Mathematics Grade 11 P2 Eastern Cape November 2025 Memo

NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 11 NOVEMBER 2025 MATHEMATICS P2/WISKUNDE V2 MARKING GUIDELINE/NASIENRIGLYN MARKS/PUNTE: 150 This marking guideline consists of 16 pages./ Hierdie nasienriglyn bestaan uit 16 bladsye.

2 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2025) QUESTION 1/VRAAG 1 Data Set/Datastel: Prizes/Pryse 3 600 4 500 5 400 6 200 x 12 300 15 800 y 43 800 87 500 1.1 x + 12300 ✓method/metode = 9700 2 x = 7100 ✓ answer/antwoord (2) 1.2 y + 186200 correct equation = 21770 10 korrekte vergelyking y = 31500  y + 186200 answer/antwoord (3) 1.3 outlier/𝑢𝑖𝑡𝑠𝑘𝑖𝑒𝑡𝑒𝑟 = 87 500 ✓ answer/antwoord (1) 1.4  = 25258,19 ✓✓ answer/antwoord (2) 1.5 Mean will increase by R200 ✓ increase by R200 Gemiddelde sal vermeerder met R200 vermeerder met R200 Median will increase by R200 ✓ increase by R200 Mediaan sal vermeerder met R200 vermeerder met R200 Standard deviation will remain the same ✓ remain the same Standaardafwyking sal dieselfde bly bly dieselfde (3) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/NOVEMBER 2025) MATHEMATICS P2/WISKUNDE V2 3 QUESTION 2/VRAAG 2 Cummulative Frequency graph(Ogive) Kumulatiewe Frekwensiegrafiek (Ogief) 150 140 130 120 110 100 Cummulative Frequency 90 80 Kumulatiewe Frekwensie 70 60 50 40 30 20 10 0 0 10 20 30 40 50 60 70 Tips in Rands Fooitjies / "Tips" in Rand 2.1 140 employees/werknemers ✓ answer/antwoord (1) 2.2 30  x  40 ✓ answer/antwoord (1) 2.3 Reading from the graph / lees vanaf grafiek ✓ Reading/lees 32 employees/werknemers  answer/antwoord (2) ✓ 118 − 24 2.4 118 − 24 = 94 ✓ answer/antwoord (2) th 2.5 25 percentile is at 35 = R25 ✓ Q1 25ste persentiel is by 35 = R25 ✓ Q3 75th percentile is at 105 = R38 75ste persentiel is by 105 = R38 (2) 2.6 𝐼𝑄𝑅/𝐼𝐾𝑊 = 𝑄3 − 𝑄1 = 𝑅38 − 𝑅25 ✓ method/metode = 𝑅13 ✓ answer/antwoord (2) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

4 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2025) QUESTION 3/VRAAG 3 y V U W(1 ; 6) S(13 ; 3) P(0 ; 2) Q x O R(3 ; 0) 3.1 Q ( 3;2)  answer/antwoord (1) 3.2 3−2 mQS =  correct substitution 13 − 3 1 korrekte vervanging =  answer/antwoord 10 (2) 3.3 1  gradient of VW mVW = mQS = 10 gradiënt van VW 1 6= (1) + c y−6 = 1 (x − 1)  substitution/vervanging 10 10 59 c= OR/OF 10 1 59 1 59 y= x+ y = x+ 10 10 10 10  answer/antwoord (3) 3.4 3 5 1 59  equating equations x− = x+ gelykstel van vergelykings 2 2 10 10 7 42 x=  simplification 5 5 vereenvoudiging xU = 6  x-value / x-waarde 3 5 13 yU = (6) − = 2 2 2  y-value / y-waarde 1 + xV 13 6 + yV 6= ; =  method/metode 2 2 2 11 = xV ; 5 = yV (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/NOVEMBER 2025) MATHEMATICS P2/WISKUNDE V2 5 3.5  13 + 1 3 + 6   correct substitution MW S ;  korrekte vervanging  2 2   9  coordinates of midpt of WS MW S 7 ;   2 koördinate van midpt van WS  11 + 3 7 + 2  M QV  ;   2 2   9  coordinates of midpt of QV M QV  7 ;   2 koördinate van midpt van QV WVSQ is a parm (Diag. bisect each other.)  reason/rede WVSQ is ’n ||m (Hoeklyne halveer mekaar) (4) 3.6 WV = (1 − 11)2 + (6 − 7)2  correct substitution korrekte vervanging = 101  answer/antwoord (2) 3.7 Let inclination of WV to be θ and QU be α Laat inklinasie van WV,  wees en van QU,  wees 1  tan  = mW V tan  = mW V = 10  = 5,71 0  size of θ / grootte van  3 tan  = mQU =  size of α / grootte van  2  = 56,31 0  answer/antwoord WUQ ˆ = UQS ˆ = 50, 60 alt s, WV QS   [verv. e, WV || QS] (4) 3.8 101  WU WU = 2 2 UQ = (6 − 3)2 +  13 − 2  = 3 13  UQ 2  2  correct subst. in to Area rule 1 3 13 101 A. of QUW =    sin ( 50, 60o ) korrekte verv. in oppervl. reël 2 2 2 = 10,50  A. of ∆. / Oppervl. van  Area of PQRO 23 = Area of ΔQUW 10 .50  correct subst. in A. of rect. korrekte verv. in Opp. van regh 4 =  answer/antwoord 7 (6) [27] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

6 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2025) QUESTION 4 / VRAAG 4 4.1.1 3 tan 45o = t  correct trig ratio t =3 korrekte trig. verhouding  t-value/waarde (2) 4.1.2 3 t 3 sin 45o = cos 45o = =  correct trig ratio k OR /OF k k korrekte trig. verhouding k =3 2 k =3 2  k-value/waarde (2) 4.2.1 8 cos  − 2 = 0 2 1 cos 2  =  simplification 4 vereenvoudiging 1 cos  = 2  answer/antwoord (2) 4.2.2 r=2 y= 3   diagram/diagram x =1 OR/OF OR/OF y = 2 2 − 12 Pyth y= 3 = 3  3 2  1 2   4 tan  − cos  = 4 2 2   −    correct substitution  1   2   korrekte vervanging   = 11  answer/antwoord (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/NOVEMBER 2025) MATHEMATICS P2/WISKUNDE V2 7 4.3.1 cos x. tan x  ( 2 sin x. cos x − 90 0 + cos x. cos( − x) )  sin x cos x. sin x cos x  = cos x 2sin x.(sin x ) + cos x(cos x )  sin x  cos x sin x =  2 sin x + cos 2 x 2   simplification sin x vereenvoudiging =  sin 2 x + cos2 x = 1 2(1) sin x  answer/antwoord = 2 (6) 4.3.2 Period/Periode = 360o  answer/antwoord (1) 4.4 cos 42o sin 48o − tan 2 ( −45o ) cos 2 132o  sin 48o = cos 42o cos 42 .cos 42 − ( − tan 45 ) o o o 2  ( − tan 45o ) 2 = ( cos (90o + 42o )) 2 cos 2 42o − ( −1) 2 =  (− sin 4 2)2 ( − sin 42o ) 2 cos 2 42o − 1  cos2 42o = 1 − sin 2 42o = sin 2 42o 1 − sin 2 42o − 1 answer/antwoord = sin 2 42o = −1 (5) 4.5.1 LHS/LK = tan  − sin  .cos  sin  sin  = − sin  .cos   cos  cos  sin  − sin  cos 2  =  simplification cos  vereenvoudiging sin  (1 − cos 2  ) =  common factor cos  gemeenskaplike faktor = tan  .sin 2  = RHS / RK (3) o 4.5.2  = −90 and / en  = 90o o   = −90 o   = 90 (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

8 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2025) 4.6 3sin  = −2 cos 2  1− sin 2  2 (1 − sin 2  ) + 3sin  = 0 2 − 2sin 2  + 3sin  = 0  standard form 2sin 2  − 3sin  − 2 = 0 standaardvorm ( 2sin  + 1)( sin  − 2 ) = 0  factors/faktore 1 sin  = − or / of sin   2  both equations 2 beide vergelykings  = 210o + 360.k  = 330o + 360.k , k    = 210o + 360.k   = 330o + 360.k , k   (6) [32] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/NOVEMBER 2025) MATHEMATICS P2/WISKUNDE V2 9 QUESTION 5/VRAAG 5 5.1 a =1  value of a / waarde van a q=0  value of q / waarde van q (2) 5.2 y   −2;0 or / of − 2  y  0  correct critical value korrekte kritieke waarde  correct notation korrekte notasie (2) 5.3 Amplitude = 1  correct answer korrekte antwoord (1) 5.4 o  answer / antwoord x=0 (1) 5.5 −90o  x  90o  correct answer korrekte antwoord (2) 5.6 h ( x ) = g ( x + 90 ) o = cos ( x + 90o )  cos ( x + 90o ) = − sin x  − sin x (2) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

10 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2025) QUESTION 6/VRAAG 6 Q R N R  S P x   T O 6.1 ˆ = + SPO alt. s, verw. e : SP OT   SP̂O =  +  ∴ ˆ =  − QOP 𝑂𝑄̂ 𝑃 = 180𝑜 − (𝜃 + 𝛼) − (𝛽 − 𝜃)[∠𝑠in𝑎𝛥/∠𝑒𝑖𝑛𝛥]  using sum of angles in  = 180o − (  +  ) gebruik van som van e in  (2) 6.2 OQ OP  application of sine rule = ˆ ˆ sin P sin OQR toepassing van sinusreël OQ x =  correct substitution sin ( +  ) sin 180 − (  +  )  o korrekte vervanging x.sin ( +  ) OQ =  reduction sin ( +  ) sin (  +  ) reduksie sin ( +  ) (3) 6.3 5 2 x o =  correct substitution sin 45 sin 90o korrekte vervanging x = 10  answer/antwoord (2) [7] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/NOVEMBER 2025) MATHEMATICS P2/WISKUNDE V2 11 QUESTION 7/VRAAG 7 H x L 7.1 L = 2x  L = 2x H = 3+ x  H = 3+ x (2) 7.2 SA = 2(2 x  x ) + 2(x(3 + x )) + 2(2 x(3 + x ))  correct substitution / korrekte vervanging = 4 x + 6 x + 2 x + 12 x + 4 x 2 2 2 simplification / vereenvoudiging = 10 x 2 + 18 x answer / antwoord (3) 7.3 10 x + 18 x = 784 2  correct substitution/ korrekte vervanging 10 x 2 + 18 x − 784 = 0  standard form/standaardvorm 5 x 2 + 9 x + 392 = 0 ( 5 x + 49 )( x − 8) = 0  factors/faktore 40  correct answer/korrekte antwoord x or x = 8 5 (4) 7.4 V = (8)(2  8)(3 + 8)  correct substitution into volume / = 1408 korrekte vervanging in volume answer/antwoord (2) [11] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

12 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2025) QUESTION 8/VRAAG 8 O A C B 8.1 Constructions: Join OA and OC  construction Konstruksies: Verbind OA en OC konstruksie Proof in ∆ AOB and ∆ COB Bewys in ∆ AOB en ∆ COB 1.AO = CO  radii  /[radiusse]  S/R  S/R ˆ = CBO 2.ABC ˆ = 90o  s on a str. line  / [e op reguitlyn] 3.OB = OB common  / [ gemeen]  S  R AOB  COB  RHS  / [90o Sk S ]  AB = BC (5) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/NOVEMBER 2025) MATHEMATICS P2/WISKUNDE V2 13 8.2 S x x+2 T U R 3 cm 4 cm Q 4 cm P 8.2.1 PQ = QR = 4 cm line from centre ⊥ to chord S R [lyn vanaf middelpunt ⊥ op koord] correct substitution/ UR = 4 + 3 2 2 Pyth. theo [ [Stelling van Pythagoras] [Pyth.theo]/ korrekte vervanging =5  answer/antwoord (4) 8.2.2 TR = 2 ( UR ) = 10  diameter  / [ middellyn] Ŝ = 900  in semi − circle /[ in semi − sirkel ]  S R 102 = x 2 + ( x + 2 )  Pyth. theo  /[ Stelling van Pythagoras ]  correct substitution/ 2 korrekte vervanging 100 = x 2 + x 2 + 4 x + 4 2 x 2 + 4 x − 96 = 0  standard form / x 2 + 2 x − 48 = 0 standaardvorm ( x − 6 )( x + 8) = 0  factors / faktore  correct value of x / x = 6 or / of x  −8 korrekte waarde van x  TS = 6 (6) [15] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

14 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2025) QUESTION 9 / VRAAG 9 G B 1 A 2 C 3 4 1 5 6 O 2 15o 1 2 D 3 F E 9.1.1 Ô1 = 30o  at centre = 2 at circumf. S R [Middelpunts  = 2 x Omtreks ] (2) 9.1.2 Ô2 = 150 o s on a str.line /[e op reguitlyn] S Â = 75o  at centre = 2 at circumf.  S/R [Middelpunts  = 2 x Omtreks ] (2) 9.1.3 D̂ = 105 o opp. s of a cyclic quad  S R [Teenoorst. e van koordevierhoek] (2) 9.1.4 ˆ = 75o ECF  tan chord theo. / [raaklyn-koord stelling]  S R (2) 9.1.5 ˆ = 90 OCG o  rad ⊥ tan  / [radius ⊥ raaklyn] S R (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

(EC/NOVEMBER 2025) MATHEMATICS P2/WISKUNDE V2 15 9.2 Q 1 P 1 2 2 4 3 M T 2 1 1 2 2 R 1 3 N S 9.2.1 Pˆ1 = Rˆ 1  tan chord theo / [raaklyn − koord stelling ] S R ˆP = MNQ ˆ corresp. s,/ ooreenk.e : MN PR  S R 1 Any 2 angles with Rˆ = PTQ 1 ˆ s same seg  / [ein dies. segment ] reasons / Enige 2 hoeke met redes (4) 9.2.2 ˆ ˆ =Q SPQ  tan chord theo /[raaklyn − koord stelling ] S R ˆ =M SPQ ˆ corresp. s,/ ooreenk . e : MN PR  S 1 ˆ =M Q ˆ 1 1 MNQP is a cyclic quad  conv.ext.  of a cyclic quad  R MNQP is ' n koordevh. [omgek . buite  van koordevh] (4) 9.2.3 Pˆ3 = N ˆ 2 alt. s, MN PR  / [verw. e : MN PR]  S/R N̂ 2 = ˆ Q1 s same seg  / [ein dies. segment ] S R ˆ = Pˆ Q1 4  tan chord theo / [raaklyn − koord stelling ] S/R  ˆP3 = ˆP4 PN bisects ˆ SPR / PN halveer ˆ SPR (4) [22] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief

16 MATHEMATICS P2/WISKUNDE V2 (EC/NOVEMBER 2025) QUESTION 10/VRAAG 10 A F 1 2 G 3 E B 2 1 3 H D C ˆ =A G ˆ corresp. s / ooreenk. e , GC AB S R 3 ˆ = Fˆ A s subt by = chords  / [e onderspan deur = koorde] S R ˆ = Fˆ G 3 R AB FE corresp. s, = / [ooreenkomstige e =] [5] TOTAL/TOTAAL: 150 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief