2024 A Level FM P2 Qns & Solns

2024 Cambridge GCE ‘A’ Level Further Mathematics Paper 2 Questions & Solutions 1 The function y satisfies 22d 2dy xyx =+ , with 1y= when 1x= . (a) Use two steps of the improved Euler method to find an approximation to the value of y when 1.1x= . Give your answer to 3 decimal places. [5] (b) State how the improved Euler method could be used to find a more accurate approximation to the value of y when 1.1x= . [1] Solution: 1 (a) ( ) 22 11 1.1 1f , 2 , 1, 1, step size 0.05 2x y x y x y h −= + = = = = . First Predictor: ( ) ( ) ( ) 22 2 1 1 1 f , 1 0.05f 1,1 1 0.05 1 2 1 1.15u y h x y= + = + = + +  = . First Corrector: ( ) ( ) ( ) ( ) ( ) ( ) 2 1 1 1 2 2 2 2 2 2 f , f ,2 0.05 1 f 1,1 f 1.05,1.152 0.05 1 1 2 1 1.05 2 1.152 1.1686875 hy y x y x u= + +  = + +  = + +  + +  = Second Predictor: ( ) ( ) ( ) 3 2 2 2 22 f , 1.1686875 0.05f 1.05,1.1686875 1.1686875 0.05 1.05 2 1.1686875 1.3603955 47 u y h x y= + = + = + +  = Second Corrector: ( ) ( ) ( ) ( ) 3 2 2 2 3 3 f , f ,2 0.05 1.1686875 f 1.05,1.1686875 f 1.1,1.360395 5472 hy y x y x u = + +  = + +  ( ) ( ) 2 2 2 20.051.1686875 1.05 2 1.1686875 1.1 2 1.3603955472 1.387325326 = + +  + +   = Thus when 1.1, 1.387xy= . (b) The approximation can be improved further by reducing the step size.

2 A curve has equation 22 e for 0 π axay x b −=   , where a and b are positive constants. The region of the plane enclosed by the curve, the y-axis and the line 22 e π abay −= is rotated through 2 radians about the y-axis. (a) Show that the volume of revolution generated is V, where ( )( ) 22221 1 e abV k a b −= − + for some constant k to be determined. [6] (b) Hence deduce the exact value of V as b→ . [1] Solution: 2 (a) y π a 22 e π abay −= O b x By ‘Shell Method’, 2 2 2 2 0 2π e e d ππ −−=−  b a x a baaV x x ( ) 2 2 2 2 2 2 2 2 2 2 00 2 2 12 π 2 e d e d 2 112 π e 1 e 22 −− −− = − − − = − − −  bb a x a b a b a b a a x x x x a ab a ( ) 2 2 2 2 22 22 22 π 1 e e π 1 1 e −− − = − − = − + a b a b ab aba aba (b) ( )  2222ππlim lim 1 1 e ab bb V a b aa −→ →= − + = .

3 The points given by the complex numbers cos i sinz r r =+ for 0 and ππr  −   lie in the z-plane. These points are mapped to points in the w-plane, where iw u v=+ , by the transformation 1wz z=+ . (a) (i) Show that the image of a circle in the z-plane, with centre 0 0i+ and of fixed radius ( )1rr  , is a conic section in the w-plane. Identify this conic section. [3] (ii) Determine the image of the unit circle, with centre 0 0i+ . [2] (b) (i) Consider points in the z-plane with a fixed value of 10, π, π2  . Describe their image points in the w-plane. [2] (ii) Determine the image of those points in the z-plane for which 1 π2 = and 01 r . Justify your conclusion fully. [3] Solution: 3 (a) (i) Given cos i sinz r r =+ where 0, 1 and ππrr   −   . ( ) ( ) ( ) 11 cos i sin cos i sin 1i cos isin cos isin w z r r r rz u v r r         − = + = + + +  + = + + − 22 22 22 11 cos , sin ------ (1) cos , sin 11 u r v r rr uv rr rr      = + = −        = =    +−       Therefore 22 22 22 cos sin 111uvrr rr + = + =   +−       which is an ellipse centred at 0 0i+ with length of semi -major axis 1r r+ and length of semi-minor axis 1r r− in the w-plane. (ii) Put 1r= in (1) above gives 2cos , 0uv == . π π 1 cos 1 2 2cos 2  −   −   −   . Thus 2 2, 0uv−   = . The image in the w-plane is the horizontal line segment joining 2 0i−+ and 2 0i+ . (b) (i) Let 00 10, π, π2  =   be fixed. From (1),

00 11 cos , sinu r v r rr    = + = −       , 0r  . ( ) ( ) 0 0 0 00 22 22 00 22 22 00 1 1 1, , sin ,cos 0 since 0, π, πcos sin 2 11 4cos sin 1 2cos 2sin uv rr rr uv rr rr uv       = + = −            − = + − − =               − = which is a hyperbola with centre 0 0i+ in the w-plane. (ii) Put 1 π2 = in (1) gives 1 π 1 π 1cos 0, sin22u r v r r r r r   = + = = − = −       . 10 1 0 0r r v r   −    . So the image in the w-plane is the vertical line starting from 0 0i+ (but excluding it) and extending below the real axis. 4 (a) Find the curved surface area of the figure generated when the line 1yx=+ , between 0x= and 9x= , is rotated through 2  radians about the x-axis. Give your answer as an exact multiple of  and verify that this area is just under 440 square units. [3] Students Annie and Bhavesh are discussing areas of surfaces of revolution. Annie claims that rotating the curves ( ) ( )f and gy x y x== about the x-axis, where ( ) ( )f g 0xx on some interval  ,pq , must yield the result fgSS on that interval (where fg and SS are the respective surface areas generated) Annie illustrates her claim with the graphs of 1yx=+ and 21 19 10 90y x x=+ , as shown on the diagram below.

(b) Use your calculator to find the surface area of the figure generated when 21 19 10 90y x x=+ is rotated through 2 radians about the x-axis, between 0x= and 9x= , giving your answer in square units to the nearest integer. You must show the definite integral that is being evaluated. [4] Bhavesh disagrees with Annie and illustrates his idea with the graphs of 1yx=+ and sin 2y x x=+ over the interval  0,9 . (c) (i) Sketch on the same diagram 1yx=+ and sin 2y x x=+ for 09 x and use your diagram to explain why Bhavesh might be correct. [2] (ii) Use your calculator to show that Bhavesh is correct. You must show the definite integral that is being evaluated. [3] Solution: 4 (a) ( ) 29 0 9 2 0 92 0 dSurface area 2 π 1 d d 2 π 1 1 1 d 2 2 π 2 99 2 π 439.85 440 yyx x xx x x =+  = + + =+   =    (b) 29 2 0 1 19 1 19Surface area 2 π 1 d 379 44010 90 5 90x x x x   = + + +         .

(c) (i) The graph of sin 2y x x=+ , though below the graph of 1yx=+ , is highly “wavy” and so the surface generated can have a larger area than that generated by 1yx=+ over the same interval  0,9 . Bhavesh may be right. (ii) ( ) ( ) 9 2 0 Surface area 2 π sin 2 1 1 2cos 2 d 446.26 440x x x x= + + +   confirming that Bhavesh is right. 5 The curve H, with equation 2224 8 18 1x xy y+ + = , is transformed from a conic section G with equation 22 0AX BY CX DY E+ + + + = . (a) By writing the points ( ),xy as the image of points of ( ),XY under the plane transformation given by cos sin sin cos xX yY   −    =         , determine for G • the lengths of the major and minor axes • the eccentricity • the coordinates of the foci [10] (b) Determine the greatest integer n for which H lies entirely inside the ellipse with equation 22 1nx y+= . [5] Solution: 5 (a) cos sin cos sin sin cos sin cos Thus substitute cos sin and sin cos into x X X Y y Y X Y x X Y y X Y             −−      ==       +       = − = + 2224 8 18 1x xy y+ + = : ( ) ( )( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 2 2 2 2 24 cos sin 8 cos sin sin cos 18 sin cos 1 24 cos 2 sin cos sin 8 sin cos cos sin sin cos 18 sin 2 sin cos cos 1 X Y X Y X Y XY X XY Y X XY XY Y X XY Y                      − + − + + + = −+ + + − − + + + =

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