TMJC 2025 H3 Mathematics Prelim Solutions

TMJC/2025 JC2 Preliminary Examination Suggested Solutions/H3 Math (9820/01)/Math Dept Page 1 of 12 2025 H3 MATH (9820) JC 2 PRELIMINARY EXAMINATION – SUGGESTED SOLUTIONS Qn Solution 1 Inequalities (i) Using Cauchy-Schwarz Inequality, ( )( ) 22 2 2 2 1 1 1 1 1 11 n n n n n i i i i i i i i i x x x n x = = = = =         =  =                     Hence, 2 2 11 1nn ii ii xx n==   (shown) (ii) Using AM-GM inequality, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 1 1 1 2 ( 1) 1 1 2 1 2 1 1 1 1 1 1 1 1 shown n in i n n nn n n a a a a na na na −− = + + + − − − + = + + + + + + + +  =+  =+  Equality holds iff ( ) ( ) ( ) 21 1 1 1 1 n a a a − = + = + = = + Therefore, a = 0 (iii) From part (i), let ( ) 1 1 i ixa − =+ Therefore, ( ) ( ) 2211 11 111 nn ii ii aa n −− ==  +  +   From part (ii), since ( ) ( ) 11 2 1 11 n ni i a n a −− = +  + ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2121 2 1 2 2 1 1 21 2 1 21 1 111 11 1 11 1 1 1 (shown) n ni i n in i n in i n in i a n a n a n a a n a a a n a −− = −− = − = + =  +  +   +  + +  + + +  +    

TMJC/2025 JC2 Preliminary Examination Suggested Solutions/H3 Math (9820/01)/Math Dept Page 2 of 12 Qn Solution 2 Number Theory (i) Let 2nk= where k + ( ) ( ) 442 44 4 4 1 4 2 4 16 2 2 8 2 2 nk k k nk k k m − + = + =+ =+ = Where 4 4 182 kmk −+= +  Therefore, 4 4nn + is even. (ii) ( ) ( )( ) ( ) 4 4 2 2 2 2 4 2 2 3 2 4 3 3 3 2 2 2 4 2 4 2 2 2 4 2 2 x y x ay bxy x ay bxy x ax y bx y a y abxy bx y abxy b x y x a y a b x y + = + + + − = + − + − + + − = + + − 2 42aa=  = (since 0a ) 22202a b b a− =  = Since 2a= , 2 42bb=  = (since 0b ) (iii) Case 1: When n is even, from part (i), 4 4nn + is even. Since 1n , 4 42nn + Hence, 4 4nn + is not prime Case 2: When n is odd, Let 21np=+ where p + ( ) ( ) ( )( ) ( ) ( ) 44 2 1 24 2 44 4 2 1 4 2 1 4 2 2 1 4 2 np p p np p p ++ = + + = + + = + + Using the expression for (ii), ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )( ) 44 2222 22 2 1 1 2 1 1 2 1 4 2 2 1 2 2 2 2 1 2 2 1 2 2 2 2 1 2 2 1 2 2 1 2 2 1 2 2 1 2 p p p p p p p p p p p p p p p p p p + + + + ++   = + + + + + + − +       = + + + + + + − +   Since p + , ( ) ( )( ) 2 2 1 12 1 2 2 1 2 1pppp +++ + + +  and

TMJC/2025 JC2 Preliminary Examination Suggested Solutions/H3 Math (9820/01)/Math Dept Page 3 of 12 ( ) ( )( ) ( ) ( ) ( ) ( ) ( ) 2 2 1 1 22 21 2 22 2 2 2 1 2 2 1 2 2 1 2 2 2 2 1 2 2 2 2 2 1 2 2 1 pp p p p p p p pp pp p p p ++ + + + − + = + − − + = + − − + = + − +  Since 4 4nn + can be written as a product of two integers more than 1, 4 4nn + is not prime when n is odd. Therefore, for any integer 1n , 4 4nn + is not prime.

TMJC/2025 JC2 Preliminary Examination Suggested Solutions/H3 Math (9820/01)/Math Dept Page 4 of 12 Qn Solution 3 Counting and Probability (a)(i) By the Bijection Principle, it can be viewed as the set of 20-digit binary sequence with five ‘1’s. Number of ways of distributing 15 identical marbles among 6 children is 15 6 1 20 15504.55 +−    ==       (ii) To ensure that each child receives at least one marble, we give each child one marble first. We are then left with 15 6 9−= marbles to be distributed among 6 children. Number of ways of distributing 15 identical marbles among 6 children such that each child receives at least one marble is 9 6 1 14 2002.55 +−    ==       (iii) To ensure that Tom, one of the children, receives exactly two marbles, we distribute the rest of the marbles among the 5 children. ( )P 13 5 1 17 44 15504 15504 2380 15504 35 228 Tom receives exactly two marbles +−          == = = (b)(i) Number of ways of distributing k distinct marbles among 6 children is 6.k (ii) Let iA denote the event that child i receive no marbles 1,2,3,4,5,6.i= ( ) 6 1 6 61 1 2 6 1 1 6 1 6 number of ways with at least one child not receiving any marbles ... 1 ... 6 6 6 6 65 4 3 2 1 11 2 3 4 5 ii i i j i j k i i j i j k k k k k k N A A A A A A A A A A = − =        = = = −  +   − + −             =  −  +  −  +  −                       0

TMJC/2025 JC2 Preliminary Examination Suggested Solutions/H3 Math (9820/01)/Math Dept Page 5 of 12 Hence required number of ways is ( ) ( ) 6 1 6 1 6 0 6 6 6 6 66 5 4 3 2 1 1 01 2 3 4 5 616 i i ii k k k k k k ik i A SA ii = = = =−          = −  +  −  +  −  +                    = − −  .

TMJC/2025 JC2 Preliminary Examination Suggested Solutions/H3 Math (9820/01)/Math Dept Page 6 of 12 Qn Solution 4 Functions/Calculus (a)(i) ( )( ) 12 12 12 12 12 12 1 11 2 11 for 3. nn nn nn nn nn nn xx nn xx xx xx yy yy yx n −− −− −− −− −− −− + + −− + ++ = − = − = = (ii) Let P( )n be the statement 01 ny for 1.n Since 1 12 201 yy = =  , (1) and (2)PP are true. Assume that ( 2), ( 1)P k P k−− are true for some , 3.kk + 12 12 2 kk kk yy k yyy −− −−++= and since all the terms are positive on the RHS (by induction hypothesis), we have also 0.ky  Also, 12 12 2 1(1) 1 10 0 2 2 kk kk yy k yyy −− −−++=  =  ++ So, ()Pk is true. Since (1) and (2)PP are true, by mathematical induction, ()Pn is true for all 1.n (iii) 12 12 1 1122 1 0 0 2 nn nn yy n n n yyy y y −− −− −−++=   = ++ since part (ii) gives 01 ny for 1.n (iv) We have 2 1 2 21 2 1 1 1 1 2 3 1 2 2 2 2 24 2 1 1 1 1 1 1 2 2 4 8 2 12 ... ... 3 ... 2 n n ny y y y y y y y y − − − + + + +  + + + + + = + + + + +  + = Since all terms in the series are positive, and the sequence of partial sums is bounded, the series converges. Alternative method: 1 1 2 3 2 1 1... lim 2 nN n N n y y y y → = + + + +  +  and since the limit on the RHS exists as it is a sum of Geometric series with ratio 0.5<1, by comparison test, the series on the LHS converges. (b)(i) ( ) ( ) ( ) ( ) ( )   44 1 00 4 0 4 0 4 0 sin 2 1 sin 2 1 d dsin sin sin 2 1 sin 2 1 d sin 2cos 2 sin d sin 2 1 1 sin 2 sin22 nn n x n xI I x x xx n x n x xx nx x xx nx nnn      − +−− = − + − −= = ==    

TMJC/2025 JC2 Preliminary Examination Suggested Solutions/H3 Math (9820/01)/Math Dept Page 7 of 12 (ii) ( )4 50 sin 11 dsin x xIx  = Using the relation in part (i), 54 43 32 21 10 15sin52 14sin42 13sin32 12sin22 11sin12 II II II II II      −=  −=  −=  −=  −=  Using MOD, ( ) ( ) ( ) ( ) ( ) 50 5 1 5 1 4 1 3 1 2 1 1sin sin sin sin sin5 2 4 2 3 2 2 2 1 2 1 1 1 1 11 0 1 0 15 4 3 2 1 4 13 15 4 II I                 − = + + + +                   = + + − + + + =+

TMJC/2025 JC2 Preliminary Examination Suggested Solutions/H3 Math (9820/01)/Math Dept Page 8 of 12 Qn Solution 5 Sequences and Series / Calculus (a) (i) Define 1n nnb a p −+=− for n + ( ) ( ) 1 11 1 1 1 1 since 2 0 since 2 nn n n n n nn nn n n n n nn n b b a p a p a a p p p p p a p a p p p − − + ++ − − + + − − − + − + − − = − − − = − − + − − + +  = −   1nnb b n + +        1,2,... 1 1,2,... is strictly increasing is strictly increasing n n n n n b ap = −+ =  − (ii) By triangle inequality, 11 1 nnn nna p a p M− + − + +−  +  +   since 1np−+ is strictly decreasing and 1 1 1 1 np p n− + − + + =   . Thus   1,2,...n nb = is bounded above by 1.M + Since   1 1,2,... n n n ap −+ = − is strictly increasing and bounded above, it converges. That is, there exists a finite number l such that