Cambridge A Level Mathematics - Further 9231 — 2022 Oct/Nov Paper 2 · Variant 2

1 (a) Find the set of values of k for which the system of equations x + 2y + 3z = 1, kx + 4 y + 6 z = 0, 7x + 8y + 9z = 3, has a unique solution. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Interpret the situation geometrically in the case where the system of equations does not have a unique solution. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1(a) 1 2 3 M1 A1 Evaluates determinant or solves system of equations (eliminate at least one variable for M1): k 4 6 = 6 k − 12 7 8 9 2 4 k − 8 x = − , y = , z = . k − 2 k − 2 3 ( k − 2 ) k  2 A1 3 1(b) Two parallel planes. Other plane not parallel. B1 SC k = 2, two parallel planes. Parallel planes not identical. B1 Accept diagram. 2

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2 A curve has equation ( x + 1 ) y + y 2 = 2 . dy 2 (a) Show that = - at the point ( 0, - 2) . [3] dx 3 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ d 2 y (b) Find the value of 2 at the point ( 0, - 2) . [4] dx ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) d B1 Differentiates ( x + 1) y correctly. ( ( x + 1) y ) = ( x + 1) y '+ y dx d 2 B1 Differentiates y 2 correctly. y = 2 yy ' ( ) dx (1) y '− 2 + 2( −2) y ' = 0 leading to 'y = − 23 B1 Substitutes (0, −2), AG. 3 2(b) ( x + 1) y ''+ 2 y ' B1 Differentiates ( x + 1) y '+ y. +2 yy ''+ 2 ( y ' ) 2 = 0 B1 Differentiates 2 yy '. 2 2 2 M1 Substitutes (0, −2) into their expression with y ''. y ''+ 2 ( − 3 ) + 2( −2) y ''+ 2 ( − 3 ) = 0 y '' = − 274 A1 Alternative method for question 2(b) − y ( x + 1 + 2 y )( − y ' ) + y (1 + 2 y ' ) B1 B1 B1 for ( x + 1 + 2 y )( − y ' ) in numerator. y ' = y '' = x + 1 + 2 y ( x + 1 + 2 y ) 2 B1 for y (1 + 2 y ' ) in numerator. 2 2 M1 Substitutes (0, −2) and y ' = − 23 into their ( 0 + 1 + 2 ( −2 ) )( 1 + 2 ( − 3 3 ) − 2 ( ) ) y '' = ( 0 + 1 + 2 ( −2 ) ) 2 expression. y '' = − 274 A1 4

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G x G 1. Find, in terms of r and e, the area of the3 (a) A curve has equation y = e x + 14 e -x , for 0 surface generated when the curve is rotated through 2r radians about the x-axis. 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(b) Using standard results from the list of formulae (MF19), or otherwise, find the Maclaurin’s series for e x + 14 e -x up to and including the term in x2. 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Mark scheme: 2 M1 23(a) x 1 − x 2 x 1 1 −2 x 2 x 1 1 −2 x 1 + ( e − 4 e ) = 1 + e − 2 + 16 e = e + 2 + 16 e Finds 1 +  dy   dx  x 1 − x 2 x 1 − x = e + 4 e ( A1 e + 4 e ) 1 x 1 − x x 1 − x M1 Correct formula for surface area with limits. 2 π d x e + 4 e )( e + 4 e ) 0 ( 1 2 x 1 1 − 2 x A1 Expanded integrand. 2π 0 e + 2 + 1 6 e d x 1 1 2 x 1 1 −2 x M1 Integrates and substitutes limits. Integrand must be 2  2 e + 2 x − 3 2 e  0 2 x −2 x of the form ae + be + c , with a and b nonzero. 1 2 1 −2 1 2 1 −2 1 A1 2π = π ( 2 e − 32 e + 32 ) ( e − 16 e + 16 ) 6 3(b)  x 2  1  ( − x ) 2  M1 Uses expansion of ex or finds first and second  1 + x + +  +  1 − x + +  x 1 x x 1 − x  2!  4  2!  derivatives e − 4 e− and e + 4 e . 5 4 + 43 x + 85 x 2 A1 2

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4 Find the solution of the differential equation 2 dx 2 ( 4t - 1 ) + 4x = 4t - 1 dt for which x = 3 when t = 1. Give your answer in the form x = f( t) . 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Mark scheme: 4 dx 4 B1 Divides through by 4t 2 − 1 . + x = 1 dt 4t 2 − 1 4 2 2 4 4 1 B1 Writes as partial fractions or a form to which the = − or = = 1 4t 2 − 1 2t − 1 2t + 1 4t 2 − 1 ( 2t ) 2 − 1 t 2 − ( 2 ) 2 formula from MF19 can be applied.  4 ( 4 t 2 −1) −1 dt ln ( 2 t −−1) ln(2 t +1) 2t − 1 M1 A1 Finds integrating factor. e = e = 2t + 1 d  2t − 1 2t − 1 2t + 1 − 2 M1 Correct form on LHS and attempt to integrate RHS.  x = = at + b dt  2t + 1  2t + 1 2t + 1 RHS must be of the form with a, b, c, d  0. ct + d 2t 1 A1 x −= t − ln(2t + 1) + C 2t + 1 1 = 1 − ln3 +C M1 Substitutes into their expression in x and t.  2t + 1    3   M1 A1 After integrating, divides their expression through x =    t + ln    by their coefficient of x. (Their coefficient must  2t − 1    2t + 1   depend on t.) 9

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5 (a) Write down the fourth roots of unity. [1] ............................................................................................................................................................ ............................................................................................................................................................ (b) Use de Moivre’s theorem to show that cos 4i = 8 cos 4 i - 8 cos 2 i + 1. 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(c) Hence obtain the real roots of the equation 16 ( 8x 4 - 8 x 2 + 1 ) 4 - 9 = 0 in the form cos ( q r ) , where q is a rational number. 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Mark scheme: 5(a) 1,  i B1 0 i π2 πi i 3π2 Accept exponential form e , e , e ,e or trigonometric form. Four values specified. 1 5(b) 4 4 2 2 4 M1 A1 Expands and takes real part. cos4= Re ( cos+ isin) = cos − 6cos sin + sin  cos 4 − 6cos 2  1 − cos 2  + 1 − 2cos 2 + cos 4  M1 Applies sin 2 = 1 − cos 2 . ( ) ( ) 8 cos 4 − 8cos 2 + 1 A1 AG. Can award final A1 without previous A1 if penultimate line seen. SC. Applying cos2 A = 2cos 2 A − 1 twice scores 1/4. 4 8 x − 8 x + 1 = 16 , x = cos5(c) ( 4 2 4 9 M1 Applies identify given in (b). ) cos4=  12 3 A1 Need  and exact value for A1. 4=  16 π + 2 kπ 4=  56 π + 2 kπ M1 Solves cos4= 12 3 or cos4= − 12 3 1 π ) A1 Gives one correct solution. cos ( 24 cos( 1124 π), cos( 1324 π),cos( 2423 π), A1 Gives other solutions. Accept 1 11 5 7  cos ( 24 π ) ,  cos( 24 π),  cos( 24 π),  cos( 24 π) cos( 245 π),cos( 247 π),cos( 1724 π),cos( 1924 π), 5

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6 y 0 1 2 3 n – 1 n x 1 The diagram shows the curve y = for x 2 0 , together with a set of ( n - 1) rectangles of unit 2 width. x + 2x By considering the sum of the areas of these rectangles, show that n + . [10] 1 ln n + 1 + n + 2n 2 + 3 3 3 - ln ` j / 1 2 1 ` j + 2 r r = 1 r 2 .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... 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Mark scheme: 6 1 1 1 n 1 M1A1 Forms sum of the areas of the rectangles. At least = Sum = + + + 2  2 three terms, including first and last for A1.    n + 2 n ( ) r = 2 r + 2 r SC Areas of these three rectangles written down but sum not clearly formed scores B1. n M1 Compares with integral.  1 x 2 + 2 x  1  ( d x ) 2 2 B1 Completes the square. x + 2 x = ( x + 1) − 1 n M1 A1 Uses formula. Correct limits for A1. n  1 −1 dx =  cosh ( x + 1)  1 ( x + 1) 2 − 1  1  2  n M1 A1 Uses logarithmic form of cosh −1 . ln x + 1 + x + 2 x  ( ) 1 2 M1 Inserts limits. ln n + 1 + n + 2 n − ln 2 + 3 ( ) ( ) n 1 2 1 A1 Adds term from r = 1.  ln n + 1 + n + 2 n − ln 2 + 3 + 3 3 AG ( )  2 ) ( r =1 r + 2 r 10

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7 (a) It is given that m is an eigenvalue of the non-singular square matrix A, with corresponding eigenvector e. Show that m-1 is an eigenvalue of A -1 for which e is a corresponding eigenvector. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The matrix A is given by 2 0 3 A = 15 - 4 3 f 3 0 2p. (b) Given that - 1 is an eigenvalue of A, find a corresponding eigenvector. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 0 1 (c) It is also given that 1 and 2 are eigenvectors of A. Find the corresponding eigenvalues. 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(d) Hence find a matrix P and a diagonal matrix D such that A -1 = PDP -1 . 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(e) Use the characteristic equation of A to show that A -1 = pA 2 + qI , where p and q are rational numbers to be determined. 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Mark scheme: 7(a) Ae = e leading to e = A -1e M1 Multiples by A -1 on LHS. −1e = A -1e A1 2 7(b) i j k  9   1  M1 A1 Uses vector product (or equations) to find     corresponding eigenvector. = −1: 3 0 3 = 36 ~ 4         15 −3 3  −9   −1  2 7(c)  2 0 3  0  0  B1      15 −4 3 1 = −4 = −4           3 0 2 0  0   2 0 3  1  5  B1      15 −4 3 2 = 10 = 5           3 0 2 1  5  2 7(d)  1 0 1   −1 0 0  M1 A1 Or correctly matched permutations of columns.    1  M1 for their eigenvectors (all three must be non- and D = Thus P = 4 1 2    0 − 4 0  zero) correctly matched to reciprocals of their    1   −1 0 1   0 0 5  eigenvalues. 2 7(e) (+ 1)(+ 4 )(− 5) = 3 − 21− 20 = 0 B1 Characteristic equation. A 3 − 21A − 20I = 0 M1 Substitutes for A. Tolerate missing I. 20 A −=1 A 2 − 21I leading to A −=1 201 A2 − 2021 I M1 A1 Multiplies both sides of equation by A − 1, need I for A1. 4

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8 It is given that y = cosh u , where u 2 0 , and 2 2 2 d u du d u -x cosh u - 1 2 + + cosh u - 2 cosh u = 4e . e dx dx o e d x o (a) Show that d 2 y dy -x + - 2y = 4 e . [4] 2 dx dx ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ du (b) Find u in terms of x, given that, when x = 0 , u = ln 3 and = 3 . 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Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... .................................................................................................................................................................................... ....................................................................................................................................................................................

Mark scheme: 8(a) dy du B1 = sinh u dx dx 2 2 2 B1 d y d u  du  = sinh u + cosh u 2 2   dx dx  dx  2 2 2 M1 Uses substitution to find y-x equation, AG. d y dy d u  du  du + − 2 y = sinh u + cosh u + sinh u − 2cosh u 2 2   dx dx dx  dx  dx 2  d 2 u du   du  2 − x A1 AG cosh u − 1  2 +  + cosh u   − 2cosh u = 4e  dx dx   dx  8(a) Alternative for 8(a) du −1 dy B1 = ( sinh u ) dx dx d 2 u  1  d 2 y d y  cosh u  d u B1 2 =   2 −  2  d x  sinh u  d x d x  sinh u  d x d 2 u  1  d 2 y  dy 2 cosh u  = − 2   2   3  dx  sinh u  dx  dx  sinh u    1  d 2 y  dy 2 cosh u  −1 dy  cosh u  dy  2 M1 Uses substitution to find y-x equation, AG. 2cosh u sinh u    2 −   3  + ( sinh u )  + 2   −   sinh u  dx  dx  sinh u  dx  sinh u  dx    d 2 y  dy 2 cosh u  dy cosh u  dy  2 − + + − 2cosh u 2   2  2   dx  dx  sinh u  dx sinh u  dx  d 2 y dy − x A1 AG + − 2 y = 4e dx 2 d x 4 8(b) m 2 + m − 2 = 0  m = 1, − 2 M1 Auxiliary equation. y = A e x + B e− 2 x A1 Complimentary function. Allow ‘ y = ’ missing. y = ke − x y ' = − ke − x y '' = ke − x B1 Particular integral and its derivatives. ke − x − ke − x − 2 ke − x = 4e − x M1 Substitutes and equates coefficients. k = −2 A1 WWW y = cosh u = Ae x + Be −2 x − 2e − x A1 Must have ‘ y = ’ or ‘cosh u = ’ . du x −2 x − x B1 y ' = sinh u = Ae − 2 Be + 2e dx A + B − 2 = 53 A − 2B + 2 = 4 A = 289 , B = 95 M1 A1 Substitutes initial conditions and forms simultaneous equations. For M1, allow substitution into their y and 'y if two linear equations in two unknowns derived. −1 28 x 5 −2 x − x A1 Substitutes for y and finds u in terms of x. u = cosh ( 9 e + 9 e − 2e ) 10

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