Mathematics Grade 12 IEB Paper 2 2016 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 2016 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 © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 2 of 11 SECTION A QUESTION 1 (a) Opp angles do not add up to 180°. (b) tan 45° =m m =1 y= x + 8 (c) (1) x=6 1 (2) B(6 ; 14) ∴ Area = (8 + 14)(6) =66 units 2 2 QUESTION 2 2sin 2 θ + 2sin θ cos θ (a) (1) M= cos 2 θ − sin 2 θ 2sin θ ( sin θ + cos θ ) M= ( cosθ + sin θ )( cosθ − sin θ ) 2sin θ M= ( cosθ − sin θ ) Therefore M=P (2) cos θ − sin θ 0 = cos θ = sin θ 1 = tan θ Reference angle: 45° θ =− { 135°;45°;225°} IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 3 of 11 (b) (1) (2) Main y 2 tan β= = x −3 Alternate: ( 11 ) ( 2) + x 2 2 = 2 x2 = 9 x = ±3 Since sin β > 0, and cos β < 0 is in the second quadrant Quadrant Two x = −3 (This mark is for the accuracy of the sign.) y= 2 r = 11 = tan β 2 = − 3 (c) (1) cos(α − 30°) − cos(α + 30°) = cos α cos30° + sin α sin 30° − (cos α cos30° − sin α sin 30°) = cos α cos30° + sin α sin 30° − cos α cos30° + sin α sin 30° 1 = 2sin α sin 30 = ° 2sin α ×   2 = sin α (2) sin α = 2sin 2 α = 0 sin α (2sin α − 1) 1 sin α = 0 OR sin α = 2 α = 0° + k180° α= 30° + k 360° α = 150° + k 360° k ∈ OR (α = 0° + k 360° or α = 180° + k .360° ) IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 4 of 11 QUESTION 3 (a) Radius of circle Q is 9 − 5 = 4 units. (If on diagram then they get the mark) x Q of the centre of the circle is 9 + 5 = 14 units. y Q of the centre of the circle is 5 units. therefore the equation of the circle is ( x − 14 ) + ( y − 5 ) = 2 2 16 ( x − p ) + y 2 − 22 y + 121 = −117 + 121 2 (b) ( x − p ) + ( y − 11) = 2 2 4 Therefore RQ is 6 units. (Note: If they use 2 + 4 they can get a max of 2 out of 3) (c) PR = (11 − 5) 2 + (14 − 5) 2 PR = 117 ∴ AB = 117 − 2 − 5 = 3,82 units (Full marks for the correct answer) QUESTION 4 (a) Draw AO and OC ˆ +D R.T.P: B ˆ = 180° Proof: Oˆ = 2× Bˆ (Angle at centre) 2 Oˆ = 2× D ˆ (Angle at centre) 1 Oˆ +Oˆ = 360° 1 2 ∴ 2B ˆ + 2Dˆ = 360° ∴Bˆ +Dˆ = 180° (b) Main: ABCˆ = 62° (tan chord theorem) AOCˆ= 124° (Angle at centre is twice the angle at the circumference) ˆ= A C ˆ = 28° (isos∆ OR OC = OA) 2 3 ∴ Ĉ1 = 37° (∆ 's in a ∆) OR Alternate: Â 3= 90° − 62° (radius ⊥ tangent) = 28° C= ˆ 2 ˆ = 28° (isos∆ OR OC = OA) A 3 ABC= 62° ˆ (tan chord theorem) ∴ Ĉ1 = 37° (∆ 's in a ∆) IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 5 of 11 (c) (1) N = Q OR M+N = M + Q (2) Dˆ =B ˆ (exterior angle of a cyclic quad) 1 D̂= 1 Aˆ +C 1 ˆ (exterior angle of triangle = sum of the two int opp angles) 2 ∴ B̂ = A ˆ +ˆ 1 C 2 QUESTION 5 (a)=y 3sin 360° + 1 y =1 A ( 360°;1) (b) 3sin x + 1 =−1 3sin x = −2 −2 sin x = 3 Reference Angle: 41,81° x = {221,81°; 318,19°} (c) k > 4 OR k < 1 QUESTION 6 (a) r = 0.9755 Very strong (b) A = 2 788, 26 B = 1 658,39 Line of best fit. =y 1 658,39 x + 2 788, 26 (c) Sub in 19 for x. = y 1 658,39(19) + 2 788, 26 y = R34 297,67 His projected income based on his line of best fit is R34 297,67. The manager would not consider this a successful day. OR NO; Link to the table using the number 17; logical argument IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 6 of 11 SECTION B QUESTION 7 (a) A = 250 B = 502 200(10) + 250(20) + 20(30) + 32(40) + 23(50) + 300(60) + 475(70) (b) (1) x≈ 1 300 x ≈ 47,14 (If they get the right answer and no working is shown, then full marks.) (2) 65 < x ≤ 75 (c) (d) (1) No. Skewed to the left as the mean is less than the median. OR Bimodel, big dip in the middle. OR Shape of Ogive is not correct. (2) No. It is not a good indicator as the majority of the people who use your product are between 65 and 75. OR Yes. It is a good indicator as the people in this age range will be looking to buy the product for their children between the ages of 5 and 25. (Many answers to be considered.) IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 7 of 11 QUESTION 8 (a) TR =3 (TP ⊥ OP or 𝑂𝑂𝑂𝑂 ⊥ 𝑅𝑅𝑅𝑅 OR drawn on diagram) OP=2 52 − 32 OR = 4 OP = OR (tangents drawn from same point) ∴ xT = 4 T(4;3) (Workings can be shown on diagram) (b) ˆ =3 tan TOR 4 =ˆ TOR 36,87° (c) ˆ = 2 × 36,87° = 73,74° (properties of kite OPTR) POR y POR = P sin ˆ 4 yP = 3,84 units QUESTION 9 (a) OC= 2 80 + 20 OC = 10 pythagoras (b) Main: ˆ = 20 tan BCO 80 = 26,57° mAC = − tan 26,57 mAC = − 0,5 OR Alternate: AO Gradient of line AC =− 1× OC (∆ABO///ΔOBC) AO BO 20 = = OC BC 80 20 1 Gradient of AC =−1× =− 80 2 OR Alternate: 20 1 = ˆ tan OCB = 80 2 ∴m = AC tan(180° − OCB) ˆ = − tan BCO ˆ 1 = − 2 IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 8 of 11 1 (c) Gradient of OB − = = 2 mAC (2k ) 2 + k 2 =20 (coordinates of point B) 5k = 20 2 B ( 2; 4) k =2 OR Alternate tan OCB ˆ =1 2 y 2 + 4 y 2 80 = y=4 x= B 10 − 2(4) xB = 2 B (2;4) (d) ˆ =θ Let COB ˆ = 90° − θ AOB ˆ = 90° − (90° − θ )= θ ∴ OAB ∴ OAB ˆ = ˆ COB ∴∆ABO / / / ∆OBC (AAA) AB BO ∴ = OB BC OB2 ∴ AB = BC IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 9 of 11 QUESTION 10 (a) = Let = AB 4k and BC 7k FE AB 4 ∴ = = ; (Proportionality theorem OR using theorem on diagram) (3) FC AC 11 (b) = Let = AG 9m and AF 17 m CD AG 9 = = (2) DF GF 8 9 4 (c) If FC =p then ED =− p p− p 17 11 20 ED = p 187 20 The length of ED in kilometres is × 374 km 40 = kilometres. 187 It will take 2 000 hours to build the track from E to D. OR Alternate: =Let = FE 4 p and EC 7 p = = FD 8m and DC 9m ∴11 p = 374 ∴ p = 34 17 m = 374 ∴ m = 22 ∴ DC = 374 − 4 p − 9m = 40 km ∴ 2 000 hours OR Alternate: 4 =FE = (374) 136 11 9 =CD = (374) 198 17 ∴ ED = 374 − 136 − 198 = 40 km ∴ 4 hours → 40 × 50 = 2 000 hours IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 10 of 11 QUESTION 11 (a) ˆ =D C ˆ (angles in same segment) 2 ˆ +C C ˆ =90° (angle in semi-circle) 1 2 ∴C ˆ +D 1 ˆ = 90° (b) Construction: Chord BF ˆ = 90° − C C ˆ = 52° 1 2 AFB = C (tan chord theorem) ˆ ˆ 1 ˆ = 52° (tan chord theorem) ABF ∴ BAF ˆ = 76° (angles of a ∆) IEB Copyright © 2016 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: MATHEMATICS: PAPER II – MARKING GUIDELINES Page 11 of 11 QUESTION 12 1 (a) Area of DADC = × 6 × 6 × sin130° 2 = 13,8 (b) ˆ = 50° (opp ∆'s cyclic quad) ABC ABD = DBC ˆ ˆ (Equal chords; subtend equal angles) ∴ DBC ˆ = 25° (c) BC = 12 (line from centre ⊥ chord) ˆ sin BDC sin 25° = (sin rule) 12 6 ∴ sin ˆ BDC = 0,845... BD � C = 57,7o ∴ BDCˆ =180 − 57,7° = 122,3° ∴θ= 180° − 25° − 122,3 (angles of ∆) = θ 32,7° (d) AC2 = 62 + 62 − 2(6)(6) cos130° AC2 = 118, 28... ∴ AC = 10,875... 9 ∴ tan TCA ˆ = 10,875... ∴ TCA ˆ =39,6° Total: 150 marks IEB Copyright © 2016