MATHS P2 GR10 MEMO NOV2019_English_hlayiso.com_.pdf

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 10 NOVEMBER 2019 MATHEMATICS P2 MARKING GUIDELINE (EXEMPLAR) MARKS: 100 This marking guideline consists of 8 pages.

Downloaded from hlayiso.com 2 MATHEMATICS P2 (EC/NOVEMBER 2019) Consistent accuracy (CA) marking, applies in ALL aspects of the marking guideline. QUESTION 1 48 50 52 59 60 68 73 76 76 76 78 79 80 81 82 82 84 91 92 98 1.1.1 76 + 78 ✓ answer Median = = 77 (1) 2 1.1.2 60 + 68 Lower quartile = = 64 ✓ lower quartile 2 Upper quartile = 82 ✓ upper quartile (2) 1.1.3 Interquartile range ( IQR ) = Q3 – Q1 ✓ substitution = 82 − 64 = 18 ✓ answer (2) 1.1.4 Min = 48 and max = 98 ✓ min and max (1) 1.1.5 ✓ min and max ✓ Q1 and Q3 ✓ Q2 (3) 1.1.6 Skewed to the left or negatively skewed ✓ answer (1) 1.2 Duration (min) No of calls (f1) Midpoint (x1) (f 1) × (x1) 2 ≤ t < 5 47 3,5 164,5 5 ≤ t < 8 139 6,5 903,5 8 ≤ t < 11 211 9,5 2004,5 11 ≤ t < 14 102 12,5 1275 14 ≤ t < 17 58 15,5 899 17 ≤ t < 20 19 A B 576 5598 1.2.1 A = 18,5 and B = 351,5 ✓ answer of A ✓ answer of B (2) 1.2.2 sum of f 1  x1 approximate mean = ✓ sum of all sum of f 1 5598 ( f1 ) (x1 ) = ✓sum of all ( f1 ) 576 = 9,7 minutes ✓ answer (3) 1.2.3 75 75th percentile lie =  576 = 432 ✓ 432 100 In the interval 11 ≤ t < 14 ✓ interval (2) [17] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2019) MATHEMATICS P2 3 QUESTION 2 2.1 A( – 2 ; 6 ) , B( 6 ; 8 ) and C( 4 ; 0 ) dAB = ( x2 − x1 ) 2 + ( y 2 − y1 ) 2 ✓ formula = (6 − ( − 2)) 2 + ( 8 − 6 ) 2 ✓ substitution ✓ distance AB = 2 17 dBC = ( x2 − x1 ) 2 + ( y 2 − y1 ) 2 = (4 − 6) 2 + ( 0 − 8 ) 2 ✓ substitution = 2 17 ✓ distance of BC  AB = BC. (5) 2.2 ABCD is a kite ✓ kite adjacent sides are equal ✓motivation (2) 2.3 A( – 2 ; 6 ) , B( 6 ; 8 ) and C( 4 ; 0 )  x + x1 y 2 + y1  Midpoint of BC =  2 ;  ✓ formula  2 2  ✓ substitution −2+6 8+6 ✓ coordinates of = ( ; ) = G( 2 ; 7 ) 2 2 G, mdpt of BC  x + x1 y 2 + y1  Midpoint of AB =  2 ;   2 2  ✓ substitution 4+ 6 0+8 = ( ; ) = H( 5 ; 4 ) ✓ coordinates of 2 2 H, mdpt of AB (5) 2.4 BAˆ D = BCˆ D (opposite ’s of a kite are =) ✓ S ✓R AEˆ H = EDˆ B (corresponding ’s , EG || DB) ✓ SR but EDˆ B = BDˆ C (diagonals of a kite)  AEˆ G = BD ˆC ✓ 3rd angle or reason ∆AEG ||| ∆CDB. (A A A) (4) [16] Copyright reserved Please turn over

Downloaded from hlayiso.com 4 MATHEMATICS P2 (EC/NOVEMBER 2019) QUESTION 3 3.1.1 x 2 = 35 2 − 28 2 ✓sub in Pythagoras ✓x = 21 x = 21 21 35  cos = 28 ✓ 21 35 35  (3) 3.1.2 2 2 2  28   21  ✓   28 sin 2  + cos 2  =   +    35   35   35  = 1 2 ✓   21 = RHS  35  ✓1 (3) 3.2 If 37 sin  + 35 = 0 35  sin  = – − 35 37 ✓ sin  = x = 37 − 35 2 2 2 37 x = 12 – 12  ✓ 3rd quadrant – 35 37 ✓ x value = – 12 24 sec − 70 cot  37 −12 ✓✓ substitution = 24( ) − 70( ) −12 − 35 ✓answer = − 74 − 24 = − 98 (6) 3.3.1 8cos( x + 10 ) = 5 5 ✓cos( x + 10 ) cos( x + 10 ) = 8 x+ 10  = 51,32 ✓ x + 10 x = 41,32 ✓ answer (3) Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2019) MATHEMATICS P2 5 3.3.2 cosec 2x = 2 1 1 sin 2x = ✓ sin 2x = 2 2 2x = 300 ✓ 2x = 30 x = 150 ✓ answer (3) 3.4 1 3 1  ✓ sin 30  tan 60 2 = 2 1 tan 30  cos 60 1 1 ✓ 3  3 2 1 ✓ =3 3 = RHS 1 ✓ 2 ✓ answer (5) 3.5.1 x ✓ using sin 55  sin 55  = 15 ✓ answer x = 15  sin 55 = 12,29 (2) OR x cos 35 = ✓ using cos 35 15 ✓ answer x = 12,29 (2) 3.5.2 4,4 tan 21  = y ✓ using tan 21 4,4 y = ✓ answer tan 21 (2) = 11,46 OR y tan 69 = 4,4 y = 11,46 OR ✓ Pythagoras y = 12,29 – 4.4 2 2 2 ✓ answer y = 11,48 (2) [27] Copyright reserved Please turn over

Downloaded from hlayiso.com 6 MATHEMATICS P2 (EC/NOVEMBER 2019) QUESTION 4 4.1 y g f ✓ intercepts ✓ turning pts x ✓ shape (3) 4.2 period of g = 3600 ✓ answer (1) 4.3 range of m(x) if m(x) = – 3f(x) + 1 range of – 3 f(x): – 3 ≤ y ≤ 3 ✓ notation range of m(x) : – 2 ≤ y ≤ 4 ✓✓ endpoints (3) 4.4 g decreasing: 900 < x < 2700 ✓ notation ✓ endpoints (2) 4.5 f ( x)  g ( x)  0 ✓ notation 90  x  180 or 270  x 360 ✓ endpoints ✓ endpoints (3) [12] Copyright reserved Please turn over

Downloaded from hlayiso.com (EC/NOVEMBER 2019) MATHEMATICS P2 7 QUESTION 5 5.1 AD̂C = 53 (s on a straight line) ✓ SR DĈB = 116  (supplementary adj s) ✓SR CB̂A = 101 (s on a straight line) ✓SR BÂD = 360 − 53 − 116 − 101  ✓ answer = 90 (s of a quad = 360 ) (4) Answer only: full marks, provided one reason is given 5.2 Let DÊB = y and FÊC = k  B̂ = 180 − 2 y and Ĉ = 180 − 2k (s of a ∆ = 180) ✓ SR In ∆ ABC: x + 180 – 2y + 180 – 2k = 1800 2y + 2k = x + 180  + 180  − 180  ✓ SR 1 y + k = x + 90 ✓S 2 1 DEˆ F = 90 – x (s on a straight line) ✓ SR 2 (4) [8] QUESTION 6 6.1.1 AP = DE and AQ = DF (given) ✓ given  = D̂ (given) ✓∆’s similar ∆ APQ  ∆ DEF (SAS) ✓ reason (3) 6.1.2 AP̂Q = Ê (∆ APQ  ∆ DEF) ✓ Statement But B̂ = Ê (given) ✓ Statement  AP̂Q = B̂  PQ || BC (a pair of corresponding s are =) ✓ Reason (3) 6.1.3 AB BC AC = = (∆ABC ||| ∆DEF) ✓ SR DE EF DF 7,5 8 = ✓ substitution 3,5 DF 8  3,5 DF = 7,5 ✓ simplification = 3,7 ✓ answer (4) 6.2.1 Converse of midpoint theorem ✓ answer (1) Copyright reserved Please turn over

Downloaded from hlayiso.com 8 MATHEMATICS P2 (EC/NOVEMBER 2019) 6.2.2 BD = 32  AD = 32 ✓ BD = AD ✓S✓R  EF = 32 (opp sides of a parallelogram) ✓ SR  CG = 2 32 (midpt theorem) ✓ answer = 8 2 (5) [16] QUESTION 7 TSA of cone = TSA of hemisphere  r 2 +  r s = 3 r 2 ✓ equating the  r s = 2 r 2 TSA s = 2x ( r = x) but s = h2 + x2 2  h2 + x2 = 4x2 ✓ use of  h = 4x − x 2 2 Pythagoras ✓ substituting = 3x s = 2x ✓ h subject of formula (4) [4] TOTAL: 100 Copyright reserved Please turn over