RI 2023 JC1 Promos 9649 (Solutions)

2023 Y5 FM Promotion Examination (Solutions) 1 Solutions Given 1 79u  , 79 5 6 A B ………..(1) Given 2 949u  , 2949 5 36A B ……....(2) Sub (1) into (2),  2 2 949 5 6(79 5 ) 56 54 7 5 0 (5 19)(5 25) 0 5 19 (NA) or 5 25 2 AA AA AA AA A           9B  . Let nP be the statement         42f1 0 2 31 0 5 9 6nn n n nnu    is a multiple of 19, where 0n  . For 0n :     2(0) 0f0 1 0 5 9 6 1 9  which is a multiple of 19. Therefore 0P is true. Assume that kP is true for some 0k  , i.e.       2f1 0 5 9 6 1 9kkkp  for some p   . To prove 1kP is true, i.e.       2( 1) 1f1 1 0 5 9 6 kkk   is a multiple of 19.               2( 1) 1 2 2 f1 1 0 5 9 6 250 5 54 6 25 10 5 9 6 171 6 25(19 ) 19 9 6 19 25 9 6 kk kk kk k k k k p p        Thus  f1k is divisible by 19. Therefore kP is true  1kP is true. Since 0P is true, by Mathematical Induction, nP is true for all non- negative integers n.

2 Solutions (a) 22 2 2 22 1 22 2( ) where 1 yx a xPF x c y ba a      and 22 .ba c     22 22 2 22 22 2 2 22 2 2 2 2 22 2 22 4 22 22 22 42 2 2 22 2 () () 1 () 1 2 1 2 1 since axxc b a axxc a c a axc a x a ca ax ac x ac a ac ax xca aa c x x ca ac xa ac x aca                         The other focus will be at  2 ,0Fc  . Similarly, 2 2 .ac xPF a  Hence 22 12 2 (Shown)ac xac xPF PF a aa    (b) 12 22 22 2 2 QF QF ac xac x aa cxa a        Since 222ac b and at the point  4, 3Q , 1QF QF k , therefore,   222 22 1 22 1616 abcQF QF a a aa     2 2 2 1616 (1) bka a  Also, 22 2 2 2 2 22 2 2 2 43 3 4 4 11 a ab b a a    ,

2 2 2 9 16 ab a   (2) Substitute (2) into (1): 2 2 2 2 2 2 2 2 2 916 16 916 1 16 14416 (Shown) 16 Therefore 144, 16 aa aka a ka a ka a AB            

3 Solutions (i) 2 0.536x  , 5 0.0346x  , 8 0.00132x  . From GC, the sequence (appears to) converge to the larger root, 0. (ii) Sketch a graph of f( )yx where f( ) 3 l n ( 1 )xx x  Note that around the origin, f( ) 3 l n ( 1 )xx x  closely “resembles a straight line”. We use a tangent approximation to 3l n (1 )yx x    at 0,  to find root of the equation 03 l n ( 1 )xx    .  0 1|3 2 10 0 202 x dy dx x         Since  is very close to zero, the root is also close to 0. OR Alternatively, consider a graphical explanation. The equation 3l n (1 ) 0xx  has a larger root 0x . Since the graph of f( )yx is a translation of the graph of 3l n (1 )yx x  by a very small amount  in the y-direction, (and f is strictly increasing around the point where x=0), the root is a small change from 0 in the x-direction. OR Alternatively, show that there is a sign change over the interval 0 to  . f( ) 3 ln( 1)xx x       22 2 f( 0 ) 3 0 l n ( 0 1 ) f ( ) 3 ln( 1) 3 ... ... 22 f( 0 ) f( ) . . . 0                           Root lies between 0 and  . Since very close to 0, so  is close to 0. (iii) (1 ) 3l n (1 )3l n (1 )g( ) 1 323 1 xx xxxx x x       

(iv)    1 2 (1 ) 3l n (1 ) 32 10 22 nn n nn n xx xxx x x           2 3 2 2 3 21 ...22 2 2 32 22 8 ...32 21 4 13 1 ...28 2 4 x                                  2 21 6    (shown)

4 Solutions (i) 3(cos sin )x   d 3( sin cos sin ) 3 cosd x        3(sin cos )y   d 3(cos sin cos ) 3 sind y         Length of the arc PQ   22 2 0 2222 0 2 0 dd ddd 9c o s s i n d 3d xy                2 0 2 2 0 2 3d w h e n 0 , 2 3 2 3 8                (ii) Area of the surface formed  22 2 0 2 0 dd2d dd 23 ( s i n c o s ) 3 d xyy               22 0 18 sin cos d       Now,   222 000 2 0 sin d cos cos d sin 1             

Or Alternatively note that dd 3(sin cos ) 3 sindd y     , so  22 00 sin d sin cos 1       22 222 00 0 2 2 cos d sin 2 sin d 2(1) (from above)4 24              Area of the surface formed 2 3 18 1 2 4 954 2      

5 Solutions (i)   22 2 2 0 1 01 0 11 1ed 2 ed e 1 e22 2 aa x a xx axIx x x            (ii) 2 2 22 22 1 0 1 2 1 2 0 00 0 e d 1 (2e ) d2 11 ee ( 1 ) d22 11 e0 e d22 x nx xn x a n n a a a n nn xa a I x xx xx x x nx x nax                        2 1 2 11 e22 an nn II n a   (shown) (iii)   22 31 4 53 2 3 3 31 1 1 e2 e22 2 aaII a aII          22 2 211 2 1 2 13 11 1 1 1 1 ee 1 e e22 2 2 2 aa a aII a I a a             22 2 2 5 4 24 2 11 121 e e e22 2 11e 1 2 aa a a Ia a aa           (iv) 2 2 0 5 0 24 5(2 ) d (2 e ) d 2 121 e 1 2 aa x a xy x x x I aa               As a , 2 42 1e1 0 2 a aa   , 2 24 11e 1 1 2 a aa   . Therefore volume required 2 units3.

6 Solutions (i) 55 i 2i34 i   The required vertex is   2 i3 132i 34 i e 2i 34 i i 22 33 32i 2 32 i i22 13 323 1 i22                   (ii) 2i * 73 i 2i 73 izz zz    41cos 10  Required area is 22 1 1 1 11 4 152 2 5 s i n 2 2 5 c o s 2 5 s i n c o s22 1 0 41 59 4125 cos 25 10 10 10 41 241925 cos 10 100                     25, 41 and 2419AB C  5 2 + i Re Im 2 + i 7 - 3i O 5

7 Solutions 1(a) [4] At point of intersection, 3 π 5π1s i n 3 = 3 ,26 6 π 5π,18 18      Polar coordinates of the points of intersection are π3 ,2 18   and 5π3 ,2 18   . (b) [3] d1 sin 3 and 3cos3d rr    Perimeter of S  5 π 5π 18 18 π π 18 18 2 2 31s i n 3 ( 3 c o s 3 )d θ d2   π1.695032 2.7423  (c) [4] Area of S =  5 18 18 5 π 18 π 18 5 π 18 π 18 2 2 2 5π 18 π 18 13 1s i n 3 d22 15 2sin 3 sin 3 d24 13 c o s 6 θ2sin3 d24 2 13 2 c o s 3 s i n 6 24 3 1 2 π 3 312 8                       O (1, 0) (1.5, 0)