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 1 ... 2! !e P e 1 ... 2! ! 1 ... ...2! ! 1 1 1 1 ... !2! 0.1 1 1 1 ... ...! 1 ! 2 !2! n t n t n n n n n t n n n t tt nt t tt n t tt n nt t t t t n n nt t t                                      0 0 e d lim e d lim ! 1 e P ! tn x n x t nt tx x x x n t n                3iii For all n with 0n t  , we have the following results.    2 2 2 2 1 1 ... 1 2 1 !1 ... 2! ! 1 ... P2! ! n n n n n n n n nt t t t n n n n n n n t n tt nn n t tt t n                                        2 2 2 2 3 11 1 ... 0 1 2! 1 1 ... 2 11 ... ... 2! ! 1 ... 2! 3! ! n n n n n n nt t t t n n tn n n n n n n n nt tt nn n t t tt P t n                                 1 P 1 n n n t t tn n       Convergence only holds for 1r . 4i 4 239 23 65 1872 2 3 13x y z      Hence, 13 | 39 23 65x y z  . Since 13| 39x and 13 | 65z, we must have 13| 23y. Since gcd 13, 23 1 , 13 |y. Finally, since y is prime, 13y . 4ii (a) Given 13y , then we have 39 65 1573x z  which yields 3 5 121x z  . (*) Taking modulo 5 on both sides of (*) gives 3 1 mod 5x , and so 5 6 2 mod 5x x x x    . Similarly, taking modulo 3 on both sides of (*) gives 5 1 mod 3z , and so 3 3 10 2 mod 3z z z z    .

4ii (b) From (a), 3 5 121x z  , 5 2x m  and 3 2z n  for some ,m n. 3 5 2 5 3 2 121 15 105 7m n m n m n           Hence, 3 2 5 2 3 5 3 7 5 21 8z x n m n m m m m            . Minimal value of z x is 3 when 3m , which in turn gives 4n . Thus, the required solution is 17x , ( 13y ) and 14z . 4iii From (i), if y is prime, then 13y which leads to 3 5 121x z  . From (ii), , 17,14x z is a solution to 3 5 121x z  . Since gcd 3,5 1 , integer solutions to 3 5 121x z  are 17 5 for 14 3 x r rz r      . When r is even, z is even; and when r is odd, x is even. It is thus impossible for x and z to be both prime unless one of them is 2. When 2x , 3r and 23z which is prime. When 2z , 4r and 37x which is prime. Hence, solutions with x, y and z prime are , , 2,13, 23 or , , 37,13, 2x y z x y z  . 5a (i) For 0x ,  1f 2f 3x x x  . 1 Since 0x , 1 x is defined and non-zero as well. Replacing x with 1 x in (1) gives  1 3f 2f xx x  . 2 5a (ii)      62 2 1 : 3f 3x xx    Hence,  2f x xx  , 0x . 5b For 0x ,    1g g gx x x x    . 3 Replacing x with x in (3) gives    1g g gx x x x     . 4    1 13 4 : g g 2 xx x    Replacing x with 1 x gives    2g gx x x   . 5      1 23 5 : 2g g x x x x    6 Replacing x with 1 x gives  1 12g g 2 x xx x   . 7      32 6 7 : 3g x x   Hence,  1g , 0x xx  .

6i For each 1n :   2 2 1 2 1 1 2 2 1 1 * n n n n n n n n n n n x x x x x dx x x x dx x               We also have the following for each 2n :     22 2 2 1 1 1 1 2 2 2 2 2 2 1 1 1 2 2 1 1 2 n n n n n n n n n n n n n n n n n n n n n n x x dx x x dx x dx dx x x d x dx x x d x dx x x x dx x                          Recursively, we have:    2 2 1 1 2 2 1 1 2 2 2 1 2 1 2 2 1 2 1 2 2 2 1 3 ... by * n n n n n n n n n n n n x x dx x x x dx x x x dx x x x dx x x x x D                        Alternative Solution (to show 2 2 1 1n n n nx x dx x D    ) Let P n be the statement 2 2 1 1n n n nx x dx x D    for n  . Since 2 2 2 1 2 1 2 2 1 3x x dx x x x x D     , 1P is true. Assume P k is true for some k  , i.e. 2 2 1 1k k k kx x dx x D    .       2 2 1 2 1 2 2 2 1 2 2 2 1 2 2 1 2 2 2 1 1 LHS of 1 for all by * (by inductive hypothesis) RHS of 1 k k k k k k k k k n n n k k k k k k k P k x x dx x x x x x x dx x x n x x x x x dx x D P k                                     Since P k is true implies that 1P k is true, and 1P is true, then by PMI, P n is true for all n  . By (*) and P n, we have 2 2 2 1 2 1 1n n n n n n nx x x x x dx x D        . 6ii Suppose 0mx  for some m  , then by (i), 2 2 1 2 1m m m mD x x x x     , which shows that D is a perfect square. 6iii Case 1: 0mx  for some m  . Then by (ii), D is a perfect square.

Case 2: 0nx  for all n  . (Equivalently, 2 1nx  for all n  .) Since there exist positive and negative terms, then k   such that kx and 1kx are of opposite signs, so that 1 0k kx x  .    2 2 1 1 1 1 1 1 2 1 and 0 k k k k k k k k D x x dx x d x x d x x d                 By cases 1 and 2, D is a perfect square or 2D d  .