# Mathematics Grade 11 P2 Gauteng November 2023 Memo (Afr)

> Grade 11 Mathematics Paper 2 memorandum from the Gauteng November Exam 2023 (Afrikaans), 14 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/gr-11-mathematics-p2-afr-2023-possible-answers/2bd82097-d7f5-4872-aad7-426c24065f42
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Document type: Memorandum
- Year: 2023
- Exam period: Gauteng November Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 14

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## About this document

Mathematics · Grade 11 · Gauteng November Exam · 2023. Memorandum, 14 pages. Read online or download the PDF.

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## Document text

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### Page 1

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             PROVINSIALE EKSAMEN
                 NOVEMBER 2023
                     GRAAD 11
                 NASIENRIGLYNE

                  WISKUNDE (VRAESTEL 2)

14 bladsye

### Page 2

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                                                         WISKUNDE
                             NASIENRIGLYNE
                                                         (VRAESTEL 2)           GRAAD 11

INSTRUKSIES EN INLIGTING

➢   A – AKKURAATHEID
➢   CA – KONTINUE AKKURAATHEID
➢   S – BEWERING
➢   R – REDE
➢   S & R – BEWERING met REDE

NOTAS:

•   Indien ʼn kandidaat ʼn vraag TWEE keer beantwoord het, merk slegs die EERSTE
    poging.
•   Indien ʼn kandidaat ʼn antwoord doodgetrek het, en dit nie weer gedoen het nie, merk
    die doodgetrekte antwoord.
•   Deurlopende akkuraatheid moet op ALLE aspekte van die nasienriglyne toegepas
    word.
•   Dit is ONAANVAARBAAR om waardes/antwoorde aan te neem om probleme op te
    los.

                                             2

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                                                                                                      WISKUNDE
                                                              NASIENRIGLYNE
                                                                                                      (VRAESTEL 2)        GRAAD 11

VRAAG 1

1.1   Klas interval                             Frekwensie               Kumulatiewe frekwensie ✓ Frekwensie waardes
      300 < x  400                                    4                                   4              ✓ Kumulatiewe
      400 < x  500                                    11                                 15                frekwensie waardes
      500 < x  600                                    22                                 37
      600 < x  700                                    13                                 50
      700 < x  800                                    7                                  57
      800 < x  900                                    3                                  60                                      (2)

1.2                              70                                                                       ✓ (300 ; 0)
                                                                                                          ✓ Eindpunte
                                 60
                                                                                                          ✓ Vorm

                                 50

        Kumulatiewe Frekwensie
                                 40

                                 30

                                 20

                                 10

                                 0
                                      0   100   200   300    400   500    600   700   800      900 1000
                                                            Daaglikse Verkope
                                                                                                                                  (3)

1.3   Q2 = R580                                                                                            Antwoord
                                                                                                             Aanvaar
                                                                                                             R560 tot R600        (2)

1.4   Op 10 dae maak hul ʼn verlies.                                                                        Antwoord
       50 dae maak hul ʼn wins.                                                                                                   (2)

1.5   Om in die top 25% van die vorige maande te wees, moet ✓ R650
      die snoepwinkel se verkope meer as R650 wees. Daarom
                                                            ✓ Gevolgtrekking
      sal R725 in die top 25% wees.                                                                                               (2)
                                                                                                                                 [11]
                                                                                      3

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                                                     WISKUNDE
                                     NASIENRIGLYNE
                                                     (VRAESTEL 2)            GRAAD 11

VRAAG 2

2.1   2k + k + 1 + k + 2 + k − 3 + 2k − 2             ✓ 7k – 2
                                          = 15
                       5                              ✓ Deel deur 5
      7k − 2
              = 15                                    ✓ Manipulasie
         5
      7k – 2 = 75
      7k = 77
      k = 11                                                                       (3)

2.2   22 ; 12 ; 13 ; 8 ; 20                            Antwoord
       = 5,22                                                                     (2)

2.3   Gemiddeld neem af met t eenhede maar die        ✓ Gemiddeld neem af
      standaardafwyking word nie geaffekteer nie.       met t eenhede
                                                      ✓ Standaardafwyking nie
                                                        geaffekteer nie            (2)
                                                                                   [7]

VRAAG 3

3.1            0 − ( −2)                              ✓ Substitusie in die
      mMQ =
              − 2 − ( −4)                               korrekte formule

                   1
      mMQ = −                                         ✓ Antwoord
                   3                                                               (2)

3.2          1                                        ✓ Loodregte lyne, produk
      mMQ  − = −1            MP ⊥ MQ                   van gradiënte
             3

      mMP = 3                                         ✓ Gradiënt van MP

      y = 3x + 6                                      ✓ Antwoord                   (3)

3.3          xR + 4            y R + ( −2)            ✓ Substitusie in die
      –2 =               0=                             middelpunt formule
               2                    2
                                                      ✓ xR
      x R = −8              yR = 2                    ✓ yR

      R (–8 ; 2)                                                                   (3)

                                                 4

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                                                        WISKUNDE
                                     NASIENRIGLYNE
                                                        (VRAESTEL 2)           GRAAD 11

3.4   PQ =       (0 − 4) 2 + (6 − ( −2)) 2                   ✓ Substitusie in die
                                                               afstandsformule
      PQ =      80
      PQ = 4 5                                               ✓ Vereenvoudigde
                                                               wortelvorm               (2)

3.5   E(–4 ; –6)                                             ✓ x E = −4
                                                             ✓ y E = −6                 (2)

3.5   Die hoeklyne van die parallelogram is loodreg op mekaar. ✓ Hoeklyne loodreg
                                                               ✓ Gegee parallelogram    (2)
                                                                                       [14]

VRAAG 4

4.1   subst. x = 0                                           ✓ x=0
      py – x – 5p = 0                                        ✓ Manipulasie
      py – (0) = 5p                                          ✓ x=5
           5p
      y=
            p
      y=5
      C(0 ; 5)                                                                          (3)

4.2   py – x – 5p = 0                                        ✓ Standaardvorm
      py = x + 5p                                            ✓ Stel gradiënte gelyk
         x
      y=    +5
         p
            1
      mAC =
            2
       1 1
       =
        p 2
      p=2

                                       OF

                                             1
      A(–10 ; 0)                   mAC =                     ✓ A (–10 ; 0)
                                             2
                                                             ✓ Vervang A
      p(0) – (–10) – 5p = 0
      10 = 5p
      p=2                                                                               (2)

                                                 5

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                                                   WISKUNDE
                                NASIENRIGLYNE
                                                   (VRAESTEL 2)        GRAAD 11

4.3   Vergelyking van AC: 2y – x – 10 = 0           ✓ Vervang p = 2
           1
      y=     x+5
           2
                                                    ✓ Vergelyking van BD
      Vergelyking van BD: y = – 2x – 10
      1                                             ✓ Vervang y in die ander
        x + 5 = – 2x – 10
      2                                               vergelyking
      x + 10 = – 4x – 20
      5x = – 30
      x = –6                                        ✓ x = –6
      y=2                                           ✓ y=2
      B(–6 ; 2)                                                                 (5)

4.4             1                                   ✓ Vervang in die korrekte
      tan  =
                2                                     formule
       = 26,57                                    ✓ Antwoord                   (2)

4.5   AĈD = 63,43      binne e van              ✓ AĈD
       = 26,56         binne e van              ✓ 
      =                                          ✓ Rede
       ABOD is ʼn koordevierhoek
      (omgekeerde e in dieselfde sirkelsegment)                                 (3)

4.6   DB is ʼn middellyn (lyn onderspan 90)         ✓ x=0

              0 + 0 5 + ( −10)                            5
      M CD =       ;                              ✓ y=–
              2          2                                2

                   5
      = M CD  0 ; − 
                   2                                                           (2)
                                                                                [17]

                                              6

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                                                                 WISKUNDE
                                        NASIENRIGLYNE
                                                                 (VRAESTEL 2)            GRAAD 11

VRAAG 5

5.1   5.1.1             2                                               ✓ Kwadrant II
              sin  =
                        3                                    y
                                                                        ✓ x=– 5
               2        2           2
              x + (2) = 3
                                                                        ✓ Vervang cos 
              x=– 5
              3cos2  – 1                                           x   ✓ Antwoord
                            2
                  5
              =3     –1
                   3  
                5
              = –1
                3
                2
              =
                3                                                                              (4)

      5.1.2 tan (– –180)                                              ✓ –tan 
            = –tan                                                     ✓ Substitusie
                  2                                                   ✓ Antwoord
              = –   
                 − 5
                 2
              =
                  5                                                                            (3)

5.2   5.2.1 t cos 15 = 4                                                                t
                                                                        ✓ cos 15 =
                                4                                                        4
              cos 15 =
                                t
                                                 t
                                                                        ✓     t 2 − 16
              sin 15                                                   ✓ t
                    t − 16
                    2                      15°
              =                                      4
                      t                                                                        (3)

      5.2.2 sin 75                                                     ✓ Komplement
            = cos 15                                                   ✓ Antwoord
                4
              =
                t                                                                              (2)

                                                         7

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                                                               WISKUNDE
                                   NASIENRIGLYNE
                                                               (VRAESTEL 2)                 GRAAD 11

      5.2.3      1 – tan2 15                                           t 2 − 16 
                                                                                        2

                       t 2 − 16 
                                       2                             ✓           
                                                                           4     
                 =1–                                                           
                             4  
                                
                                                                       t 2 − 16
                       t 2 − 16                                      ✓
                 =1–                                                      16
                          16
                                                                     ✓ Korrekte
                   16 − t 2 + 16
                 =                                                       vermenigvuldiging
                        16
                   32 − t 2                                              met KGN
                 =
                     16                                                                           (3)

5.3
       cos(90 − α ) sin( − − 540 )                                ✓ tan 45
                                                                     ✓ sin .– sin 
      tan 225 + sin . sin(180 + )
                                                                     ✓ sin .sin 
                  sin α.sin α                                        ✓ 1 – sin2 
      =
          tan 45 + (sin .)( − sin α )                                    1
                                                                     ✓
           sin2 α                                                        tan 2 α
      =
        1 − sin2 α
         sin 2 α
      =
         cos 2 α
      = tan2                                                                                     (5)

5.4   5.4.1     1 – cos = 2 sin2                                   ✓ 2 (1 – cos2)
                1 – cos = 2 (1 – cos2)
                1 – cos = 2 – 2 cos2                               ✓ standaardvorm
                2 cos2 – cos  – 1 = 0
                (2 cos + 1)(cos  – 1) = 0                                                       (2)

      5.4.2     2cos + 1 = 0     of       cos – 1 = 0                             1
                                                                     ✓ cos = –
                           1                                                        2
                cos = –          of       cos = 1
                           2                                         ✓ cos = 1
                RA = 60          of        = 0 + K.360 ; K  Z   ✓ Beide oplossings vir
                QII :  = 120 + K.360 ; K  Z                                     1
                                                                         cos = –
                QIII :  = 240 + K.360 ; K  Z
                                                                                    2
                                                                     ✓  = 0 + K.360
                                                                     ✓ KZ                        (5)
                                                                                                 [27]

                                                      8

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                                                      WISKUNDE
                                   NASIENRIGLYNE
                                                      (VRAESTEL 2)           GRAAD 11

VRAAG 6

6.1   180                                              ✓ Antwoord                    (1)

6.2       1     1                                       ✓ Eindpunte
      –     y
          2     2                                       ✓ Notasie                     (2)

      6.3.1    45  x  90                            ✓ x  45
                                                        ✓ x < 90                     (2)

      6.3.2    –90< x < – 45 of 90< x < 135         ✓ –90< x < – 45
                                                        ✓ 90< x < 135               (2)

      6.3.3    2 tan x + cos 2x = 2                     ✓ 2 tan x – 2 = –cos 2x
               2 tan x – 2 = –cos 2x                                      1
                                                        ✓ tan x – 1 = –     cos 2x
                               1                                          2
               tan x – 1 = –     cos 2x
                               2                        ✓ Waarde van x
                f (x) = g(x)
               waar x = 45                                                           (3)

6.4          sin x + cos x                              ✓ tan x + 1
      h(x) =
                 cos x                                  ✓ Op met 2 eenhede
             sin x
      h(x) =       +1
             cos x
      h(x) = tan x + 1
      h(x) = f (x) + 2
       Vertikale translasie op et 2 eenhede                                          (2)

                1                                     ✓ –2 cos 2x
6.5   p(x) = 4  − cos 2 x 
                2                                     ✓ Maks van 2
      p(x) = – 2 cos 2x
       Maksimum waarde van 2                                                         (2)
                                                                                     [14]

                                                  9

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                                              WISKUNDE
                              NASIENRIGLYNE
                                              (VRAESTEL 2)         GRAAD 11

VRAAG 7

                1                                           1
7.1   tan p =                                   ✓ tan p =
                3                                           3
      p = 18,435                               ✓ Waarde van p tot 3
                                                  desimale plekke           (2)

7.2   In BCD                                   ✓ Korrekte substitusie in
           CD         121                         die sinreël
                    =                           ✓ Manipulasie
           18,4  sin135
       sin                                    ✓ Lengte van CD
           2 
           121 x sin 9,2
      CD =
              sin135
      CD = 27,36 m                                                          (3)
                                                                            [5]

VRAAG 8

8.1   V = x2 h                                  ✓ Substitusie in die
      490 = x2h                                   volume formule
          490                                   ✓ h in terme van x
      h= 2
           x                                                                (2)

8.2   A = x2 + 4xh                              ✓ Basis van x2
                   490                              490 
      A = x2 + 4x  2                          ✓ 4x  2 
                   x                                x 
                 1960 3
      A = x2 +       cm
                   x                                                        (2)

8.3          490                                ✓ Vervang h = 10
      10 =
              x2                                ✓ x2 = 49
      x2 = 49                                   ✓ x=7
      x=7          (x > 0)                      ✓ Buite opp
                 1 960
      A = (7)2 +
                    7
                 2
      A = 329 cm                                                            (4)
                                                                            [8]

                                       10

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                                                                     WISKUNDE
                                         NASIENRIGLYNE
                                                                     (VRAESTEL 2)         GRAAD 11

VRAAG 9

9.1
              Q

                           1
                           2         O
                                               1
                                               2
                                                      P
       R

      Verbind PO en verleng                                               ✓ Konstruksie
      In ΔPOQ
                                                                          ✓ SR
      Ô1 = Q̂ + P̂1                       buite  van 
                                                                          ✓ SR
      maar Q̂ = P̂1                        e teenoor gelyke radiusse
                                                                          ✓ S
       Ô1 = 2P̂1
      Netso in  ROQ
      Ô 2 = 2P̂2                                                         ✓ S
       Ô1 + Ô 2 = 2P̂1 + 2P̂2                                          ✓ S
                       (
      Ô1 + Ô 2 = 2 P̂1 + P̂2   )
       RÔQ = 2P̂                                                                              (6)

9.2   9.2.1
      (a)       D̂ = 55°                   e in dieselfde segment        ✓ S R                (2)

      (b)       Ô 2 = 110°               middelpunt  = 2  omtreks     ✓ S R                (2)

      (c)       Ê 2 = 110° – 90          buite  van                   ✓ R
                                                                          ✓ S
                Ê 2 = 20°                                                                      (2)

                                                      11

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                                                                WISKUNDE
                                     NASIENRIGLYNE
                                                                (VRAESTEL 2)        GRAAD 11

       (d)        Ê1 = 35°              buite e van                 ✓ R
                                                                       ✓ S   ✓R
                  Ê 3 = 35°             e in ʼn halwe sirkel
                  OF
                  Ê 3 = F̂1             e teenoor gelyke radiusse    ✓ S R
                                                                       ✓ SR
                  Ê 3 = 35°             binne e van                                    (3)

       9.2.2      OE = 5                  gelyke radiusse              ✓ SR
                  HE = 4,7                loodlyn van middelpunt van   ✓ S ✓R
                                          koord
                  OH2 = 52 – (4,7)2       Pythagoras                   ✓ S
                  OH = 1,71 eenhede
                  OF
                  OE = 5            radiusse                           ✓ SR
                               OH                                      ✓✓ sin 20°
                  sin 20 =
                                5                                      ✓ OH
                  OH = 1,71 eenhede
                  OF
                  HE = 4,7          loodlyn van middelpunt na          ✓ S R
                                    koord                              ✓ tan 20°
                               OH                                      ✓ OH
                  sin 20 =
                               4,7
                  OH = 1,71 eenhede                                                       (4)
                                                                                         [19]

VRAAG 10

10.1   B̂1 = x                       verw. e, AC ‖‖ TV                ✓ S   ✓R

       V̂1 = x                       e teenoor gelyke sye             ✓ S   ✓R

       T̂2 = x                       rklyn koord stelling              ✓ S   ✓R           (6)

10.2   Â = 2 x                      teenoorst e van parm             ✓ S   ✓R
       B̂3 = 2 x                     buite  van kvh                   ✓ S   ✓R
        Â = B3                     albei gelyk aan 2x                ✓ R
       AT = BT                       sye teenoor gelyke e

                                                     12

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                                                         WISKUNDE
                              NASIENRIGLYNE
                                                         (VRAESTEL 2)            GRAAD 11

       OF
       B̂3 = 2 x             buite  van kvh                 ✓ S ✓R
                                                             ✓ SR
       T̂1 = 2 x             verw. e, AC ‖‖ TV
        T̂1 = V̂1 + V̂2   albei gelyk aan 2x
                                                             ✓ S    ✓R
        BTVC is ʼn gelykbenige trapezium
       (een paar ewewydige sye en een paar basis hoeke       ✓ Logika
       gelyk)
        CV = BT            sye van gelykbenige trap
       en CV = AT           teenoorst. sye parm
        AT = BT                                                                             (5)

10.3   In  ATB,                                             ✓ R
       5x = 180°             binne e van                   ✓ S
       x = 36°                                                                               (2)
                                                                                            [13]

VRAAG 11

11.1   D̂3 = Ĥ1 + Ê1       buite  van                    ✓ S
                                                                                             (1)

11.2   D̂3 = Ĥ1 + Ê1       buite  van 
       Ĥ1 = F̂2             rklyn koord stelling            ✓ S ✓R
                                                             ✓ SR
       en Ê1 = F̂1          rklyn koord stelling
        D̂ 3 = F̂1 + F̂2
        DEFH is ʼn kvh (buite  = teenoorst binne  )                   (
                                                             ✓ R (met D̂3 = F̂1 + F̂2   )    (4)

11.3   Ĥ1 = F̂2             rklyn koord stelling            ✓ SR
                                                             ✓ S ✓R
       Ĥ1 = F̂1             e in dieselfde sirkelsegment
        F̂1 = F̂2                                                                           (3)

                                              13

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                                                               WISKUNDE
                                   NASIENRIGLYNE
                                                               (VRAESTEL 2)           GRAAD 11

11.4   Ĥ 3 + Ĥ 4 = Ê1 + Ê 2    buite  van kvh                 ✓ S    ✓R

       Ĥ 4 = K̂ 1                 rklyn koord stelling            ✓ SR
       Ĥ 3 = K̂ 2                 e in dieselfde sirkelsegment   ✓ S    ✓R
        K̂1 + K̂ 2 = Ê1 + Ê 2
       K̂1 = Ê1                   gegee                           ✓ S
        K̂ 2 = Ê 2
       KF is ʼn rklyn              omgekeerde rklyn koord          ✓ R (as K̂ 2 = Ê 2 bewys
                                   stelling                          is)                        (7)
                                                                                               [15]

                                                                                   TOTAAL: 150

                                                   14

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## Collection

- [Grade 11 past papers](https://www.hlayiso.com/grade-11/past-papers)

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