2023 A Level H2 FM - Solutions

2023 Cambridge GCE ‘A’ Level Further Mathematics Paper 1 Questions & Solutions FM N2023 / 1 / 1 1 A matrix M has eigenvalues 1, 2 and 4, with corresponding eigenvectors 23 1,2 12         and 1 4 3    . Determine Mn , giving your answer as a single matrix. Solution: 1 Theorem: Let A be a square matrix of order n . If A has n distinct eigenvectors, then A is diagonalisable. Suppose 1,, nλλ… (multiplicity included) are the n eigenvalues of A with corresponding n distinct eigenvectors 1,, n…ee . Let D be a diagonal matrix defined as 1 2 00 00 00 D n λ λ λ   =        and ( )12Q ee e n=  (the thk column is the vector ke ). Then Q is invertible and 1A QDQ −= . Here M is a square matrix of order 3 having 3 distinct eigenvectors 23 1,2 12         and 1 4 3    with corresponding eigenvalues 1, 2 and 4. Then by the above theorem, there exists a diagonal matrix D whose diagonal consists of the eigenvalues of M and an invertible matrix Q whose columns consist of the corresponding eigenvectors of M. That is, 1 1 231 100 231 124 020 124 123 004 123 M QDQ − −        = =           .

( ) ( )( )( ) ( ) 1 111 1 M QDQ QDQ QDQ QDQ QDQ nn n products − −−− −= =    ( )( )( ) ( ) ( )( )( ) ( ) 2 111 1 1 2 3 111 1 3 1 231 1 0 0 231 124 02 0 124 123 0 0 4 123 n products n pro c n du n n ts −−− − − −−− − − − − = =        =         =  QD Q QDQ QDQ QDQ QD Q QDQ QDQ QDQ QD Q        ( ) ( ) 2 1 22 12 2 32 2 2 7 10 12 2 157 01 11 2 32 nn nn nn ++ +  − = −−   −  by GC ( ) ( ) ( ) ( ) ( ) ( )( ) ( ) ( )( ) ( )( ) 2 21 4 3 2 14 2 2 15 20 2 21 2 2 1 2 7 2 2 10 10 2 7 2 2 2 1 2 7 2 10 3 2 10 2 14 3 2 n nn n n n nn n n nn n n n ++  − + − −+ − = − + − −+ −  − + −+ −+ − . FM N2023 / 1 / 2 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 nu nαβ→ for some constants α and β . (i) Write down, correct to 4 decimal places, the value of α and the value of β . (ii) Explain why, in the case when 1x= and 2y= , the long-term behaviour of { }nu is when is even, when is odd.n nu nβα→

(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 22p q pq+= . (ii) Deduce a value for x and y such that { }nu is a constant sequence. Solution: 2 (a) (i) By GC, 2.2287, 1.8997αβ= = (to 4 d.p.). (ii) For the case 1 1ux= = and 2 1uy= = , 2 21 3 12 2uuu uu = +=  . The sequence 1,1,2, converges to the limits α (n even) and β (n odd). For the case 1 1ux= = and 2 2uy= = , the sequence 1,2, is the above sequence shifted one term to the left and so it converges to the limits α (for odd n) and β (for even n). (b) (i) Suppose when n is even, 12,,nn nu pu qu p ++→→→ as n→∞ . 2 2 3 3 221 2 1 nn n nn uu qpu p p q p qu u pq++ + = + ⇒=+ ⇒ +=  . By symmetry, if n is odd, we also have 3 3 22p q pq+= . (ii) If pq= , then 3 3 22p q pq+= becomes 342 0 or 2pp p= ⇒= . If 2xy= = , 2nu = for all 1,2,3,n=  . FM N2023 / 1 / 3

3 (a) Let ( ) 54f x px qx= + , where p and q are real constants and 0p≠ . The value of the integral ( )f d a a I xx − =∫ is to be estimated using Simpson’s Rule with four strips. Find an expression for the error obtained when using this method. (b) Let ( ) 5 432g x px qx rx sx tx u= + + + ++ , where ,,,,pqrst and u are real constants and 0p≠ . The value of the integral ( )g d a a J xx − =∫ is to be determined using Simpson’s Rule with four strips. Write down, with justification, the expression for the error obtained when using this method. Solution: 3 (a) By Simpson’s Rule with four strips, ( ) ( ) ( ) ( )11 1 1f d f 4f 2f 0 4f f32 2 2aaI xx a a a a a−    = ≈ −+ − + + +      ∫ 54 54 54 54 4 4 5 1 446 32 16 32 16 1 2 246 16 5 12 ap aq ap aqa ap aq a p aq aqa aq aq    = − + + − ++ ++ +       = +⋅   = On the other hand, ( ) ( ) 54 6 5 5 2f d d 655aa aaaaaa pqI x x px qx x x x a q −−−−  = =+= + =  ∫∫ . Error in approximating I 55 552 112 5 60aq aq aq= −= . (b) ( ) ( ) ( )32g d f d da aa a aa J x x x x rx sx tx u x − −− = = + + ++∫∫∫ . Since Simpson’s Rule gives exact value for a cubic polynomial (this can be easily proved), the error incurred for ( )32 da a rx sx tx u x − + ++∫ is zero. Therefore error in approximating J 5160 aq= which is the same as (a). FM N2023 / 1 / 4

4 (a) Exponential growth and decay can be modelled by the differential equation d ,dN Nt λ= where N is the measure of some physical quantity at time 0t≥ and λ is a parameter. Given an initial positive value, 0N , of N, (i) solve this differential equation, (ii) sketch, on one diagram, the behaviours of N according to the possible values of λ. (b) Newton’s Law of Cooling states that the rate at which the temperature of an object is changing at any instant is proportional to the difference between the temperature of the object and the temperature of its surroundings at that instant. A container of liquid is placed in a room which has a constant temperature E. Initially, the temperature of the liquid is 0B , where 0BE> . At time t later, the temperature of the liquid is B. (i) Write down and solve the differential equation which describes this situation. (ii) Sketch the solution curve for B against t. Solution: 4(a) (i) [ ] 0 0 0 0 0 d1 d dd ln ln e Nt N N N t N N NttN Nt N tN NN λ λλ λ λ = ⇒= ⇒= ⇒=   ⇒= ∫∫ (ii) 0λ > N N0 t 0λ = N0 t 0λ < N0 t (b) Newton’s Law of Cooling states that the rate at which the temperature of an object is changing at any instant is proportional to the difference between the temperature of the object and the temperature of its surroundings at that instant.

A container of liquid is placed in a room which has a constant temperature E. Initially, the temperature of the liquid is 0B , where 0BE> . At time t later, the temperature of the liquid is B. (b) (i) ( ) ( ) ( ) 0 0 0 0 0 d1 d d where 0d ln ln e Bt B B B kt B kB E B k t kt BE B E kt BE ktBE BE B E − = − −⇒ = − >− ⇒−= −  −⇒= − − ⇒=+ − ∫∫ (ii) B 0B E O t

FM N2023 / 1 / 5 5(a) Use de Moivre’s theorem to find an expression for sin 3θ in terms of sinθ . (b) A curve C has cartesian equation ( )( )2222 3 68 0xyxy y y+ +− + = . (i) Show that the polar equation of the curve is 2sin 3r θ= . (ii) Sketch C. You may assume that the curve is defined only for those values of θ for which 0r≥ . (iii) Determine an integral that gives the exact value of the total length of one loop of C. Use your calculator to evaluate this integral, giving your answer correct to 3 decimal places. Solution: F