# Mathematics Grade 11 P1 National Exemplar November 2019 Memo > Grade 11 Mathematics Paper 1 memorandum from the National Exemplar November Exam 2019 (Afrikaans), 17 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/gr11-math-p1-afrikaans-and-english-november-2019-possible-answers/e4962b19-d0cb-4874-a7fa-58cd8e95ab68 - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 11 - Language: Afrikaans - Document type: Memorandum - Year: 2019 - Exam period: National Exemplar November Exam - Paper: 1 - Curriculum: CAPS - Pages: 17 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 2019 Memo](https://www.hlayiso.com/library/gr11-math-p1-afrikaans-and-english-november-2019-possible-answers/e4962b19-d0cb-4874-a7fa-58cd8e95ab68) ## About this document Mathematics · Grade 11 · National Exemplar November Exam · 2019 · Afrikaans. 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 Downloaded from hlayiso.com testpapers.co.za NATIONAL SENIOR CERTIFICATE/ NASIONALE SENIOR SERTIFIKAAT GRADE/GRAAD 11 MATHEMATICS P1/WISKUNDE V1 NOVEMBER 2019 MARKING GUIDELINES/NASIENRIGLYNE MARKS/PUNTE: 150 These marking guidelines consist of 17 pages./ Hierdie nasienriglyne bestaan uit 17 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 2 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 2 DBE/November 2019 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne 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 guidelines. • Assuming values/answer 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 nasienriglyne van toepassing. • Dit is onaanvaarbaar om waardes/antwoorde te veronderstel om 'n probleem op te los. QUESTION/VRAAG 1 1.1.1 2 x ( x − 3) = 0 2 x = 0 or/of x = 3  x = 0  x= 3 x=0 (2) 1.1.2 2 3x − 2 x = 4 standard form/standaardvorm 3x 2 − 2 x − 4 = 0 − b ± b 2 − 4ac x= 2a substitution into correct formula/ − (− 2 ) ± (− 2)2 − 4(3)(− 4) vervanging in korrekte formule = 2(3) x =1,54 or x = − 0,87 answer/antw. answer/antw. (4) 1.1.3 ( x − 1)(4 − x) ≥ 0 critical values/kritieke waardes 1 4  1 ≤ x ≤ 4 1 ≤x≤4 (3) 1.1.4 5 − x = x +1 squaring both sides/ 5 − x = ( x +1) 2 kwadreer beide kante 5 − x = x 2 + 2 x +1 standard form/standaardvorm x 2 + 3x − 4 = 0 factors or using formula/ (x + 4)(x − 1)= 0 faktore of gebruik formule x ≠ − 4 or x =1 both solutions to/ beide oplossings x  rejecting/verwerp x = −4 (5) Copyright reserved/Kopiereg voorbehou ### Page 3 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 3 DBE/November 2019 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne 1.2 x+4 = 2y and y 2 − xy + 21 = 0 ∴ x= 2y − 4  x= 2y − 4 y 2 − ( 2 y − 4) y + 21 = 0 substitution /verv. y 2 − 2 y 2 + 4 y + 21 = 0 − y 2 + 4 y + 21 = 0 std form/stand. vorm 2 y − 4 y − 21 = 0 ( y − 7 )( y + 3) = 0 factors or using formula/ faktore of gebruik formule y = 7 or y = − 3 y-values/wrdes x = 2(7 ) − 4 or x = 2 (− 3) − 4 x = 10 or x = −10 x-values/wrdes (6) 1.3 2(x − 3) + 2 = 0 2  ( x − 3) = −1 2 (x − 3) = −1 2 conclusion/gevolgtrekking (2) ∴ roots are non - real/wortels is niereël 1.4 g ( x) = −2 x 2 − px + 3 −b − ( − p ) p  x= − p x= = = − 2a 2 ( −2 ) 4 4 2  p  p y=−2  −  − p  −  + 3  4  4 2 p  p 1  −2  − 2  p  p 1  − p− +3 =3 −2  −  − p  −  + 3 =3  4  4 8  4  4 8 p2 2 p2 1 simplification/vereenvoudiging − + = 8 8 8 p2 = 1  p = ±1 p = ±1 OR/OF 4ac − b 2 4ac − b 2 max value/maks waarde =  max value = 4a 4a 4(− 2)(3) − p 2 25 4(− 2 )(3) − p 2 25 =  = 4(− 2 ) 8 4(− 2 ) 8 − 24 − p 2 25 = −8 8 2 simplification − 192 − 8 p = −200 8p2 = 8  p = ±1 p = ±1 (4) [26] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 4 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 4 DBE/November 2019 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 2 2.1 32 x +1.15 2 x −3 27 x −1.3 x.5 2 x − 4 32 x +1.32 x −3.5 2 x −3  prime bases/priembasis =  base/basis 3 33 x −3.3 x.5 2 x − 4 adding and subtracting = 32 x +1+ 2 x −3 −3 x + 3− x . 5 2 x −3− 2 x + 4 exponents/optel en aftrek van = 3.5 eksponente = 15 answer/antw. (4) 2.2.1 1 x   = 32 2 x 1 5   =2 2   same base/dieselfde basis 2 − x = 25 −x=5 equating indices/ x = −5 gelykstelling van eksponente answer/antw. OR/OF x OR/OF 1   = 32 2 x 1 5   =2 2    same base/dieselfde basis x −5 1 1   =  simplification/vereenv 2 2 answer/antw. x = −5 (3) 2.2.2 1 3 =4 x2 −2 exp form/eksp. vorm x 3 = 22 3 x= 2 ( ) 2 −2  x = (2 2 )− 2 3 x = 2−3 1 answer/antw. x= (3) 8 Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 5 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 5 DBE/November 2019 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne 2.2.3 12 2x − = −4 2x (2 ) − 12 = −4.2 x 2 x mult by LCD/KGN (2 ) + 4.2 − 12 = 0 x 2 x ( ) 2  2 x + 4.2 x − 12 = 0 (2 + 6)(2 − 2) = 0 x x factors/faktore  2 x ≠ −6 2 x ≠ −6 or 2x = 2  x =1 no solution/geen oplossing or/of x =1 (5) OR/OF OR/OF Let 2 x = k 12 k − = −4 k 2 k − 12 = −4k  mult by LCD/KGN k 2 + 4k − 12 = 0  k 2 + 4k − 12 = 0 (k − 2)(k + 6) = 0 factors/faktore k = 2 or k = −6 2 x = 2 or 2 x ≠ −6  2x ≠ − 6 x =1 or no solution  x =1 (5) 2 4 2.3 + 2 +1 2 = 2. 2 + ( 4 2 +1 ) LCD/KGN ( ) 2 ( 2 + 1) 2 2 +1 = ( 2) +4 2 + 2 2 2 perfect square trinomial volkome vierkant drieterm 2+ 2 simplification denominator/ vereenvoudigingsnoemer = ( 2 + 2) 2 factors/faktore 2+ 2 = 2+ 2 answer/antwoord (5) [22] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 6 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 6 DBE/November 2019 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 3 3.1 -5 0 5 5 5 Tn = an + b  5n  − 10 = 5n − 10 (2) 3.2 T12 = 5(12 ) − 10 substitution/verv. = 50 answer/antw. (2) 3.3 5n −10 = 130 substitution/verv. 5n = 140 n = 28 answer/antw. 28 term(T28 ) th (2) [6] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 7 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 7 DBE/November 2019 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 4 4.1.1 13 27 45 67 14 18 22 4 4 It is a quadratic number pattern/ Dit is 'n kwadratiese getalpatrron quadratic/ kwadratiese Second difference is constant./Tweede verskil is konstant. justification/ regverdiging (2) 4.1.2 2a = 4 a=2 a =2 3(2) + b = 14 b=8  b =8 2 + 8 + c = 13  c =3 c=3 2 Tn = 2n 2 + 8n + 3  Tn = 2n + 8n + 3 (4) 4.1.3 Tn = 2n 2 + 8n + 3  substitution/vervanging T100 = 2(100 ) + 8(100 ) + 3 2 = 20803 answer/antwoord (2) 4.1.4 4n + 10 = 110  4n +10 = 110 4n = 100 n = 25  n = 25 T25 = 2(25) + 8(25) + 3 2  T25 =1453 = 1453 T26 = 2(26 ) + 8(26 ) + 3 2  T26 = 1563 =1563 (5) Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 8 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 8 DBE/November 2019 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne 4.1.5 The first and second differences are all even but the first term is odd./ argument Die eerste en tweede verskille is gelyk maar die eerste term is onewe. conclusion/gevolgtr Thus when adding an even to an odd number the answer will always be odd./Wanneer 'n ewe by 'n onewe getal gevoeg word, sal die antwoord altyd onewe wees. OR/OF OR/OF Tn = 2n 2 + 8n + 3 2n 2 has an even coeficient thus it even 8n has an even coeficient thus it even argument 3 is an odd number conclusion/gevolgtr ∴ the values will always be odd OR/OF OR/OF Tn = 2n 2 + 8n + 3 ( = 2 n 2 + 4n + 3 ) ( 2  2 n + 4n + 3 ) For all n ∈ N, ( ) 2 n 2 + 4n is even/ewe ( ) ∴ 2 n 2 + 4n + 3 is odd/onewe conclusion/gevolgtr because an even + odd will always be odd/want ' n ewe (2) en ' n onewe sal altyd ' n onewe maak 4.2 1st difference/1ste verskil: p 2 p − 4 ; p − 3; −1 2 2nd difference/2de verskil:  p p − 3 − (2 p − 4) = − 1− ( p − 3) p − 3 − (2 p − 4 ) = p − 1 − ( p − 3) 2 2 p − p +1 = − + 2 2 p − =1 2  p =− 2 p = −2 4 -4 -9 -11 -8 -5 -2 -2 -2 −2  2(− 2) − 4 = −8 2(− 2 ) − 4 ; − 2 − 3 ; −1 2  y=−9 − 8; − 5; − 2  x=− 4 x = − 4 and y = − 9 (5) [20] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 9 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 9 DBE/November 2019 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 5 5.1 1 2x + 6 f( = x) − x −3 x +3 1 2 ( x + 3) = − common factor x −3 x+3 gemene faktor 1 simplification/vereenv. = −2 x −3 (2) 5.2 x= 3 x = 3 y = −2 y = –2 (2) 5.3 1 0= −2 subst./verv. y = 0 x−3 1 2= x−3 2(x − 3) = 1 simplification/vereenv. 2x − 6 = 1 7 x= answer/antw. 2 (3) 5.4 1 =y −2 0 −3 subst/verv. x = 0 −7 = answer/antw. 3 OR/OF answer/antw  −7  (2) 0;   3  Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 10 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 10 DBE/November 2019 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne 5.5 y f x O 3 3,5 asymptotes/asimptote shape/vorm -2 x- and y- int. 7 − 3 f (3) 5.6 y= x+c m=1 − 2 =3+ c substitution of/vervanging van (3; − 2 ) c=−5  c=−5 y=x−5 (3) 5.7 Translate f 3 units to the left and 2 units up. 3 units to the left Transleer f 3 eenhede na links en 2 eenhede op. eenhede na links 2 units up/eenhede op (2) 5.8 x ∈(− ∞; + ∞ ) ; x ≠ 0  x ∈(− ∞; + ∞ ) OR/OF  x ≠0 x ∈ R; x ≠ 0 (2) [19] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 11 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 11 DBE/November 2019 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 6 6.1 y = −3 x + k 0 = − 3(5) + k  substitute/verv. ( 5 ; 0) k =15 (1) 6.2 −b x= 2a or −2 x= −3+5 2(− 1) x= 2  x=1 x=1 =1 y = −(1) + 2(1) + 15 2 substitution/vervanging y =16  y = 16 D (1;16) (3) 6.3 x<1 OR/OF  answer/antwoord x ∈ (− ∞;1) (1) 6.4 A(−3 ;0) D(1 ;16) y 2 − y1 formula/formule Ave grad / Gemidgrad = x 2 − x1 0 − 16 subst. into correct formula = /verv. in formule − 3 −1 =4 answer/antwoord (3) Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 12 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 12 DBE/November 2019 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne 6.5 D(1 ;16)  E(1;12) E(1;12)  answer/antwoord DE = 4units (2) 6.6 h(x ) = f (x − 1) − 2 = − (x − 1) + 2(x − 1) + 15 − 2 2  − (x − 1) + 2(x − 1) + 15 − 2 2 2 = − x + 2 x − 1 + 2 x − 2 + 15 − 2 = − x 2 + 4 x + 10 2  − x + 4 x + 10 ( ) = − x 2 − 4 x −10 = −(x + −4 x + 4 − 4 −10 ) 2 ( 2  − x + −4 x + 4 − 4 −10 ) = −(x − 2 ) + 14  h(x ) = −(x − 2 ) + 14 2 2 OR/OF OR/OF D(1;16 ) and a = −1 subst./vervang D(1;16 ) and a = −1 f ( x ) = −( x − 1) + 16 2 correct form/korrekte vorm h( x ) = f ( x − 1) − 2  h( x ) = −(x − 1 − 1) + 16 − 2 2 = −( x − 1 − 1) + 16 − 2 2  h( x ) = −( x − 2) + 14 2 = −( x − 2 ) + 14 2 (4) 6.7 max value of/maks wrde van f (x ) is 16  max value of f ( x ) is 16 ∴ max value of/maks wrde van f (x ) − 12 is 16 − 12 = 4  max value of f ( x ) − 12 is 16 − 12 = 4 max/ maks p ( x) = 3 4 = 81 answer/antw. (3) 6.8 x ∈ R; x ≠ 1 answer/antwoord (2) [19] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 13 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 13 DBE/November 2019 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 7 7.1 y = 3x + p − 27 \ 54 = 33+ p − 27 subs/vervanging (3 ; 54) 3+ p 81 = 3  equating indices/gelykst. eksp. 34 = 33+ p 3+ p= 4 p =1 answer/antwoord (3) 7.2 range / waardeversameling y > −27 or / of  answer/antwoord y ∈ (− 27 ; ∞ ) (2) 7.3 −y=3 x +1 − 27 g ( x ) = − 1.3 x +1 + 27 new equation/nuwe verg. 0 +1 y = − 1.3 + 27  answer/antwoord = 24 y − intercept/afsnit (0 ; 24 ) (2) [7] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 14 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 14 DBE/November 2019 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 8 8.1 A = P(1 − i) n substitution/verv. 85 000 = 200 000(1 − i ) 5 85 000 rewrite in terms of i/ i =1 − 5 skryf in terme van i 200 000 i =15,73% answer/antw. (3) 8.2  i  m 1 + ieff = 1 + nom  formula/form.  m  4  0,085  i = 0,085 1 + ieff = 1 +   4  4 4  0,085  ieff = 1 +  −1  4  ieff = 8,77% answer/antw. (3) 8.3.1 A = P(1 + i) n 2×12  0,12  substitution/verv. = 28 0001 +   12  = R 35 552,57 answer/antw. (2) 8.3.2  0,12 and n = 48 12× 4 2× 4  0,12   0,129  A = 28 000 1+  1 +  + 12  12   2  12 2× 4 2× 3  0,129 and n = 8  0,12   0,129   0,129  12 12 000 1+  1 +  − 65001+   12   2   2  = R87 267,25  12 2×4  0,12   0,129  12 000 1 +  1 +   12   2  2×3 0,129   − 65001 +   2   answer/antw. (5) OR/OF Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 15 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 15 DBE/November 2019 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne  12×3   0,12  12×3  0,12  A = 28 000 1 +  + 12 000} 28 000 1 +   12  + 12000   12  12 2 6  0,12   0,129   0,129   1 +  1 +  − 6500 1 +  12  0,12   0,129  2  12   2   2  1 +  1 +  − 6500  12   2  = R87 267,25 6  0,129   1 +   2   answer/antw. (5) [13] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 16 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 16 DBE/November 2019 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne QUESTION/VRAAG 9 9.1.1 P(A and/en B) = P( A) × P(B)  0,48 × 0,28 = 0,48 × 0,28 = 0,1248 answer/antwoord (2) 9.1.2 P(A or B) = P(A ) + P(B) = 0 ,48 + 0 ,26 substitution/verv. answer/antwoord = 0 ,74 (2) 9.2.1 10 + 6 + x + 5 = 29 method/metode x =8 value of/wrde van x 41 + 10 + 30 + 6 + 8 + 5 + 25 + y = 130 method/metode y=5 value of/wrde van y (4) 9.2.2 41 + 10 + 30 5 41 + 10 + 3 P(no nut snack or no snack ) = + adding/optel 5 130 130 43 = answer/antwoord 65 (3) 9.3.1 0,3 B (B ; B)  branch at first level tak by eerste vlak B 0,35 0,7 C (B ; C)  branches at second level/takke by tweede vlak 0,3 B (C ; B) 0,65 C outcomes/uitkomste 0,7 C (C ; C)  probabilities /moontlikhede (4) Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 17 Downloaded from hlayiso.com Mathematics/P1/Wiskunde V1 17 DBE/November 2019 CAPS/KABV – Grade/Graad 11 – Marking Guidelines/Nasienriglyne 9.3.2 P(BB) + P(CC ) = (0, 35 × 0, 3) + (0, 65 × 0, 7 )  = 0,56 = 56% (0, 35 × 0, 3) + (0, 65 × 0, 7 ) probability/moontlik- 0,56 × 200 = 112 clients/kliënte hede  answer/antwoord (3) 9.3.3 Number of clients who chose different meals/Getal kliënte wat  88 verskillende maaltye gekies het = 200 – 112 = 88 More clients preferred to make the same choice/Meer kliënte verkies om dieselfde keuse te maak.  conclusion/gevolgtr. 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