Cambridge A Level Mathematics - Further 9231 — 2021 Oct/Nov Paper 2 · Variant 2

1 It is given that y = sinh ( x 2 ) + cosh ( x 2 ) . (a) Use standard results from the list of formulae (MF19) to find the Maclaurin’s series for y in terms of x up to and including the term in x4. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ d 4 y (b) Deduce the value of 4 when x = 0 . [1] dx ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Use your answer to part (a) to find an approximation to 2 y dx , giving your answer as a rational 1y0 fraction in its lowest terms. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 1(a) ( ) ( ) 2 2 sinh cosh + x x attempt at all four derivatives of y. 4 1 2 2 1+ + x x A1 2 1(b) 1 2 4! 12 × = B1 1 1(c) 1 2 1 2 0 2 4 1 d + +  x x x M1 Substitutes their power series, must be at least 2. + a bx 1 2 523 1 1 3 10 960 0 3 5   = + + =   x x x A1 2

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2 Find the solution of the differential equation dy 4 x 3 y + = 6x 4 dx x + 5 for which y = 1 when x = 1. Give your answer in the form y = f ( x) . [7] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 ( ) 4 ln 5 4 e 5 + = + x x ( ) ( ) 4 5 d 5 6 30 d + = + y x x x x M1 Writes in correct form. 4 6 2 ( 5) 15 + = + + y x x x C A1 6 16 = + C M1 Substitutes initial conditions. 6 2 4 15 10 5 + − = + x x y x M1 A1 Division through by coefficient of y. 7

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3 y 1 0 1 2 n - 1 1 x n n n The diagram shows the curve with equation y = 1 - x 2 for 0 G x G 1, together with a set of n rectangles of width 1. n (a) By considering the sum of the areas of the rectangles, show that 1 2 ( 1 - x ) dx 1 . [4] y0 2 4n + 3n - 1 6n 2 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 1 (b) Use a similar method to find, in terms of n, a lower bound for ( 1 - x 2 ) dx . 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Mark scheme: 3(a) ( ) ( ) 2 2 2 1 ( 1) 2 1 1 1 1 01 d 1 1 − − < + − + + −   n n n n n n x x M1 A1 Forms the sum of the areas of the rectangles. Three terms, including last for A1. Allow use of sigma. 1 2 3 3 1 1 ( 1) (2 1) 1 1 6 n r n n n r n n − = − − − = −  M1 Applies 1 2 1 6 1 ( 1) (2 1). − = = − −  n r r n n n 2 2 2 2 6 ( 1)(2 1) 4 3 1 6 6 n n n n n n n − − − + − = A1 AG. [Can get this A1 without previous A1.] 4 Question Answer Marks Guidance 3(b) ( ) ( ) ( ) ( ) 2 2 2 2 2 1 1 1 0 2 1 1 1 2 1 1 1 d 1 − − − > − + − + +   n n n n n n n x x M1 A1 Forms the sum of the areas of appropriate rectangles. Three terms, including last for A1. Allow ( ) 2 2 1 1 1 . = −  n r n n r 1 2 3 3 1 1 1 1 ( 1) (2 1) 6 n r n n n n n r n n n n − = − − − − − = −  M1 Applies 1 2 1 6 1 ( 1) (2 1). − = = − −  n r r n n n 2 2 2 6 ( 1) ( 1)(2 1) 4 3 1 6 6 n n n n n n n n − − − − − − = A1 SC: 2 2 4 3 1 1 6 + −− n n n n scores 2/4. 4

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4 (a) Write down all the roots of the equation x 5 - 1 = 0 . [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Use de Moivre’s theorem to show that cos 4i = 8 cos 4 i - 8 cos 2 i+ 1. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Use the results of parts (a) and (b) to express each real root of the equation 8x 9 - 8x 7 + x 5 - 8x 4 + 8x 2 - 1 = 0 in the form coskr, where k is a rational number. 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Mark scheme: 4(a) 2 5 π ie , 0,1,2,3,4 = k k B2 OE. B1 for 1 correct fifth root of unity. B2 for exactly 5 distinct, correct roots. 2 4(b) ( ) 4 4 3 2 2 3 4 i 4 (i ) 6 (i ) 4 (i ) (i ) + = + + + + c s c c s c s c s s M1 A1 Uses binomial expansion. Can also go RHS to LHS using 1 2cos . θ − = + z z 4 2 2 4 4 2 2 2 2 cos4 6 6 (1 ) (1 ) θ − + = − + − = − c c s s c c c c M1 Applies 2 2 1 = − s c or 1 2cos . θ − = + z z 4 2 8cos 1 8cos θ θ − + A1 AG. Can get final A1 without first. SC: Using double angle formulae scores 1/4. 4 Question Answer Marks Guidance 4(c) 9 7 5 4 2 5 4 2 8 8 8 8 1 ( 1)(8 8 1) − + − + −= − − + x x x x x x x x B1 cos4 0 θ = leading to 1 2 4 (2 1)π k θ = + M1 A1 Solves cos4 0. θ = A1 for 3 5 7 1 2 2 2 2 π, π, π, π 4θ = OE. 3 5 7 8 8 1 8 8 π, cos π, cos π, cos π cos0, cos A1 Gives exactly 5 distinct, real roots. Accept 3 1 8 8 cos π, cos π. 1, ± ± 4

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5 The curve C has parametric equations + bt , for 1 G t G 2 , x = 3 t + 2t -1 + at 3 , y = 4t - 32 t -1 3 where a and b are constants. =- (a) It is given that a = 23 and b 1 2 . 2 dy 2 25 2 -2 2 dx Show that + = 4 ( t + t ) and find the exact length of C. [6] e dt o e dt o ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) It is given instead that a = b = 0 . d 2 y Find the value of 2 when t = 1. [4] dx ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a) 2 2 3 2 2 − = − + x t t 2 2 3 3 2 2 4 − = + − y t t B1 Differentiates x and y with respect to t. 4 4 2 2 4 2 2 4 4 12 23 9 12 9 4 12 1 12 2 4 4 + + − + + + + + − t t t t t t t t M1 Expands 2 2. +   x y Accept ( ) 2 25 2 2 4 25 . − + − t t ( ) 2 25 25 25 25 4 2 4 4 2 2 4 4 t t t t − − + + = + M1 A1 Factorises 2 2. +   x y AG. 2 2 3 1 5 5 85 1 2 2 3 2 12 1 1 2 d − −   + = − =   t t t t t M1 A1 Applies correct formula for arc length. 6 Question Answer Marks Guidance 5(b) 2 3 2 2 4 d d 3 2 − − + = = −   t y y x x t B1 Finds d . d y x ( )( ) ( )( ) ( ) 2 2 3 2 3 2 2 2 3 3 2 2 3 2 3 4 4 4 d d 3 2 3 2 − − − − − − − − − − +   + =     − −   t t t t t t t t B1 Differentiates d d y x with respect to t. ( ) 2 2 3 2 3 2 2 3 2 4 d 5 5 d d d 3 2 d 3 2 d 2 2 − − − −   + =     − × −   − = = − t t y t t t x t x when 1. = t M1 A1 Applies chain rule. 4

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6 The matrix P is given by 1 6 6 P = 0 2 6 f0 0 - 3p. (a) Use the characteristic equation of P to find P -1 . [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the matrix A such that 4 0 0 P -1 AP = 0 5 0 [4] f0 0 6 p. ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) State the eigenvalues and corresponding eigenvectors of A3. 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Mark scheme: 6(a) ( ) 3 1 2)( 3) 6[ 0] ( 7 λ λ λ λ λ − − − + = + = M1 A1 Finds characteristic equation. 2 1 7 6 0 − − + = P I P M1 Substitutes P and multiplies through by 1. − P 2 1 18 24 0 4 6 0 0 9     = −       P leading to 1 1 2 1 3 1 3 4 0 1 0 0 −   − −   =     −   P M1 A1 5 Question Answer Marks Guidance 6(b) 1 4 0 0 0 5 0 0 0 6 −     =       A P P M1 Applies 1. − = A PDP 1 3 1 2 4 30 36 1 3 4 0 10 36 0 1 0 0 18 0 0   − −              − −    or 5 2 1 1 0 6 2 6 4 2 16 6 0 5 0 0 3 0 0 2 − −             − −    M1 A1 Multiplies two adjacent matrices. 4 3 2 0 5 2 0 0 6     −       A1 A1 linked to first M1. 4 6(c) Eigenvectors: 1 6 6 0 , 2 , 6 0 0 3         −   B1 Accept any non-zero multiples of these. Eigenvalues: 64, 125, 216 B1 2

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7 It is given that y = x 2 w and 2 2 d w d w 2 2 x 2 + 4x ( x + 1) + ( 5x + 8x + 2) w = 5x + 4 x + 2 . d x d x (a) Show that d 2 y d y 2 + 4 + 5y = 5x + 4x + 2 . [4] 2 d x d x ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the general solution for w in terms of x. 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Mark scheme: 7(a) 2 d d 2 d d = + y w x xw x x B1 2 3 d 1 d 2 d d = − w y y x x x x 2 2 2 2 2 d d d 4 2 d d d = + + y w w x x w x x x B1 2 2 2 3 4 2 2 d 1 d 4 d 6 d d d = − + w y y y x x x x x x 2 2 2 2 2 2 2 d d d d d 4 5 4 2 4 8 5 d d d d d + + = + + + + + y y w w w y x x w x xw x w x x x x x M1 Uses substitution to find w-x equation, AG. ( ) ( ) 2 2 2 2 2 2 d d 5 4 4 5 8 2 5 4 2 d d w w x x x x x w x x x x + + + + + = + + A1 AG 4 7(b) 2 4 5 0 m m + + = leading to 2 m i = −± M1 Auxiliary equation. ( ) 2 e cos sin − = + x y A x B x A1 Complimentary function. Allow “ = y ” missing. Don’t allow e.g. ( ) 2 e cos isin . − + x A x B x 2 y px qx r = + + , ' 2 y px q = + , '' 2 y p = B1 Particular integral and its derivatives. 2 2 2 8 4 5 5 5 5 (8 5 ) (2 4 5 ) 5 5 8 5 4 2 4 5 2 + + + + + = + + + + + = + = + + = p px q px qx r px p q x p q r p p q p q r M1 Substitutes and equates coefficients. 1 = p 4 5 = − q 16 25 = r A1 ( ) 2 2 2 16 4 5 25 e cos sin − = + + − + x x w A x B x x x M1 Substitutes for y and find w in terms of x. ( ) ( ) ( ) 2 1 2 5 5 4 4 cos sin 1 ( ) − − − = + + − + x w xe A x B x x x A1 7

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8 (a) Starting from the definitions of tanh and sech in terms of exponentials, prove that 1 - tanh 2 x = sec h 2 x . [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Using the substitution u = tanh x , or otherwise, find sec h 2 x tanh 2 x d x . [2] y ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ln 3 = sec h x tanh x d x . It is given that, for n H 0 , I n n 2 y0 (c) Show that, for n H 2 , 3 n - 2 = 5 5 ( n + 1) I n 4 3 + ( n - 2) I n - 2 . [5] b l b l d [You may use the result that ( sec h x) =- tanh x sec h x .] d x ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (d) Find the value of I4. 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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) e e 2 tanh sech e e e e − − − − = = + + x x x x x x x x B1 Writes in exponential form. ( ) ( ) ( ) ( ) 2 2 2 2 2 e e e e e e 4 1 e e e e e e − − − − − − + − −   − − = =   +   + + x x x x x x x x x x x x M1 Writes over common denominator. 2 sech x A1 AG 3 8(b) 3 1 3 2 d tanh = + u u x C M1 A1 Uses substitution correctly. Allow by inspection. 2 8(c) l 2 2 n 0 2 3 sech sech tanh d − = n nI x x x x B1 Separates into correct structure. ln3 ln3 1 1 3 3 0 2 3 2 4 0 sech tanh ( 2) sech tanh d n n x x n x x x − −   + −    M1 Uses integration by parts correctly. ( ) ( ) ( ) 3 1 4 1 3 5 5 3 2 3 2 ( 2) n n n n I I − − + − − M1 A1 Uses 2 2 . 1 sech tanh − = x x ( ) ( ) ( ) 2 3 4 5 3 5 2 3 2 ( 2) n n n n I n I − − + − = + − leading to ( ) ( ) 2 3 4 5 3 5 2 ( 1) ( 2) n n n n I n I − − + = + − A1 AG 5 Question Answer Marks Guidance 8(d) ( ) 5 3 7 2 64 1 4 3 5 3 = = I B1 ( ) ( ) 2 3 4 4 2 5 5 3 (4 1) 2 I I + = + leading to 4 4928 0.105 46875 I = = M1 A1 Applies reduction formula with 4. = n 3

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