# Mathematics Grade 12 Paper 2 NSC November 2017 Memo > Grade 12 Mathematics Paper 2 memorandum from the NSC November Exam 2017, 29 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/mathematics-p2-nov-2017-memo-afr-eng/67daf7b0-80fe-45d5-9b84-58ae9bdfffb9 - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 12 - Document type: Memorandum - Year: 2017 - Exam period: NSC November Exam - Paper: 2 - Curriculum: CAPS - Pages: 29 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Mathematics](https://www.hlayiso.com/library/grade-12/mathematics) › [Mathematics Grade 12 Paper 2 NSC November 2017 Memo](https://www.hlayiso.com/library/mathematics-p2-nov-2017-memo-afr-eng/67daf7b0-80fe-45d5-9b84-58ae9bdfffb9) ## About this document Mathematics · Grade 12 · NSC November Exam · 2017. Memorandum, 29 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE 12/GRAAD 12 MATHEMATICS P2/WISKUNDE V2 NOVEMBER 2017 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 29 pages. Hierdie nasienriglyne bestaan uit 28 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 2 Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 2 DBE/November 2017 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 guidelines. 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 nasienriglyne toegepas. Hou op nasien by die tweede berekeningsfout. • Aanvaar van antwoorde/waardes om 'n probleem op te los, word NIE toegelaat nie. GEOMETRY 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 a 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 the statement AND reason are both correct. S/R Ken 'n punt toe as beide die bewering EN rede korrek is. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 3 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 1 Time for 100 m sprint (in seconds) 10,1 10,2 10,5 10,9 11 11,1 11,2 11,5 12 12 12,2 12,5 Tyd vir 100 m-naelloop (in sekondes) Distance of best long jump (in metres) 8 7,7 7,6 7,3 7,6 7,2 6,8 7 6,6 6,3 6,8 6,4 Afstand van beste sprong in verspring (in meter) 1.1 a = 14,343… = 14,34  value of a b = –0,642… = – 0,64  value of b (3) 1.2 y = 14,34 − 0,64(11,7)  substitution correctly = 6,85  answer (2) OR/OF answer y = 6,83 (calculator / sakrekenaar) (2) 1.3 The gradient increases / Die gradient neem toe  increases/neem toe The point (12,3 ; 7,6) lies some distance above the current data.  reasoning in words/ /Die punt (12,3 ; 7,6) lê bokant die huidige data. redenasie in woorde (2) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 4 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 2 12 13 13 14 14 16 17 18 18 18 19 20 21 21 22 22 23 24 25 27 29 30 36 2.1.1 472 472 x=  23 23 x = 20,52 seconds / sekonde  answer (2) 2.1.2 Q 1 = 16  Q1 Q 3 = 24  Q3 IQR/IKO = Q 3 – Q 1 = 24 – 16 = 8  answer (3) 2.2 20,52 + 5,94 = 26,46  26,46 ∴ > 26,46 ∴ 4 girls/dogters  answer (2) 2.3  whiskers ending at 12 & 36  Q 1 = 16 & Q 3 = 24 (box)  Q 2 = 20 12 14 16 18 20 22 24 26 28 30 36 (3) 2.4.1 Girls / Meisies  answer (1) 2.4.2 Five-number summary of boys: (15 ; 21 ; 23,5 ; 26 ; 38) None of the boys / Nie een van die seuns nie  answer 5 girls completed in less than 15 seconds which was the  reason/rede minimum time taken by the boys. 5 meisies voltooi in minder as 15 sekondes, wat die minimumtyd is wat die seuns geneem het. (2) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 5 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 3 y A  1 F 3 ; 3   2 G x E O C (8 ; –4) B (– 6 ; – 5) 3.1.1 y 2 − y1 mFC = x 2 − x1 1 3 − (−4) = 2  substitution of (8 ; –4) 3−8  1 3 & 3; 3  = −  2 2  gradient y = mx + c y − y1 = m( x − x1 ) 3 y=− x+c 2 3 3 − 4 = − (8) + c OR/OF ( y − (−4)) = − ( x − 8)  substitution of m and 2 2 3 (8 ; –4) c =8 y + 4 = − x + 12 2 3 3  equation of AC y=− x+8 y=− x+8 2 2 (4) OR/OF Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 6 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne  substitution of (8 ; –4) (− 4) −  3 1   1 y 2 − y1  2 & 3; 3  mFC = = x 2 − x1 8−3  2 3  gradient = − 2 y = mx + c y − y1 = m( x − x1 )  substitution of m and 1 3  1 3  1 3 = − (3) + c  y − 3  = − ( x − 3) 3; 3  2 2  2 2  2  1 3 9 c =8 OR/OF  y − 3  = − x +  2 2 2  equation of AC 3 3 (4) y=− x+8 y=− x+8 2 2 3.1.2 AC: 3x + 2y = 16 and BG: 7x – 10y = 8 15x + 10y = 80  method /metode: 7x – 10y = 8 solving simultaneously / 22 x = 88 los gelyktydig op x=4  x coordinate (x > 0) 3(4) + 2y = 16 y=2  y coordinate ∴G(4 ; 2) (3) OR/OF 7 8 BG: 7x – 10y = 8 ∴y = x− 10 10 7 8 3  method: equating ∴ x− =− x+8 [CA from 3.1.1] metode: stel vgls gelyk 10 10 2 11 44 x= 5 5  x coordinate (x > 0) x=4 3(4) + 2y = 16  y coordinate y=2 ∴G(4 ; 2) (3) 3.2 xA + 4 y +2 1  equation ito x = 3 and A =3  equation ito y 2 2 2 ∴A(2 ; 5) (2) OR/OF by translation/deur translasie:  equation ito x xA = 3 − (4 − 3) = 2  equation ito y yA = 3 12 + (3 12 − 2) = 5 ∴A(2 ; 5) (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 7 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne 3.3 The coordinates of the midpt of AB / Die koordinaat van midpt van AB is:  2 + ( −6) 5 + ( −5)   subst A & B into midpt  ;  = ( −2 ; 0 ) formula  2 2   y coordinate = 0 But the y-coordinate of E is 0 ∴ E(–2 ; 0) is the midpoint of AB  E = midpt ∴ EF || BG [midpoint theorem/middelpuntst OR/OF  Reason line divides 2 sides of ∆ in prop/lyn (4) verdeel 2 sye van ∆ in dies verh] OR/OF The coordinates of the midpt of AB / Die koordinaat van midpt van AB is:  2 + ( −6) 5 + ( −5)   subst A & B into midpt  ;  = ( −2 ; 0 )  2 2  formula AE = (−2 − 2) 2 + (0 − 5) 2 = 41  lengths of AE & EB EB = (−2 − (−6))2 + (0 − (−5))2 = 41 ∴ In ∆ABG: AE = EB and AF = FG  AE = EB or E = midpt ∴ EF || BG [midpoint theorem/middelpuntst]  Reason (4) OR/OF Equation of AB:  5 − ( −5)  y − ( −5) =  ( x − ( −6))  2 − ( −6 )  10 15 5 5 y +5= x+ ∴y = x+ 8 2 4 2  equation of AB x-intercept of AB: 5 5 0= x+ ∴ x = −2 4 2 ∴ E(–2 ; 0)  coordinates of E 1 BG: 7x – 10y = 8 3 −0 7 8 7 mEF = 2 = ∴ y= x− 3 − (−2) 10 10 10  gradient of EF 7 7 m EF = m BG = ∴ mBG = 10 10  gradient EF = gradient BG ∴ EF || BG (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 8 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne 3.4  1  1 Midpoint of AC =  5 ;    5 ;   2  2 xD + (−6) y + (−5) 1 = 5 and D = 2 2 2 ∴ D(16 ; 6)  x value  y value (4) OR/OF  method finding x by translation/dmv translasie:  method finding y D(16 ; 6)  x value  y value (4) OR/OF − 5 − (−4) 1 5 − (−5) 5 mBC = = and mAB = = −6−8 14 2 − (−6) 4 1 1 34 AD: y − 5 = ( x − 2) ⇒ y = x + 14 14 7 5 5 CD: y + 4 = ( x − 8) ⇒ y = x − 14 4 4 5 1 34  equating x − 14 = x + 4 14 7  x value ∴ x = 16  y value y=6 (4) [17] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 9 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 4 y R P ( –4 ; 5) M θ x S O K(0 ; –3) 4.1.1 5 − (−3) mPK =  substitution P & K into −4−0 gradient formula = −2  gradient of PK PK ⊥ SR [radius ⊥ tangent/raaklyn]  PK ⊥ SR OR r ⊥ tangent ∴ mPK × mRS = −1 1 ∴ mRS =  answer 2 (4) 4.1.2 1 y = x+c 2 1 5 = (−4) + c OR/OF ( y − 5) = 1 ( x − (−4))  substitution of m and P 2 2 c=7 ( y − 5) = 1 x + 2 2 1 1  equation y= x+7 y = x+7 2 2 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 10 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne 4.1.3  − 4 + 0 5 + (−3)  M ;   2 2   x value of M ∴ M(–2 ; 1)  y value of M r 2 = ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2 r 2 = (−2 + 4) 2 + (1 − 5) 2 ∴ r 2 = 20  r 2 = 20 ∴ ( x + 2) 2 + ( y − 1) 2 = 20 or ( 20 ) 2  equation (4) OR/OF  − 4 + 0 5 + ( −3)   M (– 2 ; 1) M ;  ∴ M(–2 ; 1)  2 2  ( x + 2) + ( y − 1) = r 2 2 2 r 2 = 20 (−4 + 2) 2 + (5 − 1) 2 = r 2 ∴ r 2 = 20  equation ∴ ( x + 2) 2 + ( y − 1) 2 = 20 or ( 20 ) 2 (4) OR/OF  M (– 2 ; 1)  − 4 + 0 5 + ( −3)  M ;  ∴ M(–2 ; 1)  2 2  PK = (−4 − 0) 2 + (5 − (−3)) 2 = 80 r 2 = 20 80 r= = 20 2  equation 2 2 2 (4) ∴ ( x + 2) + ( y − 1) = 20 or ( 20 ) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 11 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne 4.1.4 tan θ = mPK = −2  tan θ = −2 ∴ θ = 180° – 63,43° = 116,57°  size of θ PK̂R = 116,57° − 90° [ext ∠ of ΔMOK]  answer = 26,57° (3) OR/OF R 20 10 P 80 K In ∆RPK: PK = (0 − ( −4)) 2 + ( −3 − 5) 2 = 80 PR = ( −4 − 0) 2 + (5 − 7) 2 = 20  lengths of PK, PR & RK RK = 10 PK 2 + KR 2 − PR 2 ( 80 ) 2 + (10) 2 − ( 20 ) 2  correct values into cos PK̂R = = 2.PK.KR 2( 80 )(10) cos rule 2 5 = 5 PK̂R = 26,57°  answer (3) OR/OF 20 80 sin PK̂R = cos PK̂R =  lengths of sides 10 OR/OF 10  ratio PK̂R = 26,57° PK̂R = 26,57°  answer (3) OR/OF 20  lengths of sides tanPK̂R =  ratio 80  answer PK̂R = 26,57° (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 Downloaded from hlayiso.com Mathematics/P2/Wiskunde/V2 12 DBE/November 2017 NSC/NSS – Marking Guidelines/Nasienriglyne 4.1.5 RS || tangent at K(0 ; – 3 ) 1 ∴ m PS = m tang =  gradient 2 1 ∴ y = x −3  equation 2 (2) OR/OF 1− 5 mPK = = −2 −2+4 mPK × mtan g = −1 [radius ⊥ tangent/raaklyn] 1 ∴ mtan g =  gradient 2 1 ∴y= x−3  equation 2 (2) 4.2 t ∈ (−3 ; 7)  – 3 (A)  7 (CA from 4.1.2) OR/OF  correct inequality −3