NYJC 2021 FM CT2 P1 - modified (Solutions)

2021 NYJC JC2 CT2 9649/1 Marking Guide Page 1 of 15 Qn 1 AE: ( )( ) 2 2 30 4 4 41 3 2 2 2i kk k + += −± − = = −± CF: ( ) 2e cos 2 sin 2xy A xB x−= + PI: Let 2 2 sin cos d cos sind d sin cosd y a xb x y a xb xx y a xb xx = + = − = −− Sub into DE, ( ) ( )sin cos 2 cos sin 3 sin cos cosa xb x a xb x a xb x x−− + − + + = 220 221 ab ab −= += Solving, 1 4ab= = ∴ General solution is ( ) ( ) 2 1e cos 2 sin 2 sin cos 4 xy A xB x x x−= + ++

2021 NYJC JC2 CT2 9649/1 Marking Guide Page 2 of 15 2 2023/TJC/JC2 MYE/I/Q1 (i) cos sin cos cos cos sin sin sin cos sin sin cos cos sin rr r rr r θ θ α θα θα θ θ α θα θα −−     =     +      ( ) ( ) cos sin r r θα θα += + The transformation T rotates every vector in 2 anti-clockwise about the origin through the angle θ. (ii) det ( ) λ−=AI 0 cos sin 0sin cos θλ θ θ θλ −− =− ( ) 2 2cos sin 0θλ θ−+ = 2 22cos 2 cos sin 0θ λ θλ θ− ++ = 2 2 cos 1 0λ λθ− += For real eigenvalues, discriminant 0≥ 24cos 4 0θ −≥ ( )( )cos 1 cos 1 0θθ− +≥ cos 1 or cos 1θθ≤− ≥ Since 1 cos 1θ−≤ ≤ , cos 1 or cos 1θθ= −= 0 or θπ= When 0θ = , 10 01 = A which is the identity matrix. Eigenvalue = 1 When θπ= , 10 01 −= − A . Eigenvalue = −1

2021 NYJC JC2 CT2 9649/1 Marking Guide Page 3 of 15 3(i) Note that 12 12 ) 2(( )Te e ee+= + . Thus 12v ee= + . 12e e+ is an eigenvector of the matrix representation of T with corresponding eigenvalue 2. (Geometrically, the line ( )12 ,eeαα= +∈r  , is invariant under the transformation, and each point is transformed onto a different point on the line) 3(ii) 3 3 12 1 23) ( )22( e Te e eTe ee=− += −− + Thus 02 2 11 2 00 1 − = − A 3(iii) Since 2 is an eigenvalue of A, we have 1 )1( 0 ker  = A . Since 22 2 2 1 12 00 1 −− −= −− − AI , a basis for 1R is 22 ) 1,2ker 01 (      = −       A . Note that we are solving ( )2−=A Iv 0 . Since dim of nullspace is 1, dim of range space(i.e. column space) is 2. We can just take the 2 linearly independent columns (since 1 st 2 columns are multiples of each other) 4(i) Let ker( )∈xA . Thus =Ax 0 . Accordingly, ( ) ( )= =B A xB A x0 . Thus ker( )∈x BA Thus ker( ) ker( )⊆A BA 4(ii) Since BA is invertible, thus 0 ||| | |0| ≠⇒ ≠ BABA , both | 0|≠B and | 0≠A| , thus A and B are both invertible. 4(iii) ( ) ( ) ( ) ( )( ) ( ) ( ) 11 1 1 1 1 )( −− − − − − + = + +− = −+ = −− + + + = ++ − = + AA B B AA B B A A A AA B A A BABAB A AB AB AA BA B A

2021 NYJC JC2 CT2 9649/1 Marking Guide Page 4 of 15 5 2023/TJC/JC2 MYE/I/Q7 (i) 1 1 1 68 0.5 = 0.5 = c r rr rr rr b ca ab b + + + += ⇒ 1 1 1 0 68 0.5 0 0 0 0.5 0 rr rr rr aa bb cc + + +           =             0 68 0.5 0 0 0 0.5 0  = M Given 120 120 120  = 0x 0 6 8 120 1680 0.5 0 0 120 60 0 0.5 0 120 60          = =             1x 2 0 6 8 1680 840 0.5 0 0 60 840 0 0.5 0 60 30          = =             x The number of rabbits in first age class, second age class and third age class after 2 years are 840, 30 and 30 respectively. 5 (ii) To find λ=Mx x ( )λ−=M Ix 0 The characteristic equation is det (M – λI ) = 0 det 68 0.5 0 0 0.5 λ λ λ −  − − = 0 ⇒ ( ) ( ) 23 0.5 6 4 3 2 0λλ λ λ λ− − − − = − + += 1λ =− (repeated) or 2 λ = Taking 2λ = , (rejected the negative eigenvalue as population cannot be negative) Augmented matrix:

2021 NYJC JC2 CT2 9649/1 Marking Guide Page 5 of 15 2 6 8 0 1 0 16 0 0.5 2 0 0 0 1 4 0 0 0.5 2 0 0 0 0 0 −−     −  → −     −   RREF using GC eigenvector, x = 16 4 t t t    Smallest initial population = 16 4 1    when t = 1 6(i) If n is odd, then 21 2 1nn x x++ = + and 1 2nnxx+ = , Combining, we have 2 11 1 1 21n nn n nx xx x x+ ++ += + += + + If n is even, then 21 2nnx x++ = and 1 2 1nnx x+ = + Combining, we have 2 111 21n nn n n xx x xx+ +++ +=++= 6(ii) From GC, 11 12 1365 20 2730 2021 21 x x = > < = Thus max number of rings = 11. 6(iii) Since 1 21 321 , 2 2 , 2 1 5x xx xx= = = = += (2) )(1 nn n BCxA += − + 11 1 (2) 2 11, 1 ( 1)x B ABA CC+ + −+= ⇒+− == 22 2 (2) 4 22, 2 ( 1)x B ABA CC++ += ⇒+− == 33 2 (2) 8 55, 5 ( 1)x B ABA CC+ + −+= ⇒+− == Using GC, 21 1(2 ) ( 1)36 2 nn nx −−= − Alternatively, 2 20 2 or 1 mm m m −−= =−= Complementary solution: ( 1) (2)nn nx B A+= − Particular solution: nxC=

2021 NYJC JC2 CT2 9649/1 Marking Guide Page 6 of 15 1 2 21CC C C = ++ =− Since 1 21 1, 2 2x xx= = = Also, ( 1) 1(2) 2 nn nx B A+= −− 11 1 13(1, 2)1 (1 222)x ABA B= = −− + ⇒− + = 22 2 1(2) 52, 2 1) 2( 42xA B AB= = −− + ⇒+ = 12,63AB= −= 21 1(2 ) ( 1)36 2 nn nx −−= − 7 Source: NJC Promo 2025 Qn 3 ( ) 21 πe cos 3 y zx −− = − , ( ){ },| π π, xy x y−≤≤ ∈  . ( ) 21 πe sin 3 yz xx −−∂ = −− ∂  , ( ) ( ) 21 π2 1 e cos 3 yz yxy −−∂ = −− − ∂  . At stationary points, 0z x ∂ =∂ and 0z y ∂ =∂ . ( ) 21 πe sin 0 3 y x −− − −=  ( ) 21πsin 0 ( e 0 )3 y xy −− − = − < ∀∈  2ππ or 33x=− and

2021 NYJC JC2 CT2 9649/1 Marking Guide Page 7 of 15 ( ) ( ) ( ) ( ) 2 2 1 1 π2 1 e cos 0 3 π1 cos 0 ( e 0 )3 π1 or cos 0. 3 y y yx yx y yx −− −− −− −=  − − = − < ∀∈  = −=    When 2π 2π π, cos 1 03 33x = − − − = −≠ . When π ππ, cos 1 03 33x = −= ≠ . Therefore, the stationary points in D are 2π ,1, 13 −− and π ,1,13   . ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) 2 2 22 2 2 1 2 2 1 2 2 11 2 2 1 2 πe cos ,3 π21 e s i n 21 , 3 π4 1 e 2e cos 3 π4 1 2 e cos 3 4 1 2. y y yy y z x zx z y xyyx z yxy z x z yx y −− −− −− −− −− ∂ = − −= − ∂  ∂  − ∂=− = −−∂∂  ∂ = −− − ∂  = −− − − ∂  = − At 2π ,1, 13 −− , ( )( ) ( ) 2 11 21 0 20D= − ⋅− − ⋅− − = > . Also, 2 2 10z x ∂ = >∂ . ∴ 2π ,1, 13 −− is a local minimum point. At π ,1,13   , ( )( ) ( ) 2 11 21 0 2 0D=−⋅ −⋅ − = > . Also, 2 2 10z x ∂ = −<∂ . ∴ π ,1,13   is a local maximum point.

2021 NYJC JC2 CT2 9649/1 Marking Guide Page 8 of 15 8(i)  ( ) ( ) 22442 22 2 44 1 11 11d1 d d dd4 4 xxyAB xxxxx x ++= += = ∫∫∫ 2 4 2 1 2 3 1 11 d2 11 1 23 17 12 x xx x x += = − = ∫ 8(ii) Surface area = 22 2 44 2 11 d 312π 1 d 2π dd 62 y xxyx xx xx   +++=      ∫∫ 2 5 3 1 26 2 2 1 π3 4 d6 π3 266 2 47 π16 xx x x x x x = ++  = +− = ∫ ( ) T by 4 44 4 6 30 6 4 30 6 2 4 30 y xx y xx y xx y x− += → − − +=⇒ − + += . So Γ2 is a translation of Γ1 by 4 units in the positive direction of the y-axis. y B’ y = f(x) + 4 A’ B y = f(x) A O 1 2 x

2021 NYJC JC2 CT2 9649/1 Marking Guide Page 9 of 15 8(iii) Perimeter = 2( AB + AA’) = 2( 17 12 + 4) = 65 6 . 8(iv) ( ) 4 3 f6 xyx x += = . By the ‘shell’ method, volume ( ) ( ) 22 11 22 1 2π f 4 f d 8π d 4π 12π x x x x xx x +− = =  = ∫∫ 9(i) 21 1 0 11 21 11 21 2 0 12 22 2 1 212 21 01 1 () () () n n n n n n n n n nn p x a ax ax a x p x a ax ax