Basic Education MATHEMATICS P2 PREPARATORY EXAMINATION SEPTEMBER 2015 MEMORANDUM NATIONAL SENIOR CERTIFICATE GRADE 12

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1 Basic Educatin KwaZulu-Natal Department f Educatin REPUBLIC OF SOUTH AFRICA MATHEMATICS P PREPARATORY EXAMINATION SEPTEMBER 05 MEMORANDUM NATIONAL SENIOR CERTIFICATE GRADE N.B. This memrandum cnsists f 5 pages including this page.

2 Mathematics P September 05 Preparatry Eaminatin QUESTION. 4 pints crrect 5 9 pints crrect all pints crrect. a = 6, 58 (a = 6, ) b =,5 (b =,58...) yˆ 6,58, 5 a b equatin. yˆ 6,58,5(76) substitute 76 8,4 answer ().4 There is strng, psitive crrelatin between height and arm span. strng, psitive () [0]

3 Mathematics P September 05 Preparatry Eaminatin QUESTION.. Daily Sales Frequency Cumulative Frequency first tw cumulative frequencies crrect net tw cumulative frequencies crrect remainder crrect (ttal = 6) grunding at 0 pltting cumulative frequencies at upper limits pints crrect smth shape f curve. The median fr the data is apprimately R 87. reading f frm graph R87 ().4 The upper 5% interval is R96 t R0 (Range t accept: 94 t 0) 96 t 0 () [0]

4 Mathematics P 4 September 05 Preparatry Eaminatin QUESTION.. D D 5 D(5 ; 4) m CD tan yd 0 4 D 4 ( ) 5 6,4. m m AB CD, equal gradients AB CD D = 5 y D = 4 () substitutin int gradient frmula tan answer m AB.4 y c 0 ( ) c c y m AD 4 (0) 5 ( ) tan ( f inclinatin f AD) = f inclinatin f AD =,7 6,4,7 9,7 subst ( ; 0) answer m =,7 9,7

5 Mathematics P 5 September 05 Preparatry Eaminatin.5 AB DC 9AB DC y 0 = 5 4 = y y 0 AB // DC, y 5.() y = +.. () Substitute () in () ( + ) + ( + ) = = = 0 5 ( + ) = 0 # 0 ; = Substitute = in () y = B( ; ) substitutin substitute y = + standard frm = y = (5) [6]

6 Mathematics P 6 September 05 Preparatry Eaminatin QUESTION r ( 4) 8 ( ) ( y ) 5 m NP 4 m PT y c 5 4 c c 9 y T(9 ; 0) PT 5 (9 4) 50 Area PT tan NPT ˆ 50 NPT ˆ,8 4.6 NP = NM PT = TM ( 5) MNPT is a kite 8 (5 0) NP PT, prduct f gradients = radii radii tw pairs f adjacent sides equal in length subst int distance frmula 8 ( ) ( y ) m NP m PT subst (5; 4) c = 9 y = + 9 crdinates f T (6) substitutin int distance frmula 50 r 5 () substitutin int area frmula 57 () tan N PT ˆ,8 S/R S/R reasn 8 50 ()

7 Mathematics P 7 September 05 Preparatry Eaminatin 4.7 NTˆP NTˆM,8 NPˆT NMˆ T 90 diagnal f kite tangent perpendicular t radius NMˆ P ,6 angles in quadilateral OR 6,4 NP PT TNˆ P 68, TNˆ P MNˆ MNˆ TNˆ M P 68, P 6,4 sum Δ prp f kite (rad tan) S/R S/R S/R answer S/R S/R S/R answer []

8 Mathematics P 8 September 05 Preparatry Eaminatin QUESTION tan(80 A). cs (80 A). sin (60 cs (90 A) = tan A cs A sin A sin A sin A =. cs A cs A = sin A cs5 cs(6) cs 6 r r A) tan A, cs A, sin A, sin A sin A cs A answer (6) writing 5 in terms f 6 epansin answer 5.. sin7 tan 7 cs7 sin(45 6) cs(45 6) sin 45cs6 cs45sin 6 cs45cs6 sin 45sin 6 r r r r r r r r identity writing in terms f 6 epansins r substitutin r r r r answer (6)

9 Mathematics P 9 September 05 Preparatry Eaminatin 5. LHS sin cs sin cssin cs sin sin cssin cs sin cs tan RHS identity fr sin identity fr cs simplificatin identity OR LHS sin cs sin cssin cs sin cssin cs ( sin ) cssin cs cs cssin cs sin cs tan RHS identity fr sin identity fr cs simplificatin identity OR LHS sin cs sin cssin sin sin cssin sin cssin cs sin cs tan RHS identity fr sin identity fr cs simplificatin identity [9]

10 Mathematics P 0 September 05 Preparatry Eaminatin QUESTION 6 6. cs sin( 0) cs 90 0 cs 0 key angle = 0 using c-rati 0 0 n.60; n Z 0 n.60; n Z 40 n.0; n Z r 0 n n (0 ) n.60; n Z 40 n.60; n Z 40 n.60; n Z y (-80 ; ) (0 ; ) f g -0,5 - (-90 ; -) (-60 ; -) (90 ; ) - 60 (0 ) n n. 60 n Z (7) f -intercepts turning pints shape g intercepts turning pints shape (6) r critical values: 0 ; 80 critical values: 40 ; 90 crrect ntatin [6]

11 Mathematics P September 05 Preparatry Eaminatin QUESTION 7 7. AB tan BD h BD h BD tan 7. BC = BD CD BC BD BC.BD.cs y h h h h.cs y tan tan tan tan h h h tan tan tan h h.cs y tan tan h cs y tan h cs y tan.cs y using tan rati h BD tan using csine frmula substitutin simplificatin cmmn factr () [6]

12 Mathematics P September 05 Preparatry Eaminatin QUESTION 8 8. B D C O A E Cnstructin: Draw diameter AD. Jin D t C Prf: E ÂC DÂC 90 tan rad D ĈA 90 diameter subtends right angle A Dˆ C DÂC 90 sum f angles in triangle EÂC ADˆ C But ABˆ C ADˆ C angles in same segment EÂC ABˆ C cnstructin S / R S / R S / R S R (6)

13 Mathematics P September 05 Preparatry Eaminatin T V 4 Q S M 4 W P y R 8.. Tw tangents drawn frm the same eternal pint are equal in length. answer 8.. Ŝ angles ppsite equal sides S R (a) 8.. Rˆ tan-chrd therem S R (b) 8.. Vˆ Ŝ Rˆ eterir angle f triangle S R (c) 8.. Rˆ Qˆ Qˆ Ŝ (a) 8..5 (b) Rˆ Ŝ4 Ŝ Rˆ Ŵ 4 Ŵ Vˆ QVSW is a cyclic quad answer S R S R S R S R Tˆ Qˆ y crrespnding angles PQ RS S R Qˆ Mˆ Mˆ 4 Ŵ PMWR is a cyclic quad alt angles PQ RS angles in same segment eterir angle f triangle cnverse: eterir angle f cyclic quad Qˆ 4 y tan-chrd therem S R prven in 8.. angle at centre = twice angle at circumference cnverse: angles in same segment S R () () () () () ()

14 Mathematics P 4 September 05 Preparatry Eaminatin [8] QUESTION 9 B A P 4 Q R T 9. Qˆ Qˆ  4 Qˆ Qˆ  Rˆ 4 Rˆ vert pp angles tan-chrd therem et angle f cyclic quad S R S R S R PQ = PR sides pp equal angles R 9. In PBQ and PQA (i) Pˆ is cmmn (ii) Bˆ Qˆ tan-chrd therem (iii) PQˆ B remaining  angles in triangle PBQ PQA equiangular S S R S / R (7) 9. PA PQ PQ PB frm 9. But PQ = PR PA PR PR PB deductin PQ = PR cnclusin in full PA, PR and PB frm a gemetric sequence the rati is cnstant [4]

15 Mathematics P 5 September 05 Preparatry Eaminatin QUESTION 0 H R D F K E G 0. In Δ HKG: DG RK (RD KG) HD RH DG 9 DG = 6 units 0. Let FD = y FG = 6 y GF GE (DE HK) FH EK 6 y y (6 y) = y + y = y + y = 0 0 y = = FD S / R substitutin answer statement S / R substitutin simplificatin answer (5) [8] TOTAL: 50