Gr11 MATH P2 (ENG) June 2022 Possible Answers_hlayiso.com_.pdf

Downloaded from hlayiso.com PROVINCIAL EXAMINATION JUNE 2022 GRADE 11 MARKING GUIDELINES MATHEMATICS PAPER 2 10 pages

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (Paper 2) GRADE 11 INSTRUCTIONS AND INFORMATION: 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 deleted attempt. • Consistent accuracy applies to ALL aspects of the marking guidelines. • It is UNACCEPTABLE for candidates to assume values in order to answer questions. Marks are awarded as per the guidelines, and the following symbols are used: ➢ A – Accuracy ➢ CA – Continued Accuracy ➢ S – Statement ➢ R – Reason ➢ S/R – Statement with Reason 2

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (Paper 2) GRADE 11 QUESTION 1 y2 − y1 1.1 1.1.1 mDF = x2 − x1 ✓ substitution −3− 2 mDF = − 5 +1 ✓ answer 5 mDF = 4 (2) 4 ✓ answer 1.1.2 mDE = − 5 (1) 1.1.3 y = mx + c 4 ✓ substitution 2 = − ( −1) + c pt D( −1 ; 2) ✓ value of c 5 6 c= ✓ answer 5 4 6 y = − x+ 5 5 NOTE: Any other valid method. (3) 4 1.1.4 m=− 5 y = mx + c ✓ value of m 4 − 3 = − ( −5) + c F(−5 ; − 3) 5 c = −7 ✓ value of c 4 y = − x−7 ✓ equation 5 (3) 4 6 1.1.5 y =− x+ 5 5 4 6 y = − ( 6) + ✓ substitution 5 5 18 ✓ answer y=− 5 (2) 3

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (Paper 2) GRADE 11 1.1.6 d DF = ( y2 − y1 ) 2 + ( x2 − x1 ) 2 ✓ substitution d DF = ( −3 − 2) 2 + ( −5 + 1) 2 ✓ answer d DF = 41 ✓ substitution ✓ answer d DE = ( −2 − 2) 2 + (4 + 1) 2 ✓ conclusion d DE = 41  ∆ DEF is an isosceles ∆ (5) 1 ✓ substitution 1.1.7 areaDEF = . 41. 41 2 41 ✓ answer areaDEF = square units 2 (2) 1.2 1.2.1 Trapezuim ✓ answer (1) 1.2.2 ( x − 10) 2 + (−9 − 3) 2 = 15 ✓ substitution x 2 − 20 x + 100 + 144 = 225 ✓ simplification x 2 − 20 x + 19 = 0 ✓ standard form ( x − 19)( x − 1) = 0 ✓ factors x = 1, x  19 ✓ selection (5) 1.2.3  10 + 1 3 − 9   substitution T = ;   2 2   answer  11  =  ; −3  2  (2) 3+9 1.2.4 mSP = mRS = 10 − 1 4 mSP = mRS = ✓ value of mRS 3 4 1− y ✓ substitution  = 3 −4+7 ✓ answer 12 = 3 − 3 y  y = −3 (3) 4

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (Paper 2) GRADE 11 1.2.5 4 mSP = 3  equation SP 4 3 = (10) + c S (10 ; 3) 3 31 c = − ✓ value of c 3 4 31 y = − 3 3 ✓ x-value  x = 4 at W 4 31  y = (4) − 3 3 ✓ y-value  y = −5 W(4 ; − 5) NOTE: Point W does not have to be in coordinate form. (3) [32] 5

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (Paper 2) GRADE 11 QUESTION 2 2.1 2.1.1 sin( 90 −  ) = cos ✓ identity  5 cos = 0 3  cos = ✓ value of cos 5 4 ✓ answer  sin  = 5 (3) 2.1.2 4 k 4 tan  = =− ✓ tan  = 3 6 3  3k = −24 ✓ tan  = − k  k = −8 6 ✓ answer (3) 6

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (Paper 2) GRADE 11 2.2 2.2.1 tan 315 + cos300 ✓ − tan 45 sin150 + tan135 ✓ cos 60 ✓ sin 30 ( − tan 45 ) + ( cos 60 ) = ✓ - tan 45 ( sin 30 ) + ( − tan 45) ✓ all special angles 1 (−1) + ( ) = 2 1 ✓ answer + (−1) 2 =1 (6) 2.2.2 sin (180 + x ) cos (180 − x ) .sin 50 = tan ( 315 ) .cos 2 ( 360 − x ) .cos140 ✓ reduce numerator ✓ reduce denominator ( − sin x )( − cos x )( cos 40 ) = ✓ sin x ( − tan 45 )( cos 2 x )( − cos 40 ) ✓ -1 ✓ –cos x sin x = ✓ answer ( −1)( − cos x ) = tan x (6) 2.3 sin (x + 10°) = cos (x – 30°) ✓ identity sin (x + 10°) = sin (90° – (x – 30°)) = sin (120° – x) ✓ sin (120 − x ) Quadrant 1: = (120° – x) + 360° k; k  Z ✓ (x + 10°) = (120° – x) 2x = 110° + 360° k; k  Z ✓ 360° k; k  Z x = 55° + 180° k; k  Z Quadrant 2: = (x + 10°) = (120° – x) + 360° k; k  Z ✓ x = 55 + 180k x + 10° = 60° + x (no solution) ✓ x = – 125° Final solution: x = – 125°, x = 55° ✓ x = 55° (7) 7

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (Paper 2) GRADE 11 2.4 2.4.1 ˆ D = 40° BC [Sum of the int. angles of a triangle] ✓ answer NOTE: Candidate does not have to state the reason. (1) 2.4.2 CD 52 = sin 80 sin 60 ✓ sine ratio  ✓ answer 52. sin 80 CD = sin 60 = 59,13 m (2) 2.4.3 BC 52  sine rule = sin 40 sin 60  52. sin 40 BC = sin 60 BC = 38,59 m ✓ value of BC let AD = b (triangle ABD) ✓ correct sub into Cos  rule  b 2 = a 2 + d 2 − 2adCos116  b 2 = (38,59) 2 + (74) 2 − 2(38,59)(7 4) cos116  ✓ answer b 2 = 9468,87 b = 97,31 m (4) 2.4.4 B ˆDC = 40° (Sum of angles of BDC) ✓ B ˆDC = 40° Area of ABCD = ✓ area ABC 1 1 ✓ area CBD   74 × 52 × sin 36°   59,13 × 52 × sin 40° 2 2 ✓ answer = 2119,11 m2 (4) [36] 8

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (Paper 2) GRADE 11 QUESTION 3 3.1 Given: Circle with centre O. OM is perpendicular to PQ. RTP: PM = MQ Construction: Draw radii OP and OQ Proof: In OPM and OQM ✓ construction OP = OQ (radii) OM = OM (common) ✓ S/R OMˆ P = 90° = O ˆ MQ (given OM ⊥ PQ) ✓ S/R ✓ S/R OPM  OQM (90° ; H ; S) PM = MQ ✓ condition-congruency (5) 3.2 AB = 10 cm (given) ✓ S ✓ R AD = DB = 5 cm [Line from centre of circle perpendicular to chord bisects chord] ✓ S/R OD2 = 132 − 52 [ PT ] OD2 = 144 ✓ value of OD OD = 12  DC = 1 cm ✓ answer (5) 9

Downloaded from hlayiso.com MATHEMATICS MARKING GUIDELINES (Paper 2) GRADE 11 3.3 A = r 2 A = 3,14.( 13 )2 ✓ area of circle A = 530 ,66 cm 2  area of ΔOAB 1 Area = bh 2 1 Area = .10.12 2 Area = 60 cm 2 ✓ area of  OAB  remaining space = 530 ,66 cm 2 − 60 cm 2  remaining space = 470 ,66 cm 2 ✓ answer (3) 3.4 3.4.1 P̂2 = x (tan-chord theorem) ✓ S✓R ✓ S✓R Ŝ2 = x (angles opposite equal sides) P̂1 = x (alt angles; APQ ║SR) ✓ S✓R ˆ =x QRS (ext. angle of cyclic quad) ✓ S✓R ✓ S✓R Aˆ = x (corresp. angles; APQ ║SR) (10) 3.4.2 Aˆ = x and Pˆ1 = x [ from 3.4.1] Sˆ = 180 0 − 2 x [sup p s ] 1 ✓ S ✓R  Rˆ1 = Sˆ1 [tan/chord] chord ] ✓ S ✓R  Pˆ = Sˆ [alt Ls ; AQ║SR] 3 1 ✓ S ✓R (6) 3.4.3 R̂ 1 = P̂3 (from 3.4.2) ✓ S PS = QR (equal angles subtend equal chords) ✓ ✓ R (3) [32] TOTAL: 100 10