# Mathematics Grade 11 P2 Gauteng November 2023 Memo (English) > Grade 11 Mathematics Paper 2 memorandum from the Gauteng November Exam 2023 (English), 14 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/gr-11-mathematics-p2-english-2023-possible-answers/78699db6-4378-41d9-973a-4daae4cd46d5 - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 11 - Language: English - Document type: Memorandum - Year: 2023 - Exam period: Gauteng November Exam - Paper: 2 - Curriculum: CAPS - Pages: 14 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 P2 Gauteng November 2023 Memo (English)](https://www.hlayiso.com/library/gr-11-mathematics-p2-english-2023-possible-answers/78699db6-4378-41d9-973a-4daae4cd46d5) ## About this document Mathematics · Grade 11 · Gauteng November Exam · 2023 · English. Memorandum, 14 pages. Read online or download the PDF. Read it free online on the page above; download the PDF from there. ## Matching paper and memo - [Gr 11 Mathematics P2 (English) 2023 Question Paper](https://www.hlayiso.com/library/gr-11-mathematics-p2-english-2023-question-paper/b8f6a489-3cb4-476d-8c5a-3c2de035fcf8) — QP · 2023 · Eng ## Document text Text extracted from the PDF. Layout, diagrams and tables may not survive extraction. ### Page 1 Downloaded from hlayiso.com PROVINCIAL EXAMINATION NOVEMBER 2023 GRADE 11 MARKING GUIDELINES MATHEMATICS (PAPER 2) 14 pages ### Page 2 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 INSTRUCTIONS AND INFORMATION ➢ A – ACCURACY ➢ CA – CONSISTENT ACCURACY ➢ S – STATEMENT ➢ R – REASON ➢ S & R – STATEMENT with REASON NOTES: • 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/answers in order to solve a question is UNACCEPTABLE. 2 ### Page 3 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 1 1.1 Class interval Frequency Cumulative frequency ✓ Frequency values 300 < x  400 4 4 ✓ Cumulative 400 < x  500 11 15 frequency values 500 < x  600 22 37 600 < x  700 13 50 700 < x  800 7 57 800 < x  900 3 60 (2) 1.2 70 ✓ (300 ; 0) ✓ End points 60 ✓ Shape 50 Cumulative Frequency 40 30 20 10 0 0 100 200 300 400 500 600 700 800 900 1000 Daily Sales (3) 1.3 Q2 = R580  Answer Accept R560 to R600 (2) 1.4 On 10 days they made a loss.  Answer  50 days they made a profit. (2) 1.5 To be in the top 25% of the previous months, the tuck ✓ R650 shop would need to have sold more than R650. ✓ Conclusion Therefore, R725 would have been in the top 25%. (2) [11] 3 ### Page 4 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 2 2.1 2k + k + 1 + k + 2 + k − 3 + 2k − 2 ✓ 7k – 2 = 15 5 ✓ Divide by 5 7k − 2 = 15 ✓ Manipulation 5 7k – 2 = 75 7k = 77 k = 11 (3) 2.2 22 ; 12 ; 13 ; 8 ; 20  Answer  = 5,22 (2) 2.3 Mean would decrease by t units but the standard ✓ Mean decreases by t deviation would be unaffected. units ✓ Standard deviation unaffected (2) [7] QUESTION 3 3.1 0 − ( −2) ✓ Substitution into mMQ = − 2 − ( −4) correct formula 1 mMQ = − ✓ Answer 3 (2) 3.2 1 ✓ Perpendicular gradients mMQ  − = −1 MP ⊥ MQ 3 ✓ Gradient of MP mMP = 3 ✓ Answer y = 3x + 6 (3) 3.3 xR + 4 y R + ( −2) ✓ Substitution into –2 = 0= midpoint formula 2 2 ✓ xR x R = −8 yR = 2 ✓ yR R (–8 ; 2) (3) 4 ### Page 5 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 3.4 PQ = (0 − 4) 2 + (6 − ( −2)) 2 ✓ Substitute into distance formula PQ = 80 PQ = 4 5 ✓ Simplified surd (2) 3.5 E(–4 ; –6) ✓ x E = −4 ✓ y E = −6 (2) 3.5 The diagonals of the parallelogram are perpendicular. ✓ Diagonals perpendicular ✓ Given parallelogram (2) [14] QUESTION 4 4.1 subst. x = 0 ✓ x=0 py – x – 5p = 0 ✓ Manipulation py – (0) = 5p ✓ x=5 5p y= p y=5 C(0 ; 5) (3) 4.2 py – x – 5p = 0 ✓ Standard form py = x + 5p ✓ Equate gradients x y= +5 p 1 mAC = 2 1 1  = p 2 p=2 OR 1 A(–10 ; 0) mAC = ✓ A (–10 ; 0) 2 ✓ Substitute A p(0) – (–10) – 5p = 0 10 = 5p p=2 (2) 5 ### Page 6 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 4.3 Equation of AC: 2y – x – 10 = 0 ✓ Substitute p = 2 1 y= x+5 2 ✓ Equation of BD Equation of BD: y = – 2x – 10 1 ✓ Substitute y into other x + 5 = – 2x – 10 2 equation x + 10 = – 4x – 20 5x = – 30 x = –6 ✓ x = –6 y=2 ✓ y=2 B(–6 ; 2) (5) 4.4 1 ✓ Substitution into correct tan  = 2 formula  = 26,57 ✓ Answer (2) 4.5 AĈD = 63,43 int 's of  ✓ AĈD  = 26,56 int 's of  ✓  = ✓ Rede  ABOD is a cyclic quadrilateral (converse 's in same segment) (3) 4.6 DB is a diameter (line subtends 90) ✓ x=0  0 + 0 5 + ( −10)  5 M CD =  ;  ✓ y=–  2 2  2  5 = M CD  0 ; −   2 (2) [17] 6 ### Page 7 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 5 5.1 5.1.1 2 ✓ Quadrant II sin  = 3 y ✓ x=– 5 2 2 2 x + (2) = 3 ✓ Substitute x=– 5 3cos2  – 1 x cos   5 2 ✓ Answer =3   –1  3  5 = –1 3 2 = 3 (4) 5.1.2 tan (– –180) ✓ –tan  = –tan  ✓ Substitution  2  ✓ Answer = –  − 5 2 = 5 (3) 5.2 5.2.1 t cos 15 = 4 t ✓ cos 15 = 4 4 cos 15 = t t ✓ t 2 − 16 sin 15 ✓ t t − 16 2 15° = 4 t (3) 5.2.2 sin 75 ✓ Complement = cos 15 ✓ Answer 4 = t (2) 7 ### Page 8 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 5.2.3 1 – tan2 15  t 2 − 16  2  t 2 − 16  2 ✓    4  =1–      4    t 2 − 16 t 2 − 16 ✓ =1– 16 16 ✓ Correct 16 − t 2 + 16 = multiplication by 16 32 − t 2 LCD = 16 (3) 5.3 cos(90 − α ) sin( − − 540 ) ✓ tan 45 ✓ sin .– sin  tan 225 + sin . sin(180 + ) ✓ sin .sin  sin α.sin α ✓ 1 – sin2  = tan 45 + (sin .)( − sin α ) 1 ✓ sin2 α tan 2 α = 1 − sin2 α sin 2 α = cos 2 α = tan2  (5) 5.4 5.4.1 1 – cos = 2 sin2 ✓ 2 (1 – cos2) 1 – cos = 2 (1 – cos2) 1 – cos = 2 – 2 cos2 ✓ standard form 2 cos2 – cos  – 1 = 0 (2 cos + 1)(cos – 1) = 0 (2) 5.4.2 2cos + 1 = 0 or cos – 1 = 0 1 ✓ cos = – 1 2 cos = – or cos = 1 2 ✓ cos = 1 RA = 60 or  = 0 + K.360 ; K  Z ✓ Both solutions for QII :  = 120 + K.360 ; K  Z 1 cos = – QIII :  = 240 + K.360 ; K  Z 2 ✓  = 0 + K.360 ✓ +KZ (5) [27] 8 ### Page 9 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 6 6.1 180 ✓ Answer (1) 6.2 1 1 ✓ Endpoints – y 2 2 ✓ Notation (2) 6.3.1 45  x  90 ✓ x  45 ✓ x < 90 (2) 6.3.2 –90< x < – 45 or 90< x < 135 ✓ –90< x < – 45 ✓ 90< x < 135 (2) 6.3.3 2 tan x + cos 2x = 2 ✓ 2 tan x – 2 = –cos 2x 2 tan x – 2 = –cos 2x 1 ✓ tan x – 1 = – cos 2x 1 2 tan x – 1 = – cos 2x 2 ✓ Value of x  f (x) = g(x) where x = 45 (3) 6.4 sin x + cos x ✓ tan x + 1 h(x) = cos x ✓ Up by 2 units sin x h(x) = +1 cos x h(x) = tan x + 1 h(x) = f (x) + 2  Vertical translation up by 2 units (2)  1  ✓ –2 cos 2x 6.5 p(x) = 4  − cos 2 x   2  ✓ Max of 2 p(x) = – 2 cos 2x  Maximum value of 2 (2) [14] 9 ### Page 10 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 7 1 1 7.1 tan p = ✓ tan p = 3 3 p = 18,435 ✓ Value of p to 3 decimal places (2) 7.2 In BCD ✓ Correct substitution CD 121 into sine rule = ✓ Manipulation  18,4  sin135 sin  ✓ Length of CD  2  121 x sin 9,2 CD = sin135 CD = 27,36 m (3) [5] QUESTION 8 8.1 V = x2 h ✓ Substitute into volume 490 = x2h formula 490 ✓ h in terms of x h= 2 x (2) 8.2 A = x2 + 4xh ✓ Base of x2  490   490  A = x2 + 4x  2  ✓ 4x  2   x   x  1960 3 A = x2 + cm x (2) 8.3 490 ✓ Substitute h = 10 10 = x2 ✓ x2 = 49 x2 = 49 ✓ x=7 x=7 (x > 0) ✓ Surface area 1 960 A = (7)2 + 7 2 A = 329 cm (4) [8] 10 ### Page 11 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 9 9.1 Q 1 2 O 1 2 P R Join PO and extend ✓ Construction In ΔPOQ ✓ SR Ô1 = Q̂ + P̂1 ext  of  ✓ SR but Q̂ = P̂1 ' s opp equal radii ✓ S  Ô1 = 2P̂1 Similarly in  ROQ Ô 2 = 2P̂2 ✓ S  Ô1 + Ô 2 = 2P̂1 + 2P̂2 ✓ S ( Ô1 + Ô 2 = 2 P̂1 + P̂2 )  RÔQ = 2P̂ (6) 9.2 9.2.1 (a) D̂ = 55° ' s in the same segment ✓ S R (2) (b) Ô 2 = 110°  at centre = 2   at ✓ S R circumference (2) (c) Ê 2 = 110° – 90 ext  of  ✓ R ✓ S Ê 2 = 20° (2) 11 ### Page 12 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 (d) Ê1 = 35° int ' s of  ✓ R ✓ S ✓R Ê 3 = 35° ' s in a semi-circle OR Ê 3 = F̂1 ' s opp equal radii ✓ S R ✓ SR Ê 3 = 35° int ' s of  (3) 9.2.2 OE = 5 equal radii ✓ SR HE = 4,7 line from centre perpendicular ✓ S ✓ R to chord OH2 = 52 – (4,7)2 Pythagoras' Theorem ✓ S OH = 1,71 units OR OE = 5 radii ✓ SR OH ✓✓ sin 20° sin 20 = 5 ✓ OH OH = 1,71 units OR HE = 4,7 line from centre ✓ S R perpendicular to chord ✓ tan 20° OH ✓ OH sin 20 = 4,7 OH = 1,71 units (4) [19] QUESTION 10 10.1 B̂1 = x alt. 's, AC ‖‖ TV ✓ S ✓R V̂1 = x ' s opp equal sides ✓ S ✓R T̂2 = x tan chord theorem ✓ S ✓R (6) 10.2 Â = 2 x opp ' s parm ✓ S ✓R B̂3 = 2 x ext  cyclic quad ✓ S ✓R  Â = B3 both equal 2x ✓ R AT = BT sides opp equal ' s 12 ### Page 13 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 OR B̂3 = 2 x ext  cyclic quad ✓ S ✓R ✓ SR T̂1 = 2 x alt 's, AC ‖‖ TV  T̂1 = V̂1 + V̂2 both equal 2x ✓ S ✓R  BTVC is an isosceles trapezium (one pair parallel sides and one pair of base angles ✓ Logic equal)  CV = BT sides of isos trap en CV = AT opp sides parm  AT = BT (5) 10.3 In  ATB, ✓ R 5x = 180° int ' s of  ✓ S x = 36° (2) [13] QUESTION 11 11.1 D̂3 = Ĥ1 + Ê1 ext  of  ✓ S (1) 11.2 D̂3 = Ĥ1 + Ê1 ext  of  Ĥ1 = F̂2 tan chord theorem ✓ S ✓R ✓ SR and Ê1 = F̂1 tan chord theorem  D̂ 3 = F̂1 + F̂2 (  DEFH is a cyclic quadrilateral (ext  = opp int  ) ✓ R (with D̂3 = F̂1 + F̂2 ) (4) 11.3 Ĥ1 = F̂2 tan chord theorem ✓ SR ✓ S ✓R Ĥ1 = F̂1 ' s in the same segment  F̂1 = F̂2 (3) 13 ### Page 14 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 11.4 Ĥ 3 + Ĥ 4 = Ê1 + Ê 2 ext  cyclic quad ✓ S ✓R Ĥ 4 = K̂ 1 tan chord theorem ✓ SR Ĥ 3 = K̂ 2 ' s in the same circle segment ✓ S ✓R  K̂1 + K̂ 2 = Ê1 + Ê 2 K̂1 = Ê1 given ✓ S  K̂ 2 = Ê 2 KF is a tangent converse tan chord theorem ✓ R (if K̂ 2 = Ê 2 proven) (7) [15] TOTAL: 150 14 ## Related documents - [Gr 11 Math P2 (English) June 2023 Possible Answers](https://www.hlayiso.com/library/gr-11-math-p2-english-june-2023-possible-answers/7d433fdb-38a7-4d1e-b09f-472448486aa3) — Mathematics · Memo · 2023 · Eng - [MATHS P2 AB GR11 NOV 2023 Afri+English](https://www.hlayiso.com/library/maths-p2-ab-gr11-nov-2023-afri-english/eee869d2-4e88-4265-9665-6bcf4b2cfc27) — Mathematics · QP · 2023 · Eng - [Gr 11 Math P2 (English) June 2023 Question 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