Mathematics Grade 11 P2 Gauteng June 2024 Memo (English)

Downloaded from hlayiso.com PROVINCIAL EXAMINATION JUNE 2024 GRADE 11 MARKING GUIDELINES MATHEMATICS (PAPER 2) 11 pages

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

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

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

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

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

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

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

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

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

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