# Mathematics Grade 11 Paper 2 North West June 2019 Memo

> Grade 11 Mathematics Paper 2 memorandum from the North West June Exam 2019, 12 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/mathematics-gr-11-memo-paper-2/7d9c93f3-dc3b-4d29-bf3d-ccd86d08de1c
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Document type: Memorandum
- Year: 2019
- Exam period: North West June Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 12

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## About this document

Mathematics · Grade 11 · North West June Exam · 2019. Memorandum, 12 pages. Read online or download the PDF.

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## Document text

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### Page 1

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                                         GRADE 11

                             MATHEMATICS: MEMO
                           MID-YEAR EXAMINATION
                                         JUNE 2019

MARKS: 100
                                JUNE EXAMINATION 2012

This marking guideline consists of 12 pages.

                         Demo
                                 NW/JUNE/MATH/ EMIS/6*******

### Page 2

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Mathematics/ P2                                Mid-Year Examination                          NW/ Jun 2019

QUESTION 1

 1.1
                 y2  y1
           m
                 x2  x1
                   1 4
           mAB 
                   1  3                                                   🗸 substitution         (2)
             3
           
             4                                                              🗸 answer
 1.2
            3 a 4 2                                                         3 a
                 ;       0;1                                           🗸        0
            2       2                                                          2
             3 a
                   0
               2                                                            🗸 answer
           a  3                                                                                  (2)
                                                                                                   [4]

QUESTION 2
                                              P

                                                                            S

       Q

                                                   R                            T

 2.1
                   5  3
           mPR 
                    42
                  4                                                       ✓ m  4
           y  y1  m  x  x1 
                                                                            ✓ substitution
           y  3  4  x  2 
           y  4 x  11                                                    ✓ answer
                                                                                                   (3)

                                       Demo

                                   2          NW/JUNE/MATH/ EMIS/6*******

### Page 3

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Mathematics/ P2                                      Mid-Year Examination                       NW/ Jun 2019

 2.2
        mPQ  tan 45                                                          ✓ tan 45
             1                                                                ✓ answer               (2)
 2.3
        d      x2  x1    y2  y1 
                          2                      2

        PQ        m  2    n  3
                           2                 2
                                                                               ✓ substitution

         m 2  4m  4  n 2  6n  9
                                                                               ✓ simplification
         m 2  4m  n 2  6n  13

                   m  4    n  5
                              2              2
        QR                                                                    ✓ substitution
         m 2  8m  16  n 2  10n  25
                                                                                                      (3)
         m 2  8m  n 2  10n  41
 2.4
        PQ  QR
                                                                               ✓ PQ = QR
          m 2  4m  n 2  6n  13  m 2  8m  n 2  10n  41
        m 2  4m  n 2  6n  13  m 2  8m  n 2  10n  41
        4m  6n  13  8m  10n  41
        4m  16n  28
                                                                               ✓ m  4n  7
        m  4n  7 ................... 1

        mPQ  1
        n3
             1                                                                    n3
        m2                                                                    ✓       1
        n3  m2                                                                  m2

        m  n  1 ....................  2                                    ✓ m  n 1

        Substitute  2  in 1 :

        n  1  4n  7                                                         ✓ substitution
        3n  8
                                                                               ✓ 3n  8
               8
        n
               3
               8
        m   1
               3
            11
        
             3
                        11 8 
        Q m ; n  Q   ;  
                        3 3                                                                         (6)

                                      Demo

                                  3              NW/JUNE/MATH/ EMIS/6*******

### Page 4

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Mathematics/ P2                         Mid-Year Examination                          NW/ Jun 2019

 2.5
        mQS  mPR  1
        mQS   4   1
              1                                                                1
        mQS                                                         ✓ mQS 
              4                                                                4
        y  mx  c
          8 1  11 
           c                                                   ✓ substitution
          3 4 3 
             7                                                               7
        c                                                          ✓ c
             4                                                               4
           1   7
        y  x                                                                              (4)
           4    4                                                    ✓ answer
 2.6
             ˆ m
        tan QRT
                          8
                    5    
                         3
                         11 
                    4 
                         3
                      7
                    
                   3
                    23
                     3
                      7                                                        7
                                                                   ✓ m
                      23                                                       23
            ˆ  180  tan 1   7 
         QRT                      
                               23 
         180  16,93
                                                                     ✓ 163, 07
         163, 07
        QRS  163, 07  45                                         ✓for 45
             118, 07                                               ✓ QRS  118,07
         QPS  118, 07                                             ✓ answer               (5)
                                                                                            [23]

                                Demo

                            4          NW/JUNE/MATH/ EMIS/6*******

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Mathematics/ P2                             Mid-Year Examination                          NW/ Jun 2019

QUESTION 3

3.1

 3.1.1
         r 2  x2  y 2     [Pyth]
         OP 2   8   152
                        2
                                                                         ✓ substitution
              289
         OP  17                                                         ✓ answer               (2)
 3.1.2
 (a)             y
         tan  
                 x
                 15
                                                                        ✓ answer               (1)
                 8
 (b)
                                cos( )
         17 sin 180    
                                   8                                     ✓ sin 
                     cos                                                ✓ cos 
          17 sin  
                       8
               15  1  8 
          17                                                       ✓ substitution
               17  8  17 
           256
                                                                                               (4)
            17                                                           ✓ answer
 3.1.3
                  15
         tan   
                   8
         Ref  61,93                                                   ✓ 78, 07
           180  61,93
                                                                         ✓ answer
            118,1                                                                             (2)

                                    Demo

                                5          NW/JUNE/MATH/ EMIS/6*******

### Page 6

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Mathematics/ P2                           Mid-Year Examination                        NW/ Jun 2019

 3.2.1
         sin 40  t

         x 2  y 2  r 2 [Pyth]
                                                                       ✓ sketch
         x  1 t2

                                                                       ✓ x  1 t2

         cos 320  cos  360  40 
                                                                       ✓ cos 40
          cos 40
          1 t2                                                       ✓ answer             (4)
 3.2.2
         sin140  sin 180  40 
          sin 40                                                     ✓ sin 40
                                                                       ✓ answer             (2)
         t
 3.2.3
         tan  220    tan 220
                                                                       ✓  sin 220
           tan 180  40 
           tan 40                                                   ✓ tan 40
                t
                                                                                          (3)
               1 t2                                                   ✓ answer
                                                                                            [18]

                                  Demo

                             6           NW/JUNE/MATH/ EMIS/6*******

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Mathematics/ P2                            Mid-Year Examination                            NW/ Jun 2019

QUESTION 4

 4.1     sin  90  A        tan(A  540)
                           
        cos  540  A         cos A.sin A
           cos A           tan A
                                                                      ✓ cos A
           cos A cos A.sin A
                     sin A                                              ✓  cos A
                                                                        ✓ tan A
         1  cos A                                                      sin A
                cos A.sin A                                             ✓
                    1                                                     cos A
         1 
                cos 2 A
           cos 2 A  1
        
              cos 2 A
                                                                        ✓ sin 2 A
          sin 2 A
        
          cos 2 A
                                                                        ✓ answer                  (6)
         tan 2 A
 4.2               sin 120  .tan 330
        tan 2 x 
                         cos 240
          sin 180  60  .tan  360  30 
        
                     cos 180  60 
          sin 60( tan 30)
                                                                       ✓ sin 60
                cos 60
                                                                        ✓  tan30
            3 1                                                         ✓  cos60
              
           2      3
        
              1                                                         ✓ substitution
              2
        tan x  1
           2

        tan 2 x  1  0
         tan x  1 tan x  1  0
        tan x  1 or tan x  1                                         ✓ factors or tan x  1
        x  45 ; 45
                                                                        ✓ 45 ✓ 45              (7)
 4.3    4 cos x  3
             2

                  3
        cos 2 x 
                  4
                                                                                     3
        cos x  
                     3                                                  ✓ cos x  
                    2                                                               2
        x  150  360.k or  30  360.k                            ✓ 150  360.k
                                                                        ✓ 30  360.k
        k                                                                                        (4)
                                                                        ✓ k
                                                                                                  [17]

                                   Demo

                               7          NW/JUNE/MATH/ EMIS/6*******

### Page 8

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Mathematics/ P2                         Mid-Year Examination                    NW/ Jun 2019

QUESTION 5

 5.1                                                                 ✓answer          (1)
         supplementary

5.2

                            R                         ≫          S

                                              Q        ≫               U
                                                                 T

                                                                 P

 5.2.1
          Ŝ  90                diameter subtends right           ✓S ✓R            (2)
 5.2.2
          T̂2  90                [corr s ; QU     RS]             ✓ S/ R
          TP = 8 cm                [line from centre ⊥ to chord]     ✓ S/ R
          QT 2  82  102          [Pythagoras]
                                                                     ✓ QT = 6
          QT  102  82  6
          TU  10  6  4 cm                                         ✓ answer         (6)
                                                                                      [9]

                                Demo

                            8          NW/JUNE/MATH/ EMIS/6*******

### Page 9

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Mathematics/ P2                            Mid-Year Examination                   NW/ Jun 2019

QUESTION 6

                                              K

                                                                J

                                                  O

                    M                                                    L
                                          H

  N

 6.1.1
          ˆ  2  50  100
         MOL                             [  at centre                 ✓S    ✓R
                                          2   at circumf]                          (2)
 6.1.2
         KJ  JL                 [given]                               ✓S
         Ĵ1  90               [line from centre to
                                 midpt of chord]                       ✓ S/ R
         In  NKL                [acute s of right
         N̂  40                triangle complementary/               ✓S
                                 sum of s in  ]                                     (3)
 6.1.3
         MO  LO                          [radii]                      ✓ S/ R
         ˆ  Lˆ
         M                                [ s opp equal
                                                                       ✓ S/ R
           2   1
                                          sides]
                 180  100              [ OML is isosc]
         L̂1                 40
                      2                                                ✓S             (3)
 6.2
         Lˆ 1  N=40
                ˆ                     [calculated in 6.1.2
                                       and 6.1.3]                      ✓S
         MOLN is a cyclic              [line subt. equal s /
         quad                          converse s in the              ✓R
                                       same seg]                                      (2)
                                                                                      [10]

                                Demo

                            9            NW/JUNE/MATH/ EMIS/6*******

### Page 10

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Mathematics/ P2                      Mid-Year Examination                          NW/ Jun 2019

QUESTION 7

7.1

        Construction: Draw diameter AM and join M to B            ✓ construction

         Aˆ A
           1
              ˆ  90
                2
                                     [radius ⊥ tangent]           ✓ S/ R
         ˆB  B
              ˆ  90                [ s in a semi-circle]       ✓ S/ R
           1      2
               ˆ                     [ s in same segment]        ✓ S/ R
         B̂2  A 2

         B̂  Aˆ                                                  ✓S
           1      1                                                                      (5)

                             Demo

                        10          NW/JUNE/MATH/ EMIS/6*******

### Page 11

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Mathematics/ P2                                Mid-Year Examination                        NW/ Jun 2019

7.2

                              T

                          M
                                                    R                   B

                                                P

                                                               N

                      A                             S
 7.2.1
         P̂1  90                       [diameter subtends right  ]       ✓ S/ R
         Pˆ  Pˆ                         [vert. opp. ] s                   ✓ S/ R
          1       4
         BN is a diameter                [chord subtends 90 /
                                                                            ✓ conclusion
                                         converse s in a semi-circle]                           (3)
 7.2.2
         Nˆ  Pˆ
                3
                                         [tan chord theorem]                ✓S     ✓R
         ˆP  Pˆ                         [vert. opp. s ]                   ✓ S/ R
                                                                            ✓ S/ R
          6       3

         Mˆ  Pˆ                         [tan chord theorem]
           1    6

         AM NB                           [alt s equal]                     ✓R

         OR
         Aˆ  Pˆ                         [tan chord theorem]                ✓S     ✓R
                2

         Pˆ  Pˆ
          5       2
                                         [vert. opp. s ]                   ✓ S/ R
         ˆ  Pˆ                          [tan chord theorem]                ✓ S/ R
         B2    5
         AM NB                           [alt s equal]                     ✓R                   (5)

                                       Demo

                                  11          NW/JUNE/MATH/ EMIS/6*******

### Page 12

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Mathematics/ P2                            Mid-Year Examination                       NW/ Jun 2019

 7.2.3
                     ˆ
         M̂1  90  A               [ s in  AMP                      ✓ S/ R
         ˆ  90  B
         B          ˆ                [radius ⊥tangent]                  ✓ S/ R
           1              2
          ˆ Bˆ                      [alt s ; AM    NB]
         A      2                                                       ✓ S/ R
         MPBT is a cyclic            [ext   int opp  / converse      ✓R
         quad                        ext  of cyclic quad]

         OR
         ˆ N  ˆ                     [alt s ; AM    NB]
         M 1                                                            ✓ S/ R
         ˆ B
         N   ˆ
                  1
                                     [tan chord theorem]                ✓S ✓R
         Mˆ Bˆ
           1    1
         MPBT is a cyclic            [ext   int opp  / converse      ✓R
         quad                        ext  of cyclic quad]                                  (4)

 7.2.4
         Tˆ  Pˆ1  90               [ext  of cyclic quad]            ✓S       ✓R         (2)
                                                                                            [19]
                                    TOTAL:                                                  [100]

                                   Demo

                              12          NW/JUNE/MATH/ EMIS/6*******

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