2023 GCE A level FM P1 (Solution)

2023 A level FM Paper 1 1 A matrix M has eigenvalues 1, 2 and 4, with corresponding eigen vectors 23 1 , 2 12                   and 1 4. 3    Determine ,nM giving your answer as a single matrix. [5] 1 1 1 1 1 2 3 1 1 0 0 2 3 1 1 2 4 0 2 0 1 2 4 1 2 3 0 0 4 1 2 3 2 3 1 1 0 0 2 3 1 1 2 4 0 2 0 1 2 4 1 2 3 0 0 4 1 2 3 2 3 1 1 0 0 2 3 1 1 2 4 0 2 0 1 2 4 1 2 3 0 0 4 1 2 3 2 3(2 ) 4 n n n n n − − − − =        =                   =                    =          = M PDP M 2 7 10 1 2(2 ) 4(4 ) 1 5 7 1 2(2 ) 3(4 ) 0 1 1 4 3(2 ) 14 15(2 ) 4 20 21(2 ) 4 2 2(2 ) 7 10(2 ) 4(4 ) 10 14(2 ) 4(4 ) 2 2(2 ) 7 10(2 ) 3(4 ) 10 14(2 ) 3(4 ) n nn nn n n n n n n n n n n n n n n n  − −−  −  − − + − + − = − − + − + −  − − + − + −

2023 A level FM Paper 1 2 The sequence  nu is given by the recurrence relation 2 1 2 1 nn n nn uuu uu + + + =+   for 1n , together with positive initial values 1ux= and 2uy= . (a) In the case when 1x= and 1y= , the sequence  nu exhibits alternating convergence. That is, as n→ , when is even, when is odd, n αnu βn →  for some constants α and β . (i) Write down, correct to 4 decimal places, the value of α and the value of β . [2] (ii) Explain why, in the case when 1x= and 2y= , the long-term behaviour of  nu is when is even, when is odd. n βnu αn →  [1] (b) You are given that, for all positive values of x and y, the sequence  nu exhibits alternating convergence. (i) Suppose that, for a chosen pair of positive initial values, the alternate limits are (in either order) p and q. Show that p and q must satisfy the relationship 3 3 2 2p q p q+= . [2] (ii) Deduce a value for x and y such that  nu is a constant sequence. [2] (a)(i) Using GC, 2.2287, 1.8997 (to 4 dp)αβ== (a)(ii) Old sequence: 1 2 3 4 51, 1, 2, 2.25, 1.9151, u u u u u= = = = = New sequence: 1 2 3 41, 2, 2.25, 1.9151, u u u u= = = = This is because the terms in the sequence in (ii) starts from the second term of the sequence in (i) and in the same order. Hence, when is even, when is odd. n βnu αn →  (b)(i) Suppose when n→ , nup→ , 1nuq+ → , 2nup+ → 2 33 2 2 2 3 3 (shown) qpp pq qpp pq p q q p =+  += =+ (b)(ii) For  nu to be constant sequence, when n→ , , where nu p p q→= when n is even or odd. ( ) 2 2 3 3 43 3 20 20 0 (rejected since 0) or 2 p p p p pp pp p p p =+ −= −= =  = Hence 2xy== .

2023 A level FM Paper 1 3 (a) Let 54f ( ) ,x px qx=+ where p and q are real constants and 0.p The value of the integral f ( ) d a a I x x − = is to be estimated using Simpson’s Rule with four strips. Find an expression for the error obtained when using this method. [7] (b) Let 5 4 3