ACJC_EJC_NJC_RVHS_9649_2024_Prelim_P2_Solutions (updated)

2024 H2 FM 9649 Prelim Paper 2 (ACJC_EJC_NJC_RVHS) Section A: Pure Mathematics [50 marks] Solution P2 Q1 Let nP be the statement ( ) ( )2 7 3 5 5nn +− is divisible by 24 for all n + . When n = 1, ( ) ( ) ( )2 7 3 5 5 24 24 1+ − = = . 1P is true. Assume that kP is true for some k + , i.e. ( ) ( )2 7 3 5 5 24kk a+ − = for some integer a. We want to prove that 1kP + is true, i.e. ( ) ( ) 112 7 3 5 5 24kk b++ + − = for some integer b. ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 112 7 3 5 5 14 7 15 5 5 12 7 12 5 2 7 3 5 5 12 7 5 24 . k k k k k k k k kk a ++ + − = + − + + + −  + = =+ Since 7k and 5k are both odd, ( ) ( )75kk + is even. Thus, ( ) ( )7 5 2kk c+= for some integer c. Therefore, ( ) ( ) ( ) ( ) 12 7 5 24 12 2 24 24 where . kk a c a c a c a  + + = + = + +   1kkPP + . Since 1P is true, and 1kkPP + , by mathematical induction, nP is true for all positive integers n.

2024 H2 FM 9649 Prelim Paper 2 (ACJC_EJC_NJC_RVHS) Solution P2 Q2 Method 1 2 2 2 ddsin cos 1 sindd xxx t t x t x tt=  = = −  =− =− Then, ( ) ( ) 2 2 2 2 2 2 2 22 2 2 d d d d d d d d d d d Note: d d d d d d d d d d d d d d d d d (by product rule)d d d d d d ddd ddd dd1 d y y y x y y x t t t t x t t x x x t y x x y x x x t t x t yyx xxxt yxx x        = =  =                    =   +      = + −     = − − ( ) 2 2 2 d dd1 dd y x yyxx xx= − − Substituting into the original differential equation: ( ) 22 2 22 d d d1 4 0 4 0 (shown)d d d y y yx x y yx x t− − + =  + = Method 2 12 2 d 1 dsin sin 1 dd 1 txx t t x x xt x −=  =  =  = − − Thus, 2d d d d 1 (1)d d d d y y x y xt x t x=  = − ( ) 2 2 22 2 22 2 2 2 2 2 d d d d d d d d d d d d dd 11dd dd11 dd 1 dd1 (2) dd y y y x t t t x x t yxxxx y x yxx xx x yyxx xx    ==        = −  − = − − − − = − − Substituting (1) and (2) into the DE: ( ) 2 2 2 2 2 dd1 4 0 dd d 4 0 (shown)d yyx x y xx y yt − − + = += Method 2 ddsin cos secdd xtx t t t tx=  =  = Thus, d d d d sec (1)d d d d y y t y tx t x t=  =

2024 H2 FM 9649 Prelim Paper 2 (ACJC_EJC_NJC_RVHS) 2 2 2 2 2 22 2 d d d d d d d d d d d d dd sec secdd ddsec sec tan secdd ddsec sec tan (2)dd y y y t x x x t x x ytttt yyt t t ttt yyt t ttt    ==        =  =+  =+ Substituting (1) and (2) into the DE: ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 d d d1 sin sec sec tan sin sec 4 0d d d d d sin dcos sec sec tan 4 0d d cos d d d d tan tan 4 0d d d d 4 0 (shown)d y y yt t t t t t y t t t y y t yt t t t y t t t t y y y t t yt t t y yt − + − + =   + −  + =  + − + = += Auxiliary equation for reduced DE is 2 4 0 2imm+ =  = General solution is ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 2 2 2 2 22 e cos 2 sin 2 1 2sin 2sin cos 1 2 2 1 Since sin and 1 cos 1 2 1 , where 2 and are arbitrary constant s. t y A t B t A t B t t A x B x x x t x t A x Cx x C B A =+ = − + =