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 d