# Mathematics Grade 11 Paper 1 KZN June 2024 Memo

> Grade 11 Mathematics Paper 1 memorandum from the KZN June Exam 2024, 17 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/kzn-maths-grade-11-june-2024-p1-and-memo/95105f0d-8c1a-4872-b9d1-23d423531e17
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Document type: Memorandum
- Year: 2024
- Exam period: KZN June Exam
- Paper: 1
- Curriculum: CAPS
- Pages: 17

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### Page 1

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                                        NATIONAL
                                   SENIOR CERTIFICATE

                                             GRADE 11

                                     MATHEMATICS P1

                                        COMMON TEST

                                            JUNE 2024

MARKS: 100

TIME: 2 hours

                     This question paper consists of 6 pages including cover page

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Mathematics P1                                         2                          Common Test June 2024
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INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

1.      This question paper consists of 5 questions.

2.      Answer ALL the questions.

3.      Number the answers correctly according to the numbering system used in this
        question paper.

4.      Clearly show ALL calculations, diagrams, graphs, etc. which you have used in
        determining your answers.

5.      Answers only will NOT necessarily be awarded full marks.

6.      You may use an approved scientific calculator (non-programmable and non-graphical),
        unless stated otherwise.

7.      If necessary, round off answers correct to TWO decimal places, unless stated otherwise.

8.      Diagrams are NOT necessarily drawn to scale.

9.      Write neatly and legibly.

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Mathematics P1                                                 3                              Common Test June 2024
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QUESTION 1

   1.1     Solve for x:

              1.1.1       x(1 − 2 x) = 0                                                                          (2)

              1.1.2     2 x2 − 5x + 3 = 0                                                                         (3)

              1.1.3       x(3x − 5) = 7 (correct to TWO decimal places)                                           (4)

              1.1.4     (1 − x)( x + 3)  −5                                                                      (4)

                             x2 + 7    2       1
              1.1.5                 +     =−                                                                      (5)
                          x − 2x − 3 x +1
                           2
                                             x −3

   1.2     Given that: x − 3 = 2 x

              1.2.1    Solve for x.                                                                               (4)

              1.2.2    Hence, if      3
                                           3t − 2   3
                                                        3t = 3 , determine the value of t .                       (3)

   1.3     Solve simultaneously for x and y:
           3 y = x + 1 and ( x − y )( 5 y − 3x ) = 0                                                              (5)

                                                                        33 5
   1.4     The roots of the equation x 2 − 3x − k = 0 are x =                 . Determine the value of k.         (4)
                                                                          2
                                                                                                                  [34]

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Mathematics P1                                        4                Common Test June 2024
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QUESTION 2

   2.1     Simplify the following fully, without using a calculator:

                       24 x+1.9 x.62 x −1
              2.1.1                                                                        (4)
                        26 x.27 x.3x

                       32021 − 32017
              2.1.2
                           (
                       60. 3 36048     )                                                   (4)

   2.2     Solve for x, without using a calculator:

              2.2.1    m8 x−4 = 1                                                          (3)

                                              44
              2.2.2    4.3x − 3x − 2 + 3x =                                                (4)
                                               3

              2.2.3    x x = 22048                                                         (3)
                                                                                           [18]

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Mathematics P1                                         5                             Common Test June 2024
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QUESTION 3
                                            x
                                       1
 The function g is defined as g ( x) =   − 4
                                       2

   3.1     Write down the equation of asymptote of g .                                                    (1)

   3.2     Calculate the y - intercept of g .                                                             (2)

   3.3     Calculate the x - intercept of g .                                                             (2)

   3.4     Draw a neat sketch of g. Clearly show all the intercepts with the axes and the
           asymptote.                                                                                     (3)

   3.5     Calculate the average gradient of g between x = 0 and y = 0 .                                  (2)

   3.6     Write down the equation of k if it is given that k ( x) = g ( x) + 4 .                         (1)

   3.7     It is further given that h ( x) = 2 x+3 − 4 . Explain, in words, how graph g must be
           transformed to obtain graph h.                                                                 (3)
                                                                                                          [14]

QUESTION 4

                          a
 The function f ( x) =        + 1 has an axis of symmetry with an equation of x = 3 − y .
                         x+ p

   4.1     Write down the range of f .                                                                    (1)

   4.2     Determine the equation of the vertical asymptote of f .                                        (2)

   4.3     If f (0) = 2 , calculate the value of a .                                                      (2)

   4.4     Calculate the x-intercept of f .                                                               (3)

   4.5     Draw a neat sketch of f . Clearly show all intercepts with the axes and asymptotes.            (3)

   4.6     Write down the values of x for which f ( x)  0 .                                              (2)

   4.7     The straight line y = −2 x + 5 passes through the point of intersection of the
           asymptotes and intersects f at P(1 ; 3) and Q. Write down the coordinates of Q.                (2)
                                                                                                          [15]

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Mathematics P1                                         6                             Common Test June 2024
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QUESTION 5

 The graphs of h( x) = ax 2 + bx + c and p( x) = 8 − 2 x are sketched below.
 The x-intercepts of h are (−3 ; 0) and (1 ; 0) and the y-intercept of h is (0 ; 6). M is the
 turning point of h . D and E are the 𝑥- and y-intercepts of p respectively.

                                p               y

                                    M           E

                                                6

      (–3 ; 0)                                        (1 ; 0)                    D
                                                O                                                 x

         h

   5.1       Write down the coordinates of D.                                                            (1)

   5.2       Show that a = −2 , b = −4 and c = 6 .                                                       (4)

   5.3       Calculate the coordinates of M, the turning point of h .                                    (2)

   5.4       Write down the range of h .                                                                 (1)

   5.5       Determine the values of x for which h( x). p( x) < 0 .                                      (3)

   5.6       If h( x) = k , determine the value(s) of k for which:

                 5.6.1    roots are non-real.                                                            (1)

                 5.6.2    roots have the same sign.                                                      (2)

   5.7       Calculate how graph p must be translated so that it becomes a tangent to graph h.           (5)
                                                                                                         [19]

                                                                                         TOTAL:          100

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                                 FINAL

                            MATHEMATICS P1

                             COMMON TEST

                               JUNE 2024

                          MARKING GUIDELINES

                                NATIONAL
                           SENIOR CERTIFICATE

                                GRADE 11

 MARKS:       100

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Mathematics P1                       2         com                                  Common Test June 2024
                               GRADE 11
                             Marking Guideline

                                    These marking guidelines consist of 10 pages.

QUESTION 1
                         1                                                                       1
1.1.1   x = 0 or x =                                                       A x = 0 A x =
                         2                                                                       2
                                                                                                        (2)
1.1.2     2x − 5x + 3 = 0
             2

        (2 x − 3)( x − 1) = 0                                              A correct factors
             3
        x=        or x = 1                                                 CA answer CA answer
             2
                                                                                                        (3)
1.1.3     x ( 3x − 5) = 7
                                                                           A standard form
        3x 2 − 5 x − 7 = 0
             −(−5)  (−5) 2 − 4(3)(−7)                                     CA correct substitution
        x=
                          2(3)
        x = −0,91 or x = 2,57                                              CA answer CA answer
                                                                                                        (4)
1.1.4   (1 − x)( x + 3)  −5
        x + 3 − x 2 − 3x + 5  0
        − x2 − 2x + 8  0
        x2 + 2x − 8  0                                                    A standard form
        ( x + 4)( x − 2)  0

                                        OR       +        −         +
          −4                    2
                                                     −4        2           CA critical values

        x  −4 or x  2                                                    CA answer CA answer
                                                                                                        (4)

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Mathematics P1                       3         com           Common Test June 2024
                               GRADE 11
                             Marking Guideline
1.1.5        x2 + 7          2        1
                        +         =−
        x − 2x − 3 x +1
           2
                                     x−3
              x +7
                2
                                2     1
                           +       +     =0
        ( x − 3)( x + 1) x + 1 x − 3                 A correct factorisation
        LCD : ( x − 3)( x + 1)
        x 2 + 7 + 2( x − 3) + 1( x + 1)
                                        =0
                  ( x − 3)( x + 1)                   CA correct LCD
        x2 + 7 + 2 x − 6 + x + 1
                                     =0
               ( x − 3)( x + 1)
        x 2 + 3x + 2 = 0                             CA standard form
        ( x + 1)( x + 2) = 0                         CA factors
        x  −1 or x = −2                             CA answers with rejection
                                                                                  (5)

1.2.1   x−3 = 2 x

        ( x − 3) = ( 2 x )
                      2        2

        x2 − 6x + 9 = 4 x                            A squaring both sides
        x 2 − 10 x + 9 = 0                           CA standard form
        ( x − 9)( x − 1) = 0                         CA factors
        x = 9 or x  1                               CA answers with rejection
                                                                                  (4)
                                         OR
        x−3 = 2 x
        x −2 x −3 = 0

        ( x) −2 x −3 = 0
                 2
                                                     A standard form

        ( x − 3)( x + 1) = 0                         CA factors

         x = 3 or              x  −1                CA both equations
        x = 9                                       CA answers with rejection
                                                                                  (4)
1.2.2        3
                 3t = 9                              CA 3 3t = x from 1.2.1

        ( 3t ) = 9
                  3
         3                3

                 3t 729                              M cubing both sides
                    =
                 3     3
                  t = 243                            CA answer
                                                                                  (3)

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Mathematics P1                       4         com                                                            Common Test June 2024
                               GRADE 11
                             Marking Guideline
1.3                                                3 y = x + 1...............................(1)
                             ( x − y )( 5 y − 3x ) = 0................(2)

                                                      x = 3 y − 1...............................(3)   A rewriting in terms of x
       ( (3 y − 1) − y )( 5 y − 3(3 y − 1) )  = 0                                                 CA subst. of (3) into (2)
                          ( 2 y − 1)( 3 − 4 y ) = 0                                                   CA simplification
                                 1                3
                            y=        or y =                                                          CA both values of 𝑦
                                 2                4
                                 1                5
                            x=        or x =                                                          CA both values of 𝑥
                                 2                4                                                                                   (5)

                                                     OR                                                             OR
                          3 y = x + 1..........................................................(1)
       ( x − y )( 5 y − 3x ) = 0...............................................................(2)
             from (2)
                                                            3
         x= y                           or           y=       x.................(3)
                                                            5                                         A solving for y ito x in (2)
             sub (3) into (1)
                                                      3 
       3x = x + 1                     or            3 x  = x +1
                                                      5                                             CA subst. both into eq. (1)
                                                    9
       2x = 1                        or               x = x +1
                                                    5                                                 CA simplification
            1                                            5
         x=                or                        x=
            2                                            4                                            CA both values of 𝑥
          sub x − values into (3)
            1                        3
         y=                 or    y=
            2                        4                                                                CA both values of 𝑦
                                                                                                                                      (5)

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Mathematics P1                       5         com          Common Test June 2024
                               GRADE 11
                             Marking Guideline
1.4                    33 5
                    x=
                          2
                       3  45                        A rewriting ∆ as     45
                   x=
                           2
                   = 45
                          and
        x 2 − 3x − k = 0
                   = b 2 − 4ac
                   = (−3) 2 − 4(1)(− k )            A substitution into ∆
                   = 9 + 4k

                    9 + 4k = 45                     CA equating two ∆s
                4k 45 − 9
                    =
                 4      4
                  k =9                               CA answer
                                                                                 (4)
                                            OR                    OR

       (2 x − 3 + 3 5)(2 x − 3 − 3 5) = 0            A rewriting roots as
       (2 x − 3) − (3 5) = 0
                2          2                            factors of equation
       4 x 2 − 12 x + 9 − 9(5) = 0
       4 x 2 − 12 x − 36 = 0                         A standard form
       x 2 − 3x − 9 = 0                              CA  equation by 4
       k = 9                                        CA answer
                                                                                 (4)
                                                                                [34]

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Mathematics P1                       6         com                  Common Test June 2024
                               GRADE 11
                             Marking Guideline
QUESTION 2

2.1.1   24 x +1.9 x.62 x −1
         26 x.27 x.3x
            24 x +1.( 32 ) .( 3  2 )
                            x             2 x −1
                                                     A rewriting with prime bases
        =
                   2 .( 3 ) .3
                     6x         3 x   x

          24 x +1.32 x.32 x −1.22 x −1               CA separate prime bases
        =
                 26 x.33 x.3x
        = 24 x +1+ 2 x −1−6 x.32 x + 2 x −1−3 x− x
        = 20.3−1                                     CA simplifying: exponential laws
              1
        = 1
              3
          1
        =                                            CA answer
          3
                                                                                            (4)

2.1.2   32021 − 32017
              (
        60. 3 36048       )
          32017 (34 − 1)                             A common factor in numerator 32017
        =                                            A 32016 in denominator
            60.32016
          32017 − 2016 (80)                          A 80 in numerator
        =
                 60
        =4                                           CA answer
                                                                                            (4)
2.2.1   m8 x − 4 = 1
        m8 x − 4 = m 0                               A m0
        8x − 4 = 0                                   CA equating exponents
                   1
              x=                                     CA answer
                   2
                                                                                            (3)

        OR                                           OR

        m8 x − 4 = 1
         m8 x
                 =1
         m4
        m8 x = m4                                    A m8 x = m4
        8x = 4                                       CA equating exponents
             1
         x=                                          CA answer
             2                                                                              (3)

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Mathematics P1                       7         com           Common Test June 2024
                               GRADE 11
                             Marking Guideline

                              44
2.2.2     4.3x − 3x − 2 + 3x =
                               3
                              44
        4.3x − 3x.3−2 + 3x =
                               3
                              44              A common factor of 3x
           3x (4 − 3−2 + 1) =
                               3
                      44  44                                44
                  3x   =                    A factor of
                      9  3                                  9
                              44 9
                        3x = 
                               3 44
                        3x = 31               CA simplification
                          x =1                CA answer
                                                                                       (4)
2.2.3   x x = 22048
        x x = 22(1024)                        A rewriting the exp. as 2(1024)
        x x = 42(512)
        x x = 16(512)                         A rewriting as 16(512)
        x x = 162(256)
        x x = 256256
         x = 256                             A answer
                                                                                       (3)

        OR

        x x = 22048
        x x = 21024.21024                     A rewriting the RHS as 21024.21024
        x x = (2.2)1024
        x x = 41024
        x x = 4512.4512
        x x = 16256.16256                     A rewriting as 16256.16256
        x x = (16.16) 256
        x x = 256256
         x = 256                             A answer
                                                                                       (3)
                                                                                      [18]

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Mathematics P1                       8         com             Common Test June 2024
                               GRADE 11
                             Marking Guideline
 QUESTION 3
3.1   y = −4                                   A answer                                (1)
                     x
                 1
3.2     g ( x) =   − 4
                 2
                     0
                1
        g (0) =   − 4                        A g (0) OR substitute x = 0
                2
        g (0) = 1 − 4
            y = −3                             CA value of 𝑦                          (2)
                     x
                 1
3.3     g ( x) =   − 4
                 2
                     x
               1
           0=  −4                            A g ( x) = 0
               2
           4 = 2− x
          22 = 2− x
           x = −2                              CA value of 𝑥                          (2)
3.4.

                                               A shape

                                               CA intercepts

                                               A asymptote
                                                                                        (3)
           y2 − y1
3.5     m=
           x2 − x1
           (0) − ( −3)                         CA substitution of x- and y- intercepts
        m=                                     from 3.2 and 3.3
           (−2) − (0)
             3
        m=−                                    CA answer
             2                                                                          (2)
                     x
                1                            A answer
3.6    k ( x) =  
                2                                                                     (1)
3.7    reflection about y -axis                A reflection about y−axis
       and then translation/shift of 3 units   A translation of 3 units
       to the left                             A left                                  (3)
       OR                                                       OR
       translation/shift of 3 units            A translation of 3 units
       to the right                            A right
       and then reflection about y-axis
                                               A reflection about y−axis               (3)
                                                                                       [14]
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Mathematics P1                       9         com                 Common Test June 2024
                               GRADE 11
                             Marking Guideline
 QUESTION 4
4.1   y , y 1          or    y  ( −  ; 1)  (1 ; )   A answer                             (1)
4.2   x = 3− y
      x = 3 −1                                            A substitution
      x=2                                                 A answer
                                                                                                 (2)
                     a
4.3      f ( x) =      +1
                  x+ p
                    a
         f ( x) =      +1
                  x−2
                    a
              2=       +1                                 A substitution of a point (0; 2)
                  0−2
                   a
         2 −1 =
                  −2
              a = −2                                      CA answer
                                                                                                 (2)
                      −2
4.4          f ( x) =    +1
                     x−2
                      −2
                 0=      +1                               A    f ( x) = 0
                     x−2
                      −2
               −1 =
                     x−2
        −1( x − 2) = −2                                   CA simplification
          − x + 2 = −2
               − x = −4
                 x=4                                      CA answer
                                                                                                 (3)

4.5

                                                          CA shape based on a in 4.3

                                                          CA vertical asymptote from 4.2
                                                              but horizontal asymptote of
                                                              y = 1 must be accurate

                                                          CA x-intercept from 4.4 but y-
                                                              intercept must be accurate
                                                              (given in 4.3)
                                           f
                                                                                                 (3)
4.6     2< x  4        OR    x  ( 2;4]                  CA answer
                                                                                                 (2)
4.7     Q(3 ; − 1)                                        A x−value A 𝑦 −value
                                                                                                 (2)
                                                                                                [15]
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Mathematics P1                      10         com                                     Common Test June 2024
                               GRADE 11
                             Marking Guideline
QUESTION 5

5.1     D(4 ; 0)                                                            A answer
                                                                                                               (1)
5.2     h( x) = a ( x − x1 )( x − x2 )                                      A defining equation
        h( x) = a ( x − ( −3))( x − (1))                                    A substituting x-intercepts
        6 = a (0 + 3)(0 − 1)                                                A substituting ( 0 ; 6 )
        6 = a ( −3)
         6
            =a
        −3
        −2 = a
        h( x) = −2( x + 3)( x − 1)
        h( x) = −2( x 2 + 2 x − 3)                                          A simplification
        h( x) = −2 x 2 − 4 x + 6
                                                                                                               (4)
        a = −2 , b = −4 , c = 6
5.3     h( x) = −2 x 2 − 4 x + 6
                b
        x=−
               2a
               −4
        x=
             2(−2)
        x = −1                                                              A x−value
        h(−1) = −2(−1) 2 − 4(−1) + 6
        y =8                                                                CA 𝑦 −value
        M(−1 ; 8)                                                                                              (2)
                                           OR                                                OR
             −3 + 1
        xM =
               2
        xM = −1                                                             A x−value
        h(−1) = −2(−1) − 4(−1) + 6
                         2

        y =8                                                                CA 𝑦 −value
        M(−1 ; 8)                                                                                              (2)
5.4      y , y 8                                                          CA answer (using xM in 5.3)

       OR                                                                   OR

        y  (− ; 8]                                                        CA answer (using xM in 5.3)
                                                                                                          (1)
5.5     x < −3        or 1 < x < 4                                          A x < − 3       ACA 1 < x < 4

                                           NOTE:                            OR
       OR
                                           CA is only on 4 i.e. xD in 5.1
        x ( −  ; − 3)  (1 ; 4 )                                           x ( −  ; − 3) ACA x (1 ; 4 )
                                                                                                               (3)

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Mathematics P1                      11         com                                  Common Test June 2024
                               GRADE 11
                             Marking Guideline
5.6.1   k 8                                                               CA answer (using xM in 5.3)
           OR

        k  (8 ;  )                                                       CA answer (using xM in 5.3)
                                                                                                               (1)
5.6.2   6<k 8                                                             ACA answer
                                          NOTE:
           OR
                                          CA is only on 8 i.e. xM in 5.3
        k (6; 8]                                                          ACA answer
                                                                                                               (2)
5.7     If graph p is to now become a tangent, then this implies
        that the NEW 𝑦 −intercept is unknown
         h( x ) = p ( x )
         −2 x 2 − 4 x + 6 = −2 x + k                                       A equating equations
         −2 x 2 − 2 x + 6 − k = 0
         2 x2 + 2 x − 6 + k = 0                                            CA standard form
         =0
        b 2 − 4ac = 0
        (2) 2 − 4(2)(−6 + k ) = 0                                          CA equating ∆= 0
         4 − 8(−6 + k ) = 0
         4 + 48 − 8k = 0                 NOTE:
                                         Correct answer only:
        52 = 8k
                                         award only 1 mark
               52 13        1
         k=       = = 6 = 6,5                                              CA value of k
                8     2     2
        Graph p must shift/be translated 1,5 units downwards               CA answer
                                                                                                               (5)
        OR

        h( x ) = p ( x ) + k
        −2 x 2 − 4 x + 6 = −2 x + 8 + k                                    A equating equations
        −2 x 2 − 2 x − 2 − k = 0
        2 x2 + 2 x + 2 + k = 0                                             CA standard form
        =0
        b 2 − 4ac = 0
        (2) 2 − 4(2)(2 + k ) = 0                                           CA equating ∆= 0
        4 − 8(k + 2) = 0
        4 = 8(k + 2)                     NOTE:
                                         Correct answer only:
        1
           =k+2                          award only 1 mark
        2
               3      1
        k = − = −1 = −1,5                                                  CA value of k
               2      2
        Graph p must shift/be translated 1,5 units downwards               CA answer
                                                                                                          (5)
                                                                                                        [19]
                                                                                                   TOTAL:100

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