Mathematics Grade 12 Paper 2 Limpopo Mock 2018 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE/GRAAD 12 MATHEMATICS PAPER 2/ WISKUNDE VRAESTEL 2 MEMORANDUM SEPTEMBER 2018 MARKS/PUNTE: 150 TIME/TYD: 3 HOURS/URE This memorandum consists of 14 pages. Hierdie memorandum bestaan uit 14 bladsye Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2 /Wiskunde V2 2 LimpopoDoE/September 2018 NSC-Memorandum QUESTION/ 1 1.1 1140 1140 ̅=  22,8 ✓ 50 50 ✓ answer/ (2) 1.2 Marks Frequency Cumulative Frequency 0 4 4 10 20 24 ✓4;24;39 20 15 39 ✓44;50 30 5 44 (2) 40 6 50 1. Cumulative Frequency/𝑲𝒖𝒎𝒖𝒍𝒂𝒕𝒊𝒆𝒘𝒆 𝒇𝒓𝒆𝒌𝒘𝒆𝒏𝒔𝒊𝒆 3 Marks out of 50/ ✓(0;0) ✓upper limits/ ✓all points correct / (3) 1.4 Q1  15 (accept 14 to16)/ (aanvaar van 14 tot 16) ✓ Q1  15 Q 3  29 (accept 28 to 30)/ (aanvaar van 28 tot 30) ✓ Q 3  29 IQR/IKO = 29  15 = 14 ✓answer / (3) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2 /Wiskunde V2 3 LimpopoDoE/September 2018 NSC-Memorandum QUESTION/ 2 2.1  143  143 x  14 ,3 seconds/ ✓x  10 10 = 2,87 seconds/ . ✓14,3 ✓ = 2,87 ✓accuracy / (4) 2.2 = 17,93 ✓ =  0,07 ✓ yˆ   0,07 x  17 ,93 ✓equation/ (3) 2.3 r   0,69 ✓ r  0,69 (1) 2.4 y   0,07 (80 )  17 ,93 ✓ y  2,33 seconds. substitute/ ✓ answer/ (2) [10] QUESTION/ 3 3.1 QR = (5  3) 2  (3  3) 2 ✓substitution/ = 64  36 = 100 =10 ✓10 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2 /Wiskunde V2 4 LimpopoDoE/September 2018 NSC-Memorandum 3.2 53 33 ✓ -value/ M( ; ) 2 2 ✓ -value / (2) = M (1;0 ) 3.3 03 ✓m P (  5;3) and M (1;0 ) : m  1 5 1 = 2 1 ✓subst of m and point/ y0  ( x  1) 2 1 1 y x 2 2 ✓equation/ (3) 3.4 r  5; centre (1;0 ) : ✓r = 5 and/ (1;0 )  x  1  y  25 2 2 ✓LHS/ ✓RHS/ (3) 3.5 PM = (1  5)  ( 3) 2 2 ✓PM = 45 = 45 ✓ 25 25 ✓conclusion/ (3)  P lies OUTSIDE the circle. 3.6 S(3;9) ✓ -value/ ✓ value/ (2) 3.7 33 ✓ m PQ  3 m PQ  =-3 53 tan  =  3 ✓ tan  =  3 ✓   108 , 43    180   71,57  ✓   71,57   108 , 43  (4)   71,57  co-interior angles, PR -axis. [19] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2 /Wiskunde V2 5 LimpopoDoE/September 2018 NSC-Memorandum QUESTION/ 4 4.1.1 C (3 ;  2 ) ✓ -value/ ✓ -value (2) 4.1.2  4y  8  0 ✓ -value/ y2 ✓ -value/ (2) B( 0;2) 4.1.3 r 2  (3  0 ) 2  (  2  2 ) 2 =25 ✓ r 2  25 r5 ✓r = 5 x  3  5  2 ✓ x  2 x  35  8 ✓x 8 (4) 4.1.4 Let D ( x; y ) :  x  0 2  ( y  2) 2 =25 and ✓subst in distance formula/ 3 y x2 3 4 ✓y  x2 3 4 x2  ( x  2  2 ) 2  25 ✓replace y/ 4 9 x2  x 2  25 ✓ x 2  16 16 ✓ -value/ 25 x 2  25 16 ✓ -value/ x 2  16 x   4 or x  4 (6) 3 n/a y ( 4)  2  5 4 D ( 4;5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2 /Wiskunde V2 6 LimpopoDoE/September 2018 NSC-Memorandum 4.2 4.2.1 Let C ( x ; 2 x ) and A ( x ; 0 ) ✓coordinates of C ( x  1)  ( 2 x  2)  ( x  x)  ( 2 x  0) 2 2 2 2 and A/ x 2  2 x  1  4 x 2  8x  4  4x 2 ✓equating two r 2 / x 2  6x  5  0 r2 ( x  1)( x  5)  0 x  1 or x  5 C (1; 2) or C (5 ;10 ) ✓standard form/ ✓factors ✓ C (1; 2) ✓ C (5 ;10 ) (6) 4.2.2 r  2 or r  10 ✓r 2 ✓ r  10 (2) [22] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2 /Wiskunde V2 7 LimpopoDoE/September 2018 NSC-Memorandum QUESTION/ 5 5.1.1 y 2  12  ( 1  k 2 ) 2 ✓diagram and/ Pythagoras  11 k 2 1 k  k2 yk ✓ sin 25   k sin 25   k (2) 5.1.2 sin 50   2 sin 25  cos 25  ✓double angle expansion/  2k 1  k 2 ✓substitution (2) 5.2 tan( 207 ) sin 2 ( x  360 )  tan 333  sin( x  90 ) cos x ✓  tan 27   tan 27  sin 2 x =  ✓  tan 27   tan 27   cos x. cos x ✓ sin 2 x ✓  cos x sin 2 x  1 cos 2 x  sin 2 x cos 2 x ✓ cos 2 x cos 2 x  sin 2 x 1  ✓ cos 2 x cos 2 x 1 = ✓answer / (7) cos 2 x 1 = cos x 5.3 ✓replace/ sin2A cosA tan 2 A RHS:  cos2A sinA ✓expansion of sin 2 A = 2sinAcosA. cosA / sin 2 A sinA.cos2A 2 cos 2 A 2 cos 2 A = ✓ cos 2 A cos 2 A 1  2cos 2 A  1 LHS: ✓ replacing cos 2 A in cos2A 2 cos 2 A numerator/ = cos 2 A cos 2 A 2 cos 2 A ✓ LHS = RHS cos 2 A (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2 /Wiskunde V2 8 LimpopoDoE/September 2018 NSC-Memorandum 5.4 2 sin 2 x   cos 2 x 2 sin 2 x  cos 2 x = ✓dividing by cos 2 x / cos 2 x cos 2 x 2 tan 2 x   1 cos 2 x 1 tan 2 x   1 2 ✓ tan 2 x   2 x  153 ,43   k180  2 x  76 ,72   k .90 , k  Z ✓ 2 x  153 ,43   k180  ✓ x  76 ,72   k .90  ✓ k Z (5) [22] QUESTION/ 6 6.1 Graph done/ : ✓ intercepts/ ✓turning points/ ✓shape/ (3) 6.2.1 x  ( 120  ; 0)  ( 240 ; 270 ] ✓ (  120 ; 0 ) ✓ ( 240 ; 270 ] ✓notation / (3) 6.2.2 x  [ 180 ;  60 ]  [ 30 ;60 ]  [150 ;270 ] ✓ x  [ 180 ;  60 ] ✓ [  30 ;60 ] ] ✓ [150 ;270 ] (3) 6.3 y  sin( x  30   120 ) ✓✔ y  sin( x  30   120 )  sin( x  90 ) ✓   cos x ✔ answer/ (2)   cos x [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2 /Wiskunde V2 9 LimpopoDoE/September 2018 NSC-Memorandum QUESTION/ 7 7.1 2h ✓ratio/ tan   BE 2h ✓ BE = 2h tan  BE = tan  (2) 7.2 ED ✓ tan  tan   h ✓correct ratio / ED = h tan  (2) 7.3 BD  BE  ED  2 BE .DE . cos BEˆ D 2 2 2 ✓Correct use of cos = rule/ 2  2h   2h    h tan    2 h tan  cos 120   2   tan    tan   4h 2 =  h 2 tan 2   4 h 2  cos 60   tan  2 ✓substitution/ 4h 2  1 =  h 2 tan 2   4 h 2    tan  2  2 4h 2 ✓simplification/ =  h 2 tan 2   2 h 2 tan  2 4 h 2  h 2 tan 4   2 h 2 tan 2  = tan 2  h 2 ( 4  tan 4   2 tan 2  ) = ✓simplify / tan 2  h tan 4   2 tan 2   4 ✓ h 2 taken out as common factor/ = tan  2 h2 (5) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2 /Wiskunde V2 10 LimpopoDoE/September 2018 NSC-Memorandum QUESTION/ 8 8.1 Q̂ 1  32  (tan-chord theorem/ ) ✓S✓R (2) 8.2 P̂4  Q̂ 1  32  ( s opp equal sides/ ✓S✓R (2) ) 8.3 M̂ 1  64  ( at centre is 2 at circumference/ ✓S✓R (2)  8.4 M̂ 2  116  ( on a str line / ) ✓S✓R R̂  58 ( at centre is 2 at circumference/ ✓S✓R (4)  [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2 /Wiskunde V2 11 LimpopoDoE/September 2018 NSC-Memorandum QUESTION/ 9 9.1 Ĵ 3  90  (line from centre to midpoint/ ✓S ✓R (l ) N 2  90 (tangent to radius/ ✓S ✓R Ĵ 3  N̂ 2 ✓S ONPJ is a cyclic quad /koordevierhoek (CONVERSE of ext. of cyclic quad/ ✓R (6) ) 9.2 Ô 1  2Ĉ  2 x ( at centre =2 at circumference ✓S ✓R  Ô 4  180   2 x ( ‘s on a str line/ ) ✓S✓R P̂1  180   2 x (ext. of a cyclic quad/ ) ✓S✓R (6) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2 /Wiskunde V2 12 LimpopoDoE/September 2018 NSC-Memorandum QUESTION/ 10 10.1 h k 10.1 Construction: Join PQ and PF and draw altitudes h and k/ ✓construction/ 1 DP  h area  DPQ DP  2  area  PEQ 1 PE ✓S PE  h 2 1 DQ  k area  DPQ DQ  2  ✓S area  PQF 1 QF QF  k 2 But the area PQE  area PQF (same base and same height/ ✓S✓R ) area  DPQ area DPQ  area  PEQ area PQF ✓S DP DQ  PE QF (6) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2 /Wiskunde V2 13 LimpopoDoE/September 2018 NSC-Memorandum 10.2 10.2.1 Ĉ 3  B̂ 2 (tan-chord theorem ) ✓S✓R ✓S Ĉ 3  D̂ 2 (alternate/ ’s; BD CE ) B̂ 2  D̂ 2 BC = DC (sides opposite equal ’s/ e) ✓R (4) 10.2.2 In BAF and/ en DCF: B̂1  B̂ 2  D̂ 3 (ext. of a cyclic quad/ ) ✓S✓R ✓S Â 1  Â 2  Ĉ 3  Ĉ 4 (ext. of a cyclic quad/ ) F̂ is common/ BAF DCF ( ) ✓R (4) 10.2.3 BA AF ✓S  (similar/ ’s) DC CF BA DC  AF CF ✓S BA BC  (BC = DC) ✓Replacing/ DC AF CF BC DE But  (BD CE; prop theorem/ ✓S✓R CF EF BA DE  (5) AF EF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2 /Wiskunde V2 14 LimpopoDoE/September 2018 NSC-Memorandum 10.2.4 In ECD and EAC: Ê 1 s common/ Ĉ 3  Â 2 (tan-chord theorem ) ✓S✓R D̂ 3  Ĉ 2  Ĉ 3 ( ’s of a / ’e van ‘n ∆) ✓S ECD ||| EAC ( ) ✓R (4) 10.2.5 EC ED  (||| ’s) EA EC ✓S EC  EA.ED 2 ✓S BA.EF But ED  (from/ 10.2.3) ✓S AF EA.BA.EF EC 2  (3) AF [26] TOTAL/ : 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief