Mathematics Grade 12 Paper 2 Eastern Cape June 2014 Memo

Downloaded from hlayiso.com Mathematics/P2 1 DBE/November 2011 NSC – Memorandum MATHEMATICS P2 JUNE 2014 – COMMON TEST MEMORANDUM NATIONAL SENIOR CERTIFICATE GRADE 12 MARKS: 125 This memorandum consists of 11 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 2 Common Test June 2014 NSC – Memorandum QUESTION 1 1.1.1 AC2 = (–1 – 4)2 + (–4 – 1)2 = 9 + 9 = 45  correct substitution into AC = 18 distance formula  18 = 3 2 (3) answer 1.1.2 1  2 4  2    1  M is  ;    ; 1 (2) answer  2 2   2  1.1.3 2 4 6  mAB  2 Gradient of AB =  2  2 1  3 1  m bisectorof AB   1  gradient of  bisector is 2 2 (obviously it passes through M)  1   ; 1  2  1 y xc 2  1  1 Substitute 1       c  substitution of m  2  2  and M into 1 equation of line c 1 4 3  4 1 3 Equation is y =  x  (4) 2 4 answer 1.2 x 2  2 x  1  y 2  2 y  1 2 x  2 y  x2 – 4x + 4 x2  4x  y 2  4 y   2  y2 + 4y + 4 x2  4x  4  y 2  4 y  4   2  4  4  (x – 2)2  (y + 2)2 x  22   y  22  6  answer: centre  centre is (2; –2) and radius = 6 (6)  answer: radius [15] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 3 Common Test June 2014 NSC – Memorandum QUESTION 2 2.1.1 Since JK // LM, MJK = MLM equating gradients  4 1 02  substitution in p2 53 each gradient  2  p  2    5 2  2 p  4   10 (4) simplification  2p  6 p3 2.1.2 JK2  3  2   4  1  50  correct substitution 2 2 into distance form. LM 2  2  0  3  5  8 2 2 (5)  50 JK : LM  50 : 8  5 2 : 2 2  8 5:2  ratio answer 2.1.3 Diagonals of a parallelogram bisect each other 3 1 3 1  ;  Midpoint of JL :  ;  2 2 2 2 x 3 3 This is the same midpoint for MQ:   2 2 x 3 3 y  2 1 y  2 1 Thus  ;      2 2 2 2 2 2 x  0  y  1 answer OR answer Therefore Q is Q(0; –1) (5) only(if done by inspection) – full marks 2.1.4 Since the x coordinates of K and M are both 3, it follows the equation of KM is x = 3. answer (2) o 2.1.5 90 since KM is a vertical line (1)  answer 2.1.6 For collinearity; mJR  mJL  equating: mJR  mJL k 1 0 1 k 1 mJR = =  mJR = 1  ( 2) 5  (2) 1  ( 2) k 1 1 simplification = = answer 3 7 3  k – 1 = 7 3  k = + 1 7 4 k = (4) 7 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 4 Common Test June 2014 NSC – Memorandum 2.2.1 Q (x; 2) … (radius QR  tangent)  yQ  2 Substitute (x; 2) in 3x + 4y + 7 = 0: 3x + 8 + 7 = 0 substitution: y = 2 x = –5  x = –5  Q (–5; 2) Radius = QR = 0 – (–5) = 5  radius  Equation is (x + 5)2 + (y – 2)2 = 25 (5)  equation 2.2.2 QR = 5 units d = 2 x radius  WZ = 10 units (1) [27] answer Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 5 Common Test June 2014 NSC – Memorandum QUESTION 3 3.1 sin 15o = sin (45o – 30o) 15° as 45 – 30 = sin 45o cos 30o – cos 45o sin 30o  correct 1 3 1 1 expansion = .  . 2 2 2 2  correct special 3 1 =  angle values 2 2 2 2 3 1 2 = x 2 2 2 6 2 = (3) 4 OR sin 15o = cos 75o = cos (45o + 30o) =cos 45o cos 30o – sin 45o cos 30o 75° as 45 + 30  1   3   1   1   correct       2   2   2   2  expansion =  correct special 3 1 angle values =  2 2 2 2 3 1 2 = x 2 2 2 6 2 = (3) 4 3.2   tan 180 o   . cos 360 o          sin 180 o   . cos 90 o    cos 540 o   . cos     For each reduction : tan  . cos    tan   cos  sin   .  sin    cos  . cos   sin  – sin  sin  – cos cos   cos  cos  sin  =  tan    sin 2   cos 2  cos  sin  =   sin   cos 2  2   sin 2   cos 2  = – sin θ (9) answer Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 6 Common Test June 2014 NSC – Memorandum 3.3    sin 45o  x . sin 45o  x   12 cos 2 x LHS  sin 45  x  . sin 45  x  o o  sin 45 cos x  cos 45 sin x  sin 45 cos x  sin x cos 45 )  expansion o o    1 1   1 1    . cos x  . sin x  .  cos x  sin x  special ∠  2 2   2 2  values 1 1  cos 2 x  sin 2 x 2 2  difference of   1 squares  cos 2 x  sin 2 x 1 2  HCF of 1 2  cos 2 x 2  RHS (5) 3.4 sin 33o cos 33o  sin 11o cos 11o sin 33o. cos 11o  cos 33o .sin 11o  sin 33o. cos 11o  cos 33o . sin 11o = sin 11o .cos 11o  sin 11o.cos 11o =  sin 33o  11o    sin 33o  11o  sin 11o cos 11o sin 22o  sin 22o = sin 11o cos 11o 2 sin 11o cos 11o  2 sin 11o cos 11o = o o sin 11 cos 11 = 2 (6)  answer 3.5 tan 3x 1 tan 24 o  tan 24° tan 3x  tan 24 o  tan 3x  tan 24o 3x  24 o  k . 180 o  3x  24o  k . 180o  x  8 o  k . 60  , k   (5)   x  8  k . 60 , k   o  [28] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 7 Common Test June 2014 NSC – Memorandum QUESTION 4 4.1 F y h g Q G f H x Constr: Draw FQ  HG produced  construction sin FĜH = sin(180° – FĜQ ) = sin FĜQ  reduction FQ  sin G FQ h   sin G  FQ = h sin G h FQ FQ Also  sin H   sin H g g  FQ = g sin H  h sin G = g sin H  h sin G = g sin H s in G s in H   g h (5) 4.2.1 PW  tan α  PQ  using tan α  PW  PQ tan α  answer (2) 4.2.2 QP̂R  180o  x  y  and  RP̂Q  180o  x  y   sin QP̂R  sin 180o   x  y    sin  x  y  In Δ PQR, by the sine rule PR QR   choice of sine formula sin Q̂ sin QP̂R PR 15  correct substitution into   sin y sin 180  x  y  o  sine formula 15 sin y  sin x  y   PR  sin  x  y  (4) Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 8 Common Test June 2014 NSC – Memorandum 4.2.3 PW = PR tan α from 3.3.1 15 sin y  15 sin y . tan   PW  . tan  PW sin  x  y  sin x  y  15 sin y  . tan  sin  y  y  15 sin y  . tan  sin 2 y  simplification 15 sin y  . tan  2 sin y. cos y  sin 2y = 2 sin y. cos y 7 ,5  . tan  cos y tan α  PW = 7,5 cos y (3) [14] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 9 Common Test June 2014 NSC – Memorandum QUESTION 5 5.1 Supplementary (1) Answer 5.2.1 It is the angle between tangent and radius.  (1) Answer 5.2.2 Ŝ3  Ŝ4  90o ... Tan  radius N̂1  N̂2  90o … Tan  radius  Statement Reason Ŝ3  Ŝ4  N̂1  N̂2  90o  90o  Statement = 180°  RNOS is a cyclic quadrilateral … opp  quad supplementary  Reason (4) 5.2.3 Ŝ1  x S1  N̂2  x … Tan chord theorem Statement Reason N̂2  S3  x … base ' s of isosceles ∆ OSN Statement Reason Ŝ3  R̂ 2  x … ' s in same segment Statement Reason N̂2    ’s in same segment.  R̂  x... 1 Statement Reason (8) 5.2.4 Ô1  Q̂2  N̂2  Ŝ3  180o … sum of ' s of Δ OSN Statement with reason But Ŝ3  N̂ 2 … s opp. equal sides Statement with  Ô1  Ô2  2Ŝ3  180o reason Ô1  Ô2  O3  180o … ' s on a straight line  Statement with reason  Ô1  Ô2  2 Ŝ3  Ô1  Ô2  Ô3 2 Ŝ3  Ô3  2 Ŝ3  Ô3 1  Ŝ3  Ô3 (4) 2 OR ext. s of cyclic quad.  Ô3  SR̂N .... Statement Reason R̂ 2  Ŝ3 … ∠s in the same segment  Statement with but SO = ON … radii of a circle reason  R̂1  R̂ 2 ...  chords ;  s  Statement with 1 reason  Ŝ3  Ô3 2 (4) [18] Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 10 Common Test June 2014 NSC – Memorandum QUESTION 6 6.1 Given: ∆ABC with D A on AB and E on AC such that DE//BC h1 h2 D E B C AD AE RTP:  DB EC Construction: Join DC and BE construction Proof: 1 areas of ∆’s with the same AD.h1 Statement Reason area Δ ADE 2 AD   … height and common vertex NOTE: If area of  area BDE 1 DB are in the same ratio as their DB.h 2 found, then reason 2 bases not necessary 1 AE.h 2 area Δ ADE 2 AE    Statement area DEC 1 EC EC.h 2 2 but area ∆ DBE = area ∆ DEC ... Same base, same parallel lines Statement Reason AD AE Thus  (6) DB EC 6.2.1 In ΔPQM, GH ∥ QT QH GP  = (line parallel to one side of a Δ OR Prop Th: GH ∥ PQ) Statement Reason HM GM 1 Answer =  (3) 2 6.2.2 QH = k ; HM = 2k  RM = 3k  RM = 3k MR = QM = 3k ...M is the midpoint of QR RG RH   = (line parallel to one side of a Δ OR Prop Th: GH ∥ PQ)) Statement Reason RT RQ 5k Substitution =  6k 5 =  (5) answer 6 Copyright reserved Please turn over

Downloaded from hlayiso.com Mathematics/P2 11 Common Test June 2014 NSC – Memorandum 6.3.1 Let Ẑ2  x  α …. Tan chord theorem  Statement Reason Then AB̂X = 90o – α … sum of ' s of ∆ ABP  Statement with reason But Ẑ1  AB̂P = 90 – α ... ' s opposite equal sides:AZ = AB  Statement with reason Ẑ1  Z2  α  90o  α … adj. ' s on a straight line  Statement with = 90o reason Thus Ẑ 3 = 90o (5) 6.3.2 In ∆ AYZ and ∆ AZX 1. Ẑ2  X̂ … Tan chord theorem Statement with reason 2. Â2  Â2 ….common  Statement 3. AŶZ  AẐX (remaining angles) Statement with  Δ AYZ /// Δ AZX , ,   (3) reason 6.3.3 AZ AX statement    s ∥∣ sides in proportion  AY AZ  AZ2  AY . AX (1) [23] TOTAL MARKS: [125] Copyright reserved Please turn over