CJC-SAJC-JPJC 2025 H3 Mathematics Prelim Solutions

1 9820/01/PRELIM/2025 [Turn over Q1 Functions and Graphs (a) [3m] 1 2 0 11 1122 0 0 1 d(1 2 ) d (1 2 ) d d(1 2 ) 2 (1 2 ) d 2 (1 2 ) d 12 n x n x n n x n x n x n vI x e x u x e x ux e n x e x n x v e x nI −− − = − = − = = − + − =− − = =− +   (b)(i) [3m] 1 02 0 0 1 2 0 1 2 0 1 2 (1 2 ) d d 1 x x x I x e x ex e e =− = = =−   32 1 1 0 0 1 2 1 2 16 1 6( 1 4 ) 7 24 7 24( 1 2 ) 31 48 31 48 1 79 48 II I I I I e e =− + =− + − + =− + =− + − + =− + =− + −   =− + (b)(ii) [2m] 1 42 0 4 3 1 2 1 2 (1 2 ) d 18 1 8 79 48 633 384 xx e x I I e e − = =− + =− + − +  =− + 

2 9820/01/PRELIM/2025 [Turn over (c) [3m] 1 2 0 (1 2 ) dnx nI x e x=− 1For 0, , 0 1 2 1.2 As , (1 2 ) 0. n xx nx   −  → − → Since is continuous and bounded for 10, ,2x  1 2 0 As ,(1 2 ) 0, (1 2 ) d 0. nx nx n n x e I x e x → − →  = − → (d) [3m] 1 1 01 2 2 3 3 Given 1 2 , 1 ( 1)2 1 ( 1)2 11 ( 1) 124 1 1 1 1 1 2 4 2 4 2 1 1 1 1 1 1( 1)2 4 6 2 4 2 1 1 1 1 1 1 2 4 6 2 4 6 nn nn I nI II n II I I I I − − =− + =+ =+ = + +      = + +                 = + + +                  =+             1 1 1 2 4 2 1 1 1lim ... 2 4 2 1 1 1 1 1 1lim ... ...2 4 2 2 4 2 1As , 0, 0 2 1 1 1 1 1 1 ...2 2 4 2 4 6 nn n n In n nI n → →    ++             =                 + + + +               → → →        + + + =             1 2 1e − xe

3 9820/01/PRELIM/2025 [Turn over Q2 Numbers and Proofs (a)(i) [3m] 2 2 0 2(1 ) m mk k mxx k= +=  Let 1x= 2 2 0 2 2 22 2 2 2 2 22 ... 0 1 1 2 22since each term is positive, 2 m m k m m m k m m m m m m m m mm km = =           = + + + + + +          +                      (a)(ii) [3m]    2 (2 )(2 1) ( 1)( )( 1) (2)(1) ( )( 1) (2)(1) ( )( 1) (2)(1) 2 (2 )(2 1) ( 1) ( )( 1) (2)(1) m m m m m m m m m m m m m m m m mm  − + −= −−  −+= − This is an integer which implies all factors of m! cancel with the factors in the numerator. We note that 1,2Pmm+ is part of the numerator. Therefore 1,2Pmm+ divides 2 .m m   (b)(i) [3m] By the Cauchy-Schwarz Inequality, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 a b c b c c a a b a b c b c c a a b a b c b c c a a b a b cb c c a a b a b c a b c a b cb c c a a b a b c abcb c c a a b   + + + + + + +  + +  + + +  + + + + + + +  + + + + +  + +  + +  + + + + +  + +  + + + + +

4 9820/01/PRELIM/2025 [Turn over (b)(ii) [5m] 2 2 2 22 2 22 2 3 ( ) ( ) ( ) 11 1 ( ) ( ) ( ) 11 1 1 1 1 1 1 1 1 1 1 1 using result from (b)(i)2 113 using AM2 yz zx xy x y z y z x z x y yx z y z z x x y yz zx xy yx z y z z x x y x y z xyz +++ + + = + ++ + + = + + + + +  + +  -GM Inequality 3 since 84 xyz== Therefore the minimum value is 3 4 . It occurs when 1 1 1 1 or 22 x y zx y z= = = = = =

5 9820/01/PRELIM/2025 [Turn over Q3 Sequences and Series (a) [4m] Let nP be the statement ( ) 12 21 1 n n n nf f f + ++ − = − for 1, nn . Base case: 1n= ( ) ( )( ) ( ) ( ) ( ) 222 2 2 2 1 3 2 1 2 1 2 1 2 1 2 11 L.H.S. 1 1 1 1 1 R.H.S. f f f f f f f f f f f + = − = + − = + − = + − =− = 1P is true. Inductive Step Assume that kP is true for some 1k , i.e., ( ) 12 21 1 k k k kf f f + ++ − = − . To prove that 1kP + is true, i.e., ( ) 22 1 3 2 1 k k k kf f f + + + + − = − . ( ) ( ) ( ) 2 12 2 12 22 1 2 1 2 2 12 2 1 21 3 2 2 21 L.H.S. [by definition of Fibonacci numbers] [by definition of Fibonacci numbers] k kk kk k k k k kk k k k k k k k k kk f f f ff f ff f ff f f ff f ff f f f ++ ++ ++ + + + + ++ ++ ++ + + + =− =− = + − = =− = − − + − −( ) ( )( ) ( ) 2 1 1 2 1 [by inductive hypo thesis] 1 R.H.S. 1 k k+ + =− − =− = 1 true truekkPP + By the Principle of Mathematical Induction, nP is true for 1, nn .

6 9820/01/PRELIM/2025 [Turn over (b) [4m] Given that 1n n n fr f += . As the terms in the Fibonacci numbers nf , are positive, 1nr  As the Fibonacci sequence nf is increasing, 1 1 1 12n n n n n n n f f f f f f f + − − += = +  Since 12 nr , it is a bounded sequence. Hence subsequences 2 4 6 8, , , , ...r r r r and 1 3 5 7, , , , ...r r r r would naturally be bounded as well Consider ( ) ( ) ( ) 2 1 2 3 2 2 2 2 2 2 2 1 2 2 2 2 2 1 2 2 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 1 21 2 23 22 22 21 [by definition of Fibonacci numbers] [using the identity from1 nn nn n nn nn nn nn nn n n n nn nn n n n n n n ffrr ff ff ff ff ff f f f ff f f ff f f f f ++ + + + + + + + + ++ + + + + + + − = − −= −= −= + + −= − 2 2 2 part (a)] 1= 0 nnff +  2 2 2nnrr + The sequence 2 4 6 8, , , , ...r r r r is decreasing. Consider ( ) ( ) ( ) 2 2 2 2 1 2 1 2 1 2 1 2 1 2 21 2 21 2 1 2 2 1 2 1 2 1 2 1 2 1 2 22 2 1 1 1 2 2 2 1 2 1 2 2 12 2 [by definition of Fibonacci numbers] [using the identity1 nn nn nn nn n nn nn nn n n n n n n n n n n n n n ffrr ff ff ff f f f f ff f f f f f ff ff ff + +− +− − +− − +− − +− − + − + + + + − = − − − − = = + = = − + 2 1 2 1 from part (a)] 1= 0 nnff +−  2 1 2 1nnrr+− The sequence 1 3 5 7, , , , ...r r r r is increasing.

7 9820/01/PRELIM/2025 [Turn over (c) [6m] Since 2 4 6 8, , , , ...r r r r and 1 3 5 7, , , , ...r r r r are bounded and monotone, By the Monotone Convergence Theorem, 2 4 6 8, , , , ...r r r r and 1 3 5 7, , , , ...r r r r are convergent. Consider ( ) 2 1 2 2 2 1 2 2 1 2 2 1 2 1 2 2 2 1 2 2 1 2 2 1 2 [using the identity from part (a)] = 1 1 n nn nn nn n n n nn nn nn ffrr ff f f f ff ff ff + − − +− − − − −= = = − − − Since nf is an unbounded and increasing sequence, when the denominator of the preceding fraction is arbitrarily large and consequently 2 2 1nnrr −− is arbitrarily small. Proof by contradiction: Let ( )2lim nen rL → = and ( )2 1 0lim nn rL−→ = , where eoLL and eoLL =− . Since ( )2lim nen rL → = , there exists an 1N  such that 21 4 ner L n N −    . Likewise, ( )2 1 0lim nn rL−→ = , there exists an 2N  such that 2 1 2 4 nor L n N  − −    . There exists an 3N  such that 2 2 1 3 4 nnr r n N  −−    . Let  1 2 3max , ,N N N N= and pick any nN . Then by the triangle inequality, 0 2 2 2 1 2 1 3 4 4 4 4 e e n n n n oL L L r r r r L     −−= −  − + − + −  + + = . But this implies 3 4   , which is a clear contradiction. Therefore, eoLL= . Since eoLL= , we conclude that sequence nr converges to the same limit Now 1 1 1 1 111n n n n n n n n n f f f fr f f f r + − − − += = = + = + As n→ , nrL→ and 1nrL− → ,

8 9820/01/PRELIM/2025 [Turn over 2 11 10 1 5 1 5 or (rejected 0 )22 n L L LL L L f n =+ − − = +−= =    Q4 Counting / Inequalities (a) [1m] Number of 2-element subsets of S = 500 1247502  = (b) [4m] Case 1: Both x and y are multiples of 3 Number of multiples of 3 = 500 1663  = Thus, the number of 2-element subsets  ,xy of S = 166 2   . Case 2: Exactly one of x and y is a multiple of 3 Number of ways of choosing a multiple of 3 = 166 Number of ways of choosing a non-multiple of 3 = 500-166 = 334 Thus, the number of 2-element subsets  ,xy of S = 166 334= 55444 By addition principle, the number of 2-element subsets  ,xy of S = 166 2   + 166 334 = 69139 (c) [5m] Case 1: Both x and y are multiples of 3. From (ii), the number of 2-element subsets  ,xy of S = 166 2   Case 2: One of x and y leaves a rema