2025 A Level FM P2 Qns & Solns

2025 Cambridge GCE ‘A’ Level Further Mathematics Paper 2 Questions & Solutions FM N2025 / 2 / 1 1 The vector space V is 4  . The set S consists of the vectors a, b, c and d in 4  given by ( ) ( ) ( )1, 2,3, 4 , 2,3, 4,1 , 3, 4,1, 2= = =a b c and ( )4,1, 2,3=d . (a) Show that the set S spans V. [2] (b) (i) Express the vector ( )1,0,0,0=1u as a linear combination of a, b, c and d. (ii) Write down similar expressions for each of the vectors ( ) ( )0,1,0,0 , 0,0,1,0==23u u and ( )0,0,0,1=4u in terms of a, b, c and d. [3] [2] Solution: 1 (a) Take any vector ( ) 4, , ,x y z t   . Set ( ) ( ) ( ) ( ) ( ), , , 1, 2,3, 4 2,3, 4,1 3, 4,1, 2 4,1, 2,3x y z t    = + + + . (*) Writing in matrix form gives 1 2 3 4 2 3 4 1 3 4 1 2 4 1 2 3 x y z t                   =                M By GC, since ( )det 160 0=M , 1−M exists and thus , , ,    can be found which satisfy (*) viz ( ) ( ) 1, , , , , , x y z t    −= M . Hence  , , ,S = a b c d spans V.

Alternatively, we can show that the vectors in the set  , , ,a b c d are linearly independent. To this end, we set ( ) ( ) ( ) ( ) ( )1, 2,3, 4 2,3, 4,1 3, 4,1, 2 4,1, 2,3 0,0,0,0    + + + = . In matrix form, 1 2 3 4 0 2 3 4 1 0 3 4 1 2 0 4 1 2 3 0                    =               . Since ( )det 160 0=M where 1 2 3 4 2 3 4 1 3 4 1 2 4 1 2 3   =   M , 1−M exists and thus we can write the matrix equation as 1 00 00 00 00     −                  ==                  M which shows that  , , ,S = a b c d is a set of 4 linearly independent vectors and therefore S must span 4  . (b)(i) We want , , ,    satisfying ( ) ( ) ( ) ( ) ( )1,0,0,0 1, 2,3, 4 2,3, 4,1 3, 4,1, 2 4,1, 2,3   = + + + . That is, 1 1 2 3 4 1 1 2 3 4 1 2 3 4 1 0 2 3 4 1 0 3 4 1 2 0 3 4 1 2 0 4 1 2 3 0 4 1 2 3 0     −                                 =  =                                 . By GC, 9 1 1 11 40 40 40 40=− + + +1u a b c d .

(b)(ii) In a similar fashion, 1 1 11 9 40 40 40 40= + + −2u a b c d , 1 11 9 1 40 40 40 40= + − +3u a b c d , 11 9 1 1 40 40 40 40= − + +4u a b c d FM N2025 / 2 / 2 2 Solve the differential equation d cos sin 3 2cos 2dy x y x xx + = + , given that 51π362y=+ when 1 π3x= . Give your answer in the form ( )fyx= . [6] Solution: 2 ( ) 23 2 2cos 1d d sincos sin 3 2cos 2d d cos cos xy y x x y x x yx x x x +−+ = +  + =  d sin sec 4cosd cosyx y x xxx  + = + IF: ( ) sin d ln coscose e secx x xx x− == . Multiplying the DE by IF gives ( ) 2d sec sec 4 sec tan 4d sin 4 cos cos y x x y x x x Cx y x x x C x = +  = + +  = + + Substitute 1 π3x= , 51π362y=+ into above GS gives 5 1 1 1 1 1π 3 sin π 4 π cos π cos π6 2 3 3 3 3 3 2 1 = π2 3 2 1 π3 C C C + = + +  ++ = PS: 1sin 4 cos π cos3y x x x x= + + .

FM N2025 / 2 / 3 3 Solutions are required for the equation ( )f0 x = , for 0x , where ( ) ( )13f sin e 54 xx =+ . (a) Evaluate ( ) ( )f 1 and f 2 , and hence comment on the possible number of roots of ( )f0 x = in the interval ( )1, 2 . [2] It is given that  , the smallest positive root of ( )f0 x = , lies in the interval ( )1, 2 . (b) (i) Use the Newton-Raphson method, with initial approximation 0 1.5x = , to determine the values of 1 2 3, and x x x . Give each answer correct to 3 decimal places. [3] (ii) Explain why the values of 34 and xx are the same (correct to at least 3 decimal places). [2] (iii) Explain why the use of 0 1.5x = in the Newton-Raphson method does not give convergence to  . [1] (iv) Find the value of  to 3 decimal places. [1] (v) By considering the graph of ( )f , for 0y x x= , suggest a reason why the Newton-Raphson method should not be used to find the roots of ( )f0 x = for large values of x. [1] Solution: (a) ( ) ( ) ( ) ( ) ( ) ( ) 2 1.5 13f 1 sin e 0.5081 054 13f 2 sin e 0.8704 054 13f 1.5 sin e 0.5301 054 = + =  = + =  = + =−  By the intermediate-value theorem applied to the continuous function ( ) ( )13f sin e 54 xx =+ , ( )f0 x = has at least 2 real roots. (b) (i) ( ) ( ) ( ) ( )1 3 3f sin e f e cos e5 4 4 x x xxx = +  = .

( ) ( ) ( ) ( ) 10 13 sin ef 54 , 0,1, 2, ; 1.53f e cos e4 nnn xnn n n xxnxx x x n x x+ += − = − = =  . By GC, 1 2 3 0.810, 1.552, 19.701x x x= = = . (ii) ( ) ( ) ( ) ( ) ( ) ( ) 3 33 3 33 43 19.701 19.701 43 19.701 19.701 43 13 sin e54 3 e cos e4 1 3 1 3sin e sin e5 4 5 4 0 since e is very large.33e cos e e cos e44 to at least 3 decimal places. x xx x xx xx xx xx + =− ++  − = =  = (iii) ( ) ( )3 3 3f 0 e cos e 0 e π ln π 1.55024 2 2x x xxx  =  =  =  =   0 1.5x = in the Newton-Raphson method does not give convergence to  since its second iterate 2 1.552x = is very near a stationary point and therefore the tangent to the curve at the point where 2 1.552xx== is almost horizontal which results in it having an x-intercept that is very far from the root.

1 0.810x = 0 1.5x = 1.5502 2 1.552x = (iv) From the above sketch in (iii), choose 0 1x = . By GC, 1.227 = to 3 decimal places. (v) For large values of x, the curve ( )f , for 0y x x= become highly oscillating and therefore the tangents become very unpredictable. Moreover there are many roots close to one another and it will be difficult to identify the specific root using the Newton-Raphson method

FM N2025 / 2 / 4 4 The diagram below shows part of the curve with cartesian equation 32 20x y xy+ − = . (a) Show that the polar equation of this curve can be written as ( ) 2sec 2 tan tanr   =− . [2] (b) The value of  , where 10 tan 2 − , for which the curve is furthest from the pole, O, is denoted by  . Show that  satisfies an equation of the form 32tan tan tan 0a b c d  + + + = where a, b, c and d are integers to be determined. Hence find the value of  . Give your answer correct to 4 decimal places. [5] (c) Do not use a calculator in answering this part. Showing full working, find the area of the loop of the curve. [6] Solution: (a) cos sin 3 2 3 3 2 2 2 2 3 2 0 cos sin 2 cos sin 0 2cos sin sin since cos xr yr x y xy r r r rr           = = + − =  + − = − =  ( ) 2 2 2 0 2cos sin sin sec cos sec 2 tan tan r r         −=   = −

(b) ( ) ( ) set2 1 sec 2 tan tan 0 tan 2 tan 0 tan 0 or tan 2 0, π or tan 2 r        − = − =  − =  = =  = = for 0 π . As  varies from 0 to 1tan 2− , the parameter  traces out the loop of the polar curve. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 2 2 2 3 32 sec 2 tan tan d sec 2sec 2 tan sec sec tan 2 tan tand d 0d 2sec 2 tan sec tan 2 tan tan 0 2 1 tan 2 tan 1 tan 2 tan tan 0 3tan 4 tan 2 tan 2 0 r r r                            = =− = − + − =  − + − =  + − + + − = − + − + = That is, 3, 4, 2, 2a b c d=− = =− = . Let tanx = . Since 10 tan 2 − , 0 tan 2 , i.e. 02 x . We sketch the graph of 323 4 2 2y x x x=− + − + . By GC, the root of 323 4 2 2 0x x x− + − + = is 1.231555. Hence 1tan 1.231555 tan 1.231555 0.8888 −=  = = to 4 decimal places. =

(c) Area of the loop ( ) ( ) 1 1 1 1 1 1 11 tan 2 2 0 tan 2 222 0 tan 2 2 2 3 4 0 tan 2 ta