MEMO MATHEMATICS: PAPER II

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1 MEMO CLUSTER PAPER 2016 MATHEMATICS: PAPER II Time: 3 hours 150 marks PLEASE READ THE FOLLOWING INSTRUCTIONS CAREFULLY 1. This question paper consists of 28 pages and an Information Sheet of 2 pages(i-ii). Please check that your question paper is complete. 2. Read the questions carefully. 3. Answer ALL the questions on the question paper and hand this in at the end of the examination. 4. You may use an approved non-programmable and non-graphical calculator, unless otherwise stated. 5. All necessary working details must be clearly shown. 6. Round off your answers to one decimal digit where necessary, unless otherwise stated. 7. Ensure that your calculator is in DEGREE mode. 8. It is in your own interest to write legibly and to present your work neatly. 9. The last pages can be used for additional working, if necessary. If this space is used, make sure that you indicate clearly which question is being answered.

2 CLUSTER: PAPER II SEPTEMBER 2016 Page 2 of 28 SECTION A QUESTION 1 Refer to the sketch below: AD//BC. A is the point ( 2; 4) and D is the point (1; -5). Determine: 1.1 the coordinates of B. (2) B(1; 4) 1.2 the equation of BC in the form ax + by + c = 0. (4) m BC = m AD = m AD = 9 3 m AD = 3 y 4 = 3(x 1) AD//BC y 4 = 3x + 3 y + 3x 7 = 0 for y [0; 7]

3 CLUSTER: PAPER II SEPTEMBER 2016 Page 3 of the coordinates of C. (2) x int: Let y = 0 3x = 7 x = 7 3 C ( 7 3 ; 0) [8] QUESTION 2 Refer to the sketch below: ABC has vertices of A(0; 3), B(-3;-2) and C(5; 0). y A(0;3) C(5;0) x B(-3;-2) 2.1 Calculate the length of: (4) AB AB = (0 + 3) 2 + (3 + 2) 2 AB = AB = 34 units BC BC = (5 + 3) 2 + (0 + 2) 2 BC = BC = 68 units PLEASE TURN OVER

4 CLUSTER: PAPER II SEPTEMBER 2016 Page 4 of AC AC = (5 + 0) 2 + (0 3) 2 AC = AC = 34 units 2.2 Hence, show that ABC is a right angled isosceles triangle. (4) AC = AB = 34 units AC = AB = 34 units ABC is an isosceles triangle ABC is an isosceles triangle AC 2 + AB 2 = = 68 m AB = = 5 3 and BC 2 = 68 m AC = = 3 5 AC 2 + AB 2 = BC 2 A = 90 m AB. m AC = ( 5 3 ) ( 3 ) = 1 AB AC 5 ABC is a right angled triangle 2.3 Determine the area of ABC. (2) Area of ABC = Area of ABC = 17 unit 2 ABC is a right angled triangle Area of ABC = sin90 Area of ABC = 17 unit Given that BC is a diameter of the circumscribed circle of ABC, show that the centre of this circle is M, the point (1; 1). (2) Midpoint of BC: ( ; ) = (1; 1) 2 2 [12]

5 CLUSTER: PAPER II SEPTEMBER 2016 Page 5 of 28 QUESTION 3 Find the equation of the line that is a tangent to the circle (x + 1) 2 + (y + 2) 2 = 20 at (3; 4). (6) Centre: ( 1; 2) m r = m r = 1 2 m T = 2 y + 4 = 2(x 3) y = 2x 10 (Radius Tangent) [6] PLEASE TURN OVER

6 CLUSTER: PAPER II SEPTEMBER 2016 Page 6 of 28 QUESTION 4 If 37cosθ + 10 = 2 and θ [180 ; 360 ], determine, with the use of a diagram, the value of 1 sin 2θ 4 cos θ. (4) 12 θ 37 cosθ = y = ± (37) 2 ( 12) 2 Pyth y = ± 1225 y = ±35 but in III y = 35 1 sin 2θ 4 cos θ 1 (2sin θ. cos θ) = 4 cos θ = 1 sin θ 2 = 1 35 ( 2 37 ) = [4]

7 CLUSTER: PAPER II SEPTEMBER 2016 Page 7 of 28 QUESTION Prove: 1 cos (90 2θ)tan (180 + θ) = cos 2θ (5) LHS = 1 sin 2θ tan θ sin θ LHS = 1 (2 sin θ cos θ) cos θ LHS = 1 2sin 2 θ LHS = cos 2θ LHS = RHS 5.2 For which values of θ is this identity in 5.1 undefined? (2) θ = 90 + k. 180 ; k Z [7] PLEASE TURN OVER

8 CLUSTER: PAPER II SEPTEMBER 2016 Page 8 of 28 QUESTION 6 Evaluate without using a calculator: 6.1 cos 3. cos cos 267. sin 42 (5) = cos 3. cos 42 cos 87. sin 42 = cos 3. cos 42 sin 37. sin 42 = cos ( ) = cos 45 = 2 2 = cos 3. sin 48 sin 3. sin 42 = cos 3. sin 48 sin 3. cos 48 = sin (48 3 ) = sin 45 = tan (1 + cos 120 ). sin (5) = (tan 30 ) 2 (1 cos 60 )( sin 45 ) 2 = ( 3 3 ) 2 (1 1 2 ) ( 2 2 ) 2 = 1 3 (1 2 ) (1 2 ) = = 1 12 [10]

9 CLUSTER: PAPER II SEPTEMBER 2016 Page 9 of 28 QUESTION 7 The table gives the Olympic pole vault records for the past 12 games Olympic games. Year Height in metres , , , , , , , , , , , , Find the equation of the line of best fit for the height (y) against the year (x) in the form y = a + bx. Determine your values for a and b rounded to 2 decimal digits. (2) a = 5,45 ; b = 0, Complete the scatter plot and draw the line of best fit on the scatter plot. (3) x x x x x x PLEASE TURN OVER

10 CLUSTER: PAPER II SEPTEMBER 2016 Page 10 of Determine the value of the correlation coefficient rounded to 2 decimal digits and explain clearly what can be deduced from this result. (3) r = 0,91 Very Strong positive linear correlation. Height of Olympic pole vault records should improve over the years. 7.4 Use the model to predict the record pole vault height for the 2016 Olympic Games. (1) From Graph: y = Height for 2016 = 6,10 m OR y = 5,45 + 0,05(13) y = 6,10 m Use the model to predict the record pole vault height for the 2028 Olympic Games. (1) y = 5,45 + 0,05(16) y = 6,25 m Do you think the actual record in 2028 will be higher or lower than this prediction? Give a reason for your answer. (2) Lower This is an extrapolation and it falls outside the range, therefor it is not a reliable value. [12]

11 CLUSTER: PAPER II SEPTEMBER 2016 Page 11 of 28 QUESTION 8 In the diagram, O is the centre of the circle passing through A, B, C and D. AB//CD and B = 20. Complete the following statements and reasons to prove that AOEC is a cyclic quadrilateral. No extra steps/calculations are allowed. (5) STATEMENTS C 1 = 20 O 1 = 40 D = 20 E 1 = 40 AOEC is a cyclic quad REASONS alt. angles ; AB//CD <.. at centre = 2 < on circumference <... s in same circle segment ext. angle of line... subtend =< s on the same side [5] PLEASE TURN OVER

12 CLUSTER: PAPER II SEPTEMBER 2016 Page 12 of 28 QUESTION Complete the following statement: The line from the centre of the circle perpendicular to the chord bisects the chord. (1) 9.2 Refer to the figure below: In the circle with centre O. OT QP, OS PR, OT = 8 units, PQ = 30 units and PR = 23 units. Q T O x S P R Determine OS = x. (5) QT = TP = 15 units line form centre of circle to chord OQ = Pyth OQ = 17 units OR = 17 units Radii RS = SP = 11,5 units line form centre of circle to chord OS = x = 17 2 (11,5) 2 Pyth x = 12,5 units

13 CLUSTER: PAPER II SEPTEMBER 2016 Page 13 of 28 QUESTION 10 [6] Refer to the figure below: PQ and PS are tangents to circle QRST. ST, RS and RQ are chords and QS is joined. PQ//ST. Let S 2 = x. x 10.1 Calculate Q 1 in terms of x. (1) Q 1 = x tan chord Theorem PLEASE TURN OVER

14 CLUSTER: PAPER II SEPTEMBER 2016 Page 14 of Prove: RSQ/// RQP (4) In RSQ and RQP: S 2 = Q 1 = x Q 2 = T but T = P 1 Q 2 = P 1 Proven < s in same segment; alt < s; PQ//ST R 2 + R 3 = R 1 of < s of = 180 RSQ/// RQP <<< 10.3 Prove: QR 2 = RS. RP (2) RS = QR QR PR QR 2 = RS. RP RSQ/// RQP [7] TOTAL FOR SECTION A = 77 MARKS

15 CLUSTER: PAPER II SEPTEMBER 2016 Page 15 of 28 SECTION B QUESTION 11 The circle, with equation x 2 10x + y 2 + 8y 40 = 0, is given. A point P(x; y) on the circumference of a NEW circle, is such that it is always 3 units from the circumference of the original circle, and outside the original circle. Determine the equation of this new circle. (6) x 2 10x y 2 + 8y + 16 = (x 5) 2 + (x + 4) 2 = 81 Initial circle: Centre (5; 4) and r = 9 New circle: Centre (5; 4) and r = 12 (x 5) 2 + (x + 4) 2 = 12 2 (x 5) 2 + (x + 4) 2 = 144 [6]

16 CLUSTER: PAPER II SEPTEMBER 2016 Page 16 of 28 QUESTION 12 Solve for x correct to one decimal digit if θ [ 180 ; 180 ] sin(2θ 10 ) = cos (θ + 50 ) (8) sin(2θ 10 ) = cos (θ + 50 ) sin(2θ 10 ) = sin [90 (θ + 50 )] RA (2θ 10 ) = 40 θ 3 rd k Z 4 th 2θ 10 = θ + k θ 10 = 360 (40 θ) + k θ = k. 360 θ = k. 360 θ = 76,7 + k. 120 θ = { 163,3 ; 43,3 ; 30 ; 76,7 } [8]

17 CLUSTER: PAPER II SEPTEMBER 2016 Page 17 of 28 QUESTION 13 The company, Mega Ski, has sales of ski equipment as given by S(t) = 10[1 cos (30. t)] where t is the time, in months (t = 0 corresponds to 1 August 2015), and S(t) is in thousands of Rands Use the given graphpaper below to sketch the function of the sales on a 12-month interval [0; 12]. (4) PLEASE TURN OVER

18 CLUSTER: PAPER II SEPTEMBER 2016 Page 18 of What is the period of the function? (2) 12 months 13.3 What is the minimum amount of sales? (1) R 0, What is the maximum and when does the maximum amount of sales occur? (3) Max: R When: after 6 months 1 February 2016 [10]

19 CLUSTER: PAPER II SEPTEMBER 2016 Page 19 of 28 QUESTION 14 At a particular time during the day, a tower of height 19,2 meters casts a shadow. At the same time, a person who is 1,65 meters tall, casts a shadow 5 metres long. 19,2 m 1,65 m Shadow of tower 5 m What is the length of the shadow cast by the tower at that time? (6) Tower//Person shadow of tower = 5 19,2 1,65 shadow of tower = 58,2 m OR one side of // to other side In small triangle: tanθ = 1,65 Tower and Person is to ground; corr < s equal 19,2 In big triangle: tanθ = 5 θ = 18, Shadow of Tower Shadow of Tower = 19,2 tan18, shadow of tower = 58,2 m [6] PLEASE TURN OVER

20 CLUSTER: PAPER II SEPTEMBER 2016 Page 20 of 28 QUESTION 15 The new fitness centre offers two different fitness classes. The attendance for each class for 12 sessions is represented in the box-and-and whisker plot Determine the interquartile range of Class A. (1) IQR = 14 4 = Compare the data from Class A and Class B. How are they alike? (1) Both has the same maximum attendance of 22 OR Both skewed to the right/positive Compare the data from Class A and Class B. How are they different? (1) Class A: min = 2 and Class B: min = 10 OR Class A: range = 20 and Class B: range = 12 OR any other reason Which class has more variability? Show calculations to support your answer. (3) Class A: Range = 20 and IQR = 10 Class B: Range = 12 and IQR = 4 Because the range and IQR of Class A is more than those of Class B, Class A has a more variability 15.5 Which class has a higher mean deviation? Explain your answer. (2) Class A. Skewed positively: x M > 0 Class A will have a higher mean deviation, meaning the data items are further from x. [8]

21 CLUSTER: PAPER II SEPTEMBER 2016 Page 21 of 28 QUESTION 16 Refer to the sketch below: In the diagram O is the centre of the circle with AE and CE tangents to the circle. C 1 = 45 and O 1 = 127 A B D O E C Calculate the size of E. (5) A 1 = 90 C 1 + C 2 = 90 C 2 = 45 Radius Tangent E = 360 2(90 ) 127 Int < s of quadrilateral = 360 E = 53 Radius Tangent OR A 1 = 90 Radius Tangent C 1 + C 2 = 90 Radius Tangent AECO is a cyclic quad opp int < s of quad = 180 O 1 + E = 180 opp int < s of cyclic quad E = 53 O 1 = 127 [5] PLEASE TURN OVER

22 CLUSTER: PAPER II SEPTEMBER 2016 Page 22 of 28 QUESTION 17 In the diagram below, points R, P, A, Q and T lie on a circle. RA bisects R and AB = AQ. RA and TQ produced meet at B. Prove that: 17.1 AQ bisects PQ B. (3) R 1 = R 2 Q 3 = R 1 Q 2 = R 2 Q 2 = Q 3 AQ bisects PQ B RA bisects R Ext angle of cyclic Quad angles in same circle segment 17.2 TR = TB. (2) B = Q 3 Q 3 = R 1 Proven in 17.1 R 1 = B TR = TB angles opp = sides sides opp = angles

23 CLUSTER: PAPER II SEPTEMBER 2016 Page 23 of P = TR P (3) A 1 = 2Q 3 ext angle of ; B = Q 3 P = A 1 angles in same segment Q 3 = R 1 proven R = 2R 1 from 17.1 P = TR P OR R 2 = R 1 = Q 2 = Q 3 = B proven in 17.1 and 17.2 RP//TB = alt < s [8] Q 2 + Q 3 = P alt < s ; RP//TB but Q 2 + Q 3 = R 1 + R 2 ext < of cyclic quad PRTQ P = TR P PLEASE TURN OVER

24 CLUSTER: PAPER II SEPTEMBER 2016 Page 24 of 28 QUESTION 18 Refer to the sketch: In the circle below, AB and CD are chords intersecting at E. A C E D B If AE = 5, BE = 12 and CE = 6, what is the length of DE? (6) Constr: AC and DB In ACE and BDE: A = D < s in same circle segment C = B < s in same circle segment AE C = DE B of < s of = 180 ACE/// DBE <<< AC DB = CE BE = AE DE 6 12 = 5 DE DE = ACE/// DBE DE = 10 units OR Similarly Constr: AD and BC Prove: AED/// CEB <<< [6]

25 CLUSTER: PAPER II SEPTEMBER 2016 Page 25 of 28 QUESTION 19 Daniel decided to participate in this year s Tour de France bicycle race. One of the routes are around the Annecy Lake which is a circular lake that has a diameter of 4 km, as shown in the diagram below. Points B and D are the opposite sides of Annecy Lake and lie on a straight line through the centre of the lake, with each point 5 km from the centre. The course of the race is ABCDE, where AB and DE are tangents to the lake. A E 4 km B 5 km C 5 km D Annecy Lake Determine the length of the route. (6) Constr: AC and CE A = E = 90 radius = AC = CE = 2 km AB = DE = AB = DE = 21 km radius = 1 2 diameter Pyth Length of route = (5) Length of route 19,2 km radii Radius Tangent PLEASE TURN OVER

26 CLUSTER: PAPER II SEPTEMBER 2016 Page 26 of 28 [6]

27 CLUSTER: PAPER II SEPTEMBER 2016 Page 27 of 28 QUESTION 20 The diagram shows triangles ABC and ABD with AD parallel to BC. The sides AC and BD intersect at Y. The point X lies on AB such that XY is parallel to AD and BC. C D Y 20.1 Prove ABC /// AXY. (4) In ABC and AXY: CA B = CA B C = AY X A X B common angle corresp < s ; CB//YX B = AX Y corresp < s ; CB//YX or of < s of = 180 ABC/// AXY <<< 20.2 Hence, or otherwise, prove that 1 = XY AD BD (6) PLEASE TURN OVER

28 CLUSTER: PAPER II SEPTEMBER 2016 Page 28 of 28 AX AB = XY BC ABC/// AXY Similarly is ABD/// XBY. BX BA = XY ABC/// AXY AD AX AB + BX BA = XY BC + XY AD But AX + BX = AB AB AB = XY ( 1 BC + 1 AD ) 1 XY = 1 BC + 1 AD [10] TOTAL FOR SECTION B = 73 MARKS TOTAL: 150 marks

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