2023 A Levels H3 Math Paper 1 Soln

Solutions to 2023 GCE ‘A’ Level H3 Mathematics (9820/01) 1 (a) Prove that, for any real numbers 12, , , na a a  , 2 2 2 1 2 1 2 nna a a n a a a+ + +  + + +  . [3] (b) Prove that, for any positive real numbers x, y and z, 6x y y z z xx y z x y z x y z+ + ++ + + + + + + + . [3] (c) Hence solve the equation 3626 66xxx + += ++ . [3] Solution: (a) Applying the Cauchy -Schwarz inequality to the pair of sequences 1,1, ,1  and 12, , , na a a  gives 2 2 2 2 2 2 1 2 1 2 terms 1 1 1 1 1 1 nn n a a a a a a +  + +   + + + + + +   2 2 2 1 2 1 2 nna a a n a a a + + +  + + +  . (b) Set 1 2 33, , , x y y z z xn a a a x y z x y z x y z+ + += = = = + + + + + + in (a) above gives ( ) 3 2 3 6 x y y z z x x y y z z x x y z x y z x y z x y z x y z x y z x y z x y z + + + + + ++ +  + ++ + + + + + + + + + + + ++== ++ Equality occurs when x y y z z x x y zx y z x y z x y z+ + += =  = =+ + + + + + . (c) 3 6 3 3 62 6 6 6 6 6x x xx x x x x + + ++ = + + + + + + + From (b), putting 3yz== gives 3 3 6 6666xxxxx++ + + +++ . Equality occurs when 3x= . So solution to the equation is 3x= .

2 2 The curve C, defined for 0x , satisfies the differential equation 2 3 d d y y y x x x=− and passes through the point ( ),ab where 0a and 0b . (a) By using the substitution xu y= , find the equation of the curve C. [5] (b) State, with reasons, necessary and sufficient conditions on a and b for the curve C to have two asymptotes. Find the equations of these asymptotes. [6] Solution: (a) d d d d 11d d d dx y u y uu uy x u y u yy x x x x=  =  + =  = − . Multiply the DE 2 3 d d y y y x x x=− by u gives 2 3 3 2 d d d d d1 1 since and 1d d d d1 d 1 1 y uy uyu x x x u xy y uy uy x u yx x x x u xx uC x x Cyx =− − = − = = − = =− + =− + Substitute ,x a y b== into above gives 21a a b CCb a ab +=− +  = . ( ) 2 2 2 2 2 2 2 2 2 2 2 2 1 Note that 0 since if 0, then which has no asymptote] Thus where 0 where x a b y x ab abxy a b x ab a b a b y x ab xy a b abab x ab x aby k k x k a b +=− + = +− +  + = =−  = +  +  − + ==  −+

3 ( )( ) 2 2 3 2 x k x k kyk xk ky k x k xk ky kx k xk  + − +=  − = + + − = + + − We must have 2 0ab+ and 0k  since xk= is a vertical asymptote of C and C is defined for 0x . That is, 2 0abab + . Since 0a , if 2 0ab+ , then 0b and if 2 0ab+ , then 0b . The necessary and sufficient conditions for C to have two asymptotes are: 2 0ab+ and 0b or 2 0ab+ and 0b . Given the above conditions, the asymptotes of C are 2 and x k y kx k= = + where 2 abk ab= + .

4 3 Let n be a positive integer. (a) Write down the binomial expansion of ( )1 n x+ . [1] (b) Prove that 00 3 jn n ji nj ji==    =        . [4] (c) (i) Let 12 12 kmmm km p p p=  , where 12, , , kp p p  are distinct prime numbers and 12, , , km m m  are positive integers. Determine the number of positive divisors of m that have no square factors other than 1, justifying your answer. [2] (ii) The function  is defined on the positive integers as follows. ( )11 = and for 1m with 12 12 kmmm km p p p=  , as in part (i), ( ) ( ) 0 if 1 for some 1, 2, , 1 otherwise. i k m i k m == −  Write down the values of ( ) ( ) ( ) ( ) ( )2 , 3 , 4 , 6 and 12     . [3] (iii) Prove that, for 1m , ( ) 0 dm d = , where dm denotes that d is a positive divisor of m. [6] Solution: (a) ( ) 0 1 n n j j nxx j= +=  . (b) ( ) 0 3 1 2 2 n nnj j n j= = + =  by (a) with 2x= . ( ) 0 2 1 1 j jj i j i= = + =  by (a) with 1x= . Substituting the second equation into the first therefore gives 00 3 jn n ji nj ji==    =        . (c) (i) Any positive divisor of m is of the form 12 12 kttt kp p p  where 0 for 1,2, ,iit m i k  =  . If a positive divisor of m has no square factors, then 0 1 for 1,2, ,it i k  =  .

5 Number of square-free positive divisors of m other than 1 21k=− . (ii) ( )11 = and for 1m with 12 12 kmmm km p p p=  , as in part (i), ( ) ( ) 0 if 1 for some 1, 2, , 1 otherwise. i k m i k m == −  m im k ( )m 122= 1 1m = 1 ( ) 1 11− =− 133= 1 1m = 1 ( ) 1 11− =− 242= 1 2m = 1 0 116 2 3= 12 1mm== 2 ( ) 2 11−= 2112 2 3= 12 2, 1mm== 2 0 ( ) ( ) ( ) ( ) ( )2 3 1, 4 12 0, 6 =1    = =− = = . ( ) ( ) ( ) ( ) ( ) ( ) ( ) 12 1 2 3 4 6 12 1 1 1 0 1 0 0 d d      = + + + + + = − − + + + =  (iii) ( ) ( ) ( ) ( ) ( ) ; 1; is sq-free ; 1; is not sq-free 0 ; 1; is sq-free 1 1 d m d m d d d m d d d m d d d d d d        = + + =+      The number of square-free divisors 1d  of m is equal to the number of possible selections of the k prime numbers 12, , , kp p p  . ( ) 1 if an odd number of primes is selected 1 if an even number of primes is selectedd −= ( ) ( ) ( ) ; 1; is sq-free ; 1; is sq-free 1 0 d m d m d d d m d d dd k d     =+ =+   ( ) ( ) ( ) ( ) ( ) ( )( ) 1 2 3 4 1 1 1 1 10 1 2 3 4 11 0 k k k k k k k k k            = + − + − + − + − + + −                       = + − = 

6 4 Let n stones be placed in fixed positions on a line. Each stone is painted using one of four colours (red, white, yellow or blue) in such a way that no two adjacent stones are the same colour. Let nr be the number of ways of painting the stones such that the first and last stones are both red. Let ns be the number of ways of painting the stones so that the first stone is red but the last stone is not red. (a) Explain why 1 2 1 21, 0, 0, 3r r s s= = = = . [2] (b) Find a formula for nnrs+ and explain why 1nnrs+ = . [4] (c) Using mathematical induction, or otherwise, prove that for all 1n , ( ) 113 3 1 4 nn nr −− +−= . [6] (d) Now let n stones, where n > 1, be placed on a circle with numbered positions. Find the number of ways of painting these stones, using at most four distinct colours, in such a way that no two adjacent stones are the same colour. [2] Solution: (a) If n = 1, then the first stone is also the last stone and there is only 1 way to paint the stone red. So 1 1r = . Since the first stone is red and the last stone, which is the first stone, is red, there is no way to paint the stone such that the last (first) stone is not red. So 1 0s = . If n = 2, then the first and the last (second) stones must be painted red but in so doing they are adjacent stones painted red, thus violating the restriction that no two adjacent stones are the same colour. So 2 0r = . Since the first stone is red, the last (second) stone can be white, yellow or blue, there are 3 ways to paint the stones. So 2 3s = . (b)       paintings such that the first and last stones are red paintings such that the first stone is red and the last stone is not red painting such that the first stone is red , Since and are m nn R S T r R s S RS = = = == utually exclusive whose union is , nn T T R S R S r s=  = + = + 1 1 1 3 3 3 3 nnn n r s T − − + = =     =   . Check: 1 1 2 1 1 1 2 2 1 3 ; 3 3r s r s −−+ = = + = = .

7 Observe that if 1n+ stones are painted in such a way that the first and last stones are both red, then the nth stone must necessarily be non-red since no two adjacent stones can share the same colour . Therefore the number of ways to paint 1n+ stones such that the first and last stones are both red is equal to the number of ways to paint n stones such that the first stone is red and the last stone i