2020 H3_Solution

2020 A Level H3 Mathematics (9820/01) Qn Suggested Solution Remarks 1i Apply the AM-GM inequality to yield the following.      1 1 1 1 1 1 nn n n n n n n n x y x yn n x y x yn n x y n x y             1ii Let 2n , 1x and y a in (i) to get:   2 21 2a a  . Let 3n , 1 2x and y b in (i) to get:   23 3 11 3 2b b  . Let 4n , 1 3x and y c in (i) to get:   34 4 11 4 3c c  . In each case, equality holds only when x y . Hence,         2 32 3 4 2 3 4 4 1 11 1 1 2 3 4 2 3 4 256 1 a b c a b c abc abc                with equality only if 1a , 1 2b and 1 3c which is impossible since 1abc . Therefore,       2 3 41 1 1 256a b c    . 2i Let  1fy x ax b   , bx a .    1 1 1 1 1 1f , 0 ax b y x ba y x b xa x            2ii          3 2 2 2 2 2 3 2 2 2 2 2 2 2 2 f 1 (*) 0 1 0 (**) 0 or 1 0 p p pap ba ba b ap b a p abp a abp b abp ab p b p a b ap b a b p a b a b ap bp a b ap bp                                

Case 1: 2 0a b  . Then equation (**), and hence equation (*), holds true for all values of p for which 3f p exists. Hence, 3f x x for all x such that 3f x exists. Case 2: 2 1 0ap bp   . Then  1 fp pap b  , i.e. p is a fixed point of f. 2iii 1 1 1 11n n n n n Ax x Bx x Ax B       From (ii), 2f x x when 0b for all x such that 2f x exists; and 3f x x when 2 0a b  for all x such that 3f x exists. Period 2: Set 1A and 0B to get the recurrence relation 1 1n nx x  , with the condition that 1 1,0,1x . Period 3: Set 1A and 1B to get the recurrence relation 1 1 1n n nx x x    , with the condition that 1 0,1x . Compare A and B in (iii) to a and b in (ii) respectively. 3i Let Q n be the statement:  0 e d ! 1 e P t n x t nx x n t    , 0n  .       0 00 0 0 LHS of 0 e d e 1 e RHS of 0 0! 1 e P 1 e 1 e 0! t tx x t t t t Q x x tQ t                         Therefore, 0Q is true. Assume Q k is true for some 0k  , i.e.  0 e d ! 1 e P t k x t kx x k t    .                    1 0 1 0 0 1 0 1 1 1 LHS of 1 e d e 1 e d e 1 e d e 1 ! 1 e P (by inductive hypothesis) e1 ! 1 ! 1 e P1 ! 1 ! 1 e P 1 ! 1 ! 1 e t k x ttk x k x tk t k x k t t k k t t k kt k Q k x x x k x x t k x x t k k t tk k tk tk t k k                                                                  1 0 1 1 0 ! 1 ! 1 ! 1 e 1 ! 1 e P RHS of 1! k i kt i k it t k i t t i k tk k t Q k i                                

Since Q k is true implies that 1Q k is true, and 0Q is true, then by PMI, Q n is true for all 0n  . 3ii For any fixed 0n  ,     2 2 2 1 2 2 1 2