EJC_9649_2023_Prelim_P2_Solutions

2023 JC2 H2 Further Mathematics Preliminary Examination Paper 2 [Turn over Q1 Solution Let P( )N be the statement 1 3 1 1 2( 1)( 2)2 ( 2)2 N nN n n n n N     for integers 1N  . When 1N  : LHS  1 3 1 (1 1)(1 2)(2) 3   RHS  1 1 1 1 12 (1 2)(2) 6 3    LHS  RHS, hence P(1) is true. Assume P( )k is true for some 1k  , i.e. 1 3 1 1 2( 1)( 2)2 ( 2)2 k nk n n n n k     . Claim P( 1)k is true, i.e. 1 1 1 3 1 1 2( 1)( 2)2 ( 3)2 k nk n n n n k        . Proof: LHS 1 1 3 ( 1)( 2)2 k n n n nn     1 1 34 ( 1)( 2)2 ( 2)( 3)2 k nk n nk n n k k        1 1 1 4 2 ( 2)2 ( 2)( 3)2kk k k k k        1 ( 3)(2) ( 4)1 2 ( 2)( 3)2 k kk kk      1 1 2 6 4 2 ( 2)( 3)2 k kk kk      1 12 2 ( 2)( 3)2 k k kk    1 11 2 ( 3)2 kk   RHS Hence P( )k is true  P( 1)k is true. Since P(1) is true, and if P( )k is true then P( 1)k is also true, then by mathematical induction, P( )N is true for all positive integers 1N  . 1 3 1 1 1 lim 22( 1)( 2)2 ( 2)2nN Nn n n n N         Q2 Solution (i) The distributive axiom ()c c c  u v u v is violated. 1 1 1 1 2 2 2 2 () u v cu vcc u v cu v              uv 2 11 11 2 22 22 cu cv c u vcc cu cv c u v                uv So ()c c c  u v u v in general

2 2023 JC2 H2 Further Mathematics Preliminary Examination Paper 2 Alternative The axiom ()c d c d  u u u is violated. 1 2 ()() () c d ucd c d u    u 2 11 1 2 22 2 cu du cducd cu du cdu                uu So ()c d c d  u u u in general (ii) (a) Let 1A and 2A be matrices such that 11 A B BA and 22 A B BA  1 2 1 2 1 2 1 2 ()      A A B A B A B BA BA B A A        1 1 1 1k k k k   A B A B BA B A The set is closed under addition and scalar multiplication. Also, the set is non-empty since 0B B0 0 . Hence it is a subspace. (ii) (b) T I I I but TT(2 ) (2 ) 4( ) 4  I I I I I I The set is not closed under scalar multiplication. Hence it is not a subspace. Q3 Solution (i) 2 2 3dxIx   Let f ( ) 3 xx  and 2 ( 2) 14h  n nt f ( )nnyt  0 2 1 9 1 1 1 3 2 0 1 3 1 3 4 2 9 Let T denotes the approximation to 2 2 3dxIx   , found using trapezium rule with 5 ordinates.  0 1 2 3 42 2 22 hT y y y y y     --- (1) 88 9T  (ii) f ( ) 3 xx    f ( ) ln 3 3 xx      2 f ( ) ln 3 3 0 xx  for 22 x   f ( ) 3 xx  is concave upwards over the interval  2, 2 Trapezium rule produces an overestimate T to 2 2 3dxIx   . (iii) Let S denotes the approximation to 2 2 3dxIx   , found using Simpson rule with 5 ordinates.  0 1 2 3 4 1 4 2 43S h y y y y y     --- (2)

3 2023 JC2 H2 Further Mathematics Preliminary Examination Paper 2 [Turn over Q4 Solution (a) Differentiate (1) with