NYJC 2022 FM CT2 P1 - modified (Solutions)

2022 NYJC JC2 Common Test 2 Exam 9649/1 Solution Page 1 of 18 SOLUTIONS: Qn Suggested Solution 1 2d4 d yy xyx x− = ( ) 2 2 dd 21 yx yx y = + ( ) 2 2ln 1 4 xyc+ = + 2 2 41e x yA+= , ecA= When 0, 1xy== , 2A= 2 42e 1 x y=− Using Euler Method, 00 0, 1, 0.1x y h= = = , 0 0 (0.1)nx x nh n= + = + ( ) ( ) 21d f,d4 xyy xyxy + == ( )1 1 1hf ,n n n nyx yy − − −= +  2 1 1 1 ( 1)(0.1)(1 ,14 )0.1 n nn n ny ny yy − − − =+ −+  , 0 1y =

2022 NYJC JC2 Common Test 2 Exam 9649/1 Solution Page 2 of 18 Qn Suggested Solution n ny actualy Difference 22 2.2954 2.3889 -0.0935 23 2.4456 2.5506 -0.1050 From GC, max no of iterations = 22

2022 NYJC JC2 Common Test 2 Exam 9649/1 Solution Page 3 of 18 Qn Suggested Solution 2 (i) V0 as and trace( ) 0t==0 0 0 Let 12, VAA and k  ( )1 2 1 2 1 2 t ttk k k+ = + = +A A A A A A and 1 2 1 2trace( ) trace( ) trace( ) 0 0 0kk + = + = + =A A A A Hence 12k V+AA . Since V is closed under vector addition and scalar multiplication, V is a subspace. (ii) Note that matrices in V are of the form 1 0 0 1 0 1 1 0 xy xyyx      =+     −−      . Hence a basis for V is 1 0 0 1 ,0 1 1 0        −    . Note that matrices in W are of the form 0 0 a b   . Hence a basis for W is 0 1 0 0,0 0 0 1            . Since dim( ( )22M  ) = 4 and the 4 basis matrices are independent of each other, ( )22VW += M  . Qn Suggested Solution 3(i) 2cos sinr  = As 0,→ , r→ , thus sin 2cos 2yr =→= , if 0 → 2c 2sin osyr = = →− , if →

2022 NYJC JC2 Common Test 2 Exam 9649/1 Solution Page 4 of 18 Qn Suggested Solution (ii) 0x= (iii) When 3y= , 2cs oin s3r == , 6  = x-coordinates of the point of intersection 2cot cos 366cosr     ==        Area bounded by the curve from 6  = to 2  = 22 6 1 4cot d2   =    2 6 2 /2 /6 2 cosec d 2 cot 1       = −=− −   /2 /62 cot 2 3 3    =− =− − +  +  223 3=− Required Area ( )123 3 2 323 = − −  32 3 2= −

2022 NYJC JC2 Common Test 2 Exam 9649/1 Solution Page 5 of 18 Qn Suggested Solution (2025 VJC Promo Q9) 4 (a) 1,andn n n nF a bP cF P −= − = Sub 11nnF a bP−−=− into 1nnP cF −= ( )1 1 nn nn P P c a bP ac bcP − − =− =− General Solution: ( ) n np A bc B= +− Particular solution: Consider 1nnp p p −= = 1 ac bcP acP bc P=− = + General solution: ( ) 1 n n acA bc bcp = − + + 0:n= 0 1A ac pbc+=+ Solving, 0 1 acAp bc−= + ( ) 0 11 n n ac acp bc p bc bc  +=− +−  +   (b) For np to converge, 1,bc− so that ( ) 0, n bc− → i.e. 01 bc as b and c are positive constants. 1 e ac bcP = + (c) Since nP contains the term ( ) n bc− and 1bc , nP oscillates in sign and decays in magnitude. Thus, nP converges to 1 ac bc+ in an oscillatory manner.

2022 NYJC JC2 Common Test 2 Exam 9649/1 Solution Page 6 of 18 Qn Suggested Solution 5 (i)(a) 1 2 3 4 5 4.15 4.14740 4.14658 4.14631 4.14623 x x x x x = = = = = 4.146 = (i)(b) If the initial value is less than  , the iteration will give 1 1nx +  , then value becomes undefined as 1ln( 1)nx + − is undefined. If the initial value is more than  , the iteration will converge to  . (ii) ( )f ( ) ln 1 3x x x= − − + 1f'( ) 1 1x x=− − 1 f( ) )f '( n nn n xxx x + =− ln( 1) 3 1 11 nn n n xxx x − − +=− −− ( ) 1 ln( 1) 3 11 n n n n n x x xx x − − − +=− −+ ( )  ( ) ( )( ) ( ) 2 1 ln( 1) 3 2 2 1 ln( 1) 1 3 2 3 2 1 ln( 1) 2 n n n n n n n n n n n n n n n n x x xx x x x x x x x x x x x x − − − +=− − − − − − + − −= − − − − −= − ln( 1) 3yx= − + yx=

2022 NYJC JC2 Common Test 2 Exam 9649/1 Solution Page 7 of 18 (ii)(a) f ( )yx= has a turning point at 2x= . If we take 1 2.2x = , drawing a tangent at 1 2.2x = , it will cuts the x-axis at value bigger than  , and subsequent iterations will converge to  . Or If we start at x1 = 2.2, then the iteration will converge to  . (ii)(b) 1 2 3 4 5 1.2 1.152359 1.158448 1.158594 1.158594 x x x x x = = = = = 1.159 = ln( 1) 3y x x= − − + 3 2 ( 1)ln( 1) 2 x x xy x − − − −= − yx= x = 2

2022 NYJC JC2 Common Test 2 Exam 9649/1 Solution Page 8 of 18 Qn Suggested Solution 6(a) ( ) ( ) ( ) ( ) ( ) 2 sin 2sin d dsin sin sin sin 2 d sin 2cos 1 sin d sin 2 cos 1 d 2sin 1 (*) 1 nn nxnxI I x x xx nx n x xx n x x xx n x x nx Cn − −− = − −−= −= =− −=+ −     Integrating factor = ( ) cos dcot d ln sinsin 1e e e sin x xxx xx x −− − = = = . Multiplying the DE by IF gives d sin 5 d sin sin yx x x x  = . Integrating w.r.t x: 5 sin 5 dsin sin yx xIxx== By (*), 5 3 1 1 3 1 2 2 2sin 4 sin 4 42 2sin 2 sin 22 xxI I C C xI I C x C − = + = + − = + = + Adding gives 5 1 1 2 12 sin 4 sin 22 sin 4 sin 2 where 2 xI I x C C x x C C C C − = + + + = + + = + 51 sin 4 sin 22 1 sin sin 4 sin 2 d2 sin 1 sin 4 sin 22 xI x I C xx x x C x x x x C = + + + = + + + = + + +  Thus 11sin 4 sin 2 sin 4 sin 2 sinsin 2 2 y x x x C y x x x C xx = + + +  = + + + 

2022 NYJC JC2 Common Test 2 Exam 9649/1 Solution Page 9 of 18 Qn Suggested Solution (b) diff w.r.t 22diff w.r.t 22 dd dd d d d 2 d d d x x u y uy xy u x yx x x y y ux x x x =  =  + =  + = Dividing the last equation by x and transferring terms, 22 22 2 2 2 22 d 1 d 2 d d d d 1 d 2 1 d dd 1 d 2 1 d dd y u y x x x x x uu yx x x x x u u u x x x x x x =−  = − −   = − −  22 2 2 2 3 d 1 d 2 d 2 d d d y u u u x x x x x x=−+ which is the required equation. Substitute all the above into given DE gives 2 2 2 2 1 d 2 d 2 d 25 0d d d d 25 0d u y y u x x x x x x x u ux − + + =  += AE: 2 25 0 5ikk+ =  = Therefore GS is ( ) cos5 sin 5 1 cos5 sin 5 u A x B x y A x B xx =+  = +

2022 NYJC JC2 Common Test 2 Exam 9649/1 Solution Page 10 of 18 Qn Suggested Solution 7 ( ) ( ) 12 1 2 1 1 1 2 2 2 1 2 (i) 0 2 1 2 22 2 1 1 22 2 2 2 2 1 2 4 2 1 4 2 20 nn n n n nn n n n n nn uu u u uuu u u u u uu −− − − − − − − − =+ + = + − + − = + += 2 1 cos sin222 Auxilliary equation: 2 1 0 im m      =    +=      0 1 1General solution: cos sin 222 When 0, 1 1When 1, 2 2 2 1 cos 2sin .222 n n n n nnu A B n u A A n u B B nnu       =+           = =  = = = =  =      = +          1 1(ii) cos 2sin 222 1 5 cos where tan 222 n n n n nnu nu    −     =+           = − =   ( ) ( ) ( ) ( ) ( ) 10 1 0 1, 0 since 1 cos 1 and 0 20 2 20 2 When Since 20 2 20 the long run, amount of bacteria cannot exceed 2 2 20 2 as 5 cos , thus2 5 20 n n n n nnn n n r r n n r n n n nu x u u x u u u u u u n n n u     − =  → −  −  →     → − → = + − = + − →   − = −  00 20 5 1 billion since 1uu= − = A1