Mathematics Grade 12 IEB Paper 2 2018 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 2018 MATHEMATICS: PAPER II MARKING GUIDELINES Time: 3 hours 150 marks These marking guidelines are prepared for use by examiners and sub-examiners, all of whom are required to attend a standardisation meeting to ensure that the guidelines are consistently interpreted and applied in the marking of candidates' scripts. The IEB will not enter into any discussions or correspondence about any marking guidelines. It is acknowledged that there may be different views about some matters of emphasis or detail in the guidelines. It is also recognised that, without the benefit of attendance at a standardisation meeting, there may be different interpretations of the application of the marking guidelines. IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 2 of 10 SECTION A QUESTION 1 (a) (3; 1) 3 − ( −1) (b) mAB = =1 5 −1 y =− x + c − (3) + c 1= c=4 y =− x + 4 ( 3 − ( −1) ) + ( 5 − 1) 2 2 (c) = AB AB = 32 or 5,66 units ( x − 3 ) + ( y − 1) = 2 2 (d) 8 x 2 + y 2 − 6 x − 2y + 2 =0 ( 4 − 3 ) + ( y − 1) = 2 2 (e) 8 ( y − 1) 7 2 = y = 1± 7 (f) M is 3 units away from the y-axis. The radius of the circle is 8 units. Therefore the shortest distance of the circle from the y-axis is 3 – 8 units. QUESTION 2 (a) OMNˆ = 90o (Tangent perpendicular to line from centre) x 2 − 8 x + 16 + ( y + 4 ) =9 + 16 2 (b) (1) O ( 4; −4 ) ( x − 4) + ( y + 4) = 2 2 (2) 25 5 units (c) MN 2 + OM 2 ON = 2 (11 − 4 ) + ( −5 − ( −4 ) ) 2 2 = ON ∴ ON = 50 2 = MN 50 − 52 MN = 5 units IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 3 of 10 QUESTION 3 5 (a) 4 3 2 1 0 –1 –2 –3 –4 –5 Starting point (0o ;0o ) and Finishing point (180o ;0o ) Both Turning points correct (45o ;3) and (135o ; −3) X intercept of (90o ;0) Shape of the graph (b) 3 sin 2 x = cos 2 x 1 tan2 x = 3 Reference angle = 18,43o = 2 x 18,43o + k180o = x 9,22o + k 90o x = {9,22o ;99,22o } (c) cos 2 x = 0 { x = 45o ;135o } IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 4 of 10 QUESTION 4 (a) Look for construction on diagram or labelled BO B 1 2 Oˆ1= Aˆ + Bˆ1 Exterior angle of triangle Aˆ = Bˆ1 Isos triangle OR Radii Similarly in other triangle 1 O 2 Oˆ1= 2 × Bˆ1 A C Oˆ = 2 × Bˆ 2 2 Therefore ˆ = 2 × ABC AOC ˆ (b) (1) Bˆ1 = 73o Opposite angles of cyclic quad (2) ˆ = 40° OMB (isos triangle) ˆ = 100o MOB (angles of a triangle) Tˆ = 50o 2 (Angle at centre = two times angle at circumference) (3) Mˆ 3 = 17o (Angles in a triangle) Mˆ = 23o 2 (40 – 17) (c) (1) Pˆ1 = 56o (Tan chord theorem) Pˆ = 54o 2 (Angles on straight line) Sˆ = 54o (Angles in same segment) (2) Rˆ1 = 37o (Angles in same segment) Therefore QRSˆ = 93o ˆ = 90o But for QS to be the diameter the size of QRS Hence QS is not the diameter QUESTION 5 (a) ˆ =x ACE (Angles in same segment) ˆ = x+y AFE (Ext angle of a triangle) ˆ = CDE AFE ˆ Therefor FCDE is a cyclic quadrilateral (Converse: ext angle of cyclic quad) (b) ˆ =x AEB (Isos triangle) ˆ =x ACE (Angles in same segment) ˆ = ACE AEB ˆ Therefore AE is a tangent (Converse: tan chord theorem) IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 5 of 10 QUESTION 6 (a) LV + HV Q1 Q2 Q3 Diagram (b) 2000 households (c) Skewed to the right or positively skewed (d) (1) Standard deviation would decrease as the households using more than 100 litres would lower their consumption and hence the values will be closer to the mean. (2) The data would become more symmetrical as the mean will move much closer to the median. OR the data would become less positively skewed IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 6 of 10 SECTION B QUESTION 7 (a) Pamphlets; the correlation coefficient is closer to 1 (b) (1) It would increase as the values would be closer to line of best fit (2) Pamphlets: Gradient would increase Television: Gradient would decrease (c) Television; as the gradient is steeper and although the correlation coefficient is lower, the outliers fall on the high side (Low expenditure; High sales), so you can predict that there will be an increase in sales. QUESTION 8 (a) (1) cos334°. sin 244 = ° cos26°. ( −sin64° ) sin64°. ( −sin64° ) = − p2 (2)8 sin16°.cos16°.cos32° = 4sin32°.cos32° = 2sin64 = ° 2p (b) sin 43o =cos(90o − k )cos 23o + cos 246o sin 23o sin 43o =cos(90o − k )cos 23o − cos 66o sin 23o sin(66o − 23 = o ) sin k cos 23o − cos 66o sin 23o Therefore k = 66o 2cos 2θ .cosθ 2 (c) + 2 tanθ .sinθ = cos θ − sin θ 2 2 cosθ 2cos 2θ .cosθ 2sin2 θ + cos 2θ cos θ 2cos2 θ 2sin2 θ + cos θ cos θ 2cos2 θ + 2sin2 θ cos θ 2(cos2 θ + sin2 θ ) cos θ 2 cos θ IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 7 of 10 (d) ˆ = 135o FWG ˆ sin FGW sin135o = 6 116 ˆ FGW = 23,2o ˆ = 21,8o WFG GH sin21,8o = 116 GH = 4 units QUESTION 9 (a) CB = 10 units BA = 8 units 1 × CD × perp.ht Area of DBDC (b) .= 2 Area of DBED 1 × DE × perp.ht 2 Triangles have same perp height 5 = 4 (c) Split of CE into two 5/9 or 5:4 or 4/9 DE/CE = 4/9 DE/CF = 20/81 100 DE = 11.11 or 9 IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 8 of 10 QUESTION 10 (a) F̂1 = Sˆ3 (Tangents drawn from common point) F̂ = Sˆ 1 2 (Alternate angles DH//SG) Sˆ 2 = Hˆ 1 (Angles in same segment) Ĥ= Gˆ + Gˆ 1 1 2 (Radii; isos triangle) DDSF /// ∆OHG (A.A.A) or D̂ = Oˆ1 (Angles in a triangle) DF SF (b) = (prop theorem) OG HG OG.SF DF = HG But FH OG = 2 Therefore SF × FH 2 × DF = HG QUESTION 11 (a) AE = 4 units (Line from centre perpendicular to chord) OE = 0,8 units OA = 4,08 units EK = 4,08 + 0,8 = 4,88 units ˆ = 29 − 13 − 64 (b) cos BAC −2 × 13 × 8 ˆ = 33,69o BAC height of B sin33,69o = 13 height of B = 2 units point below B cos33,69o = 13 Point below B is 3 units IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 9 of 10 Distance from K to new point 12 + 4,882 = 4,98 units Therefore Distance from B to K once the fold has been made is 22 + 4,982 = 5,37 units QUESTION 12 (a) Line AD y= x + c –2 = 1 + c c = –3 x intercept 0= x − 3 x =3 D ( 3;0 ) OR Create a triangle by dropping a perpendicular from point D. Side lengths are 2. Therefore x coordinate of D is 1 + 2 = 3 D ( 3;0 ) 1+ 2 (b) mAB = mAB = −1 −2 − 1 mBC = 1 ˆ = 90o since m × m ABC = −1 AB BC (c) ˆ = 90o CDA (Opp angle of cyclic quad) 5 mCD = 2− x −2 mAD = 1− x 5 −2 × = −1 2 − x 1− x x 2 − 3x − 8 =0 x = 4,7 or x = −1,7 D needs to move 1,7 units to the right IEB Copyright © 2018 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 10 of 10 Alternate Solution AC is a diameter (Converse: angle in semi-circle) 3 3 Midpoint  ;  2 2 72 + 12 50 Radius of circle = = 2 2 Third side of triangle 2  50   3 2   −    2  2 = 3,2 units 3,2 + 1,5 = 4,7 units D must have coordinates of ( 4,7;0 ) Therefore it must move 1,7 units to the right Alternate Solution AC is a diameter (Converse: angle in semi-circle) 3 3 Midpoint  ;  2 2 72 + 12 50 Radius of circle = = 2 2 Circles equation is 2 2  3  3 50 x − 2 +y − 2 =     4 Find the x intercepts 2 2  3  3 50  x − 2  + 0 − 2  =     4 x = 4,7 D must move 1,7 units to the right Total: 150 marks IEB Copyright © 2018