# Mathematics Grade 11 P1 National Exemplar November 2020 Memo > Grade 11 Mathematics Paper 1 memorandum from the National Exemplar November Exam 2020, 15 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/grade-11-mathematics-2020-national-exemplar-november-exam-memorandum-paper-1/1069fa12-57c0-4875-a06a-ca2128f1197f - Updated: 2026-09-28 - Subject: Mathematics - Grade: Grade 11 - Document type: Memorandum - Year: 2020 - Exam period: National Exemplar November Exam - Paper: 1 - Curriculum: CAPS - Pages: 15 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 11](https://www.hlayiso.com/library/grade-11) › [Mathematics](https://www.hlayiso.com/library/grade-11/mathematics) › [Mathematics Grade 11 P1 National Exemplar November 2020 Memo](https://www.hlayiso.com/library/grade-11-mathematics-2020-national-exemplar-november-exam-memorandum-paper-1/1069fa12-57c0-4875-a06a-ca2128f1197f) ## About this document Mathematics · Grade 11 · National Exemplar November Exam · 2020. Memorandum, 15 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 NOVEMBER 2020 MATHEMATICS P1/WISKUNDE V1 MARKING GUIDELINE/NASIENRIGLYN EXEMPLAR/EKSEMPLAAR MARKS/PUNTE: 150 This marking guideline consists of 15 pages./ Hierdie nasienriglyn bestaan uit 15 bladsye. ### Page 2 2 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2020) NOTE/LET WEL: x If a candidate answers a question TWICE, mark the FIRST attempt ONLY. Indien ʼn kandidaat ʼn vraag TWEE keer beantwoord, merk SLEGS die EERSTE poging. x Consistent accuracy applies in ALL aspects of the marking guideline. Volgehoue akkuraatheid geld deurgaans in ALLE aspekte van die nasienriglyn. x If a candidate crossed out an attempt of a question and did not redo the question, mark the crossed-out attempt. Indien ʼn kandidaat ʼn poging vir ʼn vraag deurgetrek het en nie die vraag weer beantwoord het nie, merk die poging wat deurgetrek is. x The mark for substitution is awarded for substitution into the correct formula. Die punt vir substitusie word toegeken vir substitusie in die korrekte formule. QUESTION 1/VRAAG 1 1.1.1 (3x  2)( x  5) 0 2 3 3 answers / antwoorde ? x  or / of x 5 (2) 3 1.1.2 3x  5 x  1 0 2 b r b 2  4ac ?x 2a (5) r (5) 2  4(3)(1) 3 substitution / vervanging 2(3) 5 r 37 6 3 3 answers / antwoorde (3)  0,18 or / of 1,85 1.1.3 x 4 x2 x2 4 x 3 squaring both sides/ ( x  2) (4  x) 2 kwadreer beide kante x  2 16  8 x  x 2 3 standard form / standaardvorm x 2  9 x  18 0 3 factors / faktore ( x  6)( x  3) 0 3 both x-values / beide x-waardes 3 selection / keuse ? x 3 or / of xz6 (5) 1.1.4 2x2  5x d 3 3 standard form / standaardvorm 2x2  5x  3 d 0 3 factors / faktore (2 x  1)( x  3) d 0 -3 1 1 ?3 d x d 3 3 3 d x d 2 2 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 3 (EC/NOVEMBER 2020) MATHEMATICS P1/WISKUNDE V1 3 1.2.1 2 2 § 8 · 2 § 8 ·3 § 8 · ¨ ¸ ¨¨ 3 ¸¸ 3 ¨¨ 3 ¸¸ © 27 ¹ © 27 ¹ © 27 ¹ 2 §2· ¨ ¸ ©3¹ 3 answer / antwoord 4 (2) 9 OR/OF 2 OR/OF ª§ 2 ·3 º 3 2 3 «¨ ¸ » ª§ 2 ·3 º 3 «¬© 3 ¹ »¼ «¨ ¸ » «¬© 3 ¹ »¼ 2 §2· ¨ ¸ ©3¹ 3 answer / antwoord 4 9 (2) 1.2.2 12  2 3 1 2 32 3 1 3 2 3 3 4.3  2 3  2 3  2 2.3  2 3  2 3  2 62 3 answer / antwoord 4 (3) OR/OF OR/OF 3 36 12  2 3 1 36  12  2 3  2 3 62 3 2 3 2 3 answer / antwoord 62 3 2 3 2 (3) 4 Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 4 4 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2020) 1.3 5y  x 2 (1) x 2  3 xy  4 y 4 (2) 3 x 5x  2 x 5y  2 (3) Subst. / Vervang (3) into/in (2) : 3 substitution / vervanging ? (5 y  2)  3 y (5 y  2)  4 y  4 2 0 3 standard form / standaardvorm 25 y  20 y  4  15 y  6 y  4 y  4 2 2 0 3 method;factors / metode;faktore 10 y 2  10 y 0 3 both x-values / beide x-waardes 10 y ( y  1) 0 ?y 0 or / of y 1 x 5y  2 3 both y-values / beide y-waardes x 5 0 2 or / of x 5 1 2 (6) ?x 2 or / of x 3 OR / OF 3 y 1 5 x2 5y  x 2 1 x 2  3 xy  4 y 4 (2) 3 substitution / vervanging x2 1 y x2 3 5 5 Subst./Vervang 3 into/in 2 , 3 standard form / standaardvorm x  3x 2 1 5 ( x  2)  4 1 5 ( x  2) 4 3 factors / faktore x  53 x( x  2)  54 ( x  2) 2 4 3 both y-values / beide y-waardes 5 x 2  3 x 2  6 x  4 x  8  20 0 2 x  2 x  12 2 0 3 both x-values / beide x-waardes (6) x2  x  6 0 ( x  3)( x  2) 0 ? x 2 or / of x 3 y 1 5 2  2 or / of y 1 5 (3  2) ?y 0 or / of y 1 1.4.1 Perimeter/Omtrek 2l  2b 280 2(2 x)  2 y 3 substitution / vervanging 2 y 280  4 x 3 simplification / vereenvoudiging ? y 140  2 x Area/Oppervlakte lb 3 A 2x(140  2x) 2x u y 2 x(140  2 x) 280 x  4 x 2 (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 5 (EC/NOVEMBER 2020) MATHEMATICS P1/WISKUNDE V1 5 1.4.2 A 280 x  4 x 2  4( x 2  70 x)  4( x 2  70 x  1225  1225) 3 completing the square / vierkantsvoltooiing  4[( x  35) 2  1225]  4( x  35) 2  4900 3 4900 ? The maximum area is 4900 m 2 3 correct conclusion / korrekte gevolgtrekking Die maksimum oppervlakte is 4900 m 2 (3) OR/OF OR/OF b x  2a 3 method/metode  280 2(4) 3 35 m 35 m 3 answer / antwoord ? A 280 (35)  4(35) (3) 4 900 m 2 [31] QUESTION 2/VRAAG 2 2.1 3.2 x 1  2.4 x 3.2 x.2  2.22 x 3 3.2x.2  2.22 x 3.2 x  22 x 3.2 x  22 x 3 factorisation of numerator / 2.2 x (3  2 x ) faktorisering van teller 2 x (3  2 x ) 3 factorisation of denominator / 2 faktorisering van noemer 3 answer / antwoord (4) 2 2.2.1 5x 5 20 2 2 x 5 4 3 x5 4 5 2 2 5 x 5 4 3 x5 2 2 5 42 5 ?x 22 2 25 32 3 answer / antwoord (3) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 6 6 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2020) 2.2.2 12 x.3x 1 648 x x 12 .3 .3 648 3 12x.3x 216 x x 12 .3 216 x 12.3 216 36 x 216 3 62 x 63 62 x 63 3 equating exponents / 2x 3 gelykstel van eksponente 3 3 answer / antwoord x 2 (4) 2.3 3x  2 f ( x) x  10 x  25 2 3 x2  10x  25 0 for undefined / f is undefined when : f is ongedefinieerd wanneer vir ongedefinieerd x  10 x  25 0 2 ( x  5) 2 0 x 5 3 x 5 ? f is defined for : x  , but x z 5 3 x  ℝ 3 x z 5 f is gedefinieerd vir : x  , maar x z 5 (4) [15] QUESTION 3/VRAAG 3 3.1.1 9 ; 5 ; 1; …;−143 Tn 13  4n 3 13 3 −4n (2) 3.1.2 Tn 13  4n T7 13  4(7) 3 substitution / vervanging 15 3 −15 (2) 3.1.3 Tn 13  4n 143 13  4n 3 substitution / vervanging 156  4n ? n 39 3 answer / antwoord (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 7 (EC/NOVEMBER 2020) MATHEMATICS P1/WISKUNDE V1 7 3.2 Tn an  b 16a  b 38 1 3 16a  b 38 and/en 41a  b 113 41a  b 113 2 25a 75 3 method / metode ?a 3 16 3  b 38 3 Tn 3n  10 48  b 38 ? b 10 Tn 3n  10 3n=3  1 3n  10 9 3n ?n 3 (4) [10] QUESTION 4/VRAAG 4 4.1 −12 ; −8 ; 0 ; 12 4 8 12 4 4 3 28 3 48 28 ; 48 (2) 4.2 2a 4 3a  b 4 a  b  c 12 ?a 2 3(2)  b 4 2  2  c 12 3 a 2 3 b 2 b 2 c 12 3 c 12 3 Tn 2n 2  2n  12 ?Tn 2n 2  2n  12 (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 8 8 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2020) 4.3 For first differences: / Vir eerste verskille 4;8;12; ... 3 Tn 4n Tn 4n 3 n 48 192 4n 3 answer / antwoord ? n 48 (3) ?192 lies between T48 and T49 192 lê tussen T48 en T49 OR/OF OR/OF Tn 1  Tn 192 3 2(n  1) 2  2(n  1)  12  (2n 2  2n  12) 192 2(n  1) 2  2(n  1)  12  (2n 2  2n  12) 192 3 4n = 192 2(n 2  2n  1)  2n  2  12  2n 2  2n  12 192 2n 2  4n  2  2n  2  12  2n 2  2n  12 192 ? 4n 192 3 answer / antwoord (3) n 48 ?192 lies between T48 and T49 192 lê tussen T48 en T49 4.4 Pn  0 Ÿ Tn  168  0 Tn  168 3 2n2  2n 12  168 3 standard form / standaardvorm 2n 2  2n  12  168 2n 2  2n  180  0 3 factorisation / faktorisering 3 9  n  10 n 2  n  90  0 -9 10 (n  10)(n  9)  0 3 answer / antwoord 9  n  10 (5) but / maar : n ! 0 ? Number of terms 9 Aantal terme 9 4.5 Tn 2n 2  2n  12 3 Tn 2(n 2  n  6) 2(n 2  n  6) 3 explanation / verduideliking ? 2 u any n ! 0 is always even, so Tn will always be even 2 u enige n ! 0 is altyd ewe, so Tn sal altyd ewe wees. (2) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 9 (EC/NOVEMBER 2020) MATHEMATICS P1/WISKUNDE V1 9 QUESTION 5/VRAAG 5 5.1 b At TP/By Draaipunt: x  3 method / metode 2a 6  2(1) 3 3 x-coordinate / x-koördinaat ?y (3) 2  6(3)  7 3 y-coordinate / y-koördinaat 16 (3) OR/OF f ( x)  x2  6 x  7 ( x 2  6 x  7) 3 completing the square / [( x 2  6 x  9)  9  7] vierkantsvoltooiing [( x  3) 2  16] ( x  3) 2  16 3 x-coordinate / x-koördinaat 3 y-coordinate / y-koördinaat ? Turning point / Draaipunt : 3;16 (3) 5.2  x2  6 x  7 0 3 f ( x) 0 x2  6 x  7 0 ( x  7)( x  1) 0 ? x 7 or / of x 1 3 answers / antwoorde (2) 5.3 3 y-intercept / y-afsnit 3 x-intercepts / x-afsnitte 3 turning point / draaipunt 3 shape / vorm (4) 5.4 x 3 3answer / antwoord (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 10 10 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2020) 5.5 f ( x)  x2  6x  7 f (3) (3) 2  6(3)  7 3 f (3) 20 20 (1) 2  6(1)  7 f (1) 3 f (1) 12 12 12  ( 20) 3 substituting into gradient formula / Average/Gemiddelde m vervanging in gradiënt-formule 1  (3) 32 4 8 3 answer / antwoord (4) 5.6 f ( x)  x  6 x  7 2 ( x  3) 2  16 2 3 a 1 and/en q 16 ? h( x ) ( x  3)  4  16 3 p 1 ( x  1) 2  16 (2) [16] QUESTION 6/VRAAG 6 6.1 a f ( x) q x p a 1 3 p 33q 1 x3 (2) 6.2 a f ( x) 1 x3 a 6 1 3 substitution / vervanging 2  3 ?a 5 3 answer / antwoord (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 11 (EC/NOVEMBER 2020) MATHEMATICS P1/WISKUNDE V1 11 6.3 5 f ( x) 1 x3 5 3 y 0 0 1 x3 5 1 x3 3 x 8  x3 5 x 8 ? A(8;0) (2) 6.4 A(8;0) and/en D(3;1) y2  y1 mAD x2  x1 1 0 3 substitution / vervanging 3  (8) 1 3 mAD ?b 5 g ( x) bx  c 1 3 substitution / vervanging xc 5 1 1 0 8  c or / of 1 3  c 5 5 8 3 equation / vergelyking ?c 5 1 8 g ( x) x 5 5 (4) 6.5 x  (f ; f) but / maar x z 3 3 x  (f ; f) 3 x z 3 OR / OF (2) x  ℝ, x z 3 3 x  ℝ 3 x z 3 (2) 6.6 f ( x) g ( x) 5 1 8 5 1 8 1 x 3 1 x x3 5 5 x3 5 5 25  5( x  3) x( x  3)  8( x  3) 25  5 x  15 x 2  3 x  8 x  24 3 standard form / standaardvorm x  6 x  16 0 2 3 x-values / x-waardes ( x  8)( x  2) 0 x 8 or x 2 5 y 1 3 coordinates / koördinate 23 2 ? B (2; 2) (4) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 12 12 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2020) 6.7 8 d x  3 or x t 0 33 8 d x  3 3 x t 0 OR / OF OR / OF x >[8;  3) ‰ [0; f)@ ∈ [[−8; −3) 33 ‰ [0; ∞)] 3 (3) [19] QUESTION 7/VRAAG 7 7.1.1 y 4 3 answer / antwoord (1) 7.1.2 y  (4; f) 3 answer / antwoord (1) OR / OF OR / OF 3 answer / antwoord (1) y ! 4 7.1.3 §1· x g ( x) ¨ ¸  4 ©2¹ 0 §1· 3 y-value / y-waarde y ¨ ¸ 4 ©2¹ 3 substitution / vervanging 3 x §1· 0 ¨ ¸ 4 ©2¹ x 4 21 3 answer / antwoord 22 2 x (3) ?2 x x 2 Intercepts / Afsnitte :(0;  3) and / en ( 2;0) 7.1.4 x ! 2 3 answer / antwoord (1) 7.2 y 3 asymptote above x-axis q asimptoot bo x-as 3 y-intercept positive y-afsnit positief 3 shape / vorm x h (3) [9] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 13 (EC/NOVEMBER 2020) MATHEMATICS P1/WISKUNDE V1 13 QUESTION 8/VRAAG 8 8.1 § inom · m ieff ¨1  ¸ 1 3 formula / formule © m ¹ 12 § 0.095 · 3 substitution / vervanging ¨1  ¸ 1 © 12 ¹ 0,099247 ... ? r 9,92 % 3 answer / antwoord (3) 8.2 A P (1  i ) n R 764 050 ,60 P 1  0,08 5 3 A R764050,60 764050 .60 3 substitution / vervanging P 5 1  0.08 R 520 000 3 answer / antwoord (3) 8.3.1 ª § 0,075 · 48 º§ 0,11 ·12 A «28 000 ¨1  ¸  R 7 300 » ¨1  ¸ 48 «¬ © 12 ¹ »¼© 4 ¹ § 0,075 · 3 28 000¨1  ¸ 12 © 12 ¹ § 0,11 · R 37 760 ,78  R 7 300 ¨1  ¸ 3 −R7 300 © 4 ¹ 12 12 § 0,11 · § 0,11 · 3 u ¨1  ¸ R 30 460 ,78¨1  ¸ © 4 ¹ © 4 ¹ 3 simplification / vereenvoudiging R 42 181,59 3 answer / antwoord (5) 8.3.2 A P(1  i ) n 60 § 0, 08 · A R42181,59 ¨1  ¸ 3 substitution into correct formula © 12 ¹ vervanging in korrekte formule R62844, 06 R80 000  R62844, 06 3 R62 844,06 R17155,94 ?A P(1  i ) n 3 R17 155,964 60 § 0, 08 · R17155,94 P ¨1  ¸ © 12 ¹ 3 method / metode 17155,94 P 60 § 0, 08 · ¨1  ¸ © 12 ¹ R11515, 25 ? He needs to deposit R11 515 / Hy moet R11 515 deponeer 3 answer / antwoord (5) [16] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 14 14 MATHEMATICS P1/WISKUNDE V1 (EC/NOVEMBER 2020) QUESTION 9/VRAAG 9 9.1 If A and B are independent, then: As A en B onafhanklik is, dan: P(A and/en B) = P(A) u P(B) P(A) = 1 P(not/nie A) = 1 0,45 3 0,55 = 0,55 11 = 20 P(A or/of B) = P(A) + P(B) – P(A and/en B) 0,685 = 0,55 + 0,3 – P(A and/en B) 3 substitution / vervanging ∴P(A and/en B) = 0,165 33 = 3 answer / antwoord 200 P(A) u P(B) = 0,55 u 0,3 3 P(A) u P(B) = 0,165 33 = 200 ∴ A and B are independent events. / 3 conclusion / gevolgtrekking A en B is onafhanklike gebeurtenisse. (5) 9.2.1 S N R 9 5 c (8) 3 a = 10 12 a (10) b (13) 3 b =13 3c=8 4 3 d =14 d (14) A a = 10 ; b = 13 ; c = 8 ; d = 14 (4) 9.2.2 39 5 39 5 P A or/of (N and/en R)  3 3 75 75 75 75 44 3 answer / antwoord 75 (3) | 0,59 [12] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 15 (EC/NOVEMBER 2020) MATHEMATICS P1/WISKUNDE V1 15 QUESTION 10/VRAAG 10 10 Let the total number of balls be t. Laat die totale aantal balle t wees. 5 4 3 3and/en t t 1 Green/Groen: 5 G 4 Blue/Blou: t – 5 t 1 5 G t B G B B P(GG) = P(G) u P(G) 3 equation / vergelyking 5 4 5 u t t  1 18 20 5 3 standard form / standaardvorm t (t  1) 18 5t 2  5t 360 5t  5t  360 0 2 3 factorisation / faktorisering t 2  t  72 0 3t=9 (t  9)(t  8) 0 ? t 9 or t z 8 ? There are 9 balls. Daar is 9 balle. 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