# Mathematics Grade 11 Paper 1 Eastern Cape November 2016 Memo

> Grade 11 Mathematics Paper 1 memorandum from the Eastern Cape November 2016, 8 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/maths-p1-memo/74dcfd0f-d58b-4986-997e-6d6db1f8180b
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Document type: Memorandum
- Year: 2016
- Exam period: Eastern Cape November
- Paper: 1
- Curriculum: CAPS
- Pages: 8

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Mathematics · Grade 11 · Eastern Cape November · 2016. Memorandum, 8 pages. Read online or download the PDF.

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### Page 1

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                             NATIONAL
                        SENIOR CERTIFICATE/
                             NASIONALE
                         SENIOR SERTIFIKAAT

                             GRADE/GRAAD 11

                           NOVEMBER 2016

               MATHEMATICS P1/WISKUNDE V1
                    MEMORANDUM

MARKS/PUNTE:                 150

       This memorandum consists of 8 pages./Hierdie memorandum bestaan uit 8 bladsye.

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2                                     MATHEMATICS P1                              (EC/NOVEMBER 2016)

NOTE/LET OP:
 If a candidate answered a QUESTION TWICE, mark the FIRST attempt ONLY.
  Indien ŉ kandidaat ŉ vraag TWEE keer beantwoord het, merk SLEGS die EERSTE poging.
 Consistent accuracy applies in ALL aspects of the memorandum.
  Volgehoue akkuraatheid geld deurgaans in ALLE aspekte van die memorandum.
 If a candidate crossed out an attempt of a QUESTION and did not redo the question, mark the
  crossed-out attempt.
  Indien ŉ kandidaat ŉ poging vir ŉ vraag deurgetrek het en nie die vraag weer beantwoord het
  nie, merk die poging wat deurgetrek is.
 The mark for substitution is awarded for substitution into the correct formula.
  Die punt vir substitusie word toegeken vir substitusie in die korrekte formule.

QUESTION/VRAAG 1
1.1.1 (𝑥 + 2)2 = 1
      𝑥 + 2 = ±1                                        ±1
      𝑥 = −1 or/of 𝑥 = −3                               𝑥 = −1
      OR/OF                                             𝑥 = −3
      (𝑥 + 2)2 = 1
      𝑥 2 + 4𝑥 + 4 − 1 = 0
      𝑥 2 + 4𝑥 + 3 = 0                                  standard form/standaardvorm
      (𝑥 + 3)(𝑥 + 1) = 0                                factors/faktore
      𝑥 = −3 or/of 𝑥 = −1                               both x-values/beide x waardes           (3)
1.1.2 2𝑥 2 − 11𝑥 − 4 = 0
           −𝑏 ± √𝑏 2 − 4𝑎𝑐
      𝑥=
                 2𝑎
                                                        sub into correct formula/vervanging
           11 ± √(−11)2 − 4(2)(−4)
      𝑥=                                                 in korrekte formule
                      2(2)                              simplification/vereenvoudiging
           11 ± √153                                    𝑥 = 5,84
      𝑥=
               4                                        𝑥 = −0,34
      𝑥 = 5,84 or/of 𝑥 = −0,34                                                            (4)
1.1.3       1
      𝑥2 >     ;𝑥 < 0
                4
            2   1
          𝑥 −4>0
                1
          (𝑥 − 2) (𝑥 + 2) > 0
                              1                           factors/faktore

                                                          solutions/oplossings

                                                          critical values and method/kritiese
                          1           1                    waarde en metode
                        −
                          2           2
                                                                  1
                1                 1                       𝑥 < − 2 only/alleenlik
          𝑥 > 2 or/of 𝑥 < − 2

                    1
          𝑥 < −2                                                                                 (4)
1.1.4     𝑥 + 5 = √3 − 3𝑥
          𝑥 2 + 10𝑥 + 25 = 3 − 3𝑥                         squaring both sides/kwadrering
          𝑥 2 + 13𝑥 + 22 = 0                               beide kante
          (𝑥 + 11)(𝑥 + 2) = 0                             standard form/standaard vorm
          𝑥 = −2 𝑜𝑟/𝑜𝑓 𝑥 = −11                            both x values/beide x waardes
          𝑥 = −2                                          𝑥 = −2                         (4)

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(EC/NOVEMBER 2016)                       WISKUNDE V1                                                3

1.2.1      𝑦 2 − 9𝑥 2
           (𝑦 − 3𝑥)(𝑦 + 3𝑥)                                factors/faktore                  (1)
1.2.2      𝑦 + 3𝑥 = 2      … … … … … … … … (1)
           𝑦 2 − 9𝑥 2 = 16    … … … … … … … … … . (2)

           From/Vanaf (2)
           (𝑦 − 3𝑥)(𝑦 + 3𝑥) = 16
           Sub (1) into/in (2)
           2(𝑦 − 3𝑥) = 16                                  sub 2 for (𝑦 + 3𝑥)vervanging van 2
           From/Vanaf (1)                                   vir (𝑦 + 3𝑥)
           𝑦 = 2 − 3𝑥                                      making y the subject/maak y die
           2(2 − 3𝑥 − 3𝑥) = 16                              onderwerp.
           4 − 12𝑥 = 16                                    sub y/vervanging van y.
           12𝑥 = −12                                       x value/x-waarde
           𝑥 = −1
           𝑦 = 2 − 3(−1)                                   y value/y-waarde
           𝑦=5

           OR/OF

           From/vanaf (1) 𝑦 = 2 − 3𝑥
           Sub into (2)/vervang in (2)

           (2 − 3𝑥)2 − 9𝑥 2 = 16
           4 − 12𝑥 + 9𝑥 2 − 9𝑥 2 = 16                      𝑦 = 2 − 3𝑥
           4 − 12𝑥 = 16                                    Substitution/vervanging
           12𝑥 = −12                                       Simplification/vereenvoudiging
           𝑥 = −1                                          x value/x-waarde
           ∴𝑦=5                                            y value/y-waarde               (5)
                                                                                            [21]

QUESTION/VRAAG 2
2.1         2
       𝑎3
      ( )
        2
         𝑎6
      =                                                    answer/antwoord                  (1)
          4
2.2   2𝑥−3 − 3. 2𝑥−1
           2𝑥−2
             1   3
        2𝑥 (8 − 2)                                         2𝑥 common factor/gemene faktor.
                                                            1   3
      =        1
                                                             −2
                                                            8
          2𝑥 (4)                                               1
                                                           2𝑥 . 4
            11
      = − or/of −5 12 or/of −5,5                           answer/antwoord                  (4)
             2
2.3   10𝑥+3
      = 10𝑥 . 103                                          sub/vervang 10𝑥 = 1,5
      = 1,5 × 1000
      = 1500                                               answer/antwoord                  (2)

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4                                        MATHEMATICS P1                           (EC/NOVEMBER 2016)

2.4.1     2𝑥 = 0,125
          2𝑥 = 2−3                                           0,125 𝑎𝑠 2−3
          𝑥 = −3                                             answer/antwoord                    (2)
2.4.2                         9
          (0,5)𝑥 . √1 +           = 10
                          16
           1 𝑥       25                                        25
          (2) . √16 = 10                                     √
                                                               16
                                                                    4
           1 𝑥            4                                  10 × 5
          (2) = 10 × 5
           1 𝑥                                               2−𝑥
          (2) = 8
          2−𝑥 = 23                                           23
          ∴ 𝑥 = −3                                           𝑥 = −3                            (5)
                                                                                               [14]
QUESTION/VRAAG 3
3.1   2𝑥(𝑥 + 1) + 𝑚 = 𝑥
      2𝑥 2 + 2𝑥 + 𝑚 − 𝑥 = 0
      2𝑥 2 + 𝑥 + 𝑚 = 0                                       standard form/standaardvorm
      ∆= 𝑏 2 − 4𝑎𝑐                                           sub into ∆/vervang in ∆
      = 1 − 4(2)(𝑚)
      = 1 − 8𝑚                                               ∆= 1 − 8𝑚
      For non-real roots/vir nie-reële wortels ∆< 0          ∆< 0
      1 − 8𝑚 < 0
                                                                    1
      8𝑚 > 1                                                 𝑚>8                                (5)
           1
      𝑚>8
3.2             √𝑥+2                                         5 − 𝑥2 = 0
          𝑓(𝑥 ) =                                            𝑥+2<0
                     5−𝑥 2
          Undefined/ongedefinieerd                           𝑥 < −2
          5 − 𝑥 2 = 0 or/of 𝑥 + 2 < 0                        𝑥 = √5
          𝑥 = ±√5 or/of      𝑥 < −2                          or/of                              (5)
                                                                                               [10]
QUESTION/VRAAG 4
4.1.1 22 ; 27                                                22
                                                             27                                 (2)
4.1.2     𝑇𝑛 = 5𝑛 + 2                                        5𝑛
                                                             2                                  (2)
                          5
4.1.3     5𝑛 + 2 = 12                                        5𝑛 + 2 = 125
          5𝑛 = 125 − 2                                       125 − 2
          𝑛 = 49766                                          𝑛 = 49766                          (3)
          𝑇49766 = 125
4.1.4     General Term is multiples of 5 plus 2 therefor
          the numbers in the sequence will always end        multiples of 5 plus 2/veelvoude van
          in a 2 or a 7 as multiples of 5 end with 0 or a     5 plus 2
          5. Die algemene term is veelvoude van 5 plus       ending in 2 or 7/eindig in 2 of 7
          2 en daarom sal die patroon altyd eindig in ŉ
          2 of ŉ 5 omdat die veelvoude van 5 altyd
          eindig op 0 of 5

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(EC/NOVEMBER 2016)                      WISKUNDE V1                                                      5

4.2.1      39                                              39                                    (1)
4.2.2      2𝑎 = 2
           𝑎=1                                                𝑎=1
           3𝑎 + 𝑏 = 6                                         3𝑎 + 𝑏 = 6
           𝑏=3                                                𝑏=3
           𝑎+𝑏+𝑐 =3                                           𝑎+𝑏+𝑐 =3
           𝑐 = −1                                             𝑐 = −1                             (5)
           𝑇𝑛 = 𝑛2 + 3𝑛 − 1
4.2.3      𝑛2 + 3𝑛 − 1 = 269                                  𝑛2 + 3𝑛 − 1 = 269
           𝑛2 + 3𝑛 − 270 = 0
           (𝑛 − 15)(𝑛 + 18) = 0                               factors/faktore
           𝑛 = 15
           𝑇16 = 162 + 3(16) − 1                              𝑛 = 15
           𝑇16 = 303                                          𝑇16 = 303                          (4)
4.3        ….. ; 6 ; 2 ; x ; -18

                                                              Method/metode
                     -4   (𝑥 − 2) (−18 − 𝑥)

                 (𝑥 − 2 + 4) [−18 − 𝑥 − (𝑥 − 2)]
4.3.1      𝑥 + 2 = −2𝑥 − 16                                   setting up of equation/opstel van
           3𝑥 = −18                                            vergelyking
           𝑥 = −6                                             𝑥 = −6
           𝑇4 = −6                                            second diff/tweede verskil        (4)
           𝑑2 = −4
4.3.2      𝑑1 = 0                                             𝑑1 = 0
           𝑇1 = 6                                             𝑇1 = 6                             (3)
                                                                                                 [26]
QUESTION/VRAAG 5
5.1   𝑥=8                                                    𝑥=8
      𝑦=4                                                    𝑦=4                                 (2)
5.2   Domain     𝑥 ∈ 𝑅 ;𝑥 ≠ 8                                domain/gebied
      Range      𝑦 ∈ 𝑅 ;𝑦 ≠ 4                                range/terrein                        (2)
5.3

                              4
                                                             shape/vorm
                              3                              asymptotes/asimptote
                                    6         8              x-intercept/x-afsnit
                                                             y-intercept/y-afsnit

                                                                                                   (4)
5.4.1       8
                 ≥ −4
           𝑥−8
            8
              +4≥0
           𝑥−8
           ∴ 𝑓(𝑥) ≥ 0                                        𝑥≤6
                                                             𝑥>8
           𝑥 ≤ 6 or/of 𝑥 > 8                                 or/of                                (3)

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6                                   MATHEMATICS P1                               (EC/NOVEMBER 2016)

5.4.2     𝑓(𝑥) ≤ 3                                        𝑥≥0
          0≤𝑥<8                                           𝑥<8
                                                          notation/notasie                      (3)
5.5       𝑦 =𝑥−8+4                                        𝑦 =𝑥−8+4
          𝑦 =𝑥−4                                          𝑥−4                                   (2)
5.6       𝑔(𝑥) = 𝑓(𝑥 − 2) − 2                             𝑥 − 10
                  8
          𝑔(𝑥) = 𝑥−10 + 2                                 +2                                    (2)
                                                                                                  [18]
QUESTION/VRAAG 6
6.1   𝑦 = 𝑎(𝑥 − 𝑥1 )(𝑥 − 𝑥2 )
      𝑦 = 𝑎(𝑥 + 5)(𝑥 + 1)
      2 = 𝑎(0 + 5)(0 + 1)                           (𝑥 + 5)(𝑥 + 1)
      𝑎=5
          2                                         sub of/vervang van (𝟎; 𝟐)
              2                                     a-value/a-waarde
         𝑦 = 5 (𝑥 2 + 6𝑥 + 5)                       equation/vergelyking            (4)
              2   2   12
         𝑦 = 5𝑥 + 5 𝑥 + 2
6.2      𝑦 = 𝑘. 𝑚 𝑥
         2 = 𝑘. 𝑚0
         𝑘=2                                      𝑘=2
         𝑦 = 2. 𝑚 𝑥                               Sub of/vervang van (1;6)
         6 = 2. 𝑚                                 𝑚=3
         𝑚=3           ∴ 𝑔(𝑥) = 2. 3𝑥                                                              (3)
6.3      𝑦=0                                      answer/antwoord                   (1)
6.4.1    x-value of turning point/x-waarde van
         draaipunt. 𝑥 = −3                        𝑥 = −3
         decreasing/dalend 𝑥 < −3                 𝑥 < −3
         OR/OF (−∞ ; −3)                                                                   (2)
6.4.2    0≤𝑥≤1                                      𝑥≥0
                                                    𝑥 ≤ −1                                      (2)
6.4.3    𝑥≤0                                        answer/antwoord                             (2)
6.4.4    −5 < 𝑥 < −1                                𝑥 > −5
                                                    𝑥 < −1                                      (2)
6.5      (−5 ; 0) and 𝑦 − 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡. (0 ; 2)

            𝑦2 − 𝑦1                               formula/formule
         𝑚=
            𝑥2 − 𝑥1                               substitution/vervanging
             0−2                                  answer/antwoord
         𝑚=
            −5 − 0
            2
         𝑚=
            5                                                                                      (3)
                                                                                                  [19]
QUESTION/VRAAG 7
7.1   𝑦 = 𝑏𝑥                                            𝑦 = 𝑏 𝑥−2
      𝑦 = 𝑏 𝑥−2 + 4                                     +4
      8 = 𝑏 4−2 + 4                                     sub of/vervang van (4;8)
      4 = 𝑏2                                            𝑏=2                                       (4)
      𝑏=2                                              
7.2   𝑦 = 2𝑥−2 + 4                                     equation/vergelyking                        (1)
                                                                                                   [5]

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(EC/NOVEMBER 2016)                      WISKUNDE V1                                              7

QUESTION/VRAAG 8
8.1  𝐴 = 𝑃(1 − 𝑖)𝑛                                      formula/formule
                           25  3                        subst/vervanging
          𝐴 = 7 200 (1 −       )                        answer/antwoord
                           100
          𝐴 = 𝑅 3 037,50                                                           (3)
8.2       𝐴 = 𝑃(1 + 𝑖)𝑛
                                 𝑖 12                      formula/formule
          1 126,10 = 500 (1 + )                            subst/vervanging
                                2
          1 126,10          𝑖 12                           simplification/vereenvoudiging
                    = (1 + )                               answer/antwoord
             500            2                                                            (4)
          𝑖 = 0,14
          Interest rate/rentekoers = 14%
8.3.1                    𝑖𝑛 𝑚
          1 + 𝑖𝑒 = (1 + )                               formula/formule
                         𝑚
                      7,2 12                            subst/vervanging
          𝑖𝑒 = (1 +       ) −1                          answer/antwoord
                     1200
          𝑖𝑒 = 0,07442 …
          𝑟 = 7,442 %                                                                     (3)
8.3.2     𝐴 = 𝑃(1 + 𝑖)𝑛                                 formula/formule
          𝐴 = 120 000(1 + 7,442%)3                      subst/vervanging
          𝐴 = 148 834,46                                answer/antwoord
                                                                                          (3)
8.3.3                              48
                              7,2
          𝐴 = 120 000 (1 +        )
                             1200
          𝐴 = 𝑅159 913,20
                             7,2 30
          𝐴 = 20 000 (1 +        )                             R 159913,20
                            1200
          𝐴 = 23 931,47
          𝐹𝑖𝑛𝑎𝑙 𝑎𝑚𝑜𝑢𝑛𝑡 𝑅159913,20 − 23 931,47                  R 23931,47
          𝐹𝑖𝑛𝑎𝑙𝑒 𝑏𝑒𝑑𝑟𝑎𝑔        = 𝑅 135 981,73                  subtracting/aftrek
                                                               answer/antwoord
          OR/OF

                             7,2 18                7,2 30
          𝐴 = [120 000 (1 +      ) − 20 000] (1 +      )
                            1200                  1200
                                                                                           (4)
          𝐴 = 𝑅 135 981,73
                                                                                         [17]

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8                                              MATHEMATICS P1                              (EC/NOVEMBER 2016)

QUESTION/VRAAG 9
9.1
                                           P            PP
                                      2
                                      9                                 1st branch of tree
                                                                         diagram/1ste tak van boom
                                       2                                 diagram
                      3                                 PG
                          P            9 G                              values on first
                     10
                                                                         branch/waardes op 1ste
                                       5
                                                                         tak
                                       9 B              PB              2nd branch/2de tak
                                      3                                 values on 2nd
                       2              9 P               GP
                                                                         branch/waardes op 2de
                         G                 1
                      10                 G              GG               tak
                                       5   9
                                                                        first four outcomes/eerste
                                       9                                 vier uitkoms
                      5               3 B               GB
                                                                        last 4 outcomes/laaste 4
                     10               9 P                                uitkomste
                          B
                                           G     2
                                       4         9      BG
                                       9
                                           B            BB                                                (6)

9.2.1               5 4
          𝑃(𝐵𝐵) =     ×                                                    5       4
                   10 9                                                 10 × 9
                   20
          𝑃(𝐵𝐵) =     = 1,333 ….                                        answer/antwoord                 (2)
                   90
9.2.2           2 3        3 2
          𝑃 =( × )+( × )                                                       2       3
               10 9       10 9                                          (10 × 9)
               6    6
          𝑃=     +                                                             3       2
              90 90                                                       (10 × 9)
              12                                                          simplify/vereenvoud,
          𝑃=
              90                                                          adding/optel
          𝑃 = 0,133                                                       answer/antwoord
                                                                                                 (5)
9.3.1                              Sport       No sport      Total      labels/opskrifte
                                               Geen sport    Totaal     boys row/seuns ry
           Boys/Seuns                50            10             60    girls row/meisies ry
           Girls/Meisies             40            20             60    column totals/totale van
           Total/Totaal              90            30            120     kolomme
                                                                        120                     (5)
9.3.2            40           1                                         answer/antwoord
          (a)             =                                                                      (1)
                120           3
                 10            1                                        answer/antwoord
          (b)             =                                                                      (1)
                120           12
                                                                                                [20]

                                                                               TOTAL/TOTAAL: 150

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- [MATHS P1 GR11 MEMO NOV 2020 English+Afrikaans](https://www.hlayiso.com/library/maths-p1-gr11-memo-nov-2020-english-afrikaans/89544f20-6bf0-4ca5-b1c6-bb6abeb0c3cd) — Mathematics · Memo · 2020 · Afr
- [MATHS P1 MEMO GR11 NOV2015 EA](https://www.hlayiso.com/library/maths-p1-memo-gr11-nov2015-ea/52b2b17e-c884-403e-a909-21e71ca2ee83) — Mathematics · Memo · 2015
- [Mathematics P1 Common Paper Grade 11 2014 Memo Afr](https://www.hlayiso.com/library/mathematics-p1-common-paper-grade-11-2014-memo-afr/17a6673b-39dc-4c97-96c7-67251b603545) — Mathematics · Memo · 2014 · Afr

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## Collection

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

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Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. When citing, link the page URL above, not a PDF or /api/ URL.
