Mathematics Grade 11 Paper 2 Free State June 2024 Memo

Downloaded from hlayiso.com MATHEMATICS GRADE 11 JUNE P2 2024 MARKS: 100 TIME : 2 HOUR This question paper consists of 8 pages.

Downloaded from hlayiso.com 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.

Downloaded from hlayiso.com 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]

Downloaded from hlayiso.com 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]

Downloaded from hlayiso.com 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]

Downloaded from hlayiso.com 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]

Downloaded from hlayiso.com 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]

Downloaded from hlayiso.com 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

Downloaded from hlayiso.com Mathematics Gr 11 P2 9 FS/June 2024

Downloaded from hlayiso.com 1 GRADE 11 MATHEMATICS 2024 JUNE P2 MARKING GUIDELINE This marking guideline consists of 11 pages

Downloaded from hlayiso.com 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)

Downloaded from hlayiso.com 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]

Downloaded from hlayiso.com 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

Downloaded from hlayiso.com 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]

Downloaded from hlayiso.com 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)

Downloaded from hlayiso.com 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]

Downloaded from hlayiso.com 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]

Downloaded from hlayiso.com 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]

Downloaded from hlayiso.com 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)

Downloaded from hlayiso.com 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