2019 A Level H2 FM 9649 P1 (Qns & Solutions)(30 Sep 2024)

2019 GCE A Level H2 Further Maths 9649 Paper 1 Solutions Section A: Pure Mathematics Question 1 For a given initial value 1u = , the sequence of real numbers  nu is defined by the recurrence relation 1 1 ,1 n n n uu u + += − 1.n  Prove that 5n = gives the first value of ( ) 1nn  for which nu = , and that this is so for all but three values of  . State these three exceptional values of  . [6] [Solution] Method 1: Given 1u = 1 2 1 1 1 11 uu u   + +== −− ,  ≠ 1 1 Let 1   + =− [to check if 2u = ] 221 1 0    + = −  + = Since there is no real solution for 1 ,1   + =− 2u  1 12 3 1 2 1 11 1 11 uu u      + − + − ++= = = −−− ,  ≠ 0 Let 1 −= [to check if 3u = ] 2 10 + =  No real solution 3u  1 3 4 1 3 11 1 1 1 1 uu u     −+ −= = =− + + ,  ≠ −1 Let 1 1   − =+ 2 2 1 10 No real solution      − = +  + = 

4u  1 14 5 1 4 1 11 2 1 1 2 uu u       − + − + ++= = = =−− . Thus u1 = u5 . Hence n = 5 is the first value for nu = This implies that the sequence is periodic with period 4. Note: Need to check that 1 1   + =− , 1 −= and 1 1   − =+ has no real solution Method 2: 1 1 1 1 11 n n n n n n n uu u u u u u + + + +=  − = +−  1 1 1 1 12 1 1 11 1 () n n n n u nn u uu uu Thus 24 42 11 n n n nn u u u uu . The sequence {un} is periodic with period = 4. The first value of n is 5 for which 5u = and that this is so for all but three values of  which are 0 and ±1 Question 2 (i) Evaluate the integral 3 d b a xx . [2] (ii) Evaluate this integral using Simpson’s rule with two strips. [3] (iii) Deduce that Simpson’s rule always gives the correct value for an integral of any cubic polynomial. [2] (i) 4 3 d 4 b b a a xxx == ( ) 441 4 ba− (ii) Using Simpson’s Rule, 3 3 3 3 1d4 3 2 2 b a b a a bx x a b −+   = + +         ( ) 3 3 2 2 3 311 333 2 2 ba a a a b ab b b−   = + + + + +     3 2 2 33 3 3 312 ba a a b ab b− = + + +  ( ) 3 2 2 3 4 4 3 2 2 31 3 3 3 3 3 3 3 312 a b a b ab b a a b a b ab= + + + − − − − (iii) Let ( ) 32f x px qx rx s= + + + 3 2 3 2 d d d b b b a a a px qx rx s x px x qx rx s x+ + + = + + +   32 d d bb aa p x x qx rx s x= + + + From part (i) and (ii), using Simpson’s rule with 2 strips on 3 d b a xx gives the same exact value. Simpson’s rule also gives an exact value to 2 d b a qx rx s x++ since this is the integral of any

quadratic curve and Simpson’s rule model a curve by a quadratic approximation. Hence Simpson’s rule always gives the correct value for an integral of any cubic polynomial. Question 3 The curve C is such that d sin( )d y xyx = and 1.5y = when 1x = . (i) Use the Euler method with steps of size 1 3h = to find an approximation to the value of y when x = 2. Give all intermediate value