2018 A Level H2 FM 9649 P2 (Qns & Solutions) (29 Sep 2024)

1 2018 GCE A Level H2 Further Maths 9649 Paper 2 Solutions Section A: Pure Mathematics Question 1 (i) Let W  V . Let u 1 1 1 1 a b Wc d   =   and v 2 2 2 2 a b Wc d   =   . Then 1 1 1 1 2 2 2 2 0 and 0a b c d a b c d+ + + = + + + = u + v 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 a a a a b b b b Wc c c c d d d d +            +     = + =       +       +      ( ) ( )1 2 1 2 1 2 1 2 1 1 1 1 2 2 2 2since 0a a b b c c d d a b c d a b c d+ + + + + + + = + + + + + + + = Let  . Then u 1 1 1 1 a b Wc d       =   ( )1 1 1 1 1 1 1 1since 0 a b c d a b c d    + + + = + + + = Also W0 and Since W is closed under vector addition and scalar multiplication, W forms a linear space.

2 (ii) Let T  V such that 1a b c d+ + + = 11 00 but 200 00 TT                        . Thus T does not form a linear space. (iii) Let S  V such that a + b = c + d and a + 2b = c + 3d. Then b = 2d and c = a + d Let u 1 1 11 1 2 a d Sad d   = +   and v 2 2 22 2 2 a d Sad d   = +   u + v 1 2 1 2 1 2 1 2 1 1 2 2 1 2 1 2 1 2 1 2 2 2 2( ) a a a a d d d d Sa d a d a a d d d d d d +            +     = + =      + + + + +       +      Let  . Then u 1 1 11 1 2 () a b Sad d       = +   Also, S0 and since S is closed under vector addition and scalar multiplication, S forms a linear space. (iv) dim W = 3 1 0 0 0 1 0 0 0 1 1 1 1 aa bb abccc d a b c                              = = + +                   − − − − − −          1 0 0 0 1 0,,0 0 1 111                           −−−      is a basis for W.

3 Question 2 (i) (ii) Area = 24 0 1 tan d2     4 24 0 0 2 1 sec 1 d2 1 tan2 1 1 units24      =− =− =−   (iii) Method 1 By observation, there is a right angled isosceles triangle with a hypotenuse of length 1. Hence, 22 11 2 x x x+ =  = From the graph, the domain is 1[0, ] 2 . Method 2 cos tan cosxr   == When 4  = , 1tan cos44 2 x    ==        From the graph, the domain is 1[0, ] 2 . Common Mistake tanr =  = 0 O 1, 4   4  =

4 ( ) 22 2 2 2 2 4 2 2 sin sin cos cos 1 ryxy rx y x x y xy x  + = = = =+ = − Domain for x: 210 ( 1)( 1) 0 11 x xx x − + −  −  

5 Question 3 (i) d d y yxx−= Integrating factor is 1 d ee x x− − = e e d e e d ee e1 xx xx xx x y x x xx xc y c x −− −− −− = =− + =− − + = − −   When x = 0, y = 0, c = 1 The solution is e 1 xyx= − − When x = 0.1, y = 0.00517 When x = 0.2, y = 0.0214 (ii) d sind y yxx =+ Using Euler’s method, with x0 = 0, y0 = 0, y1 = 0 + 0.1(sin0 + 0) = 0 y2 = 0 + 0.1(sin0 + 0.1) = 0.01 (iii) Using Improved Euler’s method, wi