RI 2025 H3 Mathematics Prelim Solutions

Question 1 1(a) [2] Replace with 1 ,xx − ( ) 2f ( ) f (1 ) solving, we get 1 f ( ) (1 )f (1 ) f ( ) 1 x k x x k x k x xx k x x − − =  − = − +− − = −  Hence 2 (1 )f ( ) 1 k k xx k +−= − . (b) [2] From f ( ) f (1 ) g( ) (1)x x x− − = −−−− Replacing x with 1 – x we get f (1 ) f ( ) g(1 ) (2)x x x− − = − −−− (1)+(2) gives f ( ) f (1 ) f (1 ) f ( ) 0 g( ) g(1 )x x x x x x− − + − − = = + − . Hence g( ) g(1 ) 0xx+ − = is a necessary condition for there to be a solution. (c) [2] 1 1 1 0 1 1 1 0 Let g( ) with 0 Then g(1 ) (1 ) (1 ) (1 ) kk k k k kk kk x a x a x a x a a x a x a x a x a − − − − = + + + +  − = − + − + + − + Hence, the coefficient of kx in g( ) g(1 )xx+− is ( 1) k kkaa+− . From (b), since there is a solution f to () , ( 1) 0,k kkaa+ − = which implies that k must be odd, else we have 0, a contradiction.ka = (d) [3] Let g( ) and so g(1 ) (1 )x mx c x m x c= + − = − + We have (1 ) 2 0 20 mx m x c mc + − + = += Let 2, 1,mc= =− so g( ) 2 1xx=− is a possible function. Consider f ( ) f (1 ) 2 1.x x x− − = − We will show that there is a solution f to the equation above. In fact, it suffices to let f ( ) .xx= (e) [2] Method 1: 3 3 2g( ) (2 1) 8 12 6 1x x x x x= − = − + − 32 32 3 3 Let f ( ) , f (1 ) (1 ) (1 ) (1 ) Since f ( ) f (1 ) g( ), Compare coefficient of , 4 Compare coefficient of , 3 Compare constant term, 3 Let 0, 3, a solution is f ( ) 4 x ax bx cx d x a x b x c x d x x x xa x b c bc b c x x = + + + − = − + − + − + − − = = + =− + =− = =− = 3.x−

Method 2: Since g( ) g(1 ) 0xx+ − = and f ( ) f (1 ) g( ). ( )x x x− − =  ( )  ( ) ( ) 3 g( )If f ( ) , 2 g( ) g(1 )f ( ) f (1 ) 22 1 g( ) g(1 )2 1 g( ) g( ) g( ), satisfying .2 g( ) (2 1)Hence f ( ) is a solution.22 xx xxxx xx x x x xxx = −− − = − = − − = − − =  −== Method 3: Since g( ) g(1 ) 0xx+ − = and f ( ) f (1 ) g( ). ( )x x x− − =  ( ) ( )  ( ) 3 If f ( ) g( ), f ( ) f (1 ) g( ) 1 g(1 ) g( ) 1 g( ) g( ) g( ) g( ) g( ), satisfying . Hence f ( ) g( ) (2 1) is a solution. x x x x x x x x x x x x x x x x x x x x x x x x = − − = − − − = − − − = − + =  = = −

Question 2 2(a) [2] ( )( ) 32 2 1 1 1 0 Since 1, 1 0 (shown). z z z z z z z − = − + + =  + + = 1 3 1 3Solving, i 2 2 2z −  −= =−  . (b) [2] MC is obtained by rotating MB by 2  rad in anticlockwise direction, followed by a scaling with factor 3 . Hence, ( )i 3.c m b m− = − Thus ( )3ik = . (c) [2] ( ) ( ) i322 i3 22 i3 2 2 2 13 i22 (shown) a b a bcb b a a bc b a a bb b b a b b a ++ − = −  −+=+ −= + + − = + − − + = + − (d) [2] From (c), (1 ) 0a b c− + + = . Since 210+ + = we have 2 0a b c+ + = . Multiplying by 2 we get 3 4 2 0a b c  + + = and now using the fact 3 1 = we have the desired result 2 0a b c+ + = . (e) [4] We first note that if 2 0a b c+ + = then reversing the calculations in (d), we get the result in (c) and (b) which means that ABC is an equilateral triangle. Since XQP, YRQ and ZPR are equilateral, we have 2 0x q p+ + = , 2 0y r q+ + = and 2 0z p r+ + = . From the result given about centroids, we have 1 2 3 , , .3 3 3 x q p y r q z p rg g g + + + + + += = = Hence

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 1 2 3 2 2 2 2 2 2 3 4 2 3 2 2 2 3 3 3 1 3 1 3 1 3 0 g g g x q p y r q z p r x q p q y r p r z x q p q y r p r z x q p q y r p r z                       ++ + + + + + +   = + +        = + + + + + + + + = + + + + + + + + = + + + + + + + + = This thus implies that triangle 1 2 3G G G is equilateral. This result is also known as Napoleon’s Theorem.

Question 3 (a) [3] , 210 1 210 12 22 0 0 12 22 0 1 2, 12 ( ) d (1 ) () d(1 ) ( 1) ( 1) ( 1) d2 (1 ) 2 (1 ) 1 d2 (1 ) 2 (1 ) 1 ()2 (1 ) 2 mx mn n mx n x xmm nn mm x nn m mnn tI x t t tt tt t m t tn t n t x m t tn x n t xm Ixn x n + − + −− −− − −− = + = +  −−= − − ++ −=− + ++ −=− + +     (b) [1] We need to have 12mn− . Since m and n are integers, this means that 2mn . (c) (i) [3] We have ( ) 0,1 20 2 1lim ( ) lim d 1 x xx I x t t → → = + Using the substitution tant = , ( ) ( ) 1 1 1 20 2 tan 2 20 2 tan 2 0 tan 0 1 d 1 1 sec d 1 tan cos d cos 2 1 d2 x x x x t t      − − − + = + = +=     As x→ , 1tan 2x − → . Therefore, ( ) 2 20 2 0 1 1 sin 2lim d . 2 2 41 x x t t    → = + =+ (c) (ii) [4] The curve is symmetric about the y-axis and thus the required required exact volume is given by

4 240 4,3 2,2 0,1 2d (1 ) 2 lim ( ) 32 lim ( ) (since 1 2 ,3 6)6 312 lim ( ) (since 1 2 ,1 4)64 x x x t tt Ix I x m n I x m n      → → → + = = −   = −    Altogether, we have 42 240 12 d 2 .(1 ) 8 4 16 t tt     ==   +   

Question 4 4 (a) (i) [2] Consider a class of n students, from which we wish to choose a class committee of i students. On the one hand, this is clearly n i   . On the other hand, consider a particular student A. If student A is chosen, then we need to choose another i – 1 students from the remaining n – 1, of which there are 1 1 n i −  − ways to do so. If student A is not chosen, then we need to choose all i from the remaining n – 1, and there are 1n i −  ways to do so. By the addition principle, we have 11 1 n n n i i i −−     =+      −      . 4 (a) (ii) [4] Let nP be the statement 1 ( 1)01 nn m n m n m n m n k k n k k n − − −          − + + − =           −           . When 1n= , LHS = 1 1 1 1 1 0 1 1 m m m m m k k k k k − − −           − = − = =            −            RHS. where we have used the result in (a), with m = n. Hence 1P is true. Assume (*) holds for some positive integer n. ( )11 1 1 1 1( 1)0 1 1 12 0 0 1 1 2 ( 1) 1 n n n m n m n mn k k n k n m n n m n n m k k k nn nn ++ + − + −+       − + + −         +         −−                 = − + + + +                                      + − +   −    1 1( 1) nm n n m n k n k +− − −    +−         1 1 ( 1)01 1 2 1 ( 1)01 n n n m n m n m n k k n k n m n m n m n k k n k + −−        = − + + −                 − − − −        − + + + −                

11 .1 m n m n m n k n k n k n − − − − −     = − =     − − − −      Since 1P is true, 1nnPP + , by mathematical induction, (*) is true for all positive integers n. (b) [4] Context: Amongst m players, of which n are star players, choose a team containing k players and all star players must be included within these k players. Clearly, since all n star players are chosen for the team, we need to choose remaining kn− players from the mn− players, and there are mn kn −  − ways, which is the RHS of (*). For LHS of (*), let the star players be numbered 1 to n and let iA denote the sets of teams of k players without star player i. Hence, we need 12 nA A A   so that every star player is included. Without considering whether we include star players, we have a total of m k   ways choosing k players from m players. For the ways without any one of the star players, we have 1 1 1 n i i nmA k= −  =     and for the ways without any two of the star players, we have 2 2 ii ij nmAA k −  =      . Similarly for the rest until we have for the ways without any of the star players, which is 12 n n m nA A A nk −     =      . By principle of inclusion and exclusion, we have 12 12... ( 1) n n i i j n A A A S A A A A A A    = − +  − + −    1 ( 1)01 nn m n m n m n k k n k −−