# Mathematics Grade 12 Paper 2 Limpopo Mock 2018 Memo

> Grade 12 Mathematics Paper 2 memorandum from the Limpopo Mock Exam 2018, 14 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/2018-grade-12-math-trial-exam-paper-2-limp-memo/a3d3c686-608c-4e4b-9813-4aa1eb2786b9
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Memorandum
- Year: 2018
- Exam period: Limpopo Mock Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 14

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## About this document

Mathematics · Grade 12 · Limpopo Mock Exam · 2018. Memorandum, 14 pages. Read online or download the PDF.

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## Document text

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### Page 1

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                                        NATIONAL
                            SENIOR CERTIFICATE

                                        GRADE/GRAAD 12

                          MATHEMATICS PAPER 2/

                           WISKUNDE VRAESTEL 2

                                    MEMORANDUM

                                 SEPTEMBER 2018

MARKS/PUNTE: 150
TIME/TYD: 3 HOURS/URE

                        This memorandum consists of 14 pages.
                      Hierdie memorandum bestaan uit 14 bladsye

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### Page 2

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                  Mathematics P2 /Wiskunde V2                                           2                    LimpopoDoE/September 2018
                                                                                  NSC-Memorandum

 QUESTION/                                                         1
 1.1                                                     1140                                                                   1140
                                                    ̅=           22,8                                                      ✓
                                                          50                                                                     50
                                                                                                                            ✓ answer/

                                                                                                                                       (2)
 1.2                                                       Marks              Frequency                Cumulative
                                                                                                       Frequency
                                                       0                            4                       4
                                                      10                           20                      24               ✓4;24;39
                                                      20                           15                      39               ✓44;50
                                                      30                            5                      44                    (2)
                                                      40                            6                      50

1.

 Cumulative Frequency/𝑲𝒖𝒎𝒖𝒍𝒂𝒕𝒊𝒆𝒘𝒆 𝒇𝒓𝒆𝒌𝒘𝒆𝒏𝒔𝒊𝒆
 3

                                                                   Marks out of 50/

                                               ✓(0;0) ✓upper limits/
                                               ✓all points correct /                                          (3)

 1.4                                             Q1  15 (accept 14 to16)/ (aanvaar van 14 tot 16)                       ✓ Q1  15
                                                 Q 3  29 (accept 28 to 30)/ (aanvaar van 28 tot 30)                     ✓ Q 3  29
                                                 IQR/IKO = 29  15 = 14                                                  ✓answer
                                                                                                                         /       (3)
                                                                                                                              [10]

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 Mathematics P2 /Wiskunde V2                     3                LimpopoDoE/September 2018
                                           NSC-Memorandum

QUESTION/               2
2.1     143                                                                 143
       x     14 ,3 seconds/                                          ✓x 
          10                                                                  10
          = 2,87 seconds/              .                               ✓14,3
                                                                       ✓ = 2,87
                                                                       ✓accuracy
                                                                       /                (4)
2.2      = 17,93                                                       ✓
         =  0,07                                                      ✓
       yˆ   0,07 x  17 ,93
                                                                       ✓equation/

                                                                        (3)
2.3    r   0,69                                                      ✓ r  0,69 (1)
2.4    y   0,07 (80 )  17 ,93                                       ✓
       y  2,33 seconds.                                               substitute/
                                                                       ✓
                                                                       answer/
                                                                       (2)
                                                                                     [10]

QUESTION/               3

3.1   QR =     (5  3) 2  (3  3) 2                        ✓substitution/
          = 64  36
          = 100
          =10                                               ✓10
                                                                                (2)

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 Mathematics P2 /Wiskunde V2                            4               LimpopoDoE/September 2018
                                                  NSC-Memorandum

3.2      53 33                                                   ✓ -value/
       M(        ;   )
            2      2                                               ✓ -value /                    (2)
      = M (1;0 )
3.3                                       03                      ✓m
       P (  5;3) and M (1;0 ) : m 
                                          1 5
                                           1
                                          =
                                              2
                    1                                              ✓subst of m and point/
       y0             ( x  1)
                    2
                    1             1
            y          x
                     2            2
                                                                   ✓equation/
                                                                                               (3)
3.4    r  5; centre (1;0 ) :                                      ✓r = 5 and/        (1;0 )
         x  1  y  25
                2         2
                                                                   ✓LHS/
                                                                   ✓RHS/
                                                                                                     (3)
3.5   PM =      (1  5)  ( 3)
                              2       2
                                                                   ✓PM =      45
            =   45                                                 ✓    25
                    25                                             ✓conclusion/
                                                                                                     (3)
         P lies OUTSIDE the circle.
3.6    S(3;9)                                                      ✓ -value/
                                                                   ✓    value/                         (2)
3.7             33                                                ✓ m PQ  3
       m PQ                  =-3
                53
        tan  =  3                                                ✓ tan  =  3
                                                                   ✓   108 , 43 
         180   71,57 
                                                                   ✓   71,57 
          108 , 43 
                                                                                                     (4)
         71,57  co-interior angles, PR             -axis.
                                                                                                       [19]

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 Mathematics P2 /Wiskunde V2                                5                 LimpopoDoE/September 2018
                                                      NSC-Memorandum

QUESTION/                  4

4.1.1   C (3 ;  2 )                                                     ✓ -value/
                                                                         ✓ -value                (2)
4.1.2    4y  8  0                                                     ✓ -value/
              y2                                                        ✓ -value/           (2)
             B( 0;2)
4.1.3
        r 2  (3  0 ) 2  (  2  2 )
                                           2
                                               =25                       ✓ r 2  25
        r5                                                              ✓r = 5
        x  3  5  2                                                   ✓ x  2
        x  35  8                                                      ✓x 8           (4)
4.1.4 Let D ( x; y ) :  x  0 2  ( y  2) 2 =25 and                   ✓subst in distance formula/
              3
        y        x2                                                           3
              4                                                          ✓y        x2
                                     3                                          4
                          x2  (          x  2  2 ) 2  25             ✓replace y/
                                      4
                                     9
                          x2             x 2  25                       ✓ x 2  16
                                16
                                                                         ✓ -value/
                             25 x 2
                                           25
                                16
                                                                         ✓ -value/
                                  x 2  16
                                     x   4 or x  4                                      (6)
                                                      3
                                     n/a         y       ( 4)  2  5
                                                      4
                               D ( 4;5)

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 Mathematics P2 /Wiskunde V2                        6                 LimpopoDoE/September 2018
                                              NSC-Memorandum

4.2

4.2.1         Let C ( x ; 2 x ) and A ( x ; 0 )                          ✓coordinates of C
               ( x  1)  ( 2 x  2)  ( x  x)  ( 2 x  0)
                      2              2            2            2         and A/
               x 2  2 x  1  4 x 2  8x  4  4x 2
                                                                         ✓equating two r 2 /
               x 2  6x  5  0
                                                                                        r2
               ( x  1)( x  5)  0
               x  1 or x  5
               C (1; 2) or C (5 ;10 )
                                                                         ✓standard form/

                                                                         ✓factors
                                                                         ✓ C (1; 2)
                                                                         ✓ C (5 ;10 )

                                                                                             (6)
4.2.2          r  2 or r  10                                           ✓r 2
                                                                         ✓ r  10            (2)
                                                                                             [22]

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 Mathematics P2 /Wiskunde V2                                     7              LimpopoDoE/September 2018
                                                           NSC-Memorandum

QUESTION/                         5
5.1.1   y 2  12  ( 1  k 2 ) 2                                            ✓diagram and/
                                                                            Pythagoras
              11 k 2                                      1
                                                                    k
           k2
        yk
                                                                            ✓ sin 25   k
        sin 25   k                                                                                 (2)
5.1.2   sin 50   2 sin 25  cos 25                                       ✓double angle expansion/
                   2k 1  k          2

                                                                            ✓substitution
                                                                            (2)
5.2
            tan( 207 )              sin 2 ( x  360 )
                              
             tan 333             sin( x  90 ) cos x

                                                                            ✓  tan 27 
               tan 27                   sin 2 x
        =                                                                  ✓  tan 27 
               tan 27            cos x. cos x
                                                                            ✓ sin 2 x
                                                                            ✓  cos x
                    sin 2 x
         1                                                                        cos 2 x  sin 2 x
                    cos 2 x                                                 ✓
                                                                                           cos 2 x

              cos 2 x  sin 2 x                                                     1
                                                                           ✓
                     cos 2 x                                                     cos 2 x

                1
        =                                                                   ✓answer /                      (7)
            cos 2 x
            1
        =
          cos x
5.3                                                                         ✓replace/
                    sin2A         cosA                                      tan 2 A
        RHS:                  
                    cos2A sinA                                              ✓expansion of sin 2 A
               =
                     2sinAcosA. cosA                                        /           sin 2 A
                       sinA.cos2A
                     2 cos 2 A                                                  2 cos 2 A
                =                                                           ✓
                      cos 2 A                                                   cos 2 A

               1  2cos 2 A  1
        LHS:                                                                ✓ replacing cos 2 A in
                     cos2A
                  2 cos 2 A                                                 numerator/
              =                                                                        cos 2 A
                    cos 2 A
                                                                                  2 cos 2 A
                                                                            ✓
             LHS = RHS                                                            cos 2 A
                                                                                                        (5)

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 Mathematics P2 /Wiskunde V2                         8                        LimpopoDoE/September 2018
                                               NSC-Memorandum

5.4     2 sin 2 x   cos 2 x
        2 sin 2 x      cos 2 x
                   =                                                      ✓dividing by cos 2 x /
         cos 2 x        cos 2 x
        2 tan 2 x   1                                                                 cos 2 x
                     1
        tan 2 x                                                                         1
                     2                                                    ✓ tan 2 x  
        2 x  153 ,43   k180                                                         2
        x  76 ,72   k .90 , k  Z                                     ✓ 2 x  153 ,43   k180 
                                                                          ✓ x  76 ,72   k .90 
                                                                          ✓ k Z
                                                                                                     (5)
                                                                                                     [22]

QUESTION/                 6

6.1     Graph done/                                              : ✓ intercepts/
                                                                ✓turning points/
                                                                ✓shape/                                (3)
6.2.1   x  ( 120  ; 0)  ( 240 ; 270 ]                    ✓ (  120 ; 0 )
                                                                ✓ ( 240 ; 270 ]
                                                                ✓notation /                                 (3)
6.2.2   x  [ 180 ;  60 ]  [ 30 ;60 ]  [150 ;270 ]   ✓ x  [ 180 ;  60 ]
                                                                ✓ [  30 ;60 ] ]
                                                                ✓ [150 ;270 ]                             (3)
6.3     y  sin( x  30   120 )                          ✓✔ y  sin( x  30   120 )
           sin( x  90 )                ✓   cos x        ✔ answer/                                      (2)
             cos x
                                                                                                       [11]

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 Mathematics P2 /Wiskunde V2                              9                      LimpopoDoE/September 2018
                                                    NSC-Memorandum

QUESTION/                       7

7.1              2h                                                    ✓ratio/
      tan  
                 BE                                                               2h
                                                                       ✓ BE =
                2h                                                               tan 
      BE =
            tan                                                                                     (2)
7.2           ED                                                        ✓ tan 
      tan  
               h                                                        ✓correct ratio /
       ED = h tan                                                                                   (2)
7.3   BD  BE  ED  2 BE .DE . cos BEˆ D
          2             2        2                                      ✓Correct use of cos
         =                                                              rule/
                2
       2h                          2h 
                 h tan    2           h tan  cos 120  
                              2
      
       tan                        tan  
                 4h 2
            =           h 2 tan 2   4 h 2  cos 60  
               tan 
                   2
                                                                        ✓substitution/
                4h 2                         1
           =           h 2 tan 2   4 h 2   
              tan 
                  2
                                             2
                4h 2                                                    ✓simplification/
          =                  h 2 tan 2   2 h 2
              tan  2

              4 h 2  h 2 tan 4   2 h 2 tan 2 
          =
                            tan 2 
               h 2 ( 4  tan 4   2 tan 2  )
          =                                                             ✓simplify /
                               tan 2 
              h tan 4   2 tan 2   4                                 ✓ h 2 taken out as common factor/
          =
                              tan 
                                  2
                                                                                  h2

                                                                                                    (5)
                                                                                                   [9]

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 Mathematics P2 /Wiskunde V2                     10                        LimpopoDoE/September 2018
                                           NSC-Memorandum

QUESTION/               8

8.1    Q̂ 1  32     (tan-chord theorem/                         )            ✓S✓R                 (2)

8.2    P̂4  Q̂ 1  32  (   s opp equal sides/                                ✓S✓R                (2)
                                                               )
8.3    M̂ 1  64                ( at centre is 2        at circumference/     ✓S✓R                (2)
                                                          
8.4    M̂ 2  116                (    on a str line /                 )       ✓S✓R
        R̂  58                 ( at centre is 2        at circumference/     ✓S✓R                (4)
                                                            
                                                                                                   [10]

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 Mathematics P2 /Wiskunde V2                          11                             LimpopoDoE/September 2018
                                                NSC-Memorandum

QUESTION/                   9

9.1    Ĵ 3  90               (line from centre to midpoint/                           ✓S ✓R

       (l                                                            )

       N 2  90       (tangent to radius/                                               ✓S ✓R
       Ĵ 3  N̂ 2                                                                       ✓S
       ONPJ is a cyclic quad /koordevierhoek
                     (CONVERSE of ext. of cyclic quad/                                   ✓R                  (6)

                                                                         )
9.2
       Ô 1  2Ĉ  2 x                 ( at centre =2        at circumference           ✓S ✓R
                                                                 
       Ô 4  180   2 x               ( ‘s on a str line/                      )
                                                                                         ✓S✓R
       P̂1  180   2 x                (ext.   of a cyclic quad/
                                                                             )           ✓S✓R

                                                                                                           (6)
                                                                                                           [12]

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 Mathematics P2 /Wiskunde V2                          12            LimpopoDoE/September 2018
                                                NSC-Memorandum

QUESTION/                    10

  10.1

                                                    h

                                                          k

10.1     Construction: Join PQ and PF and draw altitudes h and k/          ✓construction/

                         1
                              DP  h
         area  DPQ                        DP
                     2                
         area  PEQ   1                    PE                              ✓S
                              PE  h
                         2

                         1
                              DQ  k
         area  DPQ                        DQ
                     2                                                   ✓S
         area  PQF   1                    QF
                              QF  k
                         2
         But the area PQE  area PQF (same base and same height/         ✓S✓R
                                                                       )
          area  DPQ         area DPQ
                         
           area  PEQ        area PQF                                     ✓S
           DP       DQ
                
           PE       QF

                                                                                                 (6)

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 Mathematics P2 /Wiskunde V2                          13                      LimpopoDoE/September 2018
                                                NSC-Memorandum

  10.2

10.2.1   Ĉ 3  B̂ 2    (tan-chord theorem                           )            ✓S✓R
                                                                                  ✓S
         Ĉ 3  D̂ 2    (alternate/                      ’s; BD CE )
         B̂ 2  D̂ 2
         BC = DC (sides opposite equal ’s/                               e)       ✓R                      (4)
10.2.2 In BAF and/ en            DCF:
         B̂1  B̂ 2  D̂ 3     (ext. of a cyclic quad/                        )   ✓S✓R
                                                                                  ✓S
         Â 1  Â 2  Ĉ 3  Ĉ 4 (ext.   of a cyclic quad/
                               )
         F̂ is common/
             BAF             DCF            (     )                               ✓R                      (4)

10.2.3   BA        AF                                                             ✓S
                                  (similar/                   ’s)
         DC        CF
         BA        DC
               
         AF        CF                                                             ✓S
         BA        BC
                                  (BC = DC)                                      ✓Replacing/              DC
         AF    CF
             BC DE
         But                      (BD CE; prop theorem/                          ✓S✓R
             CF   EF
           BA DE
                                                                                                          (5)
           AF   EF

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 Mathematics P2 /Wiskunde V2                       14                  LimpopoDoE/September 2018
                                             NSC-Memorandum

10.2.4 In ECD and EAC:
       Ê 1 s common/
         Ĉ 3  Â 2        (tan-chord theorem                 )
                                                                            ✓S✓R
         D̂ 3  Ĉ 2  Ĉ 3              ( ’s of a / ’e van ‘n ∆)           ✓S
             ECD ||| EAC                 (     )                            ✓R
                                                                                                    (4)
10.2.5     EC          ED
                                       (||| ’s)
           EA   EC                                                          ✓S
          EC  EA.ED
              2                                                             ✓S
                  BA.EF
         But ED                        (from/     10.2.3)                  ✓S
                   AF

                   EA.BA.EF
         EC 2                                                                                    (3)
                        AF
                                                                                                 [26]

                                                                           TOTAL/               : 150

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## Collection

- [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers)

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Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. When citing, link the page URL above, not a PDF or /api/ URL.