Cambridge A Level Mathematics - Further 9231 — 2025 May/June Paper 2 · Variant 2

1 Find the roots of the equation z 3 = 27 - 27i , giving your answers in the form reii, where r 2 0 and - r G i 1 r . [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 z 3 = 27 − 27i = 27 2e − 4i1 π B1 Modulus and argument of 3z (does not need to be in exponential form). 1 6 − i 121 π M1 A1 Finds one root. Cube root must be taken for M1. z1 = 3  2 e 1 6 i 127 π 16 − i 129 π A1FT A1FT Finds other two roots. FT on modulus. Withhold one z 2 = 3  2 e , z 3 = 3  2 e answer mark for omission of i. 5

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2 Let I = `1 - xj sinh x dx , where n is a non-negative integer. n 0 (a) Show that, for n H 2 , I =- 1 + n ( n - 1) I . [4] n n - 2 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the exact value of I2. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) 1 n n 1 1 n −1 M1 A1 Applies integration by parts. Appropriate choice of (1 − x ) cosh x  n (1 − x ) cosh x d x 1 − x ) sinh x d x =    0 nI = 0 ( + 0 parts for M1. n −1 1 1 n − 2 M1 Applies integration by parts again. Appropriate choice 1 − x ) sinh x dx I n = −+1 n  (1 − x ) sinh x  0 + n ( n − 1) 0 ( of parts for M1. I n = −+1 n ( n − 1) I n − 2 A1 AG, withhold for no evidence of substitution or incorrect use of limits. 4 2(b) 1 1 B1 Finds 0I . = cosh1 − 1 . x d x =  cosh x 0 I 0 = 0sinh I 2 = −+1 2 I 0 M1 Uses reduction formula from (a). I 2 = 2cosh1 − 3 A1 Accept e + e −1 − 3. 3

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3 By considering the binomial expansion of bz - l , where z = cos i + isin i , use de Moivre’s theorem z to show that a cosec 5i = , sin 5i + b sin 3 i + c sin i where a, b and c are integers to be determined. 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Mark scheme: 3 z − z −1 = 2isin B1 Use of z − z −1 = 2isin −1 5 5 −5 3 −3 −1 M1 A1 Expands and groups. No clear grouping is M1 A0. z − z = z − z − 5 z − z + 10 z − z ( ) ( ) ( ) ( ) 25 isin 5 = 2isin5− 5(2isin3) + 10(2isin) M1 A1FT Substitutes z n − z − n = 2isin n on RHS. FT on signs of coefficients only. Substituting z n − z − n = isin non RHS is M1 A0. Omission of i is M0. 5 16 A1 Cancelation of i justified.  cosec  = . sin5− 5 sin3+ 1 0 si n 6

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4 y x 0 1 2 3 n-1 n 1 x The diagram shows the curve with equation y = e for x H 1, together with a set of n - 1 rectangles x of unit width. (a) By considering the sum of the areas of these rectangles, show that n 1 r 1 n e 1 e2 + o e - 2e. [5] / r n r = 1 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ n 1 r (b) Use a similar method to find, in terms of n, a lower bound for e . 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Mark scheme: 4(a) 1 1 1 2 1 n −1 M1 Forms sum of the areas of the rectangles. e + e + + e 1 2 n −1 n 1 x A1 Compares with integral. Wrong comparison e d x 1 x (missing/wrong inequality or wrong limits) is A0. n 1 e x dx =  2e x  n M1 A1 Correct integration. Attempt at integration by parts is 1 x  1 M0 A0. n n A1 AG, must have obtained all previous marks. 1 e r   2e x  + 1 e n = 2 + 1 e n − 2e  r  1 n ( n ) r =1 5 4(b) 1 2 1 3 1 n M1 Forms the sum of the areas of appropriate rectangles. e + e + + e 2 3 n n  1 e x dx =  2e x  n A1 Compares with integral. 1 x  1 n n M1 A1 Adds term from r = 1.  1r e r   2e x 1 + e = 2e n − e r=1 4

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5 Find the particular solution of the differential equation d 2 x dx -t 6 + 3 + 6x = e , dt 2 dt dx given that, when t = 0 , x = = 0 . [10] dt .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... 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Mark scheme: 5 6 m 2 + 3m + 6 = 0  m = −14 14 i 15 M1 Auxiliary equation. M1 for correct type (complex) of roots. − 14 t 15 15 A1 Complimentary function. x = e ( A cos 4 t + B sin 4 t ) Allow ‘ x = ’ missing. Using x instead of t is A0. x = ke−t  x ' = −ke−t  x '' = ke −t B1 Particular integral and its derivatives. 6ke−t − 3ke−t + 6ke−t = e−t M1 Substitutes and equates coefficients. k = 19 A1 − 14 t 15 15 1 − t A1 FT General solution. FT on CF and their value of k. Must x = e + 9 e ( A cos 4 t + B sin 4 t ) have ‘ x = ’. 15 − 14 t 15 15 1 − 14 t 15 15 1 − t *M1 Using product rule. − 9 e x ' = 4 e ( − A sin 4 t + B cos 4 t ) − 4 e ( A cos 4 t + B sin 4 t ) 1 15 1 1 1 1 DM1 A1 Substitutes initial conditions and forms simultaneous 0 = A + 9 0 = 4 B − 4 A − 9  A = − 9 , B = 3 15 equations. 1 − 14 t 1 15 1 15 1 − t A1 Must have ‘ x = ’. + 9 e x = 3 e ( − 3 cos 4 t + 15 sin 4 t ) 10

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6 (a) Starting from the definitions of tanh and sech in terms of exponentials, prove that 1 - tanh 2 u = sech 2 u . 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It is given that x = tanh -1 t and y = t sech -1 t , for 0 1 t 1 1. d y 2 2 -1(c) Show that =- 1 - t + `1 - t j sech t . 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Mark scheme: 6(a) e u − e − u 2 B1 Writes in exponential from. tanh u = sech u = e u + e − u e u + e − u 2 2 e u + e − u − e u − e − u  e u − e − u  2 ( ) ( ) 4 M1 Writes over common denominator. 1 −  u − u  = 2 = 2 e u + e − u e u + e − u  e + e  ( ) ( ) = sech 2 u A1 Withold this mark if working with LHS and RHS simultaneously. 3 6(b) −1 du M1 A1 Differentiates implicitly. u = sech t  sech u = t − tanh u sech u = 1 dt du 1 1 M1 A1 Uses identity from (a), AG. = − = − dt tanh u sech u t 1 − t 2 First alternative method for question 6(b) −1 1 du 1 M1 A1 Differentiates implicitly. u = sech t  cosh u =  sinh u = − t dt t 2 du 1 1 1 M1 A1 Uses cosh 2 u − sinh 2 u = 1 , AG. = − = − = − d t t 2 cosh 2 u − 1 t 2 t −2 − 1 t 1 − t 2 6(b) Second alternative method for question 6(b) −1 1 −1 1 M1 A1 −1 1 u = sech t  cosh u =  u = cosh  Differentiates cosh  . t t t du 1 −2 1 M1 A1 Applies formula with chain rule and simplifies, AG. =  −t = − 2 ( ) 2 1 dt ( t ) − 1 t 1 − t Third alternative method for question 6(b) 2  M1 A1 Uses logarithmic form of sech −1 t 1  1 1   1 + 1 − t  sech u = t  cosh u = + = ln   u = ln  − 1     t t  t t 2      d u −−1 1 − t 2  t  1 M1 A1 Differentiates logarithmic form using chain and =     = − quotient rules, AG.    d t   1 + 1 − t 2  t 1 − t 2  t 2 1 − t 2  4 6(c) dy −1 1 dx 1 B1 B1 = sech t − , = d t 1 − t 2 dt 1 − t 2 dy dy dt 2 2 −1 M1 A1 Uses chain rule, AG. =  = − 1 − t + 1 − t sech t ( ) dx dt dx 4 6(d) d  dy t 1 − t 2 −1 M1 A1 Applies product rule. Might see − − 2t sech t  = 2t 2 − 1 dt  dx  1 − t 2 t 1 − t 2 − 2t sech −1 t . t 1 − t 2 3 2 2 M1 A1 Uses chain rule. 2 1 − t d y 2 ( ) 2 −1 = t 1 − t − − 2t 1 − t sech t 2 ( ) dx t 4

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7 Find the solution of the differential equation dy x + 5 - y = 1, dx x 2 + 10x + 61 given that y = 0 when x = 3 . Give your answer in an exact form. 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Mark scheme: Finds integrating factor using dx = ln f ( x ) . x + 10 x + 61 dx = 2 ln ( )  f ( x )7  x x + 5 1 2 M1 A1 f '( x ) 2 + 10 x + 61 − 12 ln ( x 2 +10 x + 61) 1 A1 e = x 2 + 10 x + 61 d   y = 1 M1 Correct form on LHS using their integrating factor.  2  2 dx  x + 10 x + 61  x + 10 x + 61 2 2 B1 Completes the square. x + 10 x + 61 = ( x + 5 ) + 36  1 −1  x + 5 *M1 A1 Integrates RHS. Attempting to integrate by parts is dx = sinh + C M0.    6  2 + 6 2  ( x + 5 ) 8 8 C = − sinh −1 ( 6 ) = − ln ( 6 + 53 ) = − ln3 DM1 A1 Substitutes initial conditions.  y −1  x + 5  −1 4 A1 y  x + 5 x 2 + 10 x + 61 = sinh   − sinh ( 3 ) Accept = ln  +  . 2  x + 10 x + 61  6  18 18  x 2 + 10 x + 61   10

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8 (a) It is given that m is an eigenvalue of the non-singular square matrix A, with corresponding eigenvector e. Show that e is an eigenvector of A3 with corresponding eigenvalue m 3 . 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The matrix A is given by -1 3 4 A = f 0 1 0 p. 0 - 2 5 (b) Show that the eigenvalues of A are -1, 1 and 5. 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(c) Find a matrix P and a diagonal matrix D such that A - 2I = PDP -1 . 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(d) Use the characteristic equation of A to show that ( A - 2 I) 3 = aA 2 + bA + cI where a, b and c are constants to be determined. 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Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. .................................................................................................................................................................. ..................................................................................................................................................................

Mark scheme: 8(a) Ae = e  A 3 e = A ( Ae ) M1 Multiples by A 2 on LHS. Trying to invert an eigenvector or missing eigenvector is M0.  A 3 e = A (e ) = 2 ( Ae ) = 3 e A1 2 8(b) −−1  3 4 M1 Forms det ( A − I ) = 0. Do not accept checking each 0 1 −  0 = 0 given value is an eigenvalue. Do not accept row operations on A. 0 −2 5 −  −−( 1 ) ( (1 − )( 5 − ) − 0 ) − 3(0) + 4(0) = 0  ( −−1 )(1 − )( 5 − ) = 0  = −1,1 and 5. A1 AG. 2 8(c) i j k  −8  1 M1 A1 Uses vector product (or equations) of rows of A − I    to find corresponding eigenvectors. = −1: 0 3 4 = 0 0       0 2 0  0  0 i j k  20  5 i j k  16  2 A1 A1       = 1: −2 3 4 = 8 ~ 2 = 5: −6 3 4 = 0 ~ 0             0 −2 4  4  1 0 −4 0  24  3  1 5 2   −3 0 0  M1 A1 Or correctly matched permutations of columns.     A column of zeros is M0. Repeated eigenvalues in D is Thus P = 0 2 0 and D = 0 −1 0     M0.      0 1 3   0 0 3  6 8(d) ( A − 2I )3 = A3 − 6 A 2 + 12 A − 8I B1 Expands. = 5A2 + A − 5I − 6A2 + 12A − 8I M1 Substitutes A3 = 5A2 + A − 5I . = − A2 + 13A − 13I A1 Alternative method for 8(d) (+ 3)(+ 1)(− 3) = 3 + 2 − 9− 9 = 0 B1 Finds characteristic equation of A − 2 I . ( A − 2I )3 = − ( A − 2I ) 2 + 9 ( A − 2I ) + 9I M1 Substitutes A − 2 I and makes ( A − 2I )3 the subject. = − A2 + 13A − 13I A1 3

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