Cambridge A Level Mathematics - Further 9231 — 2024 Oct/Nov Paper 2 · Variant 1

1 Find the set of values of k for which the system of equations x + 5y + 6z = 1, kx + 2 y + 2 z = 2, - 3 x + 4 y + 8 z = 3, has a unique solution and interpret this situation geometrically. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 1 5 6 M1A1 Evaluates determinant. 2 2 k 2 k 2 k 2 2 = − 5 + 6 = 8 − 5 ( 8k + 6 ) + 6 ( 4 k + 6 ) 4 8 −3 8 −3 4 −3 4 8 k  78 A1 Three planes intersect at a unique point. B1 4

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2 It is given that 1 -1 x = 1 + and y = cos t for 0 1 t 1 1. t dy t 2 (a) Show that = . [2] d x 2 1 - t ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ b d 2 y a 2(b) Show that = - t `1 - t j `2 - t 2j, where a and b are constants to be determined. 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Mark scheme: 2(a) dy dy dt 1 2 t 2 M1A1 Uses chain rule, AG. =  = − −t = 2 ( ) 2 dx dt dx 1 − t 1 − t 2 2(b) 2 3 2 − 12 M1A1 Applies Quotient rule. 1 − t ) d  dy 2t 1 − t + t (  = 2 dt  dx  1 − t − 12 M1A1 Uses chain rule. 2 3 2 1 − t d 2 y 2t 1 − t + t ( ) 2 3 2 − 12 5 2 − 32 = −t = −2t 1 − t − t 1 − t 2 2 ( ) ( ) ( ) dx 1 − t 3 2 − 32 2 2 3 2 − 32 2 = −t 1 − t 2 1 − t + t = −t 1 − t 2 − t ( ) ( ( ) ) ( ) ( ) 4

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3 A curve has equation y = e x for ln 4 G x G ln 12 . The area of the surface generated when the curve is 3 5 rotated through 2r radians about the x-axis is denoted by A. (a) Use the substitution u = e x to show that 12 5 2 A = 2 r 1 + u du . 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(b) Use the substitution u = sinh v to show that A = r b 904 + ln 5 l. [6] 225 3 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 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Mark scheme: 3(a) ln 125 x 2 x 125 2 1 M1A1 Correct formula and applies substitution, AG. u 1 + u ( u ) du A = 2 π ln 43 e 1 + e dx = 2π 43 2 3(b) u = sinh v  du = cosh v dv B1 ln5 ln5 2 2 M1A1 Applies substitution and cosh 2 x = 1 + sinh 2 x . A = 2 π ln3 1 + sinh v cosh v d v A = 2 π ln3 co sh v d v Need limits for A1. ln5 M1 Applies 2cosh 2 x = cosh2 x + 1. π ln3 ( cosh 2v + 1) d v ln5 A1 Correct integration. 1 = π   2 sinh2v + v  ln3 312 904 π ( 50 + ln5 − 1840 − ln3) = π ( 225 + ln 53 ) A1 AG. 6

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4 The matrix A is given by - 11 1 8 A = f 0 - 2 0p. - 16 1 13 1 (a) Show that f1p is an eigenvector of A and state the corresponding eigenvalue. 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(b) Show that the characteristic equation of A is m 3 - 19 m - 30 = 0 and hence find the other eigenvalues of A. 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(c) Use the characteristic equation of A to find A -1 . 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Mark scheme: 4(a)  −11 1 8  1  −2  M1 Multiplies matrix with eigenvector.      0 −2 0 1 = −2            −16 1 13   1   −2  = −2 A1 2 4(b) −11 −  1 8 M1 Sets determinant equal to zero. 0 −−2  0 = 0 −16 1 13 −  ( −11 − )( −−2 )(13 − ) + 128 ( −−2 ) = 0 A1 Expands determinant, AG. 2 − 2− 15 = 0 ( −−2 )( ) 3 − 19− 30 = 0 = 5, − 3 B1 3 4(c) A 3 − 19 A − 30I = 0 B1 States that A satisfies its characteristic equation. 30 A −=1 A 2 − 19I M1 Multiplies through by A −1 .  −7 −5 16   −26 −5 16  M1A1 M1 for substituting A 2 . 2   −1 1   A = 0 4 0  A = 0 −15 0       30    −32 −5 41   −32 −5 22  4

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5 Find the particular solution of the differential equation d 2 x d x 2 6 - 5 + x = t + t + 1, d t 2 d t dx given that, when t = 0 , x = 12 and = -6 . [10] dt .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... 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Mark scheme: 5 6m2 − 5m + 1 = 0  m = 12 , 13 M1 Auxiliary equation, two distinct real roots. 1 t 13 t A1 Complementary function. x = A2e + Be x = pt 2 + qt + r  x ' = 2 pt + q  x '' = 2 p B1 Particular integral and its derivatives. p = 1 −10 p + q = 1 12 p − 5q + r = 1 M1 Substitutes and equates coefficients. p = 1 q = 11 r = 44 A1 1 2 t 13 t 2 A1 General solution. Need to see ‘ x = ’. FT on CF. x = Ae + Be + t + 11t + 44 1 12 t 1 13 t M1 Differentiates. x ' = 2 Ae + 3 Be + 2t + 11 A + B + 44 = 12 12 A + 13 B + 11 = − 6  A = −38, B = 6 M1A1 Uses initial conditions. x = −38e 1 2 t + 6e 1 3 t + t 2 + 11t + 44 A1 Need to see ‘ x = ’. 10

More questions on Differential equations

6 y 1 0 x 1 2 N - 1 1 N N N x The diagram shows the curve with equation y = b 1 l for 0 G x G 1, together with a set of N rectangles 2 1 each of width . N 1 x (a) By considering the sum of the areas of these rectangles, show that b 1 l dx 2 L , where y 2 N 0 1 L = 1 . [4] N 2N `2 N - 1j ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 1 x (b) Use a similar method to find, in terms of N, an upper bound UN for y b 12 l dx . 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(c) Find the least value of N such that U - L G 10 -3 . [2] N N ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 1 1 x 1 (d) Given that b l dx = , use the value of N found in part (c) to find upper and lower bounds y 0 2 2 ln 2 for ln2. 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Mark scheme: 6(a)  1 1 x  1 1 N1 1 1 N2 1 1 NN−1 1 1 NN M1A1 Forms the sum of the areas of the rectangles. M1 ( N )( 2 ) + ( N )( 2 ) + + ( N )( 2 ) + ( N )( 2 ) for correct number of rectangles.  2 ) dx    0 (  1 N N 1 1 1 1 N M1A1 N n r r N − 1 1 ( 1 ( ( ) 2 ) 2 − 1) ( 2 )( 2 ) n 1 1 N = r = , AG. 1 = ( 2 ) = = N   N N 1 1 r − 1 n =1 n =1 2 N 2 N1 − 1 1 − N N ( N − N ( 2 ) 2 ) ( ) . Applies 4 N − 2 N 1 x N N6(b) 1 1 1 1 1 N 1 1 1 1 0 ( 2 ) d x  ( N ) + ( N )( 2 ) + + ( N )( 2 ) + ( N )( 2 ) −1 M1A1 Formsrectangles.the sum of the areas of appropriate N −1 N n 1 1 M1A1 N −1 n r N − 2 − 1 1 2 1 1 N r = 1. = 1 ( 2 ) =  N  N N 1 r − 1 1 n = 0 n = 0 2 N 2 N1 − 1 1 − N 2 N N ( 2 ) 1 − ( 2 ) ( ) Applies ) = ( 4 6(c) N1 M1 c 2 1 1 .  10 −3  2 N  103 Simplifies U N − LN to 1 − 1 = N N N 2 N 2 N 2 − 1 2 N 2 − 1 ( ) ( ) Least value of N is 500 A1 CWO. 2 N .6(d) LN  2ln21  U N  2U1 N  ln2  2 L1N M1A1FT Forms inequality. FT on their U 0.693 = 2U1 N  ln2  2 L1N = 0.694 A1A1 CWO. Must have used N = 500. 4

More questions on Summation of series

7 (a) Show that an appropriate integrating factor for 2 dy 2 x + 16 + y = x x + 16 dx is 1 x + 1 x 2 + 16 . 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(b) Hence find the solution of the differential equation 2 dy 2 x + 16 + y = x x + 16 dx for which y = 6 when x = 3 . 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Mark scheme: 7(a) d y y B1 2 + = x Divides through by x + 16 . d x x 2 + 16 1 M1A1 Finds integrating factor dx x  ) sinh −1 ( 4 x 2 +16 e = e 1 1 2 A1 AG. = 4 x + 4 x + 16 4 7(b) d 2 2 2 M1A1 Correct form on LHS and RHS. y x + x + 16 = x + x x + 16 dx ( ( ) ) 3 2 1 3 1 2 2 M1A1 Integrates RHS. RHS of the correct form. y x + x + 16 x + 16 + C = 3 x + 3 ( ) ) ( 27 25 M1 Substitutes initial conditions into their expression. 6 3 + 25 = 3 + 3 25 + C ( ) 3 2 1 3 1 2 2 8 A1 OE. y x + x + 16 x + 16 − 3 = 3 x + 3 ( ) ) ( 6

More questions on Differential equations

8 (a) By considering the binomial expansion of bz + l , where z = cos i + isin i , use de Moivre’s z theorem to show that cos 7i = a cos 7 i + b cos 5 i + c cos 3 i + d cos i , where a, b, c and d are constants to be determined. 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(c) Using the results given in parts (a) and (b), find the exact value of I9. 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Mark scheme: 8(a) z + z −1 = 2cos B1 For LHS. −1 7 7 −7 5 −5 3 −3 −1 M1A1 Expands and groups. A0 if not grouped clearly. z + z = z + z + 7 z + z + 21 z + z + 35 z + z ( ) ( ) ( ) ( ) ( ) 27 cos7 = 2cos 7+ 7 ( 2cos 5) + 21( 2cos3) + 35 ( 2cos) M1 Substitutes z n + z − n = 2cos n. cos7 = 641 cos7+ 647 cos5+ 6421 cos3+ 3564 cos A1 5 8(b) 14 π n −1 n −1 14 π 14 π n − 2 2 M1A1 Applies integration by parts. cos cosd =  cos sin cos sin d   0 nI = 0 + ( n − 1) 0 1 n −1 4 π 14 π n − 2 2 M1 Applies sin 2 = 1 − cos 2  cos  1 − cos  d ( ) I n =  cos sin 0 + ( n − 1) 0 − n2 − n2 A1 AG. I n = 2 + (n − 1) I n − 2 − (n − 1) I n  nI n = 2 + ( n − 1) I n − 2 4 8(c) − 92 1 − 12 M1A1 Applies reduction formula from (b). 2 + 8 I 7 9 I 9 = 2 + 8 I 7 = 16 ( ) 14π 1 7 21 35 M1 Applies identity from (a). Allow 64 cos7+ 64 cos5+ 64 cos 3+ 64 cosd missing/incorrect limits. I 7 = 0 14π A1 1 1 7 21 I 7 = 64  7 sin 7+ 5 sin 5+ 3 sin3+ 35sin 0 1 − 12 2 2 − 12 − 12 A1 Check their exact answer when 9I is the subject, 2 −2− 1 −2− 1 2 + 35 2 + 81 9 I 9 = 161 ( ) 7 ( ) + 75 ( ) + 213 ( ) ( ) ) ( like terms collected. ISW. 0.402237 2 2 − 1 = 128670080 2 I 9 = 50286740 ( ) 5

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