# Mathematics Grade 11 Paper 1 Gauteng June 2016 Memo

> Grade 11 Mathematics Paper 1 memorandum from the Gauteng June 2016, 11 pages. Read online or get the free PDF download, question paper linked when available.

- Page: https://www.hlayiso.com/library/gr11-math-p1-june-2016-possible-answer/8abf9959-f0c5-4918-94fd-aa3ebabec0f5
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Document type: Memorandum
- Year: 2016
- Exam period: Gauteng June
- Paper: 1
- Curriculum: CAPS
- Pages: 11

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## About this document

Mathematics · Grade 11 · Gauteng June · 2016. Memorandum, 11 pages. Read online or download the PDF.

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## Document text

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### Page 1

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           GAUTENG DEPARTMENT OF EDUCATION
                PROVINCIAL EXAMINATION
                     JUNE 2016
                      GRADE 11

                   MATHEMATICS P1

                    MEMORANDUM

11 pages

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                                                  MATHEMATICS   Grade 11
                                    MEMORANDUM    (Paper 1)

                    GAUTENG DEPARTMENT OF EDUCATION−
                          PROVINCIAL EXAMINATION

                                                 MATHEMATICS
                                                 (Paper 1)

                                                 MEMORANDUM

QUESTION 1

1.1                                          
                                4            
                     OR x  
                                3                                 (2)
1.2   1.2.1             = x+4                 Squaring
              (                                 both sides
                                              Standard form
              2–x =        + 8x +16
                2                             Factors
              x + 8x +16 + x – 2 = 0
                                              Answers
                 x2 + 9x +14 = 0
                                              x= –2 only
              (x + 2)(x + 7) =0
              x = –2 OR x = –7
                                NA                                (5)
      1.2.2   2x( x - 3) = 1                  Standard form
              2x2 - 6x – 1 =0                 Substitution
              a = 2; b = -6; c = -1           Answer
                                              answer

              x =

              x=
              x=
              x = 3,2 OR x = -0,2                                 (4)

                                       2

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                                                          MATHEMATICS           Grade 11
                                         MEMORANDUM       (Paper 1)

      1.2.3                 >0                         x 1
                                                       x3
                                                       Critical values in
                            >0                           context of an
                                                         inequality
                             >0                        correct notation

                                                      Only focus on
                   -3 < < -1 or > 1
                                                      numerator
                             OR
                   (-3 ; -1)  (1 ; )                                            (4)

1.3                                                    Divide by 2
                                                          73
                                                      
                              2                           16
                                                       Finding square root
               2                                         ( )
                                                       x = 2,89
               2 =                                     x = -1,39

                                                      Use of quadratic
      x    =                                          formula max 2/5 for the
                                                      two correct answers.
      x=
                     OR    x=

      x = 2,89 OR x = -1,39                                                       (5)
                                                                                  [20]

                                           3

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                                                             MATHEMATICS         Grade 11
                                            MEMORANDUM       (Paper 1)

QUESTION 2

2.1                                                       Expansion
                                                          33 3  4.9
      =                                                   3x 1  3
                                                          Answer
      =

      =                                                  If k-method is used
                                                         exactly the same mark
      =                                                  allocation
      = 16 OR
                                                                                   (4)
2.2   2.2.1

                                                         
                                                                  
                                                             5  10 5  2   
              
                        
                  5  10 5  2                                 5 2 5
                       52 5                              simplification
              
                  5  10 5  20
                                                         
                                                              
                                                             5 5 2 5   
                       52 5                                 5  2 55
                  25  10 5
              
                   52 5                                  answer
              
                   
                  552 5        
                5  2 55
              5

              OR

                                       +
                                                           Rationalizing the
                                                           denominator
              =             +                            Rationalizing the
                                                           denominator
              =5 - 2                                      Simplification
                                                          Answer
              =5 - 2
              =5                                                                   (4)

                                              4

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                                                               MATHEMATICS        Grade 11
                                                MEMORANDUM     (Paper 1)

      2.2.2      (                                            7 2008
                                                              7 3  71
                 =(
                                                              (7
                 =(                   +                       Answer
                 =(7 -
                 =(7
                 =343
                                  OR

                                                                2009
      2.2.2      (                                        7 2
      cont.
                                                          (7–1)
                                                          (7
                 =(               +        )2             Answer

                 =                +       )2
                = (7
                = (49)(7)
                = 343                                                               (4)
2.3   from: 2 . 4y = 1
             x
                                                          2   2y

            2x . 22 y = 20                                x + 2y = 0
            x + 2y = 0                                     OR
            x = –2y.................
                                                           x= –2y
      subst  into (4y)x =
                                                          substitution
                      (4y)–2x =                           same bases on either
                      2 4 y  2 4                        side
                           2

                   –4y2 = –4                              y = 1
                   y = 1                                 x = 2
      subst y =  1 into 
      y= 1         x = –2
      y = –1       x =2

      OR

                                                                                    (6)
                                                  5

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                                                         MATHEMATICS   Grade 11
                                            MEMORANDUM   (Paper 1)

         2 x .4 y  1                                 22 y
       2 x .2 2 y  2 0                               x + 2y = 0
       x  2 y  0.......... .1                        OR
           
         4y 
                x    1
                    16
                                                       x= –2y
                                                      22xy  24
           2  2 4
             2 xy

                                                      2 xy  4
            2 xy  4
                       2                              y = 1
                y   .......... ...... 2
                       x                              x = 2
        2
  x  2    0
        x
       x2  4  0
x  2x  2  0
                x  2 or x  2

                    2
               y
                     x
                    2               2
                           or 
                    2              2
               y  1           y 1
                                                                         [18]

                                              6

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                                                             MATHEMATICS           Grade 11
                                       MEMORANDUM             (Paper 1)

QUESTION 3

3.1   3.1.1   Roots are non-real (imaginary)              Non-real roots
              therefore  < 0.                            OR
                                                            < 0                   (2)
      3.1.2   Roots are real and equal,                    Roots are real
              = 0                                         Roots equal
                                                                                     (2)
3.2                          2
                   = (2k -1) – 4(k)(k-1)                   Substitution
                   = 4k2 – 4k +1 -4k2 +4k                  Simplification
                   =1                                      Value of 1
              1 is a perfect square, the coefficients      Perfect square
              are rational, so the roots are rational.                               (4)
                                                                                     [8]

QUESTION 4

4.1   4.1.1   -1; 8; 23; .........                        –1
                                                         8
                                                         23                         (3)
      4.1.2       2
              3k – 4 = 71                                3k2 – 4 = 71
              k2 = 25                                     k2 = 25 or
              k = 5                                        k  5k  5
              k = 5                                      k=5
                                                         No marks for k =  5        (3)
4.2   4.2.1   Quadratic number pattern                    Quadratic
                                                          second difference is
              0 ; 5 ; 12 ; 21; ....                        constant or
                                                           illustration
               5      7   9

                2 2                                        If the learner only
              Quadratic number pattern                     show the pattern
                                                           without justification
              OR/OF                                        only 1 mark
              Quadratic number pattern
              The first difference is not constant
              but the second difference is
              constant.

                                                                                     (2)
                                            7

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                                                                        MATHEMATICS                Grade 11
                                                          MEMORANDUM    (Paper 1)

        4.2.2    2a = 2
                 a = 1                                            a = 1
                 3a + b = 5
                 3(1) + b = 5
                 b = 2                                            b = 2

                 T1 = a + b + c
                  0 = 1 + 2 +c
                 c = –3                                           c = –3
                 Tn = an2 + bn + c
                 Tn = n2 + 2n -3                                   Tn = n2 + 2n -3
                                                                                                     (4)
4.3     4.3.1    Row 4                                              Row 4 = 22
                 7  6  5  4  22
                     2       2       2       2

                 Row 20                                            Row 20 = 86
                 23  22  21  20  86
                         2       2       2           2
                                                                                                     (2)
        4.3.2    (n  3)  (n  2)  (n  1)  n  4n  6  a  n  3
                             2                   2        2   2

                                                                    b  n  2; c  n  1; d  n
                                                                    Tn  4n  6
                                                                   If only the general term
                                                                   was given 1/3                     (3)
                                                                                                     [17]

QUESTION 5

5.1   x = –2 and y = 1                                               x = –2
                                                                     y=1
                                                                    Both has to be in equation
                                                                     form. If not 0/2
                                                                    If p  2 and q  1 0/2         (2)
5.2   Sub B = (0; -2) in y =                         +1              Substitution of
                                                                      (0 ; –2) and q = 1
       -2 =      +1
       -2 =     +1

      –3 =
      then k = -6
       y=      +1                                                   k value
                                                                     Answer                        (3)
                                                              8

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                                                      MATHEMATICS          Grade 11
                                      MEMORANDUM       (Paper 1)

      0 =           +1                              y=0
5.3                                                 x26
      –1 =                                          x=4
      (x +2) = 6                                    Writing Point D in
      x = 4                                          coordinate form.

      D(4 ; 0)
                                                                            (4)
5.4   C(–2 ; 0) and B(4 ; 0)                          CA from 5.3
      y = a(x+ 2)(x – 4)                            x2
      –2 = a(0+2)(0 – 4)                             x  4
      –2 = a (–8)                                   Sub. B (0; –2)
       =a                                           a =
      y = (x + 2)(x – 4)                            answer in any form
      y =            -2x – 8)
         =          – x–2                                                   (5)
5.5                                                  Shape
                                                     Coordinates of
                                                      (0 ; 4)
                                  y                  Graph not crossing
                                                      the x- axis

                                  (0; 4)

                                               x

                                                                            (3)
5.6   y=                                            y =                   (2)

5.7   y is real ,                                   y is real,
      (y R)
                                                      both condition        (1)
                                                                            [20]
                                           9

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                                                                      MATHEMATICS     Grade 11
                                                      MEMORANDUM      (Paper 1)

QUESTION 6

6.1        [0 ; 4]                                                  0
      OR                                                           4
      0 x4                                                                            (2)
6.2   h( x)  ( x2  4x  4  4)                               a = –1.
             x  2   4
                         2                                      p = –2.
                                                                q = –4.                (3)
6.3                                                              CA from 6.1
           y                   y
                             (2 ;4)      (2;4)
                                             h(x)               Shape (neg graph)
                                                                Turning point
                                                                y-intercept.
                                         2        4      x      positive y-values
                                   0
                                                                 only.
           0                                 4           x

                                                                                        (4)
6.4   h(x) = - x + 4x2
                                                                Sub x with x  5
      = –( x – 5 )2 + 4( x – 5)                                 Simplification
      = – (x2 – 10x +25) + 4x -20                               Answer
      = – x2 +10x –25 + 4x -20
      h(x – 5) = – x2 + 14x -45

                                       OR                            OR

      y = – (x – 2)2 + 4                                            (x – 2 – 5)2
      = – (x – 2 – 5)2 + 4
      = – (x – 7)2 + 4
      = – (x2 – 14x + 49) + 4                                       Simplification
      = – x2 + 14x – 49 + 4
      = – x2 + 14x – 45                                             Answer             (3)

                                                        10

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                                                 MATHEMATICS   Grade 11
                                    MEMORANDUM   (Paper 1)

6.5   k(x) = x2 – 4x                          x2
                                              –4x               (2)
6.6            9                                        9
      p(–3) = 2                               p(–3) =
                                                        2
               1                                       1
      p(–1) = 2                               p(–1) =
                                                       2
      Average gradient                        Answer

      =

      =

      Average gradient of p = –2                                (3)
                                                                 [17]

                                      11

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