Mathematics Grade 12 Paper 2 Mpumalanga Mock 2019 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS PAPER 2 September 2019 MARKING GUIDELINES MARKS: 150 This marking guidelines consist of 23 pages

Downloaded from hlayiso.com Mathematics/P2 memo 2 September 2019 Question 1 1.1 Interval Frequency Cummulative frequency  frequency 60  x  70 5 5  accumulative 70  x  80 11 16 frequency 80  x  90 22 38 (3) 90  x  100 13 51 100  x  110 7 58 110  x  120 3 61 1.2 angle point use of OGIVE 70 cumulative frequency 60  upper limits  free hand curve 50 Cummulative frequency 40 (4) (in days) 30 20 10 0 40 50 60 70 80 90 100 110 120 130 Daily sales (in Rand) 1.3 61 51  10  answer (2) 1.4 Medium = 87 (between 85 and 88)  answer (2) [11] Copyright reserved Mpumalanga Please turn over

Downloaded from hlayiso.com Mathematics/P2 memo 3 September 2019 QUESTION 2 2.1.1 By using a calculator : a = 76,60 (76,59564211)  a b = 0,1(0.09544928654)  b ∴ equation of line of least squares is y  76,60  0,10x  equation (3) 2.1.2 y  76,60  0,10(57 ) substitution  82 answer (2) 2.1.3 r = 0,1408374211  r  0,14 r  0,14 (1) 2.1.4 r value is low which means that the data is spread around the invalid regression line. Therefore prediction using the regression line will be invalid. (1) 2.2.1 22  22 (1) 2.2.2 1320 1320  60  22 22  60 (2) [10] Copyright reserved Mpumalanga Please turn over

Downloaded from hlayiso.com Mathematics/P2 memo 4 September 2019 QUESTION 3 y C(6 ; 4) B(0 ; k) E F x O A(1 ; –1) ) 3.1 AB : y  3x  2  0  y  3x  2  y  3x  2 k 2 k  2 (2) 3.2 AC  (1  6) 2  (1  4) 2  substitution  50 answer 5 2 (2) 3.3 1 2 m AB   3 1 0  m AB  3 42 1 m BC   1 60 3  m BC  3 1 m AB  m BC  3   1  m AB  mBC  1 3 (3)  AB̂C  90. 3.4 tan   m AC  1  tan   m AC  1   45  1 tan   m BC  3   18,43    45  18,43 [ext of ] method  26,57 answer (5) OR Copyright reserved Mpumalanga Please turn over

Downloaded from hlayiso.com Mathematics/P2 memo 5 September 2019 AC  (1  6) 2  (1  4) 2  50 5 2  AC = 5 2 AB  (0  1) 2  (2  1) 2  10  AB = 10 AB sin   AB AC  sin   AC 10  5 2 10  sin   5 5 2  2    26.57   answer OR (5) sin  sin B sin  sin B    AB AC AB AC 10 sin 90 sin    AC = 5 2 5 2 5  AB = 10  5 10 sin 90    26,57   sin   5 2  answer (5) Copyright reserved Mpumalanga Please turn over

Downloaded from hlayiso.com Mathematics/P2 memo 6 September 2019 3.5 BD̂C  90 altitude  BC is the diameter (line subtends 90 )  BC is diameter 06 2 4 midpoint of BC :  ;   2 2   3 ; 3  3 ; 3 x  32  ( y  3) 2  r 2 subst B0 ; 2  0  32  (2  3) 2  r 2  Substitution of (0 ; 2)  r  10 2 x  32  ( y  3) 2  10  equation OR (4) BD̂C  90 altitude  BC is diameter  BC is the diameter (line subtends 90 ) 0 6 2  4 midpoint of BC :  ;   2 2   3 ; 3  3 ; 3 x  32  ( y  3) 2  r 2 subst B0 ; 2   Substitution of (0 ; 2) 0  32  (2  3) 2  r 2  r 2  10  equation x  32  ( y  3) 2  10 (4) 3.6 1  6 2 1 7 3 M AC   ;     2 2  2 2 7 3 x7  ;  2 2  y 1 7 x0 3 2 y   2 2 2 2 (4) x7 y 1 D(7 ; 1) [20] Copyright reserved Mpumalanga Please turn over

Downloaded from hlayiso.com Mathematics/P2 memo 7 September 2019 QUESTION 4 S y P R O T x 4.1 T(6; 0)  (6; 0) (1) 4.2 radius  3 radius = 3 (1) 4.3 3 3 tan 30    tan 30   3 3 3 substitution 0 ( 3)  c 3 c  3 c 3 equation 3 (4) y x 3 3 4.4 3 gradient y xc radius  tangent 3 18 c  3 3 0 (6)  c 3 equation 18 c (3) 3 3 18 y x 3 3 Copyright reserved Mpumalanga Please turn over

Downloaded from hlayiso.com Mathematics/P2 memo 8 September 2019 4.5 3 3 18 equating x 3  x 3 3 3 simplification 9 x  54 3x  9 x-value  3 3 3 3 substitution 12 x  45 y-value 15 3 x 3 (5) 4 4 3  15  y   3 3 4 9 3 y      4   15 9 3  S ;    4 4  4.6 The distance between centres of circles: OT = 6  OT = 6 r  R  1 3  r  R  1 3 4 4 The distance between the two circles = OT –(r + R) = 6 –4 = 2 units 2 (4) [18] Copyright reserved Mpumalanga Please turn over

Downloaded from hlayiso.com Mathematics/P2 memo 9 September 2019 QUESTION 5 5.1 sin(A  180). tan(180  A).cos A   tan A cos(90  A)   sin A sin((180  A)).( tan A). cos A   sin A   sin A  identity  sin A.(  tan A). cos A answer   sin A sin A  . cos A cos A (5)   sin A 5.2.1 1 28 a  sketch 1 a2 cos 28  1  a 2 answer (2) OR cos2 28  sin 2 28  1  identity cos2 28  a 2  1 cos2 28  1  a 2 answer cos 28  1  a 2 (2) 5.2.2 1 1  b2 32  32 b cos 64 double angle  cos(2  32 ) answer (2)  2 cos 32  1 2   2b 2  1 OR Copyright reserved Mpumalanga Please turn over

Downloaded from hlayiso.com Mathematics/P2 memo 10 September 2019 cos 64  cos(2  32 )  1  2 sin 2 32 double angle   1  2 1  b2  2  1  2  2b 2  2b 2  1 answer (2) OR cos(2  32 )  cos2 32  sin 2 32  b2   1 b  2 2 double angle  b2 1 b2  2b 2  1 answer (2) 5.2.3 sin 4  sin( 32   28  ) compound   sin 32 . cos 28   cos 32 . sin 28   expansion   1  b 2  1  a 2   b.a  1  b2     1  b 1  a   ab 2 2  1 a2 (4) 5.3 1 1 2 tan x   1  sin x 1  sin x cos x LCD 1 1 1  sin x  (1  sin x) LHS :   numerator 1  sin x 1  sin x (1  sin x)(1  sin x) 2 sin x 1  sin x  1  sin x   cos 2 x 1  sin 2 x 2 sin x  cos2 x sin x 2 sin x 2 tan x  RHS:  cos x cos x cos x cos x 2 sin x 2 sin x   cos 2 x cos 2 x  LHS  RHS (5) Copyright reserved Mpumalanga Please turn over

Downloaded from hlayiso.com Mathematics/P2 memo 11 September 2019 OR 1 1 2 tan x   1  sin x 1  sin x cos x LCD 1 1 1  sin x  (1  sin x) LHS :   numerator 1  sin x 1  sin x (1  sin x)(1  sin x) simplification 1  sin x  1  sin x  2 sin x 1  sin 2 x  cos 2 x 2 sin x  sin x cos 2 x   tan x cos x 2 sin x  (5) cos x. cos x 2 tan x  cos x  LHS  RHS [18] Copyright reserved Mpumalanga Please turn over

Downloaded from hlayiso.com Mathematics/P2 memo 12 September 2019 QUESTION 6 [16] 6.1 1 sin( x  60 )  cos x 2 co-function  1  sin( x  60 )  sin 90  x   2  quadrant 1 :  1 1st quad equation x  60  90  x  360.k , k  Z 2 1 x  x  30  360.k 2 3 x  30  360.k 2  x  20  240.k x  20  240.k , k  Z quadrant2 : 1 x  60  180  (90  x )  360.k , k  Z 2nd quad equation 2 1 x  60  180  90  x  360.k 2 1 x  30  360.k 2 x  60  720.k , k  Z  x  60  720.k x  {20 ; 60; 260 }  x  {20 ; 60; 260 } OR (7) Copyright reserved Mpumalanga Please turn over

Downloaded from hlayiso.com Mathematics/P2 memo 13 September 2019 1 sin( x  60 )  cos x 2 1 cos[90  ( x  60 )]  cos x 2 quadrant 1 co-function 1 [90  ( x  60 )]  x  360.k , k  Z 2 1 90  x  60  x  360.k 1st quad equation 2 3  x  30  360.k 2 x  20  240.k , k  Z  x  20  240.k quadrant4 1 [90  ( x  60 )]  360  x  360.k , k  Z 2 1 90  x  60  360  x  360k 2 4th quad equation  1 30  x  360  x  360.k 2 1  x  330  360.k 2  x  660  720.k x  660  720.k  x  {20 ; 60; 260 } (7) x  {20 ; 60; 260 } Copyright reserved Mpumalanga Please turn over

Downloaded from hlayiso.com Mathematics/P2 memo 14 September 2019 6.2.1 g:  x  intercepts y  y  intercept 1 turning points 0.87 (3) –60° 0° 60° 120° 180° 240° 300° x 300° 6.2.2 20   x  60  or 260  x  300   20   x  60 OR  260  x  300  x  (20 ; 60 )  x  (600  ; 300  ] notation (3) 6.2.3 120   x  18 0   120   x  18 0  x   60  x   60 ;300° x  300 (3) [16] Copyright reserved Mpumalanga Please turn over

Downloaded from hlayiso.com Mathematics/P2 memo 15 September 2019 QUESTION 7 R h T C D 7.1 CR = RD (congruency OR SAS)  CR = RD CR̂D  180  2 (s in ) answer (2) 7.2 TR̂D  90    h ratio cos(90   )  RD h  RD  h sin RD  sin substitution 1  2 cos   1 2  h   h  2  h  h  2  2h2 2 CD       2   cos(180  2 ) 2 sin   sin    sin    sin   sin   2  h  2  4h 2 sin   cos 2    2 2  cos 2   2h 2  sin  2  sin   (5) 2  2h 1  cos 2  sin 2   2h 2 sin 2   1  2 cos 2   1   2h 2 sin  2  2 cos 2    4h 2 sin 2   cos 2   2h. cos   CD  sin  7.3 2h. cos  CD  . sin  substitution Copyright reserved Mpumalanga Please turn over

Downloaded from hlayiso.com Mathematics/P2 memo 16 September 2019 2h. cos 65  5,4  . sin 51  5,42 . sin 51  h . h to the nearest unit 2 cos 65  (2) h  4,96 h5 [9] Copyright reserved Mpumalanga Please turn over

Downloaded from hlayiso.com Mathematics/P2 memo 17 September 2019 QUESTION 8 8.1 A 1 2 1 1 E 2 D 1 B 2 2 3 2 O 1 C 8.1 D̂1  33 (  ' s in same circle segment) S R (2) 8.2 In Δ DAE : Ê1  90 (line from centre to midpt of chord) S R D̂1  Ĉ  33 (s in same circle segment)  Â1  57 (  ' s of  ) S (3) OR In Δ EBC : S R Ê 3  90 (line from centre to midpt of chord) Ĉ  33 ( given)  B̂1  57 (s of  ) S (3)  B̂1  Â1  57 (  ' s in same circle segment) 8.3 Ô1  2Â1  114  (  at center  2   at circumference) S R (2) Copyright reserved Mpumalanga Please turn over

Downloaded from hlayiso.com Mathematics/P2 memo 18 September 2019 8.4 D̂1  D̂ 2  Â 1  57  (s opposite  sides) S/R But D̂1  Ĉ  33 (s in same circle segment)  D̂ 2  24 S OR (2) DÔC  2  B̂1 (@centre  2  @circum) DÔC  2  57   114 S/R DÔC  Ê 2  D̂ 2 (ext  of  ) 114   90  D̂ 2  D̂ 2  24 S (2) 8.5 B̂1  B̂2  90 ( in semi  circle) S R Â 2  57 (s of  ) S (3) OR ADE  ABE (SS) S R  Â1  Â 2  57 S (3) [12] Copyright reserved Mpumalanga Please turn over

Downloaded from hlayiso.com Mathematics/P2 memo 19 September 2019 QUESTION 9 C A 1 3 2 G 2 1 1 F O 2 2 3 1 E D B 9.1 Ê  x ( at centre  2   at circum) S F̂1  180  x (opposite s of cyclic quad) R S F̂2  x ( s on straight line) (3) OR Ê  x ( at centre  2   at circum) S R F̂2  x (ext  of cyclicquad) S/R (3) 9.2 F̂2  x (ext  of cyclic quad OR straight line) B̂1  x  Ê (correspon ding s, EC || BF) S R  F̂2  B̂1  x S  DF  DB (sides oppposite  s) R (4) Copyright reserved Mpumalanga Please turn over

Downloaded from hlayiso.com Mathematics/P2 memo 20 September 2019 9.3 Â  x ( at centre  2   at circum) S S Ĉ1  Ĉ 2  x (s opposite  sides) S R Ĉ 2  Ĉ 3  x (ext.  of cyclic quad)  Ĉ1  Ĉ 2  Ĉ 2  Ĉ 3 (4)  Ĉ1  Ĉ 3 OR S Â  x ( at centre  2   at circum) R Ĉ1  Ĉ 2  Â (s opposite  sides) S R Â  F̂2 (exterior  of cyclic quad) (4) F̂2  Ĉ 2  Ĉ 3 (corresponding s, CE || BF)  Ĉ1  Ĉ 2  Ĉ 2  Ĉ 3  Ĉ1  Ĉ 3 9.4 CG  FB, CG || BF (given)  CGBF is a parm (one pair opp. sides equal and parallel) R S/R BG  FC  2 (opp. sides of parm.) S CF : FD  2 : 1 BD FD S\R  (line || to one side of  or proportionality ED CD theorem) BD 1   BD 1  ED 3 ED 3 (5) [16] Copyright reserved Mpumalanga Please turn over

Downloaded from hlayiso.com Mathematics/P2 memo 21 September 2019 QUESTION 10 D A G 1 1 H B C E F 10.1 Construction : DG  AB and DH  AC construction In ABC and DEF 1. Â  D̂ (given) 2. AB  DG (construction) 3. AC  DH (construction) S/R  ABC  DEF (SS)  B̂  Ĝ1 and B̂  Ê (given) S  Ê  Ĝ1 S/R  EF || GH (corresponding s ) S R DG DH   (line || to side of  ) DE DF but AB  DG and AC  DH (6) AB AC   DE DF Copyright reserved Mpumalanga Please turn over

Downloaded from hlayiso.com Mathematics/P2 memo 22 September 2019 10.2 K B 2 1 O 1 2 1 2 3 2 3 C 4 1 T A S 10.2.1 In CKB and AKC S/R Ĉ1  Â (tan-chord) K̂  K̂ (common  ) S KĈT  B̂ 2 (3rd  in  )  CKB ||| AKC (    ) R (3) 10.2.2 T̂2  Ô1  Â (ext  of  ) S/R KĈT  Ĉ1  Ĉ 2  Ĉ 3 But Ĉ 2  Ĉ3  90  (  s in semi-circle) S R But Ĉ1  Â (proven) KĈT  Â  90   T̂2  KĈT  Â  90 S Copyright reserved Mpumalanga Please turn over

Downloaded from hlayiso.com Mathematics/P2 memo 23 September 2019 (4) 10.2.3 COT ||| AKC TĈO  Â ( s opp equal sides; CO = AO) S/R S T̂2  KĈA ( proved ) Ô 2  K̂ (3rd  in  ) COT ||| AKC (  ) R (3) 10.2.4 CKB ||| ΔAKC BK CK ratio   KC AK  BK.AK  KC 2 S ΔCOT ||| ΔAKC KC AC   ratio OT CT  KC.CT  AC.OT  KC  AC.OT S CT (4) AC 2 .OT 2  KC 2  CT 2 [20] Copyright reserved Mpumalanga Please turn over