EJC 2025 H3 Mathematics Prelim Solutions

[Turn over 2025 EJC H3 Mathematics Prelim Suggested Solutions 1 Let a, b, c be positive real numbers such that 36abc= . (a) Show that 2 3 18a b c+ +  . [2] (b) Deduce the minimum value of 2 2 249 3 3 2 2 a b c b c c a a b+++ + + and find the values of a, b and c where this minimum is attained. [5] (c) Hence, or otherwise, prove that if p, q and r are positive real numbers such that 1 36pqr= , then ( ) ( ) ( ) 3 3 3 1 1 1 99 3 36 3 2 4 2 prp r q q r q p+ + + + + . [4] 1 Solution [11] (a) By AM-GM inequality, 3 3 2 3 3 2 3 36 3(6) 18 a b c a b c abc + +    = = = (b) By Cauchy Schwarz inequality   ( )   ( ) 2 2 2 2 2 2 2 2 2 2 2 49 ( 3 ) (3 2 ) (2 ) 2 33 3 2 2 49 4 2 6 2 33 3 2 2 4 9 2 3 since 2 3 03 3 2 2 2 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 a b c a b cb c c a a b  + + + + + + +  + + + + +  + + + +  + + + + + +++ +  + + + + + Hence 2 2 249 93 3 2 2 a b c b c c a a b+ + + + + The minimum value of 9 occurs when equality holds for AM-GM inequality and this requires 23a b c== . Substituting into 36abc= yields 2(2 ) 363a a a  = 3 27 3 a a = = 3, 6, 2a b c = = = Check against original expression 2 2 2 2 2 24 9 4(3) (6) 9(2) 3 3 3 93 3 2 2 6 3(2) 3(2) 2(3) 2(3) 6 a b c b c c a a b+ + = + + = + + =+ + + + + + (c) Consider the substitution 1 1 1,,a b cp q r= = = Then 1 1 1 1 36abc p q r pqr=   = =

2 of 20 (the “of n” is hidden for our own reference) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 3 3 3 3 3 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 22 1 1 1 9 3 36 3 2 4 2 1 1 1 369 3 36 3 2 4 2 49 3 3 2 2 1 1 1 49 1 1 13 3 2 2 11 14 9 1 1 1 13 3 2 p r q q r q p pqrp r q q r q p qr pr pq p r p q q q r q r p qr pr pq p p p p r p q q q r q r p p p p pq r pr pr pr qr q r qr p q r r r qr q r q p r +++ + + = + +  + + + = + ++ + + =  +  + + + + = + + ++ ( ) ( ) 2 2 2 2 112 49 3 3 2 2 9 pq a b c b c c a a b + = + ++ + +  by the result of the earlier part

3 of 20 (the “of n” is hidden for our own reference) 2 (a) The variables x and y are related by the differential equation 22 d2 d2 y xy x x y= − , 2x y for all ,xy  . Using the substitution y ux= , find the general solution of the differential equation. [7] (b) Describe geometrically the set of points that satisfy the general solution obtained in part (a). [4] 2 Solutions [11] (i) Differentiating the substitution with respect to x , dd dd yu xx ux=+ Substituting dd dd yu xx ux=+ and y ux= into the differential equation, we have ( ) 2 2 2 2 2 3 2 2 2 d2 d 2 d2 d 12 d 2 2 d 12 21d d 12 u x uux x x x u uuxu x u u u u ux x u uuux x u += − =− − −+= − + = − Hence, ( ) 2 2 2 2 1 2 1 dd 21 1 4 1 dd21 ln ln21 u ux xuu u uxu u x u Axu − = + −= + =+   2 2 22 , where 21 2 2 u Bx B u yy B Bxxx y By B A x == + =+ =+  (ii) Case 1: 0B . ( ) 22 22 2 2 2 22 2 11 24 16 1 0 4 1 8 11 4 y By Bx x y B x B B y B B =+ + − =  −− +=      

4 of 20 (the “of n” is hidden for our own reference) The general solution represents an ellipse centre at 10, 4B   with horizontal radius 1 8B and vertical radius 1 4B with 2x y for all ,xy  . Case 2: B = 0 The particular solution is the line 0y= .

5 of 20 (the “of n” is hidden for our own reference) 3 Consider 4-by-4 unit-square grids where each unit-square is either shaded or white. For example, the following diagram illustrates an example of such a grid. (a) (i) How many possible 4-by-4 unit-square grids are there? [1] (ii) How many 4-by-4 unit-square grids are there which contain at least one entirely shaded 3 -by-3 square? [4] (b) Determine the minimum number of shaded unit-squares in a 4-by-4 unit-square grid to guarantee that it contains at least one entirely shaded 2-by-2 square. Justify your answer. [3] (c) Define ic to be the number of shaded unit squares in the ith column from the left of a given 4-by-4 unit- square grid. For example, for the 4-by-4 unit-square above, 1 2 3 42, 0, 3, 2c c c c= = = = . How many distinct possible tuples ( )1 2 3 4, , ,c c c c are there if the given 4-by-4 unit-square grid has (i) exactly 4 shaded unit-squares, [1] (ii) at least 12 shaded unit-squares? [3] 3 Solutions [12] PIE, Pigeonhole Principle, Distribution Qns (a)(i) Total number of grids 162 65536== (a)(ii) Label the squares as shown below Let iA be the number of grids with a 3-by-3 shaded square that’s top-left square is labelled i. 72iA = , 4 1 2 1 3 2 4 3 4 2A A A A A A A A= = = = , 1 2 3 4

6 of 20 (the “of n” is hidden for our own reference) 2 1 4 2 3 2A A A A== , 2i j kA A A = and 1 2 3 4 1A A A A = . By PIE, ( ) ( ) ( ) 1 2 3 4 4 11 1 2 3 4 1 4 7 4 42 . 2 2 2 24 2 4 1 447 4 ij ik i i i i i j ik A A A A A A A A A A A A A A        = =− +− = − + =  +   −   (b) We require more than 12 from the following construct: Now we show that every grid with 13 shaded unit square will satisfy the required condition. Consider the partitioning for the grid by the four 2-by-2 squares as shown below:

7 of 20 (the “of n” is hidden for our own reference) We have 13 shaded squares (pigeons) and four 2 -by-2 squares (pigeonholes). By pigeonhole principle, there will be one 2-by-2 square with at least 13 44  = shaded squares. This would be we have at least one shaded 2-by-2 square if we have 13 shaded square in the grid. Therefore, at minimum, we need 13 shaded squares. (c)(i) Method 1: This is equivalent to 1 2 3 4 4c c c c+ + + = , where 0ic + . Hence, 4 4 1 3541 +− = − . Method 2: By listing cases, 4 3 1 column with 4 2 columns with 1,1 with 2 44 1 column with 3, 1 with 1 2 columns with 2 4 colums 2 with 1 2 4 3 1 35 C PC + ++ += (c)(ii) As the number of white squares is 16− the number of shaded squares. Counting this is equivalent to counting if the grids have at most 4 white squares. By symmetry this is the same as counting the grids that have at most 4 shaded squares. Method 1: So, we require, 4 4 1 3 4 1 2 4 1 1 4 1 0 4 1 4 1 4 1 4 1 4 1 4 1 70. + − + − + − + − + −         + + + +         − − − − −          = Method 2: This is equivalent to 1 2 3 4 5 4c c c c c+ + + + = , where 0ic + .

8 of 20 (the “of n” is hidden for our own reference) We treat 5c as the number of shaded squares not in the grid if there are less than 4 shaded squares in the grid. 4 5 1 70.51 +− = − Method 3: By listing cases: ( ) ( ) ( ) ( ) ( ) 22 4 4 4 4 3 2 2 4 4 3 1 4 4 4 4 1 70 C P C P C +  + + + + + + + + + + =

9 of 20 (the “of n” is hidden for our own reference) 4 Let S be the set of all integers from 1 to n and T be the set of all subsets of S, including the empty set  and S itself. (a) Explain why 2nT = . [1] The sets 12 and AA , which are not necessarily distinct, are chosen randomly and independently from the 2n subsets in T. (b) In this part of the question, we will calculate ( )12P AA = in 2 ways. (i) Let i be an integer such that 0 in . Show that ( )12 3P 4i A A  = . [2] (ii) Hence, find ( )12P A A = . [1] (iii) Let k be an integer such that 0 kn . Find ( )1 2 1P| A A A k = = . [1] (iv) Deduce that 0 13 24 nn nk k n k + =  =  . [3] (c) (i) Let k be an integer such that 0 kn . Find ( )12P A A k= . [2] (ii) Find ( )12E AA . [2] 4 Solution[12] (a) In constructing a subset of S, we have 2 choices for each of the n integers from 1 to n: include or exclude. By multiplication principle, there are 2n subsets of S. (b) (i) Then, ( ) ( ) ( ) ( ) 1 2 1 2 12 2 1 1 (by independence) 11 2 3 4 P i A A P i A A P i A P i A   = −   = −   =−  = Alternatively, Consider the partition: ( ) ( ) ( ) ( )1 2