Mathematics Grade 12 IEB Paper 2 2017 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 2017 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 © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 2 of 11 SECTION A QUESTION 1 (a) 0,7337 (The second mark awarded for rounding off correctly) (b) C (c) A = 0,6268 B = 0,0264 (d) No, you would be extrapolating (The second mark is for the concept of extrapolation) QUESTION 2 40 (a) mOA = = 2 20 ˆ =2 tan AOB ˆ = 63,43 AOB 1 (b) m =  2 Midpoint of OA = (1;2) 1 y=  x+c 2 Subs (1; 2) 1 2=  +c 2 1 5 y=  x+ 2 2 (c) x=3 1 5 (d) y=  (3) + 2 2 y=1  (x – 3)2 + (y – 1)2 = r2 Subs (0;0)  (0 – 3)2 + (0 – 1)2 = r2  r2 = 10  (x – 3)2 + (y – 1)2 = 10 IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 3 of 11 QUESTION 3 (a) (1) sin(31° + 22°) = k = sin 53° =k (2) cos(90° + 53°) = –sin 53° = –k (3) cos(75° – 22°) = cos 53° = 1  k 2 (There is a method mark for workings.) cos θ cos2 θ  sin2 θ (b)  2sin θ cos θ 2 sin θ 1 cos2 θ  sin2 θ   2 sin θ 2 sin θ 1 cos2θ + sin2θ  2 sin θ sin2 θ + cos2 θ  cos2 θ + sin2 θ  2 sin θ 2 sin2 θ  2 sin θ  sin θ therefore LHS = RHS (c) 3 sin2 θ  2 sin θ = 0 sin θ (3 sin θ  2) = 0 sin θ = 0 θ = 0 + k180 Alternate: θ = 0 + k360 OR θ = 180 + k 360 OR 2 sin θ = 3 θ = 41,8° + k 360 OR θ =138,2° + k 360 K  Z IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 4 of 11 QUESTION 4 (a) M(3;–1) (b) (0 – 3)2 + (y + 1)2 = 25 y2 + 2y – 15 = 0 (y + 5)(y – 3) = 0 y = –5 OR y = 3 C(0;3) 3   1 4 (c) mCM = = 03 3 3 mAC = 4 3 y = x +3 4 3 (d) 0= x +3 4 x = –4 A(–4; 0) (x – 3)2 + (0 + 1)2 = 25 (x – 3)2 = 24 x = 3 ± 24 AB = 4 units – 1,9 units AB = 2,1 units IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 5 of 11 QUESTION 5 (a) ˆ = ABC R.T.P: CAE ˆ Construction: refer to diagram for the construction Proof: F ˆ + CAE OAC ˆ = 90 (Tangent perpendicular C to line through centre) ˆ = 90 (Angles in semi-circle) FCA B ˆ + OAC OFC ˆ = 90 (Angles in triangle) O therefore ˆ = CAE OFC ˆ but D E A ˆ = ABC OFC ˆ (Angles in same segment) therefore ˆ = ABC CAE ˆ (b) Bˆ 3 = 70 (Angles in same segment) Fˆ = 52 2 ˆ +G G ˆ = 70 (Exterior angle of cyclic quad equal to the interior 1 2 opposite angle) ˆ = 52 Fˆ2 = G (tan chord theorem) 2 ˆ G = 18 1 IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 6 of 11 QUESTION 6 (a) A = 50 (b) 400 (c) P = 50 and M = 100 (d) Q3 = greater than 300 and less than 325 (approximate) Q1 = 200 IQR = 110 (ca mark based on values) (e) Mean = 250 (f) (1) It would stay the same. It is only the upper 25% of data that are affected. Or It would decrease. People leave the contract therefore less people. (2) Standard deviation would decrease. The difference between the new mean and the data would decrease. (3) It would skew the data to the left. The vales above the median are less spread out. or mean < median IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 7 of 11 SECTION B QUESTION 7 (a) mOA = 3 Equation of line OA is y = 3x Equation of line EF is 2y + x = 10 2(3x) + x = 10 7x = 10 10 x= (This represents the height of the triangle.) 7 30 y= 7 Coordinates of point E 2y + 0 = 10 y =5 E(0; 5) 1 10 Area of EBO = 5 2 7 25 Area of EBO = units2 7 (b) C(4; 0) 2(0) + x = 10 x = 10 1 18 Area of DCF = 6 2 5 54 Area of DCF = units2 5 IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 8 of 11 QUESTION 8  3  0   1   2   2 2 (a) (1) OC   18 (2) B (6; –2) 3 Alternate solution: (3) mOC = =1 3 62  18  18  2 18  cosCOB ˆ ˆ = 45  OCB COBˆ  90 ˆ  45 (Angle at centre)  CAB ˆ = 90  angles of a    COB ˆ = 45 (angle at centre)  CAB (b) Circumference = 2r Circumference = 2 18 units or 26,66 units ˆ = 60 (Angle at centre = 2 x angle at circumference) COB The size of angle θ after B moves into new position 9 1 = 18 18 sin θ (Area rule) 2 2 θ = 30° θ = 180° – 30° = 150° B needs to move 90° anti-clockwise Therefore 90° Point B needs to move 2 18  360° OR Point B needs to move 6, 66 units. IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 9 of 11 QUESTION 9 (a) Ĉ is a common angle ˆ =A D ˆ (tan chord theorem) 2 Therefore ΔADC ||| ΔDBC (A.A.A) OR Bˆ 2 = ADC ˆ (Angles in a triangle) DC AC (b) = ( ΔADC ||| ΔDBC ) BC DC DC2 = AC.BC but AC = AB + BC Therefore DC2 = BC (AB + BC) DC2 = AB.BC + BC2 AB.BC = DC2 – BC2 QUESTION 10 (a) ˆ = 90 ADL (Angle in semi-circle) (one mark for the reason) ˆ = 90 ACB Therefore DL||CB (Converse: corresponding angles are equal) (b) LC = LA (radii of the large circle) Alternative solution: SD = SL = SA (radii of small circle) AD = DC (converse: midpoint DL||BC) but In ΔACL LA = SA + SL therefore DS||CL (midpoint theorem) LC = 2SD  LC = 2SD (c) AS = SL and AL = LB; radii SL 1  = AB 4 (d) LB = 15 units (radius) 9 LM = (prop theorem) 16 15 LM = 8,44 units IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 10 of 11 QUESTION 11 (a) (1) BE = 2OA – ED (radii) (2) AE = EC (Line from centre is perpendicular to chord) BE2 = BC2 – EC2 (2OA – ED)2 = BC2 – AE2 (b) Construction DB ˆ =BDE DBE ˆ (Tangents drawn from common chord) ˆ = 180°  θ ˆ = BDE DBE 2 ˆ = 180°  θ A (tan chord theorem) 2 ˆ = 180  180°  θ (Opp angles of cyclic quad) C 2 ˆ = 180°  θ = 90 + θ C 2 2 A B C E D IEB Copyright © 2017 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 11 of 11 QUESTION 12 (a) Front view of the circles 3 6 h 3 3 h2 = 62 – 32 (Pythagoras)  h=3 3  Height of B 3 3 +6 (b) (1) sin 50° = 11,2 AB AB = 11,2 sin 50° AB = 14,62 metres (2) A view of the triangle made by the two pieces of rope and the horizontal plane sin A sin 70 = 13 14,62 ˆ = 56,68 A Bˆ = 53,32 OPTION 1 EA 14,62 = sin 53,32° sin 70° EA = 12,48 metres OPTION 2 EA2 = 132 + 14,622 – 2(13)(14,62)cos53,32° EA = 12,48 metres Total: 150 marks IEB Copyright © 2017