Mathematics Grade 12 Paper 2 Western Cape Mock 2014 Memo

Downloaded from hlayiso.com GANSBAAI ACADEMIA EKSAMEN V2 WISKUNDE September 2014 Graad 12 Totaal: 150 memo VRAAG 1 1.1 1522 ✓ 1522  101,47 ✓ 101,47 15 (2) 1.2 19,07 ✓ 19,07 ✓ 2 desimale plekke (2) 1.3 Onderste kwartiel  89 ✓ 89 Boonste kwartiel  113 ✓ 113 (2) 1.4 ✓ M by 100 ✓ Min=58 EN Max=145 ✓ Q1=89 EN Q3=113 (3) ✓✓ interval 1.5 ( x  1 ; x  1 )  (82,4 ; 120,54) ✓ antwoord  2 dae (3) [12] VRAAG 2 1

Downloaded from hlayiso.com VRAAG 3 3.1  2  3 1  0  ✓ subst M ;   2 2  ✓ antwoord vir albei  1 1    ;   2 2 (2) 3.2  1 0  3  2  ✓ subst Mid BD  ;   2 2   1 1    ;   2 2 ✓ antwoord Middelpunt van AC en BD is dieselfde ✓ gevolgtrekking Halveer AC en BD mekaar (3) 3.3 20 2 ✓ M AD M AD   03 3 1  2  3 ✓ M CD M CD   20 2 2 3 ✓ =-1 M AD  M CD    1 3 2  AD  CD  ADˆ C  90 ✓ AD ˆ C  90 (4) 3.4 Diagonale halveer mekaar (3.2) ✓ halveer mekaar DWS ABCD is parallelogram ✓ parallelogram ˆ C  90 Vanaf 1.3: Omdat AD ✓ AD ˆ C  90 DWS ABCD reghoek ✓ reghoek M AC  M BD 1 5    1 5 1 AC  BD ✓ diagonal halveer mekaar loodreg Diagonale halveer mekaar loodreg ✓ afleiding DWS ABCD is ‘n vierkant (6) 3.5 2 1  3 ✓ gradient DC tan    3 02 2 ✓   56,30993247...  180 2  123,7 ✓ antwoord (3) 2

Downloaded from hlayiso.com 3.6 OC 2  (2  0) 2  (1  0) 2 ✓ OC 2 OC 2  5  OC  2,236067977 ✓ antwoord OC  2 Buite sirkel of OC 2  (2  0) 2  (1  0) 2 OC 2  5  OC 2  4 Buite sirkel of x2  y2  4 (2) 2  (1) 2  5  4 Buite sirkel (2) VRAAG 4 4.1   8  0 1 5  ✓ subst Middelpunt AB ;   2 2   (4;3) (1) 4.2 5 1  2 ✓ subst M AD   03 1 ✓ M AD y  mx  c  5  20   c ✓ subst (0;5) c  5 ✓ antwoord y  2 x  5 (4) 4.3 AM  (5  2)  (0  4) 2 2 2 ✓ subst  4  16 ✓ vereenvoudig ✓ antwoord AM  20 (3) 4.4 ( x  4)  ( y  3)  2 2  20  2 ✓ ( x  4) 2  ( x  4) 2  ( y  3) 2  20 ✓ ( y  3)2 ✓  20  x  y  8x  6 y  5  0 2 2 ✓ antwoord (4) 4.5 AT  TK  6 ✓ AT=TK ✓ CD loodreg op AK CD  AK ✓ vlieër ACKD vlieër want diagonal CD halveer ✓ rede diagonal AK reghoekig. (4) ✓ [16] 3

Downloaded from hlayiso.com VRAAG 5 5.1 cos( ). tan(180   ).cos(90   ) ✓ sin(180   ).sin(540   ) cos( ). tan(180   ).cos(90   ).cos(90   )  sin(180   ).sin(180   ) cos .(  tan  .sin  sin ( sin ) sin cos  cos sin 1 (7) 5.2.1 y  12 12 sin A  13 (3) 5.2.2 sin 2 A =2 sin A. cos A  12   5   2    13  13   120  169 (3) 5.3 1  cos 2 x  sin x sin 2 x  cos x 1  ( 1  2 sin 2 x)  sin x  2 sin x cos x  cos x 2 sin 2 x  sin x  2 sin x cos x  cos x sin x( 2 sin x  1 )  cos x( 2 sin x  1 ) sin x  cos x  tan x (6) 5.3.2 sin 2 x - cos x = 0  cos x( 2 sin x - 1 ) = 0 1 sin x  cos x  0 2 x  90  360k of x  30  360k k  Z x  270  360k x  150  360k (8) 4

Downloaded from hlayiso.com 5.4 sin 38° = 1 a2  cos52  sin 38  1  a 2 sin 38  cos 52 cos 52. cos14  sin 52. sin 14 sin 38  cos 52  cos ( 52  14 ) sin 38  cos 52  cos 38 1  a2  1  a2  a 2 1  a2  a (5) VRAAG 6 6.1 a2 1 b 2 (2) 6.2 360 (1) 6.3 1 (1) 6.4 3 waardes van x (2) [6] 5

Downloaded from hlayiso.com VRAAG 7 7.1 CAˆ D  180  (   ) sin[180  (   ) sin    x AC x sin   AC  sin(   ) En AB sin  AC  AB  AC sin x sin  sin   sin(   ) 7.2 40sin15sin70 AB  sin120  11,25m 7.3 40sin70 AC  sin120  43,4m 1 AREA ACD  AC .CD.sin  2 1  (43,4)(40) sin 50 2  664,93m 2 VRAAG 8 8.1 supplementêr ✓ antwoord (1) 8.2.1 Hoek tussen raaklyn en raduis ✓ antwoord (1) 8.2.2 Sˆ3  Sˆ 4  90 (raaklyn  radius) ✓ Stelling Nˆ  Nˆ  90 (raaklyn  radius) 1 2 ✓✓ Stelling & Rede Sˆ3  Sˆ 4  Nˆ 1  Nˆ 2  90  90  180 ✓ Rede  RNOS koordevierhoek teenst hoeke suppl  (4) 8.2.3 Sˆ  x 1 Sˆ1  Nˆ 2  x (Raaklyn koord stelling) ✓✓ Stelling, Rede ✓✓ Stelling, Rede Nˆ  Sˆ  x (basis hoeke gelykbenige Δ) 2 3 ✓✓ Stelling, Rede Sˆ1  Rˆ 2  x (hoeke in dieselfdesegment) ✓✓ Stelling, Rede Nˆ  Rˆ  x (hoeke in dieselfdesegment) 2 1 (8) 6

Downloaded from hlayiso.com 8.2.4 Oˆ  Qˆ 2  Nˆ 2  Sˆ3  180 (Som van hoeke ΔOSN) Maar Sˆ  Nˆ (Teenst hoeke gelyke sye) 3 2  Oˆ1  Oˆ 2  2Sˆ3  180 Oˆ  Oˆ  Oˆ  180 (hoeke op reguitlyn) 1 2 3  Oˆ1  Oˆ 2  Oˆ 3  Oˆ1  Oˆ 2  2Sˆ3 2Sˆ  Oˆ 3 3 1  Sˆ3  Oˆ 3 2 (4) VRAAG 9 9.1 9.2.1 7

Downloaded from hlayiso.com 9.2.2 9.3.1 9.3.2 9.3.3 8