NYJC 2023 FM CT2 P2 - modified (Solutions)

2023 NYJC JC2 CT2 9649/2 Solutions Page 1 of 14 SOLUTIONS: Question 1 Qn Key Points 1 Consider auxiliary equation 2 10 1 ( 1 3 )2 m m m i + + = = −  2 /3 2 /3eei n i n ny A B  −=+ 2 /3 2 /3eei n i n ny A B  −=+ 00 0 2 /3 2 /3 1 3 e 3 / 3) ( )sin(2 / 3) 33 ( ) 022 1 3 1 3 1,22 3 1 3 1(cos(2 / 3) sin(2 / 3)) (cos(2 / 3) sin(2 / 3))22 3 cos(2 / 3) sin(2 / 3) 2 e 0 e e ( )co sin(2 / ( 3 s2 ii n ii n yB AB AB AB AB y n i n n i y A A B y A B n nn n AB        − =  + +− − + − = −= +−== = = + = = = + = + +− ++ + = = − = + 1 3tan ) 1 2sin(2 / 3 / 3)n − =+ Auxiliary equation Solve for P.S

2023 NYJC JC2 CT2 9649/2 Solutions Page 2 of 14 Question 2 Key Points 2(a) 321 2 2 1 3 3 33 2 23 2 1 2 2 4) 2 2 ( 141 4 1 4 4 7 0 4 4 7 0 1 2 2 0 0 2 2 8 0 4 7 14 0 41 4 1 4 aR a a RRRR R R R RR RR aaa a a a a aa a a a − +→ +→ → −+ +         ⎯⎯⎯⎯⎯ → ⎯⎯⎯⎯⎯ →     +        − + − + + −+ − ⎯⎯⎯⎯ − ⎯ + → +   20 0 ( 4 1 4) 4 71a a a        −  − + +  Since dim(ker(T))=1, 2 22 4 1 4a 16 4 3 7 0 )7 3 ( 4 4 0 a a aa a a+ − − − + = − + + = = Note that 1 3 4 0 1 1 0 0 0 0 0 0 x y z           =             Clearly, 1 1 1   − is the basis for the kernel of T, thus 1 1, 1   =   − r  , Perform correct EROs Eqn in a (b) Since 13 3 , 4 22              −    is the basis for R(T), then 13 r 3 4 22 st        =+      −    , Normal = 1 3 14 3 4 8 2 2 5            =−          −−      Equation of plane is: 14 8 5 0x y z− − = State basis or linear combination of basis

2023 NYJC JC2 CT2 9649/2 Solutions Page 3 of 14 Question 3 Key Points ( ) ( ) 2i i i i i1 e e e e 2cos e     −+ = + = 2i1 e 2cos 0 + =  since cos 0  for 11ππ22 −   . ( ) ( ) ( ) ( ) 2i i i 0 arg 1 e arg 2cos e arg 2cos arg e      +=  =+ =  ( ) ( ) ( ) ( ) ( ) ( ) 0 i 2 1 0 i i2 0 ! cos 2 1 isin 2 1!! ! e!! ee n r n r r n r r n rrr n r n r n r n r    = + = =  + + + − = − =     ( ) i i2e 1 e n=+ using ( ) 0 1 n nr r n xxr=  =+ ( ) ( ) ( ) ( ) ( ) ( ) ii i1 e 2cos e 2cos e 2cos cos 1 isin 1 by de Moivre's n n n n nn        + =  = =  + + +  Equating imaginary parts, ( ) ( ) ( ) ( ) 0 ! sin 2 1 2cos sin 1!! n n r n rnr n r    = + = +− . Exponential form Justify modulus Justify argument Consider sum Write as 0 n r r n xr=   Binomial expansion Use previous result Use de Moivre’s theorem

2023 NYJC JC2 CT2 9649/2 Solutions Page 4 of 14 Question 4 Key Points Define ( ) 1 2f tan 1 xxx x −=− + . ( )f 0 0= . ( ) ( ) ( ) ( ) 2 22 2 2 22 121f 1 1 2 0 for all 0 1 x x xx x x x x x +− =−+ + =   + Since ( )f 0 0= and ( )f0 x  for all 0x , 1 2tan 1 xx x −  + for all 0x with equality at 0x= . Required volume 1 1 20 211 1 200 2211121 220 00 21 20 1 20 1 2π tan d 1 2π tan d d 1 112π tan d d2 2 1 1 1 π32π d2 4 2 1 π 3 12π 1 d8 2 1 π32π tan82 xx x x x xx x x x x xxx x x x xx x xx xx xx − − − − =−  + =−  +  = − −  ++ =−  +  = − −  + = − −      ( ) 1 0 π 3 π2π1 8 2 4 π32π 22 π π 3     = − −  =−  =− State ( )f 0 0= Conclude ( )f0 x  State 0x= Integration by parts Writing in correct form to attempt integration

2023 NYJC JC2 CT2 9649/2 Solutions Page 5 of 14 Question 5 Key Points (a) EN is the rate at which the species of fish are caught when the population of the fish is N. State meaning (b) For equilibrium populations, d 0 1 0d NN r N ENtM =  − − = ( ) 0 0 or 0 0 or 0 since rNN r E M rNN r E M MN N r E r E r  − − =  = − − =  = = −   Thus ( )12 0, MN N r E r= = − . Obtain equation and attempt to solve Obtain 1N and 2N (c) dN/dt O M(r – E)/r N For ( )0 MN r Er  − , ( )d 0d NM N r Etr   − . For ( )MN r Er− , ( )d 0d NM N r Etr   − . Therefore 1NN= is unstable and 2NN= is stable. Sketch d d N t vs N Argue ( )MN r Er− Argue ( )MN r Er− (d) ( )2 EMY EN r E r= = − . Write Y in terms of E

2023 NYJC JC2 CT2 9649/2 Solutions Page 6 of 14 Y O r E Sketch Y vs E (e) From the quadratic graph of Y vs E, Y attains a maximum when 2 rE= and 2 24 m r M r MrYr r  = − =  .

2023 NYJC JC2 CT2 9649/2 Solutions Page 7 of 14 Question 6 (Source: HCI Promo 2025 Qn 7) (a) π 2 0 2 sin dV x x x =   ππ 22 0 0 2π cos cos dx x x x= − +     π 2 02π sinx= 2π= (bi) d cosd y xx = 2 2d cosd y xx  = 2 π 0 d1d d yLx x =+  ( ) π 2 0 1 cos d 3.8202L x x= + = (bii) 2 π 0 d2π 1 d d yA y x x =+  π 2 0 2π sin 1 cos dA x x x=+  Let dcos sin d uu x x x=−  = 1 2 1 2π 1 dA u u − =+  1 2 0 4π 1 dA u u=+  , where 4k = and cosux=− Alternatively: Let dcos sin d uu x x x=  =− 1 2 1 2π 1 dA u u − = − + 1 2 1 2π 1 dA u u − =+ 

2023 NYJC JC2 CT2 9649/2 Solutions Page 8 of 14 1 2 0 4π 1 dA u u=+  , where 4k = and cosux= (biii) Let 2dtan sec d uu  =  = π 224 0 4π 1 tan sec dA   =+  π 24 0 4π sec sec d  =    ππ 244 0 0 4π sec tan sec tan d    =−      ππ 344 0 0 4π sec tan sec sec d    = − −   π 34 0 8π sec d   ππ 44 0 0 4π sec tan sec d   =+    π 34 0 4π sec dA =  π 4 0 2π sec tan ln sec tan   =  + +  A π 4 0 2π sec tan ln sec tan   =  + +  A ( )2π 2 ln 2 1= + +

2023 NYJC JC2 CT2 9649/2 Solutions Page 9 of 14 Question 7 Key Points (a) Required probability = P( 3 draws ) + P(4 draws) + P(5 draws) + P(6 draws) = P( 111) + P(2111) + P(*2111)+ P(**2111) = 3333 1 1 1 1 1 1 1 33 1 1 1 1 (1 (1 (1 (1 3 3 ) (4 3 ))) ) p p p p p p p pp p p+ − + − + − = + − = − (b) Let X denote the time that a randomly chosen battery will last. ( ) 1 12 2 2 1 12 12 ( ) ( ) 1 2 battery is type 1) bat ()| () ( | ( | ( tery is type 2) battery is type 1) battery is type 2)| ( | s t t s t t P X s tP X s t X t P X t P X s t P X s t P X t p P X t p pp pp p e ee p e   − + − + −− + +  =   + +  +  +  + + = = Passing remark: If 12 == , the fraction simplifies to ( )| sP X s t X t e − +  = demonstrating memoryless property. Otherwise, the memoryless property does not hold in such a mixture problem.

2023 NYJC JC2 CT2 9649/2 Solutions Page 10 of 14 Question 8 Key Points (a) P(X < Y) = 0.5 P(X < Y) + P(Y < X) + P(Y = X) = 1 Thus P(X < Y) + P(Y < X) = 1 since for CRVs, P(Y = X) = 0. Since X and Y are i.i.d., P(X < Y) = P(Y < X) (b) Let max( , )M X Y= 2 P )(m ))ax( (,) P( PXY m a m a b m X m Y ab a m a ma b  =   −− −  − − − = = 2 2 32 2 3 2 3 3 2 3 2 3 2 2 2 2( )f ) E( ) f ( ) d 2 ( ) d() 2 ( 3 () () 32 2 ( ( ) (2 ) 2 3 2 2 2 3 ( ) 6 2 (2 ) ) ( ) 6 1 3 b a b a b a mam ba m am b ab a a ba b ab a ba b m a b a M m m m m a mba ba ba ba = −= =− −  −−   − − + − = = − −+= − +=  − −  =+   Find f(m) by differentiating F(m) Apply Expectation formula Integration Factorization of cubic expression