Cambridge A Level Mathematics - Further 9231 — 2024 May/June Paper 2 · Variant 2

1 Find the roots of the equation z 3 = - 108 3 + 108i , giving your answers in the form r cos i + i sin i , where r 2 0 and 0 1 i 1 2r . [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 5 5 6 6 3 π π 216cos i216sin z   B1 Finds modulus and argument of 108 3 108i   5 5 18 18 π π 1 6cos i6sin z   M1 A1FT Finds one root. FT on their modulus. For M1 must have divided their argument of 108 3 108i   by 3. 2 18 18 18 18 2 17π 17π 9π 29π 3 6cos i6sin , 6cos i6sin z z     M1 A1 Finds other two roots. 5

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2 Find the Maclaurin’s series for e 1 + x 2 + e 1 - x up to and including the term in x2 . [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 2 2 1 2 e e e e 1 x x x       B1 Changes 2 1e x  or 1e x  so that the formula given in the list of formulae (MF19) can be applied.   2 1 2 1 e e 1 x x x     B1     2 1 2 2 e 1 e 1 x x x        M1 Sums power series. 2 3 1 2 2 1 e e 2e e e x x x x         A1 Alternative method for question 2 2 1 1 f '( ) 2 e e x x x x     B1 Finds first derivative. 2 2 2 1 1 1 f ''( ) 4 e 2e e x x x x x       B1 Finds second derivative. f (0) 2e f '(0) e f ''(0) 3e    M1 Evaluates their derivatives at zero. 2 3 1 2 2 1 e e 2e e e x x x x         A1 4

More questions on Differentiation

3 It is given that x = sin -1 t and y = t cos -1 t , for 0 G t 1 1. dy 2 -1 (a) Show that = - t + 1 - t cos t . [3] dx ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ d 2 y (b) Find 2 in terms of t. 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Mark scheme: 3(a) 1 2 d cos 1 d y t t t t     , 2 d 1 d 1 x t t   2 1 1 d d dt d dt os d c y y t x t x t       M1 A1 Uses chain rule, AG. 3 3(b) 2 1 1 2 2 2 d d 1 cos cos 1 2 d d 1 1 1 y t t t t t t x t t t               M1 A1 Applies product rule. 2 2 1 2 d 2 1 cos d y t t t x     M1 A1 Uses chain rule. 4

More questions on Differentiation

= sech x d x .4 It is given that, for n H 0 , I n n 0 (a) Show that, for n H 2 , 3 n - 2 4 n - 1 I n = + n - 2 I n - 2 . [5] ` j b 5 l b 5 l ` j d [You may use the result that sec h x = - tanh x sec h x .] dx ` j ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the value of I4. 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Mark scheme: 4(a) l 2 n3 2 0 sech sech d n nI x x x   B1 Separates into correct structure. ln3 ln 0 2 2 2 0 2 2 3 ln3 0 sech sech d sech tanh ( 2) sech tanh d n n n nI x x x x x n x x x              *M1 Uses integration by parts correctly.     3 4 5 5 2 2 ( 2) n n n n I I      DM1 A1 Uses 2 2 . 1 sech tanh x x         2 2 2 2 3 3 4 4 5 5 5 5 1 2 ( 2) ( 1) ( 2) n n n n n n n I n I n I n I               A1 AG 5 Question Answer Marks Guidance 4(b) 4 2 5 I  B1  2 3 4 4 2 5 4 5 236 3 2 0.629 375 I I I      M1 A1 Applies reduction formula with 4. n  3

More questions on Integration

5 y 0 1 2 n - 1 1 x n n n The diagram shows the curve with equation y = 2x - x 2 for 0 G x G 1, together with a set of n 1 rectangles of width . n 1 2x - x 2 (a) By considering the sum of the areas of these rectangles, show that dx 1 U n , where j y0 ` 1 2 1 - [5] 1 + U n = 6n o. b n el 3 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 1 2 x - x d x . [4] j(b) Use a similar method to find, in terms of n, a lower bound Ln for y0 ` 2 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Show that lim" U n - L n = 0 . [2] n 3` j ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a)      1 1 1 1 1 2 2 1 0 2 2 2 2 2 d 2 2 2 n n n n n n n n n n n x x x                 M1 A1 Forms the sum of the areas of the rectangles, may be in summation form. M1 is for correct number of rectangles. 2 3 2 3 2 1 1 1 6 2 1 1 ( 1) ( 1)(2 1) n n n n n n r r r r n n n n n            M1 A1 Applies formulae from MF19. Summations must have (correct) limits for A1. Substituted in correctly once for M1.      2 1 1 1 1 1 2 1 2 1 1 6 3 6 3 2 6 1 1 2 1 n n n n n n n           A1 AG. This A1 requires the previous A1. 5 5(b)              1 1 1 1 1 1 2 2 2 2 1 2 2 1 0 2 d 2 2 2 n n n n n n n n n n n x x x             M1 A1 Forms the sum of the areas of appropriate rectangles. M1 is for seeing correct height of last rectangle. 2 3 2 3 1 1 2 2 1 1 1 6 1 1 ( 1)( ) ( 1)( )(2 1) n n n n n n r r r r n n n n n             M1 Applies formulae from MF19, substitutes correct limit. Substituted in correctly once for M1.    2 1 2 1 2 1 1 3 6 3 2 6 1 n n n n      A1    1 2 1 1 3 6 1 n n n    or fully expanded or fully factorised. 4 Question Answer Marks Guidance 5(c)      1 2 1 1 2 1 1 3 6 3 6 1 1 ( ) n n n n n n n U L         M1 Expresses their n n U L  in terms of 1 . n For, M1 they must taking the limit of an expression that tends to a constant. (Must be fully expanded if they do not take the limit at any stage.) 1 0 n  as n  or 2 2 0 3 3   A1 AG, CWO. Their n L must be correct. 2

More questions on Integration

1 6 (a) Show that cosh x + sinh x 2 = e 2 x . 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(b) Find the particular solution of the differential equation 1 d 2 y dy 2 + + 3 y = 5 cosh x + sinh x , 2 dx dx ` j dy 4 given that, when x = 0 , y = 1 and = . [10] dx 3 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 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Mark scheme: 6(a)       1 1 2 2 1 1 2 2 cosh sinh e e e e x x x x x x         M1 Substitutes sinh and cosh in terms of exponentials.    1 2 1 2 e e x x  A1 AG 2 Question Answer Marks Guidance 6(b) 1 2 2 1 2 3 i 0 11 m m m        M1 Auxiliary equation.   1 2 11 11 2 2 e cos sin x y A x B x    A1 Complimentary function. Allow ‘ y ’ missing. 1 1 1 2 2 2 1 1 2 4 e ' e '' e x x x y k y k y k      B1 Particular integral and its derivatives. 1 1 1 2 2 1 2 2 1 1 1 1 4 2 4 2 e e 3 e 5e 3 5 x x x x k k k k k k        M1 Substitutes and equates coefficients. PI must be correct form ( bx ae ). 4 3 k  A1   2 2 1 1 11 11 4 2 2 3 e cos sin e x x y A x B x     A1 FT Must have ‘ y ’. FT on their CF.     2 1 1 2 2 1 11 11 11 11 11 11 1 2 2 2 2 2 2 2 2 3 d e sin cos e cos sin e d x x x y A x B x A x B x x         B1 4 3 1 A   11 4 1 2 3 2 2 3 B A    1 1 3 11 , A B    M1 A1 Substitutes initial conditions and forms simultaneous equations. For M1, CF must be correct form   2 2 1 1 11 11 1 1 4 3 2 2 3 11 e cos sin e x x y x x      A1 Must have ‘ y ’. 10

More questions on Differential equations

7 (a) Use the substitution u = 1 + x 2 to find x dx . [2] y 2 1 + x ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the solution of the differential equation dy 2 -1 x - y = x sinh x , dx given that y = 1 when x = 1. Give your answer in the form y = f x [10] ` j. ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 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Mark scheme: 7(a)     1 2 2 2 1 d d 1 1 x x u u C x C u x             2 7(b) 1 s d d inh y y x x x x    B1 Divides through by .x 1 1 ln d e e x x x x        M1 A1 Finds integrating factor.   1 1 d sinh d x y x x    M1 Correct form on LHS.   d d Iy x for their integrating factor I. 1 1 2 sinh d 1 x y x x x x x       M1 A1 Integrates RHS. RHS must be of the form 1 sinh c x  . 1 1 2 sinh 1 x y x x C x       A1   1 2 sinh ln 1 x x x       1 ln 1 2 2 C     *M1 Substitutes initial conditions.     2 1 2 1 sinh ln 1 2 1 2 y x x x x x x         DM1 A1 Divides through by their integrating factor. Accept   2 1 2 1 1 sinh 1 1 2 . sinh y x x x x x x         10

More questions on Differential equations

8 (a) Find the set of values of a for which the system of equations 6x + ay = 3, 2x - y = 1, x + 5y + 4z = 2 has a unique solution. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Show that the system of equations in part (a) is consistent for all values of a. 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The matrix A is given by 6 0 0 A = 2 - 1 0 f 1 5 4 p. 2 -1(c) Find a matrix P and a diagonal matrix D such that 14A + 24 I = PDP . [7] ` j ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (d) Use the characteristic equation of A to show that 2 4 2 14A + 24I = A A + bI , ` j ` j where b is an integer to be determined. 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Mark scheme: 8(a) 6 0 2 1 0 8 24 1 5 4 a a    3 a  A1 2 8(b) If 3 a  then system has unique solution (so consistent). B1 FT If 3 a  then 6 3 3, 2 1, 11 4 7 5 4 2, x y x y x z x y z           M1 Eliminates variable using all 3 equations. Or states there are two distinct equations and three unknowns. Or gives accurate geometrical description with 3 a . So the system has infinitely many solutions (so consistent). A1 Alternative method for question 8(b) Sets 0. y  B1   3 1 2 8 ,0, is a solution so system is consistent for all values of a. M1 A1 Finds a solution with 0 y  and states conclusion. 3 Question Answer Marks Guidance 8(c) Eigenvalues of A are 6, 1  and 4. B1 Lower diagonal matrix or characteristic equation. 14 6: 2 7 0 4 1 5 2 17              i j k  M1 A1 Uses vector product (or equations) to find corresponding eigenvectors. 0 0 1: 2 0 0 10 1 1 5 5 10 1                        i j k   A1 0 0 4: 2 5 0 0 0 1 5 0 15 1                  i j k   A1 Thus 14 0 0 4 1 0 17 1 1             P and 11664 0 0 0 100 0 0 0 6400           D M1 A1 Or correctly matched permutations of columns. Column of zeros in P gets M0. 7 8(d)     3 2 6 1 4 14 24 9 0               B1 Characteristic equation. 3 2 14 24 9    A I A A M1 Substitutes for A and makes 14 24  A I the subject.       2 2 2 3 2 4 14 24 9 9      A I A A A A I M1 A1 Squares and factorises. CWO. 4

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