Mathematics Grade 11 Paper 2 Eastern Cape November 2024 Memo

NATIONAL SENIOR CERTIFICATE NASIONALE SENIORSERTIFIKAAT GRADE/GRAAD 11 NOVEMBER 2024 MATHEMATICS P2/WISKUNDE V2 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 17 pages. Hierdie nasienriglyn bestaan uit 17 bladsye. Hosted by www.ecexams.co.za.

2 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2024) QUESTION 1 / VRAAG 1 Data Set / Datastel: 16 28 41 41 42 52 54 55 58 59 60 62 64 99 1.1 Mode / Modus = 41 ✓ mode / modus (1) 1.2 Outlier / Uitskieter = 99 ✓ outlier / uitskieter (1) 54 + 55 ✓ dividing a sum by 2 1.3 Median / Mediaan = 2 deel ʼn som deur 2 Median / Mediaan = 54,5 (Answer only full marks) ✓ answer / antwoord (Slegs antwoord – volpunte) (2) 1.4 Q1 = 41 ✓ Q1 Q3 = 60 ✓ Q3 IQR / IKW = Q3 − Q1 = 60− 41 =19 ✓ answer / antwoord (3) 1.5 ✓correct min and max korrekte min en maks ✓ correct box and whisker diagram. korrekte mond-en- snordiagram (2) 1.6 The data is skewed to the left or negatively skewed. ✓ for the correct Die data is skeef na links of negatief skeef. comment vir korrekte opmerking (1) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

(EC/NOVEMBER 2024) MATHEMATICS P2/WISKUNDE V2 3 QUESTION 2 / VRAAG 2 2.1 Weight of players Frequency Cumulative frequency ✓ 4 and/en Gewig van boksers Frekwensie Kumulatiewe frekwensie 13 35  x  55 1 1 55  x  75 3 4 ✓ 19 and/en 75  x  95 9 13 22 95  x  115 6 19 115  x  135 3 22 135  x  155 1 23 (2) 2.2 The total number of boxers is 23 ✓ for Die totale aantal boksers is 23 answer/ vir antwoord (1) 2.3 1 45 + 3  65 + 9  85 + 6  105 + 3  125 + 1 145 ✓ f x xi x= 23 2155  2 155 = 23  answer/ x = 93,70 antwoord (3) 2.4 Cumulative frequency graph (ogive) Kumulatiewefrekwensiegrafiek (ogief) 25 ✓ correct grounding/ korrekte 20 Cumulative Frequency anker 15 ✓ plotting against the upper limits/ Kumulatiewe Frekwensie 10 afsteek by boonste 5 limiete 0 ✓ correct 15 35 55 75 95 115 135 155 175 shape/ Mass in kilograms / Gewig in kilogram korrekte vorm (3) 2.5 No. of boxers in the interval of 75 x  100 will be 15 − 4 = 11  15 or/of Aantal boksers in die interval 75 x  100 sal 15 − 4 = 11wees reading from the graph / lees vanaf die grafiek  answer / antwoord (2) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

4 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2024) QUESTION 3 / VRAAG 3 MARKING GUIDELINE FOR: R(3;4) 3.1 4−0  correct substitution/ mPR = korrekte vervanging 3−0 4 =  answer / antwoord 3 (2) 3.2 4 mPR = (P, O and R are collinear points) 3  gradient of/ gradiënt van PR (P, O en R is saamlynige punte) 4  substitution by R or O/ y−4= ( x − 3) vervanging met R of O 3 4  equation of PR/ = x 3 vergelyking van PR (3) 3.3 4  for equating the equations/ 6 x + 14 = x 3 vir gelykstelling van 4 vergelykings 6 x − x = − 14 3  simplification/ 14 x = − 14 vereenvoudiging 3 x = −3 y = 6 ( −3) +14 y = −4  x–value/waarde P ( −3; − 4 )  y–value/waarde (4) 3.4 0 = 6 ( x ) + 14  y=0 7 x=−  x-coordinate/ 3 x-koördinaat  7   S − ; 0   3  (1) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

(EC/NOVEMBER 2024) MATHEMATICS P2/WISKUNDE V2 5 MARKING GUIDELINE FOR: R(2;3) 3.1 3−0  correct substitution/ mPR = 2−0 korrekte vervanging 3 =  answer / antwoord 2 (2) 3.2 3 𝑚𝑃𝑅 = 2 (P, O and R are collinear points)  gradient of/ gradiënt van PR (P, O en R is saamlynige punte) 3  substitution by R or O/ 𝑦−3= (𝑥 − 2) vervanging met R of O 2  equation of PR/ 3 vergelyking van PR = 𝑥 (3) 2 3.3 3  for equating the equations/ 6𝑥 + 14 = 𝑥 2 vir gelykstelling van 3 vergelykings 6𝑥 − 𝑥 = −14 2 9  simplification/ 𝑥 = −14 2 vereenvoudiging 28 𝑥=− 9 28 𝑦 = 6 (− ) + 14 9  x–value/waarde 14 𝑦=− 3  y–value/waarde 28 14 𝑃 (− ; − ) 9 3 (4) 3.4 0 = 6 ( x ) + 14  y=0 7 x=−  x-coordinate/ 3 x-koördinaat  7   S − ; 0   3  (1) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

6 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2024) QUESTION 4 / VRAAG 4 4.1.1 −7+3 0+8 M= ; 2 2 = −2 ; 4  x-value/waarde P(− 2 ; 4)  y-value/waarde (2) 4.1.2 8−0  substitution / vervanging m= 3+ 7 4 =  answer / antwoord 5 (2) 4.1.3 4 tan  = 5 4  tan  =  = 38,66 0 5  answer / antwoord (2) 4.2 4−2 mPQ =  correct substitution/ 1 −2 + korrekte vervanging 2 4 =−  answer / antwoord 3 4 𝑚BC = 𝑚PQ = − gradients||lines =  gradient of BC/ 3 gradiënt van BC 𝑔𝑟𝑎𝑑𝑖ë𝑛𝑡𝑒||𝑙𝑦𝑛𝑒 = 4  substitute B(3 ; 8) and m 𝑦 − 8 = − (𝑥 − 3) vervang B(3 ; 8) en m 3 4 𝑦 = − 𝑥 + 12  equation / vergelyking 3 (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

(EC/NOVEMBER 2024) MATHEMATICS P2/WISKUNDE V2 7 4.3.1 4 y=0 − x + 12 = 0 3 x =9 x=9 AC = 16  AC = 16 (3) 4.3.2 AB = (3 + 7) + (8 − 0) 2 2  AB = 164 = 164 1  correct substitution in A of ΔABC / A.of/van ΔABC =  164 16sin 38, 650 2 korrekte vervanging in A van ΔABC = 63,99  63,99 1 164  correct substitution in A of ΔAPQ A. of/van Δ APQ =   8sin 38, 650 2 2 korrekte vervanging in A van ΔAPQ = 16, 00  16,00  answer / antwoord A. of/van PBCQ = 47,99 (6) [20] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

8 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2024) QUESTION 5 / VRAAG 5 5.1 − 3 tan  = 5 − 5 tan  = 3 − 5 y  tan  = 3  3 x - 5 14  diagram in the correct quadrant diagram in die korrekte kwadrant ( 3) + ( − 5 ) 2 2 r= ( Pyth)  for / vir r = 14 = 14 sin 2  − cos 2  2 2 − 5  3  =   −   14   substitution / vervanging  14  2 =−  answer / antwoord 7 (5) 5.2 r = x +y 2 2 2 ( Pyth) (1)2 = (k )2 + ( y )2 y2 = 1 − k2 y = 1 − k2 41 1 1 −k 2 49 k 5.2.1 sin 131 = sin (180 − 49) = sin 49  sin 49 = 1− k 2  answer / antwoord (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

(EC/NOVEMBER 2024) MATHEMATICS P2/WISKUNDE V2 9 5.2.2 1 − cos 2 410 = sin 2 410  for sin 2 41 = k2  answer / antwoord OR/OF OR/OF ( ) 2 1 − cos 2 410 = 1 − 1− k 2  substitution in terms of k/ vervanging in terme van k = 1 −1 + k 2 = k2  answer / antwoord (2) 5.3 tan (180 − x ) .cos ( − x ) + sin ( 360 − x ) cos ( 90 − x ) 0 2 0 sin (180 − x ) − tan x. cos x + sin 2 x .sin x  − tan x  cos x = sin x  sin 2 x  sin x sin x sin x − .cos x + sin 3 x  cos x cos x = sin x − sin x + sin 3 x =  simplification/ sin x vereenvoudiging − sin x (1 − sin 2 x ) = sin x  answer / antwoord = − cos x 2 (7) 5.4 ( ) sin − 15 . cos 75 + tan 75 . cos 75 . cos 165 0 0 0 0 0 sin 75 0 ( = − cos 75 cos 75 + 0 ) sin 75 0 cos 75 0 . cos 75 0 − sin 75 0 0 ( )  − cos 75  cos 75 0  − sin 750 = − cos 75 − sin 75 2 0 2 0 ( = − cos 2 75 0 + sin 2 75 0 )  common factor/ gemene faktor = −1  identity / identitieit cos2 750 + sin 2 750 = 1 OR/OF OR/OF ( ) sin − 15 0 . cos 75 0 + tan 75 0. cos 75 0. cos 165 0  – sin15° and/en sin15° cos 150  sin 150 , [sin15° and/ ( ) ( ) 0 cos 15 = − sin 15 0 .sin 15 0 + 0 .sin 15 0 − cos 15 0 en(− cos 1 50 )]  sin 15  common factor/ gemene faktor = − sin 2 15 0 − cos 2 15 0  identity / identiteit ( = − sin 2 15 + cos 2 15 ) cos2 150 + sin 2 150 = 1 = −1 (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

10 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2024) 5.5 3cos x 3 + 3 tan x = 1 + sin x cos x sin x 3cos x  identity / identiteit LHS / LK = + 3 tan x cos x 1 + sin x 3cos x 3sin x = + 1 + sin x cos x 3cos 2 x + 3sin x + 3sin 2 x  simplification / = vereenvoudiging cos x (1 + sin x ) 3 ( cos 2 x + sin 2 x ) + 3sin x  square identity / = vierkantsidentiteit cos x (1 + sin x ) 3 (1 + sin x )  common factor / = cos x (1 + sin x ) gemene faktor 3 = cos x (4) 5.6 sin2 𝑥 − 3 cos2 𝑥 = 0  isolating trig ratios/ sin2 𝑥 = 3 cos2 𝑥 isoleer trig. verhoudings sin2 𝑥 =3 cos 2 𝑥  tan 2 x = 3 tan2 𝑥 = 3 tan 𝑥 = ±√3  correct equations/ 𝑥 = ±60° korrekte vergelykings 𝑥 = 60° + 𝑘. 180 or/𝑜𝑓 𝑥 = −60° + 𝑘. 180°, 𝑘 ∈Ζ  𝑥 = 60° + 180°. 𝑘 or/of 𝑥 = 120° + 180° 𝑘, 𝑘 ∈ Ζ  𝑥 = −60° + 180°. 𝑘, 𝑘 ∈ 𝛧 or /of 𝑥 = 12 + 180°. 𝑘, 𝑘 ∈ 𝛧 OR/OF OR/OF sin2 𝑥 − 3 cos2 𝑥 = 0 sin2 𝑥 − 3(1 − sin2 𝑥) = 0  cos 2 x =1 − sin 2 x sin2 𝑥 − 3 + 3 sin2 𝑥 = 0 4 sin2 𝑥 − 3 = 0 standard form/ 3 standaardvorm sin2 𝑥 = 4 √3  correct equations/ sin 𝑥 = ± korrekte vergelykings 2 𝑥 = 60° + 360. 𝑘 or/𝑜𝑓 120° + 360°. 𝑘, 𝑘 ∈ Ζ  𝑥 = 60° + 360. 𝑘 or/𝑜𝑓 𝑥 = 120° + 360°. 𝑘 both 𝑥 = 240° + 𝑘. 360° or/𝑜𝑓 𝑥 = 300° + 𝑘. 360° quads/beide kwadrante  𝑥 = 240° + 𝑘. 360° 𝑥 = 300° + 𝑘. 360°, 𝑘 ∈ 𝛧 both quads / beide kwadrante (5) [30] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

(EC/NOVEMBER 2024) MATHEMATICS P2/WISKUNDE V2 11 QUESTION 6 / VRAAG 6 6.1 Period is / Periode is 360o  answer/ antwoord (1) 6.2 y  0 ; 2 or/of 0 y2  correct interval/ korrekte interval  correct notation/ korrekte notasie (2) 6.3 ( ) h( x ) = cos 60 + x + 30 + 1 0 0 = cos(90 + x ) + 1 0  cos(90° + x)  − sin x = − sin x + 1 (2) 6.4 h ( x)  x-intercept/ afsnit  y-intercept/ y-afsnit  correct shape/ korrekte vorm (3) 6.5.1 𝑥 = 150°  answer/ antwoord (1) 6.5.2 60° ≤ x ≤ 150°  correct interval/ korrekte interval  correct notation/ korrekte notasie (2) 6.5.3 𝑥 = 30°  answer/ antwoord (1) [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

12 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2024) QUESTION 7 / VRAAG 7 7.1 5 cos  = SP 5 5 SP =  SP = cos  cos  (1) 7.2 PR̂Q =  (alt s ; RQ || SP )  PR̂Q =  sin  sin  =  application of sine rule/ RP 5 5sin  toepassing van sinusreël RP = sin  answer / antwoord (3) 7.3 1 area of/van ΔRPS = (RP)(PS) 𝑠𝑖𝑛 𝛽 2  correct formula for area rule/ 1 5 𝑠𝑖𝑛 𝛼 5 korrekte formule vir oppervlaktereël = ( )( ) 𝑠𝑖𝑛 𝛽 2 𝑠𝑖𝑛 𝛽 𝑐𝑜𝑠 𝛼  substitution / vervanging 25 𝑠𝑖𝑛 𝛼 = 2 𝑐𝑜𝑠 𝛼 25 𝑡𝑎𝑛 𝛼  simplification / vereenvoudiging ∴ area of/van ΔRPS = 2 (3) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

(EC/NOVEMBER 2024) MATHEMATICS P2/WISKUNDE V2 13 QUESTION 8 / VRAAG 8 8.1 (ℎ𝑠 )2 = 𝑟 2 + ℎ2  correct formula/ (13)2 = (5)2 + ℎ2 korrekte formule ℎ2 = 169 − 25  substitution/ ℎ2 = 144 vervanging ℎ = 12 cm  answer / antwoord (3) 8.2 height of the wholeshape / hoogte van die hele vorm =r +h = 5 + 12  answer / antwoord = 17cm (1) 8.3 volume of the whole shape / volume van die hele vorm 2 1  correct formula/ =  r 3 +  r 2h korrekte formule 3 3 2 1 =  ( 5 ) +  ( 5 ) (12 ) 3 2  substitution/ 3 3 vervanging 550 =  3 = 575,96 cm3  answer / antwoord (3) 8.4 outer surface area of the whole shape/  correct formula/ 𝑏𝑢𝑖𝑡𝑒 − 𝑜𝑝𝑝𝑒𝑟𝑣𝑙𝑎𝑘𝑡𝑒 𝑣𝑎𝑛 𝑑𝑖𝑒 ℎ𝑒𝑙𝑒 𝑣𝑜𝑟𝑚 korrekte formule = 2𝜋𝑟 2 + 𝜋𝑟ℎ𝑠  substitution/ = 2𝜋(5)2 + 𝜋(5)(13) vervanging = 115𝜋 = 361,28 cm2  answer / antwoord (3) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

14 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2024) QUESTION 9 / VRAAG 9 9.1.1 Construction: Draw AO and OC  constructions 𝑂̂1 = 2𝐵̂ [∠at centre = 2 × ∠at circumf]  S/R 𝑂̂2 = 2𝐷̂ [∠at centre = 2 × ∠at circumf] S ∴ 𝑂̂1 + 𝑂̂2 = 3600 [∠𝑠 around a point] ∴ 2𝐵̂ + 2𝐷 ̂ = 3600  S/R ∴ 𝐵̂ + 𝐷 ̂ = 1800 S Konstruksie: Teken AO en OC  konstruksies 𝑂̂1 = 2𝐵̂ [𝑀𝑖𝑑𝑑𝑒𝑙𝑝𝑢𝑛𝑡𝑠∠ = 2 × 𝑂𝑚𝑡𝑟𝑒𝑘𝑠∠]  S/R 𝑂̂2 = 2𝐷̂ [𝑀𝑖𝑑𝑑𝑒𝑙𝑝𝑢𝑛𝑡𝑠∠ = 2 × 𝑂𝑚𝑡𝑟𝑒𝑘𝑠∠] S ∴ 𝑂̂1 + 𝑂̂2 = 3600 [∠𝑒 𝑟𝑜𝑛𝑑𝑜𝑚 ′𝑛 punt] ∴ 2𝐵̂ + 2𝐷 ̂ = 3600  S/R ∴ 𝐵̂ + 𝐷 ̂ = 1800 S (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

(EC/NOVEMBER 2024) MATHEMATICS P2/WISKUNDE V2 15 9.2 9.2.1 Ŝ = 1150 [ at the centre = 2×  at the circumference] S R  Middelpunts  = 2  Omtreks  (2) 9.2.2 Q̂ = 650 [opposite s of cyclic quad] S R teenoorst. e van ' n koordevierhoek  (2) 9.2.3 P̂1 + 200 = 850 [ext. of cyclic quad] S R buite.  van ' n koordevierhoek  P̂1 = 650 (2) 9.3 9.3.1 Â1 = 500 s in a str.line / [e op ' n reguitlyn] S R̂ = 1100 s in a   /[ein ' n driehoek ] S P̂ = 700 opp s of a cyclic quad  S R [teenoorst. e van ' n koordevierhoek ] OR/OF OR/OF R̂ = 1300 − 200 = 1100 ext  of a Δ / [buite  van ] S R P̂ = 700  opp. s of a cyclic quad  / [teenoorst.e van ' n kv] S R (4) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

16 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2024) QUESTION 10 / VRAAG 10 10.1 10.1.1 Â1 = 37 0  tan chord  / [raaklyn − koord stelling ] S R ˆ = 740   at centre = 2   at circumf  ˆ = 2A O S R 1 1 [ Middelpunts  = 2  Omtreks ] (4) 10.1.2 ˆ = 90 ABP 0   in a semi-circle / [ in ' n semi − sirkel ] S R (a) B̂1 = 900 s in a str.line  / [e op ' n reguitlyn] Ô3 = 900 [given] / [ gegee] S OACB is a cyclic quadrilateral OACB is ’n koordevierhoek R [converse s same seg]/[Omgekeerde e in dies segment ] (4) 10.1.2 ˆ =A C 2 ˆ = 370 1 s sameseg  / [e in dies segment ] S R (b)  OC PN alt. s = / [verw. e =] R OR/OF OR/OF ˆ = 900 APN diameter ⊥ tan  / [middellyn ⊥ raaklyn] S R Oˆ 3 = 900 given  /[ gegee]  OC PN corresp. s = /[ooreenk. e =] R (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.

(EC/NOVEMBER 2024) MATHEMATICS P2/WISKUNDE V2 17 10.2 10.2.1 Â3 = 900 [tan ⊥ rad] / [raaklyn ⊥ radius] S R Ŝ1 = 900  tan -chord theorem  / [raaklyn − koord stelling ] S R Ĉ = 900  in semi-circle / [ in ' n semi − sirkel ] S R (6) 10.2.2 Sˆ 1 = C ˆ both / beide = 90  0 MS BC correspo. s = / ooreenkomst.e = R (1) 10.2.3 AS:SC =1:1 line drawn from the centre S R [lyn getrek vanaf middelpunt ] (2) 10.2.4 MS = 3 midpoint theorem / [middelpunt stelling ] S R (2) 10.2.5 p 2 = ( p − 1) + 32  apply Pyth theo 2 Pyth.theorem / stelling toepassing van Pyth p2 = p2 − 2 p + 1 + 9 stelling 2 p = 10  simplification/ p=5 vereenvoudiging  answer / antwoord (3) [25] TOTAL / TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief Hosted by www.ecexams.co.za.