ACSI 2022 Y3EXP FYE AMath P1 Solutions

2022 Yr 3 EXP End Year Exam P1 Solutions 1 Making x the subject from Eqn (1), 75 2 yx += --- Eqn (3) Sub Eqn (3) into Eqn (2), 22 22 2 2 7512( ) 5 72 3(7 5 ) 5 7 70 210 140 0 3 20 ( 1)( 2) 0 1 or 2 y y yy yy yy yy y + −= + −= + += + += + += = −− Sub the values of y into Eqn 3, When 1y=− , 7 5( 1) 2 1 x +−= = When 2,y=− 1.5x=− 2(i) 71 10 5 2 5 ABm −= −− =− 21 ( 5)5 2 35 YX YX −= − − = −+ 2 2 12 35 5 2 15 xy y x = −+ ∴= −+ 2(ii) Sub 0X = , 3Y = 3k∴= 3(a) 2, 5ab= = 3(b)

12cos 2 2cos 5 4.5 x x =− += Insert 4.5y= , the number of solutions = 2 4 22 2 10 5() () 31 31 75 ( 3 1) 75 3 1 31 31 15 5 3 31 15 5 322 h= − −− = − += × −+ += − = + 5(a) 2 40b ac−< 2 2 [2( 3)] 4(1)(25) 0 4( 3) 100 0 [2( 3) 10][2( 3) 10] 0 ( 8)( 2) 0 28 m m mm mm m −− < −−< −− −+ < − +< ∴− < < 5(b) For 2 5 056xx >−+− ,

2 2 2 5 60 ( 5 6) 0 5 60 ( 2)( 3) 0 23 xx xx xx xx x −+ −> −−+> − +< − −< ∴<< 6(a) 3 0 0 2 nPe P = 3 2 3 ln2 ln2 3 0.231 (shown) ne n n = = = = 6(b) 7(0.23105) 00 0 7(0.23105) % increase 100% ( 1) 100% 404% (3sf) Pe P P e −= × = −× = 7(i) 1sin 5 A= 7(ii) cos( ) cos 2 5 AA−= =

7(iii) 1sec( )2 cos( )2 1 sin 1 1 5 5 A A A π π−= − = = = 8(a) Comparing coefficient of 3x , 4A= Sub 2x=− , 5C = Sub 0x= , 1 (2) 5 3 B B −= + =− 8(b) 22 35 ( 1) ( 3) ( 1) ( 1) ( 3) x ABC xx x x x − = +++− + + − 23 5 ( 1)( 3) ( 3) ( 1)x Ax x Bx Cx− = + −+ −+ − Sub 3x= , 4 16 1 4 C C = = Sub 1,x=− 84 2 B B −= − = Sub 0,x= 15 ( 3) 2( 3) 4 1 4 A A − = −+−+ =− 22 35 1 2 1 ( 1) ( 3) 4( 1) ( 1) 4( 3) x xx x x x −∴ = −+++− + + − 9(i) 16 16 16 1log ( ) log 1 log 0 qq p p = − = − =− 9(ii) 16 4 16 loglog log 4 1 2 2 qq p p = = =

9(iii) 2 16 16 4 p p p q q = = = 10(ai) 1 2 2 AB AD m m = =− Eqn of AD: 3 2( 4) 25 yx yx −= − + = −− ( 2.5,0)D∴= − 10(aii) Midpoint of AC = midpoint of BD 4 3 2.5 6 0 8( , )( , )22 2 2 xy−+ + − + += (7.5,5)C∴= 10(b) Method 1 2 83 6 8.5 2 BC BE m m =− −= − =− Since BC BEmm = and there is a common point B, B , C and E are collinear Method 2 Eqn of BC: 8 2( 6) 2 20 yx yx −= − − = −+ Sub 8.5x= , 2(8.5) 20 3 y= −+ = Since the equation is fulfilled, hence E lies on BC. Therefore, they are collinear.