Set 3 Paper 2. Set 3 Paper 2. 1 Pearson Education Asia Limited 2017

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1 Set Pper Set Pper. D. A.. D B 8. D 9. B 0. A. B. D. B.. B 6. B 7. D 8. A 9. B 0. A. D. B.. A. 6. A B 0. D.. A. D. D. A A 8. B 9. D 0. D. A. B.. A. D Sectio A. D ( ) 6. A b b b ( b) b. L.H.S.. D. b b ( b) b R.H.S. ( )( 6) By comprig the lie terms, we hve b b b ( ) ( ) ( ) 0 ( )[ ( )] 0 or 6. 7 () 8 7 () () + () : 7 () B For optio A: Both p d q < 0 p + q < 0 A is ot true. For optio B: p < q p q ( q < 0) q q p q B must be true. For optio : p q ( p < 0) q p is ot true. For optio D: p < q < 0 p > q > 0 p > q D is ot true. The swer is B. 8. D P( ) ( )( ) ( )[ ( )] ( 6)( ) Whe P() is divided by, the quotiet is B The grphs of y = f() d y = itersect t A d B, d the roots of the equtio f() = re d 6. The coordites of B re (6, ). I must be true. The roots of the equtio f() = 0 re 0 d the positive -itercept of y = f(). The solutio of the iequlity f() 0 is ot 6. II is ot true. Perso Eductio Asi Limited 07

2 Solutio Guide d Mrig Scheme The equtio of the is of symmetry of y = f(): 6 III must be true. The swer is B. 0. A Let $ d $y be the cost of two wtches respectively. ( %) y( %) 9 y 00 Gi / loss fter the two trsctios $(9 y) $( ) $90 thy loses $90.. B Let ml be the volume of wter iside the cotier origilly. The volume of wter fter decresed by 0% ( 0%) 0.7 The volume of wter fter icresed by 0% (0.7)( 0%) 0.98 The required peretge chge % %. B Number of dots i the st ptter = 9 Number of dots i the d ptter = = 7 Number of dots i the rd ptter = = Number of dots i the th ptter = + 8 = Number of dots i the th ptter = + 8 = Number of dots i the 6th ptter = + 8 = 9 The swer is B.. Let z, where is o-zero costt. y By substitutig =, y = d z = ito the equtio, we hve i.e. z y Whe = d z = 9, we hve 6 9 ( ) y y y 6. B Produce AD to meet E t F.. D y y z z y ( y) : ( z) z y z : ABD ~ AF (AAA) Are of AF Are of ABD 9 Are of AF cm 6 7. cm Areof BFD (7. ) cm 9. cm Areof EDF (60 9.) cm 0. cm AD AB AF A 7 (corr. sides, ~ s) AD : DF : ADE d EDF hve the sme height. Perso Eductio Asi Limited 07

3 Set Pper Are of ADE AD Are of EDF DF Are of ADE 0. cm cm 9. B 6. B Let AD = cm, the D = (6 ) cm. Areof ABD cm (6 ) Areof BD cm The re of ABD is smller th the re of BD by 0 cm. (6 ) I ABD, AB cm I BD, B AD BD 0 cm BD D 0 (6 0) cm 9 cm Perimeter of AB ( 9 6) cm 0cm (Pyth. theorem) (Pyth. theorem) 7. D Let r cm be the rdius of the upper prt of the frustum. 8 r r The volume of the frustum 9 (6) (8) (8 ) cm cm 8. A cos(90 ) cos(70 ) si0 t si cos[80 (90 )] si [ cos(90 )] si si si Eted AD to poit E such tht such tht AB F. I BF, BF cos B BF Bcos I DE, E side D E si( 80 ) D E Dsi AB BF E B cos Dsi AE E. F is poit o AB 0. A AD // B d AB = D ABD is isosceles trpezium. AB = DB. I is true. I AB d DB, AB = DB (proved) AB = D (give) B = B (commo sides) AB DB (SAS) I ABD d DA, BAD = DA (proved) AB = D (give) AD = DA (commo sides) ABD DA (SAS) AB DB I ABE d DE, AB = D (give) BAE = DE (corr. s, s) AEB = DE (vert. opp. s) ABE DE (AAS) II is true. I AED d EB, AED = EB (vert. opp. s) DAE = BE (lt. s, AD // B) AED ~ EB (AAA) A pir of cogruet trigles is lso pir of similr trigles. There re pirs of similr trigles. III is ot true. The swer is A. Perso Eductio Asi Limited 07

4 Solutio Guide d Mrig Scheme. D Joi AD. By solvig, we hve d 60 8 I is true. Obviously, the umber of digols of the polygo is more th 8. II is ot true. The umber of es of reflectiol symmetry of the polygo = 8. OA = OD (rdii) OAD = ODA (bse s, isos. ) I OAD, 0 OAD ODA 80 ( sum of ) ODA 60 ODA 0 AB AD 80 (opp. s,cyclicqud.) AB (70 0) 80 AB 0. B Let X, Y d Z be the mid-poit of AB, D d EF respectively. OX AB, D OY, OZ EF (lie joiig cetre to mid-pt. of chord chord ) AB = D = EF (give) OX = OY = OZ (equl chords, equidistt from cetre) OXQ OYQ (SAS) d OXP OZP (SAS) Let OPX = d OQX =. I OPQ, OPQ OQP POQ I PQR, RPQ RQP PQR 80 PQR 80 6 PQR 80 PQR 6 ( sumof ) ( sumof ). Let d be the iterior gle d eterior gle of the polygo respectively. 80 d 90 III is true. The swer is.. A For L : Slope, -itercept = b d y-itercept For L : Slope, -itercept = d d y-itercept c c c I is true. -itercept of L > -itercept of L b > d II is true. b d c Slope of L > slope of L d both d c re egtive y-itercept of L = b b 0 y-itercept of L < d c c d 0 i.e. b c d III is ot true. The swer is A. Perso Eductio Asi Limited 07

5 Set Pper. The rectgulr coordites of the imge of poit A re (, ). r ( ) r 6 d t r or (rejected) 00 The polr coordites of the imge of poit A re (, 00). 6. A The movig poit P mitis fied distce uits from L. The locus of P is pir of stright lies prllel to d t distce uits from L. The swer is A. 7. The cetre of the circle 6,, The slope of the dimeter pssig through (, 6) is. 6 ( ) P(ge 7) = = 6 9. B Mode = 70 cm = y = 70 Arrge their heights i scedig order: cm, cm, 8 cm, 70 cm, 70 cm, 70 cm 870 The medi of their heights cm 6cm 0. D Sectio B. ( )( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ). A E000AD D (6) 6 (06) i( i) i i i ( ) i The imgiry prt. D From the figure, ( ) Sice the grph psses through 0,, we hve m 0 m y Te logrithms to the bse o both sides of the epressio y, we hve log log y log y log log log y log log Referrig to the grph of y, we ow tht >. log 0 The slope of the grph of log y gist is positive, while the itercept of the verticl is is egtive. The swer is D. Perso Eductio Asi Limited 07

6 Solutio Guide d Mrig Scheme. A 0 b log b b log I is true. log b log log b II is true. The sequece: log b, log b, log b i.e. log b, log b, log b It is rithmetic sequece but ot geometric sequece. III is ot true. The swer is A. 6. Let, d d T() be the first term, the commo differece d the geerl term of the sequece respectively. = 98 d d = 9 (98) = T( ) 98 ( )() 0 Let T(m) be term of the sequece greter th 6. T ( m) 6 i.e. m 0 6 m 7 m 0 The lest vlue of m is A Refer to the figure. The swer is A. 8. B The grph of y = + f () is obtied by reflectig the grph of y = f () bout the y-is d the trsltig upwrds by uit. The swer is B. 9. D By comprig the optimum vlues of the grphs of y si( 0) b d y si( 0) b, we hve ( ) By cosiderig the me of the optimum vlues of the grph of y si( 0) b, we hve ( ) The grph of y = si ( 0 ) b is obtied by trsltig the grph of y = si ( 0 ) upwrds by uit. b b 0. D BD = 90 BD D B 7 m (Pyth. theorem) m Let N be the poit o D such tht BN D. BND B BD BN 7 7 BN AB t BN 7 7. A BEO 90 (tget rdius) BA EO // A (corr. s equl) BE = E d EO // A BO = OA (itercept theorem) I is true. Joi OD. ADO 90 (tget rdius) AB DO // B (corr. s equl) AO = OB d DO // B AD = D (itercept theorem) AD = D d BE = E DE // AB (mid-pt. theorem) DE // AO d AD // OE OEDA is prllelogrm. (by defiitio) II is true. 6 Perso Eductio Asi Limited 07

7 Set Pper OE is rdius of the circle but OA is ot. OE OA d OEDA is ot rhombus. AE is the gle bisector of DAO uless OEDA is rhombus. III is ot true. The swer is A.. B y y Substitute y ito y y 0, we hve ( y ) y ( y ) y 0 y ( ) y ( ) 0 The y-coordite of the mid-poit AB thesumof rootsof (*) ( ). The totl umber of queues c be formed 9! 8! 080. A P(llof thesme colour) (*) P(llwhite) P(llblc) 8 7. D New medi ( ) 60 New iter-qurtile rge 8 New stdrd devitio The swer is D. 7 Perso Eductio Asi Limited 07

Set 3 Paper 2. Set 3 Paper 2. 1 Pearson Education Asia Limited C

Set 3 Paper 2. Set 3 Paper 2. 1 Pearson Education Asia Limited C . D. A. C. C. C 6. A 7. B 8. D. B 0. A. C. D. B. C. C 6. C 7. C 8. A. D 0. A. D. B. C. A. A 6. D 7. C 8. C. C 0. A. D. D. D. D. A 6. A 7. C 8. B. D 0. D. A. C. D. A. D Sectio A. D ( ) 6. A + a + a a (