# Mathematics Grade 11 P1 Gauteng November 2023 Memo (English)

> Grade 11 Mathematics Paper 1 memorandum from the Gauteng November Exam 2023 (English), 18 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/gr-11-mathematics-p1-english-2023-possible-answers/990d926f-21c7-4bf0-8735-f9eace921dbf
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Language: English
- Document type: Memorandum
- Year: 2023
- Exam period: Gauteng November Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 18

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Mathematics · Grade 11 · Gauteng November Exam · 2023 · English. Memorandum, 18 pages. Read online or download the PDF.

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### Page 1

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           PROVINCIAL EXAMINATION
                NOVEMBER 2023
                     GRADE 11
            MARKING GUIDELINES

                 MATHEMATICS (PAPER 1)

18 pages

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                                                            MATHEMATICS
                             MARKING GUIDELINES
                                                            (PAPER 1)   GRADE 11

INSTRUCTIONS AND INFORMATION

A – ACCURACY
C.A. – 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 guidelines.
•   Assuming values/answers in order to solve a question is UNACCEPTABLE.

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                                                           MATHEMATICS
                                MARKING GUIDELINES
                                                           (PAPER 1)   GRADE 11

QUESTION 1

1.1 1.1.1           5    p
            K=         +
                   p+2   3

                   5     3
            K=         +
                  3+ 2   3
                                                           ✓ simplification
            K= 1 +1
                                                           ✓ answer
            K=2                                                              (2)

    1.1.2           5    p
            K=         +
                   p+2   3
            p+2>0
             p > –2                                       ✓ answer
            NOTE: Answer only, full marks.                                    (1)

1.2 1.2.1 3x2 = 4x + 2
            3x2 – 4x – 2 = 0

                 − b  b 2 − 4ac
            x=                   +1
                       2a

               − ( −4)  ( −4) 2 − 4(3)( −2)               ✓ substitution
            x=                               +1
                           2(3)
                                                           ✓ answers
             x = 1,72 or x = –0,39
            NOTE: Penalise with 1 mark for rounding-
                      off in this question ONLY. Answers
                      only, 1 mark.                                           (2)

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                                                       MATHEMATICS
                                  MARKING GUIDELINES
                                                       (PAPER 1)   GRADE 11

1.2.2 x2 + 10x > –21
        x2 + 10x + 21 > 0                              ✓ standard form
        (x + 7)( x + 3) > 0                            ✓ factors

        x < –7 or x > –3                               ✓✓ answers
        NOTE: Penalise with 1 mark for the use of
              AND instead of OR.                                               (4)

1.2.3     x2 − 5 = 2 x
                                                       ✓ squaring both sides
        ( x − 5 )2 = (2 x )2
           2
                                                       ✓ simplification
        x2 – 5 = 4x                                    ✓ standard form
        x2 – 4x – 5 = 0                                ✓ factors
        (x – 5)(x + 1) = 0                             ✓ both answers with
                                                         exclusion
         x = 5 or x = –1(N/A)
                                                                               (5)

1.2.4                   10
        3x+1 + 3x–1 =
                         9
                             10                        ✓   factorisation
        3x.31 + 3x.3–1 =
                              9
                      1 10
        3x (3–1 +      =
                      3   9
               10 10
        3x.      =
                3   9
                 1
        3x =
                 3                                     ✓ simplification
        3x = 3–1                                      ✓ same bases
        x = –1                                         ✓ answer                (4)

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                                                    MATHEMATICS
                               MARKING GUIDELINES
                                                    (PAPER 1)   GRADE 11

1.3 1.3.1                            −3
            x + y = 2 and y + 1 =        …(2)
                                    x −1
                                                    ✓ x as subject in (1)
            x = 2 – y …(1)
            sub (1) into (2)

                       −3
            y+1=                                    ✓ substitution in (2)
                    2 − y −1
                      −3
            y+1=
                    1− y
            (y + 1)(1 – y) = –3
            1 – y2 = –3
            y2 = 4                                  ✓ value of y 2
            y = 2 … sub into (1)                   ✓ values for y

            x = 2 – 2 of x = 2 – (–2)
            x=0          x=4                        ✓ values for x
            NOTE: Any other valid method.                                       (5)

    1.3.2  1 1 
           + 
          x y
            1 1 
             +                                    ✓ correct selection for x
            4 −2
               1                                       and y values
            =−
               4                                    ✓ answer                    (2)

1.4 1.4.1 2mx2 – (m – 2)x + m + 1 = 0
           = b2 – 4ac
                                                    ✓ substitution
           = [–(m – 2)]2 – 4(2m) (m + 1)
           = (m – 2)2 – 8m(m + 1)                  ✓ simplification
           = m2 – 4m + 4 – 8m2 – 8m                ✓ standard form
           = 7m2 – 12m + 4

            For real roots,   0                   ✓ condition for 
            7m2 – 12m + 4  0
            7m2 + 12m – 4  0
            (7m – 2)(m + 2)  0                    ✓ factors
                          2
            –2m                                  ✓ answers
                          7                                                     (6)

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                                                            MATHEMATICS
                                MARKING GUIDELINES
                                                            (PAPER 1)   GRADE 11

     1.4.2 For roots to have opposite signs, the value of
           ‘c’ must be positive.

             c > 0                                         ✓ condition of ‘c’
             m + 1 > 0
             m > – 1

             Real roots opposite in sign for:
              –1<m<0                                        ✓ answer              (2)
                                                                                 [33]

QUESTION 2

                       −1
2.1 2.1.1  512 x 3  3
               −3 
           64 x 
                           −1
                29 x 3  3                                 ✓ prime bases
             =  6 −3 
               2 x 
                      −1
             = {23x6} 3
             = {2–1x–2}
               1     1
             = x 2                                          ✓ simplification
               2 x
                 1
             =                                              ✓ answer
               2x2
             NOTE: Any correct alternative method.                               (3)

     2.1.2   x n+2 + x n +1 − x n − x n−1
                          x2 − 1
                x n ( x 2 − 1) + x n ( x − x −1 )
             =                                              ✓ factorisation
                             x2 − 1
                                       x2 − 1
                x n ( x 2 − 1) + x n (        )
             =                           x
                             x2 − 1
                           1
             = xn + xn )                                    ✓ simplification
                           x
                 n       n–1
             =x +x                                          ✓ answer
             NOTE: Any correct alternative method.                               (3)

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                                                                 MATHEMATICS
                                    MARKING GUIDELINES
                                                                 (PAPER 1)   GRADE 11

     2.1.3       a + 2a − 1 . a − 2a − 1
                                                                 ✓ write as A single surd
             =    (a + 2a − 1)(a − 2a − 1)

             =    a 2 − ( 2a − 1) 2
                                                                 ✓ quadratic expression
             =    a 2 − 2a + 1
                                                                 ✓ factors
             =    ( a − 1) 2
                                                                 ✓ answer
             =a–1
             NOTE:             Any correct alternative method.                              (4)

2.2 (1 –     2 )(x + y     z ) = –3 +   2)
                       ( −3 + 2 ) 1 + 2                          ✓ isolating variables
     x+y         z=             x                                 and rationalising
                         1− 2      1+ 2
                       ( −3 + 2 )(1 + 2 )
     x+y         z=
                              1− 2
                       −3− 2 2 + 2
     x+y         z=                                             ✓ simplification
                           −1
     x+y         z=1+2 2                                        ✓ simplification
     x=1
     y=2
     z=2                                                        ✓ answers                   (4)
                                                                                            [14]

QUESTION 3

3.1 3.1.1 Multiply the previous term by cos x .                  ✓ answer                    (1)

     3.1.2 sin x.cos2 x                                          ✓ answer                    (1)

3.2 3.2.1 − 1; 2; 5; 8; …

             T2 – T1 = T3 – T2
             T2 – T1 = 2 – (–1) = 3
             T3 – T2 = 5 – 2 = 3
             d = 3                                              ✓ value of d
              Tn = dn + a
              Tn = 3n + a                                       ✓ substitution
             –1 = 3(1) + a
             a = –4                                             ✓ value of a
              Tn = 3n – 4                                       ✓ answer
             NOTE: Answer only, full marks.                                                  (4)

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                                                         MATHEMATICS
                                   MARKING GUIDELINES
                                                         (PAPER 1)   GRADE 11

    3.2.2 Tn = 3n – 4
          161 = 3n – 4                                   ✓ substitution
          165 = 3n
           n = 55                                       ✓ answer              (2)
                                                                               [8]

QUESTION 4

4.1 4.1.1 25 squares                                     ✓ answer              (1)

    4.1.2 64 dots                                        ✓ answer              (1)

    4.1.3 4 ; 9 ; 16 ; 25
            5   7    9
              2   2                                      ✓ second difference

          2a = 2
          a=1                                           ✓ value of a
          3a + b = 5
          b=2                                           ✓ value of b
          a+b+c=4
          c=1                                           ✓ value of c
           Tn = n2 + 2n + 1
           Tn = (n + 1)2                                                      (4)

4.2 4.2.1 17 ; 10 ; 5 ; 2 ; 1 ; 2
            –7       –5       –3       –1       +1
                 2        2        2        2
          r=1                                           ✓ answer
          s=2                                           ✓ answer              (2)

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                                                       MATHEMATICS
                              MARKING GUIDELINES
                                                       (PAPER 1)   GRADE 11

    4.2.2   2a = 2                                    ✓ 2 nd difference
            a=1                                      ✓ value of a
            d(n) = n2 +bn + c
            d(1) = 12 +b(1) + c
            17 = 1 + b + c
            16 = b + c …(2)
            d(2) = 22 +b(2) + c
            10 = 4 +2b + c
            6 = 2b + c …(1)
            (1) – (2)
             b = –10                                 ✓ value of b
            16 = –10 + c …(2)
             c = 26                                  ✓ value of c
            NOTE: Any other valid method.                                    (4)

    4.2.3   d(n) = n2 – 10n + 26
             d(8) = 82 – 10(8) + 26                  ✓ substitution
             d(8) = 10m                              ✓ answer                (2)

4.2 4.2.4   d(n) = n2 – 10n + 26
            Minimum turning point
                   − (−10)
            n=
                     2(1)                             ✓ substitution
            n=5                                       ✓ value of n
            n>0…nN
            minimum value                             ✓ explanation and
            Mikayla is CORRECT                          choice of Mikayla
            NOTE: No marks are awarded for correct
                    name only. The explanation and the
                    choice in name must correlate.                             (3)
                                                                             [17]

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                                                               MATHEMATICS
                                  MARKING GUIDELINES
                                                               (PAPER 1)   GRADE 11

QUESTION 5

5.1 A = P(1 – i)n
      1                                                        ✓ correct substitution
        = (1 – i)2                                               into correct formula
      3
      i = 0,42%                                                ✓ answer
      NOTE: The substitution is for the correct formula
                and the interpretation of substituting for A
                and P.                                                                    (2)

5.2   5.2.1               i 
              1 + ieff = 1 + nom 
                             n                               ✓ correct substitution
                                      12
                                                                 into correct formula
                          0,075 
              1 + ieff = 1 +    
                             12 
              ieff = 0,07763…                                  ✓ simplification
              ieff = 7,76%                                     ✓ answer                   (3)

      5.2.2   A = P(1 – i)n
                                                                               0,075 12
                                0,075 12     0,078 8           ✓ 10 000(1 +          )
               A = 10 000(1 +        ) (1 +      ) +                            12
                                 12            4
                                                                          0,078 8
                              0,078 6                          ✓   x (1 +       )
                    5000(1 +        )                                       4
                                4                                            0,078 6
               A = R18 191,03                                 ✓   5000(1 +        )
                                                                               4
                                                               ✓   answer
              YES. He will have sufficient funds.              ✓   conclusion
              NOTE: The values must be substituted into
                      the correct formula. Only award the
                      conclusion mark if a valid calculation
                      has been done.                                                      (5)

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                                                         MATHEMATICS
                                MARKING GUIDELINES
                                                         (PAPER 1)   GRADE 11

5.3 Mervin:
     A = P(1 – ni)
     A = 6 000(1 + 5 x 0,085)                          ✓ correct substitution into
                                                          correct formula
     A = R8 550                                        ✓ value of A (Mervin)
     5% x R6 000 = R300
      total = R8 550 + R300                           ✓ bonus value +
     A = R8 850                                           final value of A
                                                           (Mervin)
     Haley:
     i = 0,08  4 = 0,02
     n = 5 x 4 = 20
     A = P(1 – i)n
     A = 6 000(1 + 0,02)20                             ✓ correct substitution into
                                                           correct formula
     A = R8 915,68                                     ✓ value of A (Haley)
     Haley will have the larger investment.            ✓ conclusion
     NOTE: Only award the conclusion mark if a valid
           calculation has been done.                                                 (6)
                                                                                     [16]

QUESTION 6

6.1 x                                                ✓ answer

      OR

     x (–∞;∞)                                                                        (1)

6.2 y = 0                                              ✓ answer                       (1)

6.3 f (x) = a.bx
                     2
               3
     144 = a.                                        ✓ substitution
              4
     a = 256                                           ✓ answer                       (2)

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                                                         MATHEMATICS
                                 MARKING GUIDELINES
                                                         (PAPER 1)   GRADE 11

                     −x
6.4              3
      g(x) =256  
                4                                    ✓ answer
      OR
                     x
                4
      g(x) =256  
                 3                                                       (1)

6.5 Increasing function.                               ✓ answer
    If x increases, increases f(x).                    ✓ explanation       (2)
                                                       ✓
6.6 [f (x)]2 – [f (–x)]2 = a.f (2x) – a.f (–2x)

      LHS

      [f (x)]2 – [f (–x)]2
             3             3
      [256( )x]2 – [256( )–x]2                         ✓ substitution
             4             4
                 3            3
      = [2562 ( )2x] – [2562 ( )–2x]                   ✓ answer (LHS)
                 4            4

      RHS

      a.f (2x) – a.f (–2x)
                    3             3
      = 256 [256( )2x] – 256 [256( )–2x]
                    4             4
                3          3                           ✓ answer (RHS)
      = 2562 ( )2x – 2562 ( )–2x
                4          4                                               (3)
                                                                          [10]

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                                                           MATHEMATICS
                            MARKING GUIDELINES
                                                           (PAPER 1)   GRADE 11

QUESTION 7

7.1   x = –4                                                    ✓ answer
      y = –4                                                    ✓ answer       (2)

7.2   x-intercept:
                2
      f (x) =      −1
               x+4
              2                                                 ✓ make y = 0
      0=         −1
            x+4
      x+4=2
       x = –2                                                  ✓ x- value

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

7.3                                             y
                                    f

                                                                ✓ shape
                                                                ✓ asymptotes
                                                                ✓ intercepts
                                           –2 O            x

                                           –½
                                                       y = –1

                                  x = –4

      NOTE: The mark for shape is for the graph shifting from
            quadrants 1 towards 3.                                             (3)

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                                                        MATHEMATICS
                                 MARKING GUIDELINES
                                                        (PAPER 1)   GRADE 11

7.4   (–6 ; –2)                                              ✓ x-value
                                                             ✓ y-value               (2)

7.5   y   and y  1                                        ✓ answer
      NOTE: Both conditions must be stated.                                          (1)

7.6   Reflection in the y-axis                               ✓ answer                (1)
                                                                                    [12]

QUESTION 8

8.1 y = a(x + p)2 + q
     6 = a(0 –1)2 + 8                                ✓ substitute points (1 ; 8)
    6=a+8
                                                          and (0 ; 6)
    a = –2
                                                      ✓ value for a
      y = –2(x – 1) +8
                  2

      y = –2(x2 – 2x + 1) +8
                                                      ✓ correct simplification
      y = –2x2 + 4x –2 +8
       y = –2x2 + 4x + 6
      NOTE: The mark for simplification is awarded
                either at the 3rd or 2nd last step.                                  (3)

8.2 j(x) = –2x2 + 4x + 6
     j(3) = 0 …(3;0)
     j(1) = 8 …(1;8)                                 ✓ coordinates of j(3) and
          y − y1                                        j(1)
    m= 2
          x2 − x1
          8−0                                         ✓ correct substitution
      m=
          1− 3                                        ✓ answer
       m = –4
      NOTE: Candidates do not have to write the
               coordinates.                                                          (3)

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                                                      MATHEMATICS
                               MARKING GUIDELINES
                                                      (PAPER 1)   GRADE 11

8.3 j(x) = k(x)
                      1                             ✓ equating
     –2x2 + 4x + 6 =    x–1
                      3
     –6x2 + 12x + 18 = x – 3
     –6x2 + 11x + 21 = 0                            ✓ standard form
     –6x2 – 11x – 21 = 0
     (x – 3)(6x + 7) = 0                            ✓ factors
                       7
     x  3 or x = –                                 ✓ x-values with exclusion
                       6
        7       25
     k−  = −
        6       18
         7 25 
     P  − ;−                                     ✓ y-value
         6 18 
     NOTE: Answer need not be in coordinate form.                               (5)

                          1
8.4 VW = –2x2 + 4x + 6 – x + 1
                          3                         ✓ method
                  11
    VW = –2x2 +       x+7
                   3
                   11    7                        ✓ simplification
    VW = –2  x 2 − x − 
                    6    2
                   11    121 121 7 
    VW = –2  x 2 − x +      −   −                 ✓ method
                    6    144 144 2 
                       2
                 11     625
     VW = –2  x −  +
                 12      72
                625
     VWmax =        of 8,68 units                   ✓ answer
                72
                             OR                                 OR

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                                                      MATHEMATICS
                             MARKING GUIDELINES
                                                      (PAPER 1)   GRADE 11

                         1     
    VW = –2x2 + 4x + 6 –  x − 1                   ✓ method
                         3     
                         1
    VW = –2x2 + 4x + 6 – x + 1
                         3
                11
    VW = –2x2 +    x+7                              ✓ simplification
                 3

    Turning point:

       11
    x= 3                                            ✓ method
      2( −2)
         11
    x=
         12
               625                                  ✓ answer
    VWmax =        of 8,68 units
               72                                                       (4)

8.5 8.5.1 x < 1                                     ✓ answer            (1)

    8.5.2 x  –1 … but … x  3
            –1  x < 3… or … x > 3                 ✓✓ answers
            NOTE: Both answers must be completely
                     correct.                                           (2)

8.6 k < 8                                           ✓✓ answer            (2)
                                                                       [20]

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                                                                               MATHEMATICS
                                         MARKING GUIDELINES
                                                                               (PAPER 1)   GRADE 11

QUESTION 9

ANSWERS CAN EITHER BE IN FRACTION OR DECIMAL FORM.

9.1
                                                       C
                                           0,7

                                            0,1            I             C
                                     C
                                                           0,9
                0,7                         0,2
                                                       L
                                                               0,1       I
                                                                             ✓ Tree 1
                       0,1
                                     I
                                                 0,7             C           ✓ Tree 2
                                     C
               0,2            0,9
                                             0,1                     I
                        L
                               0,1   I

                     Tree 1                      Tree 2
                                                                                                 (2)

9.2   P(C and C) = 0,7 x 0,7                                                 ✓ correct method
      P(C and C) = 0,49                                                      ✓ answer
      NOTE: Answer only, full marks.                                                             (2)

9.3   P(C) or P(L and C)                                                     ✓ 0,7
      = 0,7 + (0,2 x 0,9)                                                    ✓ +(0,2 x 0,9)
      = 0,88                                                                 ✓ answer
      NOTE:          Answer only, full marks.                                                    (3)
                                                                                                 [7]

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                                                     MATHEMATICS
                               MARKING GUIDELINES
                                                     (PAPER 1)   GRADE 11

QUESTION 10

10.1 10.1.1 P(A and C) = 0                           ✓ answer                    (1)

       10.1.2 P(B and C)
              = P(B) x P(C)                          ✓ correct formula or
              = 0,4 x 0,2                              correct substitution
              = 0,08                                 ✓ answer
              NOTE: Answer only, full marks.                                     (2)

       10.1.3 P(A and B)
              = P(A) x P(B)
              = 0,3 x 0,4                            ✓ correct substitution
              = 0,12                                 ✓ value of P(A and B)
              P(A or B) = P(A) + P(B) – P(A and B)
                         = 0,3 + 0,4 – 0,12          ✓ correct substitution
              P(A or B) = 0,58                       ✓ answer                    (4)

10.2 10.2.1 8 + 10 + x + 65 + 3 + 5 + 5 + 2 = 103
            x=5                                      ✓ answer                    (1)

       10.2.2             n(any 2)
                P(any2) =
                            n( S )                       n(any 2)
                                                     ✓             with
                          10 + 65 + 5 + 5                  n( S )
                P(any2) =
                                103                      correct substitution
                           85
                P(any2) =
                          103                        ✓ answer (any shape)
                P(any2) = 0,83                                                   (2)

10.3    9     (17 + 8) (17 + 8)   9                  ✓ correct substitution
           x          +         x
       80        79       80      79
        9     25 25      9                           ✓ simplification
           x      +    x
       80 79 80 79
       0,0712 ...
        0,07                                        ✓ answer                    (3)
                                                                                [13]

                                                                     TOTAL:     150

                                            18

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