ACSI 2020 Y3 Core Mathematics Paper 1 Solution

Confidential – for internal circulation only 2020 Y3IP Core Mathematics Final Exam P1 Worked Solutions (a) ( )( )2 10002 10002 2−+ ( )( )2 10002 2 10002 2 10002 10000 = − + =− =− By 22ab− (b) 2ab ab − =+ 22 3 3 a b a b ba a b − = + −= =− (a) 23 42mm ++− 2( 2) 3( 4) ( 4)( 2) 2 4 3 12 ( 4)( 2) 58 ( 4)( 2) mm mm mm mm m mm − + += +− − + += +− += +− (b) yzx yz += − () ( 1) ( 1) ( 1) 1 x y z y z xy xz y z xy y z xz y x z x zxy x − = + − = + − = + − = + += − (a) 2 3 5( 1) 43 x x x xx  −+   − = ( 3) 5( 1) 43 x x x xx −+     −   = 1 5( 1) 41 x+       = 5 ( 1)4 x+

(b) (i) 2 3 5( 1)05 43 x x x xx  −+   − 50 ( 1) 54 5505 44 5 5 15 4 4 4 13 x x x x  +   +  −   −   (ii) 2 (3 5)( 5 1)−+ = 3 5 3 5 5+−− = 2 5 2− (2 5 2) 48 16 5 24 8 5 51 4(6 2 5)( 5 1) 4 6 5 10 6 2 5 4( 5 1) h h h h h − = − −= − −+= = − + − =− Alternatively, ( )( ) 16(3 5) 3 5 5 1 4( 5 1) h h −= −+ =− (a) 2 6 15xx− − + = 2( 6 ) 15xx− + + = 2[( 3) 9] 15x− + − + = 2( 3) 24x− + + (b) ( 3, 24)− (c) 2 6 15 0xx− − + = (Method 1 – Complete Square) 2( 3) 24 0x− + + = 2( 3) 24x+=

2 6 3x= − 2 6 15 0xx− − + = (Method 2 – Quad. Formula) 2( 6) ( 6) 4( 1)(15) 2( 1)x − −  − − −= − 6 96 2x = − 3 2 6x=−  (d) y x (0,15) ( 3, 24)− ( 2 6 3,0)−− (2 6 3,0)− (a) 371xx+ = 371 30 3 x x x + = += =−

(b) 21 2xxe +− = 21ln( ) ln(2 ) 2 ( 1)ln 2 2 ln 2 ln 2 ln 2 2 ln 2 (1 ln 2) 2 ln 2 2 ln 2 1 ln 2 xxe xx xx xx x x +− = + = − + = − − =− − − =− − −−= − Alternatively, log 2 2log log log 2 aa aa ex e −−= − (c) 22 3 18(3 ) 7xx −−= 2 2 13 18 7 3 x x −=  Let 2 3xy= 22 2 2 2 2 7 18 0 ( 2)( 9) 0 2, 9 3 2( ),3 9 33 2 2 xx x yy yy yy rej x x − − = + − = =− = =− = = = = (a) (i) By Pythagoras’ Theorem, 2 2 2 5 12 169 13 TB TB TB =+ = = (ii) By Pythagoras’ Theorem, 2 2 2 2 5 194 194 TF TB TF TF =+ = = (b) (i) tan FBFTB TB= 5tan 13FTB=

(ii) cos FTD cos 13 194 FTB TB TF =− =− =− (a) 22(6 ( 2)) (8 2)AB= − − + − 10AB= Since AD is parallel to vertical axis, ( 2, 8)D= − − (b) Midpoint of BD = 6 ( 2) 8 ( 8),22 (2,0) + − + −  = Gradient of BD = 8 ( 8) 6 ( 2) 2 −− −− = Gradient of 1 2⊥=− Eqn of ⊥ Bisector: 1 2y x c=− + Sub 2, 0xy== 10 (2) 2 1 1 12 c c yx =− + =  =− + (c) Since DC is parallel to x axis, sub 8y=− into 1 12yx=− + , 181 2 19 2 18 x x x − =− + − =− = (18, 8)C = −

(d) 2 6 18 21 2 8 8 22 −− − 2 1 (12 144 16) ( 16 48 36)2 1 2002 100units = + + − − − + = = Alternatively, Area = 1 (10)(20)2 = 2100units (a) 2 3y x kx k= − + − Considering the discriminant >0, 2 2 ( ) 4(1)(3 ) 0 4 12 0 ( 6)( 2) 0 6, 2 kk kk kk kk − − −  + −  + −  −  (b) 4pq+= 5pq= For the new equation, () 4 kp kq k p q k + =+ = (sum of roots) 2 2 ( )( ) 5 kp kq k