# Mathematics Grade 12 Paper 2 Western Cape November 2017 Memo

> Grade 12 Mathematics Paper 2 memorandum from the Western Cape November 2017, 10 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/2017-grade-12-math-trial-exam-paper-2-wc-winelands-memo/3090fe90-9bf2-49b2-9ebd-a44bea521bb0
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2017
- Exam period: Western Cape November
- Paper: 2
- Curriculum: CAPS
- Pages: 10

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Mathematics · Grade 12 · Western Cape November · 2017. Memorandum, 10 pages. Read online or download the PDF.

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### Page 1

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SEPTEMBER 2017          MATHEMATICS PAPER 2 / WISKUNDE VRAESTEL 2
                                  MEMORANDUM

NOTE:
 If a candidate answers a question TWICE, mark only the first one.
 Consistent accuracy applies in ALL aspects of the marking memorandum.
LET WEL:
 Indien ŉ kandidaat ŉ vraag TWEE keer beantwoord, merk slegs die eerste poging.
 Volgehoue akkuraatheid is deurgaans op ALLE aspekte van die memorandum van toepassing.

              SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD       DESCRIPTORS/BESKRYWERS             Marks

QUESTION/ VRAAG 1
1.1            Percentage of   Frequency    Cumulative       frequency column/
               income (x)                   Frequency       Frekwensie kolom
               20 < 𝑥 ≤ 30     10           10
               30 < 𝑥 ≤ 40     15           25               first 3 points on
               40 < 𝑥 ≤ 50     20           45               C. f column/ 1ste 3 punte op Kf
               50 < 𝑥 ≤ 60     10           55              kolom
               60 < 𝑥 ≤ 70     5            60
                                                             last 2 points on
                                                            C. f column/ laaste 2 punte op
                                                            Kf kolom
                                                                                               (3)
      1.2.1   60 − 45                                        45
              = 15 𝑝𝑒𝑜𝑝𝑙𝑒 𝑠𝑝𝑒𝑛𝑡 𝑚𝑜𝑟𝑒 𝑡ℎ𝑎𝑛 50% 𝑜𝑛 𝑡ℎ𝑒𝑖𝑟
              𝑖𝑛𝑐𝑜𝑚𝑒 𝑜𝑛 𝑒𝑑𝑢𝑐𝑎𝑡𝑖𝑜𝑛                            Answer/ Antwoord                 (2)
      1.2.2   𝑀𝑒𝑑𝑖𝑎𝑛 = ≈ 42 − 43                            Answer/ Antwoord                 (2)
      1.3     40 < 𝑥 ≤ 50                                    Answer/ Antwoord
              OR
              𝑥 ∈ (40; 50]                                  Accept/ Aanvaar: from 40 to 50     (1)
                                                            / van 40 tot 50
                                                                                               [8]

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SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD                               DESCRIPTORS/BE         Marks
                                                                      SKRYWERS
QUESTION/ VRAAG 2
2.1                                                                    Plotting 5 - 8
&                                                                     points correct/
2.3                                                                   Afsteek van 5 – 8
                                                                      punte korrek
                                                                       Plotting 9 -10
                                                                      points correct/
                                                                      Afsteek van 9 – 10
                                                                      punte korrek

                                                                      2.3 Regression line/
                                                                      Regressielyn:

                                                                       y -int:  29
                                                                       Passing through
                                                                           (30; 56)         (2)
                                                                                             (2)
2.2   𝑎 = 29,22 (29,21542 … )                                         𝑎
                                                                      𝑏
      𝑏 = 0.89
                                                                      equation /
      ∴ 𝑦 = 29,22 + 0,89𝑥
                                                                      vergelyking            (3)
2.3   On graph/ Op grafiek
2.4   r = 0,66                                                        r
      Moderately positive relationship/ redelike (matige) positiewe    Moderately
      verwantskap                                                     positive/ Redelike
                                                                      positiewe              (2)
2.5                            559                                       559
      Mean/ Gemiddelde, 𝑥̅ = 10 = 55,90                                10
                                                                       55,90
      Standard deviation/ Standaardafwyking: σ = 11.36                 σ = 11.36
                                                                                             (4)
2.6        44,54                          67,26                        Interval
                           55,90                                      Answer/Antw
      6 scores lie within one σ/ 6 Punte val binne een σ                                     (2)
                                                                                             [15]

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               SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD                          DESCRIPTORS/BESKRYWERS          Marks

QUESTION/ VRAAG 3
3.1             4−0                                                              Subst in gradient formula/
      𝑚 𝑇𝑄 =
                0−3                                                             Vervang in gradiënt formule
               −4
          =
                3
                                                                                 Answer/Antwoord               (2)
3.2   RQ =√(10 − 3)2 + (7 − 0)2                                                  Substitute/ Vervang

         = √49 + 49
                                                                                 Answer/Antwoord
         = √98 Units/ 𝑒𝑒𝑛ℎ𝑒𝑑𝑒                                                                                   (2)
3.3          4−0  −4
      𝑚 𝑇𝑄 = 0−3 = 3          or/of                        𝑚 𝑇𝐹 = 𝑚 𝑇𝑄

          𝑦=
                      −4
                           𝑥+4                         
                                                         −8−4
                                                                =
                                                                    −4    eqn of/ vgl van TQ or gradients =
                          3                                𝑘         3
                           −4                                             Subst/ vervang (k; -8) or gradient
           −8 =                (𝑘) + 4               −4𝑘 = 3(−12)
                            3                                                                        of TF
                            −4
           −12 =                   (𝑘)                                   Simplify/ Vereenv 𝒐𝒓 cross ×
                                3
                                          𝒌= 𝟗                           Answer/Antwoord                      (4)
3.4     𝑺(𝟕: 𝟏𝟏)                                                                𝑥= 7

                                                                                 𝑦 = 11                        (2)
3.5   TŜR = TQ
              ̂R           .opp  of ⃦ / ….teenoorst.  van ⃦
                                            𝑚                        𝑚           TŜR = TQ
                                                                                          ̂R

                 −4
       𝑚 𝑇𝑄 =                         𝑚𝑅𝑄 = 1
                 3
                                                                                            −4
                                                                                 tan 𝛼 =    3
                 −4
       tan 𝛼 =                            tan β = 1                              tan β = 1
                 3

      RA/ 𝑉𝐻 = 53,13°                       β = 45°
                                                                                 𝛼 = 126, 87°
       𝛼 = 126,87°
                                                                                 β = 45°
      ∴ TŜR = 𝛼 – β

                                                                                 Answer/Antwoord               (6)
      ∴ 𝐓𝐒̂𝐑 = 81, 87°
3.6   MQ = √(2)2 + (2)2
         = √8                                                                    √8

      But/Maar RQ = √98                      … 𝑐𝑎𝑙𝑐. 𝑖𝑛 𝑞 3.2
                                                                                    √𝟖
                                                                                
                          √𝟖                                                        √𝟗𝟖
      ∴ 𝑹𝒂𝒕𝒊𝒐 =
                          √𝟗𝟖

                      𝟐
                 =𝟕                                                              Answer/ Antwoord              (3)
                                                                                                                [19]

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      SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD                                  DESCRIPTORS/BESKRYWERS        Marks

QUESTION /VRAAG 4
4.1   Radius to midpoint of chord/ radius na midpt van koord                Answer/ Antwoord                 (1)
4.2           3                                                            𝑚𝑇𝑆
      𝑚 𝑇𝑆 = −   −1
              3
       𝑚𝑃𝑁 = 𝑚 𝑇𝑆 × −1                              … … TS  NP
                                                                           𝑚𝑃𝑁
       𝑚𝑃𝑁 = 1
       𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 )                 OR/ OF      𝑦 = 𝑚𝑥 + 𝑐          Subst (-3; 8)
      ∴𝑦−8 =𝑥+3                                        8 = −3 + 𝑐
                                                       𝑐=5
                                                                            Equation/ Vergelyking            (5)
                   ∴ 𝒚 = 𝒙 + 𝟏𝟏
4.3   𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑃(0; 𝑦)     𝑦 = 0 + 11                                    𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑃(0; 𝑦)
                           𝒚 = 𝟏𝟏                                          Answer/ Antwoord
      𝑅𝑎𝑑𝑖𝑢𝑠:           11 − 5 = 6                                         𝑅𝑎𝑑𝑖𝑢𝑠
                                                                            Answer/ Antwoord
                          𝒚 = −𝟏                                                                             (4)
4.4   Substitute M (𝑥: 0)/ 𝑆𝑡𝑒𝑙 𝑀(𝑥; 0)                                     Subst (x; 0) / vervang (x; 0)
            0 = 𝑥 + 4  𝑥 = −11
         𝑴 = (−𝟏𝟏; 𝟎)                                                       Length of TO/ Lengte van TO

                                                                            Length of MO/ Lengte van MO
         𝑇𝑂 = 5 , 𝑀𝑂 = 11
        𝑇𝑀2 = 52 + 112 …pyth                                                Answer/ Antwoord
        𝑀𝑇 = √146                                                                                            (4)
4.5   (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑟 2                                            TM is the diameter of/ is die
      TM is the diameter/ is die middellyn ʘ TSM                           middellyn van ʘ TSM

            √146                 146
      𝑟 =    2
                    𝑟2 = 4                                                𝑟 =
                                                                                     √146
                                                                                      2

      Centre of / midpt ʘ TSM :
                                                                                        −11 5
                           𝑥+𝑥1        𝑦+𝑦1                                 centre (      ; )
      Midp. of TM = (              ;            )                                        2  2
                             2             2
                            −11+0          0+5
                      = (              ;         )
                              2             2
                            −11        5                                                𝟏𝟏          𝟓 𝟐
                      = (         ; 2)                                      (𝒙 +         )𝟐 + (𝒚 − 𝟐)
                             2                                                          𝟐

                      𝟏𝟏                    𝟓 𝟐       146      73              146       73
             ∴ (𝒙 +      )𝟐 + (𝒚 − 𝟐) = 4                   or 2 or 36,5            or 2 or 36,5             (5)
                       𝟐                                                       4
                                                                                                              [19]

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        SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD                           DESCRIPTORS/BESKRYWERS            Marks

QUESTION /VRAAG 5
5.1                                                                        figure 1 in correct quadrant;
                                                            3         1   𝑟 = 5 / figuur 1 in korrekte
                          -4
                               A                        𝐵                 kwadrant; 𝑟 = 5
                                                                           figure 2 in correct quadrant;
                 -3            5                            2√2
                                                                          𝑥 = 2√2 / figuur 2 in korrekte
                                                                          kwadrant; 𝑥 = 2√2

                               2
          −3     2√2                                                       substitution / substitusie
        5( )− 6(     )
          5       3
                      8
        = −3 − 6 (9)
            25                                                             Answer/ 𝐴𝑛𝑡𝑤𝑜𝑜𝑟𝑑                (5)
      =− 3
5.2.1 𝑐𝑜𝑠 (360 − 24°)                                                      𝑐𝑜𝑠 24°
      = 𝑐𝑜𝑠 24° = p                                                        Answer/ 𝐴𝑛𝑡𝑤𝑜𝑜𝑟𝑑                (2)
5.2.2    𝑐𝑜𝑠 2(24°)                                                        𝑐𝑜𝑠2(24°)
        = 2𝑐𝑜𝑠 2 24° − 1                                                  2𝑐𝑜𝑠 2 24° − 1
        = 2p2 - 1                                                          Answer/ 𝐴𝑛𝑡𝑤𝑜𝑜𝑟𝑑                (3)
5.3                                           1−𝑐𝑜𝑠2𝐴
        𝐿𝐻𝑆 = tan 𝐴                 𝑅𝐻𝑆 =      𝑠𝑖𝑛2𝐴
                                               1−(1−2𝑠𝑖𝑛2 𝐴)
                                          =     2𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴                 1 − 2𝑠𝑖𝑛2 𝐴
                                                2𝑠𝑖𝑛2 𝐴
                                          = 2𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴                     2𝑠𝑖𝑛𝐴𝑐𝑜𝑠𝐴
                                               𝑠𝑖𝑛𝐴
                                          = 𝑐𝑜𝑠𝐴                           simplify / vereenvoudig
                                           = 𝑡𝑎𝑛𝐴                          Answer/ 𝐴𝑛𝑡𝑤𝑜𝑜𝑟𝑑                (4)
5.4     sin 20° . (cos 40°) + cos 20° . sin 40°                           𝑐𝑜𝑠40° 𝑐𝑜𝑠20°
        = sin(20° + 40°)                                                   𝑠𝑖𝑛40°
        = sin 60°                                                          𝑠𝑖𝑛 60°
           √3                                                                 √3
        =                                                                                                  (5)
            2                                                                 2
5.5     2cos²θ + 5sinθ + 1 = 0
        ∴ 2(1 − sin2 θ) + 5 sin θ + 1 = 0                                  (1 − 𝑠𝑖𝑛2 𝜃)
        ∴ 2 − 2sin2 θ + 5 sin θ + 1 = 0
                                                                           standard form/
        ∴ −2sin2 θ + 5sinθ + 3 = 0                                        Standaardvorm
        ∴ 2sin2 θ − 5sin2 θ − 3 = 0                                        factors faktore
        ∴ (2sinθ + 1)(sinθ − 3) = 0                                                    1
                                                                          sin 𝜃 = − 2        sin 𝜃 ≠ 3
                  1
        ∴ sinθ = − ; sinθ ≠ 3
                  2                                                        210° and/ en 330°
        θ = 180° + 30° + 360°n or /𝑜𝑓 θ = 360° − 30° + 360°n,
                                                                           360°𝑛 , 𝑛 ϵ Z                   (7)
        ∴ θ = 210° + 360°n         or /𝑜𝑓 θ = 330° + 360°n , n ϵ Z

                                                                                                            [26]

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      SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD   DESCRIPTORS/BESKRYWERS       Marks

QUESTION /VRAAG 6
6.1   𝑎=2                                        Answer/ 𝐴𝑛𝑡𝑤𝑜𝑜𝑟𝑑            (1)
6.2
                                                 Max TPs /Maks DPe

                                                Min TP/Min DP

                                                x-intercepts / x-afsnitte

                                                                              (3)
6.3   180°                                      180°                         (1)
6.4   𝑥 = −90° or 𝑥 = 90°                        𝑥 = −90°                    (2)
                                                 𝑥 = 90°
6.5    3 − 5.2 sin 𝑥 cos 𝑥                       − 5.2 sin 𝑥 cos 𝑥
      = 3 − 5 sin 2𝑥                            sin 2𝑥
      = 3 − 5 (-1)                              Answer/ Antwoord             (3)
      =8
                                                                              [10]

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      SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD                    DESCRIPTORS/BESKRYWERS        Marks

QUESTION /VRAAG 7
7.1

                  𝑟
      𝑐𝑜𝑠 𝛼 =
                 𝐴𝐵
                                                                  expression AB /
             𝑟                                                   uitdrukking AB
      𝐴𝐵 =                                                                                     (1)
           cos 𝛼
7.2            𝑟
      𝑐𝑜𝑠 𝛼 =
             𝐴𝐶                                                   expression AC /
             𝑟
      𝐴𝐶 =                                                       uitdrukking AC
           𝑐𝑜𝑠 𝛼

      𝐵𝐶 2 = 𝐴𝐵2 + 𝐴𝐶 2 − 2𝐴𝐵. 𝐴𝐶𝑐𝑜𝑠𝐵𝐴̂𝐶                         cosine rule / cos reël

                  𝑟    2    𝑟    2      𝑟       𝑟                substitution / substitusie
      𝐵𝐶 2 = (        ) +(      ) − 2(      )(      ) 𝑐𝑜𝑠𝜃
                𝑐𝑜𝑠 𝛼     𝑐𝑜𝑠 𝛼       𝑐𝑜𝑠 𝛼   𝑐𝑜𝑠 𝛼

                2𝑟 2   2𝑟 2 𝑐𝑜𝑠𝜃
      𝐵𝐶 2 =      2  −
               𝑐𝑜𝑠 𝛼    𝑐𝑜𝑠2 𝛼
                                                                 common factor /
        2
            2𝑟 2                                                 gemeenskaplike faktor
      𝐵𝐶 =         (1 − 𝑐𝑜𝑠𝜃)
           𝑐𝑜𝑠 2 𝛼

             𝑟√2(1 − 𝑐𝑜𝑠𝜃)
      𝐵𝐶 =
                 𝑐𝑜𝑠𝛼

                                     OR
               𝑟
      𝑐𝑜𝑠 𝛼 =
              𝐴𝐶
                                                                  expression AC /
               𝑟
      𝐴𝐶 =                                                       uitdrukking AC
             𝑐𝑜𝑠 𝛼

      ∴ 𝐴𝐵 = 𝐴𝐶                                                  cosine rule / cos reël

      𝐵𝐶 2 = 2𝐴𝐵2 − 2𝐴𝐵2 𝑐𝑜𝑠 𝜃                                   substitution 𝐴𝐵 = 𝐴𝐶 /
                                                                 vervanging 𝐴𝐵 = 𝐴𝐶
        2         2
      𝐵𝐶 = 𝐴𝐵 (2 − 2𝑐𝑜𝑠 𝜃)

      𝐵𝐶 = 𝐴𝐵√2 − 2𝑐𝑜𝑠𝜃
               𝑟                                                 common factor /
      𝐵𝐶 =        √2(1 − 𝑐𝑜𝑠𝜃)                                   gemeenskaplike faktor         (4)
             𝑐𝑜𝑠𝛼

                                                             7

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7.3             50(√2(1 − 𝑐𝑜𝑠30°)                                       substituting values of BC, r
        100 =                                                          and 𝜃 / substitusie van
                      𝑐𝑜𝑠𝛼
                                                                       waardes vir BC, r and 𝜃
                 50(√2(1 − 𝑐𝑜𝑠30°)
        𝑐𝑜𝑠𝛼 =                                                                    50(√2(1−𝑐𝑜𝑠30°)
                       100                                              𝑐𝑜𝑠𝛼 =        100
        𝛼 = 75°
                                                                        𝛼 = 75°                          (3)
                                                                                                          [8]

GEOMETRY/ MEETKUNDE

S: Statement/ Bewering
R: Reason/ Rede
S/R: Both Statement and Reason/ Beide Bewering en Rede

Allow for alternative methods/ Kyk vir Alternatiewe metodes

         SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD                       DESCRIPTORS/BESKRYWERS             Marks

QUESTION /VRAAG 8

8.1.1    90°                                                      S                                      (1)
8.1.2    Supplementary/ supplementêr                              S                                      (1)
8.2.1    ̂
         E1 = 54° ... tan-chord thm/ raaklyn-koordstelling        S R                                   (2)
8.2.2    Ĉ1 = 36° ... angle in semi-circle/ hoek in halfsirkel   S R
              OR angle subt. By diameter/ hoek onderspan
                                            deur midlyn                                                   (2)
8.2.3    Ĉ3 = 35° ... alt. s; AB//CD / verw.e; AB//CD          S R                                   (2)
8.2.4    AÊD = 55° ...opp s of cyclic quad/ teenoorst e        S R                                   (2)
                                               van kvh
8.2.5    ̂
         E4 = 35° ... subt by AC/ ondersp deur AC                 S R

         Ê3 = 20°                                               S                                      (3)
8.3       ̂ C = 90° ... tangent- radius/ raaklyn- radius
         BA                                                       S R
         And/ en
          ̂C = 90° ... tangent- radius/ raaklyn- radius
         BE                                                       S

         ∴ BECA is a cyclic quad/ is ’n kvh ... opp.s suppl./
                                                                  R
                                          teenoorst e suppl.
            ̂ E = AĈE... external  of cyclic quad/
         ∴ DB                                                      R can only get this mark if BECA is   (5)
                                            Buite  van kvh       proved a cyclic quad.
                                                                                                          [18]

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        SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD                           DESCRIPTORS/BESKRYWERS          Marks

QUESTION /VRAAG 9
9.1
                                                             D
                                      A

                               1
                           S                                     *
                                             T    E                   F

                  B                           *   C

        Constr./ Konstr: Draw ST with S on AB and T on AC such              Constr./ Konstr – can be
        that AS = DE and AT = DF/ Trek ST met S op AB en T op              on diagram/ kan op diagram
        AC sodat AS = DE en AT = DF                                        wees
                                                                            S/R ; Congruency/
        AST ≡ DEF …S, , S                                               Kongruensie

        Ŝ1 = ̂
               E … from/ vanaf congruency/ kongruensie                      ̂S1 = E
                                                                                   ̂
             =̂
              B … given / gegee                                                   ̂
                                                                                 =B

         ST // BC …corresp. s =/ ooreenk. e =                            S: ST // BC        R

             AB       AC
                 =                                                               AB       AC
             AS       AT
                                                                            S:        =
                                                                                  AS       AT
             AB       AC
                 =        (Constr/ Konstr: AS = DE and/ en AT = DF        R                             (7)
             DE       DF
9.2.1   Q1 = ̂
        ̂    Q2 … given/ gegee                                              S: ̂
                                                                                Q1 = ̂Q2
           = N2 … alt. s; MN // QR / verw. e; MN // QR
             ̂                                                                       ̂
                                                                                   = N2
                                                                           R
        MN = MQ … sides opp = s / sye teenoor = e                       R                             (3)

        QMN is isosceles/ gelykbenig
9.2.2   In PMN and/ en PQR:

        1.    P is common/ P is gemeen                                     S

                ̂ ... corr./ ooreenk s; MN // QR
           ̂1 = R
        2. N                                                               S R

                                                                            R (, , ) or 3rd  given   (4)
             PMN /// PQR … (, , ) or 3rd  given

                                                       9

### Page 10

Downloaded from hlayiso.com
9.2.3    From/ vanaf /// in 9.2.2:

         PM MN PN                                                                 Proportionality/ Eweredigheid
           =  =
         PQ QR PR
                                                                                  Subst/verv PQ = 5 and/ en
           PM MN
         ∴    =                                                                     QR = 7,5
            5   7,5
                                                                                   PM     2
           PM   5   2                                                             MN = 3
         ∴    =   =
           MN 7,5 3
                                                                                   PM    2
             PM       2                                                           MQ = 3
         ∴        =        … MQ = MN from/ vanaf 9.2.1
             MQ       3

             PN       2                                                            PN    2
         ∴        =       … Prop thm/ ewer.stelling; MN // QR or/ of              NR = 3
             NR       3
                           line // to 1 side of  / lyn // aan 1 sy van                                           (5)
                                                                                                                   [19]

        SUGGESTED ANSWER/ VOORGESTELDE ANTWOORD                                 DESCRIPTORS/BESKRYWERS             Marks

QUESTION /VRAAG 10

                                                                            A

                                                       N

                                                B     1 2

                                                             2
                                                             1
                          T
                                                M                P
10.1     ̂ T = 90° = BN
        BM               ̂P                                                             S
         BMPN is a cyclic quad/ is ’n kvh … ext  = int opp / buite =
                                                      teenoorst binne                  R                         (2)

10.2    N1 = ̂
        ̂    P1 …subtended by/ onderspan deur BM                                        S R

        But/ Maar ̂    ̂ … tan-chord thm/ raakl-koordstelling
                  P1 = A                                                                S R

                                                                                        S R
        NM // AP … corresp/ ooreenk. s =

            TN       TM
        ∴        =                                                                                                 (6)
            AN       MP
                                                                                                                   [16]
                                                    Total/Totaal: [150]
                                                            10

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