Cambridge A Level Mathematics - Further 9231 — 2024 Oct/Nov Paper 2 · Variant 2

1 Find the value of dx , giving your answer in the form ln ( a + b) , where a and b are 2 - 1 6y ( x - 5) integers to be determined. [4] .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... ..........................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 7 M1A1 Applies formula. 7  1 −1 dx =  cosh ( x − 5 )  6 ( x − 5) 2 − 12   6 = cosh −1 ( 2 ) − cosh −1 (1) M1 Uses limits. = ln 2 + 3 A1 ( ) 4

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2 The curve C has equation 4 y 2 + 4 ln ( xy) = 1. 1 d y 1 (a) Show that, at the point b,2 l on C, = - . [3] 2 d x 6 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ d 2 y 1 (b) Find the value of at the point b,2 l. 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Mark scheme: 2(a) d 2 B1 Differentiates y 2 correctly. 4 y = 8 yy ' ( ) dx d xy '+ y −1 −1 B1 Differentiates ln( xy ) correctly. ( 4ln( xy ) ) = 4 = 4 y ' y + 4 x dx xy 1 1 1 B1 Substitutes (2, 12 ) into 8 ( 2 ) y '+ 4 ( 2 ) y '+ 4 ( 2 ) = 0  y ' = − 6 8 y ' y + 4 y ' y −1 + 4 x −1 = 0, AG. 3 2(b) EITHER Solution 1 (B1B1 Differentiates 8 y ' y + 4 y ' y −1 + 4 x −1 = 0. 2 + 8 y '' y and B1 for other terms. 8 ( y ' )2 + 8 y '' y − 4 ( y ' ) 2 y −2 + 4 y −1 y ''− 4 x −2 = 0 B1 for 8 ( y ' ) 2 1 1 1 2 M1) 1 8 ( − 6 ) + 8 y '' ( 2 ) − 16 ( − 6 ) + 8 y ''−=1 0 Substitutes ( 2, 2 ) . OR Solution 2 (B1B1 −1 −1 −2 −2 2 y + y 2 y '− y y ' and B1 −1 B1 for x ( ) ( ) −1 −1 y ' = − x 2 y + y ( ) −1 2 y + y −1 −2 −1 for x −2 ( ) . −1 −1 −2 −2 −1 y '' = x 2 y + y 2 y '− y y ' + x 2 y + y ( ) ( ) ( ) 1 −2 1 1 −1 M1) 1 1 and y ' = − 6 . y '' = 2 (1 + 2 ) ( − 6 )( 2 − 4 ) + 4 (1 + 2 ) Substitutes ( 2, 2 ) 12 y ''− 119 = 0  y '' = 10811 A1 4

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3 The curve C has parametric equations 1 t2 1 3 1 x = e - t - , y = 2et ( t - 1 ) , for 0 G t G 1. 2 3 2 Find the exact length of C. 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Mark scheme: 3 x = e 2 t − t 2 , y = 2e t + 2(t − 1)e t = 2te t B1B1 Differentiates x and y with respect to t. 2 2 M1A1 Factorises x 2 + y 2 . 2 2 2 t 2 2 2 t 4 t 2 2 t 4 2 t 2 x + y = e − t + 4t e = e + 2t e + t = e + t ( ) ( ) 1 2 t 2 1 2 t 1 3 e + t dt =   2 e + 3 t  0 0 1 M1A1 Applies correct formula for arc length. = 1 2e − 1 A1 2 6 7

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4 (a) Use de Moivre’s theorem to show that cot 6 i - 15 cot 4 i + 15 cot 2 i - 1 cot 6 i = . [6] 6 cot 5 i - 20 cot 3 i + 6 cot i ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Hence obtain the roots of the equation x 6 - 6 x 5 - 15x 4 + 20x 3 + 15x 2 - 6x - 1 = 0 in the form cot ( qr ) , where q is a rational number. 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Mark scheme: 4(a) 6 6 4 2 2 4 6 M1A1 Expands and takes real part. cos6= Re ( cos+ isin) = cos − 15 cos sin + 15 cos sin − sin  6 5 3 3 5 M1A1 Takes imaginary part sin6= Im ( cos+ isin) = 6cos sin− 20cos sin + 6cossin  cos 6 − 15cos 4 sin 2 + 15cos 2 sin 4 − sin 6  sin −6  M1 Divides numerator and denominator by cot6 =  6cos 5 sin− 20cos 3 sin 3 + 6cossi n 5  sin −6  sin 6 . cot 6 − 15cot 4 + 15co t 2 − 1 A1 AG cot6= 6 cot 5 − 20cot 3 + 6 cot 6 4(b) x = cot, cot 6= 1 B1 Applies identity given in (b). 6 = 14 + k M1 Solves cot6= 1 1 π ) A1 Gives one correct solution. cot ( 24 7.60 13 17 3 π ) ,cot ( 7 π ) A1 Gives other five solutions. cot ( 24 5 π ) ,cot ( 8 24 π ) ,cot ( 24 π ) ,cot ( 8 1.30, 0.41, –0.13, –0.77, –2.41 Withhold A1 mark for repeated roots. 4

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5 Find the particular solution of the differential equation d 2 y dy 2 3 + 2 + y = x , dx 2 dx dy given that, when x = 0 , y = = 0 . [10] dx .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... .......................................................................................................................................................... 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Mark scheme: 5 2 1 2 M1 Auxiliary equation. M1 for correct type of 3m + 2m + 1 = 0  m = −3 3 i roots (complex). − 13 x 2 2 A1 Complementary function. y = e A cos x + B sin x ( 3 3 ) y = px 2 + qx + r  y ' = 2 px + q  y '' = 2 p B1 Particular integral and its derivatives. p = 1 4 p + q = 0 6 p + 2q + r = 0 M1 Substitutes and equates coefficients. q = −4 r = 2 A1 − 13 x 2 2 2 A1FT General solution. Must have ‘ y = ’. FT on y = e A cos x + B sin x + x − 4 x + 2 ( 3 3 ) CF. − 13 x 2 2 2 2 1 − 13 x 2 2 *M1 Differentiates. GS must be in correct form. y ' = e − A sin x + B cos x − e A cos x + B sin x + 2 x − 4 ( 3 3 3 3 ) 3 ( 3 3 )  A = − 2, B = 5 2 A + 2 = 0 32 B − 13 A − 4 = 0 DM1A1 Uses initial conditions. − 13 x 2 2 2 A1 Must have ‘ y = ’. y = e −2cos x + 5 2sin x + x − 4 x + 2 ( 3 3 ) 10

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6 y 0 x 1 2 n - 1 1 n n n The diagram shows the curve with equation y = e 1 - x 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 e 1 - x dx 1 U , where n 0y e - 1 U = . [4] n - 1n n (1 - e ) ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 1 e 1 - x dx . [4](b) Use a similar method to find, in terms of n, a lower bound Ln for y 0 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Show that lim ( U - L ) = 0 . 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(d) Use the Maclaurin’s series for ex given in the list of formulae (MF19) to find the first three - 1z terms of the series expansion of z ( 1 - e ) , in ascending powers of 1, and deduce the value of z lim ( Un ). 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Mark scheme: 0 M1A1 Forms the sum of the areas of the rectangles.6(a)  1 1− x  1 1− n 1 1− 1n 1 1− nn−1 e dx  ( n ) e + ( n ) e + + ( n ) e M1 for correct number of rectangles.    0  r e  1 − e −1  e − 1 M1A1 Applies sum of geometric progression, AG. e n −1 − n = e = =    n n n n r = 0 n 1 − e − 1  1 − e − 1  ( ) 4 16(b) 1 1− x 1 1− n 1 1− n2 1 1− nn−1 1 1− nn M1A1 Forms the sum of the areas of appropriate 0 e d x  ( n ) e + ( n ) e + ( n ) e + ( n ) e rectangles. M1 for correct number rectangles. e − 1 e 1 e − 1 M1A1 Applies sum of geometric progression. = + = −1 n n n e1− 1 1 − e n n e 1 − 1n ( e − ) ( ) n 1 − e − 1 n e 1 − 1 ( ) − ( ) OE. E.g. or . 1 1 − − n n n 1 − e n 1 − e ( ) ( ) 4 6(c) e − 1 M1 c U n − Ln = Simplifies U n − Ln to , where c is a n n nonzero constant. e −→1 0 as n → A1 AG. CWO. n 2 6(d) 1 z M1A1 Substitutes power series for e −z1 . A1 for z 1 − e − 1 = z 1 − + 2 − 2 + ( 1 − 1z + 21z 6 z 3 = 1 − 21z + 6 1z ( ) ) ( ) enough terms. lim (U n ) = e − 1 B1 n → 3

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7 (a) Show that ( ln ( tanh x)) = 2 cosech 2x . [3] dx ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ .......................... ............................................................................................................................ (b) Find the solution of the differential equation dy sinh 2x + 2 y = sinh 2 x dx for which y = 5 when x = ln 2 . Give your answer in an exact form. 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Mark scheme: 7(a) d sech 2 x M1A1 Applies chain rule. ( ln ( tanh x ) ) = dx tanh x 1 2 A1 AG. = = . sinh x cosh x sinh2 x 3 7(b) d y 2 y B1 Divides through by sinh(2 )x . + = 1 d x sinh(2 x ) d M1A1 Correct form on LHS and RHS. ( y tanh x ) = tanh x dx y tanh x = ln(cosh x ) + C M1A1 Integrates RHS. 3 5 ) = ln 5 ( 5 4 +C M1 Substitutes initial conditions. y tanh x = ln(cosh x ) + 3 − ln 54 A1 Accept equivalent exact form. 7

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8 The matrix A is given by - 2 0 0 A = f 0 7 9p. 4 1 7 (a) Show that the characteristic equation of A is m 3 - 12 m 2 + 12m + 80 = 0 and find the eigenvalues of A. 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(b) Use the characteristic equation of A to show that A 4 = pA 2 + qA + rI , where p, q and r are integers to be determined. 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(c) Find a matrix P and a diagonal matrix D such that ( A - 3I) 4 = PDP -1 . 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Mark scheme: 8(a) −−2  0 0 B1 Sets determinant equal to zero. 0 7 −  9 = 0 4 1 7 −  49 − 14+  − 9 ( −−2 ) 2 2 M1A1 Expands, AG ( 7 − ) − 9 = 0  ( −−2 )( ) ( )  3 − 122 + 12+ 80 = 0 = −2, 4,10 B1 4 8(b) A 3 − 12 A 2 + 12 A + 80I = 0  A 4 − 12 A 3 + 12 A 2 + 80 A = 0 M1A1 Substitutes A and multiplies through by A. A 4 = 12 12 A 2 − 12 A − 80I − 12 A 2 − 80 A = 132 A 2 − 224 A − 960I M1A1 Substitutes A3. ( ) 4 8(c) i j k  72   2  *M1A1 Uses vector product (or equations) to find     corresponding eigenvectors. = −2: 0 9 9 = 36 1         4 1 9  −36   −1  i j k  0   0  i j k  0  0 A1A1        = 4: 0 3 9 = 36 ~ 3 = 10: −12 0 0 = 108 ~ 3               4 1 3  −12   −1  0 −3 9  36  1  2 0 0   625 0 0  DM1A1 Or correctly matched permutations of     columns. Thus P = 1 3 3 and D = 0 1 0          −1 −1 1   0 0 2401  Correctly matched but eigenvalues not changed scores M1 A0. M0 if a column of zeros appears in P. 6

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