# Mathematics Grade 12 Paper 2 November 2024 NSC MG Question Paper

> Grade 12 Mathematics Paper 2 question paper from the November 2024 NSC MG, 25 pages. Read online or get the free PDF download, memo linked when available.

- Page: https://www.hlayiso.com/library/mathematics-p2-nov-2024-mg-afr-eng/03cdfcc3-9099-4cae-ba6b-d59128849c75
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 12
- Document type: Question paper
- Year: 2024
- Exam period: November 2024 NSC MG
- Paper: 2
- Curriculum: CAPS
- Pages: 25

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Mathematics · Grade 12 · November 2024 NSC MG · 2024. Question paper, 25 pages. Read online or download the PDF.

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### Page 1

NATIONAL
                                  SENIOR CERTIFICATE/
                                       NASIONALE
                                   SENIOR SERTIFIKAAT

                                        GRADE 12/GRAAD 12

                            MATHEMATICS P2/WISKUNDE V2

                                         NOVEMBER 2024

                      MARKING GUIDELINES/NASIENRIGLYNE

MARKS/PUNTE: 150

                           These marking guidelines consist of 25 pages./
                            Hierdie nasienriglyne bestaan uit 25 bladsye.

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### Page 2

Mathematics P2/Wiskunde V2                          2                               DBE/November 2024
                              NSC/NSS – Marking Guidelines/Nasienriglyne

NOTE:
• If a candidate answers a question TWICE, only mark the FIRST attempt.
• If a candidate has crossed out an attempt of a question and not redone the question, mark
  the crossed-out version.
• Consistent accuracy applies in ALL aspects of the Marking Guidelines. Stop marking at the
  second calculation error.
• Assuming answers/values in order to solve a problem is NOT acceptable.

LET WEL:
• As 'n kandidaat 'n vraag TWEE KEER beantwoord, sien slegs die EERSTE poging na.
• As 'n kandidaat 'n antwoord van 'n vraag doodtrek en nie oordoen nie, sien die
   doodgetrekte poging na.
• Volgehoue akkuraatheid word in ALLE aspekte van die Nasienriglyne toegepas. Hou op
   nasien by die tweede berekeningsfout.
• Aanvaar van antwoorde/waardes om 'n probleem op te los, word NIE toegelaat nie.

                                GEOMETRY • MEETKUNDE
                A mark for a correct statement
                (A statement mark is independent of a reason)
         S
                'n Punt vir 'n korrekte bewering
                ('n Punt vir 'n bewering is onafhanklik van die rede)
                A mark for the correct reason
                (A reason mark may only be awarded if the statement is correct)
         R
                'n Punt vir 'n korrekte rede
                ('n Punt word slegs vir die rede toegeken as die bewering korrek is)

                Award a mark if statement AND reason are both correct
        S/R
                Ken 'n punt toe as die bewering EN rede beide korrek is

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### Page 3

Mathematics P2/Wiskunde V2                                               3                                           DBE/November 2024
                                                   NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 1

      Weight
                                              34   32        40   27        33   28     27    55       39   44     30     57    28     32       31
    (in kg) (x)
 Number of push-
  ups per minute                              26   21        6    20        16   26     23    7        18   14     17     8     28     25       20
        (y)

                                                                            Scatter plot
                                     30

                                     25

               Push-ups per minute
                                     20

                                     15

                                     10

                                      5

                                      0
                                          0             10             20          30             40          50           60
                                                                             Weight (in kg)

1.1         a = 39,456001…                                                                                         ✓ a = 39,46
            b = – 0,590018…                                                                                        ✓ b = – 0,59
             yˆ = 39 ,46 − 0,59 x                       CORRECT ANSWER ONLY: FULL MARKS                            ✓ equation
                                                                                                                                            (3)
1.2         r = – 0,8                                                                                              ✓(A) – 0,8
                                                                                                                                            (1)
1.3         y = 39,46 – 0,59(29)                                                                                   ✓ substitution
            y = 22,35                                                                                              ✓ answer
                                                                                                                                          (2)
            OR/OF

            y = 22,35                         (calculator)                                                         ✓✓ answer
                                                                                                                                          (2)
1.4          y = 18 ,33                                                                                            ✓(A) 18,33
                                                                                                                                      (1)
1.5         The increase in the number of push-ups will have no influence.                                         ✓ no influence OR
            The standard deviation stays the same.                                                                 standard deviation
                                                                                                                   remains the same
                                                                                                                   geen verandering /
                                                                                                                   bly dieselfde      (1)
1.6         6 is furthest y-value below the least squares regression line.                                         ✓ 6
            An increase of 10 push-ups will get the team member to                                                 ✓ difference is 10
            (40 ; 16), the minimum number of push-ups for a player
            weighing 40kg.                                                                                                                (2)
                                                                                                                                         [10]

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### Page 4

Mathematics P2/Wiskunde V2                          4                               DBE/November 2024
                              NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 2

2.1         Median = 65                                                                ✓ 65
                                                                                                         (1)
2.2         Q1 = 44                                                                    ✓ 44
                                                                                                         (1)
2.3         IQR = 84 – 44                                                              ✓ 84
                = 40                                                                   ✓ IQR
                                                                                                         (2)
2.4
                                                                                       ✓ box
                                                                                       (A) whiskers
                                                                                       ending at 5 & 120
               5                   44         65        84
                                                                                                      (2)
2.5         Number of employees who qualify = 34                                       ✓ 34
                                        34
            % of employees who qualify =  100
                                        60
                                       = 56,67% of the employees                        answer
            OR/OF                                                                                        (2)
            Number of employees who qualify = 35                                       ✓ 35
                                        35
            % of employees who qualify = 100
                                        60
                                       = 58,33% of the employees                        answer
                                                                                                         (2)
2.6         Number of intervals = 3                                                    ✓ 3
            Time allowed to work from home = 3(30 minutes)
                                           = 90 minutes OR/OF 1,5 hours                ✓ answer
                                                                                                         (2)
                                                                                                        [10]

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### Page 5

Mathematics P2/Wiskunde V2                                    5                                                  DBE/November 2024
                                        NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION / VRAAG 3                                                               y
                                                                       A

                                       C(–9 ; 0)                            O        M                                  x

                                                                   B(–1 ; k)

                                                                                           D(3 ; –9)

               −9 − 0                                         0 − ( −9 )                                correct substitution of
3.1   mDC =                          OR/OF mDC =                                                         D (3 ; − 9) & C (− 9 ; 0)
              3 − ( −9 )                                       −9 − 3
                                                                                                         into gradient formula
                    3                                              3
      mDC = −                                     mDC = −                                               answer
                    4                                              4
                                                                                                                                         (2)
3.2 Equation of DC:

    0 = − ( −9 ) + c OR/OF y − 0 = − ( x − ( −9 ) )
          3                         3
                                                                                                        correct substitution of
          4                         4                                                                    C (− 9 ; 0) or D (3 ; − 9)
         −27        3               3
     c=      or − 6           y = − ( x + 9)                                                             into equation of line
           4        4               4
          3   27                                             3   27
      y =− x−                                            y =− x−                                        answer
          4    4                                             4    4                                                                      (2)
             3              27              k − ( −9 )        −3                k −0           −3
3.3   k =−        ( −1) −           OR/OF                 =         OR/OF                  =            substitution of B ( − 1 ; k )
              4                 4               −1 − 3         4            − 1 − ( −9 )       4
          3 27                                                                        3
      k=    −                       OR/OF   k +9 =3                OR/OF        k =−    (8)                                              (1)
          4 4                                                                         4
      k = −6                                k = −6                              k = −6
                                                                                                        correct substitution of
                 ( 3 + 9 ) + ( −9 − 0 )
                            2               2
3.4   DC =
                                                                                                         D (3 ; − 9) & C (− 9 ; 0)
                                                                                                         into distance formula
      DC = 15 units                                                                                     answer
                                                                                                                                      (2)

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### Page 6

Mathematics P2/Wiskunde V2                                 6                                DBE/November 2024
                                     NSC/NSS – Marking Guidelines/Nasienriglyne

                  ( 3 − ( −1) ) + ( −9 − ( −6 ) )
                                 2                  2
3.5      DB =
                                                                                   DB = 5
         DB = 5

             DB 5 1
              = =                                                                 answer
             DC 15 3
                                                                                                                  (2)
          DM DB 1                                                                     DM DB
3.6           =    =                                                                   =
          DA    DC 3                                                                  DA DC

         Area ΔMBD
                     1
                       (DM)(DB) sin D̂         (        )
                   = 2                                                             correct use of area rule
         Area ACD   1
                     2
                       (DA)(DC) sin D̂         (        )
                     1 1                                                                       BD        DM
                   =                                                              subst. for      and
                     3 3                                                                       DC        DA
                     1                                                              into correct formula
                   =                                                               answer
                     9
                                                                                                                  (4)
3.7     y = –4x + c
         m AD = −4
                                                                  y+9
         −9 = −4 ( 3) + c                          OR/OF                = −4       correct substitution of
                                                                  x−3               m AD = −4 and D(3 ; –9)
        c=3                                                  y + 9 = −4 x + 12
          y = − 4x + 3                                           y = − 4x + 3
                                                                                   ( x − 3) + ( y + 9 ) = 612
                                                                                             2          2
         ( x − 3) + ( y + 9 ) = 612
                  2              2

          ( x − 3 ) + ( −4 x + 3 + 9 ) = ( 612 )
                      2                  2              2
                                                                                   substitution of equation
                                                                                    AD into distance formula
         ( x − 3) + ( −4 x + 12 ) = 612
                  2                  2

         x 2 − 6 x + 9 + 16 x 2 − 96 x +144 = 612
        17 x2 − 102 x − 459 = 0
         x 2 − 6 x − 27 = 0                                                        standard form

         ( x − 9 )( x + 3) = 0
         x = 9 or x = −3                                                           x values with rejection
         N/A
        y = –4(–3) + 3
        y = 15                                                                     y coordinate
         A ( − 3 ; 15 )
                                                                                                                  (6)

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### Page 7

Mathematics P2/Wiskunde V2                             7                                DBE/November 2024
                                 NSC/NSS – Marking Guidelines/Nasienriglyne

        OR/OF                                                                 OR/OF

         −9 = −4 ( 3) + c                                                      correct substitution of
                                                                                m AD = −4 and D(3 ; –9)
         c=3
         y = −4 x + 3
        N(0 ; 3)                                                              ✓ N(0 ; 3)
                   ( 3 − 0 ) + ( −9 − 3 )                                     ✓ substitution into distance
                            2               2
        ND =
                                                                                formula to calculate ND
              = 3 17
        AD = 6 17                                                                       1
                                                                              ✓ ND =      AD
               1                                                                        2
        ND = AD
               2
        N is the midpoint of AD                                               ✓ x – value ✓ y – value
        A(–3 ; 15)                                                                                            (6)
                                                                                                            [19]

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### Page 8

Mathematics P2/Wiskunde V2                               8                                    DBE/November 2024
                                   NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 4
                                                        y
                                                    S
                                                                 P(3 ; 9)

                                                                                      T

                                                            M(1 ; 3)

                                                                                R(d ; 1)

                                                        O              V    N                              x

                                                L

4.1    L(–1 ; –3)                                                                    x = –1  y = –3
                                                                                                                       (2)
               9−3
4.2     m MP =
               3 −1
        m MP = 3                                                                     m MP = 3
                     1                                                                               1
        mST = −                                                                      mST = −
                     3                                                                              m MP
              1                                       1
        9=−     ( 3) + c                     y − 9 = − ( x − 3)                      substitution of mST & P(3; 9)
              3                                       3
                                                                                      into equation of a line
                                                      1
        c = 10                 OR/OF         y − 9 = − x +1
                                                      3
             1                                        1
        y = − x + 10                             y = − x + 10                        equation of tangent ST
             3                                        3                                                                (4)
        ( x − 1) + ( y − 3) = r
                 2             2    2
4.3
                                                                                     ( 3 − 1) + ( 9 − 3) = r 2
                                                                                                2              2
        ( 3 − 1) + ( 9 − 3) = r 2
                 2             2

        r 2 = 40                                                                     value of r 2
        ( x − 1) + ( y − 3) = 40
                 2             2
                                                                                     LHS of equation of circle
        x − 2 x + 1 + y − 6 y + 9 = 40
         2                 2
                                                                                     expanding LHS
        x2 + y 2 − 2 x − 6 y − 30 = 0
                                                                                                                       (4)

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### Page 9

Mathematics P2/Wiskunde V2                             9                                DBE/November 2024
                                 NSC/NSS – Marking Guidelines/Nasienriglyne

        d 2 + (1) − 2d − 6 (1) − 30 = 0                                        d 2 + (1) − 2d − 6 (1) − 30 = 0
                   2                                                                       2
4.4
        d 2 − 2 d − 35 = 0                                                     standard form
        ( d − 7 )( d + 5) = 0
       d = 7 or d = −5
                                                                                                                 (2)
       d = 7

       OR/OF                                                                  OR/OF

        ( x − 1) + ( y − 3) = 40
               2             2

        (d − 1) 2 + (1 − 3) 2 = 40                                             (d − 1) 2 + (1 − 3) 2 = 40
       (d − 1) 2 = 36                                                          standard form
       d − 1 = 6 or d − 1 = −6
       d = 7 or d = −5
       d = 7
                                                                                                                 (2)

       OR/OF                                                                  OR/OF

         ˆ = 90 
        PRL                        ( in semi-circle)
        9 − 1 1 − ( −3)                                                       ✓ mPR  mRL = −1
                       = −1
        3 − d d − ( −1)
        d 2 − 2 d − 35 = 0                                                    ✓ standard form
        ( d − 7 )( d + 5) = 0
        d = 7 or d = −5
       d = 7
                                                                                                                 (2)

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### Page 10

Mathematics P2/Wiskunde V2                                10                                              DBE/November 2024
                                    NSC/NSS – Marking Guidelines/Nasienriglyne

                                                           y
                                                       S
                                                                    P(3 ; 9)

                                                                                                    T

                                                                           4 5
                                                           4 10

                                                               M(1 ; 3)

                                                                                        R(d ; 1)
                                                           71, 6              26, 6     116, 6

                                                           O              V           N                             x
                                                                    4 5

                                                   L

4.5     mPO = 3
              ˆ =3
        tan POV                                                                                tan PÔV = mPO
         ˆ = 71,565...
       POV                                                                                      P Ô V

                   1 − ( −3 )
        mRL =                                                                                   m RL using R(7 ; 1) & L
                   7 − ( −1)
                   1
               =
                   2
              ˆ = 1
        tan RVN
                  2
         ˆ = 26,565...
       RVN                                                                                      R V̂ N
        L̂ = 71,565... − 26.565...              [ext.  of  / buite  van ]
        L̂ = 45                                                                                answer
                                                                                                                                  (5)
       OR/OF                                                                                   OR/OF

        R̂ = 90                     [ in semi-circle / in 'n halwe sirkel]                   R̂ = 90 
        PR = ( 3 − 7 ) + ( 9 − 1)
           2              2           2

        PR = 80 = 4 5 units                                                                     PR = 80 = 4 5
        PL2 = ( 3 − ( −1) ) + ( 9 − ( −3 ) ) OR RL2 = ( 7 + 1) + (1 + 3)
                                2             2                       2           2

        PL = 160 = 4 10                                  RL = 80 = 4 5                          length of PL OR RL
                    4 5                           4 5                   4 5
        sin L̂ =                OR cos Lˆ =                 OR tan Lˆ =                         trig ratio of L̂
                   4 10                           4 10                  4 5
        L̂ = 45                                                                                answer
                                                                                                                                  (5)
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### Page 11

Mathematics P2/Wiskunde V2                            11                               DBE/November 2024
                                NSC/NSS – Marking Guidelines/Nasienriglyne
       OR/OF                                                                 OR/OF
                  ( 3 + 1) + ( 9 + 3 ) = 160 = 4 10
                         2          2
        PL =                                                                  length of PL

                  ( 7 − 3) + (1 − 9 ) =
                          2          2
        PR =                              80 = 4 5                            PR = 80 = 4 5

                  ( 7 + 1) + (1 + 3 ) =
                          2         2
        LR =                              80 = 4 5                            length of LR
                    80 + 160 − 80
        cos L =
                    2 80  160                                                substitution into the cos rule

                     2
        cos L =
                    2
        L̂ = 45                                                             ✓ answer
                                                                                                                (5)
              1− 3
4.6     mRM =
               7 −1
                 1
            =−                                                               ✓ mRM
                 3
        mRT = 3                     (tan ⊥ rad)                              ✓ mRT
                    1
        mPT = −
                    3
        mRT  mPT = −1                                                       ✓ mRT  mPT = −1
       PT ⊥ RT                                                                                                  (3)

       OR/OF                                                         OR/OF
                3 −1
       mMR =
                1− 7
                    1
              = −
                    3
                 1
       mPT = −              [proved in Q4.2]
                 3
       m PT = m MR
       PT || MR                                                      PT || MR
       M R̂ T = 90  [radius ⊥ tangent / raaklyn ⊥ radius]            M R̂ T = 90 
       P T̂ R = 90  [co-int s; PT || MR/ooreenkomst. e; PT || MR]  P T̂ R = 90 
       PT ⊥ RT                                                                                                  (3)

       OR/OF                                                                 OR/OF

        TP̂R = L̂ = 45       [tan-chord theorem/  tussen raaklyn            TP̂R = L̂
                                       en koord]
          TP = TR             [tans from common pt]
        TP̂R = TR̂P = 45  [s opp equal sides/                              TP̂R = TR̂P
                                   e teenoor gelyke sye]
       PT̂R = 90          [sum of s in ∆ / binne e van ∆]                 PT̂ R = 90 
        PT ⊥ RT                                                                                                (3)
                                                                                                              [20]
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### Page 12

Mathematics P2/Wiskunde V2                                     12                                DBE/November 2024
                                         NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 5
                                                                   y

                                                           A
                                                                           •
                                                                   O       K x

                                                      •
                                                   P(–3 ; –4)

5.1.1      r=5                                                                     r=5
                        3
            cos A = −                                                              answer
                        5                                                                                             (2)
5.1.2       cos 2A = 2cos A −1   2

                                         2
                     3                                                           substitution of cos A into
                = 2  −  −1
                     5                                                            double angle formula
                     7
                =−
                     25                                                            answer
           OR/OF                                                                                                      (2)
            cos 2A = cos A − sin A
                             2                     2

                                     2                 2
                    3  4                                                       substitution of cos A & sin A
                  =−  −− 
                    5  5                                                        into double angle formula
                      7
                  =−                                                               answer
                     25
           OR/OF                                                                                                      (2)
           cos 2A =1− 2sin 2 A
                                               2
                           4                                                     substitution of sin A into
                   =1 − 2  − 
                           5                                                      double angle formula
                       7
                   =−                                                              answer
                       25
                                                                                                                      (2)
5.1.3                                                       y
                                             (x ; 4)
                                                               B

                                                           O           x
           x = –3                                                                  x = –3
           sin ( A − B ) = sinAcosB − cosAsinB
                               4  3   3  4 
                            =  −  −  −  −                                  substitution into the
                               5  5   5  5                                   compound angle formula
                              12     12
                            =      +
                               25    25
                                24
                            =
                                25                                                 answer
                                                                                                                      (4)

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### Page 13

Mathematics P2/Wiskunde V2                              13                                           DBE/November 2024
                                  NSC/NSS – Marking Guidelines/Nasienriglyne

                                  
            cos  − 45  sin  − 45 
5.2             2            2      
                        2
                                      
              2cos  − 45   sin  − 45  
            =                                                                   multiply by
                                                                                                      2
                           2 .2                                                                       2
              sin ( − )                                                           double angle
            =
                    4
              − cos                                                                  co function
            =
                  4
              −p              1
            =        OR/OF = − p                                                      answer
               4              4                                                                                          (4)

                                                                                     OR/OF
           OR/OF

                                     
             cos  − 45  sin  − 45 
                  2          2         
                             2
                                                                          
               cos 2 cos 45 + sin 2 sin 45  sin 2 cos 45  − cos 2 sin 45     ✓ expansion
                                                                              
            =
                                               2
               2              2      2            2     
                    cos +        sin     sin −         cos                       ✓ special angles
               2        2     2     2 2        2    2       2
            =
                                       2
              1 2 1             2 
                 sin     − cos
            = 2        2 2         2
                         2
                 1                
              −  cos 2 − sin 2 
                 2       2         2
            =
                           2
                        
                 cos 2                                                             ✓ double angle
                       2
            =−
                      4
                 cos 
            =−
                   4
            =− p
                 1                                                                    answer
                 4                                                                                                       (4)
                                                                                                                        [12]

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### Page 14

Mathematics P2/Wiskunde V2                                  14                               DBE/November 2024
                                      NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 6

6.1.1       cos ( x + y ) = cos( x − (− y ))                                        ( x + y ) = ( x − (− y ))
                           = cos x cos(− y ) + sin x sin (− y )                     correct expansion
                           = cos x cos y − sin x sin y
                                                                                                                   (2)
                       cos(90  − x ) cos y + sin (− y ) cos(180  + x )
6.1.2       LHS =
                        cos x cos(360  + y ) + sin (360  − x ) sin y
                                                                                    cos ( 90 − x ) = sin x
                  =
                    (sin x ) cos y + (− sin y )(− cos x )
                      cos x(cos y ) + (− sin x ) sin y                              sin ( − y ) = − sin y
                                                                                    cos (180 + x ) = − cos x
                      sin x cos y + cos x sin y
                  =                                                                 cos ( 360 + y ) = cos y
                      cos x cos y − sin x sin y
                                                                                    sin ( 360 − x ) = − sin x
                       sin ( x + y )
                  =
                      cos ( x + y )                                                 compound angle formulae

                 = tan ( x + y )

                 = RHS
                                                                                                                   (6)
6.2           6 sin x − 11 cos(90  + x ) + 7 = 2
                   2

            6sin 2 x − 11cos ( 90 + x ) + 7 = 4                                   squaring both sides

            6sin 2 x − 11( − sin x ) + 7 = 4                                        cos ( 90 + x ) = − sin x

            6sin 2 x + 11sin x + 3 = 0
            (3 sin x + 1)(2 sin x + 3) = 0                                          factors
                         1                                                    3
           sin x = −                         OR/OF                sin x = −         both equations
                         3                                                    2
            ref  = 19 ,47                                       no solution
           x = 199,47° or x = 340,53°                                               answers
                                                                                                                   (6)
                        4 − 8sin x2
6.3.1       g ( x) =
                             3
                        4 (1 − 2 sin 2 x )                                          factors
                   =
                         3
                  4 cos 2 x                                                            4 cos 2 x
                =                                                                  
                      3                                                                    3
           Maximum value of cos 2x is 1
                                       4
           maximum value of g ( x ) =                                              answer
                                       3                                                                           (3)

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### Page 15

Mathematics P2/Wiskunde V2                          15                              DBE/November 2024
                              NSC/NSS – Marking Guidelines/Nasienriglyne
           OR/OF                                                           OR/OF

            4 − 8 sin 2 x is a maximum when sin 2 x is a minimum

           Minimum value of sin 2 x is 0                                    min of sin 2 x = 0
                                     4 − 8 (0)                                        4 − 8 (0)
            max. value of g ( x ) =                                       ✓ g ( x) =
                                         3                                                3
                                     4
                           g ( x) =                                         answer
                                     3
                                                                                                          (3)

           OR/OF                                                           OR/OF

                      − (0)                                                             − (0)
           sin x =                                                          sin x =
                       8                                                               8
                     2 −                                                             2 − 
                       3                                                               3
           sin x = 0                                                        sin x = 0
                                        4 − 8 (0)
            max. value of g ( x ) =
                                            3
                                        4
                               g ( x) =                                     answer
                                        3
                                                                                                          (3)
6.3.2      x = 180°                                                         180°
                                                                                                           (1)
                                                                                                          [18]

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### Page 16

Mathematics P2/Wiskunde V2                            16                                  DBE/November 2024
                                 NSC/NSS – Marking Guidelines/Nasienriglyne

 QUESTION/VRAAG 7
7.1     x = 90°                                                                        x = 90°
                                                                                                                  (1)
7.2     x = −180  or x  ( −90  ; 0]                                                answer
                                                                                                                  (2)
        OR/OF

        x = −180  or − 90   x  0                                                  answer
                                                                                                                  (2)
7.3.1   180°                                                                           answer
                                                                                                                  (1)
7.3.2

                                             3

                                             2                                g        turning points on
                                                                                        x-axis: x = –90° ; 90°
                                             1
                                                                                       shape
        –180°             –90°                   0°             90°       180°         turning point on y-axis
                                           –1                                           at (0 ; 2)

                                            –2
                                                                      f
                                            –3                                                                    (3)

7.4     2cos3 x − sin x = 0
        2cos3 x = sin x
                  sin x
        2 cos 2 x =
                  cos x
        2 cos x = tan x
             2
                                                                                       2 cos 2 x = tan x
        2 cos 2 x − 1 = tan x − 1                                                      2 cos 2 x − 1 = tan x − 1
        cos 2 x + 1 = tan x                                                            cos 2 x + 1 = tan x
        x = 45° + k.180°; k  Z                                                        answer
                                                                                                                  (4)
        OR/OF
                                                                                      OR/OF
        2 cos x − sin x = 0
                3

        cos x ( 2 cos 2 x − tan x ) = 0

        cos x = 0         or        2 cos 2 x = tan x                                  2 cos 2 x = tan x
        not valid                   2 cos 2 x − 1 + 1 = tan x                          2 cos 2 x − 1 + 1 = tan x
                                    cos 2 x + 1 = tan x                                cos 2 x + 1 = tan x
                                    x = 45° + k.180°; k  Z                            answer
                                                                                                                  (4)
                                                                                                                [11]

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### Page 17

Mathematics P2/Wiskunde V2                              17                                  DBE/November 2024
                                  NSC/NSS – Marking Guidelines/Nasienriglyne

QUESTION/VRAAG 8

                                                          D

                                                               16
                                                   122°
                                                         E E

                                                          C         46,85°
                                                      105,61°

                                  B                       19                   A

                         DC
8.1         tan DÂC =
                         AC
                       16
            AC =                                                                     correct subs into trig ratio
                   tan 46,85 
            AC = 15 m                                                                answer
                                                                                                                   (2)
            ( AB) = ( BC ) + ( AC ) − 2 ( BC )( AC ) cos BCA
                                                          ˆ
                   2          2            2
8.2
            (19 ) = x 2 + (15) − 2 x (15 ) cos105, 61                               correct subst. into
                 2            2

                                                                                      cosine rule
                                                                                     quadratic equation in
             x 2 + 8, 07 x − 136 = 0
                                                                                     std form

                             ( 8, 07 ) − 4 (1)( −136 )
                                       2
                  −8, 07                                                            correct subst. into
             x=                                                                      quadratic formula
                                   2 (1)
             x = 8,30 m or x  − 16 ,38 m                                            length of BC

              ˆ = 58 
            BEC                            OR/OF            ˆ
                                                           EBC = 32                           ˆ OR/OF EB̂C
                                                                                     size of BEC
                       BC                                               EC
            tan BÊC =                                     tan EB̂C =
                       EC                                               BC
                    8,3
            EC =                                           EC = 8,3 tan 32 
                 tan 58 
            EC = 5,19 m                                    EC = 5,19 m               length of EC
            DE = 10 ,81 m                                  DE = 10 ,81 m             answer
                                                                                                                   (7)

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### Page 18

Mathematics P2/Wiskunde V2                          18                              DBE/November 2024
                              NSC/NSS – Marking Guidelines/Nasienriglyne

            OR/OF                                                           OR/OF

                            ˆ
            sin105,61 sin CBA
                      =                                                      correct subst. into
                19        15                                                  sine rule
              ˆ = 49,5
            CBA
                                                                               ˆ
                                                                             BAC
             ˆ = 24,89
            BAC
                 BC         19                                               correct subst. into
                       =                                                     sine formula
             sin 24,89 sin105, 61
                                                                             length of BC
            BC = 8,3 m

              ˆ = 58                                 ˆ                                ˆ OR/OF EB̂C
                                                                             size of BEC
            BEC                         OR/OF        EBC = 32 
                       BC                                         EC
            tan BÊC =                               tan EB̂C =
                       EC                                         BC
                    8,3
            EC =                                     EC = 8,3 tan 32 
                 tan 58 
            EC = 5,19 m                              EC = 5,19 m             length of EC
            DE = 10 ,81 m                            DE = 10 ,81 m           answer
                                                                                                           (7)
                                                                                                           [9]

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### Page 19

Mathematics P2/Wiskunde V2                           19                                  DBE/November 2024
                               NSC/NSS – Marking Guidelines/Nasienriglyne
QUESTION/VRAAG 9
                                                         A
                                                        1 2

                                                          46°

                                  D

                                                                            B

                                                  2 3
                                          86° 1
                                                  C

                                      E
9.1     Â1 = 40         [ext.  of a cyclic quad / buite  van kvh]                 ✓S ✓R
                                                                                                                (2)
                            ˆ     1 ˆ                                               ✓S
9.2     B̂ = 80             A1 = 2 B 
                                      
        D̂ = 100          [opp s of cyclic quad / teenoorst. e van kvh]            ✓ S/R
         Ĉ2 = 40        [sum of s in ∆ / binne e van ∆]                          ✓ S
         Ĉ 2 = Â1 = 40 
         AD = DC           [sides opp = s / sye teenoor gelyke ]                   ✓ R
        OR/OF                                                                                                   (4)
                             ˆ     1 ˆ
        B̂ = 80              A1 = 2 B                                              ✓S
                                       
        AĈE = Â 2 + B̂ [ext  of ∆ / buite  van ∆]                                 ✓ S/R
                                                                                      ✓ S
         Ĉ2 = 40
         Ĉ 2 = Â1 = 40 
         AD = DC                                                                     ✓ R
                            [sides opp = s / sye teenoor gelyke ]
                                                                                                                (4)
        OR/OF
                            ˆ      1 ˆ                                              ✓ S
        B̂ = 80             A1 = 2 B 
                                      
         Ĉ3 = 180  − 46  − 80  [sum of s in ∆ / binne e van ∆]
                                                                                      ✓ S/R
         Ĉ 3 = 54 
         Ĉ 2 = 180  − 86  − 54  [s on a str. line / e op 'n reguitlyn]
         Ĉ 2 = 40                                                                  ✓ S
         Ĉ 2 = Â1 = 40 
         AD = DC           [sides opp = s / sye teenoor gelyke ]                   ✓ R
                                                                                                                (4)
                                                                                                                [6]
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### Page 20

Mathematics P2/Wiskunde V2                          20                              DBE/November 2024
                               NSC/NSS – Marking Guidelines/Nasienriglyne

 QUESTION/VRAAG 10

 10.1
                                                   R

                                           h2          h1
                                  A                            B

                       S                                               T

10.1    Construction: Join SB and TA and draw h1 from B ⊥ AR and            ✓ construction
        h2 from A ⊥ RB

        Konstruksie: Verbind SB en TA en trek h1 vanaf B ⊥ AR en h2
        vanaf A ⊥ RB

        Proof/Bewys:
                                                                                         1
                                                                              area ΔRAB 2 RA  h1
                   1
                     RA  h1
        area ΔRAB 2            RA                                           ✓          =
                 =           =                                                area ΔASB 1 AS  h
        area ΔASB 1            AS
                     AS  h1                                                             2
                                                                                                1

                   2
                                                                                RA
                                                                            ✓
                                                                                AS
                     1
                       RB  h2                                                  area RAB RB
        area ΔRAB 2              RB                                         ✓            =
                   =           =                                                area ABT BT
         area ΔABT 1 BT  h      BT
                             2
                     2
        area ∆RAB = area ∆RAB       [common/gemeenskaplik]
        But area ∆ASB = area ∆ABT [same base & height; AB || ST/
                                     dies. basis & hoogte; AB ||ST]         ✓ S ✓ R

            area ΔRAB area ΔRAB
                    =
            area ΔASB area ΔABT
            RA RB
             =
            AS BT
                                                                                                         (6)

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