DHS EJC RVHS 2026 H3 Math Prelim (Ans)

2026 Year 6 H3 Mathematics Preliminary Examination Solutions 1 Solution [10] Integration, Graphs and Functions (a) Let ( ) 2,9aa be a point on 29yx= . Area bounded by 216yx= , 29yx= and 29ya= 2 2 2 9 0 9 0 93 2 0 3 d34 1 d12 12 12 3 3 2 a a a yy y yy y a  =−  =  =  =   Area bounded by 29yx= , 24.5yx= and xa= 3 22 0 2 0 33 0 99d 2 9 d2 93 2 3 2 a a a x x x xx xa =−  = ==    So both areas are always equal (shown). y x

(b) Let ( )g x k x= and ( ),a k a be a point on ( )gyx= . 22 2216 ; 16yyy x x y k x x k=  = =  = Area bounded by 16yx= , y k x= and y k a= ( ) 22 220 2 22 0 3 22 0 3 3 2 22 d16 11 d16 11 16 3 11 16 3 ka ka ka yy yk yyk y k ka k =−  =− =−    =−     Area bounded by y k x= , 5yx= and xa= ( ) ( ) ( ) ( ) 0 0 3 2 0 3 2 5d 5d 25 3 25 3 a a a k x x x k x x xk ak =− =−  =−    =−    Equating the areas: ( ) 3 22 3 2 3 11 2516 2 10 016 256 2560 0 kkk kkk kk  − = − − − + = +−= By GC, ( )88k g x x=  = [Clearly, 5 8 16x x x and ( )g 0 0= so we have verified both conditions of g.] (c) Notice that h is the inverse of f, therefore g( )xx= .

2 Solutions [12] Differential Equations, Differentiation, Inequalities, Functions 2(a) d du dd y ux y uxxx = =+ ( ) ( ) ( ) 2 2 2 23du d 3 2 du 2 3 d 3 2 du 2 3 d 3 2 du 2 2 d 3 2 1 d 3 2 du 21 3 1d 2 du 1 x uxux x x ux uux xu uxu xu ux xu xu x u ux x u ++= + ++= + +=− + −= + += −  += − ( )( ) 22 2 3 14 2 1 4 5 4 1 4 5 4 1 4 5 4 3 1 1 2 2 d d 1 2 1 3 1 1ln ln ln 14 1 2 1ln ln ln 1 11 1 1 1 1 uxux u u xu uku xu uuku ux k u y x x k y x xy x x k xy x −=− −− += − − − += − + −− += − + = − + = −    

1 4 5 4 51 44 5 , where C is an arbitrary constant xyx kx xy x y k x y x y C x y += − − = + − = + 2(b)(i) Since f is convex, we have that for 120 x x a   and (0,1),t  1 2 1 2f ( (1 ) ) f ( ) (1 )f ( )tx t x t x t x+ −  + − Subtracting 1f ( )x from both sides we get: ( )  1 2 1 1 2 1 21 f ( (1 ) ) f ( ) f ( ) (1 )f ( ) f ( ) 1 f ( ) f ( ) tx t x x t x t x x t x x + − −  + − − = − − Thus, we get (for 01 t ): ( ) ( )   1 2 1 1 2 1 2 1 21 f ( 1 ( )) f ( ) ( ) f ( ) f ( )1 ( )x t x x x x x x xt x x+ − − −  −  −−− Taking limit for 1t → ,  1 2 1 2 1f ( ) ( ) f ( ) f ( )x x x x x  −  − 21 1 21 f ( ) f ( )f ( ) . xxx xx−  − (proven) 2b(ii) f ( ) ln ,x x x= 0x  f ( ) 1 lnxx =+ 1f ( )x x = Since 1f ( )x x = is positive for 0x  , f ( ) lnx x x= is a convex function. By part b(i), we have 2 2 1 1 1 21 ln ln1 ln x x x xx xx −+ − 2 1 2 1 1 1 2 2 1 1ln ln ln lnx x x x x x x x x x− + −  − 2 2 1 2 2 1 1 2 1 1 1 2 1 2 2 2 1 2 2 1 2 1 2 21 1 ln ln ln ln ln ln ln ln x x x x x x x x x x x x x x x x x xx x x x xxx x −  − − + −  − − −   2 212 1 .e x xxx x −    (shown)

3 Solutions [14] AM-GM, Cauchy Schwarz Ineqn 3a(i) Applying AM-GM Inequality four times, we have ( ) ( ) ( ) ( ) 1 2 3 3 1 2 3 1 2 4 3 1 2 4 1 3 4 3 1 3 4 234 3 234 3 3 3 3 x x x x x x x x x x x x x x x x x x x x x x x x ++  ++  ++  ++  Hence, ( ) ( ) ( ) ( )1 2 3 1 2 4 1 3 4 2 3 4 3 1 2 3 1 2 4 1 3 4 2 3 4 3 3 3 3 ( )( )( )( ) x x x x x x x x x x x x x x x x x x x x x x x x + + + + + + + +                Since 1 2 3 4, , ,x x x x are positive numbers, ( )( )( )( )1 2 3 1 2 4 1 3 4 2 3 4 1 2 3 4 81 .x x x x x x x x x x x x x x x x+ + + + + + + +  (Shown) 3(a)(ii) For each positive integer i from 1 to n, let Si = j ji x   . By AM/GM Inequality, 1 1 i n j ji S xn − −  ( ) 11 nij ji S n x −   −   Multiplying all i = 1 to n, ( ) ( ) ( ) 1 1 11 1 1 11 11 1 1 1 nn nn ij i i j i n nn nn i k ik nn n ik ik S n x S n x S n x − = =  − − == ==  −  −   −     

( ) ( ) 1 1 1 1 1 1 1 . 1 n k k nn i i n i i nn j jii x nS x nx = = = =  −  −     Maximum possible value of ( ) 1 1 1is 1 n i i nn j jii x nx = =  −   when 12 ... nx x x= = = . 3(b)(i) The Cauchy-Schwarz inequality ( ) ( )( ) 2 2 2 2 2 1 1 2 2 1 2 1 2a b a b a a b b+  + + , yields ( ) ( )( ) 2 22 11x y x y+  + + when 1 2 1 2 , 1, 1, .a x a b b y= = = = ( )( ) ( ) 22 2 .11xy xy+ ++ (Shown) 3(b)(ii) ( ) 2 2 2 2 2( ) 2( ) 3( ) 3( ) x y z x y z xy yx xz xy yz zx xy yx xz xy yx xz x y z + + = + + + + +  + + + + + = + + = + + Hence, 3.x y z+ +  (Shown) Alternative Solution ( ) 2 2 2 2 2( )x y z x y z xy yx xz+ + = + + + + + By Cauchy-Schwarz inequality, ( )( ) ( ) 2 22 2 2 2 2 2 2 2 xy yz zx xy yz zx x y z y z x x y z  + + + + + + + + + + ( ) 2 2( ) 3( ) 3( ) x y z xy yz zx xy yx xz xy yx xz x y z + +  + + + + + = + + = + + Hence, 3.x y z+ +  (Shown)

3(b)(iii) The Cauchy-Schwarz inequality ( ) ( )( ) 2 2 2 2 2 2 2 1 1 2 2 3 3 1 2 3 1 2 3a b a b a b a a a b b b+ +  + + + + , yields ( ) ( )( ) 2 2211x y z y z x y+ +  + + + + when 1 2 3 1 2 31, , , , , 1a a y a z b x b y b= = = = = = , so ( ) 2 22 11 1 yz xy x y z ++++ ++ . Similarly, ( ) 2 22 11 1 zx yz x y z ++++ ++ and ( ) 2 22 11 1 xy zx x y z ++++ ++ . Summing them up, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 222 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1 1 1 1 1 3 2( ) 2 1. (Proven) x y y z z x y z z x x y x y z x y z x y z x y z x y z x y z x y z x y z x y z xy yz zx x y z x y z x y z x y z +++ + + + + + + + + + + + + + + + + + + + + + + + + += ++ + + + + + ++ + + + + += ++ ++= ++ =

Q4 Solutions [13] Combinatorics, Pigeonhole Principle, Cauchy - Schwarz Ineqn (a) Using the relationship between lengths of chords and the subtended angle, observe that  ,ijPP is a spaced pair if and only if 120 240 ijPOP     where O is the centre of the circle. Approach 1: Pigeonhole principle or summing angles Suppose the three pairs are all spaced pairs. WLOG let 1 2 3,,P P P be in anticlockwise order around the circle. Then we must have 2 2 3 31 1, , 120POP POP POP     , but this is impossible since these will add up to more than 360 . Contradiction, therefore the three pairs cannot all be spaced pairs. Approach 2: Restrict two points to an arc Since  12,PP and  31,PP are spaced pairs, both 2P and 3P must lie on the arc measuring 120 furthest away from 1P (see diagram). Then we must have 23 120P OP   so  32,PP is not a spaced pair. (b) Have: 1f ( ) 4P = , 7f ( ) 3P = , 2 3 4 6f ( ) f ( ) f ( ) f ( ) 2P P P P= = = = , 5f ( ) 1P = Therefore ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 4 2 4 2 4 1 4 2 2 2 2 3 2 3 2 3 LHS = + + + + + + + + + + + + + + + 6 6 5 6 4 5 5 5 42= + + + + + + + = and 2 2 2 2 2 2 24 2 2 2 1 2 3 42RHS = + + + + + + = . Hence the two sides are equal. (c) (i) Each spaced pair in M appears exactly twice in the sum on the LHS, once for each point in the pair. Hence the total sum is twice the total number of spaced pairs, i.e. 2m .

(ii) From part (a), no point forms a spaced pair with both iP and jP . Therefore, Maximum no. of points that can form a spaced pair with jP = (Total no. of points) – (No. of points that form spaced pair with iP ) f( )inP=− Therefore f ( ) ) f(ji nP P− f ( ) f ( )ij nPP+  Note: Alternative method/presentation: use pigeonhole principle to argue the contrapositive. (iii) The LHS is a sum where all terms are of the form f ( )kP for some k. We will consider the number of times f ( )iP appears in the sum, for each  1, 2, ,in  . Consider any fixed i. Since the sum is over all spaced pairs, the number of times f ( )iP appears is precisely the number of spaced pairs that include iP . But the number of spaced pairs involving iP is the definition of f ( )iP . Thus, the term f ( )iP appears exactly f ( )iP times in the sum, and this is true for every i. Therefore, the point iP contributes exactly  