Mathematics Grade 12 Paper 2 NSC November 2022 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE 12/GRAAD 12 MATHEMATICS P2/WISKUNDE V2 NOVEMBER 2022 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 24 pages. Hierdie nasienriglyne bestaan uit 24 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 2 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne NOTE:  If a candidate answers a question TWICE, only mark the FIRST attempt.  If a candidate has crossed out an attempt of a question and not redone the question, mark the crossed out version.  Consistent accuracy applies in ALL aspects of the marking memorandum. Stop marking at the second calculation error.  Assuming answers/values in order to solve a problem is NOT acceptable. NOTA:  As 'n kandidaat 'n vraag TWEE KEER beantwoord, merk slegs die EERSTE poging.  As 'n kandidaat 'n antwoord van 'n vraag doodtrek en nie oordoen nie, merk die doodgetrekte poging.  Volgehoue akkuraatheid word in ALLE aspekte van die memorandum toegepas. Hou op nasien by die tweede berekeningsfout.  Aanvaar van antwoorde/waardes om 'n probleem op te los, word NIE toegelaat nie. GEOMETRY/MEETKUNDE A mark for a correct statement (A statement mark is independent of a reason) S 'n Punt vir 'n korrekte bewering ('n Punt vir 'n bewering is onafhanklik van die rede) A mark for the correct reason (A reason mark may only be awarded if the statement is correct) R 'n Punt vir 'n korrekte rede ('n Punt word slegs vir die rede toegeken as die bewering korrek is) Award a mark if statement AND reason are both correct S/R Ken 'n punt toe as die bewering EN rede beide korrek is Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 3 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 1 SCATTER PLOT / SPREIDIAGRAM 25 Number of votes / 20 15 Getal stemme 10 5 98 103 108 113 118 Intelligence quotient (IQ) / Intelligensiekwosiënt (IK) Popularity score (x) 32 89 35 82 50 59 81 40 79 65 Gewildheidspunt (x) Number of votes (y) 9 22 10 21 11 15 20 12 19 16 Getal stemme (y) 1.1.1 155 155 y 10  15,5 ANSWER ONLY: Full marks answer (2) 1.1.2 SD  4,59  answer (1) 1.2 y  SD = 15,5 – 4,59 = 10,91  value of y  SD  10 – 2 = 8 learners  answer (2) 1.3 a = 1,7709…  a b = 0,2243…  b yˆ  1,77  0,22 x  equation (3) 1.4 yˆ  1,77  0,2272  substitution = 17,61  answer  18 votes (2) OR/OF yˆ  17,92  18 votes  answer (2) 1.5.1 Points are all scattered therefore low correlation and unrealistic  R prediction./Punte is versprei daarom 'n lae korrelasie en onrealistiese voorspelling. (1) 1.5.2 r = 0,98/correlation very strong/korrelasie baie sterk  S  a reliable prediction/'n betroubare voorspelling (1) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 4 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 2 OGIVE OGIVE/ / OGIEF OGIEF 70 70 60 60 50 50 Cumulative Cumulative frequency frequency / 40 40 30 30 Kumulatiewe Frekwensie 20 20 10 10 0 0 5 10 15 20 25 30 35 5 10 15 20 25 30 35 Percentage of salary spent on fuel / Percentage Persentasie of op van salaris fuelbrandstof spent spandeer 2.1 60 employees  answer (A) (1) 2.2 20  x  25  answer (1) 2.3 60 – 34  34 = 26 employees ANSWER ONLY: Full marks  answer (2) 2.4 100  method Salary   2400 7 Salary = R34 285,71 ANSWER ONLY: Full marks  answer (2) 2.5  Ogive/Cumulative frequency graph will shift to the  answer right/will become steeper.  Ogief/Kumulatiewe frekwensie grafiek sal na regs skuif/sal steiler wees. (2) [8] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 5 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 3 y D A(4 ; 2)  α x P O F R S C(–2 ; –3) T B(6 ; –4) 3.1.1 2   4 42 mAB  OR mAB  46 64  substitution mAB  3 ANSWER ONLY: Full marks  answer (2) 3.1.2 tan  mAB  3  tan  mAB  3  108,43  answer ANSWER ONLY: Full marks (2) 3.1.3 x x y y  T 1 2 ; 1 2   2 2    2 6 3 4 T ;   2 2  7  7  xT  2  yT  T 2 ;  2  2  (2) 3.1.4 5(0)  6 y  8  xS  0 4 4 y  yS  3 3  4 S  0;   3  (2) 3.2 mCD  mAB  3  gradient  3  3 2  c OR y   3  3x   2  substitution of c = –9 y  3x  9 C(–2 ; –3) y  3x  9  equation (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 6 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 3.3.1 5x  6 y  8 5 8 y x 6 6 5 tan  mAC   tan  mAC  5 6 6   39,81    39,81 Â  108,43  39,81  68,62  Â  68,62 DĈA  68,62 [alt s ; DC||AB]  answer (4) 3.3.2 P 3; 0 and F(1,6 ; 0)  P 3; 0 Area POSC = Area FPC – Area OFS  method 1 1 1 4   4, 6  3   4, 6  3  1, 6    2 2 2 3 1 4 = 6,9 – 1,07  1, 6    2 3 = 5,83 units2  answer (5) OR/OF P 3; 0  P 3; 0 2  8 FC    2     3  0  2 3 61  5 5 Area Δ PFC  PFFCsinOF̂S 1 2 1  23   3 61  1  23  3 61      sin 39,81     sin 39,81 2  5   5  2  5  5   6,90 1  8  4  1  8  4      Area Δ OFS     2  5  3  2  5  3  = 1,07  method Area POSC = 6,90 – 1,07  answer = 5,83 units2 (5) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 7 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne P R O C(–2 ; –3)  P 3; 0 P  3;0   method Area of POSC = Area of OSCR + Area of PRC 14  1   3   2  1 3 14  1    3   2  1 3  23  2 23  2 35  6  5,83units 2  answer (5) OR/ OF P R O W 2 C(–2 ; –3) P 3; 0  P 3; 0 Area POSC = Area ROSW + Area PRC + Area  method WSC 1 4 1 1 5     2   1 3   2     1 3 3 2 2 3 2 4 1 5     2   2   35 3 2 3  6  answer  5,83units 2 (5) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 8 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne P 3; 0  P 3; 0 1 ˆ Area of PSC = (PC)(CS)sin DCA 2    1  10   361  sin 68, 62   2    1 2  10   361  sin 68, 62   = 3,833.. 1 Area of POS = (PO)(OS) 2 1 4   3   2 3  3   1 4  2 3 =2 Area POSC = 3,833… + 2  method  5,83units2  answer (5) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 9 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 4 y K x N(7 ; –2) M(3 ; –5) L P 4.1 Px; y  ; N7;2 ; M3;5 x7 y2 3 5 2 2 x  1 y8  xP  1  yP  8 P 1;  8 (2) r 2  7  3   2   5 OR/OF r 2   1  3   8   5  4.2.1 2 2 2 2  substitution into distance formula r 2  25 ( x  3) 2  ( y  5) 2  25  ( x  3)2  ( y  5)2  r 2  25 (3) 4.2.2  5   2 3  substitution mradius   37 4 3 3  mradius   mtangent   4 [radius  tangent/raaklyn  radius ] 4 4 3 4  mtangent    2   7   c y   2   x  7  4 4 3 OR 3 3  substitution of m and 22 4 22 N(7 ; –2) c y  x 3 3 3 4 22 y  x 3 3  equation (5) 4.3  subst m and P 8   4  1  c 3 28  value of c  c 3 28 22  answer  k 3 3 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 10 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 4.4.1 AB2  AM2  MB2 AB2   t  3   t  5   52 2 2    substitution into  t  6t  9  t  10t  25  25 2 2 Pythagoras  simplification (A) AB 2t 2  4t  9 (2) 4.4.2 4  substitution into t  22  correct formula  1  t  1 Minimum at t   1 AB 2 1  4 1  9 2  substitution AB 7  answer (4) OR/OF 4t  4  0  derivative = 0  t  1 t  1 Minimum at t   1 AB 2 1  4 1  9 2  substitution AB 7  answer (4) OR/OF Length of AB  2t 2  4t  9  9  2  t 2  2t    2  7  2  t  1   2  completing of the  2 square  2  t  1  7 2 Minimum at t  1  t  1 AB 2 1  4 1  9 2 AB 7 substitution answer (4) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 11 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 5 5.1.1 sin 360  x   sin x  +  sin x (2)  13    3 5.1.2 2  substitution x  coordinate  2  2 3  method tan x  2 3  2 (3) OR/OF  13   3 2 x  coordinate  2  substitution 2 3 tan x  2  method (3) 5.1.3 cos(180  x)   cos x  –  cos x (2) 5.2 cos(90   ) sin(  180)  3 sin( )  sin   – sin  sin  180     3 sin   – 3sin  sin    – sin  sin   3 sin   sin    4 sin   simplification 1   answer 4 (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 12 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 5.3 (cos x  2 sin x)(3 sin 2 x  1)  0 cos x  2sin x  0 or 3sin 2 x 1  0  both equations 1 1 1 tan x   sin 2 x   tan x   2 3 2 ref   26,565... ref  19,471...  sin 2 x  1 3 x = 153,43° + k.180° ; k  Z x = 9,74° + k.180°; k  Z  x = 153,43° OR OR/OF or x = 153,43° &333,43° x = 153,43° + k.360° ; k  Z x = 80,26° + k.180°;  x = 9,74° & 80,26°  + k.180° ; kZ k Z or x = 333,43° + k.360° ; k  Z (6) 5.4.1 LHS  cos( x  y). cos( x  y)  cos x. cos y  sin x. sin ycos x. cos y  sin x. sin y  expansion  cos x. cos y  sin x. sin y 2 2 2 2  simplification     1  sin 2 x 1  sin 2 y  sin 2 x.sin 2 y  square identity  1  sin 2 x.sin 2 y  sin 2 x  sin 2 y  sin 2 x.sin 2 y  product  1  sin 2 x  sin 2 y = RHS (4) 5.4.2 1  sin 45  sin 15 2 2  cos(45  15). cos(45  15)  identifying x and y  cos 60. cos 30  1  3      substitution  2  2  3  4  answer (3) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 13 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 1  sin 2 45  sin 2 15  identity  sin 2 15  cos2 15  sin 2 45  sin 2 15 2  2  cos 15   2   2  1  cos 2 15  2 2 cos 2 15  1  2 cos 30  2 3 1  substitution   2 2 3  answer  (3) 4 OR 1  sin 2 45  sin 2 15  cos 2 45  sin 2  45  30  2  expansion  1      sin 45 cos 30  cos 45 sin 30  2  2  substitution 2 1  1 3 1 1        2  2 2 2 2  2 1  3 1       2  2 2 2 2  2 1  3 1       2  2 2 2 2  1 3 3 1       2  8 4 8   answer 3  (3) 4 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 14 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 5.5.1 16 sin x. cos 3 x  8 sin x. cos x  8 sin x. cos x2 cos 2 x  1  factorisation  4 sin 2 xcos 2 x   4sin2x  cos2x  2 sin 4 x  double angle OR/OF (4) 16 sin x. cos 3 x  8 sin x. cos x 1  1   16cos 2 x  sin 2 x   8  sin 2 x  2  2   factorisation 1   8  2cos 2 x  1  sin 2 x  2   4sin 2 x.cos 2 x  4sin 2x  cos2x  2sin 4x  double angle (4) 5.5.2 16 sin x. cos x  8 sin x. cos x = 2sin4x 3 Minimum at x = 67,5°  answer (1) [30] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 15 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 6 y f f B g A(60° ; k) x x –180° –90° 0° 90° 180° f 6.1 180  answer (1) 6.2.1 k  3  1,73  answer (1) 6.2.2  B  120; 3   x = –120° (1) 6.3 Range of g : y   2 ; 2  y   2 ; 2 Range of 2 g ( x) : y   4 ; 4  answer (2) OR/OF ANSWER ONLY: Full marks Range of g : 2  y  2  2  y  2 Range of 2 g ( x) : 4  y  4  answer (2) 6.4 x   65 ;  5  x   65 ;  5 (2) OR/OF  65  x  5   65  x  5 (2) 6.5 sin x.cos x  p 4 sin x.cos x  4 p  2 sin 2 x  4 p 2 sin 2 x  4 p  4p  2 4p  2 1 1  answers  p =  or ANSWER ONLY: Full marks 2 2 (3) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 16 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 7 A B 2p 45° x D C 7.1 AD2  AB2  BD2   AD2  5 p  2 p  2 2  substitution in Pythagoras AD2  9 p 2 AD  3 p  answer (2) 7.2 CD 3p  correct use of sine rule  sin(135  x) sin x 3 p sin 135  x   135° – x CD  sin x 3 p sin 135 cos x  cos 135 sin x   compound angle CD  sin x 3 p sin 45 cos x  cos 45 sin x  CD  sin x  2 2  3 p  cos x  sin x   special values CD   2 2  sin x  2 3p   cos x  sin x   2   factorisation CD  sin x 3 p  sin x  cos x  CD  2 sin x (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 17 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 7.3 1  correct use of area rule Area ADC   AD  CD  sinADC ˆ 2 1  3 p sinx  cosx    (3 p) (sin45) 2  2 sin x   1 30 30sin 110  cos110  sin 45 2  2 sin 110   substitution in area rule  143,11m2  answer (3) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 18 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 8 8.1 M 2 1 64° O 1 N 2 S 1 2 1 2 R P 8.1.1 P̂  116 [opp s of cyclic quad/teenoorst. e van kvh] SR (2) 8.1.2 M̂1  64  90 [ in semi-circle/ in halwe sirkel] R M̂1  26 S (2) 8.1.3 Ô1  52 [ at centre = 2 x  at circumference/midpts.   S R = 2 x omtreks. ] (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 19 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 8.2 A E D H G B F C 8.2.1 Midpt theorem/Midpt. Stelling R (1) OR/OF Converse prop intercept theorem R (1) 8.2.2 BG = 2DE or 6x – 2 [Midpt theorem/Midpt. stelling] S R BG = 6x – 2  S R GH FC  [line || one side of  OR BG BF prop theorem; FG || CH / lyn || een sy v.  ] x 1 1   equation into x 6x  2 4 4x + 4 = 6x – 2 2x = 6 x=3  answer (6) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 20 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne BF BG = [line || one side of  OR prop theorem; FG || CH / FC GH S R lyn || een sy v. ] AE DE = [ADE ||| ABG] AG BG  S R BG = 4x + 4 1 3x  1 = 2 4x  4  equation into x 4x + 4 = 6x – 2 x = 3  answer (6) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 21 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 9 9.1 O A 1 2 D B 9.1.1 Construction: Draw OA and OB  construction In ADO and BDO OA = OB [radii/radiusse] OD = OD [common side/gemeenskaplike sy]  first pair of sides AD = DB [given/gegee]  other 2 pairs  ΔADO  ΔBDO [S;S;S]  R ADB is a straight line ˆ D ˆ  R D1 2 ΔADO  ΔBDO  OD  AB [s on a str line/e op 'n reguitlyn] (5) OR/OF Construction: Draw OA and OB  construction In ADO and BDO AD = DB [given/gegee]  first pair of sides Aˆ Bˆ [s opp; s sides /e teenoor gelyke sye]  other 2 pairs OA = OB [radii/radiusse]  ΔADO  ΔBDO [S;;S]  R ADB is a straight line ˆ D ˆ  R D1 2 ΔADO  ΔBDO  OD  AB [s on a str line/e op 'n reguitlyn] (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 22 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne 9.2 B G E T F O P S 9.2.1 OT̂G  90 [line from centre to midpt of chord/ S  R midpt. sirkel; midpt. koord] OB̂G  90 [tan  radius/raaklyn  radius] S R OT̂G  OB̂G  90 OTBG is a cyclic quadrilateral [line subtends equal s OR R converse s in the same segment/ lyn onderspan gelyke e] (5) 9.2.2 Ŝ  BT̂G [corresp s; GF || PS / S R ooreenk. s; GF || PS] But BT̂G  GÔB [s in the same segment/e in dies. S R sirkelsegment ] GÔB  Ŝ (4) [14] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 23 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 10 A S R 3 1 2 1 2 T 3 1 2 2 1 2 1 Q D3 K 2 3 1 C 1 2 P 10.1 P̂1  Q̂1 [tan-chord theorem/ tussen raaklyn  S en koord] Ŝ1  Q̂1  Q̂ 2 [ext  of cyclic quad/buite  v. kvh]  S / R  Ŝ1  P̂1  Q̂ 2 T̂2  R̂ 2  Q̂ 2 [ext  of /buite  v.  ]  S but P̂1  R̂ 2 [given/gegee] T̂2  P̂1  Q̂ 2  S  Ŝ1  T̂2  P̂1  Q̂ 2 (4) 10.2 In  ASD and ACR  identifying 's    [common /gemeenskaplike ]  S Ŝ1  T̂2 [proven/reeds bewys] T̂2  Ĉ 2 [alt s; QS || CA/verw. e; QS || CA  S / R ]  S  Ŝ1  Ĉ 2  S D̂1  R̂ 1 [sum of s in ∆/e v. ∆] ASD ||| ACR AD AS   [corresponding sides in proportion/ AR AC ooreenstemmende sy in dies. verhouding] (5) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

Downloaded from hlayiso.com Mathematics P2/Wiskunde V2 24 DBE/November 2022 NSC/NSS – Marking Guidelines/Nasienriglyne In  ASD and ACR  identifying 's    [common /gemeenskaplike ]  S Ŝ1  T̂2 [proven/gegee] T̂2  Ĉ 2 [alt s; QS || CA/verw. e; QS || CA]  S/R  Ŝ1  Ĉ 2  S ASD ||| ACR [;;]  R AD AS   [corresponding sides in proportion/ AR AC ooreenstemmende sy in dies. verhouding] (5) 10.3 AS SD  S  [ ASD ||| ACR ] AC CR AC  SD AS  CR AS CT  [line || one side of  OR prop theorem;  S  R AR CR TS || CA/lyn || een sy v.  ] AR  CT  AS  CR AC  SD AR  CT  equating   CR CR  AC  SD  AR  CT (4) [13] TOTAL/TOTAAL: 150 Copyright reserved