Mathematics Grade 11 Paper 2 North West June 2019 Memo

Downloaded from hlayiso.com GRADE 11 MATHEMATICS: MEMO MID-YEAR EXAMINATION JUNE 2019 MARKS: 100 JUNE EXAMINATION 2012 This marking guideline consists of 12 pages. Demo NW/JUNE/MATH/ EMIS/6*******

Downloaded from hlayiso.com Mathematics/ P2 Mid-Year Examination NW/ Jun 2019 QUESTION 1 1.1 y2  y1 m x2  x1 1 4 mAB  1  3 🗸 substitution (2) 3  4 🗸 answer 1.2  3 a 4 2  3 a  ;    0;1 🗸 0  2 2  2 3 a  0 2 🗸 answer a  3 (2) [4] QUESTION 2 P S Q R T 2.1 5  3 mPR  42  4 ✓ m  4 y  y1  m  x  x1  ✓ substitution y  3  4  x  2  y  4 x  11 ✓ answer (3) Demo 2 NW/JUNE/MATH/ EMIS/6*******

Downloaded from hlayiso.com Mathematics/ P2 Mid-Year Examination NW/ Jun 2019 2.2 mPQ  tan 45 ✓ tan 45 1 ✓ answer (2) 2.3 d  x2  x1    y2  y1  2 2 PQ   m  2    n  3 2 2 ✓ substitution  m 2  4m  4  n 2  6n  9 ✓ simplification  m 2  4m  n 2  6n  13  m  4    n  5 2 2 QR  ✓ substitution  m 2  8m  16  n 2  10n  25 (3)  m 2  8m  n 2  10n  41 2.4 PQ  QR ✓ PQ = QR m 2  4m  n 2  6n  13  m 2  8m  n 2  10n  41 m 2  4m  n 2  6n  13  m 2  8m  n 2  10n  41 4m  6n  13  8m  10n  41 4m  16n  28 ✓ m  4n  7 m  4n  7 ................... 1 mPQ  1 n3 1 n3 m2 ✓ 1 n3  m2 m2 m  n  1 ....................  2  ✓ m  n 1 Substitute  2  in 1 : n  1  4n  7 ✓ substitution 3n  8 ✓ 3n  8 8 n 3 8 m   1 3 11  3  11 8  Q m ; n  Q   ;    3 3 (6) Demo 3 NW/JUNE/MATH/ EMIS/6*******

Downloaded from hlayiso.com Mathematics/ P2 Mid-Year Examination NW/ Jun 2019 2.5 mQS  mPR  1 mQS   4   1 1 1 mQS  ✓ mQS  4 4 y  mx  c 8 1  11     c ✓ substitution 3 4 3  7 7 c ✓ c 4 4 1 7 y  x (4) 4 4 ✓ answer 2.6 ˆ m tan QRT  8 5       3  11  4   3 7   3 23 3 7 7  ✓ m 23 23 ˆ  180  tan 1   7   QRT    23   180  16,93 ✓ 163, 07  163, 07 QRS  163, 07  45 ✓for 45  118, 07 ✓ QRS  118,07  QPS  118, 07 ✓ answer (5) [23] Demo 4 NW/JUNE/MATH/ EMIS/6*******

Downloaded from hlayiso.com Mathematics/ P2 Mid-Year Examination NW/ Jun 2019 QUESTION 3 3.1 3.1.1 r 2  x2  y 2 [Pyth] OP 2   8   152 2 ✓ substitution  289 OP  17 ✓ answer (2) 3.1.2 (a) y tan   x 15  ✓ answer (1) 8 (b) cos( ) 17 sin 180     8 ✓ sin  cos  ✓ cos   17 sin   8  15  1  8   17      ✓ substitution  17  8  17  256  (4) 17 ✓ answer 3.1.3 15 tan    8 Ref  61,93 ✓ 78, 07   180  61,93 ✓ answer  118,1 (2) Demo 5 NW/JUNE/MATH/ EMIS/6*******

Downloaded from hlayiso.com Mathematics/ P2 Mid-Year Examination NW/ Jun 2019 3.2.1 sin 40  t x 2  y 2  r 2 [Pyth] ✓ sketch x  1 t2 ✓ x  1 t2 cos 320  cos  360  40  ✓ cos 40  cos 40  1 t2 ✓ answer (4) 3.2.2 sin140  sin 180  40   sin 40 ✓ sin 40 ✓ answer (2) t 3.2.3 tan  220    tan 220 ✓  sin 220   tan 180  40    tan 40 ✓ tan 40 t  (3) 1 t2 ✓ answer [18] Demo 6 NW/JUNE/MATH/ EMIS/6*******

Downloaded from hlayiso.com Mathematics/ P2 Mid-Year Examination NW/ Jun 2019 QUESTION 4 4.1 sin  90  A  tan(A  540)  cos  540  A  cos A.sin A cos A tan A   ✓ cos A  cos A cos A.sin A sin A ✓  cos A ✓ tan A  1  cos A sin A cos A.sin A ✓ 1 cos A  1  cos 2 A  cos 2 A  1  cos 2 A ✓ sin 2 A sin 2 A  cos 2 A ✓ answer (6)  tan 2 A 4.2 sin 120  .tan 330 tan 2 x  cos 240 sin 180  60  .tan  360  30   cos 180  60  sin 60( tan 30)  ✓ sin 60  cos 60 ✓  tan30 3 1 ✓  cos60  2 3  1 ✓ substitution 2 tan x  1 2 tan 2 x  1  0  tan x  1 tan x  1  0 tan x  1 or tan x  1 ✓ factors or tan x  1 x  45 ; 45 ✓ 45 ✓ 45 (7) 4.3 4 cos x  3 2 3 cos 2 x  4 3 cos x   3 ✓ cos x   2 2 x  150  360.k or  30  360.k ✓ 150  360.k ✓ 30  360.k k (4) ✓ k [17] Demo 7 NW/JUNE/MATH/ EMIS/6*******

Downloaded from hlayiso.com Mathematics/ P2 Mid-Year Examination NW/ Jun 2019 QUESTION 5 5.1 ✓answer (1) supplementary 5.2 R ≫ S Q ≫ U T P 5.2.1 Ŝ  90 diameter subtends right  ✓S ✓R (2) 5.2.2 T̂2  90 [corr s ; QU RS] ✓ S/ R TP = 8 cm [line from centre ⊥ to chord] ✓ S/ R QT 2  82  102 [Pythagoras] ✓ QT = 6 QT  102  82  6 TU  10  6  4 cm ✓ answer (6) [9] Demo 8 NW/JUNE/MATH/ EMIS/6*******

Downloaded from hlayiso.com Mathematics/ P2 Mid-Year Examination NW/ Jun 2019 QUESTION 6 K J O M L H N 6.1.1 ˆ  2  50  100 MOL [  at centre ✓S ✓R  2   at circumf] (2) 6.1.2 KJ  JL [given] ✓S Ĵ1  90 [line from centre to midpt of chord] ✓ S/ R In  NKL [acute s of right N̂  40 triangle complementary/ ✓S sum of s in  ] (3) 6.1.3 MO  LO [radii] ✓ S/ R ˆ  Lˆ M [ s opp equal ✓ S/ R 2 1 sides] 180  100 [ OML is isosc] L̂1   40 2 ✓S (3) 6.2 Lˆ 1  N=40 ˆ  [calculated in 6.1.2 and 6.1.3] ✓S MOLN is a cyclic [line subt. equal s / quad converse s in the ✓R same seg] (2) [10] Demo 9 NW/JUNE/MATH/ EMIS/6*******

Downloaded from hlayiso.com Mathematics/ P2 Mid-Year Examination NW/ Jun 2019 QUESTION 7 7.1 Construction: Draw diameter AM and join M to B ✓ construction Aˆ A 1 ˆ  90 2 [radius ⊥ tangent] ✓ S/ R ˆB  B ˆ  90 [ s in a semi-circle] ✓ S/ R 1 2 ˆ [ s in same segment] ✓ S/ R B̂2  A 2 B̂  Aˆ ✓S 1 1 (5) Demo 10 NW/JUNE/MATH/ EMIS/6*******

Downloaded from hlayiso.com Mathematics/ P2 Mid-Year Examination NW/ Jun 2019 7.2 T M R B P N A S 7.2.1 P̂1  90 [diameter subtends right  ] ✓ S/ R Pˆ  Pˆ [vert. opp. ] s ✓ S/ R 1 4 BN is a diameter [chord subtends 90 / ✓ conclusion converse s in a semi-circle] (3) 7.2.2 Nˆ  Pˆ 3 [tan chord theorem] ✓S ✓R ˆP  Pˆ [vert. opp. s ] ✓ S/ R ✓ S/ R 6 3 Mˆ  Pˆ [tan chord theorem] 1 6 AM NB [alt s equal] ✓R OR Aˆ  Pˆ [tan chord theorem] ✓S ✓R 2 Pˆ  Pˆ 5 2 [vert. opp. s ] ✓ S/ R ˆ  Pˆ [tan chord theorem] ✓ S/ R B2 5 AM NB [alt s equal] ✓R (5) Demo 11 NW/JUNE/MATH/ EMIS/6*******

Downloaded from hlayiso.com Mathematics/ P2 Mid-Year Examination NW/ Jun 2019 7.2.3 ˆ M̂1  90  A [ s in  AMP ✓ S/ R ˆ  90  B B ˆ [radius ⊥tangent] ✓ S/ R 1 2 ˆ Bˆ [alt s ; AM NB] A 2 ✓ S/ R MPBT is a cyclic [ext   int opp  / converse ✓R quad ext  of cyclic quad] OR ˆ N ˆ [alt s ; AM NB] M 1 ✓ S/ R ˆ B N ˆ 1 [tan chord theorem] ✓S ✓R Mˆ Bˆ 1 1 MPBT is a cyclic [ext   int opp  / converse ✓R quad ext  of cyclic quad] (4) 7.2.4 Tˆ  Pˆ1  90 [ext  of cyclic quad] ✓S ✓R (2) [19] TOTAL: [100] Demo 12 NW/JUNE/MATH/ EMIS/6*******