Grade 11 Mathematics Paper 2 Marking Guidelines - June 2024

Downloaded from hlayiso.com PROVINCIAL EXAMINATION JUNE 2024 GRADE 11 MARKING GUIDELINES MATHEMATICS (PAPER 2) 11 pages Downloaded from hlayiso.com MARKING GUIDELINES MATHEMATICS (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 1.1 BC = ( 4 − ( −4 ) ) + ( −2 − ( −6 ) ) 2 ✓ Substitution into correct formula ✓ Answer 2 BC = 80 BC = 4 5 1.2 1.3 1.4 6+4 =x 2 x=5 mBC = −2 − ( −6 ) 4 − ( −4 ) mBC = 1 2 1 2  mAD = −2 (2) y + ( −2 ) =2 2 y=6 ✓ x=5 ✓ y =5 (2) ✓ Substitution into correct formula ✓ Answer (2) ✓ mAD = −2 mBC = ✓ Substitute A(−2 ; 2) ✓ Equation AD ⊥ BC y − ( 2 ) = −2 ( x − ( −2 ) ) y = −2 x − 2 1.5 (3) tan  = −2  = 180 − 63, 434...  = 116, 57 ✓ Substitution into correct formula ✓ Answer 2 (2) Downloaded from hlayiso.com MARKING GUIDELINES 1.6 1.7 F ( 2 ; −10 ) MATHEMATICS (PAPER 2) GRADE 11 ✓ x=2 ✓ y = –10 Equation of BC: y − ( −2 ) = (2) ✓ mBC = −2 1 ( x − 4) 2 1 y = x−4 2 D is a point of intersection between AD and BC 1 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 ✓ Substitute B(−4;−6) of C (4;−2) ✓ Equation of BC ✓ Simultaneous equation substitution of y 18 = −3, 6 5  D ( 0,8 ; - 3, 6 ) y=− 1.8 AD = (4) ( −2 − ( 0,8) ) + ( 2 − ( −3, 6 ) ) 2 ✓ Length of AD ✓ Substitution into area of a parallelogram formula ✓ Answer 2 14 5 5 Area ABCE = BC  AD AD = 39, 2 = 14 5 5 2 Area ABCE = 56 units Area ABCE = 4 5  (3) [20] 3 Downloaded from hlayiso.com MARKING GUIDELINES MATHEMATICS (PAPER 2) GRADE 11 QUESTION 2 2.1 2.1.1 sin ( −20 ) ✓ − sin 20 ✓ −p = − sin 20 = −p 2.1.2 2.1.3 (2) sin160 = sin 20 =p ✓ sin 20 ✓ p (2) sin 2  + cos 2  = 1 1 − sin 2 70 ✓ cos 2 70 = cos 2 70 ✓ sin 2 20 ✓ p2 = sin 2 20 = p2 OR ✓ 1 − cos 2 20 1 − sin 70 2 ✓ sin 2 20 ✓ p2 = 1 − cos 20 2 = sin 2 20 = p2 OR 1 − sin 70 ✓ 1 − cos 2 20 = 1 − cos 2 20 ✓ 2 = 1− ( 1− p ) 2 2 ( 1 − p2 ) 2 ✓ p2 = p2 2.2 (3) tan 225 − sin (180 +  ) cos ( + 90 ) sin  .sin(360 +  ) + cos  cos ( − ) ✓ tan 225 = 1 ✓ − sin  ✓ − sin  ✓ sin  . sin  ✓ cos  . cos  1 − ( − sin  )( − sin  ) sin  .sin  + cos  .cos  1 − sin 2  = sin 2  + cos 2  = cos 2  = ✓ cos2  (6) 4 Downloaded from hlayiso.com MARKING GUIDELINES 2.3 MATHEMATICS (PAPER 2) GRADE 11 ✓ Isolate sin  2.3.1 β ✓ Isolate cos  ✓ Explanation ✓ Diagram k (1 ; −5) −5 1 and cos  = k k sin  is negative and cos  is postive in quadrant IV sin  = OR (4) In quadrant IV, y is negative and x is positive. 2.3.2 2.3.3 5 1 tan  = −5 ✓ tan  = −5 x2 + y 2 = r 2 ✓ Substitute into Pythagoras tan  = − (1) (1) + ( −5) = k 2 2 2 k = 26 2.4 2.4.1 ✓ k = 26 k 0 (2) sin  cos  cos  ✓ sin  ✓ Add fractions  sin  cos   LHS = sin 2   +   cos  sin    sin 2  + cos 2   = sin 2     sin  .cos   ✓ sin  (1) cos  = tan  = RHS = 2.4.2 ✓ sin 2  + cos2  = 1 ✓ conclusion (5) sin 2  ( tan 2  + 1) = 3 tan  ✓  tan  ✓ Substitute tan  ✓  = 60 + k.180 ✓ k Z 1   sin 2   tan  + = 3 tan     tan  = 3  = 60 + k .180 ; kk  Z (4) [29] 5 Downloaded from hlayiso.com MATHEMATICS (PAPER 2) GRADE 11 a = −2 p = −30 ✓ a = −2 ✓ p = −30 (2) −2  y  2 ✓ Interval ✓ Notation MARKING GUIDELINES QUESTION 3 3.1 3.2 y   −2 ; 2 3.3 b=6 3.4 3.4.1 3.4.2 3.5 OR (2) −60  x  120 −90  x  0 of 90  x  180 ✓ b=6 (1) ✓ Interval ✓ Notation (2) ✓ −90  x  0 ✓ 90  x  180 (2) 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 g ( 2 x ) = sin ( 2 x − 30 ) OR OR ✓ 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 Downloaded from hlayiso.com MARKING GUIDELINES MATHEMATICS (PAPER 2) GRADE 11 QUESTION 4 4.1 supplementary or (add up to 180°) 4.2 (1) A B 1 O D 4.2 ✓ Answer E 1 5 25° 1 C 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 Ê =155 4.2.4 opp ' s cyclic quad (2) ✓S✓R (2) [8] QUESTION 5 5.1 AE = 6 r =5 +6 2 2 line from centre ⊥ to chord 2 Pythagoras' theorem r = 61 5.2 OC = r = 61 radii F1 = 90 line from centre to midpt of chord x 2 = 61 − 7 2 Pythagoras' Theorem ✓S✓R ✓S ✓ Answer in surd form (4) ✓S✓R ✓S x = 2 3 = 3, 46 (3) [7] 7 Downloaded from hlayiso.com MARKING GUIDELINES MATHEMATICS (PAPER 2) GRADE 11 QUESTION 6 A E 2 1 B 2 3 1 2 1 D Q 3 2 1 C P 6.1 6.2 ✓S✓R Â = Ĉ1 alt ' s , PQ ║ AB B̂2 = Ĉ1 tan chord theorem  Â = B̂2 both equal Ĉ1 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 ✓S✓R (4) 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  D̂ 3 = D̂1 both equal Ĉ 3 ✓S✓R (5) [9] 8 Downloaded from hlayiso.com MARKING GUIDELINES MATHEMATICS (PAPER 2) GRADE 11 QUESTION 7 A 1 2 D 3 1 y 2 x C 2 1 1 3 2 E B Ĉ = x tan chord theorem Ê 2 + Ê 3 = x + y ext  of ∆ Ê 2 = x given  Ê 3 = y B̂1 + B̂2 = y ' s opp equal sides  B̂1 + B̂2 = D̂1 both equal y  ABED is a cyclic quad OR converse ext  cyclic quad ext  = to opp int  ✓ S&R ✓ R ✓ S ✓ S&R ✓ S ✓S&R (6) [6] 9 Downloaded from hlayiso.com MARKING GUIDELINES MATHEMATICS (PAPER 2) GRADE 11 QUESTION 8 B 1 2 C E 2 1 F A 1 2 3 D 8.1 G Â = Ĉ opp ' s of parm Ê1 = Ĉ ext  cyclic quad  Â = Ê1  AD = DE both equal Ĉ sides opp equal ' s ✓S&R ✓S✓R ✓R OR Ê1 = Ĉ ext  cyclic quad D̂ 3 = Ĉ alt ' s , BC ║ DG  Ê1 = D̂ 3 both equal Ĉ Â = D̂ 3 corr ' s , AB ║ DC  Â = Ê1 both equal D̂ 3 ED = AD sides opp equal ' s 10 ✓S✓R ✓S&R ✓R (4) Downloaded from hlayiso.com MARKING GUIDELINES 8.2 Ĝ = Ê1 ext  cyclic quad Â = Ê1 proven in 8.1  Â = Ĝ  AB = BG but AB = CD  BG = CD both equal Ê1 sides opp equal ' s opp sides of parm both equal AB MATHEMATICS (PAPER 2) GRADE 11 ✓S✓R ✓S ✓S✓R ✓S&R OR Ĝ = Ĉ Â = Ĉ  Â = Ĝ  AB = BG but AB = CD  BG = CD ' s in same seg opp ' s parm both equal Ĉ sides opp equal ' s opp sides of parm both equal AB ✓S✓R ✓S&R B̂2 = Ĝ alt ' s , BC ║ DG B̂ 2 = D̂ 3 ' s in the same seg ✓S&R ✓S✓R  Ĝ = D̂ 3 both equal B̂ 2 and D̂ 3 = Â corr ' s , AB ║ CD  Ĝ = Â  BA = BG but BA = DC  BG = CD both equal D̂ 3 sides opp equal ' s opp sides of parm ✓S✓R ✓S&R OR ✓S ✓S&R ✓S&R (6) [10] TOTAL: 11 100

Grade 11 Mathematics Paper 2 Marking Guidelines - June 2024

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Grade 11 Mathematics Paper 2 Marking Guidelines - June 2024

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