HCI 2025 H3 Mathematics Prelim Solutions

2025 H3 Math Prelim Mark Scheme Solution 1 Let P(n) be the statement ( ) 1 12 12 ... ...n nn x x x x x xn + + +  for   232, 2 , 2 ,...n . To prove P(2) is true Since 2 12( ) 0xx− 22 1 1 2 2 1 224x x x x x x+ +  2 1 2 1 2 12 12 ( ) 4 2 x x x x xx xx + +  Therefore P(2) is true Assume P(2 )m is true for some m , i.e ( ) 1 12 2 2 12 2 ... ...2 m m mm x x x x x x+ + +  -----------(1) To prove 1P(2 )m+ is true, i.e. ( ) 1 1 1 1 12 2 2 121 2 ... ...2 m m mm x x x x x x + + ++ + + +  -------(2) LHS = 112 2 1 ... 2 m m x x x + + + + + 112 2 2 1 2 2 2... ...1 2 2 2 m m m m mm x x x x x x ++++ + + + + +=+  112 2 2 1 2 2 2... ... 22 m m m m mm x x x x x x ++++ + + + + +            ---(3) ( ) ( )1 11 22 12 2 2 1 2 2 2... ... mm m m m mx x x x x x +++ ------------(4) ( ) 1 1 1 2 12 2 2 1 2 2 2... ... m m m m mx x x x x x + +++= Hence 1P(2 )m+ is true, by principle of mathematical induction, P(n) is true for all   232, 2 , 2 ,...n .

Solution 2(a) 1 1 1 no. of permutations where the th object i ! s in the th p ) n ( 1 ! (! o 1 sitio ) n i i n i n i i S i n nn n = = = = =− =− =    no. of permutations where the th & th object are ! in their original positio ( ( 2)! 2)!2 ! ( 2)!2!( 2)! ! 2 ns ij ij ij ij i S j S n n n n nn n     = =− =− =− − =    2(b) 12 12 ... ! ... nn n D S S S n S S S =    = −    1 1 ! ... ( 1) ... n i i j i j k i i j i j k n n n S S S S S S SS =    = − +  −   + + −      = !! ! ( 3)! ... ( 1) 132 nnnn n n − + − − + + − = !!! ! ( 3)! ... ( 1)2 3!( 3)! nnnn n n n− + − − + + −− = ! ! !! ! ... ( 1)2! 3! ! nn n nnn n− + − + + − = 1 1 1 1! 1 ... ( 1)1! 2! 3! ! nn n − + − + + −

Solution 2(c) 1 1 1 1lim lim 1 ... ( 1)! 1! 2! ! 1 1 1 11 ...1! 2! 3! 4! e nn nn D nn→ → − = − + − + − = − + − + − = Solution 3(a) ( ) 2 1 1 1i k + + ( ) ( ) 2 2 1 i 1 1i k k ++= + ( ) ( ) 2 2 2 1 i i 1i k k +−= +    ( ) 2 1 ( 1)i 1 ( 1)i 1i kk k + − + += + 2 2 2 1 1 1f ( ) 2 1 1 ... 1(1 i) (1 2i) (1 ( 1)i)n n     = + + +    + + + −     ( )( ) ( ) ( )( ) ( ) 22 1 1 2i 1 i 1 3i2 ... 1 i 1 2i + + +=   ++    ( )    ( )    ( ) 2 2 2 1 ( 3)i 1 ( 1)i 1 ( 2)i 1 i 1 ( 1)i 1 ( 1)i 1 2)i 1 ( 1)i 1 i n n n n n n n n n + − + − + − + + − + + + − + − + 2 1 ( 1)i 1 i 1 i n n ++= ++ 2 2( 1)i (1 ) ( 1)i n nn ++= − + +    22 2 2( 1)i (1 ) ( 1)i (1 ) ( 1) n n n nn + + − − += − + + ( ) ( ) 22 22 2 i11 n n n n nn + + − − =+ ++

Solution 3(b) ( ) 2 0 2 2 1 36tan arg 1 (1 i) 37 36tan arg f ( ) 37 36 2 37 8 n k k n nn nn n =  + =−  +  =− −− =−++ =  Solution 4(a) Let 1 2 3 4 , , ,x y z wa a a a x y y z z w w x = = = = + + + + and 1 2 3 4 , , ,b x y b y z b z w b w x= + = + = + = + Applying Cauchy Schwarz inequality, ( )( ) ( ) 22 2 2 2 2 2 2 2 1 2 3 4 1 2 3 4 1 1 2 2 3 3 4 4a a a a b b b b a b a b a b a b+ + + + + +  + + + Hence ( ) ( ) 2 2 2 2 2 2 2 2 2x y z w w x y z w x y zx y y z z w w x  + + + + + +  + + + + + + + --(1) 2 2 2 2 2 x y z w w x y z x y y z z w w x + + ++ + + + + + + 4(b) Using Arithmetic-Geometric inequality, ( ) ( ) 11 44 242 w x y z w x y z wxyz wxyz+ + + + + +   Hence ( ) 2 2 2 2 1 42x y z w wxyzx y y z z w w x+ + + + + + + -- (2) 4(c) 81wxyz= From (2), 2 2 2 2 6x y z w x y y z z w w x+ + + + + + +

Solution However, since for Arithmetic-Geometric Progression, equality only holds if 3w x y z= = = = . However since 27, 3wx yz== , the equality condition is not met. 2 2 2 2 6x y z w x y y z z w w x + + + + + + + Solution 5(a) G1 Correct 2y= or 1y=− with endpoints G1 Correct 1y= with endpoints G1 Correct 0y= with endpoints G1 Correct points 37 , 2 and , 244    −−       indicated. 5(b) For ( )2sin nx , period = 2 n  . From graph,  ( ) ( ) 22 00 2sin d 2sin d nnx x n nx x  =       y x O 2 1 −1 −2 y x O 2 1 −1 −2

Solution ( ) ( ) 2 0 2 0 2sin d 2sin d 7 5 11 726 6 6 6 6 6 ( 1) n nx x n nx x n n n n n n n n n k            =         = + − + − −               =  ==   5(c) From graph in (a), ( ) 2 0 sin d n nx x n n    =   =  Solution 6(a)(i) Given ( ) 12f f 3xx x −= ⎯⎯⎯⎯ (1) Replace x with 1 x in equation (1) ( )132f f xxx  −= ⎯⎯⎯⎯⎯⎯ (2) 6(a)(ii) 2 ×(1) + (2): 3f ( ) 6 1f ( ) 2 , 0 3xx x x x x x =+ = +  6(b) 1g( ) g( ) gx x x x + − − =  ⎯⎯⎯⎯⎯ (1) Replace x with x− : 1g( ) g( ) gx x x x − + − − =− ⎯⎯⎯⎯ (2) (1) − (2): 11g g 2 xxx    − − =      

Solution Replace x with 1 x : 2g( ) g( )xx x− − = ⎯⎯⎯⎯⎯⎯⎯⎯ (3) Taking (3) − (1): 122g( ) gxx xx − = − ⎯⎯⎯⎯⎯⎯ (4) Replace x with 1 x : 112g g( ) 2 xxxx  − = − ⎯⎯⎯⎯⎯ (5) Taking 2×(4) + (5): 3g( ) 3x x=− 1g( )x x=− Solution 7(a) x5 + 1 = (x + 1)(x4 – x3 + x2 – x + 1) Assume that m has an odd divisor 2k + 1 > 1. Let m = (2k + 1)r for some positive integer r. Then ( ) ( ) (2 1) 21 2 (2 1) 2 2 1 2 1 21 2 1 2 2 ... 2 2 1 m k r kr r kr k r r r + + − + = + =+ = + − + + − + This contradicts the primality of 21m+ . Hence m cannot have any odd divisor. So m must be a power of 2, i.e. m = 2k , where k  ℤ. 7(b) If p divides 221 n + , then 2 2 2 1 0 (mod ) 2 1 (mod ) n n p p + − Squaring both sides, 122 1 (mod ) n p +  Let d be the smallest positive integer such that 2 1 (mod )d p .

Solution So d divides 2n+1, but does not divide 2n (since 22 1 mod n p− ). So d = 2n+1. By Fermat's Little Theorem, 12 1(mod )p p−  Since d divides p – 1, so p – 1 = k2n+1 for some integer k. Hence p = k2n+1 + 1, where k  ℤ. 7(c) Let p and q be prime factors of 221 n nF =+ . By (ii), p = r2n+1 + 1 and q = s2n+1 + 1 for some r, s  ℤ. Since (r2n+1 + 1)(s2n+1 + 1) = (rs2n+1 + r + s)2n+1 + 1 is also of the form k2n+1 + 1. Hence any factor of 221 n nF =+ , whether prime or composite, must be of the form k2n+1 + 1, where k  ℤ. 7(d) 641 = 54 + 24  54 + 24  0 (mod 641)  54 228 + 232 = 228(54 + 24)  0 (mod 641) 641 = 5(27) + 1  5(27) + 1  0 (mod 641)  54 228 – 1 = [52 214 + 1] [52 214 – 1] = [52 214 + 1] [5(27) + 1] [5(27) – 1]  0 mod 641 Subtracting, we get 232 + 1  0 mod 641 That is, 641 divides F5 = 232 + 1.

Solution 8(a) 12 1 1 2 1 ()() ( 1) ( 1) ( 1) ( 1)( ... 1) 1 ( ... 1) nn n q nn nn n n n n qx xDx qx qx qx q q q xq q q x −− − − − − −= − −= − − + + += − = + + + ( ) 1 2 1 11 1 lim ( ) lim ... 1 f '( ) n n n n qqq n D x q q x nx x − − − →→ − = + + + = = 8(b) 1e lim 1 1ln e ln lim 1 since f( )= ln is continuous, 11 lim ln 1 n n n n n n n n xx n → → → =+  =+  =+  Let 1h n= , we have ( ) ( ) 1 0 0 1 lim ln 1 ln 11 lim h h h h h h → → =+ += 8(c) 11 1 1 ln( ) lnlim (ln ) lim ( 1) ln 1lim 1 1 lnlim 1 qqq q q qx xDx qx qx x qx q xq →→ → → −= − = − = − Let 1hq=− , we have

Solution 11 0 1 lnlim (ln ) lim 1 1 ln(1 )lim 1 f '( ) qqq h qDx xq h xh x x →→ → = − += = = 8(d)(i)   f( )g( ) f( )g( )f( )g( ) ( 1) f( )g( ) f( )g( ) f( )g( ) f( )g( ) ( 1) f( )g( ) f( )g( ) f( )g( ) f( )g( ) ( 1) ( 1) g( ) g( ) f( ) f( )f( ) g( ) ( 1) ( 1) f( ) g( ) g( ) f( ) q qq qx qx x xD x x qx qx qx qx x qx x x x qx qx qx qx x qx x x x q x q x qx x qx xqx x q x q x qx D x x D x −= − − + −= − −−=+ −− −−=+ −− =+   f( )g( ) f( )g( )f( )g( ) ( 1) f( )g( ) f( )g( ) f( )g( ) f( )g( ) ( 1) f( )g( ) f( )g( ) f( )g( ) f( )g( ) ( 1) ( 1) f( ) f( ) g( ) g( )g( ) f( ) ( 1) ( 1) f( ) g( ) g( ) f( ) q qq qx qx x xD x x qx qx qx x qx x qx x x qx qx qx x qx x qx x x q x q x qx x qx xqx x q x q x x D x qx D x −= − − + −=