JPJC 2023 JC1 Promos Further Math (Solutions)

J1 2023 Further Math Year End Exam Solutions and Mark Scheme 1(i) 1 1 1 3 22 4 22 4 4125 5 4ln ln 125 5 4ln 5 5 2ln 5 ln(2 ) ln(5 ) (2 2)ln 2 ( 4)ln 5 n n n n n nn p p nn                       2, 2, 1 and 4AB C D    1(ii) 2 22 Area of banner 130 5 650 125 625 65041 5 cm Sc m c m      Therefore the boy will never finish colouring the banner. 2(i)

2(ii) Substituting z = 0 into | z – z1 | = | z – z2 |, LHS = | 0 – z1 | = | z1 | = 1 RHS = | 0 – z2 | = | z2 | = 1, LHS = RHS Since z = 0 satisfies the equation | z – z 1 | = | z – z2 |, the locus passes through the origin. Equation of locus is y = 4tan 2 x          y = tan 8x  

3(i) 12 1 2 nn nuu u  Characteristic equation: 2 2 1 2 22 1 0 mm mm    24 4 ( 2 ) ( 1 ) 4 21 2 4 22 3 4 13 2 m      General solution: 13 13 , where and are constants22 nn nuc d c d     3(ii) 00 0 13 13When 0, 2 22 2 2 nu c d cd dc           11 0 13 13When 1, 1 22 Sub 2 , 13 13 (2 ) 122 (1 3 ) (2 )(1 3 ) 2 32 2 3 32 23 23 0 1 1 nu c d dc cc cc cc cc c c d                         13 13 22 nn nu     

3(iii) 13 13 , where and are constants22 nn nuc d c d     Since, 13 13as , 0, and 0 0 22 nn nnu c       and 0 22 2 ud c d   132 2 n nu     1 1 1321 3 2u    4(i) Since there are n rectangles between 0 and 1, each rectangle have a width of 1 n Area of 1st rectangle =  2 2 3 11 1 1aa nnn n    Area of 2nd rectangle =  2 2 3 12 1 2aa nnn n    Area of 3rd rectangle =  2 2 3 13 1 3aa nnn n    . . Area of last (nth) rectangle =  2 2 3 11 n an a nnn n             222 2 33 3 3 222 2 3 2 3 1 11 1 1Total area 1 2 3 ... 1 1 2 3 ... 1 n r an an an n annn n n an an an n ann ra nn             

4(ii)   2 22 2 33 11 22 2 3 11 1 23 3 22 2 2 2 11 2 1 2 11 (1 ) ( 21 ) 2 ( 1)62 1 (2 3 1)6 11 1 32 6 nn rr nn n rr r ra n r a n ra nnn ra n r a nn nnn n a n n ann aa nnn ann a aannn                          2 3 1 2 2 2 2 1Area lim 11 1lim 32 6 1 0003 1 3 n rn n ra nn a aannn aa aa              5 32(1 2i) (1 2i) 4 0,zz z t   Given one root is k + ki, then 32 33 22 32 33 22 32 3 2 ( i) (1 2i) ( i) (1 2i) 4( i) 0 (1 i) (1 2i) (1 i) (1 2i) 4( i) 0 ( 2 2i)(1 2i) (2i)(1 2i) 4( i) 0, using GC 62 i 42 i 4 4 i 0 (6 + 4 4 ) (2 2 4) i 0 kk kk kk t kk k k t kk k k t kk kk k k t kkk t kkk                                Compar