Cambridge A Level Mathematics - Further 9231 — 2020 May/June Paper 2 · Variant 3

1 Find the general solution of the differential equation d 2 x d x 8 t - 8 - 9x = 9e . [6] 2 d t d t .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 9 0 1,9 m m m − −=  =− M1 9 e e t t x A B − = + A1 8 8 8 8 64 t t t x ke x ke x ke =  =  =   B1 8 8 8 8 64 e 64 e 9 e 9e 9 9 t t t t k k k k − − = − = M1 1 k = − A1 9 8 e e e t t t x A B − = + − A1 6

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= 1 + 3 x e d x , where n is an integer.2 Let I n -3 x 0 (a) Show that 3I n n -3 = 1 - 4 e + 3nI n -1 . [3] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... (b) Find the exact value of I2. [3] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2(a) ( ) ( ) 1 1 1 3 3 1 3 0 0 e 1 3 e 1 3 d n n x x nI x n x x − − −   = − + + +    M1 A1 3 3 1 1 1 1 3 3 e 4 3 1 4 e 3 n n n n n nI I nI − − − − − + +  = − + A1 3 Question Answer Marks 2(b) ( ) 1 1 3 3 3 1 1 0 3 3 0 0e d e 1 e x x I x − − −   = = − = −    . B1 ( ) ( ) 3 3 3 1 1 1 3 3 1 4e 1 e 2 5e I − − − = − + − = − ( ) ( ) 3 3 1 2 3 1 16e 2 2 5e I − − = − + − M1 ( ) 3 1 2 3 5 26e I − = − A1 3

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3 The matrix A is given by 5 - 1 7 A = 0 6 0 f7 7 5 p. (a) Find the eigenvalues of A. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Use the characteristic equation of A to find A -1 . [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) 5 1 7 0 6 0 0 7 7 5 λ λ λ − − − = − B1 3 2 16 36 144 0 ( 2)( 6)( 12) 0 λ λ λ λ λ λ − + − − =  + − − = M1 2, 6, 12 λ = − A1 A1 4 Question Answer Marks 3(b) 3 2 16 36 144 0 − + − − = A A A I B1 1 2 144 16 36 −= − + − A A A I M1 2 1 74 38 70 5 9 7 1 0 36 0 0 4 0 24 70 70 74 7 7 5 − − −         =  =         −     A A M1 A1 4

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4 y x 0 1 2 3 N - 1 N The diagram shows the curve with equation y = ln x for x H 1, together with a set of ( N - 1) rectangles of unit width. (a) By considering the sum of the areas of these rectangles, show that ln N! 2 N ln N - N + 1. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Use a similar method to find, in terms of N, an upper bound for lnN!. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) + + + M1 1 ln d N x x > M1 [ ]1 1 1 ln d ln 1d ln 1 N N N x x x x x N N N = − = − +   M1 A1 ln ! ln 1 N N N N > − + A1 5 Question Answer Marks 4(b) ln1 ln 2 ln( 1) + + + −  N (or ln 2 ln( 1) + + −  N ) M1 1 ln d N x x < (or 2 ln d N x x < ) M1 ln ! ln 1 ln ( 1) ln 1 N N N N N N N N < − + + = + − + (or ln ! ( 1) ln 2(1 ln 2) N N N N < + − + − ) A1 3

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5 The curve C has parametric equations x = 12 t 2 1 - ln t , y = t2 + 1, for 2 G t G 2 . (a) Find the exact length of C. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ d 2 y (b) Find 2 in terms of t, simplifying your answer. [4] d x ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a) 1, 2 x t t y − = − =   B1 ( ) 2 2 2 2 2 1 2 4 x y t t t t − − + = −+ + = +   M1 A1 1 1 2 2 2 2 1 2 15 1 2 8 d ln 2ln 2 t t t t t −   + = + = +    M1 A1 5 Question Answer Marks 5(b) 1 d 2 d y y x x t t − = = −   B1 ( ) ( ) 2 2 1 1 2 1 d 2 d t t t t t t − − − − +  =   −   − B1 ( ) ( ) 2 2 3 2 1 1 2 1 d d 2 d d d d t y t x t t t x t t − − − − +   = × =   −   − M1 A1 4

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6 (a) Starting from the definitions of tanh and sech in terms of exponentials, prove that 1 - tanh 2 i = sech 2 i . [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ r . r for - 14 r 1 x 1 34 The variables x and y are such that tanh y = cos x + 14 b l, r with respect to x, show that (b) By differentiating the equation tanh y = cos x + 14 b l d y 1 r [4] =- cosec x + 4 d x b l. ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ r) in the form(c) Hence find the first three terms in the Maclaurin’s series for tanh -1 cos ( x + 14 b l 1 2 2 lna + bx + cx , giving the exact values of the constants a, b and c. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a) e e 2 tanh sech e e e e θ θ θ θ θ θ θ θ − − − − = = + + B1 ( ) ( ) ( ) ( ) 2 2 2 2 2 e e e e e e 4 1 e e e e e e θ θ θ θ θ θ θ θ θ θ θ θ − − − − − − + − −   − − = =   +   + + M1 2 sech θ = A1 3 Question Answer Marks 6(b) 2 1 1 4 4 d tanh cos( π) sech sin( π) d y y x y x x = +  = − + M1 A1 ( ) 2 1 1 4 4 d 1 cos ( π) sin( π) d y x x x − + = − + M1 1 4 1 4 2 1 4 sin( π) d cosec( π) d sin ( π) x y x x x + = − = − + + A1 4 6(c) 2 1 1 4 4 2 d cot( π)cosec( π) d y x x x = + + B1 1 4 '(0) cosec( π) ''(0) y y =− = 1 1 4 4 cot( π)cosec( π) M1 ( ) 1 1 1 2 2 2 2 (0) tanh 2 ln 2 2 y −   + √ = √ =   −√   M1 2 1 1 2 2 ln(3 2 2) 2 2 y x x = + √ −√+ √ M1 A1 5

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7 (a) Show that an appropriate integrating factor for 2 d y 2 2 2 ( x + 1 ) + y x + 1 = x - x x + 1 d x is x + x 2 + 1 . [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Hence find the solution of the differential equation 2 d y 2 2 2 ( x + 1 ) + y x + 1 = x - x x + 1 d x for which y = ln 2 when x = 0 . Give your answer in the form y = f ( x) . 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Mark scheme: 7(a) ( ) 2 2 2 d 1) d 1 1) y y x x x x x x + = −√( + + √( + B1 1 2 1 d sinh 1) e e x x x − √( +  = M1 A1 2 1) x x = +√( + A1 4 7(b) ( ) ( ) 2 d 1) d y x x x + √( + = 2 1 x x − + M1 A1 ( ) ( ) 2 2 1 2 2 1) d ln 1 1 x y x x x x C x + √( + = − = − + + +  M1 A1 ln 2 C = M1 ( ) ( ) ( ) 2 1 2 2 2 1 2 2 ln2 ln 1 1) ln 1) 1) x y x x x x+ x − + = = −√( + √( + √( + M1 A1 7

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i - 6 cos 4i + 15 cos 2i - 10) . [6]8 (a) Use de Moivre’s theorem to show that sin 6 i =- 321 ( cos 6 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ i + 6 cos 4i + 15 cos 2i + 10) . It is given that cos 6 i = 321 ( cos 6 1 r 3 (b) Find the exact value of 6 1 6 1 cos ( 4 x) + sin ( 4 x) d x . [4] y 0 b l ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 3 (c) Express each root of the equation 16c 6 + 16 1 - c 2 - 13 = 0 in the form coskr, where k is a ` j rational number. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................

Mark scheme: 8(a) 1 2i sin z z θ − − = B1 ( ) ( ) ( ) ( ) 6 1 6 6 4 4 2 2 6 15 20 z z z z z z z z − − − − − = + − + + + − M1 A1 ( ) ( ) ( ) 6 2isin 2cos6 6 2cos4 15 2cos2 20 θ θ θ θ = − + − M1 A1 ( ) 6 1 32 sin cos6 6cos4 15cos2 10 θ θ θ θ = − + − + A1 6 8(b) 1 3 6 6 1 1 4 4 0 cos sin d x x x π +  = 1 3 1 8 0 3cos 5d x x π +  M1 A1 [ ] ( ) 1 3 5 3 1 1 8 8 3 2 0 3sin 5 π 3 x x π + = + √ . M1 A1 4 8(c) 2 2 cos 1 sin c c θ θ =  − = B1 ( ) 3 6 2 16 16 1 13 0 6cos4 3 0 c c θ + − − =  −= M1 A1 4θ = 5 7 1 11 3 3 3 3 π, π, π, π M1 ( ) ( ) ( ) ( ) 5 7 1 11 12 12 12 12 cos π , cos π ,cos π ,cos π c = A1 5

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