# Mathematics Grade 12 Paper 2 Mpumalanga Mock 2019 Memo

> Grade 12 Mathematics Paper 2 memorandum from the Mpumalanga Mock Exam 2019, 23 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/2019-grade-12-math-trial-exam-paper-2-mp-memo/dfb26450-8c84-4036-adf5-88bf0173af39
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2019
- Exam period: Mpumalanga Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 23

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## About this document

Mathematics · Grade 12 · Mpumalanga Mock Exam · 2019. Memorandum, 23 pages. Read online or download the PDF.

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### Page 1

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                             NATIONAL
                        SENIOR CERTIFICATE

                               GRADE 12

                      MATHEMATICS PAPER 2

                            September 2019

                       MARKING GUIDELINES

MARKS: 150

                This marking guidelines consist of 23 pages

### Page 2

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 Mathematics/P2 memo                                                    2                             September 2019

 Question 1

1.1                     Interval                   Frequency          Cummulative frequency          frequency
               60  x  70                              5                         5                  accumulative

               70  x  80                              11                        16                frequency

               80  x  90                              22                        38                                  (3)

            90  x  100                                13                        51
           100  x  110                                7                         58
           110  x  120                                3                         61
1.2                                                                                                 angle point
                                                                                                    use of
                                                        OGIVE
                                    70
                                                                                                    cumulative
                                                                                                    frequency
                                    60                                                               upper limits
                                                                                                     free hand curve
                                    50

            Cummulative frequency   40
                                                                                                                     (4)

                 (in days)
                                    30

                                    20

                                    10

                                     0
                                         40   50   60   70     80      90   100   110   120   130
                                                         Daily sales (in Rand)

1.3     61 51  10                                                                                  answer
                                                                                                                      (2)
1.4     Medium = 87 (between 85 and 88)                                                              answer        (2)
                                                                                                                    [11]

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Mathematics/P2 memo                              3                                 September 2019

QUESTION 2

2.1.1      By using a calculator : a = 76,60 (76,59564211)                    a
                                   b = 0,1(0.09544928654)                     b
           ∴ equation of line of least squares is y  76,60  0,10x           equation
                                                                                               (3)
2.1.2      y  76,60  0,10(57 )                                             substitution
            82                                                              answer            (2)
2.1.3      r = 0,1408374211                                                   r  0,14
           r  0,14                                                                             (1)
2.1.4      r value is low which means that the data is spread around the     invalid
           regression line. Therefore prediction using the regression line
           will be invalid.                                                                     (1)
2.2.1      22                                                                 22
                                                                                               (1)
2.2.2      1320                                                                  1320
                 60                                                         
            22                                                                    22
                                                                              60
                                                                                                (2)
                                                                                               [10]

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Mathematics/P2 memo                              4                              September 2019

QUESTION 3

                                        y
                                                             C(6 ; 4)
                                  B(0 ; k)

                                             E       F
                                                                        x
                                         O       A(1 ; –1)
                                                 )

3.1       AB : y  3x  2  0                                     y  3x  2
           y  3x  2                                          k 2
          k  2
                                                                                                (2)
3.2       AC  (1  6) 2  (1  4) 2                             substitution

                 50                                             answer
                5 2                                                                            (2)
3.3                  1 2
          m AB             3
                     1 0                                         m AB  3
                     42 1
          m BC                                                            1
                     60 3                                        m BC 
                                                                            3
                               1
          m AB  m BC  3       1                             m AB  mBC  1
                               3
                                                                                               (3)
           AB̂C  90.
3.4       tan   m AC  1                                        tan   m AC  1
            45                                                
                           1
          tan   m BC 
                           3
            18,43                                             
            45  18,43 [ext of ]                           method
              26,57                                            answer                      (5)

                     OR

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Mathematics/P2 memo                         5                         September 2019

          AC  (1  6) 2  (1  4) 2
                  50
                5 2                                  AC = 5 2

          AB  (0  1) 2  (2  1) 2
                 10                                  AB = 10

                     AB
          sin                                                  AB
                     AC                               sin  
                                                                 AC
                      10
                 
                     5 2                                          10
                                                      sin  
                   5                                             5 2
                 
                  2
             26.57                                answer
          OR                                                                          (5)
             sin  sin B                                 sin  sin B
                                                            
              AB     AC                                   AB    AC
                     10 sin 90
             sin                                    AC = 5 2
                      5 2
                    5                                 AB = 10
                     
                   5
                                                               10 sin 90
               26,57                              sin  
                                                                5 2

                                                      answer                        (5)

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Mathematics/P2 memo                                      6                    September 2019

3.5       BD̂C  90          altitude
           BC is the diameter (line subtends 90 )              BC is diameter

                           06 2 4
          midpoint of BC :       ;   
                            2      2 
                                                                  3 ; 3
                         3 ; 3

          x  32  ( y  3) 2  r 2
          subst B0 ; 2 
          0  32  (2  3) 2  r 2                              Substitution of (0 ; 2)
           r  10
               2

          x  32  ( y  3) 2  10                              equation
                                          OR                                                    (4)

                     BD̂C  90      altitude
                                                                  BC is diameter
                    BC is the diameter (line subtends 90 )

                                      0 6 2  4
                     midpoint of BC :        ;   
                                       2       2                3 ; 3
                                       3 ; 3

                   x  32  ( y  3) 2  r 2
                   subst B0 ; 2 
                                                                  Substitution of (0 ; 2)
                   0  32  (2  3) 2  r 2
                    r 2  10
                                                                  equation
                     x  32  ( y  3) 2  10
                                                                                                (4)
3.6              1  6 2 1                                        7   3
          M AC       ;                                             
                  2      2                                         2   2

           7 3                                                 x7
           ; 
           2 2                                                  y 1
           7 x0                                3 2 y
                                                 
           2   2                                2   2
                                                                 (4)
           x7                                    y 1
          D(7 ; 1)

                                                                                              [20]

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Mathematics/P2 memo                                7                      September 2019

QUESTION 4
                                                           S
                                         y

                                        P

      R                                       O                  T                   x

4.1        T(6; 0)                                              (6; 0)
                                                                                           (1)
4.2        radius  3                                          radius = 3
                                                                                            (1)
4.3                      3                                                     3
           tan 30                                             tan 30  
                        3                                                     3

                  3                                            substitution
           0       ( 3)  c
                 3                                             c  3
           c 3                                                equation
                  3                                                                        (4)
           y       x 3
                 3
4.4                  3                                         gradient
           y          xc     radius  tangent
                      3
                                                                      18
                                                               c 
               3                                                       3
           0    (6)  c
                3
                                                               equation
                18
           c                                                                              (3)
                 3
                     3     18
           y          x
                      3     3

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Mathematics/P2 memo                            8                        September 2019

4.5          3         3     18                                 equating
               x 3     x
            3           3     3                                 simplification
           9 x  54    3x  9                                   x-value
                    
             3 3        3 3                                     substitution
           12 x  45                                            y-value
                15    3
           x      3                                                                   (5)
                 4    4

                  3  15 
           y         3
                 3 4

               9 3
           y     
                    
                 4 

             15 9 3 
           S ;     
                     
              4  4  

4.6        The distance between centres of circles: OT = 6       OT = 6
           r  R  1 3                                          r  R  1 3
                 4                                             4
           The distance between the two circles = OT –(r + R)
                                              = 6 –4
                                              = 2 units         2                       (4)

                                                                                        [18]

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Mathematics/P2 memo                                    9           September 2019

QUESTION 5

5.1        sin(A  180). tan(180  A).cos A                     tan A
                     cos(90  A)                                 sin A
               sin((180  A)).( tan A). cos A                  sin A
           
                            sin A                               identity
                sin A.(  tan A). cos A                        answer
           
                        sin A
                sin A
                    . cos A
                cos A
                                                                               (5)
             sin A
5.2.1
                             1
                             28    a                            sketch
                             1 a2

           cos 28  1  a 2                                    answer
                                                                             (2)
                                              OR
           cos2 28  sin 2 28  1
                                                                 identity
           cos2 28  a 2  1
           cos2 28  1  a 2
                                                                answer
           cos 28  1  a 2
                                                                             (2)
5.2.2

                             1
                                     1  b2
                             32 
                             32
                              b

           cos 64                                              double angle
            cos(2  32 )                                      answer        (2)
            2 cos 32  1
                     2   

            2b 2  1
           OR

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Mathematics/P2 memo                                                 10                    September 2019

           cos 64
            cos(2  32 )
            1  2 sin 2 32                                                          double angle

                     
            1  2 1  b2               2

            1  2  2b      2

            2b 2  1                                                                answer              (2)
           OR
           cos(2  32 )
            cos2 32  sin 2 32

            b2      1 b     2
                                         2
                                                                                     double angle

            b2 1 b2
            2b 2  1                                                                answer              (2)
5.2.3      sin 4
            sin( 32   28  )                                                      compound 
            sin 32 . cos 28   cos 32 . sin 28                                   expansion
             1  b 2  1  a 2   b.a                                         1  b2
                                   
               1  b 1  a   ab
                         2           2
                                                                                      1 a2
                                                                                                           (4)
5.3            1         1       2 tan x
                              
           1  sin x 1  sin x    cos x
                                                                                     LCD
                         1         1      1  sin x  (1  sin x)
           LHS :                                                                   numerator
                     1  sin x 1  sin x (1  sin x)(1  sin x)
                                                                                         2 sin x
                                                             1  sin x  1  sin x   
                                                                                        cos 2 x
                                                                  1  sin 2 x
                                                               2 sin x
                                                           
                                                               cos2 x
                           sin x
                                         2                                               sin x
                2 tan x                                                              
           RHS:          cos x
                 cos x    cos x                                                          cos x

                                                 2 sin x                                 2 sin x
                                                                                    
                                                 cos 2 x                                 cos 2 x

            LHS  RHS                                                                                     (5)

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Mathematics/P2 memo                                   11                   September 2019

                                           OR
               1         1       2 tan x
                              
           1  sin x 1  sin x    cos x
                                                                      LCD
                     1         1      1  sin x  (1  sin x)
           LHS :                                                    numerator
                 1  sin x 1  sin x (1  sin x)(1  sin x)
                                                                      simplification
                                              1  sin x  1  sin x
                                                                         2 sin x
                                                   1  sin 2 x        
                                                                          cos 2 x
                                                2 sin x
                                                                         sin x
                                                cos 2 x                         tan x
                                                                          cos x
                                                  2 sin x
                                                                                           (5)
                                                cos x. cos x
                                                2 tan x
                                            
                                                 cos x
            LHS  RHS

                                                                                           [18]

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Mathematics/P2 memo                             12                      September 2019

QUESTION 6                                                                    [16]

6.1                             1
           sin( x  60 )  cos x
                                2                         co-function
                                     1 
           sin( x  60 )  sin 90  x 
                                     2 
           quadrant 1 :
                         1                              1st quad equation
           x  60  90  x  360.k , k  Z
                           2
               1
           x  x  30  360.k
               2
           3
             x  30  360.k
           2                                               x  20  240.k
           x  20  240.k , k  Z

           quadrant2 :
                                  1
           x  60  180  (90   x )  360.k , k  Z   2nd quad equation
                                  2
                                 1
           x  60  180  90  x  360.k
                                 2
           1
             x  30  360.k
           2
           x  60  720.k , k  Z                         x  60  720.k
           x  {20 ; 60; 260 }                          x  {20 ; 60; 260 }

          OR
                                                                                         (7)

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Mathematics/P2 memo                                       13                    September 2019

                               1
           sin( x  60 )  cos x
                               2
                                      1
           cos[90  ( x  60 )]  cos x
                                      2
           quadrant 1                                             co-function
                                    1
          [90  ( x  60 )]         x  360.k , k  Z
                                    2
                               1
           90  x  60         x  360.k                        1st quad equation
                               2
               3
                x  30  360.k
               2
           x  20  240.k , k  Z

                                                                   x  20  240.k
           quadrant4
                                           1
          [90  ( x  60 )]  360         x  360.k , k  Z
                                           2
                                        1
           90  x  60  360           x  360k
                                        2
                                                                  4th quad equation
                         1
           30  x  360  x  360.k
                          2
             1
            x  330  360.k
             2
                                                                   x  660  720.k
           x  660  720.k
                                                                   x  {20 ; 60; 260 }    (7)
           x  {20 ; 60; 260 }

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   Mathematics/P2 memo                                14                               September 2019

6.2.1                                                                             g:
                                                                                   x  intercepts
                             y
                                                                                   y  intercept
                        1                                                         turning points
                             0.87
                                                                                                          (3)

         –60°           0°          60°     120°    180°        240°   300°
                                                                              x
         300°

6.2.2   20   x  60  or 260  x  300                                         20   x  60
        OR                                                                         260  x  300 
        x  (20 ; 60 )  x  (600  ; 300  ]                                   notation
                                                                                                          (3)
6.2.3   120   x  18 0                                                          120   x  18 0 

        x   60                                                                  x   60 ;300°

        x  300                                                                                          (3)

                                                                                                        [16]

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Mathematics/P2 memo                                           15                   September 2019

QUESTION 7

                                                      R

                                                      h

                                                      T

                            C                                          D

7.1     CR = RD                 (congruency OR SAS)                         CR = RD

        CR̂D  180  2                          (s in )                answer
                                                                           (2)
7.2     TR̂D  90   

                                   h                                       ratio
        cos(90   ) 
                                  RD                                                      h
                                                                            RD 
                     h                                                                  sin
         RD 
                   sin                                                    substitution

                                                                                       1  2 cos   1
                                                                                  2
              h   h 
                                  2
                                     h  h 
                                                  2
                                                                            2h2                  2
        CD                  2               cos(180  2 )
               2
                                                                               sin 
              sin    sin       sin   sin  
                   2
                       h            2                                    
                                                                                 4h 2
                                                                               sin 
                                                                                        
                                                                                      cos 2       
                    2 2  cos 2  
               2h                                                                 2
         
             sin 
                2
                        sin                                                                          (5)
                   2
         
               2h
                     1  cos 2 
             sin 2 

         
               2h 2
             sin 2 
                        
                     1  2 cos 2   1        

         
               2h 2
             sin 
                2
                        
                    2 cos 2              

         
               4h 2
             sin 2 
                        
                     cos 2       
                         2h. cos 
         CD 
                          sin 
7.3                    2h. cos 
        CD                      .
                        sin                                               substitution

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Mathematics/P2 memo                  16             September 2019

                2h. cos 65 
        5,4                 .
                 sin 51 
              5,42 . sin 51 
         h                   .                h to the nearest unit
                2 cos 65 
                                                                     (2)
         h  4,96
         h5
                                                                     [9]

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Mathematics/P2 memo                                          17                       September 2019

QUESTION 8

8.1
                                                       A
                                                       1 2

                               1                       1   E              2
                        D                                                     1   B
                                   2                   2 3

                                                       2
                                                       O
                                                   1

                                                           C

8.1         D̂1  33       (  ' s in same circle segment)                              S R
                                                                                                     (2)
8.2         In Δ DAE :
            Ê1  90                  (line from centre to midpt of chord)              S R
            D̂1  Ĉ  33               (s in same circle segment)
             Â1  57          (  ' s of  )                                          S         (3)

            OR
            In Δ EBC :
                                                                                         S R
            Ê 3  90                 (line from centre to midpt of chord)
            Ĉ  33           ( given)
             B̂1  57          (s of  )
                                                                                         S         (3)
             B̂1  Â1  57             (  ' s in same circle segment)

8.3         Ô1  2Â1  114            (  at center  2   at circumference)         S R
                                                                                                     (2)

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Mathematics/P2 memo                                   18             September 2019

8.4         D̂1  D̂ 2  Â 1  57        (s opposite  sides)        S/R

            But D̂1  Ĉ  33         (s in same circle segment)
             D̂ 2  24                                                S
            OR                                                                      (2)

            DÔC  2  B̂1        (@centre  2  @circum)
            DÔC  2  57 
                      114
                                                                        S/R
            DÔC  Ê 2  D̂ 2 (ext  of  )
             114   90  D̂ 2
             D̂ 2  24
                                                                        S
                                                                                    (2)
8.5          B̂1  B̂2  90 ( in semi  circle)                       S R

            Â 2  57        (s of  )                                S
                                                                                    (3)

            OR
            ADE  ABE (SS)                                           S R

             Â1  Â 2  57                                          S
                                                                                    (3)
                                                                                 [12]

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Mathematics/P2 memo                                       19                       September 2019

QUESTION 9

                                                 C
                          A                  1   3
                                             2

                                   G     2
                                             1                     1
                                                                           F
                                        O                              2

                                                          2
                                                      3        1
                      E                                                        D
                                                           B

9.1          Ê  x       ( at centre  2   at circum)                          S

             F̂1  180  x (opposite s of cyclic quad)                           R
                                                                                   S
             F̂2  x ( s on straight line)
                                                                                                  (3)
            OR
             Ê  x       ( at centre  2   at circum)                          S R
             F̂2  x (ext  of cyclicquad)                                         S/R
                                                                                                  (3)
9.2          F̂2  x      (ext  of cyclic quad OR straight line)
             B̂1  x  Ê     (correspon ding s, EC || BF)                        S R
             F̂2  B̂1  x                                                        S
             DF  DB          (sides oppposite  s)
                                                                                   R
                                                                                   (4)

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Mathematics/P2 memo                                       20                              September 2019

9.3          Â  x            ( at centre  2   at circum)                            S
                                                                                          S
             Ĉ1  Ĉ 2  x        (s opposite  sides)
                                                                                          S     R
             Ĉ 2  Ĉ 3  x      (ext.  of cyclic quad)
             Ĉ1  Ĉ 2  Ĉ 2  Ĉ 3                                                                  (4)

             Ĉ1  Ĉ 3

            OR
                                                                                          S
             Â  x            ( at centre  2   at circum)
                                                                                          R
             Ĉ1  Ĉ 2  Â         (s opposite  sides)
                                                                                          S     R
             Â  F̂2      (exterior  of cyclic quad)
                                                                                                        (4)
             F̂2  Ĉ 2  Ĉ 3 (corresponding s, CE || BF)
             Ĉ1  Ĉ 2  Ĉ 2  Ĉ 3
             Ĉ1  Ĉ 3
9.4          CG  FB, CG || BF                  (given)
             CGBF is a parm                (one pair opp. sides equal and parallel)      R
                                                                                          S/R
             BG  FC  2                    (opp. sides of parm.)                         S
             CF : FD  2 : 1
             BD FD                                                                        S\R
                                          (line || to one side of  or proportionality
             ED CD
                                            theorem)
             BD 1
                                                                                         
                                                                                              BD 1
                                                                                                
             ED 3                                                                             ED 3
                                                                                                        (5)
                                                                                                      [16]

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Mathematics/P2 memo                                    21                        September 2019

QUESTION 10

                                                            D

                         A

                                              G   1             1   H

                 B            C
                                         E                              F

10.1         Construction : DG  AB and DH  AC                             construction

             In ABC and DEF
             1. Â  D̂ (given)
             2. AB  DG (construction)
             3. AC  DH (construction)
                                                                            S/R
             ABC  DEF (SS)

             B̂  Ĝ1 and B̂  Ê (given)                                  S

             Ê  Ĝ1                                                      S/R
             EF || GH (corresponding s )                                 S R
                DG DH
                           (line || to side of  )
                DE DF

             but AB  DG and AC  DH
                                                                                             (6)
                AB AC
                 
                DE DF

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Mathematics/P2 memo                                      22                    September 2019

10.2

                        K

                                                     B
                                                 2
                                                     1

                                                                      O
                                                                      1
                                                                  2
                                             1
                                                 2
                                                     3     2 3
                                             C 4          1 T
                                                                               A

                                                     S

10.2.1      In CKB and AKC
                                                                          S/R
             Ĉ1  Â         (tan-chord)

             K̂  K̂        (common  )                                   S
             KĈT  B̂ 2 (3rd  in  )
             CKB ||| AKC (    )                                     R              (3)

10.2.2       T̂2  Ô1  Â      (ext  of  )                            S/R

             KĈT  Ĉ1  Ĉ 2  Ĉ 3

            But Ĉ 2  Ĉ3  90      (  s in semi-circle)
                                                                          S     R
            But Ĉ1  Â          (proven)

             KĈT  Â  90 

             T̂2  KĈT  Â  90                                       S

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Mathematics/P2 memo                                 23                 September 2019

                                                                                    (4)
10.2.3      COT ||| AKC

             TĈO  Â           ( s opp equal sides; CO = AO)   S/R
                                                                  S
             T̂2  KĈA          ( proved )

             Ô 2  K̂           (3rd  in  )
            COT ||| AKC             (  )                     R              (3)
10.2.4       CKB ||| ΔAKC
              BK CK                                               ratio
                
              KC AK
             BK.AK  KC 2
                                                                  S

             ΔCOT ||| ΔAKC
              KC AC
                                                                ratio
              OT CT
             KC.CT  AC.OT

             KC 
                         AC.OT                                    S
                          CT
                                                                                    (4)

                         AC 2 .OT 2
             KC 2 
                           CT 2
                                                                                  [20]

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