# Mathematics Grade 12 Paper 1 Northern Cape Mock 2021 Memo > Grade 12 Mathematics Paper 1 memorandum from the Northern Cape Mock Exam 2021, 19 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/nothern-c-sep-2021-trial-exam-p1-memo/3c5eb4c1-8a56-4cc4-93e1-aed176bda09d - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 12 - Document type: Memorandum - Year: 2021 - Exam period: Northern Cape Mock Exam - Paper: 1 - Curriculum: CAPS - Pages: 19 Breadcrumbs: [Home](https://www.hlayiso.com/) › [Library](https://www.hlayiso.com/library) › [Grade 12](https://www.hlayiso.com/library/grade-12) › [Mathematics](https://www.hlayiso.com/library/grade-12/mathematics) › [Mathematics Grade 12 Paper 1 Northern Cape Mock 2021 Memo](https://www.hlayiso.com/library/nothern-c-sep-2021-trial-exam-p1-memo/3c5eb4c1-8a56-4cc4-93e1-aed176bda09d) ## About this document Mathematics · Grade 12 · Northern Cape Mock Exam · 2021. Memorandum, 19 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 ————————————————————— DEPARTMENT OF EDUCATION ————————————————————— DEPARTEMENT VAN ONDERWYS ————————————————————— LEFAPHA LA THUTO ————————————————————— ISEBE LEZEMFUNDO ————————————————————— PROVINCIAL PREPARATORY EXAMINATION/ PROVINSIALE VOORBEREIDINGSEKSAMEN GRADE/GRAAD 12 MATHEMATICS P1/WISKUNDE V1 MARKING GUIDELINES/NASIENRIGLYNE SEPTEMBER 2021 MARKS/PUNTE: 150 These marking guidelines consist of 20 pages./ Hierdie nasienriglyne bestaan uit 20 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai om asseblief ### Page 2 Downloaded from hlayiso.com Mathematics P1/Grade 12 2 September 2021 Wiksunude V1/Graad 12 Marking Guidelines/Nasienriglyne NOTE:  If a candidate answered a question TWICE, mark only the FIRST attempt.  If a candidate has crossed out an attempt to answer a question and did not redo it, mark the crossed-out version.  Consistent accuracy applies in ALL aspects of the marking memorandum. Stop marking at the second calculation error.  (A) is an accuracy mark.  Assuming answers/values in order to solve a problem is NOT acceptable. LET WEL:  Indien 'n kandidaat 'n vraag TWEE keer beantwoord het, sien slegs die EERSTE poging na.  As 'n kandidaat 'n poging om 'n vraag te beantwoord, doodgetrek en nie oorgedoen het nie, sien die dood getrekte poging na.  Volgehoue akkuraatheid is op ALLE aspekte van die memorandum van toepassing. Staak nasien by die tweede berekeningsfout.  Om antwoorde/waardesom 'n probleem op te los, te veronderstel, word NIE toegelaat NIE. Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 3 Downloaded from hlayiso.com Mathematics P1/Grade 12 3 September 2021 Wiksunude V1/Graad 12 Marking Guidelines/Nasienriglyne QUESTION/VRAAG 1 ITEM SOLUTIONS/OPLOSSINGS MARK CLARIFICATION/ PUNTEVERHELDERING 1.1.1 x2  x  6  0 factors x  3x  2  0  answers x  3 or x  2 (3) 1.1.2 x x  6   1  0 standard form Penalise 1 mark for x  6x  1  0 2 incorrect rounding substitution into formula 6 62  411 x 21  x  0,17  x  0,17 or x  5,83  x  5,83  (4) 1.1.3 6x  2x  0 2 critical value(s)  2 x 3  x   0 x  0 or x  3  method If ‘and’ instead of ‘or’ penalise 1 mark  answer 0 3 (3) x  0 or x  3 1.1.4  2  x  2  x   x  2  x2   2  x2  x squaring both sides   x 2 2  x2 2 2  x2  x2 2 x2  2  0 standard form x2 1  0  x  1 x  1  0 factors  x  1 or x  1  x 1 selection (5) Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 4 Downloaded from hlayiso.com Mathematics P1/Grade 12 4 September 2021 Wiksunude V1/Graad 12 Marking Guidelines/Nasienriglyne ITEM SOLUTIONS/OPLOSSINGS MARK CLARIFICATION/ PUNTEVERHELDERING 1.2 x y 3 and x 2  3 y 2  13  x  3 y x  3 y substitution  standard form 3  y   3 y  13 2 2  factors 9  6 y  y 2  3 y 2  13  0  2y2  6y  4  0 y-values y  3y  2  0 2 x-values  y  1 y  2  0 y  1 or y  2 OR/OF  x  3  1 or x  3  2  y  x 3 x4 or x  5 substitution OR/OF  standard form x  y  3 and x  3 y  13 2 2 y  x 3  factors x 2  3 x  3  13 2 x-values   x 2  3 x 2  6 x  9  13  0 y-values x  3 x  18 x  27  13  0 2 2  2 x 2  18x  40  0 (6) x 2  9 x  20  0 x  4x  5 x  4 or x  5 y  43 or y  5  3 y 1 y2 1.3 x2  7 x 7 x 7 x 5  x 2 .x 2 .x  x 5  x 2 .x 2 . x x 5  7 .7 . 7  answer x 5  49 7 (3) [24] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 5 Downloaded from hlayiso.com Mathematics P1/Grade 12 5 September 2021 Wiksunude V1/Graad 12 Marking Guidelines/Nasienriglyne QUESTION/VRAAG 2 ITEM SOLUTIONS/OPLOSSINGS MARK CLARIFICATION/ PUNTEVERHELDERING 2.1.1 171 282 387 486 ..... answer 111 105 99 (2) 6 6 2 nd diff  6 2.1.2 2a  6  a  3  a  3  b  120  3a  b  111  c  54  3 3  b  111 answer b  111  9 b  120 (4)  3  120  c  171 c  171  117 c  54 Tn  3n 2  120n  54 2.1.3 Tn  3n 2  120n  54 and Pn  60n  2754 equating the equations  3n 2  120n  54  60n  2754 standard form 3n 2  180n  2700  0 factors n 2  60n  900  0 answer n  302  0 (4) n  30 2.2.1 1 1 1 value of a and r   ........ 8 16 32 1 1 substitution a r 8 2 answer 1   1   16    1 8  2   (3) S16    1 1 2 S16  0.2499 or 0.25 or 0.250 or 0.2500 Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 6 Downloaded from hlayiso.com Mathematics P1/Grade 12 6 September 2021 Wiksunude V1/Graad 12 Marking Guidelines/Nasienriglyne ITEM SOLUTIONS/OPLOSSINGS MARK CLARIFICATION/ PUNTEVERHELDERING 2.2.2 1023 S  Tn  substitution 4096 1 1 n 1  S  8  1  1   1023 4 1 82 4096 1 2 n 1 n 1 1 1 1 11 1023        2 512 4 82 4096 n 1 9 n 1 1 1 11 1023 1      or      2 2 82 4096 4  1  1 n 1 1 n  1  log 1      2  512  2 512 1 n 1 1 9  1   n  10       or n  1  log 1    2 2 2  512   n 1  9  n  10 (5) [18] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 7 Downloaded from hlayiso.com Mathematics P1/Grade 12 7 September 2021 Wiksunude V1/Graad 12 Marking Guidelines/Nasienriglyne QUESTION/VRAAG 3 ITEM SOLUTIONS/OPLOSSINGS MARK CLARIFICATION/ PUNTEVERHELDERING 3.1 n 19 + 25 + 31 + …  6k  13 k 1 a and d 19  25  31  ..........6n  13 a  19 d  6  substitution into correct S n  19  6n  13 n formula 2 S n  6n  32 n (3) 2 S n  3n 2  16n OR/OF n  6k  13 k 1 19  25  31  ..........6n  13 a  19 d  6 S n  219  n  16 n 2 S n  38  6n  6 n 2 S n  6n  32 n 2 S n  3n 2  16n 3.2 S n  3n 2  16 n correct substitution S34  S33  334   16 34   (333   16 33 ) answer 2 2  4012  3795 (2)  217 3.3 T1 ; T2 ; T3 ; T4 ; 120 Tk  6k  13 sequence of first 19 25 31 37 differences. T4  120  37 T4  83  T4  83  T3  83  31  T3  52 T3  52 [8] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 8 Downloaded from hlayiso.com Mathematics P1/Grade 12 8 September 2021 Wiksunude V1/Graad 12 Marking Guidelines/Nasienriglyne QUESTION/VRAAG 4 y g A(3;10) B C D x f ITEM SOLUTIONS/OPLOSSINGS MARK CLARIFICATION/ PUNTEVERHELDERING 4.1 f ( x)   x 2  4 x  5  substitution/derivative 4 x OR f / ( x)  2 x  4 2 1 x2 0  2 x  4  x-value x2 y   2   42   5  y 8 2 y 9 B 2;9  (3) 4.2 p  2 and q  9  p  2 q 9 (2) Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 9 Downloaded from hlayiso.com Mathematics P1/Grade 12 9 September 2021 Wiksunude V1/Graad 12 Marking Guidelines/Nasienriglyne ITEM SOLUTIONS/OPLOSSINGS MARK CLARIFICATION/ PUNTEVERHELDERING 4.3 The graph of f reflects over the x-axis and shifts 10 units answer up to form the graph of t x    f x   10 . (2) The roots will be non-real. OR t x    f x   10 9 8 7 6 5 4 3 2 1 -1 1 2 3 4 -1 The roots will be non-real. 4.4 The graph shifted 2 units to the right and 6 units  m  2  down  n  6   m  2 and n  6 (2) 4.5 y  8x  k derivative f ( x)   x  4 x  5 2  8  2 x  4 f /  x   2 x  4 x-value 8  2 x  4 4  2 x y-value x  2 y  (2) 2  4 2   5 (4) y  7 P 2;  7  4.6.1 2  x  3 or x  (2;3] critical value(s) notation (2) 4.6.2 0  x  4x  5 2 standard form=0 critical value(s) 0  x 2  4x  5  x  1 x  5x  1  0 x5 x  1 or x  5  x  1 and x  5 (4) [19] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 10 Downloaded from hlayiso.com Mathematics P1/Grade 12 10 September 2021 Wiksunude V1/Graad 12 Marking Guidelines/Nasienriglyne QUESTION/VRAAG 5 ITEM SOLUTIONS/OPLOSSINGS MARK CLARIFICATION/ PUNTEVERHELDERING 5.1 f x   b x substitution answer 8  bx (2) 23  b 3 b2 f x   2 x 5.2 y  2x  x  2y x  2y  y  log 2 x (2) y  log 2 x 5.3 y Answer only full marks  A(8;3) 1 x shape asymptote  point A(8;3) OR (1;0) (3) 5.4 log 2 x  4 Answer only 16 x  24 full marks answer 0  x  16 (3) Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 11 Downloaded from hlayiso.com Mathematics P1/Grade 12 11 September 2021 Wiksunude V1/Graad 12 Marking Guidelines/Nasienriglyne 5.5  2x2 h x   f x  1 4 2nits    2 2 2 x right  2x2 The graph shifted 2 units to the right (3) [13] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 12 Downloaded from hlayiso.com Mathematics P1/Grade 12 12 September 2021 Wiksunude V1/Graad 12 Marking Guidelines/Nasienriglyne QUESTION/VRAAG 6 ITEM SOLUTIONS/OPLOSSINGS MARK CLARIFICATION/ PUNTEVERHELDERING 6.1.1  substitution into correct F  x 1  i   1 n  formula i  80  0.06 420   answer 4100 1    1  4   F (3) 0.06 4 F  626 114 ,50 6.1.2 2 5  0.062   substitution A  626114,501    2   answer  849 650,68 (2) 6.2.1 P  x 1  1  i  n  substitution into correct i formula   0.11  180  x 1  1      12   0.11 660 000   0.11 12 12  0.11   answer 660 000  x  12  180  0.11  (3) 1  1    12  x  7 501 ,54 Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 13 Downloaded from hlayiso.com Mathematics P1/Grade 12 13 September 2021 Wiksunude V1/Graad 12 Marking Guidelines/Nasienriglyne 6.2.2 Outstanding balance after the 84th payment -96 substitution into correct   0.11  96  formula 7 501,54 1  1      12   477 548,81 P 0.11 12 (3) Outstandin g Balance  477 548,81 OR/OF OR/OF   0.11  84 substitution  (A) 7501,54 1    1 84  12    0.11    answer  Outstandin g Balance  660 0001     12  0 .11 (3) 12  R 477 548,77 6.2.3 substitution (A)   application of logs 0.11   n 10 000 1  1      12   64 payments  477 548,81 0.11 12 32 payments sooner  0.11  477 548,81  n  12   1  1  0.11   12  10 000  (4) n  0.11   0.5622469242  1    12   n  log  0.11  0.5622469242  1   12   n  63,10 n  64 payments  96  64  32 payments sooner [15] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 14 Downloaded from hlayiso.com Mathematics P1/Grade 12 14 September 2021 Wiksunude V1/Graad 12 Marking Guidelines/Nasienriglyne QUESTION/VRAAG 7: PENALISE -1 FOR NOTATION ONLY IN 7.1 ITEM SOLUTIONS/OPLOSSINGS MARK CLARIFICATION/ PUNTEVERHELDERING 7.1 f x   4 x 2  3  4 x 2  8xh  4h2  3 f  x  h   4 x  h   3 2  substitution f  x  h   4 x 2  8 xh  4h 2  3 simplification f x  h   f x   common factor f /  x   lim h0 h  lim   4 x  8 xh  4h 2  3  4 x 2  3 2  answer h0 h 8 xh  4h 2  lim (5) h0 h h8 x  4h   lim h0 h  lim (8 x  4h) h0  8 x  40  8x  7.2.1 y  3 x  4 5 x  2    15 x  6 x  20 x  8 2  15x2  14x  8  15 x 2  14 x  8  30 x  14  dy (2)  30 x  14 dx 7.2.2 d  2 3 x x  2   x2  dx  x  d  2 3    2 x 2   x  2x2   dx   1  3  x2  3 1 2  x 2  4 x 3 C.A only if exponent 2 is a rational   4 x 3  (4) [11] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 15 Downloaded from hlayiso.com Mathematics P1/Grade 12 15 September 2021 Wiksunude V1/Graad 12 Marking Guidelines/Nasienriglyne QUESTION/VRAAG 8 ITEM SOLUTIONS/OPLOSSINGS MARK CLARIFICATION/ PUNTEVERHELDERING 8.1.1 x3  5 x 2  7 x  3  0 factors f 1  0  A1;0 x  1 x  3  0 2  B3;0 A1 ; 0 B3 ; 0 (3) Does not have to be in a coordinate form. Accept x  1 or x  3 8.1.2 h x   x 3  5 x 2  7 x  3 Derivative h /  x   3x 2  10 x  7 0  3x 2  10 x  7 Equation derivation to 0. 0  3 x  7  x  1 at C x  7 x-value 3 3 2 7 7 7 7 y-value f       5   7   3 3 3 3 3  32 (4)  27 7 32  C ;   3 27  8.2.1 7 substitution 1 x 3 2 x-value 5 b x OR x Answer 3 3a 5  (5) (3) x x 3 3(1) 5 x 3 OR/OF OR/OF h x   x 3  5 x 2  7 x  3 Second derivative <0 h /  x   3 x 2  10 x  7 h /  x   6 x  10  0 6 x  10  0 answer 5 x  (3) 3 Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 16 Downloaded from hlayiso.com Mathematics P1/Grade 12 16 September 2021 Wiksunude V1/Graad 12 Marking Guidelines/Nasienriglyne ITEM SOLUTIONS/OPLOSSINGS MARK CLARIFICATION/ PUNTEVERHELDERING 8.2.2 hx   0 for x  3 since h x  is a reflection over y  axis answer (2)  x  3 8.3 h x   4  p  p  4 the turning points of hx   4 are A1;4 and  p0  7 76   ;  76  3 27  p 27 76 (4) p4 or p  27 OR/OF p=4 p>4 (1;4)  7 76  p = 76  ;  p > 76 27  3 27  27 [12] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 17 Downloaded from hlayiso.com Mathematics P1/Grade 12 17 September 2021 Wiksunude V1/Graad 12 Marking Guidelines/Nasienriglyne QUESTION/VRAAG 9 ITEM SOLUTIONS/OPLOSSINGS MARK CLARIFICATION/ PUNTEVERHELDERING 9.1 S . A  2.l.w  2l.h  2 w.h formula  2 x.5 x  h.5 x  x.h  substitution 720  25 x 2  6 xh simplification 360  5 x 2  6 xh 60 5 h   x 360  5 x 2  6 xh x 6 60 5 substitution h  x x 6  60 5  (5) V  5 x.x  x   x 6  25 3 V  300 x  x 6 9.2 25 3 derivative V  300x  x 6 dV 25 2  300  x equating the derivative to 0. dx 2 x-value 25 2 0  300  x 2 25 2 substitution x  300 2 x 2  24 Answer x2 6   252 2 6   2939,39  v  300 2 6  3 (5) [10] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 18 Downloaded from hlayiso.com Mathematics P1/Grade 12 18 September 2021 Wiksunude V1/Graad 12 Marking Guidelines/Nasienriglyne QUESTION/VRAAG 10 Events A and B are given such that P A or B   and P A  . 3 2 5 5 ITEM SOLUTIONS/OPLOSSINGS MARK CLARIFICATION/ PUNTEVERHELDERING 10.1.1 P A OR B   P A  PB  substitution   P B  3 2 answer 5 5 P B    3 2 (2) 5 5 P B   1 5 10.1.2 P A or B   P A  PB   P A.PB  substitution   P B   P B  3 2 2 5 5 5  simplification  P B  1 3 5 5 answer P B   1 (3) 3 10.2 P A or B   P A  PB   P A.PB  substitution 0.84  y  y  y. y standard form 0.84  2 y  y 2 substitution into y 2  2 y  0.84  0 formula/factors   2    22  410.84 y  answers 21 selection y  1.4 OR y  0.6 (5)  y  0,6 [10] Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ### Page 19 Downloaded from hlayiso.com Mathematics P1/Grade 12 19 September 2021 Wiksunude V1/Graad 12 Marking Guidelines/Nasienriglyne QUESTION/VRAAG 11 ITEM SOLUTIONS/OPLOSSINGS MARK CLARIFICATION/ PUNTEVERHELDERING 11.1 7  7  7  7  2401 codes OR 74 7777  2401 NOTE: Answer only award full marks (2) Also if left as 7 or 7  7  7  7 award full 4 marks 11.2 Case 1: The code must start with 2 and end with 0 2 0  1 5  4  1 5 4 1 way 1 5  4  1 Number of codes starting with 2 and ending with 0 is  1 5  4  1 1 5  4 1  20  60 Case 2: number of codes starting with 2 and ending with 4 2 5 4 4 OR 1 way 1 way  1 5  4  1 3 60 1 5  4 1  20 NOTE: Answer only 2 marks Case 3: number of codes starting with 2 and ending will 6 1  5  4  1  20 (4) Total number of codes = 20 + 2 0+ 20 = 60 codes OR/OF 1 5  4  1 3  60 [6] TOTAL/TOTAAL: 150 Copyright reserved/Kopieregvoorbehou Please turn over/Blaai om asseblief ## Related documents - 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