Cambridge A Level Mathematics - Further 9231 — 2025 May/June Paper 2 · Variant 3

1 Find the Maclaurin’s series for e b x + 2 l up to and including the term in x2. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 −2 ( x + 2 ) −1 B1 Finds first derivative. f '( x ) = −( x + 2) e −1 B1 Finds second derivative. −4 ( x + 2) -1 −3 ( x + 2) -1 2 x + 5 ( x + 2 ) e f ''( x ) = ( x + 2) e + 2 ( x + 2 ) e = 4 ( x + 2 ) 2 *M1 Evaluates derivatives at zero. Must have f (0) = e 1 2 f '(0) = − 14 e 1 2 f ''(0) = 165 e 1 attempted second derivative using the product rule. ( x + 2 ) −1 12 1 12 5 12 2 DM1 A1 e = e − 4 e x + 32 e x 5

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2 (a) Starting from the definitions of tanh and sech in terms of exponentials, prove that tanh 2 t + sech 2 t = 1. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) The curve C has parametric equations x = ln ( cosh t) , y = tan -1 ( sinh t) , for 0 G t G 1. Find the length of C. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) e t − e − t 2 B1 tanh t = sech t = e t + e − t e t + e − t 2 t − t 2 t −2 t M1 Writes in a single fraction over common  e − e  4 e + e + 2  t − t  + 2 = 2 = 1 denominator, AG. e t + e − t e t + e − t  e + e  ( ) ( ) A1 Withold A1 mark if they used mixed variables within a single line of working. 3 2(b) d x B1 = tanh t d t dy cosh t M1 A1 = = sech t dt 1 + sinh 2 t Applies 1 + sinh 2 t = cosh 2 t. 1 1 2 2 M1 Substitutes their derivatives into a correct  0 tanh t + sech t d t =  01d t 2 2  dx   dy  formula for arc length eg.   +   dt  dt   dt  = 1 A1 5

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3 The curve C has equation 9y 2 - 3 sinh -1 ( xy) = 1 - 3 ln 3 . 1 dy 1 (a) Show that, at the point ( 4 , ) on C, =- . 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Mark scheme: 3(a) d 2 B1 Differentiates y 2 correctly. 9 y = 18 yy ' ( ) d x d −1 xy '+ y M1 A1 Differentiates sinh −xy1 ( ) correctly. −3sinh ( xy ) = −3 ( ) 2 2 dx 1 + x y 1 18 A1 Substitutes( 4, 3 ) and shows at least one 3 y '− 3 4 y5' + 1 3 = 0  y ' = − 12 3 intermediate step, AG. 4 3(b) 2 B1 Differentiates 18 yy ' 18 yy ''+ 18 ( y ' )  2 2 2 2 2 − 12  B1 B1 xy '+ y 1 + x y . Differentiates −3 xy ) 1 + x y ( xy ''+ 2 y ' ) − ( xy '+ y ) ( ) (   2 2 −3 2 2  1 + x y 1 + x y     2 5 5 3 4 1 M1 1 ( ) y = − 3 ( 4 y ''− 1) − 3 ) ( 5 )( 3 18 Substitutes ( 4, 3 ) . and = 0 2 3 y ''+ 184 − 3 25 9 45 ( 4 y ''− 1) − 60 18 18 3 y ''+ 4 − = 0 25 − 65 y ''+ 1087 = 0  y '' = 294 A1 5

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4 Find the particular solution of the differential equation d 2 x d x 2 + - 2x = 2t + t - 1, dt 2 d t dx given that, when t = 0 , x = = 0 . 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Mark scheme: 4 m2 + m − 2 = 0  m = 1, −2 M1 Auxiliary equation. M1 for correct type of roots. x = Aet + Be−2t A1 Complementary function. Allow “ x = ” missing. Using x instead of t is A0. x = pt 2 + qt + r  x ' = 2 pt + q  x '' = 2 p B1 Particular integral and its derivatives. p = −1 2 p − 2 q = 1 2 p + q − 2 r = −1 M1 Substitutes and equates coefficients. p = −1 q = − 32 r = − 54 A1 x = A e t + B e− 2 t − t 2 − 32 t − 54 A1 FT General solution. FT on CF and their values of p, q and r. Must have “ x = ”. x ' = Ae t − 2 Be −2 t − 2t − 32 *M1 Differentiates. A + B − 54 = 0 A − 2 B − 32 = 0  A = 43 , B = − 121 DM1 A1 Uses initial conditions. x = 43 e t − 121 e−2 t − t 2 − 23 t − 54 A1 Must have “ x = ”. 10

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5 (a) Use de Moivre’s theorem to show that sec 5 i sec 5 i = . 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(b) Hence, obtain the roots of the equation 3 x 5 - 10x 4 + 40 x 2 - 32 = 0 in the form sec ( q r ) , where q is rational. 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Mark scheme: 5(a) ( cos5+ isin 5) = ( c + is )5 M1 Uses binomial theorem. cos5= c5 − 10s 2 c3 + 5s 4 c A1 2 2 2 5 2 3 2 M1 Uses s = 1 − c . = c − 10 1 − c c + 5 1 − c c ( ) ( ) = 16c5 − 20c3 + 5c A1 1 sec 5 M1 Divides simplified numerator and sec 5=  16c 5 − 20c 3 + 5c sec 5 denominator by c5. sec 5 A1 AG. = 5 sec 4 − 20 sec 2 + 16 6 5(b) 5 2 4 2 x 5 2 B1 Relates with equation in part (a). Can be x = 5 x − 20 x + 16  = 3 ( ) 4 2 3 implied by M1 mark. 5 x − 20 x + 16 sec 5 = 2  cos5 = 23 M1 Solves cos5= 23 . 3 1 π ) A1 Gives one correct solution. x = sec ( 30 13 11 sec ( 25 π ) A1 Gives four other solutions. Allow different 30 π ) ,sec ( 30 π ) ,sec ( 30 23 π ) , sec ( 30 values of q as long as all five solutions are found. -1.35, -1.15, 1.01, 2.46, 4.81 4

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6 y 0 x 1 2 n - 1 1 n n n 1 The diagram shows the curve with equation y = for 0 G x G 1, together with a set of n rectangles x 2 + 1 of width 1. n (a) By considering the sum of the areas of these rectangles, show that n n 1 r . [5] 2 21 n + r 4 =/r 1 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ n n (b) Use a similar method to find a lower bound for . Give your answer in terms of n n 2 + r 2 =/r 1 and r. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ n n (c) Deduce the exact value of lim . [1] n " 3 n 2 + r 2 =/r 1 ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a) M1 A1 Forms sum of the areas of the rectangles.   1 1 1 1 1  n 2 n 2 n 2    + + + = + + + 2 2   1 2 2 n   n n 2 2 + n 2 n 2 + n 2 2 + 1 2 + 1 2 + 1  12 + n 2  n n n   1 M1 Evaluates integral. 1 1 −1 1 d x =  tan x  = 4 π 2   0 0 x + 1 A1 CAO 1 n n 2 n n 1 A1 AG. Clear and convincing comparison with =  2 2  2 2  4 π the integral. n r =1 n + r r =1 n + r 5 6(b)   2 2 2 *M1 A1 Forms the sum of the areas of appropriate 1 1 1 1 1  n n n   + + + 2  =  2 2 + 2 2 + + 2 2  rectangles. M0 for more than n rectangles. 12 ( n −1) 2 n  0 + 1 + 1 + 1  n  0 + n 1 + n ( n − 1) + n  n 2 n 2   1 n n 1 1 n n 1 1 DM1 A1 Compares with integral. Correctly adjust for +  2 2 −  4 π   2 2  4 π − 0th and last rectangles. (Allow sign errors n r =1 n + r 2n r =1 n + r 2n only for M1). Limits must match area covered by chosen rectangles. 4 6(c) 1 4 π B1 1

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7 Find the solution of the differential equation dy 2x + 6 - y = 4 , dx x 2 + 6x + 5 given that y = 0 when x = 0 . Give your answer in an exact form. 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Mark scheme: 7 2 x + 6 2 M1 A1 Finds integrating factor using dx = ln x + 6 x + 5  x 2 + 6 x + 5 ( ) f '( x ) dx = ln f( x ) .  f( x ) − ln ( x 2 + 6 x + 5 ) 1 A1 e = x 2 + 6 x + 5 d   2 y = 2 4 M1 Correct form on LHS. dx  x + 6 x + 5  x + 6 x + 5 4 1 1 B1 Partial fractions. Or completing the square. = − x 2 + 6 x + 5 x + 1 x + 5 4  x + 1  *M1 A1 Integrates RHS. (RHS must be of the correct dx = ln   + C 2 form.)  x + 6 x + 5  x + 5  0 = ln 15 + C DM1 Substitutes initial conditions. y  x + 1  1  5 x + 5  A1 2  5 x + 5  x + 6 x + 5 ln   . 2 = ln   − ln 5 = ln   Accept y = ( ) x + 6 x + 5  x + 5   x + 5   x + 5  9

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8 (a) Find the values of a for which the system of equations 3 x + 3 y + 8 z = 1 , 2 ax + 3y + 4z = 2 , ay - z = 3, does not have a unique solution. 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The matrix A is given by 3 3 8 2 A = f 0 3 4 p. 0 0 -1 (b) Given that B = A -1 , use the characteristic equation of A to show that B 2 = p I + q A , where p and q are constants to be determined. 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(c) Find a matrix P and a diagonal matrix D such that A -1 = PDP -1 . 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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) 3 2 3 8 M1 A1 Sets determinant equal to zero and forms 3 2 9 term quadratic equation. a 3 4 = 0  8 a − 3a − 2 = 0 0 a −1 3 A1 1 17 a = 16 ( ) 3 38(b) (− 2 ) (− 3)(+ 1) =  3 − 72 2 + 92 = 0 B1 Finds characteristic equation. 9 2 I = 72 A 2 − A 3  92 B = 72 A − A 2 M1 Using C-H Theorem and replaces  with A. B 2 = 79 I − 92 A M1 A1 Multiplies by B 2 = A −2 CAO 4 8(c) Eigenvalues of A are 32 , 3 and −.1 B1 Upper diagonal matrix or characteristic equation. i j k  − 92  1 M1 A1 Uses vector product (or equations) to find 3    corresponding eigenvectors. = 2 : 0 3 8 =  0  0 3 5    0 2 2  0  0 i j k  12  2 i j k  −20   2  A1 A1 3    5     = 3: − 2 3 8 =  6  ~ 1 = −1: 2 3 8 =  −10  ~  1         0 0 4  0  0 0 4 4  10   −1   1 2 2   23 0 0  M1 A1 Or correctly matched permutations of    1  columns. Column of zeros or repeated 0 1 1 Thus P = and D =  0 3 0        column in P, or repeated eigenvalues in D is  0 0 −1   0 0 −1  M0. 7

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