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                                Comparing real and imaginary parts, 32 32 2 32 6+ 4 4 0 224 0 2 0 (1 ) (2 ) 0 1 2 (rej since is negative) When 1, 6 then (1 2i) (1 2i) 4 6 0 kkk t kkk k k kk ko r k k kt z z z                   

Using long division, 32 2 (1 2i) (1 2i) 4 6 [ ( 1 i)][ (1 2i) (2 5i) ( 3 3i)] zz z zz z           Using quadratic formula, 2(2 5i) (2 5i) 4(1 2i)( 3 3i) 2(1 2i) (2 5 i ) ( 4 ) 2(1 2i) 391o r i 55 z i              6(a) To find intersection, solve cos 3 3 0  . Using GC, 0.2463617  . 0.2463617 0.2463617 22 06 2 11A r e a ( c o s 3)d ( 3)d22 0.211549 0.212 units (3 s.f.)         6(b) 21 0 1 0 1 1 2 0 13 2 0 33 22 dSurface area 2 π(2 1) 1 d d 14π (1 ) 1 d 1 4π (2 ) d 8π (2 )3 8π (3) , where 3 and 23 8π 33 223 2. yx a xx xx x b xx x                   

7(i) d sind y att  and d 1c o sd x att  ds i n d1 c o s yt x t   7(ii) Distance = arc length between t  to 3t  22 3 dd ddd yx ttt        3 22 sin 1 cos dat t t     3 22 c o s dat t    3 2s i n d 2 tat     23 2 2s i n d s i n d22 ttat t     23 2 4c o s c o s22 tta                41 0 0 ( 1 )a    8a

8(i) 10 0 0 0.6 10, 50, 0 (0.6 ) where 1 10 2510 . 6 Sub 0, (0.6 ) 25 50 25 25 25(0.6 ) 25, 0 nn n n n n CCC n bCc d d a d nC c c c Cn               8(ii) As , 0.6 0, 25 n nnC   Eventually the number of bacteria will stabilise at 25 million and hence the antibiotic is NOT effective. 8(iii) 15 10 151 21 3 1 3 1 100% 33.333%3 d a a a      The antibiotic needs to eliminate 100% 33.333% 66.667% 67% of the bacteria every month in order for the number of bacteria to be reduced to 15 million eventually. 8(iv) Assumption: The bacteria do not multiply on their own OR No bacteria will live forever unless eliminated by the antibiotic OR any logical/equivalent answer.

9(i) 1 21 2 2 21 3 22 2 221 21 22 2 tan (e ) de e( 1 e )d1 e d e( 1 e ) 2 e ( 1 e )d e( 1 e ) 2 e e( 1 e ) dd d 2e (shown)dd d x x xx x xx xx xx x xx x y y x y x yy y xx x                 232 2 32 2 dd d d d 2e 2 edd d d d xxyy y yy xx x x x          2 2 3 3 0, 4 d1 d2 When d 0d d1 d2 xy y x y x y x     3 42 1 2 x xy  9(ii) 30.05 0.05 1 00 0.0524 0 tan (e ) d d 42 1 2 44 4 8 0.03989 (5 dp) x xxx x xxx          9(iii) Using GC, 0.05 1 0 tan (e ) d 0.03989 (5 dp)x x  9(iv) The 2 answers are the same up to 5 decimal place. The approximation by the Maclaurin series is accurate as the interval [0, 0.05] is close to 0.

10(a) 1 1111cos cos 2 cos3 cos 4 cos52345y xx xxx    10(b) 2 1111cos cos3 cos5 cos 7 cos93579y xxx xx   