# Mathematics Grade 11 Paper 1 Gauteng June 2017 Memo

> Grade 11 Mathematics Paper 1 memorandum from the Gauteng June 2017, 11 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/gr11-math-p1-june-2017-possible-answers/19ca7dfb-d358-4c42-bbfc-c68d580cfe21
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Document type: Memorandum
- Year: 2017
- Exam period: Gauteng June
- Paper: 1
- Curriculum: CAPS
- Pages: 11

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## About this document

Mathematics · Grade 11 · Gauteng June · 2017. Memorandum, 11 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 1

                   MEMORANDUM

11 pages

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                                                                    MATHEMATICS
                                         MEMORANDUM
                                                                    (Paper 1)            GRADE 11

                           GAUTENG DEPARTMENT OF EDUCATION
                                   PROVINCIAL EXAMINATION

                                                                  MATHEMATICS
                                                                  (Paper 1)

                                                                  MEMORANDUM

INFORMATION

A – Accuracy

CA – Continued Accuracy

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 marking memorandum.

   •   Assuming values/answers in order to solve a problem is UNACCEPTABLE.

                                                  2

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                                                              MATHEMATICS
                                            MEMORANDUM
                                                              (Paper 1)              GRADE 11

                                 QUESTION 1                                  MARKS: 19

                                       𝑥 2 − 2𝑥 = 15
                                𝑥 2 − 2𝑥 − 15 = 0                ✓ std. form
1.1.1                            (x – 5)(x + 3) = 0              ✓ correct factors
                                     x = 5 or x = -3             ✓ both solutions
                                                                                                (3)

                                  4𝑥 2 − 𝑥 − 5 < 0
                              (4𝑥 − 5)(𝑥 + 1) < 0                ✓ factors

1.1.2                           +           -        +
                                     -1         1¼

                                    −1 < 𝑥 < 14
                                                     1           ✓ critical values
                                                                 ✓ correct inequality
                                                                                                (3)

                              √2𝑥 − 1 + 2 = 𝑥
                               (√2𝑥 − 1 )2 = (𝑥 − 2)2            ✓ squaring both sides
                                   2𝑥 − 1 = 𝑥 2 − 4𝑥 + 4
                                         0 = 𝑥 2 − 6𝑥 + 5        ✓ std. form
1.1.3
                                         0 = (𝑥 − 5)(𝑥 − 1)      ✓ factors
                                         𝑥 = 5 or 𝑥 = 1          ✓ both solutions
                                    Test: x = 5 only             ✓ accepting x = 5
                                                                                                (5)

                                     2
                                 2𝑥 3 – 8 = 0
                                           2
                                      2𝑥 3 = 8                        2
                                           2                     ✓ 𝑥3 = 4
                                          𝑥3 = 4
                                          𝟐 𝟑            𝟑       ✓ raising both sides to the
                                    (𝒙𝟑 )𝟐 = ± (𝟒)𝟐                power
                                                                           3
                                                                             2
                                           𝑥 = ± 23 OR
1.1.4                                                            ✓x = ±23 𝐎𝐑 𝑥 = ±8
                                           𝑥 =±8
        OR                                                       OR

            2                                                         2
        𝑥3 = 4                                                  ✓ 𝑥3 = 4
         x   4
        3       2
                    3
                          3                                     ✓ raising both sides to the
                                                                  power of 3
        x2 = 64
        x=±8                                                     ✓ x = ±23 𝐎𝐑 𝑥 = ±8
                                                                                                (3)

                                                         3

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                                                         MATHEMATICS
                                        MEMORANDUM
                                                         (Paper 1)             GRADE 11

                               𝑎+𝑏+𝑐 =0
                                     𝑏 = −𝑎 − 𝑐            ✓ 𝑏 = −𝑎 − 𝑐
                             2
                          𝑎𝑥 + 𝑏𝑥 + 𝑐 = 0
                           −𝑏 ± √𝑏 2 − 4𝑎𝑐
                       𝑥=
                                  2𝑎
                     −(−𝑎 − 𝑐) ± √(−𝑎 − 𝑐)2 − 4𝑎𝑐
                   =                                        ✓ sub. in corr. formula
                                  2𝑎
                     𝑎 + 𝑐 ± √𝑎2 + 2𝑎𝑐 + 𝑐 2 − 4𝑎𝑐
                   =
                                  2𝑎
 1.2
                          𝑎 + 𝑐 ± √𝑎2 − 2𝑎𝑐 + 𝑐 2           ✓ simplification to
                      =
                                    2𝑎                       √𝑎2 − 2𝑎𝑐 + 𝑐2

                              𝑎 + 𝑐 ± √(𝑎 − 𝑐)2
                          =                                 ✓ (𝑎 − 𝑐)2
                                     2𝑎

                                𝑎 + 𝑐 ± (𝑎 − 𝑐)
                           =
                                      2𝑎
                                             𝑐              ✓x=1
                              ∴ 𝑥 = 1 or 𝑥 =
                                             𝑎                                                 (5)

                                QUESTION 2                             MARKS: 15

                          3𝑦 + 𝑥 = 2 … … … … . . (1)
                          𝑦 2 + 𝑥 = 𝑥𝑦 + 𝑦.........(2)
       from (1)                   𝑥 = 2 − 3𝑦                ✓ 𝑥 = 2 − 3𝑦
                          subst. 𝑥 = 2 − 3𝑦 in (2)
                   2
                  𝑦 +   (2  − 3𝑦) = 𝑦(2 − 3𝑦) + 𝑦           ✓ sub. 𝑥 = 2 − 3𝑦 in........ (2)
                    𝑦 2 + 2 − 3𝑦 = 2𝑦 − 3𝑦 2 + 𝑦
2.1                        4𝑦 2 − 6𝑦 + 2 = 0                ✓ std. form
       [÷ 2]                 2𝑦 2 − 3𝑦 + 1 = 0
                         (2𝑦 − 1)(𝑦 − 1) = 0                ✓ factorisation
                                     1
                               𝑦 = or 𝑦 = 1                 ✓ both y values
                                     2
                                    1
                           𝑥=         or 𝑥 = −1            ✓ both x values
                                    2
                                                                                               (6)

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                                                 MATHEMATICS
                                    MEMORANDUM
                                                 (Paper 1)             GRADE 11

                    L.H.S: √𝑏√𝑎 − 𝑏 . √𝑏√𝑎 + 𝑏
                      = √(𝑏√𝑎 − 𝑏)(𝑏√𝑎 − 𝑏)         ✓ (𝑏√𝑎 − 𝑏)(𝑏√𝑎 − 𝑏)
                      = √𝑏 2 𝑎 − 𝑏 2
2.2                                                 ✓ simplification √𝑏 2 𝑎 − 𝑏 2
                      = √𝑏 2 (𝑎 − 1)
                                                    ✓ factorisation
                                                    ✓b
                     = 𝑏√𝑎 − 1 = R.H.S                                              (4)

                 3𝑎 = 21𝑏
              (3𝑎 )𝑐 = (21𝑏 )𝑐                     ✓ intro. power c on both sides

                3𝑎𝑐 = (3𝑏 × 7𝑏 )𝑐                  ✓ (3𝑏 × 7𝑏 )𝑐

2.3             3𝑎𝑐 = 3𝑏𝑐 × 3𝑎𝑏                    ✓ deducing that 7𝑐 = 3𝑎
                3𝑎𝑐 = 3𝑏𝑐 + 𝑎𝑏

                 𝑎𝑐 = 𝑏𝑐 + 𝑎𝑏                      ✓ equating the indices

                𝑎𝑐 = 𝑏(𝑐 + 𝑎)                      ✓ factorisation
                     𝑎𝑐                                                             (5)
                ∴𝑏=
                      𝑎+𝑐

                            QUESTION 3                         MARKS: 18

3.1.1   13                                          ✓ answer
                                                                                    (1)

               𝑇𝑛 = 𝑎𝑛 + 𝑏
3.1.2         −3 = 4(1) + 𝑏
              −7 = 𝑏                                ✓ b = −7
               𝑇𝑛 = 4𝑛 − 7                          ✓ answer
              ANY other valid method                                                (2)

            394 = 4𝑛 − 7                            ✓ equation
            401 = 4𝑛
3.1.3    100,25 = 𝑛                                 ✓ 100,25 = 𝑛
        since 𝑛 ∉ ℕ, 394 is NOT a term              ✓ any valid explanation
                                                                                    (3)

                                         5

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                                                                     MATHEMATICS
                                    MEMORANDUM
                                                                     (Paper 1)             GRADE 11

                          𝑇𝑛 = 𝑛2 + 2𝑛 + 𝑐
                           0 = (1)2 + 2(1) + 𝑐                          ✓ corr. sub.
                           0=3+𝑐                                        ✓ simplification
                          −3 = 𝑐
3.2.1
        OR                                                              OR

                              T1 =a + b + c                             ✓ corr. sub.
                              0=1+2+c                                   ✓ simplification
                                0=3+c
                                 –3 = c                                                                 (2)

                           𝑇𝑛 = 𝑛2 + 2𝑛 + 𝑐
3.2.2                     𝑇10 = (10)2 + 2(10) − 3                       ✓ corr. sub.
                              =117                                      ✓ answer
                                                                                                        (2)

                            𝑛2 + 2𝑛 − 3 > 360
                          𝑛2 + 2𝑛 − 363 > 0                             ✓ 𝑛2 + 2𝑛 − 363 > 0

                              (−2)±√22 −4(1)(−363)
3.2.3           C.V:    𝑛=                                              ✓ corr. sub. in corr. formula
                                         2(1)

                         𝑛 = −20,08 or 𝑛 = 18,08                        ✓ both critical values

                       ∴ 𝑛 = 19                                         ✓ n = 19
                       𝑇19 > 360                                                                        (4)

                                 a = 10                                 ✓ answer
3.3.1
                                                                                                        (1)

         b = 99 + 98 + 97 + 96 ........+ 50 +..........+ 3 + 2 + 1      ✓ sequence
                           (99 + 1) = 100
                            (98 +2) = 100
3.3.2
                                    :
                       ∴ 𝑏 = (100) × 49 + 50                            ✓ (100) × 49
                         b = 4950                                       ✓ answer
                                                                                                        (3)

                                                6

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                                                     MATHEMATICS
                          MEMORANDUM
                                                     (Paper 1)           GRADE 11

               QUESTION 4                                           MARKS: 6

                 𝑥−2≥0                                  ✓𝑥−2≥0
4.1
                   𝑥≥2                                  ✓ answer
                                                                                    (2)

4.2                       2                             ✓ answer
                                                                                    (1)

                              √3 − 2
              𝑘 =2+
                                4                               1
                      1
                = 2                                     ✓𝑘=2
                      4                                         4

4.3                   1 2
         𝑔(𝑘) = (2 ) − 1
                      4                                 ✓ substitution
                          1         65
                 = 4           OR          OR 4,06
                          16        16
                                                        ✓ answer

                                                                                    (3)

                                                                    MARKS: 12
               QUESTION 5

         for x- intercept, make y = 0
                  3𝑥 − 1 = 0
                       3𝑥 = 1
                       3 𝑥 = 30                         ✓ 1 = 30
5.1                      x=0                            ✓𝑥=0
         for y- intercept, make x = 0
                     𝑦 = 30 − 1
                      y=0                               ✓y=0
                                                                                    (3)

                                       7

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                                      MATHEMATICS
                    MEMORANDUM
                                      (Paper 1)            GRADE 11

               y

                                         ✓ shape
                                  x
5.2.            0                        ✓ passing thr. origin

                                         ✓ asymptote at y = -1
               -1                                                     (3)

               (−1 ; ∞)
5.3
              OR 𝑦 > −1                  ✓ answer
                                                                      (1)

5.4.1          𝑥 = −2                    ✓ answer
                                                                      (1)

5.4.2          𝑥 > −1                    ✓ answer
                                                                      (1)

              3ℎ(𝑥) = 726

              ℎ(𝑥) = 242                 ✓ 242
5.5
               3𝑥 – 1 = 242

               3𝑥 = 243

                3 𝑥 = 35                 ✓ 35

                𝑥=5                      ✓ answer
                                                                      (3)

                              8

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                                                         MATHEMATICS
                                      MEMORANDUM
                                                         (Paper 1)            GRADE 11

                            QUESTION 6                                MARKS: 10

                                   𝑥=1                      ✓𝑥=1
6.1                                𝑦 = −2                   ✓ 𝑦 = −2

                                                                                         (2)

                      for x-intercept, make y = 0
                                    3                       ✓ y=0
                             0=         −2
                                 𝑥−1
                                   3
                            2=
                                   𝑥−1
                            𝑥 = 212                         ✓ 𝑥 = 2½
6.2
                     for y -intercept, make x = 0
                                      3
                                𝑦=      −2
                                    0−1
                                𝑦 = −5                      ✓ 𝑦 = −5

                                                                                         (3)

                                     y

                                                            ✓ shape

                                                            ✓ x and y-intercepts

                                                            ✓ asymptotes
6.3                                                  1
                                         1       2
                              -2                     2

                               -5
                                                                                         (3)

                                                            ✓ reflection
6.4   Reflection about the y-axis.                          ✓ y-axis
                                                                                         (2)

                                             9

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                                               MATHEMATICS
                                  MEMORANDUM
                                               (Paper 1)             GRADE 11

                             QUESTION 7                      MARKS: 20

                    −𝑏
              𝑥=
                    2𝑎
                    −(8)                         ✓ corr. sub. in corr. formula
                =
7.1.1                2(2)
                                                 ✓ x -co-ord. - 2
                = −2                             ✓ y -co-ord. - 5
                                                                            2
                                                   ( if not ordered pair)
              D(- 2 ; - 5)                                                  3
                                                                                 (3)

             𝑥 ≥ 0 or 𝑥 ≤ −4                     ✓𝑥≥0
7.1.2                                            ✓ 𝑥 ≤ −4
                                                                            1
                                                  ( incorrect notation)
                                                                            2
                                                                                 (2)

             LM = 𝑘(𝑥) − 𝑝(𝑥)
                = 2𝑥 2 + 8𝑥 + 3 − (2𝑥 − 4)
                = 2𝑥 2 + 6𝑥 + 7                  ✓ simplification

                 4𝑎𝑐 − 𝑏 2                       ✓ formula
         MinLM =
                    4𝑎
                                                 ✓ corr. sub. in corr. formula
                   4(2)(7) − 62
                 =
                       4(2)                      ✓ answer
                    1
                 = 22

        OR                                       OR
7.1.3        LM = 𝑘(𝑥) − 𝑝(𝑥)
                = 2𝑥 2 + 8𝑥 + 3 − (2𝑥 − 4)
                = 2𝑥 2 + 6𝑥 + 7                  ✓ simplification

                     −𝑏
               𝑥=
                     2𝑎
                     −(6)                        ✓ corr. sub. in corr. formula
                 =
                     2(2)

                     3
                 =−
                     2
                       3        3
         MinLM = 2(− 2)2 + 6 (− 2 ) + 7          ✓ sub. x in eqn.
                                                 ✓ answer
               = 212                                                             (4)

                                          10

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                                                 MATHEMATICS
                            MEMORANDUM
                                                 (Paper 1)            GRADE 11

                            𝑘(−2) − 𝑘(3)
           Ave. grad =
                               −2 − 3
                                                   ✓ −5
7.1.4                       −5 − 45                ✓ −45
                          =
                            −2 − 3
                                                   ✓ answer
                          = 10                                                       (3)

               2𝑥 2 + 8𝑥 + 3 = 2𝑥 + 𝑡
                2𝑥 2 + 6𝑥 + 3 − 𝑡 = 0              ✓ std. form

                    ∆ = 𝑏 2 − 4𝑎𝑐
                      = (6)2 − 4(2)(3 − 𝑡)         ✓ corr. sub. in formula
                      = 39 − 24 + 8𝑡
                      = 12 + 8𝑡                    ✓ value of ∆
7.1.5
                   for = roots, ∆ = 0
                         12 + 8𝑡 = 0               ✓ ∆= 0
                               8𝑡 = −12
                       1       3
               𝑡 = −1 or − or -1,5
                      2          2                 ✓ answer

                                                                                     (5)

                                 y

                                                   ✓ shape

                                        1½   x
7.2
          –4                                       ✓ x-intercepts

                                     1;–5

                                                   ✓ turning point in 3rd quadrant

                                                                                     (3)

                                       11

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