# Mathematics Grade 12 Paper 2 Eastern Cape Mock 2016 Memo

> Grade 12 Mathematics Paper 2 memorandum from the Eastern Cape Mock Exam 2016, 16 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/maths-p2-memo/24553409-0f7d-4afb-86a9-bfba72dd0d1c
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2016
- Exam period: Eastern Cape Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 16

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## About this document

Mathematics · Grade 12 · Eastern Cape Mock Exam · 2016. Memorandum, 16 pages. Read online or download the PDF.

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## Document text

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### Page 1

NATIONAL
                   SENIOR CERTIFICATE

                       GRADE/GRADE 12

                     SEPTEMBER 2016

          MATHEMATICS P2 / WISKUNDE V2
                MEMORANDUM

MARKS / PUNTE:      150

                   This memorandum consists of 16 pages.
                 Hierdie memorandum bestaan uit 16 bladsye.

### Page 2

2                                              MATHEMATICS P2                                 (EC/SEPTEMBER 2016)

    QUESTION 1 / VRAAG 1
           Day/Dag        0               1       4      6      9     12     17     19
        Weight / Gewig   124             121     103    90     71     50     27     16

    1.1                                                                        2-4 correct points
                                                                               5-7 correct points
                                                                               plotting all points

                                                                               2 – 4 punte korrek
                                                                               5 – 7 punte korrek
                                                                            alle punte afgesteek

                                                                                                        (3)
    1.2 a 124,84                                                     A
        b     5,83                                                   B
                                                                     equation / vergelyking
         y 124,84 5,83x                                                                                 (3)
    1.3 x ; y     8;5 ; 75,25                                          8;5 ; 75,25 and/en
        y-int 124,84                                                   y-int / y-afsnit 124,84
                                                                    regression line /
                                                                    regressielyn                        (2)
    1.4        124,84 5,83x 80                                        substitution / vervanging
                       5,83x 44,84
                                                                     answer / antwoord
                            x 7,69
          On the morning of the 8th day the bar of soap will
          be less than 80 grams. Op die oggend van die 8ste
          dag sal steen seep minder as 80 gram wees.                                                    (2)
    1.5 r    0,998                                                             answer / antwoord        (1)
    1.6 Very strong negative correlation. Baie sterk negatiewe                 answer / antwoord
        korrelasie.                                                                                      (1)
                                                                                                        [12]

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### Page 3

(EC/SEPTEMBER 2016)                                WISKUNDE V2                                              3

   QUESTION 2 / VRAAG 2
    5 7
    4 344
    3 111468
    2 24899
    1 5567
    0 35
   2.1 21 learners / leerders                                    answer / antwoord         (1)
   2.2 3 pages / bladsye                                         answer / antwoord         (1)
   2.3 x 28,19                                                    answer / antwoord
                                                                                           (2)
   2.4        13,12                                              answer / antwoord
                                                                                           (1)
   2.5    28,19 13,12; 28,19 13,12                               interval / interval
          15,07;41,31
                                                                 8 learners / leerders
           8 learners are outside one standard deviation
           8 leerders is buite een standaardafwyking             28,57%

            8
               100 38,10%
            21
                                                                                           (3)
                                                                                           [8]

   Copyright reserved                                                                    Please turn over

### Page 4

4                                           MATHEMATICS P2                           (EC/SEPTEMBER 2016)

    QUESTION 3 / VRAAG 3

    3.1        5 1                                   subst. P and R into correct formula
          mPR                                        verv. van P en R in korrekte formule
               4 4
               1                                            1
                                                     mPR
               2                                            2                                    (2)
    3.2    mSQ   2          SP   PT                  mSQ     2
          y     4    2x 1                            subst. m and Q into correct formula
                                                     verv. van m en Q in korrekte formule
                                                      y    2x 2
          y    2x 2                                                                              (3)
    3.3   Equation of PR / Vergelyking van PR
                 1
          y 1       x 4                              substituting m and P into equation of a
                 2
                 1                                   str. line / vervanging van m en P in die
             y     x 3                               vergelyking van reguitlyn
                 2                                   equation of PR / vergelyking van PR
                     1
            2x 2       x 3
                     2                               equating PR and SQ / gelykstel van PR
            4x 4 x 6                                 en SQ
                5 x 10
                                                     x-value / x-waarde
                     x 2                             y-value / y-waarde
          y         2 2 2   2

                                 OR/OF
                                                     substituting m and R into equation of a

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### Page 5

(EC/SEPTEMBER 2016)                                              WISKUNDE V2                                                  5

                  1                                                   str. line / vervanging van m en R in die
              y 5     x 4
                  2                                                   vergelyking van reguitlyn
         2 y 10 x 4
                                                                      equation of PR / vergelyking van PR
             2y x 6
                    1
               y        x 3
                     2                                                equating PR and SQ / gelykstel van PR
                   1                                                  en SQ
          2x 2         x 3
                    2
          4x 4 x 6                                                    x-value / x-waarde
              5 x 10                                                  y-value / y-waarde
                x       2
                                                                                                                   (5)
          y       2       2   2       2
   3.4        x 1                    y 4                              substituting into correct formula
              2                       2
                 2                    2                               vervanging in korrekte formule
           x     5            and/en    y 8                           x-value / x-waarde
         S (−5 ; 8)                                                   y-value / y-waarde                           (3)
   3.5                            2              2                    subt. into correct form / verv. in
         SQ               5 1             8 4
                                                                      korrekte formule
                  6 5                                                 SQ = 6 5
                                  2              2
         PT               2 4              2 1
                                                                      PT =     5
                      5
                              1
         Area PQS                6 5 5
                              2                                       Subt into correct form. / verv. in
                              15 unit 2 / eenhede2                    korrekte formule
                                                                      15 units2 / eenhede2

                              OR/OF
                                                                     SQ = 6 5
              area of ∆ PQS = ST PT                                  ST = 3 5
                             3 5  5                                  PT = 5
                                          15 unit 2 / eenhede2       subst into form / verv. in formule
                                                                     15 units2 / eenhede2
                                                                                                                   (5)
                                                                                                                  [18]

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### Page 6

6                                                  MATHEMATICS P2                           (EC/SEPTEMBER 2016)

    QUESTION 4 / VRAAG 4

    4.1   x2       8 x 16   y2    6y 9      20                completing square
                    2         2                               kwadraatsvoltooiing
           x 4  y 3               20                                 2          2
                                                               x 4        y 3       20
            M 4;3                                             x-coordinate / x-koördinaat
                                                              y-coordinate / y-koördinaat            (4)
    4.2    0 4
                    2
                        y 3
                              2
                                  20                          subst. x = 0 into circle equation
                    2
                                                              verv. x = 0 in sirkel vergelyking
           y 3        4                                              2
                                                               y 3       4
          y 3 2
            y 1                                               y     1
          Q (0; 1)                                                                                   (3)
    4.3              3 1          1                                       1
          mradius                                             mrad
                      4 0         2                                       2
          mtan/ rkl 2                                         mtan      2 / mraaklyn = 2
          y 1 2 x 0                   [tangent radius]        subst. mtan/ rkl 2 and Q into
                                        [raaklyn              correct form. / verv. mtan/ rkl 2
                                        radius]
               y   2x 1                                       en Q in korrekte formule.
                                                              equation / vergelyking
                                                                                                     (4)
    4.4   y 6                                                 answer / antwoord                      (1)

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### Page 7

(EC/SEPTEMBER 2016)                            WISKUNDE V2                                               7

   4.5    6 2x 1                                         6 = 2x + 1
            5                                                5
          x                                              x
            2                                                2
            5
         U    ;6
            2                                                                             (2)
   4.6         11 6
         mAU
                  5
               0
                  2
                2                                                     mAU        2
                   6 11
          mAD                                                                 1
                   10 0                                               mAD
                                                                              2
                 1
                 2
                            1
          mAU mDA       2
                            2                                         mAU mDA         1
                        1
              AU   DA
               Â 90º                                                 Â    90
          DQ̂U 90º [tangent radius] / [raaklyn radius]                DQˆ U      90
          QUAD is a cyclic quad.[opp. s add up to 180º]               R
         QUAD is ʼn koordevierhoek [ teenoorst. e se som is 180º]                           (6)
                                                                                          [20]

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### Page 8

8                                                   MATHEMATICS P2                           (EC/SEPTEMBER 2016)

    QUESTION 5 / VRAAG 5

    5.1.1    sin    52          sin 52                           − sin 52°
                                                                   answer
                                1 t2                                                                 (3)
    5.1.2    cos 2.19             cos 38                             cos 2.19   cos 38
                   2
             2 cos 19       1 cos 38                                 cos 2.19   2 cos 2 19   1
                                   cos 38     1
                   cos 19                                        simplification /
                                        2                        vereenvoudiging
                                       1 t2   1
                       cos 19
                                         2
                                                                 answer / antwoord                   (4)
    5.2        2 cos(180 x) .sin(180 x).sin 74
             sin( x 360 ).sin 37 .sin 53 .sin( x 90 )
                                                                 − cos x
                  2( cos x) .sin x.sin 74                          sin x
               sin x.sin 37 .cos 37 .( cos x)                      sin x
                  2 sin 74                                        cos37°
                  1                                              − cos x
                    sin 74                                       1
                  2                                                 sin 74
                 4                                                2
                                                                 answer / antwoord
                      OR/OF

               2 cos(1800 x).sin(1800 x).sin 740
             sin( x 3600 ).sin 37 0 .sin 530 .sin( x 900 )
                       2( cos x).sin x.sin 2 37 0                − cos x
                                                                  sin x
                    sin x.sin 37 0 .cos 37 0 .( cos x)            sin x
                         2sin 2 37 0                             cos37°
                                                                 − cos x
                        sin 37 0.cos 37 0
                                                                 2sin 2 370
                        4sin 37 0.cos 37 0
                         sin 37 0.cos 37 0
                                                                 answer / antwoord
                        4
                                                                                                     (7)

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### Page 9

(EC/SEPTEMBER 2016)                        WISKUNDE V2                                          9

   5.3.1     1 cos 2 x 0                             1 cos 2 x        0
             cos 2 x 1
                                                     2x       0    360 .k
                 2 x 0 k .360
                   x 0 k .180 ; k                    x    0       180 .k ;

                                                     k                         (4)
   5.3.2                          2sinx
             L.H.S/LK
                             2 1 cos2x
                            2 sin x
                         2 2 cos 2 x
                               2 sin x               1 2 sin 2 x
                         2 2 1 2 sin 2 x
                              2 sin x
                         2 2 2 sin 2 x               removing brackets
                         2 sin x                     verwyder hakies
                        2 sin 2 x
                          1                          2 sin 2 x
                        sin x                                                 (3)
                                                                             [21]

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### Page 10

10                                                    MATHEMATICS P2                                    (EC/SEPTEMBER 2016)

     QUESTION 6 / VRAAG 6

     6.1   sin x 600           sin 900 2 x                                                       co-ratio / ko-
                                                                                                           funksie
             x 600          900 2 x 3600.k or/of x 600 1800              900 2 x      3600.k
                                                                                                 both gen. solns
           3x 300 3600.k                      or/of        x       300 3600.k                    beide algemene
           x 100 1200.k                                                 ;k   Z                   oplossings
                                                                                                 -30°
             x           300 ; 100 ; 1300
                                                                                                 10°
                                                                                                 130°
                               OR/OF
                                                                                                 co-ratio / ko-
           cos 2 x cos 30 x                                                                                funksie
           2 x 30 x 360 .k               or    2x     30       x   360 .k ; k
           3x       30     360 .k        or      x    30       360 .k                            both gen. solns
                                                                                                 beide algemene
                x        30 ; 10 ; 130                                                           oplossings

                                                                                                 -30°
                                                                                                 10°
                                                                                                 130°                  (5)
     6.2                                                                                 g:
                                                                                                 x-intercept.
                                                                                                 x-afsnit
                                                                                                 y-intercept.
                                                                                                 y-afsnit
                                                                                                 shape / vorm
                                                                                         f:
                                                                                                 x-intercept.
                                                                                                 x-afsnit
                                                                                                 y-intercept.
                                                                                                 y-afsnit
                                                                                                 shape / vorm

                                                                                                                       (6)
     6.3 240°                                                                    answer / antwoord                     (1)
     6.4 h x cos 2 x 90              1 sin 2 x 1                                 substitution / vervanging
                                                                                 sin 2x – 1                           (2)
                                                                                                                     [14]

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### Page 11

(EC/SEPTEMBER 2016)                                                 WISKUNDE V2                                          11

   QUESTION 7 / VRAAG 7

  7.1      NP 2        ON 2          OP 2
                        103
                                 2
                                      145
                                                2                             using Pyth theorem correctly
                                                                              korrekte gebruik van stelling
                                                                              van Pythagoras
           NP          177,86                                                 answer / antwoord                    (2)
  7.2                                                                         subst. into cosine rule
         PQ   2
                       103
                             2
                                     103
                                            2
                                                    2 103 103 .cos(120 )      verv. in cosinus formule

                                                                              answer / antwoord
          PQ           178,40
                                                                                                                   (2)
  7.3                    177,86
                                       2
                                   177,86     178.40
                                                         2          2         substitution / vervanging
           cos N̂
                               2(177,86)(177,86)
                                                                              N̂
                  N̂   60.20                                                                                       (2)
                                                                                                                   [6]

   Copyright reserved                                                                                     Please turn over

### Page 12

12                                                      MATHEMATICS P2                 (EC/SEPTEMBER 2016)

     QUESTION 8 / VRAAG 8
     8.1 Bisects the chord / Halveer die koord                           answer / antwoord       (1)
     8.2

     8.2.1   Q̂ 90 [           in semi circle.] / [   in halwe sirkel]    S R
                                                                          S/R
             L̂1   90 [corresp. s , QN║LO]
                   Q̂
                       [ooreenk. e , QN║LO]
           QL = LP [line from centre perp. to chord]
                   [lynstuk vanaf middelpt is loodreg op koord]           R                        (4)
     8.2.2 MF 8ML                                                                                  (1)
     8.2.3 OP 2 OL2 LP 2                                                  using Pyth correctly
                      2             2                                     korrekte gebruik van
             4M L          3M L              72                           Pythagoras
             16M L2       9M L2         49
                                                                          simplification /
             7M L2        49
                                                                          vereenvoudiging
               ML          7                                              ML                       (3)
                                                                                                   [9]

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### Page 13

(EC/SEPTEMBER 2016)                                    WISKUNDE V2                                  13

   QUESTION 9 / VRAAG 9

   9.1        D̂3       30   [ s opp equal sides] /[   e
                                                           teenoor gelyke sye]   S R
                                                                e                S/R
             Ô1 120         [sum of   s of a ∆.] / [som v/d        van ʼn ]              (3)
   9.2    Â 60     0
                             [ at centre = 2 at circumf.] /                      S   R
                              [middelpunts    = 2 omtrekshoek]                           (2)
   9.3    Ĉ 120             [opp. s of cyclic quad.] / [teenoorst. e van k.v]   S   R   (2)
   9.4    AD̂B 70            [tan chord theorem.] / [raaklyn koord stelling]     S   R   (2)
                                                                                         [9]

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### Page 14

14                                                  MATHEMATICS P2                                (EC/SEPTEMBER 2016)

     QUESTION 10 / VRAAG 10
     10.1                                                                               constr.
                                                                                        konstr.

              Û 2    90   Û1     [tan      radius.] / [raaklyn     radius]                 S/R
              Ẑ 90                [      in semi circle] / [    in halwe sirkel]            S/R

               Ŝ 180      (90   Û1 ) 90          [sum of s of a ∆]                         S/R
                                                   [som van die e van ʼn ]
               Ŝ    Û1
                                                            e
              Ŝ Ŷ          [ s in same segment] / [           in dieselfde segment]        S/R
               Û1    Ŷ                                                                                (5)

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### Page 15

(EC/SEPTEMBER 2016)                                     WISKUNDE V2                                                  15

   10.2

   10.2.1    Â 2   x          [ s opp. = sides] / [    e
                                                            teenoor gelyke sye]                S/R
                                                                                               S/R
             Ĉ 2   Â 2    x [tan chord theo.] / [raaklyn koord stelling]
                                                                                               S/R
             Â1    Ĉ1        [tan chord theo.] / [raaklyn koord stelling]
             Ĉ 2   Ŝ1         [alt.    s , CB ║TS] / [verw.    e
                                                                     , CB ║TS]                 S/R
                                                                        e
             Ĉ1    T̂        [corresp     s , CB║TS] / [ooreenk.           , CB║TS]           S/R       (5)
   10.2.2      Ŝ1 T̂ x [proven in 10.2.1] / [bewys in 10.2.1]                                 S
              CS = CT   [sides opp. = s ] / [sye teenoor gelyke hoeke]                         R         (2)
   10.2.3    AR AT                                                                     S/R
                       [line ║to one side of a ∆]
             BR CT
                       [lynstuk ║ aan een sy van ʼn ]
                                                                                       CS = CT
            CS = CT            [proved in 10.2.2] / [bewys in 10.2.2]
                                                                                       substitution
                      3
             AT         4                                                              vervanging
                      2
                                                                                       AT
             AT 6 cm                                                                                     (4)
                                                                                                        [16]

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### Page 16

16                                              MATHEMATICS P2                                     (EC/SEPTEMBER 2016)

     QUESTION 11 / VRAAG 11
                                                F
                                                          R           Z
                                                                  C
                                                       B
                                                                               P
                                                                      D
                                                      A

                                                              N

     11.1     Â1    Ĉ 2      [tan chord theo.] / [raaklyn koord stelling]                  S     R
                     Ẑ         [tan chord theo] / [raaklyn koord stelling]
                                                                                             S/R
              Ĉ 2 Ẑ           [both = x] / [beide = x]
              BC ║ RZ           [corresp. s =] / [ooreenk.            e
                                                                          =]                 R
                                                                                                                    (4)
     11.2   Ẑ P̂        [ s in same segment] / [ e in dieselfde segment]                    S/R
                                                                                             S
               Ĉ 2      [corresp. s ; BC ║ RZ] / [ooreenk. e ; BC ║RZ]
              BC is a tangent to circle ACP [conv. of tan chord theorem]
              BC is ʼn raaklyn aan sirkel ACP [ omgekeerde van raaklyn
                                               koord stelling]                               R                      (3)
     11.3     B̂1   D̂ 2        [ext.   of a cyclic quad.]                         S     R
                                [buite van koordevierhoek]
                                                                                   S/R
              R̂    B̂1         [corresp. s , BC║RZ]
                                [ooreenk. e , BC║RZ]                               S/R
              R̂    D̂ 2
                                                                                   R
              Ẑ P̂             [ s in same segment]
                                [ e in dieselfde segment]
             Â 2 Ĉ 4          [3rd ] / [3de ]
             ∆ RZA ||| ∆ DPC [equiangular or              ]
                                [gelykhoekig of           ]                                                         (5)
     11.4   ZA RA
                                      [similar ∆s] / [gelykvormige e]              S/R
            PC DC
                   ZA DC                                                                     ZA DC
             RA                                                                    RA
                             . . . .(1)                                                        PC
                       PC
            AR AZ
                                           [line ║ to one side of a ∆]             S/R
            AB AC
                                      [lynstuk ║ aan een sy van ʼn ]
                   AZ AB
             AR             . . .(2)
                       AC                                                                    AZ AB
                                                                                   RA
            ZA DC AZ AB                                                                       AC
              PC          AC
             DC AC                                                                 simplification
                         1
             PC AB                                                                                                  (5)
                                                                                                                   [17]
                                                                                       TOTAL/TOTAAL:               150

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