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, with x0 = 0, y0 = 0 u1 = 0 y1 = 0 + 1 2 (0.1)[(sin0 + 0) + (sin 0 + 0.1)] = 0.005

6 u2 = 0.005 + 0.1(sin 0.005 + 0.1) = 0.0154999979 y2 = 0.005 + 1 2 (0.1)[(sin 0.005 + 0.1) + (sin 0.01549999979 + 0.2)] = 0.0210 (iv) When y is small, sin y  y. Thus the solution to d sind y yxx−= is an approximation to the solution of d d y yxx−= . The answers obtained in part (iii) are closer to those obtained in (i). Hence Improved Euler’s method gives a better approximation than the Euler’s method.

7 Question 4 (i) 32 2 9 18 0  + − − = Using GC,  = −3, −2, 3 Thus M has only one positive eigenvalue 3. (ii) Given M 33 2 3 2 11        =      −−    M 3 3 9 2 3 2 6 1 1 3 −−           − =− = −          −      M2 2 3 1 3 1  =− 1 3 M (M 3 2 1   − ) = 1 3 M 3 32 1       − = M 39 26 13        =      −−    Since M 33 2 3 2 11        =      −−    , M 6 3 18 4 6 2 12 2 1 6            ==          − − −      Thus the solution is 6 4 2 x y z        =      −   

8 (iii) Method 1: By Cayley-Hamilton Theorem Note: The Cayley-Hamilton theorem states that every matrix satisfies its own characteristic equation, Characteristic eqn: 32 2 9 18 0  + − − = By Cayley-Hamilton Theorm, 32 2 9 18 0+ − − =M M M I 32 2 9 18=− + +M M M I ------- (*) 4 3 2 2 9 18=− + +M M M M ( ) 222 2 9 18 9 18=− − + + + +M M I M M (from (*)) 224 18 36 9 18= − − + +M M I M M 213 36=− MI Hence 13, 0, 36a b c= = =− Method 2 42 a b c= + +M M M I ( ) ( ) 421 1 1 1 a b c− − − −= + +PDP PDP PDP PP 4 1 2 1 a b c−− = + +PD P P D D I P 42 a b c= + +D D D I By comparing the diagonal entries, 81 9 3 16 4 2 81 9 3 a b c a b c a b c = − + = − + = + + Hence, by GC, 13, 0, 36a b c= = =− . Method 3 42 a b c= + +M M M I 42 a b c= + +M x M x Mx x 42 a b c  = + +x x x x 42 a b c  = + + Using the eigenvalues found in (i), 81 9 3 16 4 2 81 9 3 a b c a b c a b c = − + = − + = + + Hence, by GC, 13, 0, 36a b c= = =− .

9 Question 5 (a)(i) PQ = 223 5 34+= 5 5 3tan sin and cos 3 34 34   =  = = Let z = i34e and z = i 434e 34 cos isin 44    +     = + + +         ( ) 2 2 2 234 cos sin i sin cos2 2 2 2 2 2 8i2 2 4 2i      = − + +   = − + =− + Complex number representing R is ( ) ( )1 2i + 2 4 2i 1 2 2 4 2 i+ − + = − + + (ii) z = ( )i 34e + For R to lie on the imaginary axis, the real part of z is zero. 1 1 1 3cos 5sin 0 534 cos tan 1 3 5tan 1.3984457373 Smallest positive value of 0.713     − − + − =  + =− += = (b)(i) Let arg(z) =  ,  0, 2 Then 2arg 2arg 2zz == ,  2 0, 4 For Re(z2) < 0, Re(z) Im(z) O

10 3 5 72 , or ,2 2 2 2             Hence 3 5 7, or ,4 4 4 4             (ii) For Im(z3) > 0, ( ) ( ) ( )3 0, or 2 ,3 or 4 ,5      2 4 50, or , or ,3 3 3 3                     Im(z) Re(z) O 3 