# Mathematics Grade 11 P1 Gauteng June 2022 Memo (English)

> Grade 11 Mathematics Paper 1 memorandum from the Gauteng June 2022 (English), 12 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/gr11-math-p1-eng-june-2022-possible-answers/b06146e9-ad56-4ff5-a8dd-a4f68e5a7a66
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Document type: Memorandum
- Year: 2022
- Exam period: Gauteng June
- Paper: 1
- Curriculum: CAPS
- Pages: 12

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## About this document

Mathematics · Grade 11 · Gauteng June · 2022. Memorandum, 12 pages. Read online or download the PDF.

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### Page 1

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           PROVINCIAL EXAMINATION
                   JUNE 2022
                   GRADE 11
            MARKING GUIDELINES

                   MATHEMATICS
                     PAPER 1

12 pages

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                                                     MATHEMATICS
                     MARKING GUIDELINES
                                                     (Paper 1)              GRADE 11

QUESTION 1

1.1     x  −3 ;    x  −1                                        answer
                                                               ✓ exclusion         (2)

1.2.1   ( x + 4)( x − 1) = 0
        x = −4 or x = 1                                        ✓ answers

                                                                                   (1)

1.2.2   3x 2 − 2 x = 14
        3x 2 − 2 x − 14 = 0                                     standard form
             −(−2)  (−2) 2 − 4(3)(−14)                         correct
        x=
                       2(3)                                        substitution
            1  43
        x=
               3
        x = −1,85 or x = 2,52                                     answer
                                                                  answer
        NOTE: Penalise 1 mark for incorrect rounding in this
                   question ONLY.                                                  (4)
                   Any other valid method.
1.2.3                       45
        x 2 − 2 x = 18 − 2
                         x − 2x
        restrictio n  x 2 − 2 x  0
                                                                method
        let : k = x 2 − 2 x
                      45
         k = 18 −
                      k
         k = 18k − 45
             2

         k 2 − 18k + 45 = 0
         (k − 15)( k − 3) = 0
                                                                values of k
         k = 15 or  k = 3
        substitute x 2 − 2 x  back  int o  k .
         x 2 − 2 x = 15 or  x 2 − 2 x = 3                    equating

         x 2 − 2 x − 15 = 0 or  x 2 − 2 x − 3 = 0
         ( x − 5)( x + 3) = 0 or  ( x − 3)( x + 1) = 0
         x = 5 or  x = −3 or  x = 3 or  x = −1           all values of x   (4)
        NOTE: Any other valid method.

                                               2

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                                            MATHEMATICS
                     MARKING GUIDELINES
                                            (Paper 1)             GRADE 11

1.2.4     5 − x +1 = −x
          5 − x = −x −1
        5 − x = x2 + 2 x + 1                         method

        0 = x 2 + 3x − 4                             standard form
        0 = ( x + 4)( x − 1)                         factors
         x = −4 or x  1
                                                     critical values
                                                     rejection
                                                                         (5)
        NOTE: Any other valid method.

1.2.5   32 x +1 − 4.3x = −1
        3.32 x − 4.3x + 1 = 0                        standard form
        (3.3 − 1)(3 − 1) = 0
             x       x
                                                     factors
             1
        3x =     or 3x = 1                           equating
             3
        3x = 3−1 or 3x = 30
        x = −1        x=0                            answers

        OR

        32 x +1 − 4.3x = −1
        let 3x = k
         3k 2 − 4k = −1
        3k 2 − 4k + 1 = 0                            standard form
        (3k − 1)(k − 1) = 0
                                                     factors
             1
        k=      or k = 1                             equating
             3
        3x = 3−1 or 3x = 30                          answers
        x = −1           x=0
                                                                         (4)

                                        3

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                                                    MATHEMATICS
                       MARKING GUIDELINES
                                                    (Paper 1)           GRADE 11

1.2.6(a)   x 2 − 3x  0
           x( x − 3)  0                                     factors
           CVs 0 ; 3
           0 x3
                                                             answer
                                                                                (2)

1.2.6(b)   3 x  6                                         ✓ critical values

           NOTE: Answer can be written as separate          ✓ answer            (2)
                 inequalities.

1.3        x + y = −10                  xy = −600...(2)      setting up
           x = −10 − y...(1)                                  equations
           sub (1) in (2)                                    substitution
           (−10 − y ) y = −600
           −10 y − y 2 = − 600
           y 2 + 10 y − 600 = 0                              factors
           ( y − 20)( y + 30) = 600
           y = 20 or of y = −30                              values of y

           as y = 20       as y = −30
           x = −30          x = 20                           values of x

           NOTA: Enige ander geldige metode.                                    (5)
                                                                                 [29]

                                            4

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                                                          MATHEMATICS
                           MARKING GUIDELINES
                                                          (Paper 1)           GRADE 11

QUESTION 2

2.1.1   2 8 − 4 32 + 3 50
                                           Mark
        = 2 4.2 − 4 16.2 + 3 25.2          allocated        simplify surds
        = 2.2 2 − 4.4 2 + 3.5 2            to 1st step.
                                                            simplification
        = 4 2 − 16 2 + 15 2
                                                            answer
        =3 2                                                                      (3)
             1
2.1.2    −
        3 2 [ 12 + 3 (3 3) ]
             1         1             1 1
         −
        3 2 [( 2 2.3) 2 + (3.3 2 ) 3 ]
                                                            simplify surds
             1     1           3 1
         −
        3 [21.3 + (3 ) ]
             2     2           2 2

             1    1        1
         −
        3 2 [2.3 2 + 3 2 ]
        2.30 + 30                                          ✓ simplification
        2 +1                                               ✓ simplification
        3                                                                         (4)
                                                              answer
2.1.3   52006 − 52004 + 24
             52004 + 1
        52004(52 − 1) + 24
             52004 + 1                                      factorise
        52004(24) + 24
                                                            simplify
           52004 + 1
        24(52004 + 1)                                       factorise
          52004 + 1
                                                            answer
        24                                                                        (4)

2.2.1
        x = 6 + 6 + 6 + .........
                                                              method
         x 2 = 6 + 6 + 6 + ...
         x2 = 6 + x                                          answer             (2)

2.2.2   x2 − x − 6 = 0                                     ✓ standard form
        ( x − 3)( x + 2) = 0                               ✓ factors
        x = 3 x  −2
                                                           ✓ answer
                                                           ✓ exclusion            (4)

                                                 5

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                                                 MATHEMATICS
                   MARKING GUIDELINES
                                                 (Paper 1)         GRADE 11

2.3   d 2 = ( 8 − 1)2 + ( 8 + 1)2   Pythag         method
      d 2 = 8 − 2 8 +1+ 8 + 2 8 +1
      d 2 = 18                                     value of d 2
      d =3 2                                       answer
                                                                       (3)

                                                                       [20]

                                             6

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                                                 MATHEMATICS
                          MARKING GUIDELINES
                                                 (Paper 1)                   GRADE 11

QUESTION 3

3.1.1   -7       0         9        20
             7        9        11
                 2        2
                                                          ✓ second difference
        2a = 2
        a = 1                                               value of a
        3(1) + b = 7
        b = 4                                               value of b
        1 + 4 + c = −7
         c = −12                                                                   (4)
                                                             value of c
        Tn = n + 4n − 12
                  2

3.1.2   Tn = n 2 + 4n − 12
        128 = n 2 + 4n − 12                               ✓ equating
        0 = n + 4n − 140
             2

        0 = (n − 10)( n + 14)                                factors
         n = 10  n  −14                                  selection of n = 10   (3)

3.1.3   Tn = 2n + 5
        599 = 2n + 5                                      ✓ equating
        594 = 2n
        n = 297
                                                             value of n
        This difference will be between term number 297      conclusion
        and 298.                                                                    (3)

3.2.1   93                                                   answer                (1)

3.2.2   Tn = 10n − 3                                      ✓   Tn = 10n − 3
        Tn = (10n − 2) − 1
                                                          ✓ method
        Tn = 2(5n − 1) − 1
        Thus :
        (10n − 2) = 2(5n − 1)                              equating
        10n − 2 EVEN for any value of n.                   explanation
        Tn is ODD when :
        Tn = (10n − 2) − 1                                                          (4)

                                                                                    [15]

                                          7

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                                                       MATHEMATICS
                        MARKING GUIDELINES
                                                       (Paper 1)               GRADE 11

QUESTION 4

4.1     1 − p;2 p − 3; p + 5
        T2 − T1 = T3 − T2                                  ✓ method
         (2 p − 3) − (1 − p) = ( p + 5) − (2 p − 3)
         2 p − 3 −1+ p = p + 5 − 2 p + 3
        3 p − 4 = − p + 8                                  ✓ simplification
         4 p = 12                                          ✓ answer
        p=3                                                                        (3)

4.2.1   T1 = 1 − p
        T1 = 1 − 3
                                                            ✓   answer
        T1 = −2
                                                                                    (1)
        NOTE: Answer only: full marks
4.2.2   T2 = 2 p − 3
        T2 = 2(3) − 3
        T2 = 3
        d = T2 − T1
        d = 3 − (−2)                                        ✓   answer
        d =5
                                                                                    (1)
        NOTE: Answer only: full marks
4.3     1- p ; 2p - 3 ; p + 5
        = 1 - 3 ; 2(3) - 3 ; 3 + 5
        = -2 ; 3 ; 8 ; 13 ; 18; …
        All the terms except T1 end in either a 3 or an 8    explanation
        while perfect squares end on 0; 1; 4; 5; 6; 9 .

                                                                                    (2)

                                                                                    [7]

                                                  8

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                                              MATHEMATICS
                      MARKING GUIDELINES
                                              (Paper 1)               GRADE 11

QUESTION 5

5.1   −3 x  2                                        ✓ answer

      NOTE: Candidates can present the answer as
            separate inequalities.                                           (1)

5.2   f ( x) = a( x − x1 )( x − x2 )
      f ( x) = a( x + 3)( x − 2)
      − 8 = a(1 + 3)(1 − 2) (1;−8)
                                                       ✓ sub. point D and
      − 8 = a(4)( −1)
                                                         x-intercepts
      a = 2
      f ( x) = 2( x + 3)( x − 2)                       ✓ value of a

      f ( x) = 2( x 2 + x − 6)                         ✓ simplification of
                                                         binomial
      f ( x) = 2 x 2 + 2 x − 12
       a = 2 b = 2 c = −12                        ✓ values of a, b, c

                                                                             (4)

5.3          b
      x=−
            2a
              2
      x=−
            2(2)
            2
      x=−
            4
            1                                          ✓ value of x
      x=−
            2
             1     1        1
       f (− ) = 2( ) 2 + 2( ) − 12                    ✓ sub for x into
             2     2        2
                                                         f(x)
           1 1
      f (− ) = − 1 − 12
           2 2
           1       1
      f (− ) = −12                                     ✓ value of y
           2       2
                           1      1
       turning  po int( − ;−12 )
                           2      2

      NOTE: Answer does not have to be in coordinate                         (3)
           form.

                                        9

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                                            MATHEMATICS
                  MARKING GUIDELINES
                                            (Paper 1)            GRADE 11

           1
5.4   x = − +7
           2
           1
      x=6
           2                                      ✓ answer

      NOTE: Answer only: full marks.                                  (1)

5.5   In  g  m = 6
                          1                       ✓ gradient of q
       In  q  m = −
                          6
       y = mx + c
            1
      0 = − (−3) + c
            6
          1
      0= +c
          2
            1
      c=−                                         ✓ value of c
            2
              1    1                              ✓ answer
      y =− x−
              6    2                                                  (3)

      NOTE: Answer only: FULL marks
                                                                      [12]

                                       10

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                                            MATHEMATICS
                       MARKING GUIDELINES
                                            (Paper 1)               GRADE 11

QUESTION 6

6.1

                                                 ✓ shape of p with
                                                   y-intercept = 1
                                                 ✓ shape of h
                                                 ✓ x- and y-intercepts
                                                   of h
                                                 ✓ turning point of h

                                                                         (4)

6.2   See sketch in Q6.1.                        ✓ shape of f(x) with
                                                   y-intercept = 1       (1)

6.3   y = 2( x − 1 + 2) 2 − 8
      y = 2( x + 1) 2 − 8                        ✓ answer                (1)

6.4   p( x) = 4 x
                           1
                1      ( x+ )
       p( x + ) = 4 2
                2
              1          1                       ✓ expansion
      p ( x + ) = 4 x.4 2
              2
                             1
              1
      p ( x + ) = 4 x.(22 ) 2                    ✓ simplification
              2
              1
      p ( x + ) = 4 x.2                          ✓ simplification
              2
                1
       p ( x + ) = 2 p( x)
                2                                                        (3)

                                                                          [9]

                                      11

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                                                MATHEMATICS
                     MARKING GUIDELINES
                                                (Paper 1)               GRADE 11

QUESTION 7

7.1   p=4                                               value of p

                                                        value of q
      q=2

                    a
       f ( x) =       +2
                   x−4
           a                                            sub. point (5 ; 3)
      3=       + 2 T (5;3)
          5−4
      3=a+2
      a = 3− 2
      a =1                                              answer               (4)

7.2                 1                                ✓ answer
      h( x) = −        +2
                   x−4                                                        (1)

7.3   𝑦∈ℝ; y2                                          answer

      NOTE: Both conditions must be stated.                                   (1)

7.4   Axis of Symmetry of f is:

      y =x-p+q                and         y=-x-p+q

      y =x-4+2                        y=-x+4+2
                                                     ✓ method
      y =x-2                          y=-x+6

              Selection of: y = - x + 6
                                                     ✓ selection
                            c=6

                                                                              (2)

                                                                              [8]

                                                                      TOTAL: 100

                                           12

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