Set 1 Paper 2. 1 Pearson Education Asia Limited 2017

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Mavis Byrd
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1 . A. A. C. B. C 6. A 7. A 8. B 9. C. D. A. B. A. B. C 6. D 7. C 8. B 9. C. D. C. A. B. A. A 6. A 7. A 8. D 9. B. C. B. D. D. D. D 6. D 7. B 8. C 9. C. D. B. B. A. D. C Section A. A (68 ) [ ( ) n ( n 6n n A 6 ( )( ). C. B. C ( )( )( ) 8 ) ] g() g( ) () () c [ ( ) ( ) c] c ( c) f The required reminder is. 6 ( ) hs equl roots. ()( ) 8 ( )( ) or or Set Pper Set Pper 6. A The grph of y = f() intersects the -is t two distinct points. The discriminnt of the eqution f() = is greter thn zero. I is true. According to the grph, the solution of the inequlity f() < is < < b. II is not true. The eqution of the is of symmetry of y = f(): b ( ) b III is not true. The nswer is A. 7. A or 8 or or The solution of the compound inequlities is. 8. B Let $ be the cost of cn of Potto Ct Food. The selling price of cn of Potto Ct Food $ ( %)( %) $. The profit percentge $(. ) % $ % 9. C : : 7 Person Eduction Asi Limited 7

2 Solution Guide nd Mring Scheme. D Averge usge before p.m. = vehicles per minute = ( 6) vehicles per hour = vehicles per hour The verge usge from 8.m. to 6 p.m vehiclesper hour 9vehiclesper hour. B. C Join BE.. A y is prtly constnt nd prtly vries directly s., where,. y The grph of is stright line with non-zero y-intercept. The nswer is A.. B y. A Mimum bsolute error of the mesured weight of solid metl g. g The minimum weight of the solid metl (.) g. g Mimum bsolute error of the mesured weight of smller solid metl g. g The mimum weight of ech smller solid metl (.) g. g.g The smllest possible vlue of n The smllest possible vlue of n is 77. Obviously, BCDE is squre. BE = BAE 9 nd AB =AE AB AE BE (Pyth. theorem) AB BE AB BE 7 Are of ABCDE = Are of ABE + re of BCDE ( AB)( AE) ( BC)( CD) ( 7) 8 6. D With the nottions in the figure, ABC ~ ADE (AAA) AB (corr. sides, ~ s) AB 8 AB AC (Pyth. theorem) nd AE 6 8 (Pyth. theorem) The totl surfce re [( 8 ) 8 ] Person Eduction Asi Limited 7

3 Set Pper 7. C ADG ~ FEG (AAA) AD : BC :, BF CF nd AD BE (opp. sides of //grm) AD EF Are of ADG Are of FGE Are of ADG 7 Are of ADG 7 8 ABF ~ GEF (AAA) BE EF BF : EF : Are of ABF Are of GEF Are of ABF 9 7 Are of ABF Are of ABED [(6 7) 8] Let AD 8, then height of ABED 8 8 Are of ABCD ( ) B In BCD, CD tncbd BC tn6 BC BC BDC CBD BCD 8 ( sum of ) BDC BDC In CDE, DE coscde CD DE cos DE AD BC (property of rectngle) Are of 9 C When ADE sin6 8 6, sin. ( sin ) ttins its mimum when sin ttins its minimum. The required lrgest vlue. D [ ( )] 9 In ABE nd DBE, AB BD (given) AE DE (given) BE BE (common side) ABE DBE (SSS) I is true. AB BD nd ABE DBE AD BE (prop. of isos. ) i.e. BDE is right-ngled tringle. BDE CBD (lt. s, AD // BC) In BDE, BDE DBE BED8 DBE 9 8 DBE ABE DBE ABE DBE (corr. s, s) II is true. In BDE, DBE BDE BE DE (sides opp. equl s) III is true. The nswer is D. Person Eduction Asi Limited 7

4 Solution Guide nd Mring Scheme. C CBD = CAD (s in the sme segment) ACB = CAD (lt. s, AD // BC) ACB = ABD (rcs prop. to s t ce ) BAD ABC8 (int. s, AD // BC) BAC CAD ABD CBD 8 8 ACB 8 ACB In BCE, CED EBC ECB (et. of ) ACB 66. A Join AB. ABC ADC 8 ( ABO8) 6 8 (opp. s,cyclicqud.) ABO OA = OB (rdii) BAO = ABO = (bse s, isos. ) In OAB, AOB 8 ( sum of ) AOB Alterntive Solution Join OC. OB = OC (rdii) OCB = OBC = 8 (bse s, isos. ) In OBC, BOC ( sum of ) BOC AOC ADC ( t centre twice t ce ) AOB 6 AOB. B As shown in the figure below: The nswer is B.. A The polr coordintes of the imge of A re (, ). The rectngulr coordintes of the imge of A (cos,sin). A (, ) The locus of P is the perpendiculr bisector of AB. The locus of P psses through the centre of the circle. ( bisector of chord psses through centre) Centre of the circle,, ( 6) By substituting, into y 8, we hve () 8 6. A Slope of L, slope of L, -intercept of L b, c b -intercept of L d, y-intercept of L d nd y-intercep of L. c Slope of L < < I is true. Slope of L slope of L c c II is true. -intercept of L > -intercept of L b > d III is true. For IV: y-intercept of L > y-intercept of L b d c bc d d bc IV is not true. The nswer is A. Person Eduction Asi Limited 7

5 Set Pper 7. A 6 ( 6) Centre of C, (, ) Rdius of C PQ ( 6) [ ( 6)] The eqution of C is ( ) [ y ( )] y y 8. D Refer to the tble below: 6y 9 6y Totl number of possible outcomes = nd totl number of fvourble outcomes = 8 The required probbility B According to the stem-nd-lef digrm, we hve h nd 7. If = 7, IQR = ( + 7) ( + h) = 7 h IQR 7 h h i.e. h I is not true. IQR = ( + ) ( + h) = + ( h) ( h) h h i.e. 7 h II is true. If h = nd =, IQR ( ) () h III is not true. The nswer is B.. C Men of A ( 8) ( ) ( ) ( ) Men of B ( b ) ( b ) ( b ) ( b ) b my not be greter thn b. I my not be true. Mode of A = + Mode of B = b + + > b + II is true. Medin of A = ( b ) ( b ) Medin of B b > b III is true. The nswer is C. Section B. B The H.C.F. of the epression y nd P is y. is one of the fctor of P. The L.C.M of the epression 8 y nd P is ndy re the fctors of P. P y. D y. Person Eduction Asi Limited 7

6 Solution Guide nd Mring Scheme. D n y n log y log( ) n log y log log n log y log log log is the y-intercept of the grph.. D log 6 log y...() (log y)...( ) By substituting () into (), we hve ( ) ( ) When, log y y When, log y y or or. D Let L cut the -is t (, ), where >. Then, we hve AC log nd BC log. AC > BC log log h log log logh log logh log log logh h h I is not true. According to the shpe of the grph, we hve h > nd >. h > II is true. AC logh BC log 6. D log log logh log log logh log h III is true. Answer is D. Obviously, nd re the roots of eqution 6. ( 6) nd ( ) 7. B ( i)( i) i i i ( ) ( ) i ( i)( i) is purely imginry. 8. C cos 7 cos ( cos )(cos ) cos cos When cos,.96 or or i.e.. (cor.tod.p.) When cos, or 6 ( 6 ) cos cos or 6.96 or 8.6 (cor.tod.p.) For 6, the eqution cos 7cos hs roots. 6 Person Eduction Asi Limited 7

7 Set Pper 9. C The ngle between AC nd the plne ABD is CAD. In ABD, AD BD AD AB AB BD m (Pyth. theorem) m In ACD, AD coscad AC CAD 8 (cor.to thenerest degree) The ngle between AC nd the plne ABD is 8.. D ABC 9 ( in semi-circle) ACD 9 (tngent rdius) ABC = ACD I is true. OBE 9 (tngent rdius). B ABC ABE ABO ABO OBC ABE OBC II is true. Alterntive method ABE ACB ( in lt. segment) OB = OC (rdii) OCB OBC (bse s, isos. ) ABE = OBC II is true. Let DCB =. Then DBC =. (tngent properties) BDC8 ( sum of ) BAC DCB ( in lt. segment) OA = OB (rdii) OAB OBA (bse s, isos. ) AOB8 ( sum of ) AOB = BDC III is true. The nswer is D. ( ) y 9...() y m...( ) By substituting () into (), we hve ( ) ( m) 9 m 9 ( m ) 6...(*) The circle nd the stright line intersect. of (*) ( m 6m )(6) 9 6m 6 ( m)( m) m The rnge of vlues of m is. B 8 6 Mid-point of OA, (, ) nd m. 8 6 Mid-point of OB, (, ) The eqution of the perpendiculr bisector of OA y ( ) 6 8 y nd the eqution of the perpendiculr bisector of OB y 6 [ ( )] 8 y y By solving, we hve = nd y. y The coordintes of the circumcentre of OAB re,. Alterntive Solution 8 ( 8) The -coordinte of the circumecentre Let (, y) be the coordintes of the circumcentre of OAB. Then (, y) is the centre of the circle pssing through O, A nd B, with rdius equl to y. Rdius is equl to the distnce between centre nd A. y y 6 y y 6 y y ( 8) ( y 6) The coordintes of the circumcentre of OAB re,. 7 Person Eduction Asi Limited 7

8 Solution Guide nd Mring Scheme. A The number of different groups cn be formed 6 C C7 99. D P(consists of both se) P(ll mle) P(llfemle ) 9 6 C C C C 8. C The required vrince ( ) 6 8 Person Eduction Asi Limited 7

Set 1 Paper 2. 1 Pearson Education Asia Limited 2014

Set 1 Paper 2. 1 Pearson Education Asia Limited 2014 . C. A. C. B 5. C 6. A 7. D 8. B 9. C 0. C. D. B. A. B 5. C 6. C 7. C 8. B 9. C 0. A. A. C. B. A 5. C 6. C 7. B 8. D 9. B 0. C. B. B. D. D 5. D 6. C 7. B 8. B 9. A 0. D. D. B. A. C 5. C Set Pper Set Pper