Mathematics Grade 11 Paper 2 Gauteng June 2017 Memo

Downloaded from hlayiso.com GAUTENG DEPARTMENT OF EDUCATION PROVINCIAL EXAMINATION JUNE 2017 GRADE 11 MATHEMATICS PAPER 2 MEMORANDUM 10 pages 1

Downloaded from hlayiso.com Mathematics Paper 2 MEMORANDUM GRADE 11 GAUTENG DEPARTMENT OF EDUCATION PROVINCIAL EXAMINATION MATHEMATICS Paper 2 MEMORANDUM INSTRUCTIONS AND INFORMATION A – Accuracy CA – Consistent Accuracy S – Statement R – Reason SR – Statement / Reason NOTE: • 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 memorandum. Stop marking at the second calculation error. • Assuming values/answers in order to solve a problem is UNACCEPTABLE. 2

Downloaded from hlayiso.com Mathematics Paper 2 MEMORANDUM GRADE 11 QUESTION 1 MARKS: 25 1.1 MMR = ✓ sub. into the gradient (3) formula MMR = ✓ gradient of line MR equation of line MR is: ✓ equation of line MR 1.2 (4) y = 5x -14 ✓ mMS = 5 MS || PR  mPR = 5 ✓mPR = 5 equation of line PR: ✓sub. (2 ; 4) ✓answer 1.3 mPR = 5 (5) ✓ ✓ mMR = ✓ ✓ )..(sum of angles of ∆) ✓ 1.4 and (4) ✓equating ✓ ✓ ✓ R(4 ; 6) 3

Downloaded from hlayiso.com Mathematics Paper 2 MEMORANDUM GRADE 11 1.5 = (2) = ✓sub. into dist. formula = ✓answer = 1.6 = (5) = ✓ sub. into dist. formula ✓ OR = or Area of PMR ✓ OR = = OR (✓ rounding-off) answer ✓ = 28 1.7 S(2 ; 4) ✓x=2 (2) ✓ y = 4 QUESTION 2 MARKS: 26 2.1 ✓diagram correct in correct quad. (4) y ✓ 0 x ✓ 25 ✓answer  (-7: y) 2.2 (6) ✓ (-sinx) (-sinx) ✓ ✓ ✓ ✓ = ✓ answer 4

Downloaded from hlayiso.com Mathematics Paper 2 MEMORANDUM GRADE 11 2.3 ✓ (6) ✓ Ref. angle ✓ or ✓ ✓ 2θ = 180° + 77,2909° + k.360° ✓ 2.4 (5) ✓ ✓ ✓ ✓ ✓ 2.5 ✓ (5) ✓sin = ✓ = ✓ ✓answer 5

Downloaded from hlayiso.com Mathematics Paper 2 MEMORANDUM GRADE 11 QUESTION 3 MARKS:14 3.1 (6) ✓ y-intercepts of f ✓ y-intercepts of g ✓✓ both turning points of g ✓ both x-intercepts of f ✓ both turning points of f 3.2 2 ✓✓ answer (2) 3.3 2 cos x +1 = 1 – sin x ✓ tan x = -2 (6)  sin x = -2 cos x ✓ 63, 4349°  tan x = -2 ✓ x = 116, 57° +  Ref. angle = 63, 4349....° k.180° with k є Z  x = 180° - 63, 4349....° + k. 180° ✓ -63,43°  x = 116, 57° + k.180° with k є Z ✓ 116, 57°  x = -63,43° OR 116,57° OR 296,57° ✓ 296,57° 6

Downloaded from hlayiso.com Mathematics Paper 2 MEMORANDUM GRADE 11 QUESTION 4 MARKS: 13 O ✓ construction 1 2 (radii, OP and OR) P Q R 4.1 Construction: Join PO and OR In ∆POQ and ∆ ROQ 1) OQ = OQ..... common 2) PO = OR ......radii ✓SR ie (PO = OR ..... radii) 3) Q̂1  Q̂2  90 .. given ✓S ie (Q̂1  Q̂2  90) 90 HS ✓SR ✓ PQ =RQ (5) Â1  90 ..line from centre to midpt. of chord ✓SR 4.2.1 B̂  180  (Ô1  Â1 ) .... sum of angles of a Δ ✓ answer = 50° (2) 7

Downloaded from hlayiso.com Mathematics Paper 2 MEMORANDUM GRADE 11 Ĉ  90 ... angle in a semi-circle ✓SR 4.2.2 D̂  180  (Ĉ  B̂) ... sum of angles of Δ ✓ answer = 40° (2) ......Th. of Pyth. = = 50 ✓ BD = 50 BO = .....radii 4.2.3 = 25 ✓ BO = 25 OA = ...Midpt. Th. = 20 ✓ SR ie (OA = 20 .. Midpt.Th.) AE = OE – OA = 25 – 20 = 5 units ✓ AE = 5 (4) 8

Downloaded from hlayiso.com Mathematics Paper 2 MEMORANDUM GRADE 11 QUESTION 5 MARKS: 10 L̂  Ĵ1  x ... angles opp. = sides ✓SR 5.1 ✓S N̂1  L̂  x ...= chords subtend = angles ✓R (3) M̂  Ĵ 3  y ........... angles opp. = sides ✓SR ✓SR K̂1  M̂  y ........= chords subtend = angles Q̂2  K̂1  Ĵ1 ... ext. angle of Δ ✓SR 5.2 = x +y P̂2  N̂1  Ĵ 3 ... ext. angle of Δ ✓SR =x+y  Q̂2  P̂2  x  y (4) P̂4  P̂2 .... vert. opp. angles ✓ S R for P̂4  P̂2 OR 5.3 Q̂4  Q̂2 .... vert. opp. angles Q̂4  Q̂2 but Q̂2  P̂2 ........proved in Q5.2  P̂4  Q̂4 ✓ P̂4  Q̂4  JQ = JP...... sides opp = angles OR isos. ∆ ✓R (3) 9

Downloaded from hlayiso.com Mathematics Paper 2 MEMORANDUM GRADE 11 QUESTION 6 MARKS: 12 K̂ 2  T̂1 ...... angles opp. = sides OR equal radii ✓SR T̂1  Ĉ ......... angles in same segment ✓ T̂  Ĉ 6.1 ✓R  K̂ 2  Ĉ (3) T̂2  Â ...... angles opp. = sides OR equal radii ✓SR ✓SR P2  T̂2  Â ... ext. angle of 6.2 = 2 T̂2 ✓ K̂1  P̂2 K̂1  P̂2 ........... angles in same segment ✓R K̂1  2T̂2 (4) P̂4  2T̂ ........ angle at centre = twice angle at ✓ P̂4  2T̂ circumference ✓R T̂  T̂1  T̂2 T̂1  Ĉ ....... angles in same segment ✓SR 6.3 1 1 T̂2  K̂1 ..... proved in Q6.2 ✓ T̂2  K̂1 2 2 1 1 P̂4  2(Ĉ  K̂1 ) ✓ P̂4  2(Ĉ  K̂1 ) 2 2 = 2Ĉ  K̂1 ) (5) TOTAL: 100 10