RI 2022 H3 Test 3(Questions and Solutions)

RI 2022 RAFFLES INSTITUTION 2022 Year 6 H3 9820 Lecture Test 3 Time allocated: 2 hours Total Marks: 60 Instructions: Write your name and CT group on all the work you hand in. Answer all questions. 1 A clothes shop sells a particular make of T-shirt in four different colours. The shopkeeper has a large number of T-shirts of each colour. (a) A customer wishes to buy eight T-shirts. (i) In how many ways can he do this? [2] (ii) In how many ways can he do this if he buys at least one of each colour? [2] (b) The shopkeeper places eight T-shirts in a line. (i) In how many ways can she do this? [1] (ii) In how many ways can she do this if no two T-shirts of the same colour are to be next to each other? [2] (iii) Use the principle of inclusion and exclusion to find the number of ways in which she can do this if she has to use at least one T -shirt of each colour but with no other restriction. [4] (c) The shopkeeper is left with r T-shirts of one colour, where 4r and another 2 T-shirts of another colour. She wishes to store them in 2 identical boxes so that no box is empty. In how many ways can she do this? [4] 1 (a) (i) [2] Let 1 2 3 4, , ,x x x x be the respective number of T -shirts the customer buys in the 4 different colours. Then the problem is equivalent to the number of integer solutions to 1 2 3 4 8x x x x+ + + = , with 1 2 3 4, , , 0x x x x  . (Identical objects into distinct boxes) The number of ways is thus 11 165.3  =

RI 2022 (a)(ii) [2] Let 1 2 3 4, , ,x x x x be the respective number of T -shirts the customer buys in the 4 different colours. Then the problem is equivalent to 1 2 3 4 8x x x x+ + + = , with 1 2 3 4, , , 1x x x x  . Let 1iiyx=− (i.e. ensure he has one of each different col our). Then we need to find the number of integer solutions to 1 2 3 4 4y y y y+ + + = with 1 2 3 4, , , 0y y y y  . The number of ways is thus 7 353  = . (b)(i) [1] The number of ways is 84 65536= . (b)(ii) [2] There are 4 ways to choose the first shirt, and subsequently, 3 ways for the next shirt so that no two shirts are of the same colour. The number of ways is thus 74(3) 8748= . (b)(iii) [4] Let iA denote the event in each colour i is not used. Then required answer = __ __ __ __ 8 1 2 3 4 1 2 3 4 4A A A A A A A A   = −    We have 1 2 3 4 1 2 3 4A A A A A A A A   = + + + 1 2 1 3 3 4A A A A A A−  −  − −  1 2 3 1 2 4 2 3 4A A A A A A A A A+   +   +   1 2 3 4A A A A−    ( ) 8884 4(3) 6 2 4 40824= − + − = .

RI 2022 (c) [4] Consider the 2 T-shirts of one colour first. They can be split into the 2 boxes in the following manner: (0, 2) or (1, 1). Let the number of T-shirts contained in each of the two boxes be a and b respectively, where a b r+= . Case 1: 1 and 1. Here the 2 boxes are essentially still identical. If r is even, since ( )1 22 ra a b + = , a can only take a value from 0, 1, 2, …, 2 r when r is even. There are 12 r + ways. 2 r is not an integer when r is odd. So 1 22 rraa −   . Similarly, there are 11 122 rr−++= ways if r is odd. Case 2: 0 and 2. Here the 2 boxes are now distinct. Let Box A be the one with 0 of the other, and Box B be the one with 2. Box B can contain 0, 1, 2, …, r – 1 (cannot contain all r or else Box A is empty) of the shirts. So there are r ways. Hence in total, if r is even, there are 32122 rr r ++ + = ways. If r is odd, there are 1 3 1 22 rr r+++= ways.

RI 2022 2 Let ( ) sin d . 1 2 sin n x nxIx x  − = + (a) Show that 0 sin d .sin n nxIx x  = [4] (b) Hence show that for 2n , 2nnII −= . [2] (c) Hence evaluate nI for all nonnegative integers n. [3] 2 (a) [4] ( ) ( ) ( ) 0 0 sin sin sin d d d 1 2 sin 1 2 sin 1 2 sin n x x x nx nx nxI x x x x x x  −− = = + + + +   Using the substitution tx=− in the first integral, we get ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 0 0 00 00 0 sin( ) sin ( d ) d 1 2 sin( ) 1 2 sin sin( ) sin d d (since sin is an odd function) 1 2 sin( ) 1 2 sin 2 sin( ) sin d d 1 2 sin( ) 1 2 sin 1 2 sin d 1 2 sin n tx tx t tx x x nt nxI t x tx nt nx tx tx nt nx tx tx nx x x      − − −= − + + − + =+ ++ =+ ++ + = +    0 sin dsin nx xx  =   (b) [2]   2 0 0 0 20 sin sin( 2) dsin 2cos( 1) sin dsin 2cos( 1) d 2 sin( 1) 01 nn nn nx n xI I x x n x x xx n x x n x I In     − − −−−= −= =− = − =  =−    (c) [3] Thus we split into the odd and even n. We have 1 0 sin dsin xIx x  == and 0 0I = . Hence if is odd 0 if is even n nI n =  .

RI 2022 3 Let 0 1 2, , ,...b b b be a sequence of positive real numbers such that 0 1b = , 112 2 1 .n n nb b b −−= + − + Let 1nnab=+ for 0n . (a) Show that 1nnaa −= for 1n . [2] (b) Hence express na in terms of n. [2] (c) Show that ( ) ( ) 11 0 1 2 1 2 1 2 . N nN nN n b a a + = = − − − [3] (d) Use a sketch to explain why 0 21lim ln 2 x x x→ − = . [2] (e) Hence calculate 1 2n n n b  =  . [2] 3 (a) [2] Note that 11nnab= +  . We also have ( ) ( ) 11 11 2 1 1 1 1 1 1 1 2 1 1 1 2 11 11 1 1 since 1 n n n nn n n nn n a b b aa a a aa a −− −− − − − − = + + + − + = + + − = + − = + − = + −  =

RI 2022 (b) [2] Hence ( ) ( ) ( ) ( ) 2 1 21 1 1 2 22 1 22 1 220 = ... 2 n n nn n n aa a a a − − − − =   = = == (c) [3] ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) ( ) 2 11 21 1 1 1 1 1 11 12 11 0 2 1 2 2 2 2 1 2 1 2 1 2 1 2 1 2 1 2 . NN nn nn nn N n n n nn n N nn nn n NN nn nn nn N N ba aa aa aa aa == + = + − = + −− == + =− = − + = − − − = − − − = − − −     (d) [2] Consider the function f ( ) 2 xx = and the gradient of the function at 0x= .We see that 00 2 1 f ( ) f (0)lim lim f (0) ln 2 x xx x xx→→ −− = = = . (e) [2] Hence ( ) ( ) 11 11 0 2 0 2 lim 2 lim 1 2 1 2 212 2lim 2 212 2lim 2 2ln 2. N N nn nn nnn N Nn Nn x x bb aa x −  →== + → −→ → = = − − −  −=−   −= − = −  

RI 2022 4 Let 1, 2,..., .nSn= For any subset X of nS , define the capacity of X , ()cX to be the sum of all the elements of X . If the capacity of X is odd, we say that X is an odd subset of nS and similarly, if X is even, we say that X is an even subset of nS . For example, if 3n= , then 3( ) 1 2 3 6cS = + + = , ({ 1,2}) 1 2 3c = + = , ({ }) 0c = and thus 3S is an even subset of 3S , and  1, 2 is an odd subset of 3S and the empty set is an even subset of 3S . Let A be the set of all subsets of nS that do not contain 1, and B be the set of all subsets of nS that contains 1. For example, if 3n= ,         , 2 , 3 , 2,3A= and        1 , 1, 2 , 1,3 , 1, 2,3B= . (a) Show using a bijection between A and B that the number of odd subsets and even subsets of nS are the same. [3] (b) If 3n , show that the number of odd subsets and even subsets in A are also the same. [2] (c) Hence show that if 3n , the sum of all capacities of odd subsets of nS is equal to the sum of all capacities of even subsets of nS . [2] (d) Determine the sum of all capacities of odd subsets. [3] (a) [3] Consider the map f: AB→ , defined by f ( ) 1XX= . If  f ( ) f ( ) 1 1X Y X Y X Y=   =   = thus f is injective. For any set YB , removing 1 from it clearly gives an element of A. Hence f is bijective. Hence if the subset X is odd (resp. even), then 1X  is even (resp. odd). Hence the number of odd s