Greybound and Elto is 5

e The probability of a randomly selected person liking to travel by Greybound and Translucky is z
1

e The probability of a randomly selected person liking to travel by Translucky only is io’

e The probability of a randomly selected person liking to travel by Elto is >

The partially completed Venn diagram drawn below represents the given information.

ELTO GREYBOUND
& 7
TRANSLUCKY
12.1 Use the given information to determine the values of a, b, c, d and ep, U (5)
12.2. How many people liked to travel by Greybound bus company? (1)

12.3. Calculate the probability of a randomly selected person liking to travel by only
one bus company. (2)

[8]

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QUESTION 13

Nonhle who is a Grade 12 learner has 8 textbooks from eight different subjects: Mathematics,
English, Accounting;History, Tourism, Afrikaans, Geography and Drama which she wants
to arrange in a lin¢/oma shelf.

13.1 In how rhanyways can the textbooks be arranged? (1)

13.2 In how many ways can the textbooks be arranged if the Mathematics textbook and
the Accounting textbook must be on each end of the shelf? (3)

13.3 If the Mathematics textbook and the Accounting textbook must be on each end of
the shelf, what is the probability that the History textbook and the Tourism textbook
are not next to each other? (4)

[8]
TOTAL MARKS: 150

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INFORMATION SHEET: MATHEMATICS

INLIGTING BLADSY

_—b+Vb7 4ae

2a
A= Pl +ni) 24 = P(l—ni) A= P(I-i)" A=P(+i)"
T, =a+(n—Nd 8, = 5 (2a +(n-1)d)
T, =ar"" s _ alr" =I 3 r#l 8, =“; -1er<1

" r-l I-r

peslisi"-1 pawlcd+)")

1 7 i

. Ff (x+h)— f(x)

(x) = lim

PC) h—>0 h

xX) +X. +y,
d=\(X)—%)? +0 -94)" wo a2 21 Ye x)

y=mxt+e y-y, =m(x-x,) m= m=tand

(=a) (pa =r

In AABC: 4 = b = £ a* =b* +07 —2be.cos A area AABC = ab.sin C
sn A sn B snC 2

sin(a + B) = sin a.cos 8+ cosa@.sin 3 sin(a —f) = sin @.cos 3 —cosa@.sin B

cos(a +B)= cos a.cos #—sin a.sin & cos(a— #)= cos @.cos 8 +sin a.sin B
cos’ a—sin* a

cos 2a =41-2sin* a sin 2a@ =2sin a@.cosa
2cos* a1

>
n 2

Sfx dens)

n n
P(A) = m8) P(A or B) = P(A) + P(B)LB(A and B)
A
4 (x-x)\y-y)

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KWAZULU-NATAL PROVINCE

EDUCATION
REPUBLIC OF SOUTH AFRICA

NATIONAL
SENIOR CERTIFICATE

MARKS: 150

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Mathe:

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September 2023 Preparatory Examinations

QUESTION 1
1.1.1 x=0 or x=4 AAY ¥ answer (2)
1.1.2 2x? +3x-7=0 A” standard form
CAY substitution
2
x= 3) = 42-7) into formula
2(2)
x=1,27or x =—2,77 CAY 1,27
va (4)
Penalise 1 mark for CAY 2,77
incorrect rounding off.
1.1.3 (x-1)(x-4)>0 AY factors
i |
1 T
x<lorx>4 CAY x<1
CAY x>4
OR OR
(x-1)(x-4)>0 AY factors
. -
1
x<lorx>4 CAY x
CAY x: G3)

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September 2023 Preparatory Examinations

1.1.4 (3"-3)(3°-7)=0

3°=3 or x =7
ih log, 7 =x

1 1,77=x

*
I

OR

Let 3° =m 2. 3° =m?
m -10m+21=0
(m-3)(m-7) =0

m=3 or m=7

3*=3 or 3° =7
log, 7=x

x=1 or 1,77=x

AY factors

CAY 3* =3 or
3 =7

CAY using logs
CAY x=1
CAY x=1,77

OR

AY m=3 or m=7

CAY 3° =3 or
3°=7

CAY using logs (5)
CAY x=1
CAY x=1,77

1.2 X= 2= Yorn)

(2-y) +? +6(2=9)-4y+4=0
4-4) 4y? ty’ +12—6y—4y+4=0
2y°-14y 1 20=0

y’ -Ty+10=0

(y-2)(v-5)=0

y=2 or y=5

x=2-2=0
x=2-5=-3

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AY equation (3)

CAY substitution

CAY standard form

CAY x values (6)

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1.3 For equal roots A=0
25-n’ =0 AV 25-n’ =0
(5-n) =n)=0 AY factors
n=t5 CAY answer
OR OR

For equal roots A=0

25-n’ =0 AY 25-n’ =0
=n Av 089
=+ y
ns answer only = full marks CAY answer @)

[23]

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QUESTION 2

September 2023 Preparatory Examinations

2.1.1 3; 73 11 ;------ ------ 399.
a=3
d=4
n=20
T, =a+¥("-1)d
Ty =3+(20-1)4 AY substitute into
* correct T,
formula
Ly =79
OR CAY answer
3; 7; WW ;------ ------ 399 OR
4,454; 0,
T, =4n-1
T,, =4n-1
T,, =4(20)-1 AY substitute into
correct T,
Ty =79 formula (2)
CAY answer
2.1.2 T, =399
T, =a+(n-l)d
399 =3+(n—-1)4 AY substitute into
correct T, formula
396 =(n-1)4
99-n-1
100=n CAY answer
OR
OR
T, =399
T, =4n-1
399 =An-1 AY substitute into
correct T, formula
400 =4n
100=n

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September 2023 Preparatory Examinations

CAY answer (2)
2.2.1 T, =a
T,; =a+24
T,, =a 443-1)
a+24 at 12d AY a+24=a+12d
24 = 12d
2=d AVE, (2)
222 | sg =$[2a+(n-1)d]
200 CAY substitute into
Soop =| 2(a)+(200-1)2
an) [ (2) ( ) ] the correct
formula
Sry = 100(2a +398)
Sy = 200a +39 800
CAY answer
OR OR
T, =a+(n-ld
Typ =a + (200—1)(2)
= at+398
2
8, = 5a +1) ”
Sooo — 20044 (a +398)] CAY substitute into
2 the correct
formula
Sy) = 100(2a +398)
= 200a+39 800
CAY answer
[8]

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QUESTION 3

3.1 1311; 27; 49; 77; 111; 151; ...

10; 16; 22; 28; 34; 40; wo.
Ce a a a

Loar
The thés is 27 AY answer (1)
3.2 Secorpiifference is 6

2a=6

az=3 AY a=3
10=3a+b

10=3(3)+b

1=b CAY b=1
1=3+l+e

3=c CAY c=-3
T, =3n? +n-3 CAY answer

OR OR

-; 16; 22; 28; 34; 40; wu.
6;.6; 6; 63.002.

2a=6
a=3
AY a=3
‘T =an’+bntc
49 =3(4) +(4) +e
77 =3(5) +.b(5) +0

49 =484 4D +6 veeeeeseee (1)

TIA TSASDHC vocceceseee (2) ae

QW=2THBD eee Q)-() danny

a cay b=

49 =48+4(l) +c

CAY c=-3
3=c

CAY answer (4)

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Mathe: September 2023 Preparatory Examinations

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T, =3n +n-3
3.3 First difference 10; 16; 22; 28; 34; 40...

T =6n AY T.=6n+4

0 = OF IG
418=
414
69=n CAY value of n
W=n+l CAY between T,,

between T;, and T, and Try
~. between 7,, an

69 70 OR
OR
v = -
418=T,., -T, AY 418=T,,, -T,
418 =3(n+1) +(2+1)—3-Gn? +n—3)
418 =3(n? +2n+l)+n+1-3-3n —n4+3
418 =3n? +6n+3+n—-2-3n’ -—n+3
418 =6n+4
414=6n
69=n CAY value of n
70=n+1 CAY between Ty (3)
“. between 7;, and 7, and T,,
[8]

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September 2023 Preparatory Examinations

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QUESTION 4
4.1.1
_xtl AY answer qd)
x
4.1.2 T,=T,xr
xtl
K=(x+t2 CAY (x41)x——
(x+ 1) CAY answer (2)
T, =
x
4.1.3
(2+1) 3 CAY value of r
r= ==
2 2
“no, sinee r> CAY answer with (2)
OR motivation
No, for the series to converge: — 1 <r<1
4.2 For AP:
7-1, =T,-T,
a-—6=b-a
20 = 6 = Deessssseesseesssseees (0) AY equationiGh)
For GP:
rt
7, 7,
b_16
a b

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September 2023 Preparatory Examinations

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16d =D veeeecrrseee (2) AY equation (2)

CAY standard form

CAY values of a

CAY values of b

b=2(9)-6=12

b=2()-6=-4 (5)

[10]

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Mathemayiss wn i loa d le d fr om St ‘a Ye } S, i G Sa ¢ ‘om September 2023 Preparatory Examinations

QUESTION 5

5.1 x=3 AY answer
ye-l AY answer (2)
=>
5.2 x — int :
0-5 i AY sub y=0
j=
-x
3-x=-3
x=6 CAY x=6
y intercept = ——-1=-2 AY y=-2
OR
OR
_ 3 _
*=@=3)
3
=—-]
y x-3
— intercept:
3 AY sub y=0
0=—--1
x-3
1-3
x-3
x-3=3
CAV x=6
x=6
—intercept = 3 ja v
y P 0-3 AY y=-2 QB)

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Mathempayiss nl loa d le d fr om St ‘a Ye i S, i G Sa ¢ ‘om September 2023 Preparatory Examinations

53 ;
L AY shape
\ CAY intercepts
= 3 en? CAV asymptotes (3)
\
5.4 yr-xt+e
-[=-34e CAY sub (3;-1)
2=c
yeoxt2 CAY answer
OR OR
3
a —3-x -
yey 1 wernt
yeord2 CAY answer
OR OR
yer]
ye—(x—3)-1 AY -(x-3)-1
ys-xt+3-1
yeerd2 CAY answer (2)
[10]

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September 2023 Preparatory Examinations

QUESTION 6
6.1 m= tan135° AY m=tan135°
m=-l AY m=-1 (2)
“y=-xt2
6.2 0=-*+2 AY equating to zero
x=2
“S (2;0) AY answer (2)
Answer only = full marks
6.3

y=a(x—x )(x—x)

ymafx+h)(0-2)

-12=a( 143 (1-2)

y =8x? -12x-8

Accept method of simultaneous equations

AY sub x= and

2

x=2
AY sub (1;-12)

» 3
y=a(? 3-1] (4)

6.4

-b
x=—
2a

_-(-12)
~ 2(8)

x

Alw

AY x value

CAY substitution

CAY y value

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September 2023 Preparatory Examinations

g(x) =16x—-12
0=16x-12
12=16x

OR

x-intercepts S(2;0) and x{-40]

midpoint x =

OR

“ahe"B8wnloaded from Stanmorepliysics,.com
OR

AY 0=16x-12

CAY x value

CAY y value

OR

AY subst into
correct
midpoint
formula

CAY x value

CAY y value

OR

AY completing the
square

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September 2023 Preparatory Examinations

( 3) 25
y=8| x-—| -—
4 2 CAY V x value and
y value
23) @)
4 2
_ v
6.5 k> -25) CAY answer (1)
6.6 CAY answer
x>H
OR OR
(-1)(16x—12)<0
-l6x <-12
- 3 AV answer (1)
4
6.7 { 1 } AY x-—coordinate
T| -L=
2 AY y-—coordinate (2)
[15]

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September 2023 Preparatory Examinations

QUESTION 7
71 (4:-1) AV answer qd)
7.2 y=log,x
-l=1 AY sub(4;-1)
a’
AY a'=4
1
a=—
4
(2)
73 ly
an
4 AY answer qd)
7A A(-L4) and B(1;0) AY B(1;0)
2 2 CAY sub points A
AB= ,(-1-1) +(4-0
(-I-l) +(4-0) and B
AB= V4+16 = 20
CAY answer
AB = 25 =4,47 @)
75 x>4 AY answer qd)
8]

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QUESTION 8
81 | 4=P(1-i)’
12)’ AY sub into correct
10767, 26 ff oof 1-15 | formula
10767, 26 =(0.88)"
15800
CAY correct use of
logs
10767, 26
log cose) 75800 =
2,999998...=n
CAY answer
n=3years G)
8.2 _\m
100 m
r 7,.64Y AY subst into correct
(1 )=(14 et) formula
2
ro 0 ie) |
200 (2)
v
r=7,79% CAY answer
8.3.1
To : z 4 i
R500 R500 R500 R500 7
oon
x|(1+i)"-1 100
pte)

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### Page 31

“athe"BSwnloaded from Stanmoreplivaias,.com

September 2023 Preparatory Examinations

5.8 \4 AY values of i & 7
500 [ 2 | -1 .
Ke 1200 CAY subst into
~ 5,8 correct
1200
ia
afl CAY answer
GB)
8.3.2 | R368 100 - R12 691,25 = R355 708,75 CAY
R355 708,75
x|1—(1+i)"
P= [ ( ) ] AY n=72
i
10,4)” CAY sub into
x] 1—| 1+ correct
355708,75--- ~ "J formula
aoa 10,4
1200
R6661,78 =x CAY answer (4)
833 xf(iesy'-1]
Bal Outstanding = P(1 +i)’ -+——~_+
alance Outstanding = P(1 +i) ; CAY
10.4. 355708, 751 +14)"
6661,78 [1 2 } 1 1200
355708, 75} 1 10.4 " 1200
= + _
, ( ve | 10,4 CAY
6
1200 6661,78 (+104) -1
° 1200
10,4
1200
Balance outstanding = R99 128,46 CAV R99 128,46
oR in
Remaining instalments = 72—56=16 donot
conn
10,4" AvLn=16
6661,78 1-{1+ } .
Bal _ 1200 CAV’ subst into correct
alance © 10,4 formula
1200

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### Page 32

Mathenis5 wnloaded from Sta Ye is sigs,.com September 2023 Preparatory Examinations

= R99128,52

CAY R99128, 52

(3)

[15]

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MathenpSisswnloaded from Sta nenlivsias.com September 2023 Preparatory Examinations

QUESTION 9
Penalise ONCE for incorrect notation, either in 9.1 or 9.2.
9.1 f(x) =2x° +9
+h) +9
+4xh+2h? +9 AY calculating f(x + h)
(x+h)- f(x)
h
2x? +4xh+2h? +9-2x° -9
f (x)= lim h CAY sub into formula
2
f(x) =lim 4xh+2h
oh CAY simplifying
- h(4x+2h)
=lim So”?
£O) io CAY factors
f(x= lim 4x +2h
: CAY answer
f (x) =4x
(5)
9.2.1] y=2x? +x AY product
dh )
Y _ aya] CAY answer (2)
dx
9.2.2 Vytx =xt3
rey 2 AY squaring both
( y+x) =(x+3) sides
pox ax? 46x49 AY correct product
y= x 45x49 —
nl
* =2x45 CAY ans 3)
9.2.3 l -l
a 4x7 + 3x ? AY 4x1 43x 2
dx
= 4x2 3 3 CA CAYY each term (3)
= Ae x

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Mathenis5 wnloaded from Sta repliysiGs..com September 2023 Preparatory Examinations

9.3.1) Swat] AY A Y answer (2)
9.3.2 | AtPx=2..y=241=9

P(2;9) CAY y-—coordinate

m=12 CAY m=12

y=l2x

9=12(2)+e

-l5=c

y=12x-15 CAY answer (3)

[18]
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“atheBSwnloaded from Stanmoreplivaics,.com

September 2023 Preparatory Examinations

QUESTION 10

10.1 | f(x) =ax* + bx?
f() = all +0(IY =atb Ay aust _5
_ 2

FOR)= a2 (2) =8a+4b AY correct subst into

goon 2-7 average gradient
Ave entra ( A ) formula
Ave grad = —
5,5 —TA43D cceveveeseseseevees (1)
S'(x) =3ax? +2bx
S'(6) =3a(6)° +26(6) Av equation (1)
18 = 1080 412D.veesrsescsseeee (2) AY equation (2)
(x42 22528041 2B occccersecseeeenre (3)

AY solving

40 = 80 Ao csosesestssseseen 3)-Q) simultaneously
-1
ea
2

-l
5,5= {F) 3b

2
5,5=-3,5+3b
9=3b (5)
3=b

10.2 3x? AY f'

f (x)= 4 6x Le

2
0= 23x" 46x Cv legiating

; Sid
0 =-3x? +12x Jonny
ry

0=x° -4x
0=x(x-4)

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September 2023 Preparatory Examinations

“atheBSwnloaded from Stanmoreplivaics,.com

x=0 or x=4
xsd CAY x=4 (4)
AtN x=4.. y= F(a) +3(4) =16
N(4;16) CAV y=16
10.3 f(x) =-3x+6 CAY f(x)
-3x+6>0 CAY -3x+6>0
-3x>-6
x<2
k=2 CAY answer (3)
OR
f(x) =-3x+6
0 =-3x+6
x=2
wk=2
OR
Midpt x = —— =2
x=2
=2
[12]

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September 2023 Preparatory Examinations

QUESTION 11

11.1 Volume = Ibh

25=(I,5x)(x)h

AY substitution

AY answer (2)

11.2 Area = 4€55x)h + 3(1,5x)x + 2xh

AY substitution in
to SA formula

0 =-300+27x* —100
0 =27x° - 400

400 = 27x°

[400
x= 3/—
27

x=2,46

woe 2
Area = 4(1, 5x) x + HL, 5x?) + 2% Av substituting h 2)
1,5x° 1,5x
Area = 100 4 542 + 50
x 15x
11.3 1
A= 10027! 44,522 + 20%
1,5
F-10087 +910 av 9
dx
= 9-0
x x

CAY simplification

CAY value of x

(3)

[7]

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September 2023 Preparatory Examinations

“ahe"BSwnloaded from Stanmoreplivaics,.com

QUESTION 12

121 | a=4 AY
b=13 AY
c=5 AY
d=4 CAY
e=6 CAY (5)
12.2 | Greybound bus = 23 people CAY answer qd)
Y addi
23 P(only one bus) = Orae?2 12 3 CAY adding (2)
40 4010 CAY answer
[8]

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### Page 39

Mathe:

"Béwnloaded from Stanmoreniiysics,.com

September 2023 Preparatory Examinations

QUESTION 13
13.1 | 8!ways =40 320 ways AY 8! OR 40 320 dd)
13.2 1x6k1+ 1=1440 ways AY 6!
OR AY 1x61
2x61 Oways Av 1440 (3)
13.3 1-[ SS OE) | 2 AY subtracting from1
1440 1440 3 CAY denominator
OR Av 215!
1440 —(2!5!42!5!) CAY answer
1440
4
2 “
3
[8]
TOTAL MARKS: 150
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