# Mathematics Grade 12 Paper 1 Limpopo Pre June 2024 Memo

> Grade 12 Mathematics Paper 1 memorandum from the Limpopo Pre June Exam 2024, 13 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/mathematics-maths-p1-memo-may-june-2024-limpopo/7314f523-dd00-4b79-a088-0c2cc9236655
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2024
- Exam period: Limpopo Pre June Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 13

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## About this document

Mathematics · Grade 12 · Limpopo Pre June Exam · 2024. Memorandum, 13 pages. Read online or download the PDF.

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## Document text

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### Page 1

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yea LIMPOPO
PROVINCIAL GOVERNMENT

REPUBLIC OF SOUTH ARRICA

DEPARTMENT OF

EDUCATION

VHEMBE WEST DISTRICT

MARKS: 120

This question paper consists of 7 pages including the cover page

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Mathematics P1 2
Marking Guidelines/Nasienriglyne

NOTE:
. If a candidate answers a question TWICE, only mark the FIRST attempt.
. Consistent Accuracy applies in all aspects of the marking memorandum.

LET WEL:
. Indien ’n kandidaat ’n vraag TWEE keer beantwoord, merk slegs die EERSTE poging.
. Volgehoue akkuraatheid is DEURGAANS op ALLE aspekte van die memorandum van

toepassing.
QUESTION/VRAAG 1
1.1.1 | x2-5x-6=0
(x — 6)(x +1) =0
x=6 or x=-l v factors
Y both answers
(2)
OR/OF OR/OF
xo Sx 6=0 Y correct subst into correct
5 — | ul
x= St yy -4MC9) formula
2(/)
ya 5t V49
2
x=6 or x=-l Y both answers
(2)
1.12 | Gx-1)(r-4)=16 y /
3x2 —13x 12-0 standard form
x= 14) C13) 4@)C12) Y correct subst into correct
2(3) formula
xe 13313
6
x =5,12 or x=-0,78
v Vv answers
(4)
OR/OF OR/OF
3x? -13x-12=0 Y standard form
> 13
x*—-__x=4
3, 2 3 2
eo yy _13) -a{-2] v adding (- 2) both
3 6 J 6 } 6
13 2 313 sides
(.-2) 28
6) 36
_ 13+V313
6
x=5,12 or x=-0,78 V ¥ answers (4)

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Mathematics P1 3
Marking Guidelines/Nasienriglyne
11.3 | x(4-x)20 Y factorisation
-x(x-4)20 or -x(x-4)>0
x(x - 4) <0
0 4
0 4
O<x<4 or xe [0:4] vv OSxS4
(3)
1.1.4 52x —]
Sé+1
v .
(+=) _ 4 factors in numerator
5* +1
5*-1=4 VY 5*-1=4
5* =
x=l Y answer
(3)
OR/OF OR/OF
521
S*4+l
5 —1=4.5* +44
5 —4.5*-5=0 Y standard form
5*—5 )(5*+1)=0
( ¢ ) Y¥ factors
S*=5 or S*#-1
x- v answer
(3)
D2 | 2A BY ccccccccccsvvssvsvssssssvsvvssssvsssvsssssssee () x=2—3y
x? AY 5 HO cece eecteceeteceeee (2)
Substitute (1) in (2):
2 =
(2—3y)y +4y(2—3y)-5=0 Y correct subst into correct
4-12y+9y? +8y-12y? -5=0 formula
—3y? -4y-1=0
3y? +4y +1=0 v either standard form
By+1l(y+1)=0
1
=—~ or y=-l
y 3 y v y—values
x=3 orx=5 v¥ x—values
(5)
OR/OF OR/OF

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Mathematics P1

Marking Guidelines/Nasienriglyne

4

eC.
x2+ax}?—*|_5=0
F
3x? + 8x—4x7 -15=0
—x? +8x-15=0
x?-8x+15=0
(«—5)(x-3)=0
x=3 orx=5

Substitup (1) a (2):

1
y=- or y=-1
3

Y correct subst into correct
formula

Y either standard form

v x—values

v y—values (5)

1.3

ab =2V10
be =3V2
ac = 6V5
ab.bc.ac = 2V10.6V5.3v2
(abc) =36V100
abc = 1360 = 6V10

OR/OF
ac =6,/5 anew
be =3,/2 ob = 37
ab = 2/10

(85 3.2) 9
Le oe )
18V10 =2/10.c?
e=9

c=3

Volume = abc = 2V10.3 = V360 = 6V10

0 volume = abc

| ab.be.ac=2 40.6530
(abe)? =36100

answer
65)
OR/OF

a= oP
bai

Cc

value of c
0 Volume = abc
answer (5)
[22]

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Mathematics P1 5

Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 2
2.1.1 | 59 answer
e)
2.1.2 15 29 41 51
\ f \ sf \ fF
14 12 10
\ Ss 0\
2 2
2a=-2 ¥ second difference of — 2
a= -l Ya
3(-1)+b=14
b=17 Vb
(-1)+(17)+e=15
c=-l Ye
T, =n? +17n-1 (4)
2.13 | 7 =-(27)'+17(27)-1 substitution
27 answer 0)
221 | -18 1
"36.22 v answer
(1)
2.2.2 nl nl
1 (1)
T, =36| —— T, = 36| — —|
2 2)
9 1 n-1
—_ =36{ ——
4096 2)
1 ( a. 1 yy
16384 \ 2) 16384 (2)
aaa
L2J 4 2)
14=n-l answer
n=15 (3)
OR/OF OR/OF
36;-18:9; 228s
2 4 8 4096
If you look only at the denominator: 2; 4; 8; ... ; 4096
2* = 4096 2* = 4096
pk — 912
k=12 k=12
“.n=15 terms answer
(3)

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Mathematics P1
Marking Guidelines/Nasienriglyne
2.2.3 S = a
2 Jop .
36 Y correct subst into correct
= 1 formula with -—1<r<1
1-/-—
2
=24 v answer if—1<r<1
(2)
2.2.4 250
-1s\(1)> -1 1
q v r=_ andn=250
So50even = 1 4
~-]
4
=-24 v S50 even = —24
250
36f(1Y" J
4
so aN J
250 odd 1 1
4 v Sys0.0dd = 48
=48
Soa _ 48
Seven — 24
=2 v answer
(4)
OR/OF OR/OF
T, +7; +T; +T, +...+ Tago
T, +1, +15 +71, +...+T 500
_ atar? tart +...+ar%8 Y atar2 +ar4+..+ar8
art+ar> +ar>+...+ar*® Vart+ar +ar+...+ar
_ atar? +ar*+...+.ar%8
(atar?+ar*+..+ar%) “4
I a+ar? +ar4+..+ar®*)
oF
=2
v answer (4)
[17]

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Mathematics P1 7

Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 3
3.1.1 | p+6-(2p+3)=p-2-(p+6) Y equating i.t.o p
—p+3=-8 v simplifying
p=ll (2)
3.12 | 7, =25+(n-1)(-8) =33-8n Y subst into 7, formula
33-8n <-55
—8n <-88
nol von>il
“. Term 12 will be the first term smaller than —55 vn=12 3
«. Term 12 sal die eerste term kleiner as — 55 wees. @)
320 |g =" ta+n= (x3) +(x-18)]
6 2 2
= 6x - 63 ¥ 6x—63
s =" a+r = [e—-3)+@-2)
5 2 2
= 9x -135 v 9x -135
6x — 63 = 9x -135
3x = 72 y
mas is
“ S = atl = > [(x—-3)+(x—45)] V7 [en 3) +r 45)]
2 2
= oy — 4g]
2
=b 2(24) -48]= 0 = RHS ¥ substitution of x
[2(24) — 48}
2 (5)
OR/OF OR/OF
9
Y'-3k)=0
a Y expansion
(x-21)+(x—24)+ (x-27)=0 V 3x_72=0
“.3x-72=0 ~
3x = 72
x=24 v 24
15
Y'(24-3k)
k=l
= 21418415 4.....+-21. Y substitution of x
Ss ="[a+)]
"2
15 v sum of 15 terms
=—}21-21
at ] (S)
=0=RHS
OR/OF OR/OF

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Mathematics P1 8
Marking Guidelines/Nasienriglyne

(x —3)+(x- 6) + (x—9) + («-12) + («—-15) + (-18) Vv expansion
=(x—3)+(x-6)+(x—9) + (x—12) + (x -15) + (x -18)
+(x—21)+(x—-24)+(x-27)
v —72=
-3x-72=0 3x—72=0
3x =72 V
x=24 24
15
(24 - 3k)
k=1
= 21418415 +.....4-21. Y substitution of x
S,="[a+]]
2
15
=—|21-21
2 [ ] ¥ sum of 15 terms (5)
=0=RHS [10]
QUESTION/VRAAG 4
41 y>0 v answer
qd)
OR/OF OR/OF
ye(0; 0) v answer
Q)

42 6
g:y=| =
2
1. fl y
ere(3)

y=log i or y= —log | x or ys log, | ¥ equation (2)
2 x
43 Yes. The vertical line test cuts g? once v yes
Ja. Die vertikale lyn toets sny g' slegs eenkeer. Y valid reason (2)
OR/OF OR/OF
Yes. For every x-value there is a unique y-value v yes
Ja. Vir elke x-waarde is daar ‘n unieke y-waarde Y valid reason
(2)
OR/OF OR/OF
Yes. g is a one-to-one function / Ja. g is ‘n een-tot-een funksie | ¥ yes
Y valid reason
(2)
OR/OF OR/OF
Yes. The horizontal line cuts g only once v yes
Ja. Die horisontale lyn sny g slegs een keer Y valid reason
2)

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Mathematics P1 9
Marking Guidelines/Nasienriglyne
441 y =-logo x
2=-log: a ¥ correct subst into correct
a1 ay 1 formula (a ; 2)
=2 = = =
“ 4 “4 Lo 4 v answer (2)
442° |ulo:'l or mM '(2;a) Y answer
a) ()
45 M/l_4-71 v4
a 9
ry Vv2
4
(3)
[1]
QUESTION/VRAAG 5
5.1.1 x=-2 v answer
= v answer
y=3
(2)
5.1.2 ( 0: 7)
[ 3 5 v answer
Gr Q)
5.1.3 43=0
x+2 vy=0
143(x+2)=0
3x=—-7
xa
~ 3 v answer
x-intercept (_ 1 ; 0) Q)
Ls)
3.14
4 y
|
v asymptotes at
y=3and
=-2
3,5 *
a _. Y intercepts at
a Bp y=3,5 and
x=-2,3
¥ shape
x (reasonable
—2,3\ -2 0) representation in
correct quadrants)
| (3)
,

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Mathematics P1 10

Marking Guidelines/Nasienriglyne

521 | —2x+4=0 vy=0
2x=4
x=2 vx=2
“.S(2; 0)
(2)
$.2.2 Equation of k:
y=a(x+1) +18 Y y=a(x+l) 418
O=a(241V +18 = or O=a(-4+1) +18 7 substitute (2,0) or
9a=-18 €4;9)
a=2 Ya
y=-2(x +1) 418 GB)
5.2.3 — 2x? — 4x +16 =-2x+4 Y equating
—2x? -2x+12=0
x? +x-6=0 Y standard form
(: * 3)(« 7 2) =o 3 =) v choosing x =-3
¥= >? OF X= v answer
y=-2(-3)+4=10
T(-3 ; 10) (4)
5.2.4 x<-3 or x>2 vv answer
(2)
OR/OF OR/OF
vv
(0; -3)U(2; w) answer
(2)
5.2.5(a) | x<-l ¥| answer
(2)
OR/OF OR/OF
(-o ; -1) v| answer
(2)
5.2.5(b)

0 shape of cubic with
local min tp moving to
local max tp

turning points at x=2

and x =-4
O point of inflection at
x=-l
(3)
[24]

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Mathematics P1

Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 6

61 | f@=x +2

fe th) =(x+hy +2
=x 4+2xh+h? +2

= 2xh+h*

ho

=lm
h0 h

hoo h

hod

=2x

OR/OF

f (x) = lim Let = £0)

=li

foe+h)— f(x) = + 2xh + #2 4+2—(2 +2)
£1 () = tim LD = £0)

2xh +h?

=lim h(2x +h)
= lim(2x + A) Vv

x eben th b2-(x, +2)
m2 2 2

hoo h
li 2xh + h?
=limst oo"
h0 h
= fim Hx +4)
h>0 h
= lim(2x + h)
hv
=2x

V x? +2xh+h? +2

2xh + I?
v lim
hoo h

v lim 222 +4)

h0 h

v answer

6)
OR/OF

Y x 42xh+h? +2

2
v im eth

hoo h

v lim 222+)

h>0 h

v answer (5)

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Mathematics P1
Marking Guidelines/Nasienriglyne
6.2.1 y=4x5 + 2x7! VY 42x71
dy _ 2 _ 94-2
e =12x 2x VY 1x?
v —2x7
(3)
6.2.2 y=4¥x + 333)?
1
= axe + 9x8 V 4x3 v 9x8
dw 42 4 5
nis Ste vot v 54x
(4)
6.3 Point of contact: (1; 5)
m=2 ¥ m=2
y-y,=m(x-x,) or y=2x+e Y substitution of (1 ; 5)
y—S5=2(x-1) S=2+e¢
c=3
y=2x+3 y=2x+3 Y answer
(3)
[14]
QUESTION/VRAAG 7
7A 3
Ix) =-2+ )r—1)Gr+3) vv¥—2¢ +e —De +3)
Y correct simplification
A(x) = -(2x +3)(2 + 2x-3 P
()=-2r+9(e +2x-3) 6)
h(x) = -2x3 —7x? +9
OR/OF
OR/OF
A(x) = —(2x + 3)(x -1)(x + 3) VV —(2x + 3)(x-1)(x + 3)
Y correct simplification
= 2
W(x) =-(0x+3)(2 +2x-3) 6)
h(x) = -2x3 — 7x? +9
7.2 h'(x) =—6x? -14x y
2 _ first derivative
—6x* -14x =0 v=0
—2x(3x+7)=0
x=0 or x=—~
3 Y both answers
(3)
73 x<—l or x>0 vv answer
3 (2)
OR/OF OR/OF
ences T Yu(0s)
— > vv answer
( 3 } (2)
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Mathematics P1 13

Marking Guidelines/Nasienriglyne

TA y=4x+7 v y=4x+7
— 6x? -14x = 4 v W(x)=4
0 = 6x? +14x+4
Q=3x2 + 7x42 0 standard form
0=(3x+1)(x +2)
x=-_ or x=-2 Y both answers
3 (5)
[12]
QUESTION/VRAAG 9
8.1 Volume of Sphere
-4 a(8)° = ABT = 214,66
3 3 v answer
e)
8.2 rt? =8? (Pythagoras) Y substitution or reason
r?=64—x? Pythagoras
()
83 Veone = 1 ah
3
1 (6a )(g ) Vh=8+x
=" 64-x J(8+x
3 v 1 (64-22 )(8+x)
z 2 3) 3
= 3 512 +64x—8x° —x Y expansion
av 040 lon, 32 yi _O4n lor 3a,
ax 3 3 3 a 3 3 3
0 = 64 -16x — 3x7
0=(8-3x)(x+8)
x= 8 x#-8 Vx= 8
3 3
1 of 512 32)
Veone 3 L. he ) v volume of the cone
Venere 2048
8 3
“3 Vv > oop
7 7
(7)
[9]

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