Mathematics Grade 12 IEB Paper 2 2019 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 2019 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 © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 2 of 13 SECTION A QUESTION 1 (a) A (0 ; 3) (b) −2 x + 3 = 3 x − 7 −5 x = −10 x=2 y = −1 ( 3 − ( −1) ) + ( 0 − 2) 2 2 (c) = d d = 20 ( x − 2 ) + ( y + 1) = 2 2 (d) 20 (e) ( 4; −5 ) (f) 0= −2 x + 3 3 x= 2 3 =0 3  + c 2 9 = 3x − y 2 QUESTION 2 (a) (1) Method of finding third side 1− m2 (2) 2cos2 25° − 1 or cos2 25° − sin2 25° = 2m 2 − 1 or m 2 −( 1 − m 2 )2 (3) cos30° cos 25° − sin30° sin 25° 3m 1−m 2 − 2 2 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 3 of 13 sin ( 21° + w ) sin2 β (b) − .cos2 β sin (w + 21° ) cos β 2 = 1 − sin2 β = cos2 β (c) B 5 units A C 5 units 1 Area ∆ABC = × 5 × 5 × sin36° 2 Area ∆ABC = 7,3 units2 QUESTION 3 (a) p=2 (b) 180° (c) 3 sin 2 x = 2 sin x 6 sin x cos x = 2 sin x 2sin x ( 3cos x − 1) = 0 1 cos x = 3 =x 70,5° B ( 289,5°; −1,9 ) (d) k>3 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 4 of 13 QUESTION 4 M (a) Construction AM and MC EAM= 90° ˆ (Line from centre perpendicular to tangent) ˆ = 90° MCA (Angle in semi-circle) ˆ + MAC CMA ˆ =°90 (Angles in a triangle) Therefore ˆ = EAC AMC ˆ but ˆ = ABC AMC ˆ (Angles in same segment) Hence ˆ = ABC EAC ˆ Alternate solution (there are a number of others) M Construction AM and MB ˆ = 90° (Angle in semi-circle) MBA ˆ = 90° MAE (Line from centre perpendicular to tangent) MBC = MAC ˆ ˆ (Angles in same segment) ˆ = CAE CBA ˆ IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 5 of 13 (b) Fˆ= 1 55° (Tan chord theorem) Fˆ= 35° 2 (Angles in semi-circle) = 35° Eˆ1 (Angles in same segment) (c) = Cˆ 110° 1 (CF = CG radii; Isos triangle) D= 80° ˆ 1 (Angle @ centre) Dˆ= 3 80° (Vert opp) Dˆ + Bˆ = 180° 3 Therefore DHBJ is a cyclic quad (Converse: Opp. angles of a cyclic quad are supplementary) Alternate solution = Cˆ 110° 1 (CF = CG radii; Isos triangle) D= 80° ˆ 1 (Angle @ centre) = ˆ 100° D 2 (Angles on a straight line) D = Bˆ ˆ 2 Therefore DHBJ is a cyclic quad (Converse: Exterior angle of a cyclic quad is equal to the interior opposite angle) IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 6 of 13 QUESTION 5 (a) (1) = x 180° + k .360 (2) Shape of graph Points on ( −90° ; –1) and ( 90° ; 1) Understanding of Asymptotes at −180° and 180° cos θ sin2 θ + cos3 θ (b) = − 2sin2 θ cos 2θ cos θ = 1 − 2 sin2 θ cos θ (sin2 θ + cos2 θ ) − 2sin2 θ cos θ = 1 − 2sin θ (5) 2 QUESTION 6 (a) 16 (b) (1) Look at the diagram (2) IQR = 38 – 20 = 18 (c) (1) A = –2,6 and B = 2 (2) Extrapolation or prediction is way outside HV used (d) (1) (2) The values of those to the right of median are more spread out thus the mean is pulled to the right of the median. (e) Yes; it needs a service as one standard deviation from the mean shows that a large percentage of the logs are between 9,2 and 10,8 metres. This shows that the machine is inaccurate. IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 7 of 13 SECTION B QUESTION 7 (a) –0,8 (b) (1) Get closer to –1 or decrease (2) Decrease or get steeper (3) A person who plays professional sport and studies via the internet (c) (1) More of the data is closer to the mean OR 4 low and 3 high hence further apart (2) Mean would decrease and standard deviation will decrease (3) Decrease the price on any day from Monday to Saturday, increase in sales and data will be more dispersed OR any intervention that would increase the sales on a day between Monday and Saturday. QUESTION 8 CN DG 2 (a) (1) = = NE GE 3 DG = GK Therefore EK 1 = KG 2 DH 4 (2) = HF 1 DE = 5m and DF = 5k 1 × 2m × 4k × sin Dˆ 2 1 × 5m × 5k × sin Dˆ 2 8 = 25 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 8 of 13 (b) (1) Ê2 = x (Angle at centre) N̂3 = x (Alternate angles // lines) Ĝ1 = x (Tan chord theorem) ALT: E= ˆ N= 2 ˆ Gˆ 3 1 (With the reasons above) Therefore GN = NE (Converse: isos triangle) (2) ˆ= 180° − 2 x GNE (Angles in a triangle) Hˆ = 2 x 2 (Opp. angles of cyclic quad) Eˆ1= Nˆ 1= 90° − x (Angles in isos triangle) ∆GON|||∆GHE (A.A.A) (3) Working with HE HE GE = (Similar triangles) ON GN HE HB = (Prop theorem) EP BC QUESTION 9 (a) FH = 50 HG = 65 102 = 50 + 65 − 2 × 50 × 65 cos Hˆ Hˆ = 82,4° ALT: Cosine rule for any two sides and angle 1 Area ∆FGH =× 50 × 65 × sin82,4° 2 Area ∆FGH =28,3 m 2 (b) (1) Coordinates of C x 2 − 6x + 8 = 0 ( x − 4 )( x − 2 ) = 0 C(4; 0) IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 9 of 13 (2) x 2 − 6 x + 9 + y 2 − 4 y + 4 =−8 + 13 ( x − 3 ) + ( y − 2) = 2 2 5 Centre of circle (3; 2) 1 Gradient of line AB × =−1 2 Gradient of line AB = −2 3 − = 2 (5) + c c = 13 y= −2 x + 13 Coordinates of B 0= −2 x + 13 x = 6,5 B ( 6,5;0 ) Therefore CB = 2,5 units IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 10 of 13 QUESTION 10 2 8 (a) = y x+ 3 3 ˆ =2 tan FEA = ˆ FEA 33,7° 3 ˆ = −1 tan EFA ˆ = 45° EFA Therefore =ˆ EAF 101,3° AF 52 (b) = sin33,7° sin 45° AF = 5,66 CF = 2,83 (Line from centre perpendicular to chord) 2 = CB 40 − 2,83 2 2 CB = 5,7 Alternate solution T AT = sin33,7° 52 AT = 4 TF = 4 units (Isos Triangle) AF = 32 CF = 2,83 (Line from centre perpendicular to chord) 2 = CB 2 40 − 2,832 CB = 5,7 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 11 of 13 QUESTION 11 Construction MD ˆ =x MDE (Tan chord theorem) ˆ =x DME (Tan chord theorem) OR (isos triangle; tangents from common point) ˆ = 180° − 2 x DEM 180° − 2 x= x + 45° + 2 x − 40° (Ext angle of triangle) −5 x =−175° = 35° x QUESTION 12 (a) 5 x + 12y = 60 A(0; 5) and B (12;0 ) AB= 2 52 + 122 AB = 13 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 12 of 13 (b) DD ADC = : BCD 8 : 18 8 = 13 × AC 26 AC = 4 units ( 5k ) + (12k ) 4 2 2 = 2 169k 2 = 16 16 k2 = 169 4 k= 13 Coordinates of C  48 45  C ;   13 13  Alternate solution DD ADC = : BCD 8 : 18 8 AC= 13 × 26 AC = 4 units If C(a; b) then: a 4 = (line parallel to side of ∆) 12 13 Coordinates of C  48 45  C ;   13 13  IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 13 of 13 Alternate solution DD ADC : BCD 8= : 18 8 = 13 × AC 26 AC = 4 units = ˆ CBO 22,62° y sin 22,62° = 9 y = 3,5 units m cos 22,62° = 9 m = 8,3 x =12 − 8,3 =3,7 C (3,7;3,5) Total: 150 marks IEB Copyright © 2019