# Mathematics Grade 11 P2 Gauteng June 2024 Memo (English) > Grade 11 Mathematics Paper 2 memorandum from the Gauteng June Exam 2024 (English), 11 pages. Read online or download the free PDF. - Page: https://www.hlayiso.com/library/gr-11-mathematics-p2-english-june-2024-possible-answers/a434fdfa-63d5-4142-bd70-67e2e579c02e - Updated: 2026-09-27 - Subject: Mathematics - Grade: Grade 11 - Language: English - Document type: Memorandum - Year: 2024 - Exam period: Gauteng June Exam - Paper: 2 - Curriculum: CAPS - Pages: 11 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 June 2024 Memo (English)](https://www.hlayiso.com/library/gr-11-mathematics-p2-english-june-2024-possible-answers/a434fdfa-63d5-4142-bd70-67e2e579c02e) ## About this document Mathematics · Grade 11 · Gauteng June Exam · 2024 · English. Memorandum, 11 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 PROVINCIAL EXAMINATION JUNE 2024 GRADE 11 MARKING GUIDELINES MATHEMATICS (PAPER 2) 11 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. QUESTION 1 ✓ Substitution into correct 1.1 ( 4 − ( −4 ) ) + ( −2 − ( −6 ) ) 2 2 BC = formula BC = 80 ✓ Answer BC = 4 5 (2) 1.2 6+4 y + ( −2 ) ✓ x=5 =x =2 2 2 ✓ y =5 x=5 y=6 (2) 1.3 −2 − ( −6 ) ✓ Substitution into correct mBC = 4 − ( −4 ) formula ✓ Answer 1 mBC = 2 (2) 1.4 1 ✓ mAD = −2 mBC = 2 ✓ Substitute A(−2 ; 2)  mAD = −2 AD ⊥ BC ✓ Equation y − ( 2 ) = −2 ( x − ( −2 ) ) y = −2 x − 2 (3) 1.5 tan  = −2 ✓ Substitution into correct  = 180 − 63, 434... formula  = 116, 57 ✓ Answer (2) 2 ### Page 3 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 1.6 F ( 2 ; −10 ) ✓ x=2 ✓ y = –10 (2) 1.7 1 ✓ mBC = −2 Equation of BC: y − ( −2 ) = ( x − 4) 2 ✓ Substitute B(−4;−6) of C 1 (4;−2) y = x−4 2 ✓ Equation of BC D is a point of intersection between AD and BC ✓ Simultaneous equation 1 substitution of y y = x − 4 (1) and y = −2 x − 2 (2) 2 (1) = (2) 1 x − 4 = −2 x − 2 2 x − 8 = −4 x − 4 5x = 4 4 x = = 0,8 5 y = −2 ( 0,8 ) − 2 18 y=− = −3, 6 5  D ( 0,8 ; - 3, 6 ) (4) ✓ Length of AD ( −2 − ( 0,8) ) + ( 2 − ( −3, 6 ) ) 1.8 2 2 AD = ✓ Substitution into area of a 14 5 parallelogram formula AD = 39, 2 = 5 ✓ Answer Area ABCE = BC  AD 14 5 Area ABCE = 4 5  5 Area ABCE = 56 units 2 (3) [20] 3 ### Page 4 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 2 2.1 2.1.1 sin ( −20 ) ✓ − sin 20 = − sin 20 ✓ −p = −p (2) 2.1.2 sin160 ✓ sin 20 = sin 20 ✓ p =p (2) 2.1.3 sin 2  + cos 2  = 1 1 − sin 2 70 ✓ cos 2 70 = cos 2 70 ✓ sin 2 20 = sin 2 20 ✓ p2 = p2 OR ✓ 1 − cos 2 20 1 − sin 70 2 ✓ sin 2 20 = 1 − cos 20 2 ✓ p2 = sin 2 20 = p2 OR 1 − sin 70 2 ✓ 1 − cos 2 20 = 1 − cos 2 20 ( ) 2 ✓ 1 − p2 ( 1− p ) 2 = 1− 2 ✓ p2 = p2 (3) 2.2 tan 225 − sin (180 +  ) cos ( + 90 ) ✓ tan 225 = 1 sin  .sin(360 +  ) + cos  cos ( − ) ✓ − sin  ✓ − sin  1 − ( − sin  )( − sin  ) = ✓ sin  . sin  sin  .sin  + cos  .cos  ✓ cos  . cos  1 − sin 2  = ✓ cos2  sin 2  + cos 2  = cos 2  (6) 4 ### Page 5 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 2.3 2.3.1 ✓ Isolate sin  β ✓ Isolate cos  ✓ Explanation k ✓ Diagram (1 ; −5) −5 1 sin  = and cos  = k k sin  is negative and cos  is postive in quadrant IV OR In quadrant IV, y is negative and x is positive. (4) 2.3.2 5 ✓ tan  = −5 tan  = − 1 tan  = −5 (1) 2.3.3 x2 + y 2 = r 2 ✓ Substitute into Pythagoras (1) + ( −5) = k 2 2 2 ✓ k = 26 k = 26 k 0 (2) 2.4 2.4.1  sin  cos   sin  LHS = sin 2   +  ✓  cos  sin   cos   sin 2  + cos 2   cos  = sin 2   ✓  sin   sin  .cos   ✓ Add fractions sin  (1) = ✓ sin 2  + cos2  = 1 cos  ✓ conclusion = tan  = RHS (5) 2.4.2 sin 2  ( tan 2  + 1) = 3 tan  ✓  tan  ✓ Substitute tan   1  ✓  = 60 + k.180 sin 2   tan  + = 3  tan   ✓ k Z  tan  = 3  = 60 + k .180 ; kk  Z (4) [29] 5 ### Page 6 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 3 3.1 a = −2 ✓ a = −2 p = −30 ✓ p = −30 (2) 3.2 −2  y  2 ✓ Interval OR ✓ Notation y   −2 ; 2 (2) 3.3 b=6 ✓ b=6 (1) 3.4 3.4.1 −60  x  120 ✓ Interval ✓ Notation (2) 3.4.2 −90  x  0 of 90  x  180 ✓ −90  x  0 ✓ 90  x  180 (2) 3.5 g ( 2 x ) = sin ( 2 x − 30 ) ✓ cos (120 − 2x ) = cos ( 90 − ( 2 x − 30 ) ) ✓ t = −120 = cos (120 − 2 x ) = cos ( 2 x − 120 )  t = −120 OR OR g ( 2 x ) = sin ( 2 x − 30 ) ✓ sin ( 2 ( x − 15 ) ) = sin ( 2 ( x − 15 ) ) ✓ t = −120 cos ( 2 x + t ) must be translated 120 to the right  t = −120 (2) [11] 6 ### Page 7 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 4 4.1 supplementary or (add up to 180°) ✓ Answer (1) 4.2 A 1 B O 1 E 5 25° 1 C D 4.2 4.2.1 D̂1 = 25 ' s opp equal radii ✓S&R (1) 4.2.2 Ô1 = 50 ext  of ∆ ✓S✓R (2) 4.2.3 Â1 = 25 ' s in same segments ✓S✓R OR  at centre = 2 × at circumference (2) 4.2.4 Ê =155 opp ' s cyclic quad ✓S✓R (2) [8] QUESTION 5 5.1 AE = 6 line from centre ⊥ to chord ✓S✓R r =5 +6 2 2 2 Pythagoras' theorem ✓S ✓ Answer in surd r = 61 form (4) 5.2 OC = r = 61 radii ✓S✓R F1 = 90 line from centre to midpt of chord x 2 = 61 − 7 2 Pythagoras' Theorem ✓S x = 2 3 = 3, 46 (3) [7] 7 ### Page 8 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 6 A E 2 1 1 1 2 B D 2 3 Q 2 1 3 C P 6.1 Â = Ĉ1 alt ' s , PQ ║ AB ✓S✓R ✓S✓R B̂2 = Ĉ1 tan chord theorem  Â = B̂2 both equal Ĉ1 (4) 6.2 D̂ 3 = B̂1 + Â ext  of ∆ ✓S✓R D̂ 3 = B̂1 + B̂ 2 B̂2 = Â ✓S D̂1 = B̂1 + B̂2 ext  cyclic quad ✓S✓R  D̂ 3 = D̂1 both equal B̂1 + B̂2 OR D̂1 = B̂1 + B̂2 ext  cyclic quad ✓S✓R Ĉ 3 = B̂1 + B̂ 2 alt ' s , AB ║ PQ ✓S&R  D̂1 = Ĉ 3 both equal B̂1 + B̂2 en D̂ 3 = Ĉ 3 tan chord theorem ✓S✓R  D̂ 3 = D̂1 both equal Ĉ 3 (5) [9] 8 ### Page 9 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 7 A 1 2 D 3 1 2 y x 2 C 3 1 1 2 E B Ĉ = x tan chord theorem ✓ S&R Ê 2 + Ê 3 = x + y ext  of ∆ ✓ R ✓ S Ê 2 = x given ✓ S&R  Ê 3 = y ✓ S ✓S&R B̂1 + B̂2 = y ' s opp equal sides  B̂1 + B̂2 = D̂1 both equal y  ABED is a cyclic quad converse ext  cyclic quad OR ext  = to opp int  (6) [6] 9 ### Page 10 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 QUESTION 8 B 1 2 C E 2 1 F A 2 1 3 D G 8.1 Â = Ĉ opp ' s of parm ✓S&R Ê1 = Ĉ ext  cyclic quad ✓S✓R  Â = Ê1 both equal Ĉ ✓R  AD = DE sides opp equal ' s OR Ê1 = Ĉ ext  cyclic quad ✓S✓R D̂ 3 = Ĉ alt ' s , BC ║ DG  Ê1 = D̂ 3 both equal Ĉ Â = D̂ 3 ✓S&R corr ' s , AB ║ DC ✓R  Â = Ê1 both equal D̂ 3 ED = AD sides opp equal ' s (4) 10 ### Page 11 Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (PAPER 2) GRADE 11 8.2 Ĝ = Ê1 ext  cyclic quad ✓S✓R Â = Ê1 proven in 8.1 ✓S  Â = Ĝ both equal Ê1 ✓S✓R  AB = BG sides opp equal ' s but AB = CD opp sides of parm ✓S&R  BG = CD both equal AB OR Ĝ = Ĉ ' s in same seg ✓S✓R Â = Ĉ opp ' s parm ✓S&R  Â = Ĝ both equal Ĉ  AB = BG sides opp equal ' s ✓S✓R but AB = CD opp sides of parm ✓S&R  BG = CD both equal AB OR B̂2 = Ĝ alt ' s , BC ║ DG ✓S&R B̂ 2 = D̂ 3 ' s in the same seg ✓S✓R  Ĝ = D̂ 3 both equal B̂ 2 and D̂ 3 = Â corr ' s , AB ║ CD ✓S  Ĝ = Â both equal D̂ 3 ✓S&R  BA = BG sides opp equal ' s ✓S&R but BA = DC opp sides of parm  BG = CD (6) [10] TOTAL: 100 11 ## Related documents - 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