# Mathematics Grade 12 MP March 2021 Memo > Grade 12 Mathematics memorandum from the MP March 2021, 22 pages. Read online or get the free PDF download, question paper linked when available. - Page: https://www.hlayiso.com/library/maths-april-qp-memo-mp-2021-gr12/bce1992a-1659-4444-a434-9e73df51831e - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 12 - Document type: Memorandum - Year: 2021 - Exam period: MP March - Curriculum: CAPS - Pages: 22 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 MP March 2021 Memo](https://www.hlayiso.com/library/maths-april-qp-memo-mp-2021-gr12/bce1992a-1659-4444-a434-9e73df51831e) ## About this document Mathematics · Grade 12 · MP March · 2021. Memorandum, 22 pages. Read online or download the PDF. 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Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com GRADE 12 MATHEMATICS Date: 13 April 2021 Time: 2 hours Marks: 100 Instructions: Read the following instructions carefully before answering the questions. o This question paper consists of 7 questions in Section A and one question in Section B o Answer ALL the questions in SECTION A and SECTION B is optional. o Clearly show ALL calculations, diagrams, graphs, etc. that you have used in determining your answers. o Answers only will not necessarily be awarded full marks. o You may use an approved scientific calculator (non-programmable and non- graphical), unless stated otherwise. o If necessary, round off answers to TWO decimal places, unless stated otherwise. ### Page 2 Downloaded from hlayiso.com 13 April 2021 Page 2 of 12 Term 1 Test 1 o Diagrams are NOT necessarily drawn to scale. o An information sheet, with formulae, is included at the end of the question paper. o THE diagram sheet that is included at the end of the paper must be handed in with your test, with construction lines added to the diagrams where necessary. o Number the answers correctly according to the numbering system used in this question paper. o Write legibly and present your work neatly. ### Page 3 Downloaded from hlayiso.com 13 April 2021 Page 3 of 12 Term 1 Test 1 SECTION A QUESTION 1 Given the sequence –5; 4; 21; 46; …… 1.1 Determine the general term of the above sequence. (4) 1.2 Determine T15 (1) 1.3 Which term in the sequence will be equal to 364? (3) [8] QUESTION 2 m 2.1  32 (2) i =2 5−i  500 2.1.1 Determine the value of m for which the above-mentioned statement is true, by using the correct sum formula. (4) 2.1.2 Determine the value for S  − S 4 (3) [7] QUESTION 3 2 x; x + 1; 6 − x; . . . are the first three (3) terms of an arithmetic sequence. 3.1 Determine the value for x . (2) 3.2 If x = 4 , how many terms in the sequence add up to − 575 . (4) [6] QUESTION 4 n The sum of the first n terms of a series is given by : Sn = (14 − 4n) 8 4.1 Determine the sum of the first 25 terms of this series. (1) 4.2 Determine the value of term 25. (3) 4.3 Determine the general term of the series (5) [9] ### Page 4 Downloaded from hlayiso.com 13 April 2021 Page 4 of 12 Term 1 Test 1 Question 5 5.1 If cos26 = q , write the following in terms of p : 5.1.1 cos334 (1) 5.1.2 sin 52 (3) 5.1.3 sin 86 (2) [6] Question 6 6.1 Given in the diagram below ABC and PQR with  = P̂, B̂ = Q̂ and Ĉ = R̂. A B C AB AC Prove the theorem that states that if ABC |||PQR then = . (6) PQ PR ### Page 5 Downloaded from hlayiso.com 13 April 2021 Page 5 of 12 Term 1 Test 1 6.2 Given in the diagram below, VR is the diameter of the circle with centre O. S is a point on the circumference. P is the midpoint of RS. The circle with RS as diameter intersects VR at T. ST, OP and SV are drawn. 6.2.1 Give a reason why OP ⊥ RS. (1) 6.2.2 Prove that ROP ||| RVS . (4) 6.2.3 Prove that RVS ||| RST . (3) 6.2.4 Prove that ST 2 = VT. TR (5) [19] ### Page 6 Downloaded from hlayiso.com 13 April 2021 Page 6 of 12 Term 1 Test 1 QUESTION 7 7.1 In the diagram below, O is the centre of the circle and P is a point on the circumference of the circle. Arc AB subtends AÔB at the centre of the circle and AP̂B at the circumference of the circle. Use the diagram to prove the theorem that states that AÔB = 2AP̂B (5) ### Page 7 Downloaded from hlayiso.com 13 April 2021 Page 7 of 12 Term 1 Test 1 7.2 In the diagram, O is the centre of the circle and P, Q, S and R are points on the circle. PQ = QS and QR̂S = y .The tangent PT at P meets SQ produced at T. OQ intercepts PS at A. 7.2.1 Give a reason why P̂2 = y . (1) 7.2.2 Prove that PQ bisects TP̂S. (4) 7.2.3 Determine P Ô Q in terms of y . (2) 7.2.4 Prove that PT is a tangent to the circle that passes through P, O and A. (2) 7.2.5 Prove that OÂP = 90 . (4) [18] Total Section A: 73 marks ### Page 8 Downloaded from hlayiso.com 13 April 2021 Page 8 of 12 Term 1 Test 1 SECTION B: OPTIONAL QUESTION 8 8.1 Calculate the following without using calculator: 81.1 sin 236.cos169 + sin 371.cos(−124) (4) − cos10 + sin 2 190 8.1.2 (6) cos(−145).cos235 8.2 Prove the following identities: cos2A + sin A 2 sin A + 1 8.2.1 = (5) cos2 A 1 + sin A sin( x + 45 ) sin 2 x + 1 8.2.2 = (6) cos(x − 45 ) (sin x + cos x) 2 8.3 Determine the general solution for: 8.3.3 2 sin(3x − 15) + 1 = 0 (4) 8.3.4 Hence determine all possible values for x , If x [−270; 90] (2) [27] Total Section B: 27 marks ### Page 9 Downloaded from hlayiso.com 13 April 2021 Page 9 of 12 Term 1 Test 1 INFORMATION INFORMATION SHEET SHEET: MATHEMATICS − b  b 2 − 4ac x= 2a A = P(1 + ni) A = P(1 − ni) A = P(1 − i ) n A = P(1 + i ) n Tn = a + (n − 1)d Sn = n (2a + (n − 1)d ) 2 Tn = ar n −1 Sn = ( a r n −1 ) ; r 1 S = a ; −1  r  1 r −1 1− r F=  x (1 + i ) − 1 n  P= x[1 − (1 + i ) − n ] i i f ( x + h) − f ( x ) f ' ( x) = lim h→ 0 h  x + x2 y1 + y 2  d = ( x 2 − x1 ) 2 + ( y 2 − y1 ) 2 M  1;   2 2  y − y1 y = mx + c y − y1 = m( x − x1 ) m= 2 m = tan x 2 − x1 ( x − a ) 2 + ( y − b )2 = r 2 a b c 1 In ABC: = = a 2 = b 2 + c 2 − 2bc. cos A area ABC = ab. sin C sin A sin B sin C 2 sin ( +  ) = sin  . cos  + cos . sin  sin ( −  ) = sin  . cos  − cos . sin  cos( +  ) = cos  . cos  − sin  . sin  cos( −  ) = cos  . cos  + sin  . sin  cos2  − sin 2   cos 2 = 1 − 2 sin 2  sin 2 = 2 sin . cos 2 cos2  − 1  n 2  fx  (x − x ) i x=  = 2 i =1 n n n( A) P( A) = 𝑃(𝐴 𝑜𝑟 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) – 𝑃(𝐴 𝑎𝑛𝑑 𝐵) n(S ) yˆ = a + bx b=  (x − x )( y − y ) (x − x) 2 ### Page 10 Downloaded from hlayiso.com 13 April 2021 Page 10 of 12 Term 1 Test 1 Diagram Sheet Question 6 6.1 A B C 6.2 ### Page 11 Downloaded from hlayiso.com 13 April 2021 Page 11 of 12 Term 1 Test 1 7.1 ### Page 12 Downloaded from hlayiso.com 13 April 2021 Page 12 of 12 Term 1 Test 1 7.2 ### Page 13 Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 MATHEMATICS TEST TERM 1 MARKING GUIDELINE 2021 Time: 1,5 hours Marks: 73 Section A Time: 2 hours Marks: 100 Section A and B ### Page 14 Downloaded from hlayiso.com QUESTION 1 1.1 − 5; 4; 21; 46;..... ✓a= 4 9 17 25 ✓ b = −3 8 8 2a = 8 ✓ c = −6 a=4 ✓ Tn 3a + b = 9 3(4 ) + b = 9 (4)  b = −3 a + b + c = −5 4 − 3 + c = −5  c = −6 Tn = 4n 2 − 3n − 6 1.2 T(15 ) = 4(15) − 3(15) − 6 2 ✓answer (1) = 849 364 = 4n 2 − 3n − 6 ✓ 360 = 4n − 3n − 6 1.3 2 4n 2 − 3n − 370 = 0 − 37 n= (4n + 37)(n − 10) = 0 ✓ 4 , NA − 37 ✓n = 10 n= or n = 10 4  n = 10 (3) [8] QUESTION 2 2.1.1  1m  m  32 (2) 5−i  500 2561 −  i =2   2   500 256 + 128 + 64 + ... 500 1 Sn = ( a 1− rm ) ✓ 1− 2 1− r  1m  ✓correct use of logs 2561 −    2   500 ✓ m  5.4 1 1− ✓m= 6 2  1m  5121 −   (4)  2   500 1 ### Page 15 Downloaded from hlayiso.com m 1 125 1−  2 128 m 1 3  2 128 3 m  log 1 2 128 m  5 .4 m=6 S − S4 = a − ( a 1− rn) 1− r 1− r 2.1.2  14  2561 −  = 256 −  2  ✓ S − S4 1 1 1− 1− 2 2 ✓substitution = 512 − 480 = 32 ✓answer (3) [7] QUESTION 3 3.1 2 x; x + 1; 6 − x; . . . x + 1 − 2 x = 6 − x − ( x + 1) ✓ T2 − T1 = T3 − T2 − +1 = −2 x + 5 ✓answer (2) x=4 3.2 8; 5; 2; … n ✓ substitution of S n − 575 = [2(8) + (n − 1)(−3)] 2 ✓substitution of a and d − 1150 = n(19 − 3n) 3n 2 − 19n − 1150 = 0 ✓standard form (3n + 50)(−23) = 0 50 ✓ n = 23 only n = 23 or n  − (4) 3 [6] ### Page 16 Downloaded from hlayiso.com QUESTION 4 4.1 25 S 25 = [14 − 4(25)] ✓answer (1) 8 3 = −268 4 4.2 T 25= S 25 − S 24 ✓method ✓substitution 3  24 3 ✓answer T 25= −268 −  [14 − 4(24 )] = 22  (3) 4  8 4 4.3 1 5 S1 = T1 = 5 S1 = T1 = [14 − 4(1)] = ✓ 4 8 4 2 5 1 T2 = [14 − 4(2)] − = T = 1 8 4 4 ✓ 2 4 3 3 3 3 T3 = [14 − 4(3)] − = − ✓ T3 = − 4 8 2 4 T 5; 1; − 3; → n = 9 − 4 n 9 − 4n ✓ Tn = 9 − 4n ; ; − ; → Tn = 5 1 3 4 9 − 4n 4 4 4 Tn = ✓ 4 (5) [9] QUESTION 5 5.1.1 cos334 = cos(360 − 26) ✓answer = cos 26 =q (1) 5.1.2 1 26o q sin 52 = sin 2(26) ✓diagram = 2 sin 26 cos 26 ✓identity = 2. 1 − q 2 (q ) = 2q 1 − q 2 ✓answer (3) 5.1.3 sin 86 = sin(60 + 26) = sin 60 cos26 + cos60 sin 26 ✓identity 3 1 = .q + 1 − q2 2 2 ✓answer (2) ### Page 17 Downloaded from hlayiso.com [6] QUESTION 6 6.1 Constructions: Draw PQ on AB such that PQ =AD ✓Construction Draw PR on AE such that PR=AE In ΔADE and PQR 1. AD = PQ [construction] 2. AE = PR [construction] 3. 𝐴̂ = 𝑃̂ [given] ΔADE  PQR [ SS ] ✓S/R D̂1 = Q̂ [from congruency] But D̂1 = B̂ ✓S/R  D̂1 = B̂  DE//BC [ corrsponding S equal] . ✓ D̂1 = B̂ ✓S/R AB AC = [line || side of ] AD AE . But AD = PQ and AE = PR [construction] ✓S/R AB AC = PQ PR (6) ### Page 18 Downloaded from hlayiso.com 6.2.1 Line from centre to midpoint of chord ✓Reason (1) 6.2.2 In ΔROP and RVS 1. R̂ = R̂ [common] ✓S/R 2. Ŝ1 = 90  [angle in semi-circle] ✓R P̂ 2 = 90  (proven) Ŝ1 =P̂ 2 ✓S 3. V̂ = Ô 2 (angles in  ) ✓R (4) ΔROP ||| RVS [∠;∠;∠] 6.2.3 In ΔRVS and RST 1. R̂ = R̂ (common) 2. T̂1 = Ŝ1 = 90  (angle in semi circle) ✓S ✓R 3. TŜR = V̂ (s in ) ✓R ΔRVS ||| RST (∠;∠;∠) (3) 6.2.4 In ΔSTV and RST RT̂S = VT̂S = 90  (Angles on straight line) ✓S ✓R R̂ = 90 − TŜR ✓S = TŜV TŜR = V̂ (angles in  ) ΔRST ||| STV ([A,A,A) RT TS ✓R = (from similarity) ST VT ✓S (5) ST 2 = VT. TR [19] ### Page 19 Downloaded from hlayiso.com QUESTION 7 7.1 1 2 1 2 1 Construction: Draw PO extended ✓Construction OP = OA (radii) P̂1 =  (angles opp. equal sides) ✓S/R But Ô1 = P̂1 +  (ext. angle of triangle) ✓S/R Ô1 = 2 P̂1 Similarly ✓S Ô 2 = 2P̂2 ✓S (5) AÔB = 2AP̂B 7.2.1 Angles in the same segment ✓answer (1) 7.2.2 P̂2 = Sˆ1 = y ✓S ✓R ( angles opp equal sides) Ŝ1 = P̂3 = y (tan cord theorem) ✓S ✓R (4) P̂2 = P̂3 PQ bisects TP̂S 7.2.3 PÔQ = 2Ŝ1 = 2 y (  at centre = 2  at circumference) ✓S ✓R (2) 7.2.4 TP̂A = P̂2 + P̂3 (proven) TP̂A = PQ̂O (proven) ✓S PT is a tangent (converse theorem tan cord) ✓R (2) 7.2.5 OP̂Q + OQ̂P = 180 -2 y ✓S ✓R (angles of triangle) OQ̂P = 90 -y (angles opp equal sides) ✓S/R 90  − y + y + QÂP = 180  ✓S (4) QÂP = 90  [18] ### Page 20 Downloaded from hlayiso.com OPTIONAL: QUESTION 8 8.1.1 sin 236.cos169 + sin 371.cos(−124) ✓ − sin 56(− cos11) = − sin 56(− cos11) + sin 11.(− cos56) ✓ sin 11.(− cos56 = sin(56 − 11) = sin 45 ✓ sin 45 1 = 1 2 ✓ 2 (4) 8.1.2 − cos2 10 + sin 2 190 cos(−145).cos 235 − cos2 10 + sin 2 (180 + 10) = cos(180 − 35).cos(270 − 35) ✓ + sin 10 2 − cos 2 10  + sin 2 10  = − cos 35 .( − sin 35 ) ✓ − cos35 = ( − cos 2 10  − sin 2 10  ) ✓ − sin 35 cos 35  sin 35  − cos 2  10 = cos35 sin 35 − cos 20 = ✓ − cos20 cos35 sin 35 − 2 cos 20 = 2 cos35 sin 35 − 2 cos 20 = sin 2  35 − 2 sin 70 ✓ − 2 sin 70 = sin 70 ✓ sin 70 = −2 (6) 8.2.1 cos 2A + sin A 2 sin A + 1 = cos2 A 1 + sin A LHS: ✓ 1− 2 sin A 2 cos 2A + sin A 1 − 2 sin 2 A + sin A = 1 − sin 2 A ✓ 1− sin A 2 cos 2 A 1 + sin A − 2 sin 2 A = 1 − sin 2 A ### Page 21 Downloaded from hlayiso.com (1 + 2 sin A)(1 − sin A) ✓factorise numerator = ✓factorise denominator (1 − sin A)(1 + sin A) (4) 1 + 2 sin A = (1 + sin A)  LHS = RHS 8.2.2 sin( x + 45 ) sin 2 x + 1 = cos(x − 45 ) (sin x + cos x) 2 LHS: ✓ sin x cos45 + cosc sin 45 sin( x + 45 ) sin x cos 45  + cos c sin 45  = ✓ cos x cos45 + sin x sin 45 cos(x − 45 ) cos x cos 45  + sin x sin 45  2 2 2 sin x. + cos x ✓Substituting 2 = 2 2 cos x 2 + sin x 2 ✓1 2 2 =1 RHS: sin 2 x + 1 sin 2 x + 1 ✓simplifying denominator 2 = (sin x + cos x) sin x + sin x cos x + cos2 x 2 ✓square identity sin 2 x + 1 ✓1 = (7) 1 + 2 sin x cos x sin 2 x + 1 = 1 + sin 2 x =1  LHS = RHS 8.3.1 2 sin(3x − 15) + 1 = 0 1 sin(3x − 15) = − 2 1 Ref angle: x = 30 sin(3x − 15) = − ✓ 2 3rd 3x − 15 = 180 + 30 + k.360; k  3x = 225 + k.360; k   ✓ x = 75 + k.120 x = 75 + k.120; k   ✓k  4th 3x − 15 = 360 − 30 + k.360; k  3x = 345 + k.360; k   ✓ x = 115 + k.120 (4) ### Page 22 Downloaded from hlayiso.com x = 115 + k.120; k  8.3.2 x  − 245  ; − 165  ; − 125  ; − 45  ; − 5 , 75  ✓ three correct ✓six correct (2) ## Related documents - [MATHS P1 GR12 MEMO SEPT 2021 English IsiXhosa](https://www.hlayiso.com/library/maths-p1-gr12-memo-sept-2021-english-isixhosa/6e14ca2d-44f7-417b-b314-e40962f91676) — Mathematics · Memo · 2021 · Eng - [Maths-April-QP Memo-EC-2021-Gr12](https://www.hlayiso.com/library/maths-april-qp-memo-ec-2021-gr12/e1469b3f-b9fe-40c2-9c23-509439908f72) — Mathematics · Memo · 2021 - [Maths-April-QP Memo-Gauteng-2021-Gr12](https://www.hlayiso.com/library/maths-april-qp-memo-gauteng-2021-gr12/e8fb9e00-d9bb-4e33-a21d-a354cdc802ad) — Mathematics · Memo · 2021 - [MATHS P1 QP GR12 SEPT 2021 Afrikaans](https://www.hlayiso.com/library/maths-p1-qp-gr12-sept-2021-afrikaans/ecc6ca73-048f-4257-8f9a-db12f35059e1) — Mathematics · QP · 2021 · Afr - [MATHS P1 QP GR12 SEPT 2021 English](https://www.hlayiso.com/library/maths-p1-qp-gr12-sept-2021-english/12f7db49-5bd7-414f-af58-65ed09d4c3e4) — Mathematics · QP · 2021 · Eng - [Mathematics P2 Nov 2020 ANSBK Afr Eng (2) final](https://www.hlayiso.com/library/mathematics-p2-nov-2020-ansbk-afr-eng-2-final/199754f9-9f9b-4bd9-8890-beebee25863f) — Mathematics · QP · 2021 · Eng ## More from Grade 12 Mathematics - [MATHS P1 ANSWER BOOK GR12 SEPT 2025 English Afrikaans watermark](https://www.hlayiso.com/library/maths-p1-answer-book-gr12-sept-2025-english-afrikaans-watermark/d78c3fd3-dcd5-4f3f-911b-e5635b4686cc) — Addendum · 2025 · Afr - [Maths Grade 12 NSC P2 QP September 2025 Gauteng](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-gauteng/1d6bf7fd-d288-4a01-99f7-2c87e247760c) — QP · 2025 - [Maths Grade 12 NSC P2 QP September 2025 Afr Limpopo](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-afr-limpopo/f2f8d481-5d87-4e89-bb21-6077bbcf403d) — QP · 2025 · Afr - [Maths Grade 12 NSC P1 QP September 2025 Afr Northern Cape](https://www.hlayiso.com/library/maths-grade-12-nsc-p1-qp-september-2025-afr-northern-cape/f5b3a4ce-fa91-4475-96e0-0654a87e1405) — QP · 2025 · Afr - [Maths Grade 12 NSC P2 QP September 2025 Eng Northern Cape](https://www.hlayiso.com/library/maths-grade-12-nsc-p2-qp-september-2025-eng-northern-cape/4ede8207-ca82-4419-bae1-fae734498cc7) — QP · 2025 · Eng - [MATHS P2 ANSWER BOOK GR12 JUNE 2024 Afr+English](https://www.hlayiso.com/library/maths-p2-answer-book-gr12-june-2024-afr-english/4a71e975-4ac4-4f55-8e23-0b8155a5b0c6) — Addendum · 2024 · Eng ## Collection - [Grade 12 past papers](https://www.hlayiso.com/grade-12/past-papers) --- Hlayiso: free CAPS and IEB past papers, memos, notes and study guides for South African learners. 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