# Mathematics Grade 11 Paper 2 Gauteng June 2017 Memo

> Grade 11 Mathematics Paper 2 memorandum from the Gauteng June 2017, 10 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/gr11-math-p2-june-2017-possible-answers/1b7113c4-1c6c-4a95-a99d-05b3d3d421bd
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Document type: Memorandum
- Year: 2017
- Exam period: Gauteng June
- Paper: 2
- Curriculum: CAPS
- Pages: 10

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## About this document

Mathematics · Grade 11 · Gauteng June · 2017. Memorandum, 10 pages. Read online or download the PDF.

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## Document text

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### Page 1

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           GAUTENG DEPARTMENT OF EDUCATION
               PROVINCIAL EXAMINATION
                     JUNE 2017
                      GRADE 11

                    MATHEMATICS

                       PAPER 2

                    MEMORANDUM

10 pages

                          1

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                                                                         Mathematics Paper 2
                                         MEMORANDUM
                                                                                GRADE 11

                            GAUTENG DEPARTMENT OF EDUCATION
                                   PROVINCIAL EXAMINATION

                                                             MATHEMATICS Paper 2

                                                             MEMORANDUM

INSTRUCTIONS AND INFORMATION

A – Accuracy
CA – Consistent Accuracy
S – Statement
R – Reason
SR – Statement / Reason

NOTE:
   •   If a candidate answered a question TWICE, mark only the first attempt.
   •   If a candidate CROSSED OUT an answer and did not redo it, mark the crossed out answer.
   •   Consistent accuracy applies to ALL aspects of the memorandum. Stop marking at the second
       calculation error.
   •   Assuming values/answers in order to solve a problem is UNACCEPTABLE.

                                                  2

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                                                             Mathematics Paper 2
                                    MEMORANDUM
                                                                  GRADE 11

QUESTION 1                                                                 MARKS: 25
1.1   MMR =                                 ✓ sub. into the gradient            (3)
                                            formula

      MMR =                                 ✓ gradient of line MR

      equation of line MR is:               ✓ equation of line MR
1.2                                                                             (4)
                       y = 5x -14           ✓ mMS = 5
      MS || PR  mPR = 5                    ✓mPR = 5
      equation of line PR:

                                            ✓sub. (2 ; 4)
                                            ✓answer
1.3      mPR = 5                                                                (5)
                                            ✓
                                            ✓
       mMR =

                                            ✓
                                            ✓                 )..(sum of
                                            angles of ∆)
                                            ✓
1.4             and                                                             (4)

                                            ✓equating

                                            ✓
                                            ✓
                                            ✓
      R(4 ; 6)

                                        3

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                                                             Mathematics Paper 2
                                  MEMORANDUM
                                                                   GRADE 11

1.5       =                                                                        (2)
          =                                    ✓sub. into dist. formula
          =                                    ✓answer
          =
1.6      =                                                                         (5)
          =                                    ✓ sub. into dist. formula
                                               ✓        OR
          =      or
      Area of PMR
                                               ✓                           OR
      =
      =                 OR
                                               (✓ rounding-off) answer ✓

      = 28
1.7   S(2 ; 4)                                ✓x=2                                (2)
                                               ✓ y = 4

QUESTION 2                                                                 MARKS: 26
2.1                                            ✓diagram correct in correct quad.  (4)
                                   y
                                               ✓
                                  0        x   ✓
                             25
                                               ✓answer
                             
                       (-7: y)

2.2                                                                                (6)

                                               ✓ (-sinx) (-sinx)
                                               ✓
                                               ✓
                                               ✓
                                               ✓              =
                                               ✓ answer

                                       4

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                                                     Mathematics Paper 2
                            MEMORANDUM
                                                         GRADE 11

2.3                                   ✓                                    (6)
                                      ✓
      Ref. angle                      ✓

                             or

                                      ✓
                                      ✓ 2θ = 180° + 77,2909° + k.360°
                                      ✓

2.4                                                                        (5)

                                      ✓
                                      ✓

                                      ✓
                                      ✓
                                      ✓

2.5                                   ✓                                    (5)
                                      ✓sin   =

                                      ✓          =

                                      ✓
                                      ✓answer

                                  5

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                                                 Mathematics Paper 2
                                    MEMORANDUM
                                                       GRADE 11

QUESTION 3                                                             MARKS:14

3.1                                                                         (6)

                                                 ✓ y-intercepts of f
                                                 ✓ y-intercepts of g
                                                 ✓✓ both turning points
                                                   of g
                                                 ✓ both x-intercepts of f
                                                 ✓ both turning points
                                                   of f

3.2   2                                          ✓✓ answer                  (2)
3.3     2 cos x +1 = 1 – sin x                   ✓ tan x = -2               (6)
             sin x = -2 cos x                   ✓ 63, 4349°
             tan x = -2                         ✓ x = 116, 57° +
       Ref. angle = 63, 4349....°                 k.180° with k є Z
       x = 180° - 63, 4349....° + k. 180°       ✓ -63,43°
       x = 116, 57° + k.180° with k є Z         ✓ 116, 57°
       x = -63,43° OR 116,57° OR 296,57°        ✓ 296,57°

                                             6

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                                                             Mathematics Paper 2
                                         MEMORANDUM
                                                                   GRADE 11

                               QUESTION 4                               MARKS: 13

                                         O

                                                             ✓ construction
                                     1       2                 (radii, OP and OR)
                       P                 Q               R
4.1

                       Construction: Join PO and OR

                            In ∆POQ and ∆ ROQ
                              1) OQ = OQ..... common
                              2) PO = OR ......radii         ✓SR ie (PO = OR ..... radii)
                              3) Q̂1  Q̂2  90 .. given    ✓S ie (Q̂1  Q̂2  90)

                                                 90 HS       ✓SR
                                                             ✓ PQ =RQ
                                                                                            (5)

        Â1  90 ..line from centre to midpt. of chord      ✓SR

4.2.1
        B̂  180  (Ô1  Â1 ) .... sum of angles of a Δ   ✓ answer
           = 50°                                                                            (2)

                                                  7

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                                                       Mathematics Paper 2
                                       MEMORANDUM
                                                             GRADE 11

        Ĉ  90 ... angle in a semi-circle            ✓SR
4.2.2
        D̂  180  (Ĉ  B̂) ... sum of angles of Δ
                                                       ✓ answer
          = 40°                                                                    (2)

                             ......Th. of Pyth.
            =
            = 50                                       ✓ BD = 50

        BO =        .....radii
4.2.3       = 25                                       ✓ BO = 25

        OA =       ...Midpt. Th.
            = 20                                       ✓ SR ie (OA = 20 .. Midpt.Th.)

        AE = OE – OA
           = 25 – 20
           = 5 units                                   ✓ AE = 5
                                                                                   (4)

                                                  8

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                                                       Mathematics Paper 2
                                         MEMORANDUM
                                                             GRADE 11

                               QUESTION 5                            MARKS: 10

      L̂  Ĵ1  x ... angles opp. = sides             ✓SR

5.1                                                    ✓S
      N̂1  L̂  x ...= chords subtend = angles        ✓R
                                                                                  (3)

      M̂  Ĵ 3  y ........... angles opp. = sides    ✓SR
                                                       ✓SR
      K̂1  M̂  y ........= chords subtend = angles

      Q̂2  K̂1  Ĵ1 ... ext. angle of Δ              ✓SR
5.2       = x +y

      P̂2  N̂1  Ĵ 3 ... ext. angle of Δ             ✓SR
          =x+y

              Q̂2  P̂2  x  y                                                  (4)

      P̂4  P̂2 .... vert. opp. angles                 ✓ S R for P̂4  P̂2   OR
5.3
      Q̂4  Q̂2 .... vert. opp. angles                   Q̂4  Q̂2

      but   Q̂2  P̂2 ........proved in Q5.2
              P̂4  Q̂4                               ✓ P̂4  Q̂4

       JQ = JP...... sides opp = angles OR isos. ∆    ✓R
                                                                                  (3)

                                                  9

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                                                                Mathematics Paper 2
                                           MEMORANDUM
                                                                         GRADE 11

                                QUESTION 6                                 MARKS: 12

      K̂ 2  T̂1 ...... angles opp. = sides OR equal radii     ✓SR
      T̂1  Ĉ ......... angles in same segment                ✓ T̂  Ĉ
6.1
                                                               ✓R
                   K̂ 2  Ĉ                                                            (3)

      T̂2  Â ...... angles opp. = sides OR equal radii       ✓SR

                                                               ✓SR
      P2  T̂2  Â ... ext. angle of
6.2       = 2 T̂2
                                                               ✓ K̂1  P̂2
      K̂1  P̂2 ........... angles in same segment             ✓R
      K̂1  2T̂2                                                                         (4)

      P̂4  2T̂ ........    angle at centre = twice angle at   ✓ P̂4  2T̂
                            circumference                      ✓R

      T̂  T̂1  T̂2

      T̂1  Ĉ ....... angles in same segment                  ✓SR
6.3
              1                                                          1
      T̂2      K̂1 ..... proved in Q6.2                       ✓ T̂2      K̂1
              2                                                          2

                   1                                                         1
      P̂4  2(Ĉ  K̂1 )                                       ✓ P̂4  2(Ĉ  K̂1 )
                   2                                                         2
          = 2Ĉ  K̂1 )
                                                                                         (5)
                                                                                 TOTAL: 100

                                                  10

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