2021 A Level H2 FM - Solutions

2021 GCE ‘A’ Level Further Math Paper 1 1 (a) Given ( ) 1ef, xyxy xy++= with 11 1, 1xy= = . Step size h = 0.5. Euler’s Method: ( ) 2 2 1 11 1ef , 1 0.5 5.194528 5.1951y y h xy  += += + ≈≈  . Improved Euler’s Method: Predictor: 2 5.194528u ≈ from Euler’s method above Corrector: ( ) ( ) 2 1.5 5.194528 2 1 11 22 0.5 1 e 1 ef , f , 1 29.0532 2 1 1.5 5.194528hy y xy xu +++= + + ≈+ + ≈  × (b) The large discrepancy between these two answers could be due to the high positive concavity of the solution curve in the neighbourhood of 1.5x= . 2 (a) (b) 2 12 022 (1 )( )40 60 2 or -3 2 λ λ λλ λλ λ − − =−− −− = − = + − = When 2, 2 0xyλ = −+ = , associated eigenvector = 21 When 3,4 2 0xyλ = − += , associated eigenvector = 12− Lines that are parallel to the eigenvectors will remain invariant under transformation given by matrix M.

3 Pn : 112 21nF n < + , n +∈  . 1 1 1 11 12 2 1! 2 3 21 1F  = = <= ⋅+ so 1P holds. Assume Pk holds. That is, 112 21kF k < + , k +∈  or ( ) ( ) ( ) 11 1 112 1 122 2 2 ! 21 11 1 1 1 112 1 2122 2 2 2 2 1! 2 1121 k k k kk k k kk kk     + + +−         < +      + + +− + +      +     ⇒ <=++ ++   ( ) ( ) ( )( ) ( ) 2 1 2 12 31 21 2123 2 1 23 k kkk kk kk +− + ++−= ++ ++ ( ) ( )( ) ( ) ( ) 2 12 1 2 12 3 2 1 1 0 4k kk k k +− + + = +− + − >  since ( ) ( ) 22 11 11 4k kk+ −< + =+ . Hence ( ) 21 1 21 23 k k k + <+ + and we conclude ( ) 11 1 1 112 1 122 2 2 2 1! 23 kk k k      + + +− +           <+ + . Thus 1Pk+ must hold. Hence Pn holds by induction.

4 (a) 22 112 200 1 1 11dd d 1b b abu axxuV yx x uax a u a+= + −    = π= π = π     +    ∫∫ ∫ 1 32 1 211 dab ua uu +π = −+ ∫ ( ) 1 3 1 3 12ln 11 2ln 1 1 ab uuau ab aba ab + π =−−  π = +− + − + (b) ( ) ( ) ( ) 2 2 2 2 2 22 d1 1d 1 1d d1 1 xyy ax x ax yy x xxx ax ax = ⇒=+ + = = = + + 2 22 2 d dddyy xy yxxx= ⇒= . 1 22 00 dd b bab W xy yxV + = π = π=∫∫ . 5 (a) y 1e2xyx x −= +− O 1 x -2

From the sketch, we see that the graph of 1e2xyx x −= +− is strictly increasing on  and so it cuts the x-axis at only 1 point. [ ( )11 de 2 1 e 1 0 dxx yyx x x xx−−= +−⇒ = + +>∀ ∈  ] Since the graph of ( )fyx= is a vertical translation of the graph of 1e2xyx x −= +− by a very small amount δ , ( )fyx= also cuts the x-axis at only 1 point. Since δ is very close to zero, ( ) 1fe 2 xxx x −≈ +− so the root of ( )f0 x = , α, is approximately equal to the root of 1e 20xxx − +−= which is 1, whence 1≈α . (b) To approximate α, the root of ( )f0 x = , we employ the N-R iteration formula: ( )( )1 ff nnn nxxx x+ = − ′ where ( ) ( ) ( ) 11f e 2 f 1e 1xxxx x x x−− ′= +−−⇒ = + +δ . ( ) ( ) ( ) ( ) 0 10 0 f f1 1 11f f1 3 3xxx x −= − = − = −= +′′ δδ . ( ) ( ) 3 1 21 1 3 f1 1 e 1 2f 3 3311f3 3 f1 2 e13 3xxx x δ δ δ δδ δδδ δ δ  + + ++ −−  =− = +− = +−′  ′ + ++   Using 2 e1 2!x xx=++ +  gives

2 2 2 2 2 2 2 1211 13 3 2! 3 3 1 3 121 13 3 2! 3 1211 3 2! 3 3 9 3 1 3 221 3 3 39 x δδδ δ δ δδδ δ δ δδ δ δ δ δ δδ   + + + −−      ≈+ −   + ++ +       ++ ++ − −= +− + + ++ + ( ) 2 2 122 22 1 2 61 23 3 9 12 11 3 6 3 3 27 2 1 1 us ing 1 1 ...3 18 3 27 1 3 18 xx δ δ δδ δδ δ δ δδ δ δ δδ − − = +− ++  = +− + +     ≈+ − − + + =−+  ≈+ − where terms in 3δ and higher powers of δ have been ignored. 6 (a) ( ) 222 3 2 cos , sin 4 33 3 2 5 5 cos cos sin xr yr x y x xy rr r θθ θ θθ = = += + ⇒= + ( ) ( ) 32 2 5cos cos 1 cos cos 4cos 1 r θθ θ θθ ⇒= + − = +

(b) ( ) ( ) ( ) 1 11ππ22 1 00 d 1cos ; cos 1 πd π 21 222 0 0 1 π2 22 0 cos d cos cos d si n cos 1 sin cos d 1 1 cos cos d n nn vu nn n n n θθ θ θθ θ θθ θ θ θ θθ θ θθ − − = = −− − = = +− = −− ∫∫ ∫ ∫ ( ) ( ) 11ππ22 2 00 11ππ22 2 00 11ππ22 2 00 2 1 cos d cos d cos d 1 cos d 1cos d co s d 1 nn nn nn nn n nn n n nII n θθ θθ θθ θθ θθ θθ − − − −  = −−  ⇒= − −⇒=  −⇒=  ∫∫ ∫∫ ∫∫ where 1 π2 0 cos dnnI =∫ θθ and 1 π2 0 0 11d π2I = =∫ θ . (c) Note: E is symmetrical about 0=θ and ( )22cos cos 4cos 1 0xr= = +≥θθθ . Total area enclosed by E

( ) 1 π 22 22 0 1 11π ππ2 22 6 42 0 00 6 42 0 00 0 12 cos 4cos 1 d2 16 cos d 8 cos d cos d 16 8 531 31 116 8642 42 2 17 2 17 π4 I II I II I = ×+ = ++ = ++   =⋅ ⋅+ ⋅+     = = ∫ ∫∫∫ θ θθ θθ θθ θθ 7 (a)(i) For (*) to describe a circle, 1BA= = . (a)(ii) Put 1, 0AB C= = = in (*) gives 22 0x y Dx Ey F++ + += . (*)' Rewriting (*)' gives 22 22 22 22 02 2 44 (*)''2 2 44 D E DExy F D E DExy F   + ++ − − +=       + ++ = +−     If (*)'' is the equation of a circle, then 22 2 044DErF + −> 22 4DE F⇒+> • If 22 4DE F+= , then 0r = and so (*)'' has solution ,22DExy= −= − i.e. a single point ,22DE−− . If 22 4DE F+< , then 2 0r < which is impossible and so no point will satisfy (*)'', i.e. the curve does not exist.

(b) We are given a transformation T where ': 'xxT yy  →     such that '1'1xk xy ky    =    −     . We wish to find the possible values of k in terms of the coefficients of (*) such that T transforms 22 0Ax By Dx Ey F+ + + += (standard conic with xy term zero) to ( ) ( ) 22 0A x B y x yD xE yF′ ′ ′′ ′ ′+ + + + += (non-standard conic with 1C = ) (*)''' where '1 ' and ''1x k x kx y x kx y y x kyy k y x ky +   = = ⇒ = + = −+   − −+    Substitute the above expressions for 'x and 'y into (*) ''' gives ( ) ( ) ( )( ) ( ) ( ) 22 0 A kx y B ky x kx y ky x D kx y E ky x F ++ −++ − + ++ −+= The coefficient of xy is 222 1Ak Bk k− +− . Since coefficient of xy is zero, 22 2 10Ak Bk k− + −= ( ) ( ) ( ) ( ) 2 2 2 2 2 2 10 2 10 244 2 1 Ak Bk k k A Bk BA BAk k BA BA − + −= + − −= −± − += =−± − +

Alternative Interpretation: We are given a transformation T where ': 'xxT yy  →     such that '1'1xk xy ky    =    −     . We wish to find the possible values of k in terms of the coefficients of (*) such that T transforms 22 0Ax By xy Dx Ey F+ ++ + += (Put C = 1 in (*)) to ( ) ( ) 22 0A x B y Dx Ey F′ ′ ′ ′ ′′ ′′ ′+ + + += (*)''' where '1 ' and ''1x k x kx y x kx y y x kyy k y x ky +   = = ⇒ = + = −+   − −+    22 ,11kx y x kyxy kk′′ ′ ′−+⇒= = ++ Substituting x and y above into 22 0Ax By xy Dx Ey F+ ++ + += gives 22 2 2 22 22 1 1 11 011 kx y x ky kx y x kyAB k k kk kx y x kyDEFkk ′′ ′ ′ ′′ ′ ′− + −+   ++   + + ++    ′′ ′ ′−+  + + +=  ++   The coefficient of xy′′ is ( ) 2 22 22 1 1 Ak Bk k k − + +− + . Since coefficient of xy′′ is zero,

( ) ( ) ( ) ( ) 2 2 2 2 2 2 10 2 10 2 44 2 1 Ak Bk k k B Ak AB BAk k AB BA − + + −= + − −= −± − += =−± − + 8 (a)(i) ( )1 11 14 54 0n n nn n n nH H HH H H H+ −+ −−= − ⇒ − + = for n +∈  . which gives a second order homogeneous recurrence relation. AE: ( )( )2 5 4 0 1 4 0 1, 4mm m m m− +=⇒ − − =⇒ = . GS: ( ) ( ) ( )14 4nn nnH A B AB=+= + . ( ) 0 0 3000 3000 4H AB AB=⇒= + = + . ( ) 1 1 4500 4500 4 4H AB A B=⇒= + = + . Solving, A = 2500, B = 500. Thus ( )2500 500 4 , 0,1, 2,nnHn= +=  . (a)(ii) ( )11 911 2500 500 4 2097154500 2.10 10H = + = ≈× . (b) This is a modified model with the same initial conditions. In the absence of harvesting, the rate of growth of the population of hornet per year, ( )11 4n