Cambridge A Level Mathematics - Further 9231 — 2021 May/June Paper 2 · Variant 2

1 (a) Given that a is an integer, show that the system of equations ax + 3y + z = 14, 2 x + y + 3 z = 0, - x + 2y - 5z = 17, has a unique solution and interpret this situation geometrically. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the value of a for which x = 1, y = 4 , z = -2 is the solution to the system of equations in part (a). [1] ................................................................................................................................................... ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ................................................................................................................................................... ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 1(a) 3 1 2 1 3 11 26 1 2 5 = − + − − a a M1 A1 Finds determinant. If solving equations x, y, z in terms in a. det 0 ≠ A leads to unique solution. A1 States that det 0. ≠ A (Since a is an integer.) (M1 A0 A0 if one of x, y, z found in terms of a.) (M1 A1 A0 if two of x, y, z found in terms of a.) The three planes intersect at a single point. B1 4 1(b) 4 = a B1 1

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2 The variables x and y are related by the differential equation d 2 y dy + 3 + 2 y = 2 x + 1. 2 dx dx (a) Find the general solution for y in terms of x. [6] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) State an approximate solution for large positive values of x. [1] ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................................

Mark scheme: 2(a) 2 3 2 0 m m + + = leading to m = -1, -2 M1 Auxiliary equation. 2 e e x x y A B − − = + A1 Complimentary function. leading to ' leading to '' 0 y p qx y q y = + = = B1 Particular integral and its derivatives. 3 2( ) 2 1 q p qx x + + = + M1 Substitutes and equates coefficients. 1 1 q p = = − A1 2 e e 1 x x y A B x − − = + + − A1 Must see ‘y =’. 6 Question Answer Marks Guidance 2(b) 1 = − y x B1 FT Accept 1 ≈ − y x . 1

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3 y 0 1 2 n - 1 1 x n n n The diagram shows the curve with equation y = x3 for 0 G x G 1, together with a set of n rectangles of width 1. n 1 (a) By considering the sum of the areas of these rectangles, show that x 3 dx 1 Un , where y0 2 n + 1 U n = . [4] e 2n o ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ x 3 dx . [4](b) Use a similar method to find, in terms of n, a lower bound Ln for 1y0 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Find the least value of n such that U n - L n 1 10 -3 . 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Mark scheme: 3(a) ( )( ) ( )( ) ( )( ) ( )( ) 1 3 0 3 3 3 3 1 1 1 1 2 1 1 d n n n n n n n n n n x x − < + + + +   Need last term for A1. 2 2 3 4 4 1 2 1 ( 1) 1 2 4 n r n n n r n n n = + + = =        M1 A1 Applies 3 1 4 1 2 2 ( 1) , = = +  n r r n n AG. 4 3(b) ( )( ) ( )( ) ( )( ) 1 3 0 3 3 3 1 1 1 1 2 1 d n n n n n n n x x − > + + +   M1 A1 Forms the sum of the areas of appropriate rectangles. Need the last term for A1. 2 2 3 4 4 2 1 1 1 ( 1) 1 2 4 n r n n n r n n n − = − − = =        M1 A1 Applies 3 1 4 1 2 2 ( 1) . = = +  n r r n n Accept 2 1 1. 2 n n n + −       4 3(c) 2 2 3 3 1 1 1 10 leading to 10 2 2 n n n n n n − + − − = < >             M1 Simplifies their − n n U L to . k n Least value of n is 1001. A1 2

More questions on Integration

4 Find the solution of the differential equation d y 1 sin i i , + y = tan 2 di r . Give your answer in the form y = f( i) . [9] where 0 1 i 1 r , given that y = 1 when i = 12 [You may use without proof the result that 1 i .] cosec i d i = lntan 2 y .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... 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Mark scheme: 4 1 2 d cosec tan sin d θ θ θ θ = + y y θ 1 2 ln tan 1 osec d 2 c e e tan θ θ θ θ = =  M1 A1 Finds integrating factor. ( ) 1 1 1 2 2 2 1 1 1 2 2 2 2 1 2 1 2 3 2 s d tan sin tan ec sin cos tan d y θ θ θ θ θ θ θ θ   = =   =   M1 Correct form on LHS and uses an appropriate identity. 2 1 1 1 2 2 2 tan tan y C θ θ = + ( ) 1 1 2 2 2 or sec C θ + M1 A1 Integrates RHS. 1 1 2 C = + M1 Substitutes initial conditions. ( )[ ] 2 1 1 2 2 1 tan cose cot c y θ θ θ = + = M1 A1 Divides through by their integrating factor. 9

More questions on Differential equations

5 (a) State the sum of the series z + z 2 + z 3 + ... + zn , for z ! 1. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Given that z is an nth root of unity and z ! 1, deduce that 1 + z + z 2 + ... + z n - 1 = 0 . [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ i + i sin i) , use de Moivre’s theorem to show that (c) Given instead that z = 13 ( cos 3 -m 3 cos i - 1 3 cos m i = . [7] 10 - 6 cos i =/m 1 ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... 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Mark scheme: 5(a) 1 1 + − − n z z z or 1 . 1 − n z z z B1 1 5(b) 1 n z = and 1 z ≠ leading to 2 3 ... 0 n z z z z + + + + = leading to 2 1 1 ... 0 n z z z − + + + + = M1 A1 Must see 1. = nz 2 5(c) 1 cos isin 1 3 cos isin m m z z z θ θ θ θ ∞ = + = = − − −  M1 A1 Applies sum to infinity and substitutes for z. ( )( ) ( ) 2 2 i c s isin sin os 3 cos 3 cos in θ θ θ θ θ θ − + + − + M1 A1 Rationalises denominator. ( ) 2 2 2 2 s 3 c cos sin isin cos i in 3 cos s cos 9 6cos in os θ θ θ θ θ θ θ θ θ θ − − + + − + − + M1 Applies 2 2 1 sin cos θ θ + = or i i . e e 2 cos θ θ θ − + = 1 3cos Re 10 6 1 cos θ θ ∞ =  =   − −    m m z M1 A1 Takes the real part, AG. 7

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6 The matrix A is given by 5 - 223 8 A = 0 - 6 0 f0 0 1p. (a) Find a matrix P and a diagonal matrix D such that A 2 = PDP -1 . 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(b) Use the characteristic equation of A to find A3. 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Mark scheme: 6(a) −and 1. B1 Upper diagonal matrix or characteristic equation. 22 3 88 1 5: 0 8 0 0 0 11 0 0 0 λ       = − =       −     i j k 154 3 22 3 2 6: 11 8 77 3 0 0 7 0 0 λ   −     = − − = −           i j k 22 3 56 2 1: 4 8 0 0 0 7 0 28 1 λ         = − =         − − −      i j k M1 Uses vector product (or equations) to find corresponding Eigenvectors. A1 A1 A1 A1 for each correct Eigenvector. Thus 1 2 2 0 3 0 0 0 1     =     −   P and 25 0 0 0 36 0 0 0 1     =       D M1 A1 Or correctly matched permutations of columns. (Accept scalar multiples of the eigenvectors shown here.) P must have at least two non-zero columns. 7 6(b) ( )( )( ) 3 5 6 1 31 30 0 λ λ λ λ λ − + − = − + = =0 B1 Finds characteristic equation. 3 3 3 30 31 1 30 = − +  − = I 0 I A A A A M1 Finds 3 A in terms of A. Allow missing I. 2 3 68 3 125 248 0 216 0 0 0 1   −   = −       A M1A1 Substitutes for A. 4

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7 (a) It is given that y = sec h -1 x + 12 . dy 1 Express cosh y in terms of x and hence show that sinh y = - 2 . [3] d x 1 x + 2 b l ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the first three terms in the Maclaurin’s series for sec h –1 x + 12 in the form b l lna + bx + cx2 , where a, b and c are constants to be determined. 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Mark scheme: 7(a) ( ) 1 1 1 2 2 sech cosh cosh y x y x y − = = +  = + B1 Relates to cosh ,y d sinh d y y x B1 Differentiates LHS. ( ) 1 2 2 x − − + B1 Differentiates RHS. AG. 3 7(b) 2 3 2 2 d d 1 sinh cosh 2 d 2 d y y y x x x −     + = +         M1 A1 M1 A1 for LHS. B1 for RHS. B1 ( ) ( ) 1 1 1 2 (0) sech cosh (2) ln 2 3 − − = = = + y M1 A1 Relates to 1 cosh− and uses logarithmic form. 4 '(0) 3 = − y 16 ''(0) 3 3 = y M1 Evaluates derivatives at 0. = x ( ) 2 4 8 ln 2 3 3 3 3 = + − + y x x A1 Alternative method for question 7(b) ( ) ( ) ( ) 2 2 2 3 1 1 1 2 4 2 2 d 1 1 d 1 y x x x x x x − = − = − + − − + + − B1 ( ) ( ) ( ) ( ) ( ) 3 1 2 2 2 1 2 2 3 3 1 1 1 2 2 4 2 2 4 2 d 1 2 d y x x x x x x x x − − − − = + − − −− + − − + M1 A1 ( ) ( ) 1 1 1 2 (0) sech cosh (2) ln 2 3 − − = = = + y M1 A1 Relates to 1 cosh− and uses logarithmic form. Question Answer Marks Guidance 7(b) 4 '(0) 3 = − y 16 ''(0) 3 3 = y M1 Evaluates derivatives at 0. = x ( ) 2 4 8 ln 2 3 3 3 3 = + − + y x x A1 7

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8 The curve C has parametric equations - x = 2 cosh t , y = 32 t 1 4 sinh t2 , for 0 G t G 1. dx dy 2 (a) Find and show that = 1 - sinh t . [3] dt dt ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The area of the surface generated when C is rotated through 2r radians about the x-axis is denoted by A. 1 t - 4 sinh 2t ( 1 + cosh 2t)dt . [4] (b) (i) Show that A = r y 32 1 0 b l .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... (ii) Hence find A in terms of r, sinh2 and cosh2. 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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) d 2sinh d = x t t B1 ( ) 3 1 1 1 2 2 2 2 2 d cosh 2 1 cosh 2 1 sinh d = − = − − = − y t t t t M1 A1 Applies 2 2 sinh cosh 2 1, = − t t AG. 3 8(b)(i) 2 2 2 2 2 2 4 2 2 2 d d 4 ( 1) 4 2 1 ( 1) d d     + = + − = + − + = +         x y s s s s s s t t M1 Factorises 2 2 d d . d d     +         x y t t 4 cosh t A1 ( ) ( )( ) 1 1 2 3 3 1 1 2 4 2 4 0 0 2π sinh 2 cosh d π sinh 2 cosh 2 1 d t t t t t t t t − = − +   M1 A1 Correct formula for surface area, AG. A0 if limits missing. 2 2 d d d d     +         x y t t does not need to be simplified for M1. 4 Question Answer Marks Guidance 8(b)(ii) ( ) ( ) 1 1 1 1 1 1 4 16 16 4 2 0 2 0 sinh2 cosh2 1 d (1 cosh2 ) 1 cosh2   + = + = + −    t t t t M1 A1 Integrates. ( ) ( ) 1 1 1 3 3 3 1 1 2 2 2 2 2 0 0 0 cosh2 1 d sinh2 sinh2 d   + = + − +     t t t t t t t t t M1 A1 Integrates by parts. ( ) ( ) 1 3 3 3 1 1 1 1 1 2 2 4 2 2 2 4 4 2 0 sinh 2 cosh 2 sinh 2 cosh 2 t t t t t   + − − = − +   A1 ( ) 8 2 3 3 1 6 11 4 8 1 π sinh 2 cosh 2 (1 cosh 2) − + − + A1 OE. Must be exact. (Decimal answer is 3.980131435…) 6

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