2020 HCI H2 Prelim P2 Solution

2020 HCI H2 Mathematics Preliminary Examination P2 Solution Page 1 of 15 No Suggested Solution 1(i) 1(ii) 2(a) ( )OP OP OQ OR++          = ( ) 2OP OP OP PQ PQ +++            = 23OP OP PQ +        = ( ) 2 2 2 3 5 cos 60OP +°   = ( ) ( ) 22 125 35 2 +  =87.5 Alternatively, y x y x

2020 HCI H2 Mathematics Preliminary Examination P2 Solution Page 2 of 15 No Suggested Solution 30 30 60 POQ QOR POR = ° = ° = °    ( ) 2 22 5 3 755 10 5 cos30 25 3 22 5 10cos 60 25 87.5 OP OP OP OQ OP OR OP OP OQ OR OP OP OP OQ OP OR = = × − °= =  = × °= ++ =++ =                                       2(b) 253 0 2233 0 22 33 23 OE OF OG OE OF OF OG OE OF OF OG FE GF −+= −−+= −=− =                           Since //FE FG     and F is the common point, E, F and G are collinear. :EF EG = 3:5 Since they are parallel, 0EF EG×=      3(i) 1 1 1 f ( 1) f ( ) ( 1)2 ( 2)2 2 (2( 1) 2) 2 12 (shown)2 rr r r r r rr r rr r r − − − +− =− −− = −−+ = =  1 2a= 2 F E G 3

2020 HCI H2 Mathematics Preliminary Examination P2 Solution Page 3 of 15 No Suggested Solution 3(ii) [ ] 11 1 0 12 f( 1 ) f( )2 1 2 f(2 ) f( 1 )2 f( 3 ) f(2 ) ... f( ) f( 1 ) f( 1 ) f( ) f ( 1) f (1) ( 1)2 ( 1)2 ( 1)2 1 nn r rr n r r n n r rr r nn nn n n n = = =  = +− = − +− + + −− ++− = +− = − −− = −+ ∑∑ ∑ 1 1 2 2 ( 1)2 1 ( 1)2 2 n rn r n rn n = + = −+ = −+ ∑ 3(iii) 1 3 33 2 4 2 11 1 2 1 ( 4 2 (2 1 1 2)2 2 2 22 2 ) 21 (( )2 2) 10 ( 16) ( 1)2 2 n nn r rr rr n r n rr rr n n n rr r n r n + = = = = = + − + + += + =−+ = −+ − = − + − + − − ∑ ∑∑ ∑∑ Alternatively, 3 2 2 23 ( 2)2 2 n r r rn r r r r = −= − −= + = ∑ ∑ 2 2 5 2 2 rrn r r = + = = ∑ 2 5 1 24 n r r r + = = ∑ 24 11 35 3 1 1 224 1 ( 1)2 2 (3(2 ) 2)4 1 ( 1)2 964 ( 1)2 24 n rr rr n n n rr n n n + = = + + + = − = + +− + =+− = +− ∑∑ 4(i) 1cosyx x −= − ( ) 1 22 2 d1 1 11d 1 y xx x − = + = +− −

2020 HCI H2 Mathematics Preliminary Examination P2 Solution Page 4 of 15 No Suggested Solution ( ) ( ) 2 3 2 3 2 2 2 2 2 2 d d 1 1 d d y xxx yx x x − − = − = Since ( ) 1 22d 11d y xx − −= − , 3 2 2 d 1d d d y x yx x −  − = 32 2 dd 1dd yy xxx = − (shown) ( ) 1 1 cos cos cos yx x xy x xy x − − = − −= −= Differentiating throughout w.r.t. x, ( ) ( ) d1 sin 1d d 1 sin 1d y xyx y xyx −− −=  − −= Differentiating throughout w.r.t. x, ( ) ( ) ( ) ( ) 2 2 22 2 22 2 2 2 d ddsin 1 1 cos 0d dd dd 10dd dd 10d sin 1 d 1d 1 d 1d c d dd 1d d os y yyxy xyx xx yy xx yy xxx xy yy y y x xy x x x −     −    −+ − − −=     −− =   −− =   = −       − −  2 x 32 2 dd 1dd yy xxx = − (shown)

2020 HCI H2 Mathematics Preliminary Examination P2 Solution Page 5 of 15 No Suggested Solution 4(ii) When 0x= , 2y π=− , d 2d y x = and 2 2 d 0d y x = 4(iii) 32 2 3232 32 dd 1dd d d dd 13 1d d dd yy xxx y y yy xx x xx = −  = −+ −     When 0x= , 3 3 d 1d y x = Series expansion is ( ) 23012 ...2 2! 3!y xx xπ  = −+ + + +    3 2 ...26 xyx π= −+ + + 4(iv) 32 2 d 1d 1 2 ... 2 ...d d2 6 2 1 y xxxxx x π= + = −+ + + =+ +−  5(i) Note: 2r Im Re M r 2r O

2020 HCI H2 Mathematics Preliminary Examination P2 Solution Page 6 of 15 No Suggested Solution 2i 2i 2 ba ar = − = −= ( ) ( ) ( ) ( ) arg arg 2i arg 2i arg 2 ba a π θ = − = −+ = −+ 5(ii) ( ) by previous part2AOB π= BOC θ= 5(iii) Since the length of adjacent sides of OBDC are equal, it is a rhombus. bisects . cos 2 2 cos 2 2 4 cos 2 OD BOC OMCOM r OM r b c OD OM r θ θ = = += = =   ( )arg 2 22 b c COM π πθ += −+ = −+  i 224 cos e2bc r θπ θ  −+= 6(a) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) P' P P P PP P PP * P 1P P 'P A B B AB B AB B AB BA AB ∩= − ∩ = − = −×  = −  = × O B M 2r O B D 2r 2r

2020 HCI H2 Mathematics Preliminary Examination P2 Solution Page 7 of 15 No Suggested Solution Alternative ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) P' P P PPP P P PP * P 1P P 'P A B AB A ABA BA B AB BA AB ∩= ∪− = + − ∩− = −×  = −  = × 6(b) 30.7 0.15 0 14bb− ≥⇒≥ ------(1) When 0m= , 0.3 0.7 0.8b+≤ 50.7 0.5 7bb≤ ⇒≤ Or Use 0.5 0.7 0b−≥ 3 59 3 0.314 7 140 14bb≤≤⇒ ≤ ≤ ( )93 P140 14AB≤ ∩≤ 7(a) Required probability 13 2 13! 2!13 15! 15 C×× = 6 7= or 0.857 (3 s.f.) Alternative Total numbers of ways = ( )15 1 ! 14!−= Number of ways leaders seated together 0.15 A B C 0.2

2020 HCI H2 Mathematics Preliminary Examination P2 Solution Page 8 of 15 No Suggested Solution ( )14 1 ! 2 13! 2= −×= × Probability = 13! 2 1 611 14! 7 7 ×− = −= 7(b) Let X be the event that the 3 students from the same class are seated separately. Let Y be the event that both leaders are seated separately. Required probability ( )P P( ) XY Y ∩= ( ) ( ) ( ) PP 'X XY PY −∩= 12 11 33 12! 11!3! 2! 3!12 11 15! 15 67 CC× ×−×× × = 95 156= or 0.609 ( 3 s.f.) 7(c) Required probability 13 6 15 7 C C= 4 15= or 0.267 (3 s.f) 8(i) { }( ) { } { } { } { } { }( ) ( ) ( ) ( ) 21 98 21 21 22 98 98 98 1 36 7 36 1 7 P(match (sum 5))P(match (sum 5)) P(sum 5) , 2 P 1,1 P 1,1 or 1 2 or 2,1 or 1,3 or 3,1 2 = × = = ×+ ×+ × = << = ∩ < Method 1 P(match) { } { } { }( ) ( )21 1 9 8 12P 1,1 or 3,3 or 5,5 3= ×== From above, 1 7P(match (sum 5)) =< Since 11 7 12P(match (sum 5)) P(match)<= ≠= , ∴ the 2 events are not independent. {1,2} or {2,1} {1,3} or {3,1} {1,1}

2020 HCI H2 Mathematics Preliminary Examination P2 Solution Page 9 of 15 No Suggested Solution Method 2 From above, 7 36P(sum 5)<= and 1 36P(match (sum 5)) =∩< P(match) { } { } { }( ) ( )21 1 9 8 12P 1,1 or 3,3 or 5,5 3= ×== Since 711 36 36 12P(match (sum 5)) P(match)P(sum 5)( )( )∩ =<< ≠= , ∴ the 2 events are not independent. 8(ii) Let n be least no. of rounds such that P(match with at most n rounds) 0.75≥ Method A: P(match with at most n rounds) 0.75≥ P({match round 1}or{match round 2}or…or{match round n}) 0.75≥ ( )( ) ( ) ( ) ( ) ( ) 211 11 1 11 1 11 1 12 12 12 12 12 12 12 ... 0.75 n− + + ++ ≥ ( )1 11 12 12 11 12 1 0.75 1 n− ≥− ( )11 12 0.25 n ≤ Method B: (method of complementation) P(match with at most n rounds) 0.75≥ 1− P(no match in all n rounds) 0.75≥ P(no match in all n rounds) 0.25≤ ( )11 12 0.25 n ≤ To solve ( )11 12 0.25 n ≤ , Method 1: Method 2: ( ) ( )11 12 ln 0.25 ln n≥ 15.93n∴≥ Least n = 16 Using GC, least n = 16 8(iii) Method 1: P(match on 3rd draw) { } { }( )eP orsa ime, fdiff rerent t,same d f e en ,same,sam= ( ) ( )7 21 78 721 798 933= ×× ×× + 1 6= 1 or 3 or 5 to be repeated Numbers other than the one to be repeated For 3 cases of 1 or 3 or 5 Either 1 or 3 or 5 drawn Same 1 or 3 or 5 drawn earlier Remaining numbers not drawn n ( )11 12 n 15 0.271 0.25> 16 0.249 0.25< 17 0.228 0.25< Sum of n terms of a GP with 1 12a= , 11 12r =

2020 HCI H2 Mathematics Preliminary Examination P2 Solution Page 10 of 15 No Suggested Solution Method 2: P(match on 3rd draw) { } { }( )eP orsa ime, fdiff rerent t,same d f e en ,same,sam= ( ) ( )36 6 1 41 77 67 1 7 8 9898 9= × +××× ×+ × 1 6= Method 3: P(match on 3rd draw) nd 9 3 37 1 12 1 6 P(match in 3 draws) P(match on 2 draw) C × = − = − = 9(i) Each bottle of hand sanitiser produced by Cleanser has an equal chance of being chosen. The event of one bottle of hand sanitiser being