2024 A Levels H3 Math Paper 1 Soln

Solutions to 2024 GCE ‘A’ Level H3 Mathematics (9820/01) 1 Let a, b and c be positive real numbers. (a) Using the fact that ( ) 2 0ab− , prove that 1 1 4 a b a b+ + . [3] (b) Prove that ( ) 1 1 1 9abc abc+ + + +  . [4] (c) By using the results in parts (a) and (b), prove that 9 1 1 1 1 1 12a b c a b a c b c a b c + +  + ++ + + + + . [4] Solution: (a) ( ) 2 2 2 2 20 2 0 2a b a b ab a b ab−   + −   +  . ( ) 2 22 2 2 2 4a b a b ab ab ab ab+ = + +  + = since 22 2a b ab+ ( ) 2 44 aba b ab ab a b++    + since , 0 , 0a b ab a b  +  1 1 4 a b a b +  + . (b) Apply C-S inequality to the pair of sequences 1 1 1, , & , ,abc abc ( ) ( ) ( ) 2 2 22 2 2 1 1 1 1 1 1 a b c a b c abc abc      ++                 + +  + +            Squaring gives ( ) 1 1 1 9abc abc+ + + +  . (c) From (b), 1 1 1 9 a b c a b c+ +  ++ . ( ) 1 1 1 9 1 1 1 9 22a b a c b c a b c a b a c b c a b c + +   + +  + + + + + + + + + +  .

2 From (a), 1 1 4 a b a b+ + . 1 1 1 1 1 1 1 1 1 1 1 1 2 2 2a b c a b a c b c      + + = + + + + +           1 4 1 4 1 4 2 2 2 1 1 12 a b a c b c a b a c b c       + +     + + +      = + + + + + 2 (a) Prove, by induction, that for all positive integers n, ( ) 21 2 1 2 n AB − − = − for some positive integers A and B with 2221AB−= . [5] (b) Hence, show that for all positive integers n, ( ) 2 2 1 2 n CD− = − for some positive integers C and D with 22 21CD−= . [3] (c) Deduce that for all positive integers n, there exists a positive integer N such that ( )2 1 1 n NN− = − − . [2] (d) Hence show that for all positive integers n, there exists a positive integer N such that ( )2 1 1 n NN+ = + − . [2] Solution: (a) P(n): ( ) 21 2 1 2 n AB − − = − 22, , ,2 1n A B A B++   − =  . For 1n= , ( ) 21 2 1 2 1LHS RHS − = − = − = where 1AB== satisfying 2221AB−= . So P(1) holds. Assume P(k) holds so that ( ) 21 2 1 2 k AB − − = − for some 22, , ,2 1k A B A B++  − =  . ( ) ( ) ( ) 2 1 2 2 1 2 1 2 1 2 1 kk+− − = − −

3 2024 GCE ‘A’ Level H3 Mathematics 9820/01 ( )( ) ( ) ( ) 3 2 2 2 3 2 2 4 3 AB A B A B = − − = + − + Clearly 3 2 ,4 3A B A B ++ +   and moreover, ( ) ( ) ( ) ( ) 22 2 2 2 22 3 2 4 3 2 9 4 12 16 9 12A B A B A B AB A B AB+ − + = + + − + + ( ) ( ) 2 2 2 2 22 2 9 4 12 16 9 12 2 1 A B AB A B AB AB = + + − + + =− = Thus P(1) holds and P(k + 1) holds if P(k) holds, P(n) holds by induction. (b) ( ) ( ) ( )( ) ( ) 21 2 2 1 2 2 1 2 1 2 2 2 n n AB AB A B A B − − = −  − = − − = + − + Clearly 2,C A B D A B ++= +  = +   and moreover, ( ) ( ) 2222 2 2 2C D A B A B− = + − + ( ) ( ) ( ) 22 2 2 2 2 22 22 4 4 2 2 2 1 A B A B A B AB A B AB AB = + − + = + + − + + =− = (c) From (a), ( ) 21 2 1 2 n AB − − = − 2 2 2 1AA= − − since 2 2 22 1 2 1A B B A− =  = − 22 2 2 2 1 1 where 2 AA N N N A + = − − = − − =   ( ) 2 22 2 2 1 2 2 1 2 1 where 2 n CD DD N N N D + − = − = + − = + − =   Therefore ( )2 1 1 n NN− = − − for all positive integers n where N +  . (d) ( ) 112 1 1 2 1 1 nn NN NN + = = = + − − − − .

4 3 Let ( ) ( ) 1 1 f1 n rr n r n xx r r+= =−  . (a) State the value of ( )f0n . [1] (b) Show that ( ) ( )11f n n xx x−− = . [3] (c) Deduce that ( ) ( ) 1 1 1f f 0 d 1 nnn x yxy y− −−= − . [4] (d) Hence show that ( ) 1 1 1f 1 1 23n n= + + + +  . [2] (e) By considering ( ) ( )1f 1 f 1nn+ − , deduce that, if n is even, then ( ) 1 1 0 110 1 n r r n r r − + = −=  + . [3] (f) Determine an expression for the value of ( ) 1 1 0 11 1 n r r n r r − + = −  + when n is odd. [2] Solution: (a) ( ) ( ) 1 1 0f 0 1 0 n rr n r n r r + = = − =  . (b) ( ) ( ) ( ) ( ) 11 1 11 f 1 f 1 nn rrr r nn rr nn xx x x rr r++ −==     = −  = −       . ( ) ( ) ( ) ( ) ( ) ( ) ( ) 23 1 1 1 11 1 1 1 1 1 1 2 3 11 1 1 1 f n n n r nr r n nn rr rr n rr n n n n nx x x x x x nr nnx x x xrrxx = + − ==          − = − + − + + − = + −                   −−      =− − = − =           (c) Integrating w.r.t. x from 0 to x gives ( ) ( ) ( ) ( )integrate w.r.t 0 01 1 1 1f f d nn xx xnn xxx x x xx− − − − =  =  

5 2024 GCE ‘A’ Level H3 Mathematics 9820/01 ( ) ( ) ( ) ( )111 0 1 1 11 11f f 0 d 1 d d 11 n nnxx yx nn x x yyx x y y x y y −=− −−− −−− = = − = −−   . (d) Since ( )f 0 0n = from (a), the above gives ( ) 1 1 1f d 1 nn x yxy y− −= − ( ) ( ) 1 21 1 f 1 d nn x x y y y y − − = + + + +  since 21 11 1 nn yy y y y− −+ + + + = −  ( ) ( ) 1231 21 0 0 1 1 1f 1 1 d 1 2 3 2 3 nnn y y yy y y y y nn− = + + + + = + + + + = + + + +  (e) ( ) ( )1 1 1 1 1 1 1 1 1f 1 f 1 1 1 2 3 1 2 3 1nn n n n n+ − = + + + + + − + + + + =++   . Using the definition ( ) ( ) 1 1 f1 n rr n r n xx r r+= =−  , the above yields ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 11 11 f 1 f 1 1 2 1 11 11 11 1 1 1 111 1 1 1 1 1 11 1 1 ------ (*) 11 1 111 1 if nn nn rr rr nn r n r rr nn rr rr nn rr r n r nn rr r n n r nn nrr rr + + ++ == + + + == ++ == +   − − = −    +    +   − + − − = −    ++    +   − = −           ( ) ( ) ( ) ( ) ( ) ( ) 2 11 11 1 1 1 1 11is even so that 1 11 1 111 1 0 1 110 1 1!1 0 since 1 1! n nn rr rr n r r n r r nn nn rr rr nn rr r n n n n r r r r r n r + ++ == + = + = −= ++ +   − − − =         +    − − =           +      − = − =      − − − +         ( ) ( ) 1 2 0 11! 11 0 ( 1) 1 n k k n r n krk k− += = −  − = = −  + Replacing the dummy index k with r and dividing by 1− gives ( ) 1 1 0 110 1 n r r n r r − + =  −=  + . (f) If n is odd, (*) gives

6 ( ) ( ) ( ) 11 11 1 1 0 1 1 1 1 111 11 121 11 nn rr rr n r r nn rr r n n r n r rn ++ == − + = +   − − − = −    ++      − =−  ++   using the calculations in (e). 4 (a) A curve satisfies the differential equation 22d 2ed xyy yxx=− and is such that 1 when 1yx c== . Use the substitution xy z= to show that the curve has equation 2 eex xy c = −+ . [6] (b) Showing all necessary working, determine the values of c for which the curve (i) has exactly two vertical asymptotes [3] (ii) has exactly two stationary points. [6] Solution: 4 (a) 2 d d d d zzxxy xy zx z−=  = . Substitute into given DE gives 2 2 2 2 2 2 2 2 d 1d 2e d 2ed d 2ed e ( is an arbitrary constant) since e x x x x x zzx xx zzz zz x z xx z xx z A A xxyy zA − =−  − = − = =+ == +

7 2024 GCE ‘A’ Level H3 Mathematics 9820/01 Since 1 when 1yx c== , 11 ee AccA=  = −+ . Thus equation of solution curve is 2 eex xy c = −+ . (b) (i) For the solution curve to have two vertical asymptotes, 22 e e 0 e exx cc− + =  = − must have exactly two distinct solutions. Therefore ( )ln e 0 e 1 e 1c c c−   −    − . (ii) ( ) ( ) 22 2 2 set 2 e e 2 ed 0dee eexxx x c x xxyy xc c− + −=  = =−+ −+ ( ) 221 2 e e xxc − = − The function ( ) ( ) 22f 1 2 e xxx=− is even and has a maximum point at the point x given by ( ) ( )( ) ( ) 2222f 0 1 2 2 e 4 e 0 2 1 0 0 xxx x x x x x x =  − − =  + =  = . So maximum value of f is ( )f 0 1= . The graph of ( ) ( ) 22f 1 2 e xy x x= = − is shown below: Thus for