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    