Cambridge A Level Mathematics - Further 9231 — 2023 May/June Paper 2 · Variant 3

1 (a) Find the Maclaurin series for sin -1 x up to and including the term in x3. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 1 5 1 (b) Deduce an approximation to d u , giving your answer as a fraction. [1] 2 y0 1 - u ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 1(a) 1 2 2 '( ) 1 f x x    B1 Correct first derivative.   3 2 2 ''( ) 1 f x x x    B1 Correct second derivative.     5 3 2 2 2 2 2 '''( ) 3 1 1 f x x x x       M1 Differentiates their ''( ) f x using product rule. (0) 0 '(0) 1 ''(0) 0 '''(0) 1 f f f f     M1 Evaluates their derivatives at 0 x  . Must have attempted all three derivatives. 3 1 6 1 sin x x x    A1 CWO. Alternative method for question 1(a) sin '( ) sec y x f x y    (B1) Finds first derivative.   2 ''( ) tan sec f x y y  (B1) Finds second derivative.   2 2 3 4 5 '''( ) 2tan sec sec sec 3sec 2sec f x y y y y y y     (M1) Differentiates   2 tan sec y y using product and chain rule. (0) 0 '(0) 1 ''(0) 0 '''(0) 1 f f f f     (M1) Evaluates their derivatives at 0 x  . Must have attempted all three derivatives. 3 1 6 1 sin x x x    (A1) CWO. 5 Question Answer Partial Marks Guidance 1(b) 151 750 B1 1

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2 The variables x and y are related by the differential equation d 2 x dx 2 6 2 + 5 + x = t + 10 t + 13 . dt dt (a) Find the general solution for x in terms of t. [6] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) State an approximate solution for large positive values of t. [1] ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) 2 3 1 6 , 5 1 1 0 2 m m m      M1 Auxiliary equation. 1 1 3 2 e e t t x A B     A1 Complementary function. Allow with ‘x =’ missing. 2 ' 2 '' 2 x p qt rt x q rt x r         B1 Particular integral and its derivatives. 2 2 12 5 10 10 13 r q rt p qt rt t t         M1 Substitutes and equates coefficients. 1, 0, 1 r q p    A1 1 1 3 2 2 1 e e t t x A B t       A1 Must have x . 6 2(b) 2 1 x t   B1FT Must have ‘x =’. Accept ‘x is approximately’ but do not accept ‘ x ‘. FT on their PI. 1

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3 By considering the binomial expansions of z + and z - , where z = cos i + i sin i , use de z z Moivre’s theorem to show that cos 4 i + a cos 2i + b cot 4i = , cos 4 i - a cos 2i + b where a and b are integers to be determined. [7] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 3 1 2isin z z     and 1 2cos z z     B1 Use of 1 2isin z z     and 1 2cos z z           4 1 4 4 2 2 4 6 z z z z z z          M1 A1 Expands and groups.       4 1 4 4 2 2 4 6 z z z z z z          M1 A1 Expands and groups. Only withhold one A1 mark (this or the previous) for no clear grouping.     4 4 4 4 4 2cos2 6 4 2cos2 6 2cos4 2 cos 2cos4 2 sin            M1 Substitutes 2cos n n z z n    once in LHS. 4 3 cos4 c 4cos2 3 4 c cos o 2 t os4           A1 7

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4 The curve C has equation 4y 3 + ( x + y) 6 = 109 . dy 1 (a) Show that, at the point ( - 4 , 3) on C, = . [3] dx 17 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ d 2 y (b) Find the value of 2 at the point ( - 4 , 3). [5] dx ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a)   3 2 d 4 12 ' d y y y x  3 y correctly.     6 5 d 6( ) (1 ') d x y x y y x     B1 Differentiates   6 x y  correctly.      2 1 12 3 ' 6 1 1 ' 0 ' 17 y y y       B1 Substitutes ( 4,3),  AG. 3 Question Answer Partial Marks Guidance 4(b)   2 2 2 '' 4 y y y y  B1 Differentiates 2 . y y      5 4 2 5 1 0 x y y x y y         B1 B1 Differentiates   5 ( ) 1 . x y y       2 2 4 0 1 1 1 7 12 5 1 18 96 ( 1) 0 17 7 17 7 y y          M1 Substitutes ( 4,3).  96 289 y A1 5

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5 (a) Starting from the definitions of cosh and sinh in terms of exponentials, prove that 2 cosh 2 x = cosh 2 x + 1. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the solution of the differential equation dy + 2y tanh x = 1 dx for which y = 1 when x = 0 . Give your answer in the form y = f ( x) . [8] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 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Mark scheme: 5(a) 1 2 cosh e e x x x    B1     2 2 1 2 2 2 4 e e e e 2 cosh 2 1 x x x x x         M1 A1 Expands, AG. 3 Question Answer Partial Marks Guidance 5(b) d 2lncosh 2 tanh 2 e e cosh x x x x    M1 A1 Finds integrating factor.   2 2 d cosh cosh d y x x x  M1 Correct form on LHS,   d d yI x for their integrating factor I, and attempt to integrate their RHS.   2 2 1 1 1 cosh cosh d cosh2 1d sinh2 2 2 2 y x x x x x x x C         M1 A1 Uses   2 1 cosh cosh 2 1 . 2 x x   M1 A1 is for RHS. 1 C  M1 Finds C.   2 1 1 sech sinh2 1 4 2 y x x x    M1 A1 Divides through by their coefficient of y. 8

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6 y 1 O 1 2 n - 1 1 x n n n The diagram shows the curve with equation y = ( 1 - x) 2 for 0 G x G 1, together with a set of n rectangles of width 1. n 1 (a) By considering the sum of the areas of these rectangles, show that ( 1 - )x 2 dx 1 U n , where y0 2n 2 + 3n + 1 U n = 2 . [5] 6n ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 1 ( 1 - )x dx . [4](b) Use a similar method to find, in terms of n, a lower bound Ln for y0 2 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ lim(c) Show that ( U n - L n ) = 0 . [2] "3 n ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a)   2 2 2 2 1 0 0 1 1 1 1 1 1 d 1 1 1 n x x n n n n n n            M1 A1 Forms the sum of the areas of the rectangles. Must have first two terms and last term for A1. Accept   2 1 1 n n n  for the last term. Accept written in summation form with correct limits.     2 2 2 3 1 3 1 1 1 1 1 1 1 2 n n r r n r n nr r n n n n             3 1 1 2 1 2 1 2 1 1 n n r r r r n n               M1 A1 Setting up correct series so that formulae from MF19 can be applied. Can also recognise sum as 1 2 3 1 n r r n      2 2 2 2 3 3 3 2 2 1 1 1 ( 1)(2 1) 1 2 3 1 2 3 1 6 6 6 n n n n n n n n n n n n n n n n n n               A1 AG. Must have gained previous accuracy mark. 5 Question Answer Partial Marks Guidance 6(b)         2 1 1 1 1 1 1 0 2 2 2 2 1 d 1 1 1 n n n n n n n x x            M1 A1 Forms the sum of the areas of appropriate rectangles. Must have first two and last terms for A1. Accept   1 2 1 n n n  for the last term. Accept written in summation form with correct limits.       1 2 2 2 2 3 3 3 1 3 1 1 1 ( 1)(2 1) 2 6 n r n n n n n n n n nr r n n n n             1 2 2 3 1 2 1 1 n n r r r r n n             M1 Setting up correct series so that formulae from MF19 can be applied. Recognising sum as 1 2 2 3 1 3 1 1 1 1 n n r r r r n n n        without wrong working scores M1 A1 M1. 2 2 2 3 1 6 n n n   A1 4 6(c) 1 n n U L n   M1 Simplifies n n U L  to . c n l 0 n  as n  A1 AG. 2

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= ( 1 + x ) d x .7 The integral In, where n is an integer, is defined by I n 3 2 21 n 0 (a) Find the exact value of I -1 giving your answer in the form lna, where a is an integer to be determined. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ d 2 2 (b) By considering x ( 1 + x ) 1 n or otherwise, show that dx b l, n ( n + 1) I n 4 5 = nI n - 2 + 3 3 . [5] b l ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) A curve has equation y = x2 , for 0 G x G 23 . The arc length of the curve is denoted by s. Use the substitution u = 2x to show that s = 12 I 1 and find the exact value of s. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 7(a)   1 4 4 3 3 2 2 1 1 0 0 1 d sinh ln3 I x x x             M1 A1 Recognises integral. 2 7(b)       1 1 1 2 2 2 2 2 1 2 2 d 1 1 1 d n n n x x nx x x x             M1 A1 Uses the product rule to differentiate.      1 1 2 2 2 2 1 2 1 1 1 1 n n n x x x       M1* Uses 2 2 1 1 x x  .   4 1 3 2 2 0 2 1 n n n n x x nI nI I            DM1 Integrates both sides using the limits given. Requires previous method mark.       2 2 5 5 4 4 1 1 3 3 3 3 n n n n n n n I nI n I nI          A1 Substitutes limits and rearranges. AG. 5 7(c) 2 3 2 0 1 4 d s x x    M1 Forms correct integral with correct limits. 4 3 1 0 2 1 1 1 d 2 2 u u I    A1 AG.  1 1 5 4 2 3 3 I I   M1 Applies reduction formula with 1 n . 5 1 ln3 4 9 s   A1 4

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8 The matrix A is given by a - 6a 2 a + 2 A = 0 1 - a 0 f0 2 - a - 1 p where a is a constant with a ! 0 and a ! 1. x 1 (a) Show that the equation A y = 2 has a unique solution and interpret this situation geometrically. f z p f 3 p [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Show that the eigenvalues of A are a, 1- a and - 1. 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(c) Find a matrix P and a diagonal matrix D such that A 4 = PDP -1 . 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(d) Use the characteristic equation of A to find A4 in terms of A and a. 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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) 6 2 2 0 1 0 ( 1) 0 2 0 1 a a a a a a a         or 2 2 ( 1) 12 2 7 1 1 2 1 1 1 2 1 2(2 ) 1 3 1 1 a a a a ax a a a y a a a z a a                   equations. Three planes intersect at a single point. B1 Must be clear that there is just one point where all 3 planes intersect. 3 8(b) 6 2 2 0 1 0 0 0 2 1 a a a a a            M1 Equates determinant to zero or working to find characteristic equation.   )( 1 ) 0 , (1 1 , 1 a a a a              A1 AG. Factorisation must be clear. 2 Question Answer Partial Marks Guidance 8(c) 2 2 1 : 0 1 2 0 0 0 0 2 1 0 0 1 a a a a a a                       i j k  M1 A1 Uses vector product (or equations) to find corresponding eigenvectors. (2 )(2 2) 2 1: 1 6 2 2 0 0 0 2 0 1 ( 1)(2 ) a a a a a a a a                                 i j k  A1 2 2 2 4 2 1 : 2 1 6 2 2 4 2 10 5 2 5 2 2 2 1 0 2 1 a a a a a a a a a a a a                                   i j k  A1 1 2 2 0 0 1 0 1 1            P and 4 4 0 0 0 1 0 0 0 (1 ) a a            D M1 A1 Correctly matched permutations of columns. Their eigenvectors must be non-zero for M1. 6 8(d)   3 2 2 )( 1 ) 0 (1 ( ( 1) ) a a a a a a                 B1 Finds characteristic equation multiplied out. 2 4 2 2 1) ( ( ) a a a a      A A A M1 A1 Multiplies through by . A 3

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