# Mathematics Grade 12 Mpumalanga March Test 2025 Memo

> Grade 12 Mathematics memorandum from the Mpumalanga March Test 2025, 6 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/mathematics-maths-grade-12-nsc-memo-march-2025-mpumalanga/a515e02a-e730-4551-adb1-a1472fe4f9bc
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2025
- Exam period: Mpumalanga March Test
- Curriculum: CAPS
- Pages: 6

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## About this document

Mathematics · Grade 12 · Mpumalanga March Test · 2025. Memorandum, 6 pages. Read online or download the PDF.

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

## Document text

Text extracted from the PDF. Layout, diagrams and tables may not survive extraction.

### Page 1

education

MPUMALANGA PROVINCE
REPUBLIC OF SOUTH AFRICA

NATIONAL SENIOR CERTIFICATE

MATHEMATICS
TERM 1
MARCH 2025

MEMO

MARKS: 58

TIME: 70 minutes

This marking guidelines consists of 6 pages.

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### Page 2

QUESTION 1

17 March 2025

Yele'ly Y substitution
BES 222 Rivce ¥ answer (2)
a=8)d =7:n =36
T,=at(n-ld
Tg =8+(36-1)(7)
Ti, =953
1.1.2. ¥ substitution
a=8;d =7:n =36 Y answer (2)
36
Sy = = (28) + (36-1)(7)]
Sig = 4698
£13: ¥ value of T72
T, =at(n-ld ¥ substitution in T72.m
Ty =8+(72~1(7) = 505 ee (3)
1 5_,=84+(72—1 -1\(7)
15 + Tyg = 786
505484 497 —7in = 786
—7Fm = ~224
m=32
12.1 | r=x=-2 v answer (1)
1.2.2. |-1<x-2<1 ¥ substitution
1<x<3 ¥ answer (2)
[10]

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### Page 3

17 March 2025

QUESTION 2
2.1.1 Y value of a
16 0 -12 29 ¥ value of b
LMA4 Y value of ¢
¥ answer
-16 -12 -8
Waa °
4 4
2a=4 -16=302)+6 16=2-22+¢
a=2 b= -22
T, =2n* -22n + 36
21.2 ¥ substitution
32; 28; 24; 20; .... v answer
a=32 (2)
d=-4
T, =a+(n-l)d
T, =32+(n-I)(4)
T, =-4n+36
2.1.3 | 2n? -22n +36 =—4n +36 ¥ equating
Y factors
2 -
2n° —18n=0 Yn=9
n-9=0 ¥ answer (4)
n(n—9)=
n=0 or n=9
-.T,, and T,, are the terms
22 | =-54, 7, =162, T, =—486 Y series
a=52,r=-3,n=7. ¥ substitution
v¥n=7
7 v answer
s, -2%4 = I ta)
=-29538
[14]

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### Page 4

QUESTION 3

17 March 2025

341.1 | 3-9) Vx=1
Vy=-9 (2)
3.1.2 | x=l ¥ answer (1)
3.1.3 | ye[-9; 0) ¥ answer ()
3.1.4 | (0;-8) v VY answer (2)
3.2.1 | The graph shifted 9 units upwards and then ¥9 units upwards
shifted 1 unit to the left v1 unit to the left (2)
3.2.2 |x>0orx<0 v answer (1)
3.2.3 | x=y? Y interchange x and y
¥ equation
y= tvx. x20 Y¥ restriction (3)
[12]
QUESTION 4
4.1 27 _ 3 ¥ substitution
reel ¥ cube root
3/— =a
8
3
a
2
4.2
h:
¥ shape
Y any point
v x-intercept (3)
43 3Y ¥ swapping x and y
x= (3) Y answer (2)

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### Page 5

17 March 2025

g(x) = log, x
4.4 ve(o . 33) v answer (1)
8
[8]
QUESTION 5
Bish
¥ correct diagram
¥x=V1l-m’
¥ answer
(3)
l-m?
cos36°=V1—m?
OR
cos” 36°+sin* 36°=1 g ai
2260 _1_ cin2 260 correct diagram
oa Ae = lait 26 Fi ¥ substitution
=l-m ¥ answer (3)
cos36°=V1—m?
5.2 —2sin 15° .cos 15° ¥ —sin 30°
cos(45° — x) cos x — sin(45° — x)sin x ¥ compound angle
: ¥ compound angle
_ —sin 30 1
[cos xcos 45° + sin x cos 45°] cos x —[sin 45° cos x — sin x cos 45%| sin;x
1
aust 7 |
¥ cos 45° = sin 45 = —
"| I 1 ; 1 I v2
cosx.—=+ sin x.—= |cosx—| —=cosx-—sin x.—= |sin x ¥ common factor
| 2 v2 E V2 (6)
lV
= 2
A iggst ee coe) gs eorex
v2 2 v2 v2

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### Page 6

17 March 2025

_ —sin 30
cos[(45° — x) +x]
1

¥ —sin 30°
¥ ¥ compound angle

-= 1
ao 2 v=—
cos45° Y cos45°
_l 1
<r v2
v2 (6)
_-v2
Z
5.3.1 AC? = (12)? + (20)? — 2(12)(20) cos 110° ¥ substitution
AC* =708,17 answer &
“AC = 26,61
5.3.2 | sin DAC _ sin ADC ¥ ar ratio
= substitution
' io . wi ¥ answer (3)
sin DAC _ sin 71°
7 26,61
sin DAC = 0,2487
DAC =14,4°

[14]

TOTAL MARKS: 58

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