Mathematics N1 — Memo

August 2021 — N1 — Mathematics

N1 2021 Nated Engineering

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🎯 100 marks 📄 9 pages 📝 31 items found ∑ 8 math symbols 📚 1 topics

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Q1— Section header —

1.21 Instruksies: − 5 log 5 25 • Ken punte toe soos aangedui −2 = − 5 log 5 5  • (1 punt vir antwoord alleen, want = ( − 5)( − 2) log 5 5  vraag sê sonder ŉ sakrekenaar) = 10(1) 

1.3log 5 ( 5 𝑥𝑥 − 15 ) = 2 Instruksies: 2 (5 𝑥𝑥 − 15) = 5  • Ken punte toe soos aangedui 5 𝑥𝑥 = 25 + 15  • 1 punt vir oorskrywing in 5 𝑥𝑥 = 40 eksponentvorm 5 𝑥𝑥 40 • 1 punt vir skuiwing van 15 =  5 5 • 1 punt vir deling deur 5

Q2— Section header —

2.33 ( 𝑥𝑥 + 1 ) ( 𝑥𝑥 − 2 ) Instruksies: = (3 𝑥𝑥 + 3)( 𝑥𝑥 − 2) • 3 punte vir die finale antwoord 2 = 3 𝑥𝑥 − 6 𝑥𝑥 + 3 𝑥𝑥 − 6 2 = 3 𝑥𝑥  − 3 𝑥𝑥  − 6 

Q3— Section header —

3.13.1.1 7 𝑎𝑎 + 14 − 2 𝑏𝑏 − 𝑎𝑎 𝑏𝑏 Instruksies: = (7 𝑎𝑎 + 14) + ( − 2 𝑏𝑏 − 𝑎𝑎 𝑏𝑏 )  • Ken punte toe soos aangedui = 7( 𝑎𝑎 + 2) − 𝑏𝑏 (2 + 𝑎𝑎 ) 

3.23 1 5 Instruksies: + − 2 2 4 𝑥𝑥 𝑥𝑥 8 𝑥𝑥 • 1 punt vir noemer/KGV 8 𝑥𝑥 3(2) + 1(8 𝑥𝑥 ) − 5( 𝑥𝑥 ) • 1 punt vir die teller 6 = 2 8 𝑥𝑥 • 1 punt vir die teller 8 𝑥𝑥

Q4— Section header —

4.112 ( 𝑦𝑦 − 3 ) − 3 ( 𝑦𝑦 − 9 ) = − 5 ( 𝑦𝑦 + 6 ) Instruksies: 12 𝑦𝑦 − 36  − 3 𝑦𝑦 + 27  = − 5 𝑦𝑦 − 30  • Ken punte toe soos aangedui 12 𝑦𝑦 − 3 𝑦𝑦 + 5 𝑦𝑦 = 36 − 27 − 30

4.3𝑦𝑦 = √𝑐𝑐 − 𝑎𝑎 𝑥𝑥 Instruksies: • 1 punt vir substitusie 𝑦𝑦 = � 33 − ( − 2)33  • 1 punt vir die antwoord 𝑦𝑦 = 9,950  • (1 punt vir antwoord alleen) (2)2

4.5Laat die getal 𝑥𝑥 wees. Instruksies: 15 𝑥𝑥 − 3000 = 0  • 1 punt vir vergelyking 15 𝑥𝑥 = 3000  • 1 punt vir skuiwing van 3000 3000 • 1 punt vir deling deur 15 15 𝑥𝑥 =  15 • 1 punt vir die antwoord

Q5— Section header —

5.1.23 Instruksies: 𝑚𝑚 =  2 • Moenie penaliseer as antwoord as 𝑚𝑚 = 1,5 geskryf is nie (1)1

5.1.3𝑐𝑐 = 3  Alternatiewe antwoord: 𝑐𝑐 = ( 0 ; 3 )  (1) Kopiereg voorbehou Blaai om asseblief NASIENRIGLYN -7- WISKUNDE N1

5.25.2.1 𝑥𝑥 𝑦𝑦 = − 3 Instruksies: 3 • 1 punt vir antwoord alleen 𝑦𝑦 = −  𝑥𝑥 (1) de de 5.2.2 2  en 4  kwadrant (2)2

5.2.3Hiperbool  (1)1

5.3 Instruksies: 𝑦𝑦 Tabel:   𝑥𝑥 − 𝟑𝟑 − 2 − 1 0 1 𝟐𝟐 𝑦𝑦 𝟏𝟏𝟏𝟏 8 6 4 2 𝟏𝟏 • 1 punt vir die korrekte beginwaardes op die tabel ( − 3; 10) • 1 punt vir die korrekte eindwaardes op

Q6— Section header —

6.1• Sy, sy, sy (S, S, S) Instruksies: • Sy, ingeslote hoek, sy (S, ∠ , S) • 2 punte vir enige TWEE • Hoek, syhoek of hoek, hoek, sy ( ∠ , ∠ , S ) antwoorde • Regte hoek, skuinssy, sy (90°, hypotenuse, S )

6.22 2 2  AC = AB − BC Instruksies: 2 2 2 • 1 punt vir stelling AC = (60) − (20)  • 1 punt vir substitusie 2 2 AC = � (60) − (20) • 1 punt vir die antwoord AC = 56,569 cm  • (2 punte vir antwoord

6.3Twee sye is ewe lank.  Instruksies: Twee hoeke is ewe groot.  • Ken punte toe soos aangedui (2) [7]7

Q7— Section header —

7.16 Instruksies: cos 3°  =  𝑥𝑥 • 1 punt vir cos 3° 𝑥𝑥 cos 3° = 6 6 • 1 punt vir 𝑥𝑥 6 𝑥𝑥 =  6 • 1 punt vir cos 3° cos 3° 𝑥𝑥 = 6,008 m  • 1 punt vir die finale antwoord

7.2cos 30° Instruksies: sin 60° • 2 punte vir korrekte breuke = cos 30° ÷ sin 60° • 1 punt vir die finale antwoord √3 √3 • (1 punt vir antwoord alleen) =  ÷  2 2 = 1  (3) [7]7

Q8— Section header —

8.1A = b × ⊥ h Instruksies: A = 2,5 × 1,8  • 1 punt vir substituering van waardes 2 A = 4,5 m  • 1 punt vir die finale antwoord • (2 punte vir antwoord alleen) • Moenie penaliseer as die eenheid nie

8.28.2.1 P = S + S + S Instruksies: P = 4 + 20,616 + 21  • 1 punt vir substituering van waardes P = 45,616 m  • 1 punt vir die finale antwoord • (2 punte vir antwoord alleen) • Moenie penaliseer as die eenheid nie

8.2.2A = 𝑙𝑙 × b Instruksies: A = 12 × 21  • 1 punt vir substituering van waardes 2 A = 252 m  • 1 punt vir die finale antwoord • (2 punte vir antwoord alleen) • Moenie penaliseer as die eenheid nie

8.2.31 Instruksies: V = b × h × 𝑙𝑙 2 • 1 punt vir substituering van waardes 1 • 1 punt vir die finale antwoord V = (20,616)(4)(12)  2 • (2 punte vir antwoord alleen) 3 V = 494,784 m  • Moenie penaliseer as die eenheid nie

8.3Die hoeveelheid geld bespaar: Instruksies: = R 12 588 − R 10 999 • 1 punt vir R 1 589 = R 1 589  • 1 punt vir berekening • 1 punt vir persentasie Die persentasie geld bespaar: (moenie punt toeken as die %-simbool

Mathematical Content

°√÷×∠⊥∴trig

Topics Identified

Trigonometry

Mark Allocation

Total100 marks

Sections15 + 13 + 18 + 13 + 7 + 7 + 11 = 84 marks

Per-question sum100 marks

Paper information

TitleMemo

SubjectMathematics

LevelN1

Document typeMemorandum

Year2021

MonthAugust

Pages9

Downloads3

Examination context

Decade2020s

Years ago5y

SeasonWinter (SH)

Start daySunday

Exam periodAugust 2021

Days since exam1,898

About Mathematics examination preparation

TVET Mathematics at N-level combines algebra, trigonometry, geometry, and calculus. The examination tests both computational accuracy and the ability to apply mathematical methods to engineering and technical contexts.

How to use this resource for revision

A TVET past examination paper is most useful when it is treated as an active revision exercise. Try the questions yourself before checking the corresponding memorandum. Where you make an error, identify the reason — was it a knowledge gap, a calculation mistake, an interpretation issue, or an examination technique problem?

For stronger examination practice, complete the paper without unnecessary interruptions and keep track of the questions that take the most time. Afterwards, review the memorandum and separate knowledge gaps from mistakes caused by interpretation, calculation, structure, or examination technique.

algebraic manipulation

trigonometric problem solving

calculus techniques

geometric reasoning

applying maths to engineering contexts

A practical revision approach

Practice showing full working on every question — the marking memorandum allocates substantial method marks. For N4–N6 levels, focus on calculus and geometry as these carry the heaviest mark weightings.

File details

File size367.9 KB

Pages9

Archived on8 September 2026

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