Set 6 Paper 2. Set 6 Paper 2. 1 Pearson Education Asia Limited 2017

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1 Set 6 Pper Set 6 Pper. C. C. A. D. B 6. D 7. D 8. A 9. D 0. A. B. B. A. B. B 6. B 7. D 8. C 9. D 0. D. A. A. B. B. C 6. C 7. A 8. B 9. A 0. A. C. D. B. B. B 6. A 7. D 8. A 9. C 0. C. C. D. C. C. D Section A. C 00 ( ( ( ( [( ( ]. C (. A su tu sv tv sw tw u( s t v( s t w( s t ( s t( u v w. D k k k k 0 The qurtic eqution hs equl roots. Δ = 0 i.e. ( k (( k 0 k k 8 0 k k 0 ( k ( k 0 k or k. B L.H.S. m( m m ( m m R.H.S. n( n n n ( n Compring the coefficients of like terms on oth sies, we hve m n m n Solving the ove two equtions, we hve m = n n =. m + n = + = 6. D is fctor of f (. f ( 0 ( ( k 0 k When f ( is ivie, reminer f D The grph of = + opens upwrs. > 0 I is true. = -intercept of the grph of = + > 0 II is true. The grph of = + oes not intersect with the -is. 0 i.e. 8. A ( ( 0 III is true. The nswer is D. Person Euction Asi Limite 07

2 Solution Guie n Mrking Scheme 9. D A Let $ e the cost price of ech phone. Selling price of Dvi s phone $ ( 0% $. Selling price of Mr s phone $ ( 0%( 0% $0.99 From the question, we hve B Gin $( $78 Peter gine $78. An two equilterl tringles re similr. Are of new tringle perimeter of new tringle Are of originltringle perimeter of originltringle ( 0% 0.6 ( 6% The re of the tringle is ecrese 6%.. B z 6 z n z re negtive. < z < < z z k z, where k 0 B sustituting =, = n z = into the eqution, we hve k( k z When = n = 6, ( z 6. B Mimum solute error 0. g 0.g Lest possile weight of coins (00 0. g 99.7g 99.7 Lest possile weight of coin g 9.9g. B The numer of ots in the n pttern ( 6 The numer of ots in the r pttern ( ( 0 The numer of ots in the th pttern ( ( ( The numer of ots in the th pttern ( ( ( ( The numer of ots in the 6th pttern ( ( ( ( ( 8. A 9 8 n re negtive. < 6. B Let cm n h cm e the slnt height n the height of the circulr cone respectivel. Curve surfce re = 80 cm ( 80 h 9 Volume of the cone ( (9 cm cm Person Euction Asi Limite 07

3 Set 6 Pper 7. D AFD ~ CFE (AAA DF AD EF CE (corr. sies, ~ s CDF n CFE hve the sme height. Areof CDF DF Are of CFE EF Areof CFE cm cm AFD ~ CFE Are of AFD DF Are of CFE EF Are of AFD 6cm 6cm Are of the qurilterl ABEF (6 cm 9cm 6 8. C (6 80 Size of n interior ngle 6 ( sum of polgon 0 BAF = 0 AF = AB AFB ABF (se s, isos. AFB ABF0 80 ( sum of AFB 60 AFB 0 Similrl, FAE = 0 AGB 0 0 (et. of D In AEB, ABE 8 (et. of 6 BCD BAD ( s in thesme segment In BCF, AFC 6 (et. of 0. D For A: 60 n 8 0 For B: The size of ech interior ngle = 80 0 = 60 Ech interior ngle is greter thn its eterior ngle 0. For C: The polgon hs 8 sies. However, the numer of igonls of polgon is NOT 8. 8(8 The numer of igonls of the polgon = = For D: The sum of the interior ngles of the polgon = 80 (8 = 880 The nswer is D.. A lies in qurnt II. sin > 0 n cos 0 sin > cos I is true. For 90 < < 80, cos ecreses s increses. For 90 < < < 80, cos > cos II is true. For 90 < < 80, tn increses s increses. For 90 < < < 80, tn < tn III is not true. The nswer is A.. A Construct AM where M is point on BE such tht AM BE. M Are of ABE = 0 cm cm AM 0cm AM cm AB = AE ABC = AED (se s, isos. BC = ED ABC AED (SAS AC = AD AC = AD n AM CD CM = MD (prop. of isos. i.e. CM CD BE cm In ACM, AM tnacm CM cm cm ACM ACB (j. s on st.line (cor.to sig.fig. Person Euction Asi Limite 07

4 Solution Guie n Mrking Scheme. B. B The slope of n c re n c Epecte vlue of csh coupon got $ $ respectivel, while the -intercepts re n respectivel. Since the two stright lines hve no intersection, the must e prllel lines ut not equl lines. i.e. Their slope re equl ut -intercepts re ifferent. c n c Onl II must e true. The nswer is B.. C Let the coorintes of P e (,. The locus of P: ( ( 0 0 ( ( C The locus of P is prol. ( ( Coorintes of the centre, (, Rius ( k k Distnce etween the origin n the centre 7. A ( 0 ( 0 rius ( 0 k The origin lies outsie the circle. The nswer is C. P( green ll 0 P( lue ll 0 P( ellow ll B There re possile three notes comintions: {$0, $0, $0}, {$0, $0, $00}, {$0, $0, $00}, {$0, $0, $00}. The require proilit 9. A The inter-qurtile rnge of the istriution A The men of n is 6. 6 Men of the whole set of t I is true. For II n III: Arrnge the set of t (ecept n in scening orer: 8,,, 6, 6, 8, 9, 9, 0 The mein of the ove set of t is 6. Cse : If = = 6, then the mein of the whole set of t is lso 6. Cse : If > 6 n < 6, then the mein of the whole set of t is lso 6. Cse : If < 6 n > 6, then the mein of the whole set of t is lso 6. The mein is 6 whtever n re. II is true ut III is not true. The nswer is A. Section B. C From the H.C.F., we know tht the smllest constnt n the lowest egree of mong three epressions re n respectivel. From the L.C.M., we know tht the highest egrees of n c mong three epressions re n 6 respectivel. 6 The thir epression is c. Person Euction Asi Limite 07

5 Set 6 Pper. D From the grph, (. B ( B For I, i i i, which is rel. Since m not necessril e rtionl numer, e rtionl numer. I m not e true. For II, i ( i i i i ( i( i i The rel prt of is. i ( i i i i ( i( i i m not i i The rel prt of is. The rel prt of is equl to the rel prt of. II is true. For III, i = The imginr prt of is. ( i i = i ( i( i The imginr prt of is. III is not true. The nswer is B.. B The eqution of BC is = +. The coorintes of C re (0,. B sustituting = 0 into = + 8, we hve 0 8 The coorintes of A re (, (...( B sustituting ( into (, we hve 8 B sustituting = into (, we hve The coorintes of B re (,. At A(, 0, P ( (0 8 At B(,, P ( ( At C(0,, P (0 ( At O(0, 0, P (0 (0 0 The gretest vlue of + is. 6. A n ( n n 69...( ( B solving, we hve n 6. 9 (86 0 I is true. n II is true D ( [( 8(6] [( 8(6] 0 III is not true. The nswer is A. 9 9 ( The grph of = g( is otine reflecting the grph of = f( long the -is. 9 Person Euction Asi Limite 07

6 Solution Guie n Mrking Scheme g( f ( log log log log log log log log 8. A When = 0, =. tn[(0] 0 I is true. Compring the grph of tn, we know tht the grph of tn cn e otine the following trnsformtions: (i Trnslte ownwrs units. (ii Reuce long the -is to times the originl. (iii Enlrge long the -is to times the originl. The grph of tn hs no reflection out the -is. > 0 II is true. Sustitute into tn, we hve tn If tn, we hve The lowest point of the grph is. tn III is not true. The nswer is A. 9. C In ACD, AD AC AD CD (6 cm cm 0. C AP AD cm cm In APM, PM PM In MBF, MF MF In PEF, PF (propert of squre AP ( MB AM 9 cm 8 PE [( cm BF 8 cm 8 PF 77 cm In MPF, cosine formul EF cm ] cm MF PF PM cosmfp ( MF( PF ( ( 77 ( 9 ( ( 77 tnmfp ( Join CD n CE. C is the incentre of OAB. C is the centre of the circle. CD OA n CE OB (tngent rius 6 Person Euction Asi Limite 07

7 Set 6 Pper OC OD = CE = rius = cm cm cm OF ( cm OF is n ngle isector of OAB. 90 AOF OF AB (tngent rius In AOF, OF cos OA ( cm OA OA. C 0 ( cm (6 cm...( B sustituting ( into k 0, we hve ( 9 ( k k 0 0 k 0...(* The circle n the stright line interest t onl one point. of (* 0 ( 0 (( k k 0 k. D AOB 90 AB is imeter of the circumcircle OAB. (converse of in semi-circle The coorintes of the circumcentre of OAB 0 0,, B sustituting, into 6 : : 6, we hve. C The require numer of -igit even numers P 7. C P(Vick wins the gme D For I n II: From the grph, moe of A < moe of B A n B re in norml istriutions. Men of A = moe of A n men of B = moe of B men of A < men of B Both I n II re true. A is less isperse thn B. Vrince of A < vrince of B III is true. The nswer is D Person Euction Asi Limite 07

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