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 =+ = − + = − + − = − =  = − + − =

2024 H2 FM 9649 Prelim Paper 2 (ACJC_EJC_NJC_RVHS) Solution P2 Q3 (a) ( ) ( )( ) 2 211d 1 2et 1 3 3      −− = = − − − = + −−IM so the eigenvalues of M are –2 and 2. Let 1 2 =− , 2 1 1 13 1 33 1 0 0 1 3 1 0 0 00 RR R −  ⎯⎯⎯ →   . Corresponding eigenvector 1 1 3 =− e Let 2 2 = , 21 1 31 1 0 1 1 0 3 3 0 0 0 0 RR R + −    −− ⎯⎯⎯→   −    . Corresponding eigenvector 2 1 1 = e So the eigenvalues of M are –2 and 2 and their corresponding eigenvectors are 1 3  − and 1 1   . (b) Hence, we have 1 1 1 2 0 1 1 3 1 0 2 3 1 − −   =   −−    M Thus 1 1 1 2 0 1 1 3 1 0 2 3 1 n n − −   =   −−    M 1 1 1 2 0 1 1 3 1 0 2 3 1 n − −    =    −−     ( )1 1 1 1 20 3 1 3 14 02 1 n n  −    −=    −    ( ) ( ) ( ) ( ) ( ) ( ) 3 2 2 2 2 4 33 1 2 2 2 3 2 nnnn nnnn  + − − −=  − − + − (c) Method 1 ( ) ( ) ( ) ( ) 2 3 1 1 12 3 3 1 1 3 1 nn nn nn −  + − − −=  − − + − M . As ( ) 1 if is odd1 1 if is even n n n −−=   , ( ) ( )3 3 1 1 1 0 nn − − = − − = when n is even So nM is a diagonal matrix for all even values of n. Method 2 As ( )22 n n−= when n is even, 1 1 1 1 1 20 3 1 3 1 02 n n n −    =    −−    M for even values of n 12n −= QIQ 2n= I which is a diagonal matrix. So nM is a diagonal matrix for all even values of n.

2024 H2 FM 9649 Prelim Paper 2 (ACJC_EJC_NJC_RVHS) Solution P2 Q4 (a) By the reflective property of the parabola, an y incoming light ray parallel to its axis will reflect off the parabola and pass through the focus of the parabola. If this is also a focus of the ellipse, then by the reflective property of the ellipse, the light ray will reflect off the ellipse and pass though the other focus of the ellipse, which is the point S. Hence, incoming light rays which are parallel to the axis will all reflect off both mirrors and pass through point S. (b) ( ) 2 37.6 4 9.4y x x== . Hence the focus F of the parabola is at ( )9.4,0 . (c) The centre K of 2C is at the midpoint of SF, which is ( )4.6,0 . The semimajor axis is 9.6 4.6 5a= − = The distance of the foci from the centre, 9.4 4.6 4.8c= − = . Since 2 2 2c a b=− , we have 2 2 2 2 5 4.8 1.4b a c= − = − = Hence, the equation of the ellipse 2C is ( ) 2 2 22 4.6 15 1.4 x y− += . (d) Consider a light ray travelling along the line 4.2y= . When it touches the parabola, 2 2 4.237.6 0.46914937.6y x x=  = = . Hence, the point where it touches the parabola is ( )0.469149,4.2 . Since it reflects off the parabola and passes through its focus at ( )9.4,0 , the reflected light ray has equation ( ) 4.2 9.4 0.469149 9.4 0.470280 9.4 y x yx =−− =− − Substitute this into the equation of the ellipse: ( ) ( ) 22 2 22 2 4.6 0.470280 9.4 15 1.4 0.15284 2.48936 9.81680 0 xx xx −− += − + = Using GC, 9.5898x= or 6.6977x= (reject since 9.4x ). Then ( )0.470280 9.4 0.0893yx=− − =− . By symmetry, the light ray travelling along the line 4.2y=− will reflect off the parabola and intersect the ellipse at 0.0893y= . Hence, the range of y-coordinates of the ellipse must include 0.0893 0.0893y−   .

2024 H2 FM 9649 Prelim Paper 2 (ACJC_EJC_NJC_RVHS) Solution P2 Q5 (a)(i) 2 πi e k n kz = , where 1, 2, ...,kn= . (a)(ii) Method 1 Note that ( )2 π 2 π 2 πi i i i2 πe e e e nk kk n n n n k kzz − −−  − = = = = , so 2 1k n k kz z z− == . If n is odd, 1 2 n +−  . Then for all 1,1 2 nmm + −   , ( )( ) ( )( ) 1122 21 1 1 1 22 1 22 2 2 m n m m n m m n m m n m m n m m n m m n m m n m zz z z z z zz z z z z zz zz − −− − −− − −  + + ++= + + + +   ++= + + +  ++=  ++ = So ( ) 11 2 2 2 1 2 2... 2 11 1 1 2 1 1 1 n n n n nnz z z z − −+ + + = + = − + =+ + + + +  Method 2 2 πi111 2 3 1 2 2 2 2 2 2 2 ...1 1 1 1 1 1 1e nn k kkn n k nz z z z z z ==− + + + + + = =+ + + + + + +  ( ) ( ) πi ππii1 1 1 even number of terms since is od ππ2 cos isin2e π1 i tanπ2cosee 2 π 1 ππ 2πi tan tan ... tan tan k n n n n kk k k k nn n kk nn k k n n nnn n n n n − −= = =     −        = = = −  +   −   −    = − + + + +                  ( ) ( ) d πtan 1 π 2 ππ 2πi tan tan tan tan ... tan π π π 2π 2πi tan tan tan tan ... 0 i 0 0 ... n n nnn n n n n n n n n n n  +     −   −    = − + + + + +                       = − − + − + +                = − + + ( )0 .n + =

2024 H2 FM 9649 Prelim Paper 2 (ACJC_EJC_NJC_RVHS) (b)(i)&(ii) (b)(iii) 6 23456 0 7 1 1 (using the sum of a geometric series)1 r r z z z z z z z z z = = + + + + + + −= −  By re-writing the numerator in terms of its factors, we have ( )( )( )( )( )( )( ) ( )( )( )( )( )( ) 6 1 2 3 4 5 6 0 1 2 3 4 5 6 1 1 (shown) r r z z z z z