NATIONAL SENIOR CERTIFICATE EXAMINATION EXAMINATION 05 MATHEMATICS: PAPER II MARKING GUIDELINES Time: 3 hours 50 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 05
NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page of 4 SECTION A QUESTION (a) 5+7 5+5 () K ; = K 6; 0 () () m=5 Therefore, y5=5x+5 y=5x+40 (3) (3) For J: 5x + 40 = 0 x = 8 J( 8;0) 0 0 5 m= 6 ( 8) 7 (3) 5 (4) (i) tan J ˆ ˆ = J = 35,5 () 7 tan J ˆ + Jˆ 5 J ˆ + J ˆ = 78,7 () (ii) (iii) Ĵ = 43, () IEB Copyright 05
NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page 3 of 4 (b) () OTR ˆ = 90 ; line from centre bisects chord () () 60 3 m OT = 40 m PR = 3 y + 6 = x 4 3 3y = x 6 (3) (3) For R: 0 = x 6 x = 3 R(3; 0) Therefore, radius = 3 (3) [] IEB Copyright 05
NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page 4 of 4 QUESTION (a) cos P Q cos 33 0,68... cos P cos Q 0,569... cos P Q cos P cos Q () (b) cos θ sin θ cos 80 θ cos 90 θ cosθ sin θ cosθ sin θ = (5) (c) () a b () () On graph (Accept A as shown) () (3) On graph (Indicated by B and C ) () (d) cos 7.sin98 sin8sin8 t (3) IEB Copyright 05
NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page 5 of 4 (e) () Kˆ is acute. K K x + 5 = 3 x =44 x = a = a =36 (4) 39 3 () Kˆ is obtuse a 39 3 a 36 () [] QUESTION 3 (a) IQR = 4 000 7 000 = R7 000,5 IQR = R0 500 Q 3 +,5 IQR = R4 500 Q,5 IQR = R3 500 Therefore, R5 000 is an outlier. (5) (b) for maximum * (c) Skewed to the right. Or positively skewed. () [0] (4) IEB Copyright 05
NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page 6 of 4 QUESTION 4 (a) STATEMENT REASON Cˆ Cˆ Fˆ Ext of cyclic quad Dˆ Eˆ 80 Opp 's of cyclic quad B ˆ Dˆ Tan/chord thm Bˆ Bˆ Dˆ Dˆ Not correct Dˆ Â Not correct (5) (b) () In the diagram below, circle centre M intersects a second smaller circle at A and B. A, B, C and T are points on circle centre M. AB is the diameter of the smaller circle. ˆ AMB = 90 ; angle in semi circle ˆT = 45 ; angle at centre Ĉ = 35 ; opp. angles of cyclic quad (6) () M ˆ + Cˆ 80 ; opp 's do not add up to 80 () IEB Copyright 05
NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page 7 of 4 (d) () () AB 4 = = PT 6 3 AC 5 = = PQ 7,5 3 BC TQ 3 AB AC BC = = = PT PQ TQ 3 ABC PQT; sides in prop. (4) ˆ AB AC sin A height from K Volume of Pyramid PQTM 3 = Volume of Pyramid ABCK PQ PT sin P ˆ height from M 3 Since ΔABC PQR, we have A ˆ = P ˆ Therefore, sin A ˆ = sin P ˆ, Volume of Pyramid PQTM Therefore, Volume of Pyramid ABCK = PQ PT AB AC 3 = = 9 4 (4) [0] 75 marks IEB Copyright 05
NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page 8 of 4 SECTION B QUESTION 5 (a) () Very strong and positive. () () (i) y = 0,6x 38,4 y = mx + c (5) (ii),74 million () (3) Not reliable because prediction far from interval of values in data given. Too few points to model on. () [0] IEB Copyright 05
NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page 9 of 4 QUESTION 6 (a) () sin B sin0 0 30 sin B 0,886... ˆB 6,8 A 80 0 6,8 A 43, OB 30 sin 43, sin 0 OB = 3,7 cm Therefore B moves 35,4 3,7,7 cm (6) () 50 30 0 cm. () (b) sin sin cos sin cos sin.cos sin.cos cos sin cossin cos cossin cos sin cos cos p (6) (c) () cos 3 sin 3 cos 30 cos 3 sin 3 cos.cos30sin.sin 30 3 cos 3 sin 3 cos. sin. cos 3 sin 3cos 3.sin 3sin cos tan (4) 3 () tan 3 30 k.80 ; k (3) [0] IEB Copyright 05
NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page 0 of 4 QUESTION 7 (a) () 6 6 x y y x x y y 3 9 6 3 5 Therefore, B(0;3) is centre of one circle. For M: m rad x 03 5 x =6 x=4 M(4;0) 3 0 3 0 4 4 4 m tan = 3 M tan OR M tan = 3 9 4 0 9 3 = slope of line joining centres 4 y0 x4 3 4 6 y x 3 3 () A( 9 9) B(0; 3) AB = ( 9 0) + ( 9 3) = 8 + 44 = 5 AB = 5 MN = 5; MNAB is a rectangle (5) (7) IEB Copyright 05
NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page of 4 (b) () ˆK = 90 ; radius tangent W ˆ = K ˆ ; tan/chord theorem Ŵ = 90 KR is a diameter; converse of angle in semi-circle (5) () ˆ KTL = 90 ; angle in semi circle KTL ˆ = W ˆ = 90 ; proven TL//WR ; corres angles equal KL KT = LR TW ; prop int thm (6) (c) ˆT =P+S; ˆ ˆ ext of S=R; ˆ ˆ angles in same segment P ˆ = R; ˆ alt 's PS // QR Pˆ Ŝ T ˆ = Ŝ (5) [8] IEB Copyright 05
NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page of 4 QUESTION 8 (a) () Statement Cˆ Aˆ Bˆ Statement  Bˆ Deduction C=B ˆ ˆ () (3) Statement Cˆ Aˆ Bˆ Statement Ĉ Dˆ Bˆ Deduction Aˆ Dˆ Statement Cˆ Aˆ Bˆ Statement  Pˆ Deduction Cˆ Pˆ Bˆ (3) (b) No/Not necessarily. We do not know whether GO = OE. (If "OG is a diameter" is stated, give a mark.) () IEB Copyright 05
NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page 3 of 4 (c) () B ˆ = P ˆ ; alt<'s PQ // BA ˆB = C ˆ 3; OB = OC C ˆ ˆ 3 = C ; vert. opp<'s ˆP ˆ = C PQ = QC; (6) IEB Copyright 05
NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II MARKING GUIDELINES Page 4 of 4 () The Cartesian plane is introduced in the diagram above so that A ;,B 6;3,C ;3 and P 4;3. 0+8 3 3 m OC = 0 4 y x 3 (i) O ; = O4; (ii) mid-point of PC = ; 3. x Q ; PQ = PC and QC is horizontal yq 3 3 7 Q ; (4) (4) [9] 75 marks Total: 50 marks IEB Copyright 05