Cambridge A Level Mathematics - Further 9231 — 2020 May/June Paper 2 · Variant 1

1 Find the solution of the differential equation d y -7 x + 5y = e d x for which y = 0 when x = 0 . Give your answer in the form y = f ( x) . [6] ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ........................................................................................................................................................................... ...........................................................................................................................................................................

Mark scheme: 1 5d 5 e e =  x x M1 A1 ( ) 5 2 d e e d − = x x y x M1 5 2 1 2 e e− = − + x x y C A1 1 2 0 = −+ C M1 5 7 1 1 2 2 e e − − = − x x y A1 6

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2 It is given that y = 2x . d y x (a) By differentiating lny with respect to x, show that = 2 ln 2 . [3] d x ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... d2 y (b) Write down 2 . [1] d x ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... (c) Hence find the first three terms in the Maclaurin’s series for 2x. [3] ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ...................................................................................................................................................................

Mark scheme: 2(a) 1 d ln ln 2 ln 2 d =  = y y x y x M1 A1 d ln2 2 ln2 d = = x y y x AG A1 3 2(b) ( ) 2 2 2 d 2 ln2 d = x y x B1 1 Question Answer Marks 2(c) ( ) 2 (0) 1, '(0) ln2, ''(0) ln 2 = = = y y y B1 ( ) ( ) 2 2 2 ln 2 ''(0) 2 (0) '(0) 1 ln 2 2! 2 = + + + = + + +   x y y y x x x x M1 A1 3

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3 (a) Find the roots of the equation z 3 = - 1 - i , giving your answers in the form r e ii, where r 2 0 and 0 G i 1 2 r . [5] ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... Let w = z 31 k 3 k 3 k 3 + z 2 + z 3 , where k is a positive integer and z 1 , z 2 , z 3 are the roots of z = - 1 - i . (b) Express w in the form Re ia, where R 2 0 , giving R and a in terms of k. [3] ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ...................................................................................................................................................................

Mark scheme: 3(a) 5 1 2 4i π 3 1 i 2 e z = −−= B1 1 5 6 12 i π 1 2 e z = M1 A1 1 13 6 12 i 2 2 e π = z , 1 21 6 12 i π 3 2 e z = A1 A1 5 3(b) ( ) 5 1 2 4i π 3 3 3 1 2 3 3 2 e k k k k k z z z + + = M1 ( ) 1 2 | | 3 2 = = k R w A1 5 4 π k α = A1 3

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4 y 1 x 0 1 2 n - 1 1 n n n The diagram shows the curve with equation y = x2 for 0 G x G 1 , together with a set of n rectangles of width 1. n (a) By considering the sum of the areas of these rectangles, show that 1 2 x d x 1 2 . [4] y 2 2n + 3n + 1 0 6n ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 1 (b) Use a similar method to find, in terms of n, a lower bound for x 2 d x . [4] 0y ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ................................................................................................................................................................... ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) ( )( ) ( )( ) ( )( ) ( )( ) 1 2 2 2 2 2 1 1 1 1 2 1 1 0 d − < + + + +   n n n n n n n n n n x x M1 A1 2 2 3 3 2 1 1 ( 1)(2 1) 2 3 1 6 6 n r n n n n n r n n n = + + + + = =  M1 A1 4 4(b) ( )( ) ( )( ) ( )( ) 1 2 2 2 2 1 1 1 1 2 1 0 d − > + + +   n n n n n n n x x M1 A1 1 2 2 3 3 2 1 1 ( 1)( )(2 2 1) 2 3 1 6 6 − = − − + − + = = =  n r n n n n n r n n n M1 A1 4

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5 The curves C 1 : y = cosh x and C 2 : y = sinh 2x intersect at the point where x = a . (a) Find the exact value of a, giving your answer in logarithmic form. 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(b) Sketch C1 and C2 on the same diagram. [2] (c) Find the exact value of the length of the arc of C1 from x = 0 to x = a . [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a) 1 2 cosh 2sinh cosh sinh =  = a a a a M1 A1 ( ) 1 1 1 2 2 1 4 sinh ln 1 − = = + + a M1 ( ) 1 1 2 2 ln 5 = + √ a A1 4 Question Answer Marks 5(b) (B1 for C1 correct, B1 for C2 correct and intersecting C1 in the first quadrant) B1 B1 2 5(c) 2 0 1 sinh d +  a x x M1 2 0 0 cosh d cosh d =   a a x x x x M1 A1 [ ]0 sinh sinh a x a = M1 1 2 A1 5

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= 1 - x d x .6 The integral In, where n is an integer, is defined by I n 2 2 2 0 (a) Find the exact value of I1. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ d 2 - 21 n (b) By considering x 1 - x or otherwise, show that d x e ` j o, nI n+2 n - 1 - 21 n = 2 3 + ( n - 1) I n . [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Find the exact value of I5 giving the answer in the form k 3, where k is a rational number to be determined. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a) ( ) 1 1 2 2 1 2 2 1 1 0 6 0 1 1 d sin π I x x x − −   = − = =    M1 A1 2 6(b) ( ) ( ) ( ) 1 1 1 2 2 2 2 2 2 2 1 d 1 1 1 d − − − −   − = − + −     n n n x x nx x x x M1 A1 ( ) ( )( ) ( ) 1 1 2 2 1 2 2 2 1 1 1 1 − − − = − − − + − n n n x x x M1 ( ) 1 1 2 2 2 2 0 1 + −   − = − +     n n n n x x nI nI I M1 ( ) 1 2 1 2 3 1 2 2 4 1 2 ( 1) 2 3 ( 1) − + − − + = − −  = + − n n n n n n n nI n I nI n I A1 5 6(c) 1 2 3 3 − = I B1 3 2 5 10 7 2 3 2 5 3 2 3 2 3 − = +  = √ I I I M1 A1 3

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7 It is given that x = t 3 y and 3 d 2 y 3 2 d y 3 2 2 + 4t + 6t + 13t + 12t + 6 t y = 61e . t 2 1 t d t ` j d t ` j (a) Show that d 2 x d x 21 t + 4 + 13x = 61e . [4] 2 d t d t ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the general solution for y in terms of t. 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Mark scheme: 7(a) 3 2 d d 3 d d = + x y t t y t t B1 2 2 3 2 2 2 d d d 6 6 d d d = + + x y y t t ty t t t B1 2 2 3 2 3 2 3 2 2 d d d d d 4 13 6 6 4 12 13 d d d d d + + = + + + + + x x y y y x t t ty t t y t y t t t t t 1 2 61e = t M1 A1 4 7(b) 2 4 13 0 2 3 + + =  = −± m m m i M1 ( ) 2 e cos3 sin3 − = + t x A t B t A1 1 1 1 2 2 2 1 1 2 4 e e e =  =  =   t t t x k x k x k B1 1 4 2 13 61 4 + + =  = k k k k M1 A1 ( ) ( ) 1 1 2 2 3 2 3 2 3 e cos3 sin3 4e e cos3 sin3 4 e − − − − = + +  = + + t t t t t y A t B t y t A t B t t M1 A1 7

More questions on Differential equations

8 (a) Find the values of a for which the system of equations 3x + y + z = 0, ax + 6y - z = 0, ay - 2z = 0, does not have a unique solution. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The matrix A is given by 3 1 1 A = 0 6 - 1 f0 0 - 2p. (b) Use the characteristic equation of A to find the inverse of A2. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Find a matrix P and a diagonal matrix D such that A 5 = PDP -1 . [7] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 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) ( ) ( ) 2 3 1 1 6 1 0 3 12 2 0 0 2 − =  − + −− + = − a a a a a M1 2 5 36 0 4, 9 + − =  = − a a a M1 A1 3 8(b) 3 2 3)( 6)( 2 0 ( 7 36 ) λ λ λ λ λ − − + + = = − B1 ( ) 2 2 3 1 36 7 36 7 − = −  = − I A A A I A M1 ( ) 2 1 4 1 1 1 0 1 1 36 0 0 9 − − −     =       A M1 A1 4 Question Answer Marks 8(c) Eigenvalues of A are 3, 6 and 2 −. B1 4 1 λ 3: 0 1 1 0 0 0 3 1 0 0 −       = =       −    i j k  M1 A1 1 6: 3 1 1 3 0 0 1 0 λ −     = − = −     −   i j k 9 2: 5 1 1 5 0 8 1 40 λ −     = − =     −   i j k A1 A1 Thus 1 1 9 0 3 5 0 0 40 −     =       P and 243 0 0 0 7776 0 0 0 32     =     −   D or 5 5 5 3 0 0 0 6 0 0 0 2     =       −   D M1 A1 7

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