Mathematics Grade 12 Paper 2 Gauteng Mock 2014 Memo

Downloaded from hlayiso.com 1 10612/14 GAUTENG DEPARTMENT OF EDUCATION PREPARATORY EXAMINATION PROOFREAD DATE ___________ MATHEMATICS SIGNATURE _________________ (Second Paper) PROOFREAD DATE ___________ SIGNATURE _________________ MEMORANDUM NOTE:  If a candidate answered a question TWICE, mark the FIRST attempt ONLY.  Consistent accuracy applies in ALL aspects of the memorandum.  S/R refers to STATEMENT as well as REASON QUESTION 1 1.1.  Negatively skewed negatively skewed /  Negatief skeef negatief skeef (1) 1.2 1.3 1.2 shape grounding at 10 accuracy of points Subtract a mark if grounded at (0; 0) and/or at the end (4) 1.3  Q2 shown on the graph, linked to cumulative frequency (1) [6] 1

Downloaded from hlayiso.com 1 10612/14 QUESTION 2 2.1 points plotted correctly -1 for every 2 mistakes -1 if points are joined (3) 2.2 y  26,88 x  161,24 value of A = 161,24 value of B = 26,88 equation of the line (3) 2.3 y  26,88(24)  161,24 substitute in formula y  806,36 answer as Natural Number th 806 people treated on 24 of June (2) answer only CANNOT USE GRAPH – NO LINE OF BEST FIT IS(using calculator) yˆ  806,42 DRAWN YET full marks 2.4 See grid straight line graph passes through ( x ; y) graph passes through y-intercept from regression line formula (3) 2.5 r  0,98 correct answer This is a very strong positive, linear correlation. interpretation As the days increase the number of patients increase (3) Hierdie is ’n baie sterk positiewe lineêre korrelasie. Soos wat die dae vermeerder, vermeerder die getal pasiënte 2.6 This point is clearly an outlier/Die punt is ’n duidelike outlier/ uitskieter uitskieter justification The 16th of June is a public holiday and people might (2) have decided to wait for the next day or certain medical [16] facilities might have been closed. Die 16de Junie is ’n publieke vakansiedag en mense het dalk besluit om te wag tot die volgende dagof sekere mediese fasiliteite kon dalk gesluit gewees het. 2

Downloaded from hlayiso.com 1 10612/14 QUESTION 3 3.1 QR  x  x   y  y  Q R 2 Q R 2  1  7  2  3  0 2 substitution in correct  45 formula  3 5  a 3  answer for QR a  3 OR (3) QR  x  x   y  y  Q R 2 Q R 2 a 5  1  7  2  3  0 2 substitution in correct a 5  45 formula 5a 2  45 a2  9  5a 2  45 Squaring both a  3 ; a0 side OR a 3 ; a 0 (3) QR 2  x  x   y  y  Q R 2 Q R 2 5a 2  1  7  2  3  0 2 5a 2  45 substitution in correct a 2  9 formula a  3 ; a0  5a 2  45 Squaring both sides a 3 ; a 0 (3) 3

Downloaded from hlayiso.com 1 10612/14 3.2 3x  t y  2 3 3  t y  2 ty  7........1  3 3  t y  2 2 y  x2 3  ty  7 y    3  2 2 3 y  4.........2  y4 ty  7 t 4   7 7 value of t t 4 (4) OR 2 y  x2 3 3x  t y  2 3 2 3 2  y   x y x t t t t 3 2 2 equations equal  x  x2 t t 3 3( 3) 2 2( 3) substitute x = –3    2 t t 3 7 value of t t (4) 4 3.3.1 tan   m EB 30 1 tan    gradient of EB =   5 1 2 1 tan    1  tan    2 2 OR   26,565...  180   180   26,565    153,43   153,43   153,43 (3) 4

Downloaded from hlayiso.com 1 10612/14 3.3.2 Equation of EF: 93 y  y1  m( x  x1 ) mBC  mEF  3  mBC  3 35 y  0  3( x  1)  mBC  mEF  3 y  3x  3 substitute point (1 ; 0) equation of line (4) 3.3.3 tan F Eˆ A  3 (From line in 1.3.2)  FÊA  71,57  FÊA  71,57  BÊF  153,43  71,57 ' s on a straight line/ ' e op rt lyn BÊF  81,86  BÊF  81,86 EB̂C  180   81,86  EB̂C  98,14  (co-interior angles /ko-binnehoeke; EF  BC)  EB̂C  98,14  (3) [17] QUESTION 4 4.1.1 Q̂  90 ( in semi-circle/ in halfsirkel)  Q̂  90 mPQ.mQR  1 10  4 10  2   1 9a 93 subs into mPQ.mQR  1 6 8   1 9  a 12 4  1 9a simplification 4  9  a a  13 (3) 5

Downloaded from hlayiso.com 1 10612/14 4.1.2 midpoint of PR   3  13 4  2  C ;   2 2  C 5 ; 3  C 5 ; 3 r 2  (13  5) 2  (4  3) 2  substitute in distance formula r= (13  5) 2  (4  3) 2 or r 2  65  r  65 / r 2  65  65 Eq of circle: ( x  a ) 2  ( y  b) 2  r 2 ( x  5) 2  ( y  3) 2  65  ( x  5) 2  ( y  3) 2  65 (4) 4.2 rA  3 and A(2 ; 3) rB  4 and B(1 ; -1) radius and centre of circle A radius and centre of circle B  rA  rB  7 sum of two radii Distance between A and B AB  ( x B  x A ) 2  ( y B  y A ) 2 calculating distance between centres AB  (1  2) 2  (1  3) 2  1 16  17 AB = 17  rA  rB  17 correct conclusion The circles intersect each other (6) 4.3 149 x 2  4x  y 2  2 y  4 149 4.3.1 ( x  2) 2  4  ( y  1) 2  1  completing the square 4 169 ( x  2) 2  ( y  1) 2  standard form 4 N(2 ;  1)  N(2 ;  1) (3) 6

Downloaded from hlayiso.com 1 10612/14 4.3.2 169 r2  4 13 1 r OR r 6 OR r  6,5 answer of radius 2 2 (1) 4.3.3 NC  AB  AC  CB  6 (line from centre  chord/ lyn van midpt van sirkel  op koord)  AC  CB  6 x A  2  6  4 Substitute x  4 in  x A  4 149 (4) 2  4(4)  y 2  2 y  substitute x  4 4 149 32  y 2  2 y  4 128  4 y  8 y  149 2 4 y 2  8 y  21  0 (2 y  3)(2 y  7)  0 3 7 y or y 2 2 invalid y  0  7  7 coordinates of A A   4 ;   and B  8 ;   coordinates of B  2  2 (5) [22] QUESTION 5 5.1.1 b  tan 110  tan 110 a (1) 5.1.2 b  sin 110  sin 110 a 2  b2 answer  0,94 (2) 5.2.1 cos 146  cos  90  56    sin 56   sin 56  answer q OR (2) cos 180  34    cos 34   sin 56   sin 56  q  answer (2) 7

Downloaded from hlayiso.com 1 10612/14 5.2.2  x  1  q2 y x r  y 2 2 2 1  12  q 2 q 56o  x  1  q2 0 x sin 112  2sin 56 cos 56  2sin 56 cos 56 answer (3)  2q 1  q 2 OR  cos 56   1  q 2 cos 56   sin 56   1 2 2 cos 56   1  sin 2 56   2sin 56 cos 56  1 q2 answer sin 112 (3)  2sin 56 cos 56 [8]  2q 1  q 2 QUESTION 6 6.1 cos  P  Q   cos P cos Q  sin P sin Q  correct formula (1) 6.2 cos 2 P  cos P cos P  sin P sin P  cos P cos P  sin P sin P  cos P  sin P 2 2 replacing  cos 2 P  1  cos 2 P  sin 2 P with 1  cos 2 P  2 cos 2 P  1  2 cos 2 P  1 (3) 6.3.1 sin 2 x  cos 2 x  1 LHS  sin 2 x  cos 2 x  1 2 sin x cos (1  2 sin 2 x)  1   2sin x cos x 2 sin x cos (2 cos 2 x  1)  1  1 2 sin 2 x 2 sin 2 x  2 sin x cos x  2 cos 2 x  1  2 cos 2 x  2 sin x cos x 2 sin x(sin x  cos x) factors  2 cos x(cos x  sin x) sin x sin x   cos x cos x  tan x (5)  RHS 8

Downloaded from hlayiso.com 1 10612/14 OR sin 2 x  cos 2 x  1 LHS   2sin x cos x sin 2 x  cos 2 x  1  cos 2 x  sin 2 x 2sin x cos x   cos 2 x  sin 2 x   1  2sin x cos x   cos 2 x  sin 2 x   1 2sin x cos x  cos 2 x  sin 2 x  1  simplification 2sin x cos x  cos 2 x  sin 2 x  1 2sin x cos x  2 sin 2 x factors  2sin x cos x  2 cos 2 x sin x 2sin x  cos x  sin x    cos x 2 cos x  sin x  cos x  sin x (5)  cos x  tan x  RHS 6.3.2 sin 45  cos 45  1 tan 22,5  sin 45  cos 45 1  tan 22,5  sin 45  cos 45  1 sin 45  cos 45 1 2 2  1 2  2 2 substitution of 2 2 2  1 2 2 1  2 2 1 2 1  2 1 1 2 1   answer 2 1 2 1 (3)  2 1 OR 9

Downloaded from hlayiso.com 1 10612/14 tan 22,5  sin 45  cos 45  1  tan 22,5  sin 45  cos 45  1 sin 45  cos 45 1 sin 45  cos 45 1 1 1  1  2 2 1 1 1  1 substitution of 2 2 2 1  11 1 2 1  2 1 2 1  2 2  1 2 2 1  2 1 1 2 1   answer 2 1 2 1 (3)  2 1 6.4.1 sin 2 x. cos 2 x 1 1  .2 sin 2 x. cos 2 x  .2 sin 2 x. cos 2 x 2 2 1  sin 4 x 2 1  max of sin 4 x  1  .1 2 1 1   2 2 (3) 6.4.2 sin 4 x  1  sin 4 x  1 4 x  90  k .360  k    4 x  90  k.360  k   x  22,5  k .90 k    x  22,5  k.90  x  67,5 x  67,5 (4) [19] 10

Downloaded from hlayiso.com 1 10612/14 QUESTION 7 7.1 f: shape intercepts  turning points g: shape intercepts  turning points (6) 7.2 180  answer (1) 7.3 1 h  x    sin  x  90  correct equation 2 OR (2) 1 h( x)   sin    90  x   1 2   sin   90  x  1 2     sin  90  x   2 1 1  h  x   cos x  cos x (2) 2 2 7.4 cos 2 x  1 2 x  0  k .360 , k    2 x  0  k.360 , k    x  0  k .180  x  0  k .180 , k   (2) OR cos 2 x  1  2 x  k.360 , k   2 x  k .360 , k    x  k .180  x  k .180 , k   (2) [11] QUESTION 8 8.1 In MPQ : PM d  correct substitution into sine  sin z sin 180   y  z   rule d sin z  PM   PM  d sin z ......1 sin  y  z  sin  y  z  In MPN : MN MN  sin x   sin x PM PM MN  PM sin x .....2 substituting for PM into d sin z. sin x equation 2 and final answer  MN  (4) sin  y  z  11

Downloaded from hlayiso.com 1 10612/14 8.2 1 000sin y sin  90  y   correct replacement of angles MN  sin 2 y  2sin y cos y 1 000 sin y cos y  2 sin y cos y Answer  500 m (3) [7] QUESTION 9 9.1 …perpendicular / …loodreg  perpendicular /loodreg (1) 9.2.1 AB  CD (line from centre to midpoint of chord) / S/R (lyn van die midpt van sirkel na midpt van koord) [S/R Statement DB  9 cm (AD  DB / Given) and reason] DB = 9 cm In DCB (CD) 2  (24 ) 2  (9) 2 (pythagoras ) using Theorem of CD  576  81 2 Pythagoras(S/R)  495  CD  3 55 CD  3 55 (4) 9.2.2 CD 3 DE 3 55  3 3 55 DE 3 DE DE  55  DE  55 (2) [7] QUESTION 10    E1  90º 10.1 E1  90º (  in a semi  circle /in halwe sirkel )  in a semi  circle (2) 10.2 Ê1  90 (  in a semi  circle /in halwe sirkel )  Ê  Â2  180  Ê  Â 2  180  (opp ' s cyclic quad / oorst ' e van kvh )  opp ' s cyclic quad  Â 2  180   (90  28)   62  A 2  62 º (3) OR 12

Downloaded from hlayiso.com 1 10612/14   F A C  90 (  in a semi  circle /in halwe sirkel )  F A C  90 º S/R   A3  28º ( s in the same segment subt by DC /  A3  28º S/R e in dieselfde segment ondersp deur DC)   A2  90 º 28º   A 2  62 º  62 º (3) 10.3   F 2  30 º (equal chords subtend equal ' s) /  F 2  30 º ( gelyke koorde onderspan gelyke hoeke)  equal chordssubtend eq ' s (2) [7] QUESTION 11 11.1 1 2 2 1 Construction: Draw a diameter/ middellyn AOE and join CE construction Proof: Â1  Â 2  90 ( tan/rk lyn  radius )  Â1  Â 2  90 Ĉ1  Ĉ2  90 º ( in a semi circle / in halwe sirkel )  tang  radius but Ĉ1  Â 2 (s in the same segment)  Ĉ1  Ĉ2  90º (S/R)    A1  C2  Ĉ1  Â2 (S/R) (5) 11.2.1   C1  90º ( in a semi circle /in halwe sirkel )  C1  90º (S/R)   D1  90º (OE  BC)  D1  90 º (S/R)     C1  D1  C1  D1 (S)  OE || AC (corr s are equal / ooreenk e is gelyk )  corr  s are equal (4) OR  C1  90º ( in a semi circle / in halwe sirkel )    C1  90º (S/R) D 2  90º (OE  BC)    D 21  90 º (S/R) C1  D̂ 2  180    C1  D̂ 2  180  (S)  OE || AC (co interior s are equal / ko  binnee is gelyk)  co interior  s are equal (4) 13

Downloaded from hlayiso.com 1 10612/14 11.2.2   Ax ( tan chord / rk lyn koord ) A x  tan chord   O1  x (corr  s /ooreenk e OE || AC )  O1  x S/R (3) 11.2.3 In PAC   (sum of s of a triangle/ P  180 º  x  90 º  x  P  90 º 2 x som van e van driehoeke)  sum of  s of a triangle  90 º 2 x (2) [14] QUESTION 12 12.1 proportionally  proportionally (1) 12.2.1   C1  C 2 (given)     C1  D (corr s /ooreenk e AD || CG)  C1  D (S/R)     C 2  A1 (alt s /verwisselende e AD || CG)  C2  A1 (S/R)      A1  D  A1  D  CA  CD (sides opp equal s / sye teenoor gelykee )  CA  CD (S/R) (4) 12.2.2 BC BG BC BG  (line || to one side of Δ / lyn || een sy van Δ )   CD GA CD GA but CA  CD (proved)  BC BG   AC AG replace CD  CA (3) 12.3.1 In ΔKMT and ΔKON   for any two of the M1  O (ext. of a cyclic quad / buite  kvh) three statements   T2  N (ext. of a cyclic quad/ buite  kvh)   KK (common/ gemeensk)  ΔMTK ||| ΔONK ( ) equiangular / gelykhoekig /  (3) 12.3.2 MT TK MK MT TK MK      ON NK OK ON NK OK  KM.KN  KT.KO (1) 14

Downloaded from hlayiso.com 1 10612/14 12.3.2 MK KT  OK NK 5 KT 5  KT   6  KT 8 6  KT 8 KT  6KT  40 2 KT 2  6KT  40  0  KT 2  6KT  40  0 (KT  10)(KT  4)  0 factors KT  10 or KT  4  KT  4 units answer (4) [16] TOTAL: 150 15