RI 2023 JC1 MYE 9649 (Solutions)

2023 H2 FM Y5 Common Test 1(a ) 222 1 1 11 1 1 11 21 121 2 11 13 11 24 11111 3542 1 ...1 21 112 31 11 2 11 11 11 1122 2 nn rrrr n n rrr nn nn nn                                    11 11 21 51 1 1 42 22 ( 1 ) n nn nn       (b) 22 11 21 51 1 1lim 42 22 ( 1 ) 5 4 rr n n r nn              

2 Intercepts:   0, 0 , , 0 a Vertical Asymptotes: xa 22 2 2 2 22 2 2xa x xa xa a a xaxa xa xa       Oblique Asymptotes: 2yx a Turning points:  2 2 2102dy a xa adx xa      12 , 3 2 2aa and  12 , 3 2 2aa Set of values is    \ 3 22 , 3 22 aa  . Accept  32 2 o r 32 2ya ya  x y O

3 22 241 1 4xy y x   2Area of rectangle 2 2 1 4Ax yx x  2 2 22 d( 8 ) 2 1 622 1 4d 21 4 1 4 Ax x xxx xx     When d 0d A x  , 222(1 4 ) 8xx 1 22 x Thus the x-coordinates for the bottom corners of the rectangle are 11 and 22 22  but we only take the positive value of x for the area of the rectangle. Using first derivative test: 2 22 d2 1 6 2 ( 1 2 2 ) ( 1 2 2 ) d1 4 1 4 Ax x x xx x    x 1 22  1 22 1 22  d d A x  0  Thus the area of the rectangle inscribed is maximum at 211 12 1 4 units 8222  OR Using second derivative test: x y

        2 2 22 2 2 22 22 22 332 22 22 d2 1 6 d 14 814 3 2 21 6 d 21 4 d1 4 14 3 2 4 21 6 8 38d d 14 14 Ax x x xxx x A x xx x xx x x xA x xx             Since 2x is the length of the rectangle, 0x When x = 1 22 , 2 32 2 1183 8 8d 22 0d 114 8 A x           Thus, area is a maximum when 1 22 x Thus the area of the rectangle inscribed is maximum at 211 12 1 4 units 8222 

4(a) [5] 5 25 2 55 5 2222 2( 9 ) ( 13 ) 9 1 ( 13 ) 9 xx xx     5 2 23 23 53 53 1 5 22 22 21 1 ...92 9 2 ! 9 3 ! 9 55 51. . .18 216 11664 xxx x x xx                               5 22 22 51 5(1 3 ) 1 3 ... 1 22x xx     5 25 2 55 5 2222 2 23 2 22 3 3 23 ( 9 ) ( 13 ) 9 1 ( 13 ) 9 55 5 1 5243 1 ... 1 ...18 216 11664 2 51 5 5 5 2 5243 1 ...18 2 216 11664 12 135 14625 24305243 ... 28 4 8 xxx x x xx x x xx x x xx x                          (b) [2] Expansion of 5 2 1 9 x is valid for 199 x x  . Expansion of 5 2 2(1 3 )x is valid for 2 131 3 xx   . Hence, expansion of 55 222(9 ) (1 3 )x x is valid for 1 3 x  . This further implies 11 . 33 x  Range of validity for which expansion is valid is 11,. 33 

5[8] Use de Moivre’s theorem to show that 55 3sin 5 cos ( 10 5 )tt t  where tant  . 5 54 3 2 23 4 5 53 2 4 42 3 5 cos5 isin 5 (cos isin ) cos 5i cos sin 10cos sin 10i cos sin 5cos sin isin (cos 10cos sin 5cos sin ) i(5cos sin 10cos sin sin )                            Compare imaginary part, 42 3 5 53 5 55 3 sin 5 5cos sin 10cos sin sin cos (5 tan 10 tan tan ) cos ( 10 5 )tt t                where tant  . (shown) Let 5   and 22tanx t 55 3sin 5 cos ( 10 5 )tt t  55 3sin cos ( 10 5 )5 tt t   54 20c o s (1 0 5 )5 tt t  Since 5cos 05   and tan 05   , 42 10 5 0tt   2 10 5 0xx  2tan 5  is a root of 2 10 5 0xx  . (deduced) 2 10 80tan 5 2 552    Since 0t a n t a n 154   , 2tan 15   so 2tan 5 2 55  

6[2] Perpendicular bisector of the line segment from the points represented by 23 i and p on Argand diagram. Im (b) [6] (i) To find line equation passing through (1, 8)and (2 ,1 ) , 88 ( 1 ) 3511 ( 2 ) y yxx      Represent 23 i from part (a) as point (2 , 3 ) To equation of line passing through (2 , 3 ) and perpendicular to 35yx ,  1 3 1732 33 17 33 yx c cc yx          To find point of intersection between 35yx and 17 33yx  , 1735 33 10 8 33 4 5 41 335 55 xx x x y        Point of intersection is 41 3,55  . Represent (, )px y , (-2,3) p Re

41 3 (2 ) 3,,55 2 2 21 1(, ) , 55 xy xy          21 1 i55p  the least possible value of 22 2 2 24 1 1 1 3 6 2 2 1 0 units55 5 5 5 5 5zp                           Alternatively, From previous part, the line equation passing through (1, 8)and (2 ,1 ) , 35yx the least possible value of 22 11 23555 42 1 0 units51013 zp    

7(a) Let nP be the statement 2 124 6 2 n nunn     , for all integers 2n . When 1n  , 1 17246 22u  (as given) Since LHS = RHS, 1P is true. Assume kP is true for some 1k  , i.e. assume 2 124 6 2 k kukk     . To prove that 1kP is true, i.e. to prove 1 2 1 12( 1) 4( 1) 6 2 k kuk k        Now, LHS 2 1 1 (1 )2kkuu k  2211 24 6 ( 1 )22 k kk k       1 22 123 (1 ) 2 k kk k      1 22 1 22 1 2 121 44 6 (1 ) 2 1(1 )4 (1 ) 6 (1 ) 2 12( 1) 4( 1) 6 RHS 2 k k k kk k k kk k kk                     1 is true is truekkPP  . Since P1 is also true, by Mathematical induction nP is true for all positive integers n. (b) 98 32 2 12 2 2    12 22 (proven) Let nP be the statement 11 11 ... 2 1 2 23 n n     , for all positive integers n. From previous part, 1P is true. Assume kP is true for some 1k  , i.e. assume: 11 11 ... 2 1 2 23 k k     . To prove that 1kP is true,

i.e. to prove: 11 11. . .2 2 2 23 1 k k       . Now, LHS 11 1 11 ... 23 1 kk       121 2 1 k k     . We want to show that:   121 2 22 2 1 kk k      1i.e. 2 1 2 2 0 1 2 3 2 ( 1)( 2) 0 kk k kk k         22 31 3 323 2 23 2 23 2 0 24 2 2kk kk k k                 is true.kP Since 1 is trueP and 1P is true is truekk P , by mathematical induction, is true for allkP positive integers.