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 values of y correct to 4 decimal places. [4] (ii) (a) What feature of C do these values of y suggest occurs at some point P between x = 1 and x = 2? [1] (b) Use the results of part (i) and the differential equation d sin( )d y xyx = to estimate the x-coordinate of P. [2] (i) d sin( )d y xyx = ( )1 11.5 sin 1 1.5 1.832498 1.83253y = +    2 141.832498 sin 1.832498 2.046794 2.046833y = +     3 152.046794 sin 2.046794 1.957970 1.958033y = +     Hence 1.9580y  when 2x = (ii)(a) The y values increases and decreases suggesting that at some point P between x = 1 and x = 2, there is a maximum turning point. *(b) Let d sin( )d y xyx = = 0 sin( ) 0xy = xy = When 2.046794y  , 1.532.046794x 

Question 4 (i) Determine the eigenvalues and corresponding eigenvectors of the matrix 23 .65 =  M [5] (ii) By expressing M in the diagonalised form 1−QDQ , find nM for positive integers n. [4] [Solution] (i) det (M − I) = 0 det 23 65   −  − = 0  (2 − )(5 − ) – 18 = 0 2 – 7 − 8 = 0 ( − 8)( + 1) = 0 . So  = 8 or −1 are the eigenvalues When  = 8, 6 3 0 6 3 0 26 3 0 x x y y xy −     =  − + =  =    −     A corresponding eigenvector is 1 2   When  = −1, 3 3 0 3 3 06 6 0 x x y x yy      =  + =  = −         A corresponding eigenvector is 1 1  − (ii) Let Q = 11 21   − then its corresponding diagonal matrix is D = 80 01   − Then M = 11 21   − 80 01   − 1 11 21 −   − Mn = 11 21   − 80 01 n   − 1 11 21 −   − = 11 21   − 80 0 ( 1) n n   − 111 213   − = 1 8 ( 1)1 3 2(8 ) ( 1) nn nn +  −  − 11 21   − = 1 8 ( 1)1 3 2(8 ) ( 1) nn nn +  −  − 11 21   − = 1 12 8 2( 1) 8 ( 1)1 3 2(8 ) 2( 1) 2(8 ) ( 1) n n n n n n n n + ++  + − + −  + − + − Recall: 11 nn−−=  =M QDQ M QD Q

Question 5 The sequence  nP is given by 01 5, 10PP== and ( )11 1 32 n n nP P P+− = + + for 1n  . (i) By considering the sequence  nQ , where 2nnP Q n=+ for 0,1,2,n = , find an expression for nP as a function of n. [9] (ii) Describe the long-term behaviour of nP . [1] [Solution] (i) ( )11 1 32 n n nP P P+− = + + for 1n  . 2nnP Q n=+  11 2( 1)nnP Q n−− = + − and 11 2( 1)nnP Q n++ = + + ( )11 1 32 n n nP P P+− = + + 11 12( 1) 2 2 2 32 n n nQ n Q n Q n 112 4 4 2 2 2 6n n nQ n Q n Q n  1120 n n nQ Q Q Characteristic equation: 2m2 – m – 1 = 0 (2m + 1)(m − 1) = 0. So m = 1 or 1 2 General solution is 1 ( 1)(1) ( ) 22 n nn n n BQ A B A 01 5, 10PP==  Q0 = 5 and Q1 = 8 So 5 = A + B ---- (1) and 8 = A – 1 2 B ---- (2) Solving (1) and (2): A = 7 and B = −2 Thus 12 7 2( ) 2 2 n nnP Q n n= + = − − + (ii) As 1, 0, 2 n n →  − → 27nPn→+ (a arithmetic sequence) and thus nP → in terms of its value. (Do not just state that nP → )

Question 6 The function f is given by ( ) 2f 1 cos 1x x x= − + − for 01 x . It is known, from graphical work, that the equation ( )f0 x = has a single root x = . (i) Express ( )g x in terms of x, where ( ) f ( )g. f ( ) xxx x=−  [2] A student attempts to use the Newton -Raphson method, based on the form ( )1 gnnxx+ = , to calculate the value of  correct to 3 decimal places. (ii) (a) The student first uses an initial approximation to  of 1 0x = . Explain why this will be unsuccessful in finding a value for  . [1] (b) The student next uses an initial approximation to  of 1 1x = . Explain why this will also be unsuccessful in finding a value for  . [1] (c) The student then uses an initial approximation to  of 1 0.5x = . Investigate what happens in this case. [1] (d) By choosing a suitable value for 1x , use the Newton-Raphson method, based on the form ( )1 gnnxx+ = to determine  correct to 3 decimal places. [Solution] (i) ( ) 2f 1 cos 1x x x= − + − f’(x) = 22 2 sin sin 2 1 1 xx xx xx −− − = − −− ( ) 22 22 f ( ) 1 cos 1 1 cos 1g f ( ) sin sin 11 x x x x xx x x x xxx xx xx − + − − + −= − = − = + − −+ −− or 22 2 (1 ) 1 (cos 1) 1 sin x x xx x x x − + − −+ +− (ii) (a)Given f(x) = 0 has a root . x1 = 0, f’(0) = 0, so g(0) is not defined. The iterative formula ( )1 gnnxx+ = is not defined at x1 = 0. (b) x1 = 1, f’(1) is undefined as it has a vertical asymptote at x = 1. (c) 1 0.5x = , x2 = g(0.5) = 22 2 (1 0.5 ) 1 0.5 (cos0.5 1) 0.64398338870.5 0.5 1.204 1 0.91519469570.5 1 0.5 sin 0.5 − + − −+ = +