# Mathematics Grade 11 P2 Gauteng November 2021 Memo (Afr)

> Grade 11 Mathematics Paper 2 memorandum from the Gauteng November 2021 (Afrikaans), 17 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/gr11-mathematics-p2-afr-nov-possible-answers/08b12143-a599-4a7c-b0c3-25e32248330e
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Document type: Memorandum
- Year: 2021
- Exam period: Gauteng November
- Paper: 2
- Curriculum: CAPS
- Pages: 17

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## About this document

Mathematics · Grade 11 · Gauteng November · 2021. Memorandum, 17 pages. Read online or download the PDF.

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### Page 1

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             PROVINSIALE EKSAMEN
                 NOVEMBER 2021
                    GRAAD 11
                NASIENRIGLYNE

                 WISKUNDE (VRAESTEL 2)

17 bladsye

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                                                                  WISKUNDE
                                         NASIENRIGLYNE
                                                                  (Vraestel 2)           GRAAD 11

INSTRUKSIES EN INLIGTING:

A – Akkuraatheid

C.A. – Kontinue Akkuraatheid (volg op)

S – Bewering

R – Rede

S en R – Bewering en Rede

NOTA:

•    Indien ’n kandidaat ’n vraag TWEE keer beantwoord het, merk slegs die EERSTE poging.

•    Indien ’n kandidaat ’n 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                  ∑𝑛
                      𝑖=1 𝑥𝑖     408                                  ✓ antwoord
      Gemiddeld =              = 19 = 21,47
                       𝑛                                                                          (1)

1.2   Standaardafwyking = 7,81                                        ✓✓ antwoord                 (2)

1.3   Die limiete vir die een standaardafwyking is                    ✓ interval
      (𝑥̅ − 1𝜎 ; 𝑥̅ + 1𝜎)
                                                                      ✓ 13 mense
      = ( 21,47 – 7,81; 21,47 + 7,81) = (13,66 ; 29,28)

       13 mense lê binne 1 standaardafwyking van die
      gemiddeld af.                                                                               (2)

1.4   5 12 13 15 18 18 18 19 20 21                                    ✓ Q1 = 18
      21 22 23 23 26 29 33 35 37
      IKVW = 26 – 18                                                  ✓ Q3 = 26
      =8
                                                                      ✓ IKVW = 8                  (3)

1.5                                                                   ✓✓ houer / mond

                                                                      ✓ punt / snor
                                                                                                  (3)

1.6   Daar is ’n merkbare verskil tussen die laagste waarde (5) en    ✓ rede
      die volgende laagste waarde (12) terwyl die verskille tussen
      al die ander data waardes hoogstens binne 3 waardes is          ✓ 5 is ‘n uitskieter

       5 is ’n uitskieter                                                                        (2)
                                                                                                 [13]

                                                     3

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                                                                                                   WISKUNDE
                                                                                NASIENRIGLYNE
                                                                                                   (Vraestel 2)       GRAAD 11

VRAAG 2

2.1                            Lengte (cm)             Aantal leerders               Kumulatiewe   ✓ 89
                                                       (f)                           frekwensie
                         145  ×  150                         6                              6    ✓ 215
                         150  ×  155                        23                             29    ✓ 247
                         155  ×  160                        60                             89
                         160  ×  165                        74                            163
                         165  ×  170                        52                            215
                         170  ×  175                        32                            247
                         175  ×  180                         3                            250
                                 Totaal                      250                                                            (3)

2.2                                        Ogief wat die verspreiding van punte toon               ✓ Vorm
                                                                                                   ✓ grond op x- as
                                                                                                 ✓ plot die punte
                                                                            

      Kumulatiewe frekwensie
                                                                   

                                                          

                                                 
                                                    Q1       Q2       Q3                                                   (3)
                                 
2.3   Q2  162                                                                                     ✓
                                                                                                   ✓
      Q1  158                                                                                     ✓
                                                                                                   ✓
      Q3  167

       IKVW  162                                                                                                          (4)

2.4   165  ×  170                                                                                ✓ antwoord               (1)

2.5    35
                               = 14%                                                               ✓ antwoord
      250                                                                                                                   (1)

2.6    = 6,43                                                                                     ✓✓ antwoord              (2)
                                                                                                                           [14]

                                                                                           4

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                                                                     WISKUNDE
                                                 NASIENRIGLYNE
                                                                     (Vraestel 2)          GRAAD 11

VRAAG 3

3.1      3+12
      E( 2          ;
                        0+3
                              )                                      ✓ substitusie in
                         2                                             middelpuntformule
           1            1
      = (7 2 ; 1 2 )                                                 ✓ Antwoord                  (2)

3.2   MBC= 12−3
                3−0                                                  ✓ substitusie in
                                                                       gradiëntformule
            1
         =3                                                          ✓ Antwoord                  (2)

                                  1
3.3   tan θ = mBC =                                                  ✓ tan θ = mBC
                                  3

                1                                                    ✓ antwoord
      θtan-1(3 ) = 18,43̊                                                                        (2)

3.4                           1                                               1
      MAD = MBC = 3                       AD||BC, gelyke gradiënte   ✓ MAD = 3
                6−0
      MAB = 1−3 = –3
                                      1                              ✓ MAB = –3
       MAD × MAB = 3 × –3 = –1

      AD ⊥ AB                                                       ✓ MAD × MAB = –1            (3)

3.5   Inklinasie van nuwe lyn = 45̊ + 18,43̊ = 63,43̊                ✓ 18,43̊
       tan 63,43̊ = 2 = mlyn                                        ✓ 63,43̊
      y – 6 = 2(x – 1)                                              ✓m=2
           y = 2x + 4                                                ✓ vergelyking               (4)
                                                                                                [13]

                                                          5

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                                                           WISKUNDE
                                           NASIENRIGLYNE
                                                           (Vraestel 2)          GRAAD 11

VRAAG 4

4.1   MQP = MOS = 6           QP||OS, gelyke gradiënte      ✓ MQP = 6

      y – 17 = 6(x + 3)                                     ✓ vergelyking

      y = 6x + 35                                                                       (2)

4.2   6x + 35 = –x                                          ✓ opstel van vergelyking

           7x = –35                                         ✓x=–5

             x = –5
                                                            ✓ koördinate van Q
             y = – (–5) – 5

      Q(–5;5)                                                                          (3)

4.3   OQ2 = (–5 – 0)2 + (5 – 0)2                            ✓ substitusie in
                                                               afstandsformule
           = 50
                                                            ✓5 2
      OQ = 50 = 5 2 eenhede                                                             (2)

4.4   𝑀𝑜𝑠 = 6                                               ✓ 80,54°

      ∴ inklinasie van OS is – tan–1 (4) = 80,54°           ✓ 135°

      𝑀𝑂𝑄 = −1                                              ✓ 54,46°

      ∴ inklinasie van QO is 180° –tan–1 (1) = 135°

      𝛼= 135° – 80,54° …

      = 54,46°                                                                          (3)

4.5   QS2 = OS2 + OQ2 – 2OS.OQ.cos                         ✓ korrekte gebruik van
                                                              cosreël
          = 148 + 50 – 2( 148 )( 50 .cos 54,46)
                                                            ✓ substitusie in formule
       QS = 9,90 eenhede
                                                            ✓ 9,90                      (3)
                                                                                       [13]

                                                     6

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                                                              WISKUNDE
                                              NASIENRIGLYNE
                                                              (Vraestel 2)                 GRAAD 11

VRAAG 5

5.1   5.1.1 sin 28° = cos 62°                                 ✓✓ Antwoord

                =k                                                                               (2)

      5.1.2 cos 242° = –cos 62°                               ✓✓ Antwoord

                = –k                                                                             (2)

5.2                         sin 𝑥   cos 𝑥
      = -tan x . cos x = cos 𝑥 × sin 𝑥                        ✓ tan x . cos x
           –sin x
                                                              ✓ –sin x
      =1                                                         sin𝑥
                                                              ✓ cos𝑥

                                                              ✓1                                 (4)

5.3   cos 50° − cos 𝜃    −(cos 𝜃 + cos 50°)                          cos 50°−cos 𝜃
                        = cos 𝜃 + sin 40°                     ✓✓ sin 40 + cos 𝜃
      sin 40 + cos 𝜃

      cos 50° = sin 40°                                       ✓✓ –(cos  + cos 50°)

                = −1
                                                              ✓ antwoord                         (5)

5.4   = √22sin 20° . 23.−tan 45°                              ✓ √22sin 20° . 23.−tan 45°

                1                                                       1
      = √22.2 . 2−3                                           ✓ √22.2 . 2−3

            1
      =√4                                                        1
                                                              ✓ 2
        1
      =2                                                                                         (3)

5.5   5.5.1 sin2 x = 3                                        ✓ antwoord
                     4

                            √3
                sin x = ± 2
                                                                                                 (1)
      5.5.2 Verwysingshoek = 60°                              ✓✓✓ antwoorde

                60°; 120°; 240°                                                                  (3)
                                                                                                [20]

                                                    7

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                                                           WISKUNDE
                                           NASIENRIGLYNE
                                                           (Vraestel 2)                GRAAD 11

VRAAG 6

                                       5 y
                                       4
                                       3
                                       2
                                       1
                                                                                         x
        –90                –45                    45       90           135              180
                                      –1
                                      –2
                                      –3
                                      –4
                                      –5

6.1   f ( x) = tan x − 1                                   ✓ asimptote
                                                           ✓ afsnitte met asse
                                                           ✓ vorm
      g ( x) = cos 2 x                                     ✓ periode
                                                           ✓ vorm met oop punt/kol
                                                             by x = −90
                                                                                                (5)

6.2   6.2.1   cos 2 x + 1 = tan x                          ✓ x = 45
               cos 2 x = tan x − 1
               x = 45                                                                         (1)

      6.2.2   2 cos 2 x = 1                                ✓ cos 2 x = 0
                                                           ✓ oplossings
               2 cos 2 x − 1 = 0
               cos 2 x = 0
               x = 45 or x = 135                                                            (2)

      6.2.3 −90° < 𝑥 < −45° of 90° < 𝑥 < 135°              −90° < 𝑥 < −45°
                                                            90° < 𝑥 < 135°                     (2)

      h( x) = cos 2( x − 30  )                            ✓ h( x) = cos 2( x − 30 )
6.3                                                                               

                                                                                                (1)
                                                                                               [11]

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                                                                        WISKUNDE
                                               NASIENRIGLYNE
                                                                        (Vraestel 2)             GRAAD 11

VRAAG 7

7.1
           R

                                   p
       h                   q

                        180 − P
           M                       P            r           Q
      Konstruksie: Trek RM ⊥ PQ of QP verleng                             konstruksie
      RM = p sin Q
                                                                          RM = p sin Q
               CM = q sin (180  − P)                                     RM = q sin (180 − P)
                  = q sin P
       p sin Q = q sin P
                                                                           p sin Q = q sin P
      Deel deur sinP.sinQ gee dan
                                                                          Deel deur sinP.sinQ
        p     q
           =
      sin P sin Q
      OF                                                                 OF

                R (qcos P; qsin P)         y

                                                                          R(qcosP; qsinP)

                                       q            p
                                                                          konstruksie
                                                                          RM = q sin P

                           M               P        r           Q   x
      P is in standaardposisie soos in die diagram getoon.                RM = p sin Q
      Konstruksie: Trek RM ⊥ PQ of QP verleng
      RM = q sin P = p sin Q                                              Deel deur sinP.sinQ
      Deel deur sinP.sinQ dan kry ons
       𝑝          𝑞
              = 𝑠𝑖𝑛 𝑄                                                                                  (5)
      𝑠𝑖𝑛 𝑃

                                                        9

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                                                               WISKUNDE
                                            NASIENRIGLYNE
                                                               (Vraestel 2)           GRAAD 11

7.2
              L

                       1

               h
                                                                   I
                                                           d
                                                  J

                        K

7.2   7.2.1   In  LKJ :
               h
                  = sin y
              LJ
              h = LJ sin y            −−−      (1)               h substitusie in formule

              In  DAB :
               LJ     d
                   =
              sin x sin L 1

              maar Lˆ1 = y − x
                          d sin x                                 sinreël
               LJ =                   − − − (2)
                       sin ( y − x)
                                                                       
              sub (2) in (1) :                                    L1 = y − x
                  d sinx siny
               h=
                   sin( y − x)                                    LJ

                                                                  substitusie                 (5)

      7.2.2              70. sin 12. sin 30                     substitusie
                     h=
                               sin 18                            antwoord
                      = 23, 55 m                                                               (2)
                                                                                              [12]

                                                      10

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                                                       WISKUNDE
                                       NASIENRIGLYNE
                                                       (Vraestel 2)            GRAAD 11

VRAAG 8

8.1   Volume van silinder:

                V1 = r2h
                                                       ✓ Substitusie
                = (12) .(19)
                       2

                                                       ✓ antwoord
                = 2736  m      3

      Volume van hemisfeer (halwe sfeer):

                    14 3
            V2 =      r                                 1
                    23                               ✓
                                                           2
            Totale Volume = V1 + V2
                                                       ✓ Substitusie
                 1  4 3     3                      ✓ antwoord
            =        r (12) 
                 2  3        

            = 1152 m3

             Totale Volume = V1 + V2                   ✓ Totale Volume in 𝜋

             = 2736 + 1152
                                                       (5 uit 6 indien numeriese
             = 3888 m      3                          waarde van pi gebruik is )    (6)

                                             11

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                                                                WISKUNDE
                                           NASIENRIGLYNE
                                                                (Vraestel 2)             GRAAD 11

8.2   Buiteoppervlak van silinder:

                A1 = 2r 2 + 2rh                          ✓ Substitusie

                                                           ✓ antwoord
                     = 2 (12 ) + 2 ( 12 )( 19 )
                               2

                     = 2337 ,34 m 2                        ✓ Substitusie

      Buiteoppervlak van hemisfeer (halwe sfeer):
                                                           ✓ antwoord
                 1
             A2 = (4r 2 )
                 2

                =
                    1
                    2
                     (4r 2 )(12) 2   
                = 452.39 m2
                                                           ✓ Totale buiteoppervlak
                                                           ✓ antwoord, korrek afgerond
      Totale Buiteoppervlak = A1 + A2

                             = 2337,34 + 452.39

                             = 685.73 m2  686 m2                                              (6)
                                                                                              [12]

                                                    12

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                                                                       WISKUNDE
                                                NASIENRIGLYNE
                                                                       (Vraestel 2)          GRAAD 11

VRAAG 9

9.1   9.1.1 Die teenoorgestelde hoeke van ’n sikliese vierhoek is        antwoord
            supplementêr.                                                                          (1)

      9.1.2 Die hoek tussen die raaklyn en koord is gelyk aan die  antwoord
            hoek in die teengestelde segment.                                                      (1)

9.2

                                   F

                                                   S

                               O
                                                    2
                                                  1 D
                                       2
                                   1
           B                       A                              T

      Konstrueer AF(middellyn) en FD (koord)                            ✓ Konstruksie

                                                                       ✓ Bewering ✓ Rede
      D1 = 90      
                                           (  in ’n halwe sirkel )

       
                                                                        ✓ Bewering en Rede
      A1 = 90                             ( rad ⊥ raaklyn )

              
      A2 = D2                          (∠𝑒 in dieselfde sirkelseg. )    ✓ Bewering en Rede

                        
      A1 + A2 = D1 + D2
                      
                                                                        ✓ Gevolgtrekking
       BAS = D                                                                                    (6)

                                                           13

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                                                                                    WISKUNDE
                                                         NASIENRIGLYNE
                                                                                    (Vraestel 2)           GRAAD 11

9.3                                                F                G
                                                     1     2

                                                                         1
                                                                         2      H
                                                                          3
                                             1                     70º
                                 E           2

                                                       1   2 3   30º

               P                                           Q                                   R

      9.3.1                                                                        raaklyn – koordstelling
              P Q E = H 3 = 70                   (raaklyn − koordstell ing )
                                                                                      
              Q1 = 70                                                              Q1 = 70                   (2)

                                                                                      
      9.3.2
              E1 = 70                           (verw. e, EF / / QH )              E 1 = 70
                                                                                   verw. e =
              G =180 − E1                   (teenoorst. e kvh EFGH )                            
                   =180 − 70                                                       G =180 − E1
                                                                                   R
              G =110                                                                                            (4)

                                                                                                   
      9.3.3
              Q2 =180 − (Q1 + Q 3 )                       (e op rtlynPQR )         180 − (Q1 + Q3 )
                   =180 − (70 + 30)
                                                                                       
                   = 80                                                             Q2 = 80

                                                                                     teenoorst e van kvh
                                        
              E F H =180 − Q2                    (teenoorst. e kvh EFHQ )            
                                                                                    F1 =100
                       F1 =180 − 80
                           =100                                                                                 (4)
                                                                                                                [18]

                                                                 14

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                                                                     WISKUNDE
                                               NASIENRIGLYNE
                                                                     (Vraestel 2)        GRAAD 11

VRAAG 10

10.1

       10.1.1       O1 = 2 (middelpunt  = 2x omtrek  )           beweringrede            (2)

       10.1.2       W =        ( e in dieselfde sirkel segment)   beweringrede            (2)

       10.1.3       Z = 90  (  in ’n halwe sirkel)                bewering & rede
                    Y2 = 90 −  (binne e van ʼn  )                 bewering & rede          (2)

10.2   WY 2 = WZ 2 + YZ 2 (Pythagoras stelling)                       bewering & rede
       YZ = WU = UZ (gegee)                                           bewering & rede
        OW = OY = r (gegee)                                          bewering & rede
                     4r 2   4r                                             4r 2
        YZ =             =                                          YZ =          OF
                      5      5                                               5
                                                                          4r
                                                                     YZ =    
                                                                           5                   (4)

10.3   OW=OV= r =OU+OV                  ( radiusse )                 beweringrede
       r = 3 + 4 (Pythagoras stelling)
        2       2    2
                                                                     r 2 = 32 + 4 2 
       r = 5 cm                                                      r =5 cm
       UV=5 – 2                                                     antwoord
            =3 cm                                                                              (2)
                                                                                              [12]

                                                          15

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                                                                       WISKUNDE
                                             NASIENRIGLYNE
                                                                       (Vraestel 2)        GRAAD 11

VRAAG 11

           
11.1                                                                    bewering & rede
       S = Q2 = x                (verw. hoeke TQ‖PS)

           
       S = W1 = x                (∠𝑒 in dieselfde sirkel segment)       bewering & rede

           
       S = Q4 = x                (  tussen raaklyn en koord)           bewering & rede
                                                                                                 (3)

11.2   11.2.1                                                        bewering & rede
                R1 = S + Q3              (buite  van  )

                           
                L3 = W1 + Q1             (buite  van  )               bewering & rede

                   
                S = W1 = x        (  e in dieselfde sirkel segment)    bewering
                                                                      gevolgtrekking
                Q1 = Q3                ( = koorde onderspan =  )

                         
                   R1 = L3                                                                      (4)

                                                       16

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                                                                                  WISKUNDE
                                                           NASIENRIGLYNE
                                                                                  (Vraestel 2)         GRAAD 11

11.2.2                                                                            bewering  rede
         W1 = Q4                          ( raaklyn koordstelling )

                 
         Q1 = Q3                  ( = koorde onderspan =  )                        bewering & rede

                                                    
          Q4 + Q3 = W1 + Q1
                                                                                    bewering & rede
                                                 
         netso            L3 = W1 + Q1                     (buite  van      )
                                                                                   bewering & rede
                                  = R3            ( ooreenk. e ,QT ||RS )
                                        
                R3 = Q4 + Q3
         PRKQ is ʼn koordevierkant.
         (buite  = teenoorst. binne  )

         Alternatiewelik:
                                           
         Q1 + Q3 = Q3 + S

         OF

                     
         W1 = Q4                                   ( raaklyn koordstelling)
                 
         Q1 = Q3                                  ( = koorde onderspan =  )

                                                
                                                                                                             (5)
          Q4 + Q3 = W1 + Q1
                                                                                                            [12]

                                                                                                 TOTAAL:    150

                                                                    17

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