2021 A Level H2 FM Paper 1 (Solutions)(updated 3 Oct 2024)

1 2021 Further Mathematics Paper 1 (9649/1) [Solution] (a) Given 00 1, 1, 0.5x y h= = = and ( ) 1ef, xy xy xy ++= . By Euler method, ( )1 0 0 0 0.5f , 5.194528 5.195y y x y= + =  (to 3 dp) By Improved Euler method, ( )1 0 0 0 0.5f , 5.194528 5.195u y x y= + =  ( ) ( )1 0 0 0 1 1 0.5 f , f , 29.0532y y x y x u= + +  (to 3 dp) (b) The Euler method uses the slope at the point (x0, y0) to obtain the estimate of y1, while the Improved Euler method uses the average of the slope based on the initial point and the point (x1, u1). Since the values of f(x0, y0) and f(x1, u1) differ greatly, there is a large discrepancy between the two answers. [Solution] (a) Let  be the eigenvalue of the matrix M Then 12 022   − =−− (1 − )(−2 − ) – 4 = 0 −2 −  + 2 + 2 – 4 = 0  2 +  − 6 = 0  ( + 3)( − 2) = 0  = 2 or −3

2 If  = 2, 1 2 0 2 4 0 x y −     =    −      −x + 2y = 0 and 2x – 4y = 0  x = 2y A corresponding eigenvector is 2 1   If  = −3, 4 2 0 2 1 0 x y      =          4x + 2y = 0 and 2x + y = 0  y = −2x A corresponding eigenvector is 1 2  − (b) Points on the line x = 2y and y = −2x under M will be map to another point on the same line respectively – that means the lines are x = 2y and y = −2x are invariant under the transformation represented by M. [Solution] Let P(n) be the statement “ 1 2 1F ( ) 21 n n  + for n = 1, 2, 3, …. where 1 1 1 2 2 21 2 ( 1)...( 1)F ( ) ! n n n + + −= ” n = 1, 1 21 1 2 11F ( ) 1! 2 3 = =  as 2 > 3 . Thus P(1) is true. Assume P(k) is true for some positive integer k, that is 1 2 1F ( ) 21 k k  + . When n = k + 1, to prove 1 1 2 1F ( ) 23 k k +  + LHS = 1 1 1 1 2 2 2 21 1 2 ( 1)...( 1) ( )F ( ) !1 k kk kk + + + − += + < 1 21k+ 1 2() 1 k k + + = 1 21k+ 21 22 k k + + = (2 1)(2 3)21 22 (2 2) 2 3 kkk k kk +++ =+ ++ 222 (2 2)4 8 3 4 8 4 (2 2) 2 3 (2 2) 2 3 (2 2) 2 3 kk k k k k k k k k k ++ + + +=  = + + + + + +

3 1 23k  + So P(k) is true  P(k + 1) is true, by Mathematical Induction, 1 2 1F ( ) 21 n n  + for all positive integers n . [Solution] (a) By disc method, 2 0 d1 b xVx ax =  + 2 220 d21 b x xa x ax= ++ ( )22 221 2 1 220 21 d21 b aaa a x ax x xa x ax + + − − = ++ (or use long division) 2 2 21 1 220 d21 b a a a x xa x ax +=− ++ ( )32 2 211 1 2200 22 dd 21 bb aa a a x a xx a x ax +− =− ++ ( ) ( ) 2 1 21 320 0 111ln 1 b b a axx ax a a a − += − + −  − ( ) ( ) 2 3 3 3 2 1 1ln 1 1 b aba a a ab a = − + + −   +  ( )3 12ln 1 1 1ab aba ab  = − + − + + (b) ( ) ( ) ( ) ( ) ( ) 22 2 2 2 22 1d 1 1 d 1 1 1 ax x ay x y x x x ax ax ax +−= = = = + + +  (shown)

4 By disc method, 1 2 0 d b aby y W x y += = =  2 0 d dd xb x yxx x = = =  2 0 d xb x yx = = =  V= (shown) [Solution] (a) Graph of y = xex – 1 + x – 2 We note that when x =1, y = 1 + 1 – 2 = 0 and there is only one zero. f(1) = − f(1.1) = 1.1e0.1 + 1.1− 2 −  = 0.316 −  If  is very small and positive, f(1) < 0 and f(1.1) > 0  1 <  < 1.1 f(0.9) = 0.9e−0.1 + 0.9 – 2 −  = −0.286 −  If  is very small and negative, f(1) > 0 and f(0.9) < 0  0.9 <  < 1 In both cases,  is close to 1. (b) f’(x) = xex – 1 + ex – 1 + 1 = (x + 1)ex – 1 + 1 Using the Newton-Raphson method 1 1 1 f ( ) 2 f '( ) ( 1) 1 n n x n n n n n n x n n x x e xx x x x xe − + − + − −= − = − ++ = 1 1 12 1 [ 2 ] ( 1) 1 n n xn n x x x n n n n n x n x e x e x x e x xe − − − − + + − + − − ++ = 12 1 2 ( 1) 1 n n x n x n xe xe − − ++ ++ x0 = 1 x1 = 3 3 13 + =+ Since 2 2 d d yy xx=

5 223 2 3 2 22 3 3 9 3 18 2 2 3 3 183 (1 ) 2 (1 )(1 ) 2 (2 )(1 ) 1(2 ) 1 ex e           + + + + + + + + +== + + + +++ = 2 2 2 2 2 2 2 22 3 18 3 9 18 3 22 3 18 3 9 6 1 2 3 2 2 1 3              + + + + + + + + + =+ + + + + + + = 22 11 3 3 2(3 2 )3 [1 ( )]   −−+ + + + = 2 2 2 2 3 3 18 3 18 1 (3 2 )[1 ( ) ( ) ...]3     + + − + + + + = 2 2 2 3 3 18 9 1 (3 2 )[1 ...]3    + + − − + + = 22 3 3 18 1 (3 2 )[1 ...]3   + + − + + = 2 2 2 23 3 3 18 1 (3 2 ...)3   + + − − + + = 23 18 1 (3 ...)3 + − + = 2 1 ...3 18 + − + [Solution] (a) ( ) 22 2 3 2 5x y x xy+ = + 4 3 3 3 25 cos cos sinr r r   =+ ( ) 22cos 5cos sinr   =+ ( ) 2cos 1 4cos=+ 34cos cos=+ (b) 11 22 1 00 cos d cos cos dnn      −= ( ) 1 221 2 2 0 0 sin cos 1 sin cos dnn n       −−= + −  ( ) ( ) 1 2 22 0 1 1 cos cos d nn     −= − −  ( ) 11 22 2 00 1 cos d cos dnnn     −= − −  ( ) ( ) 1 1 1 2 2 2 2 0 0 0 cos d 1 cos d 1 cos dn n n nn          −+ − = −   ( ) 11 22 2 00 cos d 1 cos dnnnn      −=−

6 11 22 2 00 1cos d cos dnn n n      −−= (shown) (c) Area enclosed ( ) 2 23 0 12 4cos cos d2    =  + ( ) 2 6 4 2 0 16cos 8cos cos d     = + + Using the result in (b),   11 22 22 000 11cos d 1d 2 2 4     = = = 11 22 42 00 3 3 3cos d cos d 4 4 4 16      = = =  11 22 64 00 5 5 3 15cos d cos d 6 6 16 96      = = =   Area enclosed 15 316 8 96 16 4        = + +           17 4 = [Solution] (a) A = 1 and C = 0 (i) For (*) to be a circle, B = A = 1 (ii) A = B = 1 and C = 0 x2 + y2 + + Dx + Ey + F = 0 2222 2 2 4 4( ) ( ) 0D E D Ex y F+ + + − − + = 2222 2 2 4 4( ) ( )D E D Ex y F+ + + = + − For (*) to be a circle, 22 22 44 0 4 0DE F D E F+ −   + −  If 22 40D E F+ − = then 2 Dx=− and 2 Ey=− thus (*) is the point ( 2 D− , 2 E− ) If 22 40D E F+ −  then no solution

7 (b) Let 1' 1' k x x k y y      =    −      1 2 1 ' 1 ' 1 1 ' 1 ' 1 x k x k x y k y k y k − −        ==        − +         Substitute ( )2 1 ''1x kx yk=− + and ( )2 1 ''1y x kyk=+ + into the original equation with C = 1 We have: ( ) ( ) ( )( ) ( ) ( ) ( ) 22 2' ' ' ' ' ' ' ' ' ' ' ' 1 0A kx y B x ky C kx y x ky D kx y E x ky F k− + + + − + + − + + + + = Making the coefficient of the ''xy term = 0 That is: ( ) 2 2 1 0k B A k+ − − = So 2 22( ) 4( ) 4 1 ( )2 B A B Ak A B B A− −  − += = −  + − Note: if we do by actual rotation matrix cos sin sin cos    − for clockwise rotation of angle , we need to “normalise” the given matrix and work out the angle – a bit more troublesome. [Solution] (a)(i) nH denote the population of hornets n years after the start of 2019 0H = 3000| 1H = 4500 (1 year after the start of 2019 that is at the end of 1st year or start of 2020) ( )11 4n n n nH H H H+−− = − 11 5 4 0n n nH H H+−− + = Auxiliary equation: m2 – 5m + 4 = 0

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