# Mathematics Grade 11 P1 Gauteng June 2024 Memo (English)

> Grade 11 Mathematics Paper 1 memorandum from the Gauteng June Exam 2024 (English), 11 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/gr-11-mathematics-p1-english-june-2024-possible-answers/6a96bfec-c2fb-499b-b231-9b94ad3095db
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Language: English
- Document type: Memorandum
- Year: 2024
- Exam period: Gauteng June Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 11

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## About this document

Mathematics · Grade 11 · Gauteng June Exam · 2024 · English. 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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           PROVINCIAL EXAMINATION
                     JUNE 2024
                     GRADE 11
            MARKING GUIDELINES

                 MATHEMATICS (PAPER 1)

11 pages

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                                                              MATHEMATICS
                                MARKING GUIDELINES
                                                              (PAPER 1)                  GRADE 11

INSTRUCTIONS AND INFORMATION

A – ACCURACY
CA. – CONTINUED ACCURACY

NOTE:

•   If a candidate answerd 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 guidelines.
•   Assuming values/answers in order to solve a problem is UNACCEPTABLE.

                                                 2

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                                                         MATHEMATICS
                                    MARKING GUIDELINES
                                                         (PAPER 1)               GRADE 11

QUESTION 1
 1.1   1.1.1   x9                                         ✓ answer
                                                                                      (1)

       1.1.2   x=9                                         ✓ answer
                                                                                      (1)

       1.1.3     (1) 2 + 4
                   9 −1
                5                                          ✓ answer
                8
                Irrational                                                           (1)

       1.1.4   x=0                                         ✓ answer
                                                                                      (1)

 1.2   1.2.1   2 x 2 − 3x − k = 0
               2 x 2 − 3x − 5 = 0
               (2 x − 5)( x + 1) = 0
                                                           ✓ factors
                       5
               x =         of x = −1                      ✓ answer
                       2
                                                                                      (2)

       1.2.2   2 x 2 − 3x − 4 = 0
                   −(−3)  (−3) 2 − 4(2)(−4)               ✓ substitution
               x=
                              2(2)
                x = 2,35 of x = −0,85
                                                           ✓✓ answer
               NOTE: Penalise 1 mark for incorrect
               rounding in this question ONLY.                                        (3)

 1.3   1.3.1   x−4= − x−2
               16 − 8 x − x 2 = x − 2                      ✓ square both sides
               x 2 − 9 x + 18 = 0                          ✓ standard form
               ( x − 6)( x − 3) = 0
                                                           ✓ factors
               x = 3 of x  6
                                                           ✓ critical values
                                                           ✓ rejection                (5)

                                               3

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                                                           MATHEMATICS
                                      MARKING GUIDELINES
                                                           (PAPER 1)            GRADE 11

      1.3.2      x 2 + 4  3x + 2
                 x 2 − 3x + 2  0                            ✓ standard form
                 ( x − 2)( x − 1)  0                        ✓ factors
                 x  1 or x  2
                 OR
                                                             ✓✓ answers
                 x  (−;1)  (2; )
                                                                                     (4)

1.4   y + 2 = 2x
       y = 2x − 2         (1)
                                                             ✓ y as subject
      2x2 = 2 − y 2        (2)
       2 x 2 = 2 − (2 x − 2) 2
                                                             ✓ substitution
      2 x 2 = 2 − (4 x 2 − 8 x + 4)
      2 x2 = 2 − 4 x2 + 8x − 4
       6x2 − 8x + 2 = 0
       3x 2 − 4 x + 1 = 0                                   ✓ standard form
      (3 x − 1)( x − 1) = 0
              1                                              ✓ factors
      x =         of x = 1
              3                                              ✓ both x- values
                4
      y =−           of y = 0
                3                                            ✓ both y- values

                                        OR
      y + 2 = 2x                                                         OR
              y+2
      x =               (1)
                2                                            ✓ x as subject
      2x2 = 2 − y 2       (2)
            y+2 2
       2(        ) = 2 − y2                                 ✓ substitution
              2
         y2 + 4 y + 4
      2(              ) = 2 − y2
               4
          y + 4y + 4
            2
      (               ) = 2 − y2
                2
       y + 4 y + 4 = 4 − 2 y2
          2

      3y2 + 4 y = 0                                         ✓ standard form
      y (3 y + 4) = 0
                                                             ✓ factors
                               4
      y =0     of      y=−
                               3                             ✓ both y-values
                           1
      x =1    of      x=                                    ✓ both x-values         (6)
                           3
1.5   1.5.1 For unequal roots:   0                         ✓ condition of 
                                                4

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                                                               MATHEMATICS
                                          MARKING GUIDELINES
                                                               (PAPER 1)               GRADE 11

                    2 x 2 − 4 x + (3 − p) = 0
                    = b 2 − 4ac
                     = (−4) 2 − 4(2)(3 − p)                    ✓ substitution into 
                    = 16 − 24 + 8 p                                                        (4)
                     = −8 + 8 p                                ✓ expression for 
                    −8 + 8 p  0
                   8p  8
                    p 1
                                                                 ✓ answer

      1.5.2        From 1.5.1
                   3 − p  0 (a  0 and  b  0)              ✓   c0
                    − p  −3
                                                                 ✓ answer
                   1  p  3
                   NOTE: Any other valid method                                             (2)

1.6   ( p 2 + 1) x 2 = −2 pqx − q 2
                                                                 ✓ standard form
       ( p 2 + 1) x 2 + 2 pqx + q 2 = 0
       = (2 pq) 2 − 4( p 2 + 1)q 2                              ✓ substitute into 
       = 4 p q − 4 p q − 4q
               2    2      2   2      2
                                                                 ✓ simplification
        = −4q 2                                                ✓ expression for 
       q 2  0, q  ; q  0
       −4q 2  0                                                ✓ explanation
       Roots non-real                                                                      (5)
                                                                                            [35]

                                                    5

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                                                                     MATHEMATICS
                                                MARKING GUIDELINES
                                                                     (PAPER 1)               GRADE 11

QUESTION 2
 2.1   2.1.1   (2 −1 + 3−1 ) 2
                   1 1                                                 ✓ converting exponents
               = ( + )2
                   2 3
                   3+ 2
               =(        )
                     6
                   5                                                   ✓ simplification
               = ( )2
                   6
                  25
               =                                                       ✓ answer
                 36                                                                                (3)

       2.1.2   2 .4 x + 2 − 4 x − 3
                    2 x .2 x
                  2 .4 x .4 2 − 4 x .4 − 3
               =
                             4x
                                1
                  4 x (32 − )
                               64                                      ✓✓ factors
               =            x
                          4
                           1
               = 32 −
                          64
                  2 048 − 1
               =
                        64
                  2 047
               =                                                       ✓ answer
                    64
                                                    63
               NOTE: Accept 31                                                                     (3)
                                                    64

                    1
       2.1.3    −
               3 2 [ 12 + 3 (3 3)]
                            1                   1
                        −
               = 3 2 [2 3 + (3 3) 3 ]                                  ✓ simplify brackets
                          1         1   1   1
                        −
               = 3 [2.3 + 3 .3 ]
                          2         2   3   6                          ✓ surds as rational
                            1       1   1
                        −                                                  exponents
               = 3 [2.3 + 3 ]
                            2       2   2

                            1   1
                        −
               = 3 .3 (2 + 1)
                            2   2
                                                                       ✓ factors
               = 30 (3)
               =3
                                                                       ✓ answer                    (4)

                                                          6

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                                                          MATHEMATICS
                                     MARKING GUIDELINES
                                                          (PAPER 1)             GRADE 11

      2.2.1      3
              2 x = 16
                 4

                  3
              x4 = 8
                  3 4        4
                   
               x 4 3 = 83                                  ✓ method

               x = (3 8 ) 4                                ✓   x as subject
               x = 16                                      ✓ answer
                                                                                      (3)
              NOTE: Any other valid method.

      2.2.2   5 x +1 + 5 x = 150
              5 x.51 + 5 x = 150
               5 x (51 + 1) = 150                          ✓ factors
               5 (6) = 150
                  x

              5 x = 25                                      ✓ simplification
              5 = 5
                  x      2
                                                            ✓ 25 as base of 5
              x = 2
                                                            ✓ answer                 (4)
              NOTE: Any other valid method.

2.3   9 − 54
        6 2
          (9 − 9.2.3 )   2
      =                                                    ✓   
                                                                    2
              6 2        2                                          2
        9 2 − 3 2 . 3. 2
      =
               6.2
        9 2 − 3.2 3                                         ✓ simplification
      =
            12
        3(3 2 − 2 3 )                                       ✓ factorisation
      =
             12
        3 2 −2 3
      =
            4                                                                         (3)
                                                                                     [20]

                                               7

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                                                                  MATHEMATICS
                                             MARKING GUIDELINES
                                                                  (PAPER 1)               GRADE 11

QUESTION 3

 3.1   T4 = −17                                                     ✓ answer                   (1)

 3.2   Tn = −7n + 11                                                ✓ answer                   (1)

 3.3   Tn = −7n + 11
        −227 = −7n + 11                                            ✓ substitute Tn
       − 238 = −7n
                                                                    ✓ answer
        n = 34                                                                                 (2)

 3.4   2a = −7
            7                                                       ✓ value for a
       a=−
            2
                                                                    ✓ conclusion
        maximum
                                                                                               (2)
       NOTE: Any other valid method.
                                                                                                [6]

QUESTION 4

 4.1   69          0          − 63          T4
        − 69           − 63          − 57
               6              6
       next term (T4 ) is: − 120                                    ✓ answer                   (1)

 4.2   Tn = an2 + bn + c
       T2 = a(2) 2 + b(2) + c = 4a + 2b + c
       T1 = a(1) 2 + b(1) + c = a + b + c
       T2 − T1 = 3a + b
       2a = 6
                                                                    ✓   2 nd difference
       a = 3
                                                                    ✓ value for a
       3a + b = −69
       3(3) + b = −69
       9 + b = −69
        b = −78                                                    ✓ value for b
       a + b + c = 69
       3 + (−78) + c = 69                                           ✓ value for c
        c = 144
       Tn = 3n 2 − 78n + 144                                                                   (4)

                                                       8

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                                                                MATHEMATICS
                                         MARKING GUIDELINES
                                                                (PAPER 1)                GRADE 11

 4.3   Smallest term (minimum) is at the turning point of the
       quadratic equation:
                − b − (−78) 78                                    ✓ substitute into A.O.S
       n =        =       =   = 13
                2a    2(3)   6                                    ✓ value for n
       Tn = 3n − 78n + 144
                2

       T13 = 3n 2 − 78n + 144
       T13 = 3(13) 2 − 78(13) + 144
                                                                  ✓ substitution
       T13 = −363
                                                                  ✓ answer                     (4)
                                                                                               [9]

QUESTION 5
 5.1   The points A( − 4 ; 0) and B(0 ; 8) lie on g.

                8−0                                               ✓ substitute for mAB
       m AB =
              0 − (−4)
                                                                  ✓ value of mAB
       m AB = 2
       y = mx + c
        8 = 2(0) + c  A(0;8)                                   ✓ value of c
       c = 8
        g ( x) = 2 x + 8
       NOTE: Any other valid method.                                                          (3)

 5.2    f ( x) = a( x − x1 )( x − x2 )
        f ( x) = a( x + 4)( x − 2)                                ✓ substitute points A
       subst (0;8)                                                  and C
       8 = a (0 + 4)(0 − 2)
       8 = a (−8)
       a = −1                                                     ✓ value of a
        f ( x) = −( x + 4)( x − 2)
        f ( x) = −( x 2 + 2 x − 8)
       f ( x) = − x 2 − 2 x + 8
                                                                  ✓ standard form (A)
        a = −1
       b = −2                                                     ✓ value of b
       c =8                                                       ✓ value of c                (5)

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                                                       MATHEMATICS
                                  MARKING GUIDELINES
                                                       (PAPER 1)                GRADE 11

 5.3   f ( x) = − x 2 − 2 x + 8
              − b − (−2)                                 ✓ correct substitution
       x =        =
               2a 2(−1)
        x = −1
        f (−1) = −(−1) 2 − 2(−1) + 8                    ✓ correct substitution
        f (−1) = 9
       NOTE: Any other valid method.                                                 (2)

 5.4   y9                                               ✓ answer                    (1)

 5.5   mg = 2
                 1                                       ✓
        mp = −                                              mp
                 2
       y − y1 = m( x − x1 )
                  1                                      ✓ substitute point C
        y − 0 = − ( x − 2)
                  2
               1
        y = − x +1
               2                                         ✓ answer                    (3)

 5.6   f (−1) = 9                                        ✓   f (−1) = 9 and
       g (−1) = 6                                            g (−1) = 6
        f (−1) − g (−1)
        PQ = 9 − 6
        PQ = 3                                          ✓ answer
                                                               OR
                OR

       DP(−1;9)                                          ✓ using TP and g (−1)
       g (−1) = 6
        PQ = 9 − 6                                      ✓ answer
        PQ = 3                                                                      (2)

 5.7   x  2 ; x  −4                                    ✓✓ answer (A)               (2)
                                                                                     [18]

QUESTION 6
 6.1   x = −3                                            ✓ answer
       y =1                                              ✓ answer
                                                                                     (2)

                                            10

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                                                    MATHEMATICS
                               MARKING GUIDELINES
                                                    (PAPER 1)                GRADE 11

6.2            −4
      f ( x) =      +1
               x+3
           −4
      0=         +1
          x+3
             −4
      −1 =
            x+3
      − x − 3 = −4
      − x = −1
      x =1
                                                      ✓ x-intercept
           −4
      y=         +1
          0+3
          −4
      y=       +1
            3
               1                                      ✓ y- intercept
      y =−
               3                                                                  (2)

6.3   y = −4                                          ✓ answer                    (1)

6.4

                                                      ✓ shape of k

                                                      ✓ k and h intercepts

                                                      ✓ asymptotes k and h

                                                      ✓ shape of h

                                                                                  (4)

6.5   p ( x) = 3h( x)
       p( x) = 3(2 − x − 4)
       p( x) = 3.2 − x − 12
      asymptote: y = −12                              ✓ answer
      NOTE: Answer only, full marks.                                              (1)

6.6   q( x) = 2 x − 4                                 ✓ answer                    (1)

6.7   x=0                                             ✓ answer                    (1)
                                                                                  [12]

                                                                         TOTAL: 100
                                         11

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