Cambridge A Level Mathematics - Further 9231 — 2021 Oct/Nov Paper 2 · Variant 1

1 Find the Maclaurin’s series for ex tanx from first principles up to and including the term in x2. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 ( ) 2 e tan sec + x x x B1 Finds first derivative. ( ) 2 2 e tan 2sec 2sec tan + + x x x x x M1 A1 Finds second derivative. (0) 0, = y '(0) 1, = y ''(0) 2 = y M1 Evaluates derivatives at 0. = x 2 = + y x x A1 5

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2 The matrix A is given by - 1 2 12 A = 0 1 0 f 0 0 3p. Use the characteristic equation of A to show that A 4 = pA 2 + qI , where p and q are integers to be determined. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 ( )( )( ) 3 2 1 1 3[ 0 3 3 ] λ λ λ λ λ λ − − − + = + = − 2 3 4 2 3 3 3 0 3 3 0 − − + =  − − + = A A A I A A A A M1 A1 Substitutes A and multiplies through by . A ( ) 2 4 2 2 3 3 3 3 10 9 = + − + − = − A A A I A A A I M1 A1 Substitutes 3. A 6

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3 The curve C has equation xy 3 - 4x 3 y = 3 . d y (a) Show that, at the point ( - 1, 1) on C, = 11. [3] d x ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ d 2 y (b) Find the value of 2 at the point ( - 1, 1) . [5] dx ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) 2 3 3 '+ xy y y B1 Differentiates 3. xy ( ) 3 2 4 ' 3 0 − + = x y x y B1 Differentiates 3 . x y ( ) 3 3 2 3( 1)(1) ' 1 4 ( 1) ' 3( 1) (1) 0 y y − + − − + − = leading to ' 11 y = B1 Substitutes ( 1,1), − AG. ( ) 2 3 2 3 ' 3 4 12 − = − y xy x x y y 3 Question Answer Marks Guidance 3(b) ( ) ( ) 2 3 2 2 3 4 '' ' 6 ' 3 12 − + + − xy x y y xyy y x B1 B1 Differentiates ( ) 2 3 3 4 '. − xy x y ( ) 2 2 3 ' 12 ' 2 0 + − + = y y x y xy B1 Differentiates 3 2 12 . − y x y ( ) 2 2 3 2 2 3 '' 4 '' 6 ' 6 ' 24 ' 24 0 − + + − − = xy y x y xy y y y x y xy ( )( ) ( )( ) ( ) 2 3 2 2 2 3 2 2 2 2 3 3 4 12 ' 24 3 ' 12 6 ' 3 12 '' 3 4 − + − − − + − = − xy x x y xy y y x y y xyy y x y xy x '' 11( 75) 3(11) 12(9) 0 + − + − = y M1 Substitutes (–1, 1) '' 900 = y A1 5

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4 y 0 1 2 3 4 N – 1 N x ln x The diagram shows the curve with equation y = 2 for x H 2 , together with a set of ( N - 2) rectangles x of unit width. (a) By considering the sum of the areas of these rectangles, show that N ln r 2 + 3 ln 2 1 + ln N 1 - . [7] 2 r 4 N =/r 1 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ N ln r(b) Use a similar method to find, in terms of N, a lower bound for 2 . [3] r =/r 1 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) 2 1 3 2 ln ln2 ln 4 = = = +   N N r r r r r r M1 A1 Compares with sum of the areas of the rectangles. M1 for writing out sum, A1 for considering 3 2 ln . = N r r r < 2 2 ln 2 ln d 4 + N x x x M1 Compares with integral. 2 2 2 ln ln 1 d +   = −      N N x x x x x M1 A1 Finds integral. 1 2 ln ln2 ln 1 ln2 1 2 3ln2 1 ln 4 2 4 = + + + +   < + − + = −      N r r N N N N r M1 A1 Inserts limits, AG. A1 requires second M1. 7 4(b) 2 2 2 2 2 1 2 1 2 ln ln ln ln ln d − = = = + > +    N N N r r r r N x N x r r N x N M1 A1 Compares with integral. Lower limit of 1 scores M0. Accept 2 1 2 1 2 ln ln . + = >   N N r r x r x 2 ln 2 1 ln 1 ln 2 + + = − + N N N N A1 Accept ln2 1 ln( 1) 1. 2 1 + + + − + N N 3

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5 Find the particular solution of the differential equation d 2 y dy - 2 + y = 4 cos x, 2 dx dx dy given that, when x = 0 , y =- 4 and = 3 . [11] dx .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... 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Mark scheme: 5 2 2 1 0 1 − + =  = m m m ( ) e = + x y Ax B A1 Complementary function. Accept ‘y =’ missing. sin cos y p x q x = + , ' cos sin y p x q x = − , '' sin cos y p x q x = − − M1 A1 Particular integral and its derivatives. 2 0 − + + = p q p 2 4 −− + = q p q M1 Substitutes and equates coefficients. 2 = − p 0 = q A1 ( ) e 2sin = + − x y Ax B x A1 ( ) ' e e 2cos = + + − x x y A Ax B x M1 Differentiates. 4 = − B 2 3 9 + − =  = A B A M1 A1 Forms simultaneous equations using initial conditions. ( ) e 9 4 2sin = − − x y x x A1 11

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6 (a) Use de Moivre’s theorem to show that cosec 5 i cosec 5i = 4 2 . [6] 5 cosec i - 20 cosec i + 16 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Hence obtain the roots of the equation x 5 - 10x 4 + 40x 2 - 32 = 0 in the form cosec ( qr ) , where q is rational. 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Mark scheme: 6(a) ( ) ( ) 5 isin cos5 i 5 θ θ + = + c s M1 Uses binomial theorem. 5 2 3 4 sin5 10 5 θ − + = s s c c s A1 ( ) ( ) 2 5 2 3 2 10 1 5 1 s s s s s − − + − M1 Uses 2 2 1 = − c s or 2 2 cosec 1 cot θ θ = − after dividing numerator and denominator by 5. s 5 3 16 20 5 s s s − + A1 5 5 3 5 cosec cosec 16 20 5 cose 5 c 1 θ θ θ × + = − s s s M1 Divides simplified numerator and denominator by 5. s 4 2 5 6 c 2 osec cosec co c 5 0 se 1 θ θ θ − + A1 AG 6 6(b) ( ) 5 4 2 2 5 20 16 x x x = − + leading to 5 4 2 2 5 20 16 x x x = − + M1 Relates with equation in part (a). 2 cose 5 c θ = leading to 1 2 sin5θ = M1 Solves 1 2 sin5 . θ = ( ) 1 30 π cosec x = A1 Gives one correct solution. ( ) ( ) ( ) ( ) 5 7 13 11 30 30 30 30 c π osec ,cosec ,cosec ,cos π ec π π − − A1 Gives four other solutions. Allow different values of q as long as all five solutions are found. 9.56, 2, –1.49, –1.09, 1.02 4

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7 (a) Show that an appropriate integrating factor for 2 dy 2 2 x - 1 + y = x - x x - 1 dx is x + x 2 - 1 . [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Hence find the solution of the differential equation 2 dy 2 2 x - 1 + y = x - x x - 1 dx = f ( x). [7] for which y = 1 when x = 54 . Give your answer in the form y ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 7(a) 2 2 2 2 d 1 1 d 1 1 − − + = − − y x x x y x x x B1 Divides through by 2 1. − x 1 2 1 d cosh 1) e e − √( − =  x x x M1 A1 Finds integrating factor. M1 for correct form 2 1 d 1) e , √( −  a x x where a is a non-zero constant. 2 1 x x + − A1 AG 4 7(b) ( ) ( ) 2 2 d 1 d 1 + − = − x y x x x x M1 A1 Correct form on LHS and simplifies RHS. ( ) 2 2 1 1 + − = −+ y x x x C M1 A1 Integrates RHS. For M1, needs to integrate non-zero multiple of 2 . 1 − x x 3 4 2 C = + leading to 5 4 C = M1 Substitutes initial conditions. 5 2 4 2 1 1 −+ = + − x y x x M1 A1 Divides through by coefficient of y. 7

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8 (a) Starting from the definition of cosh in terms of exponentials, prove that 2 cosh 2 A = cosh 2A + 1. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The curve C has parametric equations - 4t , for - 2 G t G 2 . x = 2 cosh 2t + 3t, y = 32 cosh 2t 1 1 The area of the surface generated when C is rotated through 2r radians about the y-axis is denoted by A. 1 + 3t) cosh 2t dt . [4] (b) (i) Show that A = 10 r 21 ( 2 cosh 2 t -y 2 .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... (ii) Hence find A in terms of r and e. 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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) ( ) 1 2 cosh e e− = + A A A B1 Writes in exponential form ( ) ( ) 2 2 2 1 2 2 2 2 1 2cosh e 2 e e e 1 cosh2 1 − − = + + = + + = + A A A A A A M1 A1 Expands, AG. 3 8(b)(i) d d 4sinh 2 3 3sinh 2 4 d d = + = − x y t t t t B1 ( ) ( ) 2 2 2 2 4sinh 2 3 3sinh 2 4 25(sinh 2 1) 25cosh 2 + + − = + = t t t t M1 A1 Expands and applies 2 2 cosh sinh 1 = + A A ( )( ) 1 1 2 2 1 1 2 2 2 2 c 2cosh2 3 2 30 c 2 π 5cosh 2 d π 20 o o h sh s d t t t t t t t t − −   =   +   +   A1 Correct formula for surface area, AG. 4 8(b)(ii) 1 1 2 2 1 1 2 2 2 20 cosh 2 d 10 cosh 4 1d − − = +   t t t t M1 Applies ( ) 2 2 1 cosh cosh2 1 . = + A A ( ) 1 2 1 2 1 1 4 2 10 sinh 4 10 sinh 2 1 − + = +     t t M1 A1 Integrates. [ ] 1 1 1 2 2 2 1 1 1 2 2 2 1 1 2 2 30 cosh2 d 30 sinh 2 sinh2 d − − −   = −       t t t t t t t M1 A1 Integrates by parts. 1 2 1 2 1 2 15 sinh 2 cosh 2 0 − − =     t t t A1 Accept 1 2 1 2 cosh 2 d 0 − = t t t since cosh 2 t t is odd. ( ) ( ) 2 2 1 4 1 e π 0 e 1 − − + A1 64.82457... Question Answer Marks Guidance 8(b)(ii) Alternative method for question 8(b)(ii) 10 ( ) 1 2 1 2 π 2cosh 2 3 cosh 2 d − +  t t t t = ( ) 1 1 2 2 1 1 2 2 3 2 1 ) π cosh 2 sinh 2 ( sinh 2 5π sinh 2 4si h 2 3 d 0 n t t t t t t t − −   + − +    M1 A1 Integrates by parts. 2cosh 2 3 u t t = + , ' 4sinh 2 3 u t = + ' cosh 2 v t = , 1 2 sinh2 v t = 1 1 1 2 2 2 1 1 1 2 2 2 2 1 2 1 d π sinh 4 20π sinh 2 15π sinh 0 2 d t t t t t − − −   − −           = 1 1 1 2 2 2 1 1 1 2 2 2 1 2 d 1 π sinh4 10π cosh4 1 i 0 15π s nh2 d t t t t t − − −   − − −           M1 A1 Applies 2 2sinh 2 cosh 4 1 = − t t ( ) 1 2 1 2 5 10 15 2 4 2 π sinh 4 sinh 4 10 cosh 2 t t t t −   − + −   M1 A1 Integrates ( ) 1 2 1 2 5 15 2 2 π sinh 4 10 cosh 2 t t t −   + −   = ( ) π 5sinh2 10 + = ( ) ( ) 2 2 1 4 1 e π 0 e 1 − − + A1 64.82457... 7

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