2024 A Level FM P1 Qns & Solns

2024 Cambridge GCE ‘A’ Level Further Mathematics Paper 1 Questions & Solutions 1 The matrix A is such that 3 1 4 16 14 7 3 6 5 3 2 7 7 18   −  −= −  − A . (a) In the case when 2 = , determine the rank of A, showing all necessary working. [3] (b) In the case when 1 = , determine whether the four column vectors of A form a basis for the space of four-dimensional vectors. [1] Solution: 1 (a) 21 32 41 2 43 32 8 3 0.1 2 3 1 4 2 3 1 4 16 14 7 3 0 10 15 35 6 5 3 2 0 4 6 14 2 7 7 18 0 4 6 14 2 3 1 4 0 1 1.5 3.5 0 4 6 14 0 0 0 0 2 3 1 4 0 1 1.5 3.5 0 0 0 0 0 0 0 0 RR RR RR R RR RR − − − − + + −−        − − −   =→    − − −     −−    −  −→  −−   −  −→    A Since there are two non -zero rows in the ref of A, the dimen sion of the row space is 2 and therefore rank(A) = 2. (Alternatively, since there are 2 pivot columns in the ref of A, rank(A) = 2.) (b) By GC, since 2 3 1 4 16 14 7 3 147 06 5 3 2 2 7 7 18 −  −= =−  −  − A , 1−A exists. Therefore the four column vectors of A form a basis for 4  . Explanation:

Given any 4 x y z w        , we ask if there exist real numbers a, b, c and d such that 2 3 1 4 16 14 7 3 (*)6 5 3 2 2 7 7 18 2 3 4 16 14 7 3 6 5 3 2 2 7 7 18 2 3 1 4 16 14 7 3 6 5 3 2 27 x ya b c d z w a b c d x a b c d y a b c d z a b c d w −                    −         + + + = −−−−          −          −          + − +       + + −   =   + + −    + − +    − − − 1 7 18 (#) ax by cz dw ax by cz dw ax by cz dw −                =         −                =                         = −−−−            A A From (#), real numbers a, b, c and d can be found which satisfy (*) and we conclude that the 4 column vectors of A span 4  . Moreover, if 0x y z w= = = = , then 0a b c d= = = = . Therefore the 4 column vectors of A are also linearly independent. We conclude that the four column vectors of A form a basis for 4  .

2 A population of size P, at time t, is modelled by the logistic growth equation ( )d 0.3 1 0.0075dP P P ht = − − where h (h > 0) is the harvesting term, which represents the rate at which individuals are removed from the population in order to maintain a constant size. Determine the number of equilibrium points of the population for the possible values of h. [4] Solution: 2 For equilibrium populations, ( ) 2d 0 0.3 1 0.0075 0 0.00225 0.3 0dP P P h P P ht =  − − =  − + = . Discriminant D = (–0.3)2 – 4(0.00225)h = 0.09 – 0.009h. 0 0.09 0.009 0 10D h h  −    . So Number of equilibrium population = 0 if 10 1 if 10 2 if 10 h h h   =  

3 Let 2 0 1 2 n ny a a x a x a x= + + + + +  , where the coefficients satisfy the recurrence relation 2 12 for 2n n nn a a a n−−= +  . By writing 2n in the form ( )1n n n−+ , or otherwise, find functions ( ) ( )f ,gxx and ( )h x , and a constant k (which depends upon the initial terms 01 and aa ), such that ( ) ( ) ( ) 2 2 ddf g h dd yyx x x y k xx + + = . [6] Solution: 3 23 0 1 2 3 0 2 1 1 1 2 3 1 d 23d nn nn n nn nn n y a a x a x a x a x a x y a a x a x na x na xx  =  −− = = + + + + + + = = + + + + + =     ( ) ( ) 2 22 232 2 d 2 3 2 1 1 d nn nn n y a a x n n a x n n a x x  −− = = +  + + − + = −   . ( ) 2 1 2 2 d 1 d n n n yx n n a x x  − = =− . ( ) ( )( ) ( ) 2 11 2 21 1 1 2 21 1 2 1 1 1 2 2 11 1 1 2 22 dd 1dd 1 nn nn nn n n n n n n n nn n nn nn nn yyx n n a x na xxx a n n n a x a n a x a a a x a a x a x  −− ==  − =  − =  − −− =  −− −− == + = − + = + − + =+ = + + = + +      ( ) 12 1 1 0 2 12 10 nn nn nn a a x a x a x a y a xy  −− −− == = + − + = + − +  ( ) 2 102 dd 1dd yyx x y a a xx + − + = − .

4 Find all possible solutions of the system of equations 22 22 1 kx y z k x ky z x y z + + = + + =− + + =− identifying the different cases that arise according to the value of k. [10] Solution: 4 Consider the augmented matrix 212 2 1 2 1 1 1 1 M kk k  =− − . 2 1 2 1 1 1 1 1 1 1 1 2 1 2 0 2 1 0 0 2 1 0 1 1 1 1 0 2 1 3 0 0 3 kk k k k k k k k k −−           − → − − → − −          − − − −      If 0k = , 1 1 1 1 0 2 1 0 0 0 0 0 −  −− The system has infinitely many solutions. Let z =  , 1 2y =− , 1111 22x   =− − + =− − If 2k = , 1 1 1 1 1 1 1 1 0 0 1 0 0 0 1 0 0 0 2 6 0 0 0 6 −−       −→      −    The last row implies 0 = -6 which is inconsistent, hence the system has no solutions. If 2 and 0kk , 1 1 1 11 1 1 1 10 2 1 0 0 1 0 20 0 3 0 0 1 3k kkk −−  − − →  −  −  − The system has a unique solution where 3z=− , 13 22yz kk= =−−− 331 1 3 2 22x z y kk=− − − =− + + = +−−

5 The strength, x, of an electric current in a circuit at time t is modelled by the differential equation d d xL Rx Et += . (a) In the simplest model, L and R are positive constants and E is a constant. Show that, when t is large, the strength of the electric current in the circuit is approximately E R . [4] (b) In a more sophisticated model, L and R are positive constants but cosE F t= , where F is a constant. Show that, when t is large, the strength of the electric current in the circuit is approximately ( )22 cosF tRL −+ where tan LR = . [7] Solution: 5 (a) dd dd x x R EL Rx E xt t L L+ =  + = . IF = d ee RRttLL = . Multiply by IF gives d e e e e d ed R R R R Rt t t t tL L L L LE E Ex x t Ct L L R =  = = +  e since e 0 for large RRttLLEEx C tRR −− = +   . (b) dd cos cosddx x R FL Rx F t x tt t L L+ =  + = . Multiply by IF = e R tL gives d e e cos e e cos dd R R R Rt t t tL L L L FFx t x t tt L L =  =  ------ (*) de , cosd e cos d e sin e sin d R tL vutR R R tt t tL L L RI t t t t t L=== = − ------ (1) de , sind e sin d e cos e cos d e cos R tL vutR R R R tt t t tL L L L RRt t t t t t I LL== = − + =− + ------ (2) Substitute (2) into (1) gives 2 22e sin e cos sin cos e R R Rt t tL L L R R L RI t t I I t t C L L L RL = − − +  = + +  +  .

From (*), 2 22e e cos d sin cos e R R Rt t tL L L F F L Rx t t t t CL L L RL = = + +  +  ( )22 sin cos e RtLF FC L t R t LRL= + ++ For large t, ( ) ( ) ( ) 2 2 2 2 22 22 sin cos e sin cos cos RtLF FC Fx L t R t L t R t LR L R L F R L t RL  − = + +  + ++ = + − + ( )22 cosF tRL =− + , tan LR = . 6 For ( ) 3f2 xx=− , the equation ( )f0 x = has a single real root, r, where 3 2r = . The Newton-Raphson process, using the iterative formula ( ) ( ) 1 f f n nn n xxx x+ =−  , with the initial value 0xh= , can be used to find a decimal approximation for r. (a) Show that 1 2 21 3xh h=+  . [2] For each iterate, nx , in the Newton-Raphson process the corresponding error term, n , is defined by nn xr =− . (b) Calculate, correct to 2 decimal places, the values of 0 and 1 in each of the cases • 10h= • 1h= • 0.1h= . [3] (c) Let c be the value of h, where 0 hr , for which 10= . (i) Find the exact value of c. (ii) Verify numerically that 10 for 0 hc and 10 for c h r . [4] [1] Solution: 6 (a) ( ) ( ) 32f 2 f 3x x x x = −  = . ( ) ( ) 3 0 10 2 2 2 0 f 2 1 2 2 1 f 3 3 33 x hx x h h h h x h h h−    = − = − = − − = +         . (b)