# Mathematics Grade 9 Paper 2 Free State November 2022 Memo > Grade 9 Mathematics Paper 2 memorandum from the Free State November Exam 2022, 17 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/grade-9-mathematics-2022-free-state-november-exam-memorandum-paper-2/0e2151b6-c32e-44c0-9b3d-cc45277062e5 - Updated: 2026-09-28 - Subject: Mathematics - Grade: Grade 9 - Document type: Memorandum - Year: 2022 - Exam period: Free State November Exam - Paper: 2 - Curriculum: CAPS - Pages: 17 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 9](https://www.hlayiso.com/library/grade-9) › [Mathematics](https://www.hlayiso.com/library/grade-9/mathematics) › [Mathematics Grade 9 Paper 2 Free State November 2022 Memo](https://www.hlayiso.com/library/grade-9-mathematics-2022-free-state-november-exam-memorandum-paper-2/0e2151b6-c32e-44c0-9b3d-cc45277062e5) ## About this document Mathematics · Grade 9 · Free State November Exam · 2022. Memorandum, 17 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 NATIONAL SENIOR CERTIFICATE NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 11 MATHEMATICS P2/WISKUNDE V2 NOVEMBER 2022 MEMORANDUM MARKS/PUNTE: 150 This memorandum consists of 17 pages. Hierdie memorandum bestaan uit 17 bladsye. MST GROUP Please turn over/Blaai om asseblief ### Page 2 Mathematics/P2/Wiskunde V2 2 MST/November 2022 CAPS/KABV – Grade/Graad 11 – Memorandum NOTE: • If a candidate answered a question TWICE, mark only the FIRST attempt. • If a candidate crossed out an answer and did not redo it, mark the crossed-out answer. • Consistent accuracy applies to ALL aspects of the marking memorandum. • Assuming values/answers in order to solve a problem is unacceptable. LET WEL: • As 'n kandidaat 'n vraag TWEE keer beantwoord het, sien slegs die EERSTE poging na. • As 'n kandidaat 'n antwoord deurgehaal en nie oorgedoen het nie, sien die deurgehaalde antwoord na. • Volgehoue akkuraatheid is op ALLE aspekte van die memorandum van toepassing. • Dit is onaanvaarbaar om waardes/antwoorde aan te neem om 'n probleem op te los. QUESTION/VRAAG 1 56 68 69 71 71 72 82 84 85 88 89 90 92 93 94 96 97 99 102 103 127 128 134 135 137 144 156 1.1 Range/Omvang = 156 – 56  max – min = 100 kg  answer/antw (2) 1.2 Mode/Modus = 71 kg  answer/antw (1) 1.3 Median/Mediaan = T14 = 93 kg  answer/antw (1) 1.4 Q 1 = T 7 = 82  Q 1 = 82 Q 3 = T 21 = 127  Q 3 = 127 IQR = Q 3 – Q 1 = 127 – 82 = 45 kg  answer/antw (3) 1.5  box/mond  whiskers/snor 50 60 70 80 90 100 110 120 130 140 150 160 (2) 1.6 SD = 25,838 ≈ 25,84 kg  answer/antw (2) 1.7 x = 98,59  x = 98,59 x + 1σ = 98,59 + 25,84 = 124,43 kg  124,43 127 > 124,43 ∴ I agree with this person/Ek stem met die persoon saam.  conclusion/ gevolgtrekking (3) [14] MST GROUP Please turn over/Blaai om asseblief ### Page 3 Mathematics/P2/Wiskunde V2 3 MST/November 2022 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION/VRAAG 2 WEIGHT LOSS OVER 4 WEEKS FREQUENCY CUMULATIVE GEWIGSVERLIES FREKWENSIE KUMULATIEWE IN 4 WEKE f f (IN GRAMS/GRAM) 1 000 < x ≤ 1 500 2 2 1 500 < x ≤ 2 000 3 5 2 000 < x ≤ 2 500 3 8 2 500 < x ≤ 3 000 4 12 3 000 < x ≤ 3 500 5 17 3 500 < x ≤ 4 000 7 24 4 000 < x ≤ 4 500 2 26 4 500 < x ≤ 5 000 1 27 2.1 1250 × 2 + 1750 × 3 + .... + 4750 × 1 Average/Gemiddelde = 27 81250 = 27  81 250 = 3009,259 ≈ 3009,26 g  answer/antw (2) 2.2 OGIVE/OGIEF  grounded 28  upper limits  cumulative Cumulative Frequency/Kumulatiewe frekwensie 26 frequency 24  smooth 22 curve 20 18 16 14  geanker 12  boonste 10 limiete 8  kumulatiewe frekwensie 6  gladde 4 kurwe 2 0 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 Weight loss (in grams) over 4 weeks/ Gewigsverlies (in gram) in 4 weke (4) 2.3 (3200 ; 14)  14 27 – 14 = 13 participants/deelnemers (accept/aanvaar: 12 – 14)  13 (2) [8] MST GROUP Please turn over/Blaai om asseblief ### Page 4 Mathematics/P2/Wiskunde V2 4 MST/November 2022 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION/VRAAG 3 y C(10 ; 11) F α A(–2 ; 3) x G O D(5 ; –1) E 3.1 11 − 3  subst into/in m = gradient form/ AC 10 − (−2) gradiëntvorm 8 2  answer/antw = = 12 3 (2) 3.2 2 m =m = [DE | | CA]  gradients equal/ DE AC 3 y − y1 = m( x − x1 ) y = mx + c gradiënte gelyk 2 2 y − (−1) = ( x − 5) − 1 = (5) + c  subst m & 3 3 (5 ; –1) into eq 2 13 13 of straight line/ y = x− OR/OF − =c 3 3 3 in vgl v rt lyn 2 13  equation/vgl y = x− 3 3 (3) 3.3 2  tan FĜO = m tan FĜO = mAC = AC 3  33,69° FĜO = 33,69° ∴ α = 90° + 33,69° [ext/buite ∠ of ∆] = 123,69°°  answer/antw (3) MST GROUP Please turn over/Blaai om asseblief ### Page 5 Mathematics/P2/Wiskunde V2 5 MST/November 2022 CAPS/KABV – Grade/Graad 11 – Memorandum 3.4.1 midpoint of BE = midpoint of AD [diagonals of rectangle] R midpt v BE = midpt v AD [hoeklyne van reghoek]  − 2 + 5 3 + (−1)  M ;   x-value/waarde  2 2  3   y-value/waarde = M  ; 1 2  (3) 3.4.2 BE = AD [diagonals of rect /hoeklyne van regh] R BE = (−2 − 5) 2 + (3 − (−1)) 2  subst into distance = 49 + 16 formula/subst in = 65 units/eenh afstandsformule  answer/antw (3) [14] MST GROUP Please turn over/Blaai om asseblief ### Page 6 Mathematics/P2/Wiskunde V2 6 MST/November 2022 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION/VRAAG 4 y P R x O S  2 T0 ;− 5   3 4.1 x + ay – a = 0 0 + ay – a = 0 x=0 y=1 y=1 ∴ P(0 ; 1) (2) OR/OF x + ay – a = 0 x y = − +1 x=0 a ∴ P(0 ; 1) y=1 (2) 4.2 OP = 1 unit/eenh ∴ SP = 3 units/eenh ∴ S(3 ; 0)  S(3 ; 0) 3 + a(0) – a = 0 ∴a=3  answer/antw OR/OF (2) PO 1 mPS = − =− 1 OS 3  m =− PS 3 1 y=− x +1 a 1 1 − =− a 3 ∴a=3  answer/antw (2) MST GROUP Please turn over/Blaai om asseblief ### Page 7 Mathematics/P2/Wiskunde V2 7 MST/November 2022 CAPS/KABV – Grade/Graad 11 – Memorandum 4.3 PO 1 1 mPS = − =−  mPS = − OS 3 3 m ×m = −1 (RT ⊥ PS) PS TR  mTR = 3 ∴ mTR = 3 2  equation/vgl ∴ y = 3x – 5 3 (3) 4.4 2 x + 3y = 3 and/en y = 3x – 5 3 2 x + 3(3x – 5 )=3  subst of eq of TR 3 into/subst v vgl v x + 9x – 17 = 3 TR in x + 3y = 3 10x = 20  simplify/vereenv x =2 2 1 x=2 ∴ y = 3(2) – 5 = 3 3 1 1 y= ∴ R(2 ; ) 3 3 (4) 4.5 1 Area of/Opp van ΔPRT = PT× ⊥ h 2 2 6 1 2 3 = (6 )(2) ⊥h 2 3 2 =6 sq units/vk eenh  answer/antw 3 (3) 4.6 PT is the diameter of the circle passing through P, R and T. [PT subtends 90° at circumference]  S/R PT is die middellyn vd sirkel wat deur P, R en T gaan. [PT onderspan 90° op omtrek] 1 2 r = ×6 2 3 1 =3  answer/antw (2) 3 [16] MST GROUP Please turn over/Blaai om asseblief ### Page 8 Mathematics/P2/Wiskunde V2 8 MST/November 2022 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION/VRAAG 5 y 5.1 P(x ; 24) •T 25 β • x O R 5.1.1 2 2 x +y =r 2 2 2 2 x + (24) = 25  subst 2 x = 49 x = −7  answer/antw (2) 5.1.2(a) 24  answer/antw sin β = 25 (1) 5.1.2(b) cos(180° − β ) = − cos β  reduction/reduksie 7 = −( − ) 25 7 = 25  answer/antw (2) 5.1.2(c) tan(− β ) = − tan β  reduction/reduksie 24 = −( ) −7 24  answer/antw = 7 (2) 5.1.3 24 y T sin β = =  equating ratios/ 25 15 stel verh's gelyk 2 ∴ y T = 14  y 5 T 7 x cos β = − = T  equating ratios/ 25 15 stel verh's gelyk 1 ∴ xT = −4 5  x T 1 2 ∴ T( − 4 ; 14 ) 5 5 (4) MST GROUP Please turn over/Blaai om asseblief ### Page 9 Mathematics/P2/Wiskunde V2 9 MST/November 2022 CAPS/KABV – Grade/Graad 11 – Memorandum 5.2 sin 2 x 2 sin x. cos x.(1 + ) sin x cos 2 x  tan x = sin x cos x cos x cos 2 x + sin 2 x cos x = 2 sin x. cos x.( )×  identity/identiteit cos 2 x sin x 1 = 2 cos 2 x( )  simplify/vereenv cos 2 x  answer/antw =2 (4) OR/OF sin 2 x sin x 2 sin x. cos x.(1 + )  tan x = cos 2 x cos x sin x cos x sin 2 x cos x = 2 sin x. cos x.(1 + )× cos 2 x sin x sin 2 x  simplify/vereenv = 2 cos 2 x(1 + ) cos 2 x = 2(cos 2 x + sin 2 x)  identity/identiteit =2  answer/antw (4) 5.3.1 1 − cos 2 A 4(−sinA)  –sin A sin 2 A = − 4sinA  identity/identiteit 1 = − sinA 4  answer/antw (3) 5.3.2 1  correct equation/ − sin 2 x = 0,21 korrekte vgl 4 sin 2 x = −0,84  – 0,84  ref/verw ∠ ref/verw ∠ = 57.14° = 57,14° ∴ 2x = 237,14° + k.360° or 2x = 302,86° + k.360°  237,14° & 302,86° x = 118,57° + k.180° or x = 151,43° + k.180° ; k ∈ Z  118,57° & 151,43°  k.180° ; k ∈ Z (6) [24] MST GROUP Please turn over/Blaai om asseblief ### Page 10 Mathematics/P2/Wiskunde V2 10 MST/November 2022 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION/VRAAG 6 y 1 f f f 2 x –90° –45° 0° 45° 90° 135° –1 –2 6.1.1 b = 2  answer/antw (1) 6.1.2 45°< x ≤ 67,5° OR/OF x ∈ (45° ; 67,5°]  extreme values/ grenswaardes  notation/notasie (2) 6.1.3 The graph of f has to shift 55° to the left to form the graph of h.  x = –10° Die grafiek f het 55° na links geskuif om grafiek h te vorm.  x = 80° x = –10° and/en x = 80° (2) 6.2.1 y 1  TP(–90° ; 1) k  TP(90° ; -1) 1 2 g 1  (–150° ; ) and/en 2 (120° ; –0,87) x –150° –90° –60° 0 30° 90° 120°  y-intercept/afsnit − 12 –1 (4) MST GROUP Please turn over/Blaai om asseblief ### Page 11 Mathematics/P2/Wiskunde V2 11 MST/ November 2022 CAPS/KABV – Grade/Graad 11 – Memorandum 6.2.2 min value/waarde = –1 – 3 = – 4  answer/antw (2) 6.2.3 cos( x + 60°) = − sin x = cos(90° + x)  co-ratio/ko-verh  correct equations x + 60° = −(90° + x) or x + 60° = 90° + x korrekte vgls 2 x = −150° + k .360° no solution x = – 75° + k.180°  x = –75° + k.180° ∴ x = –75° or/of 105°  – 75°  105° (6) cos( x + 60°) = − sin x = − cos(90° − x)  co-ratio/ko-verh x + 60° = 180° − (90° − x) or x + 60° = 180° + (90° − x)  correct equations x = 90° + x + k .360° 2 x = 210° + k .360° korrekte vgls no solution/geen oplossing x = 105° + k.180°  x = 105° + k.180°  – 75°  105° ∴ x = –75° or 105° (6) OR/OF cos( x + 60°) = − sin x  co-ratio/ko-verh sin[90° − ( x + 60°)] = − sin x  correct equations 150° – x = 180° + x or 150° – x = 360° – x korrekte vgls  x –2x = 30° + k.360° no solution/geen oplossing = – 15° – k.180° x = – 15° – k.180°  – 75°  105° ∴ x = –75° or 105° (6) 6.2.4 –75°< x < 105° OR/OF x ∈ (–75° ; 105°)  extreme values/ grenswaardes  notation/notasie (2) 6.2.5 y = cos( x + 60°) = sin[90° − ( x + 60°)]  co-ratio/ko-verh = sin(30° − x) = − sin( x − 30°) ∴ θ = 30°  answer/antw OR/OF (2) By inspections/Deur inspeksie: θ = 30°  answer/antw (2) [21] MST GROUP Please turn over/Blaai om asseblief ### Page 12 Mathematics/P2/Wiskunde V2 12 MST/November 2022 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION/VRAAG 7 P y 2y θ 2y T R Q 7.1 2 2 2 PQ = QR + PR − 2QR.PR. cos θ 2 2 2 y = (2 y ) + (2 y ) − 2(2 y )(2 y ). cos θ  subst into cosine 2 2 2 rule/in cos-reël 4y + 4y − y  cos θ as subject/ cos θ = 2(2 y )(2 y ) as onderwerp 2 7y =  simplify/vereenv 2 8y 7 =  answer/antw 8 (4) 7.2 PT ⊥ QR [∠ in semi-circle/halfsirkel] R PT PT  correct ratio/ sin θ = = PR 2 y korrekte verh ∴ PT = 2y sin θ  answer/antw (3) OR/OF PT ⊥ QR [∠ in semi-circle/halfsirkel] R θ Q̂ = 90° – [∠s opposite = sides/∠e to = sye] 2 θ PT  correct ratio/ sin(90° − ) = 2 y korrekte verh θ  answer/antw PT = y. cos (3 2 [7] MST GROUP Please turn over/Blaai om asseblief ### Page 13 Mathematics/P2/Wiskunde V2 13 MST/November 2022 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION/VRAAG 8 16 2,9 FIGURE 1 FIGURE 2 8.1 Surface area(SA) = 2πrh + πr2 + 2π r2  SA of tin/OA v blik Buite-opp = 2πrh + 3πr2  SA of hemisphere/ BO v hemisfeer = 2π(2,9)(16) + 3π(2,9)2  SA of base/ = 118,03π BO v basis = 370,80 cm2  subst  answer/antw (5) 8.2 1 4 3 Volume of hemisphere/v hemisfeer = × πr  V of hemisphere 2 3 1 4 = × π(2,9) 3 2 3 = 16,26π  subst 80% of/van volume = 0,8 × 16,26π  answer/antw = 40,86 cm3 (3) OR/OF 1 4 3  vol of hemisphere/ Volume of hemisphere/v hemisfeer = × πr 2 3 vol van hemisfeer 1 4 3 80% of/van volume = 0,8 × × πr 2 3 1 4 = 0,8 × × π(2,9) 3  subs 2 3 = 40,86 cm3  answer/antw (3) [8] MST GROUP Please turn over/Blaai om asseblief ### Page 14 Mathematics/P2/Wiskunde V2 14 MST/November 2022 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION/VRAAG 9 L 1 2 58° K 1 2 O N R 9.1 LK̂R = 90° [∠ in semi-circle/halfsirkel] SR (2) 9.2 R̂ = 180° − (90° + 58°) = 32° [∠s/e of/van ∆] SR (2) 9.3 N̂ = 32° [∠s/e in same segment/dieselfde segment] SR (2) [6] MST GROUP Please turn over/Blaai om asseblief ### Page 15 Mathematics/P2/Wiskunde V2 15 MST/November 2022 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION/VRAAG 10 10.1 C O 1 2 x y A B 10.1.1 AO and/en CO are radii/is radiusse  = Ĉ1 = x [∠s opp equal sides/∠e to gelyke sye] S Ô1 =  + Ĉ1 = 2 x [ext/buite ∠ of/van ∆]  S R (3) 10.1.2 B̂ = Ĉ 2 = y [∠s opp equal sides/∠e to gelyke sye] Ô 2 = B̂ + Ĉ 2 = 2 y [ext/buite ∠ of/van ∆] S AÔB = 2 x + 2 y S = 2( x + y ) S = 2(Ĉ1 + Ĉ 2 ) = 2AĈB (3) MST GROUP Please turn over/Blaai om asseblief ### Page 16 Mathematics/P2/Wiskunde V2 16 MST/November 2022 CAPS/KABV – Grade/Graad 11 – Memorandum 10.2 P 2 1 x S 1 2 1 M R 2 1 1 2 N Q 10.2.1 ext ∠ of cyc quad/buite ∠ v koordevh R (1) 10.2.2 MP = QM [radii] Q̂1 = x [∠s opp equal sides/∠e to gelyke sye]  SR (2) 10.2.3 M̂1 = 180° − 2 x [∠s/e of/van ∆]  S R̂ = 90° − x [∠ at centre = 2× ∠ at circumference/  SR midpts∠ = 2 × omtreks∠]] (3) 10.2.4 In ∆NSR: R̂ = 90° − x and N̂ 2 = x Ŝ 2 = 180° − (90° − x + x) [∠s/e of/van ∆] = 90°  S PS = SR [line from centre ⊥ chord/lyn v midpt ⊥ kd]  SR (3) [15] MST GROUP Please turn over/Blaai om asseblief ### Page 17 Mathematics/P2/Wiskunde V2 17 MST/November 2022 CAPS/KABV – Grade/Graad 11 – Memorandum QUESTION/VRAAG 11 B 2 1 T 3 F 2 2 E 1 1 2 3 1 A C 11.1 B̂1 =  [tangent-chord theorem/rkl-kdst] SR  = T̂3 [corresp ∠s/ooeenk ∠e ; TE | | AC] SR ∴ B̂1 = T̂3 (4) 11.2 BE = CE [tangents from same point/rklyne v dieselfde pt] R B̂1 = Ĉ 3 [∠s opp equal sides/∠e to gelyke sye] S Ĉ 3 = T̂3 [ B̂1 = T̂3 ] S ∴TBEC a cyclic quad [converse ∠s in the same segment] R TBEC is koordevh [omgekeerde ∠e in dieselfde segment] (4) 11.3 B̂1 = T̂2 [∠s in the same segment/∠e in dieselfde segment] SR B̂1 = T̂3 [proven/bewys in 11.1] ∴ T̂2 = T̂3 ∴ ET bisects/halveer BT̂C (2) 11.4 B̂ 2 = Ê 2 [tangent-chord theorem/rkl-kdst] SR Ĉ 2 = Ê 2 [∠s in the same segment/∠e in dieselfde segment] R ∴ TB = TC [sides opposite equal ∠s/sye to gelyke ∠e] R (4) 11.5 Ĉ1 = T̂2 [alternate/verwiss ∠s/e ; TE | | AC] S ∴ Ĉ1 =  ∴ AT =TC [sides opposite equal ∠s/sye to gelyke ∠e] S T is a point that is equidistant from A, B and C on the circle/ S T is 'n punt wat ewever is vanaf A,B en C op die sirkel. ∴ T is the centre of the circle/T is die middelpunt van die sirkel. 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