2025 A Level FM P1 Qns & Solns

2025 Cambridge GCE ‘A’ Level Further Mathematics Paper 1 Questions & Solutions FM N2025 / 1 / 1 1 For positive constants a and b, the sequence  nq is given by 1 2qa= and 2 1 2 for 1n n bq a n q+ = +  . Given that a and b are such that  nq converges, determine the limit of the sequence in terms of a and b. [4] Solution: 1 Suppose  nq converges to q. Then as 1,, nnn q q q +→ → . 2 1 2n n bqa q+ =+ becomes 2 222 2 0bq a q aq bq= +  − − = 22 22 2 4 4 2 a a bq q a a b +=  =  + 22 1 2 3 12 2 0 2 2 0 2 2 0 bbq a q a a q a a qq=   = +    = +   and so on …. Inductively, 20nqa for all 1n . Thus 20qa . Hence 22q a a b= + + since 22 0q a a b= − +  .

FM N2025 / 1 / 2 2 The complex numbers 12 and zz are given by 12 5 i and 239 izz= + = + . (a) (i) Write down, in their simplest exact forms, the moduli of 12 and zz . State also, in inverse tangent form, the arguments of 12 and zz . [2] (ii) Use the answers to part (a)(i) to deduce the modulus and argument of 3,z where 4 1 3 2 zz z= . [2] (b) Deduce that 111 1 1π tan 4 tan4 239 5 −−+= . [2] Solution: 2 (a) (i) 12 26 and 169 2zz == . 11 12 11arg tan and arg tan5 239zz −−== . (ii) ( ) 444 11 3 22 26 22 169 2 zzz zz= = = = and 4 111 3 1 2 2 11arg arg 4arg arg 4 tan tan 5 239zz z z z −−= = − = − . (b) ( ) ( )( ) ( )( ) ( ) ( ) 44 1 3 2 5 i 476 480i 239 i 114244 1 i476 480i 2 1 i239 i 239 i 239 i 239 i 57122zz z + + − + += = = = = = ++ + + − . Thus ( )3 1arg arg 2 1 i π4z = + = . Hence 1 1 1 11 1 1 1 1 1π 4 tan tan π tan 4 tan4 5 239 4 239 5− − − −= −  + = .

FM N2025 / 1 / 3 3 The circle C has centre at the origin and radius r. (a) Write down the cartesian equation of the circle T, obtained when C is translated by the vector 0 a  − , where 0 ar . [1] A spindle is used in hand spinning. A 3D printer is used to produce a spindle. To generate the spindle, the arc of T which lies on or above the x-axis is rotated completely about the x-axis. (b) Show that the surface area of this spindle can be expressed in the form 22 2 2 1 sin rakr r a a r− − −− where k is a constant whose exact value is to be determined. [7] Solution: 3 (a) ( ) 22 2 2 2 2 y y ax y r x y a r →++ = → + + = . Cartesian equation of T is ( ) 222x y a r+ + = . Alternative, the centre ( )0,0 of C is mapped under the translation vector 0 a  − to the centre ( )0, a− of T. So cartesian equation of T is ( ) 222x y a r+ + = . (b) y T b a x (0, –a)  r By Pythagoras’ theorem, 22b r a=− . Differentiate ( ) 222x y a r+ + = w.r.t. x:

( ) ( )( ) 2 22 2 2 d d d2 2 0 1 1 d d d x y ay y x y xx y a x x y a x y a ya+++ + =  =−  + = + − = ++  + . Surface area of spindle 2 0 d22 π 1 d d b yyx x =+   ( ) ( ) 22 2 0 4π d b x y ayx ya ++= + ( ) ( ) 2 222 2 0 4π d since b ry x x y a r ya = + + = + ( ) 0 0 22 2 2 2 220 1 0 1 22 2 2 1 4π d since 0 4π 1 d 14π d since 0 4π sin 4π sin 4π sin b b b b yr x y aya arx ya r b a x x y a r y a r x rx xr b a r br b a r rar r a a r − − − = +  + =−  + = − + + =  + = −  −   =−     =−    −= − −     so 4 πk   =   Alternatively, Surface area 2 0 d22 π 1 d d ra xyy y − =+   0 22 π d ra ryy x − =   since ( ) 2 22 2 22 dd 1dd x y ax y a x ry x y xx+++=−  + = =  ( ) 2 2 220 4π d 4π d r a r u y a a y u ar y r u rur y a − =+  −= = =  −−+ 

FM N2025 / 1 / 4 4 A curve has equation siny x x= for 02 πx . The total length of the curve is denoted by L. A student decides to approximate L using four straight line segments OP, PQ, QR and RS, where ( ) ( ) ( )1 1 3 30,0 , π, π , π,0 , π, π and 2π,02 2 2 2O P Q R S    = = = = − =       . The curve and the four line segments are shown on the diagram below. (a) Calculate the value of the student’s approximation to L, giving your answer correct to 4 decimal places. [2] (b) (i) Write down the integral which gives the exact value of L. [2] (ii) Use the trapezium rule with four strips to obtain a second approximation to L. Give your answer correct to 4 decimal places. [3] (iii) Use your graphing calculator to obtain a third approximation to L. Give your answer correct to 4 decimal places. [1] (c) Comment on the accuracy of these three approximations to L. [2] Solution: 4 (a) First approximation to L is 2 2 2 2 2 2 1 1 1 1 3 1 32 π 2 π π π π π2 2 2 2 2 2 22 π π 10π 14.377522 A            = + + + + +                       = + + 

(b) (i) ( ) 2π 2π 2 2 00 d1 d 1 sin cos ddyL x x x x x x= + = + + . (ii) Let ( ) ( ) 2 f 1 sin cosx x x x= + + . By trapezium rule, second approximation to L is ( ) ( ) ( ) ( ) 2 1 π π 3πf 0 f 2 π 2 f f π f2 2 2 2 π 1 6.362265 2 1.414214 3.296908 1.414214 15 .40404 A       = + + + +             = + + + +  (iii) By GC, third approximation to L is 3 15.3746A  . (c) 1A gives a large underestimation of L since it uses line segments to approximate a curve with high curvature. On the other hand, 2A gives a slight overestimation of L. FM N2025 / 1 / 5 5 The surface S has equation 2tanz x y= , for all values of x and 11ππ22 y−   . The point 12, π, 24P lies on S. The tangent-plane to S at P is denoted by T. (a) Express the equation of T in the form 2πax by cz+ + = , where a, b and c integers to be determined. [3] (b) State a vector which is normal to T. [1] Let z x =+ u i k and z y =+ v j k , evaluated at P. (c) Show that uv is also normal to T. [1] (d) Describe the geometrical significance of each of u and v in relation to S at P. [3] Solution:

5 (a) 2 2 2tan tan , 2 tan seczzz x y y x y y xy=  = =  . Equation of T is ( ) ( ) 2 2 21 1 1 1 12 tan π tan π 2 2 2 tan πsec π π4 4 4 4 4 12 2 8 π4 82 π z x y z x y x y z     − = − + −           − = − + −   + − = (b) A normal vector to T is 8+−i j k . (c) At P, z x = + = +u i k i k and 8=+v j k . ( ) ( )88 = +  + =− − +u v i k j k i j k which is parallel to 8+−i j k . (d) u is a direction vector of the tangent line to S at P in the direction of i on the xy-plane; v is a direction vector of the tangent line to S at P in the direction of j on the xy-plane. FM N2025 / 1 / 6 A business takes out a loan of $1000 000 from a bank. At the end of each month, interest is charged by the bank and a repayment is made by the business. The amount owed by the business at the end of month n is nM , where ( )10 1.008 for 0, and 1000 000nnM M F Gn n M+ = − +  = . (a) State the annual interest rate charged by the bank, and the significance of the constants F and G. [2] (b) Given that the solution of the recurrence system is of the form 1.008nnM A Bn C=  + + , determine A, B and C in terms of F and G. [6] (c) Given that the loan is repaid after exactly 60 months, and that the value of F is at least 2500, determine the maximum possible value of G. [3] v T P u

Solution: 6 (a) Removing the repayment made by the business, we have 1 1.008nnMM+ = for 0,1,2, ,11n=  and 0 1000 000M = . Interest paid at the end of month 12 without monthly repayment ( ) 12 12 0 0 0 12 1.00