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

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1 . 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 ( a + ) a. C L.H.S.. C a a + + a( ) + a a R.H.S. ( )( + 6) + By comparig the like terms, we have a a f ( ) + f ( a) ( a) + ( a a + ) + a a + Set Paper. C ( ) a( a ) a( a ) 0 ( a)[ + ( a )] 0 a or a 6. A ( )( + 7) k k k 0 0 The equatio has distict real roots. Δ > 0 i.e. 6 ()( k 0) > 0 6 k + 0 > 0 k < 7. B It is give that p < q < 0. For optio A: Both p ad q < 0 p + q < 0 A is ot true. For optio B: p < q p q > ( q < 0) q q p > q B must e true. For optio C: p > q < ( p < 0) q p C is ot true. For optio D: p < q < 0 p > q > 0 p > q D is ot true. The aswer is B. Set Paper 8. D P( ) ( + )( + ) ( + )[ ( )] ( 6)( ) Whe P() is divided y, the quotiet is 6. Pearso Educatio Asia Limited 0

2 Solutio Guide ad Markig Scheme. B The graphs of y f() ad y k itersect at A ad B, ad the roots of the equatio f() k are ad 6. The coordiates of B are (6, k). I must e true. The roots of the equatio f() 0 are 0 ad the positive -itercept of y f(). The solutio of the iequality f() 0 is ot 6. II is ot true. The equatio of the ais of symmetry of y f(): + 6 III must e true. The aswer is B. 0. A Let $ ad $y e the origial prices of two watches respectively. ( + %) 700 y( %) y 00 Gai / loss after the two trasactios $( y) $( ) $0 Cathy losses $0.. C Let m e the umer of male studets i the school. The the umer of female studets i the school ( + 0%) m. m m +.m 0 m 0 The umer of female studet D y y z z + y ( + y) : ( z) z + y z + :. B f() is partly costat ad partly varies iversely as. k f ( ) k +, where k, k 0 f() k + k...() f() k k +...() () (): k k 0 By sustitutig k 0 ito (), we have k + ( 0) k 7 0 f ( ) 7 0 f () 7. C Maimum asolute error of each measuremet m 0. m Percetage error of the perimeter 0. 00% (8 + ).7% (cor. tod.p.). C Produce AD to meet EC at F. ABD ~ ACF (AAA) Pearso Educatio Asia Limited 0

3 Set Paper Area of ACF + Area of ABD 7 Area of ACF cm 7. cm Area of BCFD (7. ) cm. cm Area of EDF (60.) cm AD AB AF AC 7 AD : DF : Area of ADE Area of EDF Area of ADE 0. cm (corr. sides, ~ s ) AD DF (0.) cm cm 6. C With the otatios i the figure, y + (7 ) (Pyth. theorem) The total surface area 8 cm ( ) 0 + (7 + ) cm 7. C Let r ad h e the ase radius ad the height of the coe respectively. Volume of the sphere volume of the coe πr π r h r h Height of the coe r Base radius of the coe r 8. A cos (0 θ ) cos (70 θ ) si 0 ta si θ cos[80 + (0 θ )] si θ + [ cos (0 θ )] si θ si θ si θ. D I ABD, FD AB 0 m 0 m taα AF 0 AF m taα CAF ECA (alt. s, AF // EC) β I AFC, AF cosβ AC AF AC cosβ 0 m taα cosβ 0 The distace etwee A ad C is m. taα cos β 0. A AD // BC ad AB DC ABCD is a isosceles trapezium. ABC DCB. I is true. I ABC ad DCB, ABC DCB (proved) AB DC (give) BC CB (commo sides) ABC DCB (SAS) I ABD ad DCA, BAD CDA (proved) AB DC (give) AD DA (commo sides) ABD DCA (SAS) ABC DCB I ABE ad DCE, AB DC (give) BAE CDE (corr. s, s) AEB DEC (vert. opp. s) ABE DCE (AAS) II is true. Pearso Educatio Asia Limited 0

Solutio Guide ad Markig Scheme I AED ad CEB, AED CEB (vert. opp. s) DAE BCE (alt. s, AD // BC) AED ~ CEB (AAA) A pair of cogruet triagles is also a pair of similar triagles.

4 Solutio Guide ad Markig Scheme I AED ad CEB, AED CEB (vert. opp. s) DAE BCE (alt. s, AD // BC) AED ~ CEB (AAA) A pair of cogruet triagles is also a pair of similar triagles. There are pairs of similar triagles. III is ot true. The aswer is A.. D Joi AD. OA OD (radii) OAD ODA (ase s, isos. ) I OAD, 0 + OAD + ODA 80 ( sum of ) ODA 60 ODA 0 ABC + ADC 80 (opp. s, cyclic quad.) ABC + (70 0 ) 80 ABC 0. B Let X, Y ad Z e the mid-poit of AB, CD ad EF respectively. OX AB, OY CD, OZ EF (lie joiig cetre to mid-pt. of chord chord ) AB CD EF (give) OX OY OZ (equal chords, equidistat from cetre) OXQ OYQ (SAS) ad OXP OZP (SAS) Let OPX α ad OQX β. I OPQ, OPQ + OQP + POQ 80 ( sum of ) α + β α + β 6 I PQR, RPQ + RQP + PQR 80 α + β + PQR PQR 80 PQR 6 ( sum of ). C Let α ad β e the iterior agle ad eterior agle of the polygo respectively. α + β 80 ad α β 0 By solvig, we have α ad β 60 8 I is true. Oviously, the umer of diagoals of the polygo is more tha 8. II is ot true. The umer of aes of reflectioal symmetry of the polygo 8. III is true. The aswer is C.. A The equatio a + y ca e rewritte as a y +. From the figure, the y-itercept of the graph > 0 > 0 i.e. > 0 From the figure, the graph slopes upwards. a > 0 a > 0 ( > 0) a < 0 Oly optio A satisfies > 0 ad a < 0. The aswer is A.. A AOC 0 ad OC OA (prop. of square) Whe C(, ) is rotated clockwise aout O through 0, the image is A. i.e. Coordiates of A are (, ). With the otatios i the figure, joi OB ad AC. Pearso Educatio Asia Limited 0

5 Set Paper ( ) Slope of AC OB AC (prop. of square) Slope of OB 0. A Refer to the figure: 6. D The movig poit P maitais a fied distace uits from L. The locus of P is a pair of straight lies parallel to ad at a distace uits from L. The aswer is D. 7. C The cetre of the circle 6 k, k, The slope of the diameter passig through (, 6) is. k 6 ( ) k k 8. C 7 P(age 7) 6 The aswer is A. Sectio B. D a a a + a a a( a ) ( a + ) a + ( a + )( a ) ( a )( a + ) a a a a + a a a a. D C F C Mode 70 cm y 70 Arrage their heights i ascedig order: cm, cm, 8 cm, 70 cm, 70 cm, 70 cm The media of their heights cm 6 cm. D i( i) + i i + i ( ) + + i + + The imagiary part + + ( ) Pearso Educatio Asia Limited 0

6 Solutio Guide ad Markig Scheme. D From the figure, Sice the graph passes through 0,, we have m 0 m y Take logarithms to the ase o oth sides of the epressio y, we have log log y log y log log log y log log Referrig to the graph of y, we kow that >. log > 0 The slope of the graph of log y agaist is positive, while the itercept of the vertical ais is egative. The aswer is D.. A a 0 loga loga I is true. log a log a + log II is true. The sequece: log a, log a, log a i.e. log a, log a, log a It is a arithmetic sequece ut ot a geometric sequece. III is ot true. The aswer is A. 6. A Refer to the figure. 7. C Let a, d ad T() e the first term, the commo differece ad the geeral term of the sequece respectively. a 8 ad d ( 8) T ( ) 8 + ( )() 0 Let T(m) e a term of the sequece greater tha 6. T ( m) > 6 i.e. m 0 > 6 m > 7 m > 0 The least value of m is B The graph of y + f ( ) is otaied y reflectig the graph of y f () aout the y-ais ad the traslatig upwards y uit. The aswer is B.. D ( ) a + ( ) The graph of y si ( 0 ) is otaied y traslatig the graph of y si ( 0 ) upwards y uit. 0. D CBD 0 BD CD BC 7 m Let N e the poit o CD such that BN CD. BN CD CB BD BN 7 7 BN AB taθ BN m (Pyth. theorem) 7 7 The aswer is A.. A BEO 0 (taget radius) BCA 6 Pearso Educatio Asia Limited 0

7 Set Paper EO // CA (corr. s equal) BE EC ad EO // CA BO OA (itercept theorem) I is true. Joi OD. ADO 0 (taget radius) ACB DO // CB (corr. s equal) AO OB ad DO // CB AD DC (itercept theorem) AD DC ad BE EC DE // AB (mid-pt. theorem) DE // AO ad AD // OE OEDA is a parallelogram. (y defiitio) II is true. OE is a radius of the circle ut OA is ot. OE OA ad OEDA is ot a rhomus. AE is the agle isector of DAO uless OEDA is a rhomus. III is ot true. The aswer is A.. A P(all of the same colour) P(all white) + P(all lack) D New media ( ) 60 New iter-quartile rage 8 New stadard deviatio The aswer is D.. C y + k y + k Sustitute y + k ito + y + y 0, we have ( y + k) + y + ( y + k) y 0 y ( k + ) y + ( k + k ) 0 (*) The y-coordiate of the mid-poit AB the sum of roots of (*) ( k + ) k +. D Referrig to the Ve s diagram, the set X Y is equivalet to the set Y. 7 P(Y ) P(X Y ) P( Y ) P( Y ') 7 P( X Y ) P( X ) + P( Y ) + 7 Pearso Educatio Asia Limited 0

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

Set 3 Paper 2. Set 3 Paper 2. 1 Pearson Education Asia Limited 2017 Set Pper Set Pper. D. A.. D. 6. 7. 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 7. 8. 9. B 0. D.. A. D. D. A 6. 7. A 8. B 9. D 0. D. A. B.. A. D Sectio A. D ( ) 6. A b b b ( b)