DHS EJC RVHS H3 Math 2024 Prelim Questions

1 (a) Given a differential equation of the form d gd xx yy =  , use the substitution x vy= to show that 11 d dg( ) vyv v y =− . [2] (b) Show that the differential equation ( ) 22 2 2 2dee d xx yy xxy y y x y             = + + , ,0xy  can be written in the form d gd xx yy =  . [2] (c) It is given that the curve of the differential equation (b) passes through the point ( )ln 9, 2− . Solve the differential equation given in (b), leaving your answer in the form h xy y =  . [7] 2 (a) Show that 1 1 n k k n n k −  − =   for any positive integers n, k where nk . [2] It is given that 12, , ... , na a a + and s is a positive integer where sn . Let S be 12 ... si i ia a a , where the sum is taken over all 1, ..., { 1,2,..., }si i n  such that 121 ... si i i n     . (b) Find, in terms of n and s, the number of terms in S. [1] (c) Let m be a fixed integer where 1 mn . Find, in terms of n and s, the number of terms in S such that  12, ,..., siim i . [1] (d) Let ( ) 1 12 nng a a a= . Prove that ( )( ) ( ) ( )121 1 1 1 n na a a g+ + +  + . [6]

2 3 In this question, we will examine different ways of dividing circles. (a) Consider the number of regions formed when a circle is cut n times. The cases where the maximum number nm is achieved for 2n= and 3 is illustrated below with the values of nm stated. Illustrate the case where the maximum number nm is achieved for 4.n= Hence deduce, with justification, an expression for 1nnmm −− in terms of n and solve for .nm Explain briefly why this represents the maximum number of regions. [7] (b) Suppose a circle is divided into 21n+ congruent sectors, with n of them randomly coloured black and the other 1n+ randomly coloured white. A smaller concentric circle is placed on the larger circle and also divided into 21n+ congruent sectors, with 1n+ of them randomly coloured black and the other n randomly coloured white. A possible case for 7n= is illustrated below. Prove that, for all n, we can always find 1n+ sectors with matched colours by suitably rotating the smaller circle if necessary. [5]

3 5 An ice-cream shop sells single scoop cones with k different flavours available. You may assume that the shop does not allow mixing of flavours in any single scoop. (a) An order of n single scoop cones is made where n k . Given that all k flavours are bought and the order includes an odd number of cones for each flavour, state a condition between n and k and find the number of such possible orders. [3] Choosing from the k possible flavours, a group of n kids orders one single scoop cone each. (b) By considering all possible orders made by the n kids, explain why ( ) 0 S, k nk r r k n r P = = where S(n, r) denotes the number of ways to partition n distinct objects into r disjoint, non- empty subsets and ( 1)( ) (2 1).k r kP k k k r=  − +−− [4] (c) Apply the principle of inclusion and exclusion to enumerate the number of possible orders which include all k flavours and show that 0 1S( , ) ( ) , ! k r n rn k c k rk = =−  where rc are expressions, in terms of r and k, to be determined. [5] 4 The Fibonacci series is defined by 1 2 2 1 1, or .f 1 n n nF F F F F n ++= = + = (a) Show that 2 2 1 1 1 n in i F F F + = =− , and state a similar result for 21 1 n i i F + =  . [3] Let S be the set of all integers that can be written in the form 12 ... tn n nF F F+ + + , where in is a sequence of positive integers such that 1 2n  and 1 1iinn+ + for all 11 it  − . For example, 5 is in S because, 55 F= . 54 is in S because, 3 5 7 954 2 5 13 34 F F F F= + + + = + + + . 190 is in S because, 2 4 6 9 12190 1 3 8 34 144 F F F F F= + + + + = + + + + . (b) Show that both 55 and 191 are in S. [2] (c) (i) Suppose k is in S, where 232 ... su u uk F F F F= + + + + . Show that 1k+ is also in S. [3] (ii) Hence, use mathematical induction to prove that all positive integers are in S. [6]

4 6 The functions f and g are defined as follows. f ( ) 2 g( ) 1 xx xx = =+ An arrangement of functions f and g is a composition of functions f and g, where each function can be composed any number of times, but at least once each. For example, fgfgfgfgf and 3gf g are both arrangements of f and g. (a) Given that h( )x ax b=+ describes a function that is equal to an arrangement of f and g, find the set of possible values of a and b. [2] (b) (i) Show that 2g f ( ) fg( )xx= . [1] (ii) List all the arrangements of f and g that are equal to the function 44x+ . [2] (c) (i) Find an expression for the arrangement g fg fg ( )i j k x in terms of x, where i, j and k are non-negative integers. [1] (ii) Hence, or otherwise, show that for all positive integers m, the number of arrangements of f and g that are equal to the function 44xm+ is 2( 1)m+ . [5] (iii) By using a suitable bijection, show that the number of arrangements of f and g that are equal to the function 44xm+ is equal to the number of arrangements of f and g that are equal to the function 4 4 1xm++ , where m is a positive integer. [3] 7 In this question, all variables represent positive integers. The greatest common divisor of x, y and z, written gcd( , , )x y z , is the largest positive integer that divides each of x, y and z. We say that ( , , )x y z is a Py thagorean triple if 2 2 2x y z+= . If, in addition, gcd( , , ) 1x y z = , we say that ( , , )x y z is a primitive Pythagorean triple. Examples of primitive Pythagorean triples are ( ) ( )3,4,5 , (5,12,13), (7,24,25) & 9,40,41 . (a) (i) Find consecutive integers a and b such that (11, , )ab is a Pythagorean triple. [1] (ii) By an appropriate generalisation, show that there exist infinitely many primitive Pythagorean triples. [2] (b) (i) Find integers x, y and z satisfying gcd( , , ) 1x y z = , and gcd( , , ) 1xy yz zx  . [1] (ii) Show that if ( , , )x y z is a primitive Pythagorean triple, then gcd( , , ) 1xy yz zx = . [5] (c) Let ( , , )x y z be a Pythagorean triple. Show that ( ) ( ) ( ) ( ) 24 4 4 4 2 2xy yz zx z x y+ + = − . [2] (d) Deduce carefully that the equation 4 4 4 2u v w t+ + = has infinitely many integer solutions such that gcd( , , ) 1u v w = . [2]

5 8 A Gaussian integer is a complex number where the real and imaginary parts are both integers. (a) Given any complex number z, show that there is a Gaussian integer w such that 1 2 zw−  . [3] (b) Suppose that s, t are Gaussian integers with 0t . By considering the complex number s t , deduce that there are Gaussian integers q, r such that rt and s qt r=+ . [2] (c) Let s and t be the Gaussian integers 5 4i+ and 1 2i+ respectively. By considering Gaussian integers near 5 4i 1 2i + + , show that there are exactly 3 pairs of Gaussian integers ( , )qr such that rt and s qt r=+ , and find these pairs. [4] (d) Let s, t be Gaussian integers such that s t is not a Gaussian integer, 0t . Prove that there are always at least two pairs of Gaussian integers (q, r) such that rt and s qt r=+ . [5] End of Paper