JPJC 2024 J2 H2 FM Prelim P1 Solutions

1 2024 J2 H2 FM Prelims P1 (Solutions) Qn Solution 1 Let Pn be the statement that 2 1 1 1 2 n r r n   for all integers 2n . To show P2 is true: LHS = 2 2 1 1 5 4r r  . RHS = 1 3 52 2 2 4   . Hence P2 is true. Suppose Pk is true for some integer 2k  , i.e. 2 1 1 1 2 k r r k   . We want to show Pk + 1 is also true, i.e. 1 2 1 1 1 2 1 k r r k      .         1 22 2 1 1 2 2 2 2 2 1 1 1 1 1 12 1 12 1 2 1 12 1 k k r r r r k k k k k k k k k k k k                       Hence Pk + 1 is true. Since P2 is true and Pk implies Pk + 1, by the Principle of Mathematical Induction, Pn is true for all integers 2n . 2(i) Using GC,   1 0 2 2 rref 0 1 0 2 0 0 0 0        A or   7 201 2 3 3 ref 0 1 0 2 0 0 0 0              A TDim( ) Rank( ) 2R A  By the Rank-Nullity Theorem, TDim( ) 4 2 2K   

2 2(ii) Basis for TR 1 2 1 , 1 3 7                       Let 1 0 2 2 0 0 1 0 2 0 0 0 0 0 0 w x y z                       . Then 2 2 0 2 0 w y z x z       2 2 2 w y z x z    2 2 2 2 2 0 2 1 0 0 1 w y z x z y zy y z z                                                     Basis for TK 2 2 0 2,1 0 0 1                              As T 3 1 11 R        , 3 1 11        Ax for some 4x  . Observe 3 1 2 1 1 1 2 1 11 3 7                             , so a possible solution 1 2 0 0            x The general solution of 3 T( ) 1 11        x is 1 2 2 2 0 2 , ,0 1 0 0 0 1                                            x 

3 3(i) Since the daily removal of concentration level is 7%, i.e. the concentration remaining end of each day is 93%. The recurrence relation is   7 1 00.93 2.5, 0n nu u u    i.e. 1 00.6017 2.5, 0n nu u u    Hence, 7 7 7 0 2.5(0.93 ) 1 0.93 (0.93 ) 6.2767, 0 n n n u c c u       When 0, 0 6.2767 6.2767n c c     Therefore,   6.28 1 0.602 (to 3 s.f.), 0, n nu n n     3(i) Alternatively:             7 1 7 7 2 27 7 2 27 7 7 3 3 27 7 7 3 0.93 2.5 0.93 0.93 2.5 2.5 0.93 (0.93 )2.5 2.5 0.93 0.93 2.5 (0.93 )2.5 2.5 0.93 0.93 2.5 (0.93 )2.5 2.5 n n n n n n u u u u u u                      Therefore,          1 27 7 7 7 0 7 07 0.93 0.93 2.5 0.93 2.5 .... (0.93 )2.5 2.5 1 0.93 2.5 since 01 0.93 6.28 1 0.602 (to 3 s.f.), 0, n nn n n n u u u n n                    3(ii) Since    76.2767 1 0.93 6.2767 n nu    for all 0n , the concentration level will always be less than 7 mg/l. So the Local Authority should allow the factory to go ahead with the discharge 4 Since 2 is an eigenvalue of A,     2 2 det 2 0 2 3 3 2 7 3 0 2 8