# Mathematics Grade 11 Paper 2 Free State June 2024 Memo

> Grade 11 Mathematics Paper 2 memorandum from the Free State June Exam 2024, 20 pages. Read online or download the free PDF.

- Page: https://www.hlayiso.com/library/fs-maths-grade-11-june-2024-p2-and-memo/694b2de0-ade8-42fc-83be-c24c3250f506
- Updated: 2026-09-27
- Subject: Mathematics
- Grade: Grade 11
- Document type: Memorandum
- Year: 2024
- Exam period: Free State June Exam
- Paper: 2
- Curriculum: CAPS
- Pages: 20

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## About this document

Mathematics · Grade 11 · Free State June Exam · 2024. Memorandum, 20 pages. Read online or download the PDF.

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## Document text

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                MATHEMATICS
                     GRADE 11
                JUNE P2                   2024

MARKS: 100

TIME : 2 HOUR

                This question paper consists of 8 pages.

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Mathematics Gr 11 P2                          2                                  FS/June 2024

INSTRUCTIONS AND INFORMATION

Read the following instructions carefully before answering the questions.

1.     This question paper consists of 6 questions.

2.     Answer ALL the questions in the SPECIAL ANSWER BOOK provided.

3.     Clearly show ALL calculations, diagrams, graphs, etc. which you have used in
       determining your answers.

4.     You may use an approved scientific calculator (non-programmable and non-
       graphical), unless otherwise stated.

5.     If necessary, round off answers to TWO decimal places, unless stated otherwise

6.     Information sheet with formulae is attached at the end of the question paper.

7.      Diagrams are NOT necessarily drawn to scale.

8.      Write neatly and legibly.

9.     Answers only will NOT necessarily be awarded full marks.

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Mathematics Gr 11 P2                            3                                   FS/June 2024

QUESTION 1

 PQRS is a quadrilateral with vertices P(– 5; 0), Q(0; t), R(3; – 2) and S(– 2; – 5).
 P is on the x-axis and Q is on the y-axis.

 1.1      Determine the lengths of PS and RS.                                             (4)

 1.2      Calculate the coordinates M, the midpoint of PR.                                (2)

 1.3      Show that the value of t = 3 if it is given that PR and QS bisect each other.   (1)

 1.4      Determine the gradient of PQ.                                                   (2)

 1.5                             ˆ .
          Calculate the size of PQR                                                       (4)

 1.6      What type of a quadrilateral is PQRS? Justify you answer.                       (3)
                                                                                          [16]

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Mathematics Gr 11 P2                            4                                   FS/June 2024

QUESTION 2

 In the diagram below, A(– 2; 4), B(– 2; – 1) and C ( p; −4 ) are the vertices of
 ABC . Q is the x- intercept of the line AC. θ is the angle of inclination of AC and
  ˆ
 BAC=37o .

 2.1       Determine the size of θ.                                                         (2)

 2.2       Calculate the equation of line AC.                                               (3)

 2.3       Calculate the coordinate of Q, the x-intercept of AC                             (2)

 2.4       Determine the integral value of p if the points A, Q and C are collinear.        (3)

 2.5       Determine the equation of the line parallel to BC and passing through            (5)
           point A.
                                                                                          [15]

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Mathematics Gr 11 P2                            5                                 FS/June 2024

QUESTION 3

 3.1     If θ = 89,4o and α = 122o, evaluate the following (round off answers correct
         to three decimal places):

         3.1.1      sin α                                                                 (1)

         3.1.2         tan 
                    cos2 ( +  )                                                         (2)

 3.2     If 3tan β + 2 = 0 and sin β > 0, determine without the use of a calculator
         and with the aid of a sketch, the value of the following:

         3.2.1      cos β                                                                 (3)

         3.2.2        13 cos  − 13sin 2                                                 (2)

 3.3     If sin 4o = m, determine tan 94o in term of m.                                   (3)

 3.4     Simplify without the use of a calculator:

          cos ( 360o − x ) tan ( x − 180o )
              sin ( − x ) cos ( 90o − x )                                                 (5)

 3.5     Evaluate without using a calculator:

          tan120o.sin115o
          cos 240o cos 205o                                                               (6)

 3.6     Prove the identity:

             1         1       2
                  +         = 2
         1 − cos x 1 + cos x sin x                                                        (3)

 3.7     Determine the general solution:

         5 tan ( θ + 10o ) = 4
                                                                                         (4)
                                                                                        [29]

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Mathematics Gr 11 P2                                6                             FS/June 2024

QUESTION 4

 Given: f ( x ) = sin 2 x and g ( x ) = 3cos x for x   −90o ;90o 

 4.1       Sketch the graphs of f(x) and g(x) on the same system of axes in the given   (6)
           interval. Show all coordinates of the intercepts and the turning points.

 4.2       Write down the period of f(x).                                               (1)

 4.3       Write down the range of g(x).                                                (2)

 4.4       For which value(s) of x is f ( x ) .g ( x )  0 ?                            (2)
                                                                                        [11]

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Mathematics Gr 11 P2                           7                                   FS/June 2024

QUESTION 5
 5.1     Complete the following theorem:

         A line drawn from the centre of the circle, perpendicular to the chord,
         ……… the chord.                                                                  (1)

 5.2     In the diagram below, O is the centre of the circle and AC is the diameter.
         AC is produced to meet ED produced at G. OE CD and B̂ = 72o .

         5.2.1   Calculate, giving reasons, the size of:

                 (a)     Ĉ1                                                             (2)

                 (b)     Ô1                                                             (2)

                 (c)     D̂4                                                             (2)

                 (d)     AED                                                             (2)

                 (e)     D̂1                                                             (2)

         5.2.2   Prove that AF = FD                                                      (3)

         5.2.3   Is OEDC a cyclic quadrilateral? Give a reason for your answer.          (2)
                                                                                         [16]

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Mathematics Gr 11 P2                         8                                  FS/June 2024

QUESTION 6
 6.1     Complete the theorem:

         The tangent is ……………to the radius drawn from to the point of contact.         (1)

 6.2     Use the diagram below, A, B and C are on the circle with centre O. PA is a
         tangent to the circle at A. Use the diagram to prove the theorem which states
                ˆ =ˆ
         that: BAP    C                                                                (6)

 6.3     In the figure below, PQRS are points on the circle. PQ is produced to meet
                                                                    ˆ = Pˆ
         RU at U. TP is the tangent to the circle at P. PR = SP and U     1

         Prove that: RU = RQ.                                                      (6)
                                                                                   [13]
                                                                              TOTAL: 100

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Mathematics Gr 11 P2     9                FS/June 2024

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                        1

                 GRADE 11

            MATHEMATICS

               2024 JUNE P2

       MARKING GUIDELINE

    This marking guideline consists of 11 pages

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                                          2

QUESTION 1

1.1    PS= (x2 − x1 ) 2 + ( y2 − y1 ) 2

       PS= ( −2 + 5 ) + ( −5 + 0) 2           ✓subt in corr form
                        2

                                              ✓answer
       PS= 34

       RS= (3 + 2) 2 + ( −2 + 5) 2            ✓subt in corr form
                                              ✓answer
       RS= 34                                                   (4)
1.2      x + x y + y2 
      M 1 2 ; 1         
         2          2 
         −5 + 3 0 − 2 
      M        ;      
         2        2                         ✓x-value
      M(– 1; – 1)                             ✓y-value
                                                                (2)
1.3   t −5
            = −1                              ✓Method
        2
      t − 5 = −2
                                                                (1)
      t =3
1.4           y −y
      mPQ = 2 1
              x2 − x1
             3−0
       mPQ =                                  ✓subt in corr form
             0+5
             3
       mPQ =
             5                                ✓answer
                                                                (2)

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                                          3

1.5         −2 − 3
      mQR =                                                 ✓subt in corr form
             3−0
              5
      mQR = −                                               ✓answer
              3
                   3 5
      mPQ  mQR =  − = −1
                   5 3                                      ✓Method
         ˆ
       PQR= 90  o                                          ✓answer

              OR                                            OR
                                 3
      Inclination of PQ = tan −1   = 30,96o               ✓inc PQ
                                 5
             −2 − 3
      mQR =
              3−0
               5
      mQR = −                                               ✓gradient QR
               3
                                       5                  ✓inc QR
      Inclination of QR = 180 − tan −1   = 120,96o
                                       3
       PQRˆ = 120,96o − 30,96o = 90o                       ✓answer
                                                                                (4)
1.6   Square                                                ✓square
      Diagonals bisect each other/diagonals have the same
      midpoint.                                             ✓diagonals
      All sides are equal.                                  ✓any of the other
      Interior angles = 90o                                 two reasons.
                                                                             (3)
                                                                            [16]

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                                            4

QUESTION 2

2.1   θ = 90o +37 o (ext  of a triangle)       ✓Method
      θ = 127 o                                 ✓answer
                                                                  (2)
2.2   m = tan127   o

                                                ✓gradient
      m = −1,33
      y − y1 = m ( x − x1 )
       y − 4 = −1.33 ( x + 2 )                  ✓Subt of A
       y = −1,33 x − 2, 66 + 4
       y = −1,33 x + 1,34                       ✓answer
                                                                  (3)
2.3   −1,33 x + 1,34 = 0                        ✓y = 0
      −1,33x = −1.34
                                                ✓x = 1
      x =1
                                                                  (2)
      Q(1; 0)
2.4   y2 − y1
              =m
      x2 − x1
       4+4                                      ✓Method
               = −1.33
      −2 − p                                    ✓simplification
      2, 66 + 1.33 p = 8
      1.33 p = 5.34
                                                ✓answer
      p=4

      OR

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                              5

      y2 − y1
              =m
      x2 − x1
                                  ✓Method
      −4 − 0
             = −1.33
       p −1                       ✓simplification
      −1.33 p + 1.33 = −4
      −1.33 p = −5,33
      p=4                         ✓answer
                                                     (3)
2.5          y2 − y1
      mBC =
             x2 − x1
             −4 + 1
           =                      ✓subt
              4+2
               1                  ✓mBC
           =−
               2
                       1          ✓mnew
      mnew = mBC = −
                       2
      y − y1 = m ( x − x1 )
                                  ✓subt of point A
               1
      y − 4 = − ( x + 1)
               2
            1    1
      y =− x− +4
            2    2
            1    7                ✓eq.
      y =− x+
            2    2                                   (5)
                                                    [15]

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                                          6

QUESTION 3

3.1.1   sin122o = 0,848                       ✓answer
3.1.2        tan 
          cos2 ( +  )
                 tan122o
        =                                     ✓subt
          cos 2 ( 89.4o + 122o )
               tan122o
        =
            ( cos 211, 4 )
                        o 2

        = −2,197                              ✓answer
                                                             (2)
3.2.1           2
        tan  = −
                3
        r = x + y2
         2   2                                ✓diagram

        r 2 = ( −3) + ( 2 )
                    2         2
                                              ✓value of r
        r = 13
                         3
         cos  = −
                         13                   ✓cos β
                                                             (3)
3.2.2       13 cos  − 13sin     2

                                      2       ✓subt
               3         2 
        = 13  −    − 13     
               13        13 
        = −3 − 4
                                              ✓answer
        = −7                                                 (2)
3.3

        r 2 = x2 + y2
        1 = x2 + m2
        x2 = 1 − m2                           ✓ x = 1 − m2
        x = 1 − m2
        tan 94o = tan (180o − 86o )
                  = − tan 86o                 ✓reduction
                         1 − m2
                  =−                          ✓answer
                           m                                 (3)

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                                                  7

3.4   cos ( 360o − x ) tan ( x − 180o )
                                                                             ✓cos x
           sin ( − x ) cos ( 90o − x )                                       ✓tan x
         cos x. ( tan x )                                                    ✓– sin x
      =                                                                      ✓sin x
        ( − sin x ) .sin x                                                     sin x
           cos x sin x                                                       ✓
      =− 2 .                                                                   cos x
          sin x cos x
             1                                                                                  (5)
      =−
          sin x
3.5   tan120o.sin115o
      cos 240o cos 205o                                                      ✓ − tan 60o
          − tan 60o.sin 65o                                                  ✓sin 65o
      =                                                                      ✓ – cos 60o
        − cos 60.(− cos 25o )                                                ✓– cos 25o
           3.sin 65o
      =−                                                                     ✓sin 65o
          1
            sin 65o
          2
                                                                             ✓answer
      = −2 3                                                                                    (5)
3.6       1              1         2
                  +           =
      1 − cos x 1 + cos x sin 2 x
               (1 + cos x ) + (1 − cos x )                                   ✓numerator
      LHS =                                                                  ✓denominator
                 (1 − cos x )(1 + cos x )
                     2
             =
                1 − cos 2 x                                                  ✓simplification
                    2
              =
                 sin 2 x
                                                                                                (3)
              = RHS
3.7   5 tan ( θ + 10o ) = 4
                         4                                                   ✓std form
      tan( + 10o ) =
                         5
                      4
      ref = tan −1     = 38, 66o
                      5                                                      ✓ref angle

              First Quad                                Third quad
      ( + 10 ) = 38, 66o + k.180o
               o
                                               ( + 10 ) = 218o + k.180o
                                                      o

                                                                             ✓answer
                = 28, 66 + k.180o                     = 208, 66 + k .180   ✓answer
                                          k                                                    (4)
                                                                                               [29]

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                                      8

QUESTION 4

4.1                                         f:
                                            ✓shape
                                            ✓turning points
                                            ✓(0; 0)

                                            g:
                                            ✓shape
                                            ✓turning point

                                            ✓end points of f and
                                            g.
                                                              (6)
4.2   Period = 180o                         ✓answer
                                                              (1)
4.3    y  0;3                            ✓✓answer
                                                              (2)
4.4    −90o ;0o    or   −90o  x  0o   ✓notation
                                            ✓critical values
                                                              (2)
                                                             [11]

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                                              9

QUESTION 5

5.1        bisect                                                ✓answer
                                                                            (1)

5.2.1(a)   Ĉ1 = 72o (angles in the same segment)                ✓S
                                                                 ✓R
                                                                            (2)
5.2.1(b)   O1 = ˆ
           ˆ    C1 = 72 (corr angles, OE//CD)
                                    o                            ✓S
                                                                 ✓R
                                                                            (2)
5.2.1(c)   ˆ = 2 D
           O      ˆ (angle at centre = 2 × angle at the cirm.)   ✓S
            1      4
                                                                 ✓R
           D̂4 = 36o                                                        (2)
5.2.1(d)   N1 + ˆ
           ˆ    N2 = 108 (Opp angles of a cyclic quad)
                                        o                        ✓S
                                                                 ✓R
                                                                            (2)
5.2.1(e)   ˆ + (D
           D    ˆ +D
                   ˆ )+D
                       ˆ = 180o (sum of angles of a triangle)    ✓S/R
            1    2  3   4

           D̂1 + 90o + 36o = 180o (angle in the semi-circle)     ✓S/R
            D̂1 = 54       o
                                                                 ✓S
                                                                            (3)
5.2.2      Eˆ 2 = D
                  ˆ = 54o (corr angles OE//CD)
                   1
                                                                 ✓S/R
           ˆF = Eˆ + D
                     ˆ (ext. angle of a triangle)
            2       2           4                                ✓S/R
           F̂2 = 90     o

            AF = FD (perpendicular from centre to chord)        ✓R
                                                                            (3)
5.2.3      No                                                    ✓No
           Ê2 + ˆ
                C1o  180o (opp angles are not suppl)            ✓S/R
                                                                            (2)
                                                                           [16]

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                                      10

QUESTION 6

6.1    bisect                                   ✓answer
                                                                (1)
6.2

                                                ✓construction
       ˆ +C
       C  ˆ = 90 (angle in the semi-circle)
                                o               ✓S/R
        1   2

        Ĉ2 = 90o                              ✓S
       And Aˆ +A ˆ = 90o (tan ⊥ radius)         ✓R
                    1       2

       A1 = 90 − ˆ
        ˆ        A2     o
                                                ✓S/R
       But ˆ
           C = ˆ
               A (angles in the same segment)
                2           2                   ✓S/R
       A1 = ˆ
       ˆ    C2                                                  (6)

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                                        11

6.3   P̂1 = ˆS (tan chord thrm)                ✓S
                                               ✓R
      and Sˆ = Qˆ (ext angle of cyclic quad)
                  1                            ✓S
      but ˆP1 = ˆ
                U (given)                      ✓R
              ˆ                                ✓S
       Û = Q1
                                               ✓R
       RU = RQ (angles opp sides are equal)              (6)
                                                         [13]
                                                    TOTAL: 100

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