Mathematics P2 1 Preparatory Examination September 2016 NSC-MEMORANDUM. Education. KwaZulu-Natal Department of Education REPUBLIC OF SOUTH AFRICA

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1 Mathematics P Preparatry Examinatin September 06 Educatin KwaZulu-Natal Department f Educatin REPUBLIC OF SOUTH AFRICA MATHEMATICS P PREPARATORY EXAMINATION SEPTEMBER 06 MEMORANDUM NATIONAL SENIOR CERTIFICATE GRADE MARKS : 50 This memrandum cnsists f 5 pages.

2 Mathematics P Preparatry Examinatin September 06 SECTION A QUESTION. min 9 ; maximum 9 ; upper quartile 75 Lwer quartile 8 and medium 55 Therefre five number summary is (9; 8; 55; 75; 9). A 9 and 9 A 8 A 55 A 75 () CA crrect Q & Q CA median crrectly shwn CA bth crrect whisers. Data is sewed t the left /Data is negatively sewed CA CA Answer. mean mar fr M cee-nai High ,8 5 OR 57 x 50,8 5.5 Bee Vee High Schl. Bee Vee High Schl perfrmed better because half f the learners gt abve 60% whilst half f M cee-nai learners gt mre than 55%. The median f Bee Vee High was higher than that f M cee-nai High. A sum () () CA answer () [Answer nly full mars] CA Bee Vee High CA Reasning () []

3 Mathematics P Preparatry Examinatin September 06 QUESTION. (6; 60) A 6 A 60 (). y,6 x + 7,5 AA gradient AA y - intercept (). r 0, AA answer () [8] QUESTION. AC ( 5 7) + ( ( ) + () ()) A crrect Subst + 9 5,7 CA answer (). M BC 6 () y y m (x x ) y 6 ( x ) y 8 x + y x + A CA crrect subst. f (;6) And (7;-) CA equatin in any frm ()

4 Mathematics P Preparatry Examinatin September 06. Bˆ θ α β... Ext tanα mbc α 6,9 5 tan β m AB 6 β 9,8 θ α β θ α β 6,9 9,8 87, Aˆ BC 87, OR Distance AB DistanceBC DistanceAC csˆ a B 0 0,05 ˆ B 87, ( + 5) + ( 6 ) c b ac + ( 7) + ( 6+ ) 00 ( 5 7) + ( + ) 5 ( 6) ( 5) 0 ( )( 6) CA tan α CA α 6,9 5 A tan β 6 CA β 9,8 CA A BC ˆ 87, A Distance AB A Distance BC A Distance AC CA substitutin in csine rule CA answer P ; P 7 ; AA bth c-rdinates ()

5 Mathematics P 5 Preparatry Examinatin September 06.5 m AC 7+ 5 thrugh ( ; ) equatin: y ( x + ) y x y x+ r y+ x m AB ; 6x + 5y 8 6 5y 6x + 8 y m m AB m AB 6 8 x m 6 x + 5y 8 y x+ r A CA subst. (-;) CA equatin.in any frm A m A m.m AB 6 5 () A - () [8]

6 Mathematics P 6 Preparatry Examinatin September 06 QUESTION.. At W, y x + () W( 5;) x 5 x 5 r 5 (x + 5) + (y ) 5.. VZ r x 5 0 units.. m GZ + 0+ A subst y CA x 5 CA c -rdinates f W CA r 5 CA equatin f the circle. CA answer () A substitutin int frmula CA answer.. Midpint f GZ is ; A crdinates..5 m GZ () () m CA gradient f y x + perpendicular bisectr CA substitutin int frmula y x + CA answer ()

7 Mathematics P 7 Preparatry Examinatin September W ( 5; ) int x + y 0 LHS ( ( 5) + 6() RHS W is n the line that bisects GZ perpendicularly and W n GZ. lines intersect at W. OR x+ x+ 6 x 5 Thisis OR thex- value f the crdinate fw. Equatin f WZ: y y + y+ ( x+ ) 5+ y+ ( x+ ) 7 y x 7 x x+ 9x x+ 5x 5 x 5 y m( x x) ( 5) + This is the crdinate f W. A substitutin A 0 A equating eq. f () perpendicular bisectr t the hrizntal line y A x 5 A equatin f WZ A x 5 () ().. circle M: M( ; ) ; r 5 circle N: N(;) ; r r + r 8 and r r MN ( ( )) + ( + 9+ r,6 r + r > MN > r r The tw circles intersect at tw distinct pints. ) A r 5 A r A r +r 8 A MN,6 A cmparing A cnclusin (6)

8 Mathematics P 8 Preparatry Examinatin September 06.. circle M circle N (x + ) + (y ) 5 (x ) + (y ) 9 x + x + + y 6x+ y y + 5 x The equatin f the cmmn chrd is : 6x + y x+ + y 6y+ 9 9 M equating A simplifying CA equatin f the chrd () [] QUESTION tan α A crrect rati α β r CA α 0 CA β 0 tan (-β) -tanβ ( ) tan β A crrect rati CACA β 0 () β OP ( ) + ( ) + 6 OP OP 5.. csβ OQ A using distance fr frmula CA answer () OP OQ csβ cs0 8 CA cs 0 8 CA

9 Mathematics P 9 Preparatry Examinatin September 06 8 Q ; 0 r CA c-rdinates OP cs0 OQ OQ cs0 CA cs 0 8 OQ 8 Q ; 0 CA 8 CA c-rdinates () 5. csα+ sinα [ sin0 csα+ cs0 sinα] sin ( 0 + α) ; β 0 M fr intrducing A intrducing 0 special angle A sum cmpund frmula A calculating A calculating β OR csα + sinα sin + ( α β) sin0 csα + cs0 sin α sin ( α + β) sin ( 0 + α ) sin ( α + β), OR β 0, β 0 A fr intrducing A intrducing 0 special angle A sum cmpund frmula A calculating A calculating β

10 Mathematics P 0 Preparatry Examinatin September 06 csα+ csα+ csα+ csα+ sinα sinα sinα sin( α+ β) ( sinαcsβ + csαsinβ) sinα sinαcsβ + csαsinβ ( sinβ) csα+ ( csβ) sinα A expansin sinβ and csβ sin β cs β A cmparing cefficients A values f trig ratis β β sinβ β 0 A calculating A calculating β []

11 Mathematics P Preparatry Examinatin September 06 QUESTION 6 6. A asympttes A shape f f A shape f g A crrect x- intercept f f ( 90 ; 90 ; 70 ) A crrect x intercept f g (0 ; 60 ) A (0; ) and (80 ; ) ( 80 ; ) and (60 ; ) CA fr any tw () () < x < 0 r 80 < x < 60 OR 80 < x < 0 80 < x < 60 CA 80 < x < 0 CA 80 < x < 60 [penalize ne mar fr incrrect ntatin] () 6.5 y cs (x 5 ) A cs (x 5 ) () []

12 Mathematics P Preparatry Examinatin September 06 QUESTION 7 7. cs 5. cs x + sin 5. sin x sin x cs(5 x) sin x cs (90 x) A cs (5 x) A cs (90 x) x x n. 60 r 5 x 60 (90 x) n. 60 x n. 60 r x 6 n x n. 60 r x 7 n x 6 + n. 60 r x 7 + n. 0, n Ζ 7.. E FˆD FBˆ C α crrespnding ' s AD // BC EFˆD AÊF θ + AEˆ F... ext f ΔAEF In Δ ABE : α θ sinaêb sineâb AB BE ABsin BE sin ( 90 + θ) ( α θ) ABcsθ sin ( α θ) CA x 6 + n. 60 CA x 7 + n.0 A n Ζ A AÊF α θ A sine rule applicatin A substitutin x x Area f Δ BCE. ( 8 ) sin A(x) x. ( 8 x).. x.8 x. x 8 9 x x x x A area rule A sin 50 A simplifying ()

13 Mathematics P Preparatry Examinatin September At maximum area: A ( x ) 0 A 9 ( x). x 9 0 x x 9 x M A ( x ) 0 A derivative CA x () 7.. BC CE 8 () A fr bth BC and CE 9 BE BC + CE BC. CE cs 50 () + (9) x x 9 ( cs 0 ) cs 0 CA applying csine rule and substitutin ,765 BE,695,69 CA answer () [9]

14 Mathematics P Preparatry Examinatin September 06 QUESTION 8 8. PT TQ cm (line frm center perpendicular t chrd PQ) PQ cm + cm cm 8. OT OQ QT.. pythagras 69 5 OT 5 TR OR OT cm 5cm 8cm A R A answer () A OT 5 CA TR 8cm In Δ PTR, PR TR + PT cm CA PR 08 PR 08 cm r cm r, cm CA PR r, () [6]

15 Mathematics P 5 Preparatry Examinatin September 06 QUESTION 9 9. Interir ppsite angle A S () 9. Q N O P x K L M Cnstructin : Draw diameter LOQ and jin QP r Jin OL and OP STATEMENT REASON Let P LˆM L ˆ x Pˆ P + 90 angle subtended by the diameter Lˆ 90 x LM OL, tan radius Qˆ x Sum f the angles f a triangle Nˆ x Subtended by the same chrd LP P LˆM Nˆ 9.. Â 80 AÊD c interir ' s, AB//ED A cnstructin A S/R A S/R A S A S/R A 0 () 9.. Bˆ ext cyclic quad ABDE A R A 70 () 9.. Dˆ Bˆ 70...( alt s; DE//CA) CA 70 () 9.. Bˆ Dˆ ( s pp sides) CA 70 ()

16 Mathematics P 6 Preparatry Examinatin September ( Bˆ + Dˆ )...( sumf Δ) Ê Dˆ Ê 0... tanchrdtherem CA CA A R E ˆ 0 D ˆ 0 () [7] QUESTION 0 0. Pˆ Bˆ x... alt s; SP//BC Pˆ Pˆ x... given Q P x... tan chrd therem 0. PC BC Pˆ Bˆ x prved abve (Δ PCB) 0. Qˆ Bˆ x... prved RCQB is a cyclic quad cnverse ' sin the same segment 0. Bˆ... crresp 'ssp BC Ŝ Rˆ... ' s in the same segment, cyclicquadrcqb In Δ PBS and Δ QCR Pˆ Qˆ x... prved Ŝ Rˆ prved A S A R () A P ˆ B C x ˆ A reasn () A S A R () Remaining sequal Δ PBS Δ QCR 0.5 In Δ PBQ and Δ PCR ˆP is cmmn PQˆ B Rˆ... ext f cyclic quad RCQB Δ PBQ Δ PCR ( rd Δ) PB QB ( Δ s) CP CR PB. CR QB. CP A R A S A S () [7]

17 Mathematics P 7 Preparatry Examinatin September 06 QUESTION In Δ KLM LD 8 9 (LM//DE; prprtinality therem) 6 LD DML MDˆ E x alt s, LM DE LM LD (sides pp s) A S/R A LD A S A answer A R [5] TOTAL: [50]