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 = −