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 2g( ) ,x px qx rx sx tx u= + + + + + where p, q, r, s, t and u are real constants and 0.p The value of the integral g( ) d a a J x x − = is to be estimated using Simpson’s Rule with four strips. Write down, with justification, the expression for the error obtained when using this method. [1] (a) 54 54 5 0 5 ( ) d d d 0 2 5 2 5 a a aa aa a I px qx x p x x q x x xq qa − −− =+ =+ =+  =   x a− 2 a− 0 2 a a f ( )x 54pa qa+ 54 32 16 pqaa−+ 0 54 32 16 pqaa+ 54pa qa+ By Simpson’s rule, 5 4 5 4 5 4 5 4 4 5 ( ) ( ) 4( )6 32 16 32 16 5 62 5 12 a p q p qI pa qa pa qa a a a a a qa qa  − + + + + − + + + =  = Error = (b) The error is 51 60 qa . This is because Simpson’s Rule is exact (i.e. no error) for cubic functions. 555 2 1 12 5 60 qa qa −=

2023 A level FM Paper 1 4 (a) Exponential growth and decay can be modelled by the differential equation d ,d N Nt = where N is the measure of some physical quantity at time 0t  and  is a parameter. Given an initial positive value, 0,N of N, (i) solve this differential equation, [2] (ii) sketch, on one diagram, the behaviours of N according to the possible values of  . [3] (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 0,B where 0 .BE At time t later, the temperature of the liquid is B. (i) Write down and solve the differential equation which describes this situation. [3] (ii) Sketch the solution curve for B against t. [1] (a) (i) d d 1 d 1 d ln ( 0) N Nt NtN N t C N    = = = +    e e tC t N NA   += = 000,t N N A N= =  = . 0e tNN = 0 0= 0 t N 0N O

2023 A level FM Paper 1 (b) d ( ), where is a positive constantd 1 d d ln( ) ( ) B k B E kt B k tBE B E kt C B E =− − =−− − =− +    e e kt kt B E A B A E − − −= =+ 00 0 0, ( )e kt t B B A B E B B E E − = =  = −  = − + 0( )e ktB B E E −= − + 0B B t O E

2023 A level FM Paper 1 5 (a) Use de Moivre’s theorem to find an expression for sin3θ in terms of sinθ . [3] (b) A curve C has cartesian equation ( )( ) 2 2 2 2 3 6 8 0x y x y y y+ + − + = . (i) Show that the polar equation of the curve is 2sin3r θ= . [2] (ii) Sketch C. You may assume that the curve is defined only for those values of θ for which 0r . [2] (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. [3] (a) By de Moivre’s theorem, ( ) 3 cos isin cos3 isin3θ θ θ θ+ = + ( ) ( ) ( ) ( ) ( )( ) ( ) ( ) ( ) 3 3 2 2 3 3 2 2 3 3 2 2 3 cos isin cos 3 cos isin 3 cos isin isin cos 3icos sin 3cos sin isin cos 3cos sin i 3cos sin sin θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ θ + = + + + = + − − = − + − Comparing imaginary part: ( ) 23 23 3 sin3 3cos sin sin 3 1 sin sin sin 3sin 4sin θ θ θ θ θ θ θ θθ =− = − − =− (b)(i) ( )( ) 2 2 2 2 3 6 8 0x y x y y y+ + − + = Since 2 2 2x y r+= , cosxr θ= , sinyr θ= ( ) ( ) ( ) 322 4 3 3 3 33 3 6 sin 8 sin 0 6 sin 8 sin 0 6sin 8sin 0 0 or 6sin 8sin 0 (rej 0) r r r θ r θ rr θ r θ rr θθ rr θθ r − + = − + = − + = = − + =  ( ) 3 3 3 6sin 8sin 0 6sin 8sin 2 3sin 4sin 2sin 3 (shown) r θθ r θθ θθ θ − + = =− =− = (b)(ii) 2sin3r θ= 0θ= O

2023 A level FM Paper 1 (b)(iii) 2sin 3 d 6cos3d r θ r θθ = = ( ) 23 2 0 223 0 dTotal length of l loop of = d d 4sin 3 6cos3 d 4.455 (3dp) π π rCr θθ θ θ θ + =+ =    

2023 A level FM Paper 1 6 Complex numbers iz x y=+ are represented by points ( ),P x y in the complex plane. (a) On separate Argand diagrams, sketch the loci 1L and 2L given by 12 2i 3 3 2z− − = + and ( ) 1arg 2 i 3z π+ − = . [4] (b) By finding cartesian equations for these loci, prove that 1L and 2L meet tangentially. [7] (a) (b) For 12 2i 3 3 2z− − = + , cartesian equation is ( ) ( ) 222 12 2 3 3 ---(1) 2xy − + − = +  For ( ) 1arg 2 i 3z π+ − = , cartesian equation is ( )11 tan 2 3y πx− = +  ( ) ( ) 1 3 2 1 3 2 ---(2) yx yx − = + = + + Substitute (2) into (1), we have ( ) ( )( ) ( ) ( ) ( ) 222 222 22 2 12 1 3 2 2 3 3 2 12 1 3 3 2 14 4 1 2 3 3 3 3 4 74 2 3 4 3 0 4 xx xx x x x x xx − + + + − = +  − + + = +  − + + + + = + + + − + − = ( ) ( ) 2 7Discri