Cambridge A Level Mathematics - Further 9231 — 2020 Oct/Nov Paper 2 · Variant 3

1 (a) By differentiating e - x2 , find the Maclaurin’s series for e - x2 up to and including the term in x2. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 1 5 e - x 2 d x , giving your answer as a rational fraction in its lowest (b) Deduce an approximation to y terms. 0 [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 1(a) 2 f '( ) 2 e− = − x x x B1 Finds first derivative. 2 2 2 f ''( ) 4 e 2e − − = − x x x x B1 Finds second derivative. f (0) 1 f '(0) 0 f ''(0) 2 = = = − M1 Evaluates derivatives at zero. 2 2 e 1 − = − x x M1 A1 5 1(b) 3 1 1 5 2 5 0 0 1 74 1 d 3 375   − = − =      x x x x M1 A1 Substitutes 2 1−x or better. 2

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2 The variables x and y are related by the differential equation d 2 y dy 2 9 2 + 6 + y = 3x + 30x. 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 1 9 6 1 0 3 + + =  = − m m m ( ) 1 3 e − = + x y Ax B A1 Complimentary function. 2 ' 2 '' 2 = + +  = +  = y p qx rx y q rx y r B1 Particular integral and its derivatives. 2 2 18 6 12 3 30 + + + + + = + r q rx p qx rx x x M1 Substitutes and equates coefficients. 3, 6, 18 = = − = − r q p A1 ( ) 2 1 3 e 3 6 18 − = + + − − x y Ax B x x A1 6 2(b) 2 3 6 18 = − − y x x B1 FT 1

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3 (a) Show that the system of equations x - 2y - 4z = 1, x - 2 y + kz = 1, - x + 2 y + 2 z = 1, where k is a constant, does not have a unique solution. 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(b) Given that k =- 4 , show that the system of equations in part (a) is consistent. Interpret this situation geometrically. 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(c) Given instead that k =- 2 , show that the system of equations in part (a) is inconsistent. Interpret this situation geometrically. 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(d) For the case where k !- 2 and k !- 4 , show that the system of equations in part (a) is inconsistent. Interpret this situation geometrically. 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Mark scheme: 3(a) 1 2 4 1 2 4 2 2(2 ) 4(0) 0 1 2 2 − − − = −− + + − = − k k k M1 A1 Shows that determinant is zero. 2 3(b) 2 4 1, 2 4 1, 2 2 1, − − = − − = −+ + = x y z x y z x y z 1, 2 3  = − − = − z x y (or 2 2 1 −+ + = x y z is not parallel to 2 4 1 − − = x y z ) B1 Derives one equation with two unknowns or states that third plane is not parallel to the repeated one. Two of the planes are identical (or coincident). B1 There is a line of intersection with the other plane. B1 3 3(c) 2 4 1, 2 2 1, 2 2 1, − − = − − = −+ + = x y z x y z x y z 1 1 −= B1 Derives contradiction. Two parallel planes, not identical. B1 2 3(d) 2 4 1, 2 1, 2 2 1, − − = − + = −+ + = x y z x y kz x y z ( 4 ) 0, 2 2 0 2 −− = − =  = k z z B1 Derives contradiction. The three planes form a triangular prism. B1 2

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4 y 1 x 0 1 2 n - 1 1 n n n The diagram shows the curve with equation y = 1 - x 3 for 0 G x G 1, together with a set of n rectangles of width 1. n (a) By considering the sum of the areas of the rectangles, show that 1 2 ( 1 - x ) dx G 2 . [4] y 3 3n + 2n - 1 0 4n ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 1 (b) Use a similar method to find, in terms of n, a lower bound for y ( 1 - x 3 ) dx . [4] 0 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) ( ) 3 1 1 3 3 3 0 1 1 1 1 d 1 ... 1 1   −     − + − + + −          n n x x n n n n  M1 A1 Forms the sum of the areas of the rectangles. ( ) 4 2 2 1 3 4 1 1 1 1 1 4 − = − = − = −  n r n n r n n M1 Applies ( ) 1 2 2 1 4 1 3 1 . − = = −  n r r n n ( ) 2 2 2 2 2 4 1 3 2 1 4 4 − − + − = = n n n n n n A1 AG 4 4(b) ( ) 3 3 3 3 3 3 1 0 1 1 1 1 2 1 1 1 1 . . 1 d .     −     − + − + + −             −  n n n n n x x n n  M1 A1 Forms the sum of the areas of appropriate rectangles. ( ) 4 2 2 1 3 4 1 1 1 1 1 4 − = − − − = − = −  n r n n n n r n n n n M1 Applies ( ) 1 3 1 2 2 1 1 . 4 − = = −  n r r n n 2 2 2 2 4 ( 1) ( 1) 3 2 1 4 4 − − − − − = = n n n n n n n A1 4

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5 It is given that x = sinh -1 t , y = cos -1 t , where - 1 1 t 1 1. dy 1 (a) By differentiating cosy with respect to t, show that =- . [4] dt 1 - t 2 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ d 2 y (b) Find 2 in terms of t, simplifying your answer. [5] dx ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a) d sin 1 dt − = y y M1 A1 Differentiates both sides with respect to t. 2 d 0 1 cos 1 n dt si 0 π < < − −  >  = y y y y M1 Applies 2 2 sin cos 1. + = y y 2 d 1 dt 1 = − − y t A1 AG, justifies taking positive square root. 4 5(b) 2 d 1 dt 1 = + x t B1 2 2 d 1 d 1 + = − − y t x t B1 Finds first derivative. ( ) ( ) ( ) 2 1 1 1 1 2 2 2 2 2 2 2 2 2 2 1 1 (1 ) 1 d 1 dt 1 1 − −   − + + + − +   − = −   − −   t t t t t t t t t M1 Differentiates d d y x with respect to t. 2 2 2 2 d d 1 d dt d d 1   +   = − ×   −   y t t x x t M1 Applies chain rule. ( ) ( ) ( ) 1 1 2 2 3 2 2 2 2 2 2 1 (1 ) 1 2 1 1 −     − + + −         = − = −   −     −   t t t t t t t A1 OE (simplified). 5

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i - 4 cos 2i + 3) . [5]6 (a) Use de Moivre’s theorem to show that sin 4 i = 18 ( cos 4 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the solution of the differential equation d y 3 + y cot i = sin i di r . [6] for which y = 0 when i = 12 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a) 1 2isinθ − − = z z B1 Use of 1 2isinθ − − = z z . ( ) ( ) ( ) 4 1 4 4 2 2 4 6 − − − − = + − + + z z z z z z M1 A1 Expands and groups. ( ) ( ) ( ) 4 2isin 2cos4 4 2cos2 6 θ θ θ = − + M1 Substitutes 2cos θ − + = n n z z n ( ) 4 1 cos4 4cos2 3 sin 8 θ θ θ = − + A1 AG 5 6(b) cot lnsin sin θ θ θ θ = =  d e e M1 A1 Finds integrating factor. ( ) ( ) 4 1 cos4 4cos2 3 d sin sin 8 d θ θ θ θ θ = − = + y M1 Correct form on LHS and uses identity given in (a). 1 1 sin 4 2sin 2 3 8 4 sinθ θ θ θ   = − + +     C y A1 0 π 1 3 8 2 C   = +     M1 Substitutes initial conditions. 1 1 3 sin 4 2sin 2 3 in π s 8 4 2 y θ θ θ θ   = − + −     A1 OE 6

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7 The matrix P is given by 1 4 2 P = 0 - 1 1 f0 0 2p. (a) State the eigenvalues of P. [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Use the characteristic equation of P to find P -1 . 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The 3 # 3 matrix A has distinct eigenvalues b, - 1, 1 with corresponding eigenvectors 1 4 2 0 - 1 1 f0p, f 0p, f2p, respectively. (c) Find A in terms of b. 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Mark scheme: 7(a) − B1 1 7(b) 3 2 2 2 0 − − + = P P I P B1 States that P satisfies its characteristic equation. 1 2 2 2 −= + − P I P P M1 Multiplies through by 1 − P . 1 2 1 4 3 1 0 10 1 0 1 1 0 1 2 0 0 4 1 0 0 2 −     −          = −               = P P M1 A1 4 Question Answer Marks Guidance 7(c) 1 0 0 0 1 0 0 0 1 −     = −       b A P P M1 Applies 1 − = A PDP 1 4 3 4 2 1 0 1 1 0 1 2 0 0 2 1 0 0 2     − −       = −            b M1 A1 Multiplies two adjacent matrices. 4 4 3 1 0 1 1 0 0 1 + − −     −       b b b A1 4

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8 (a) Sketch the graph of y = coth x for x 2 0 and state the equations of the asymptotes. [2] (b) Starting from the definitions of coth and cosech in terms of exponentials, prove that coth 2 x - cosech 2 x = 1. 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The curve C has equation y = ln coth 12 x for x 2 0 . a k dy (c) Show that =- cosechx . [3] dx ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (d) It is given that the arc length of C from x = a to x = 2a is ln4, where a is a positive constant. Show that cosha = 2 and find, in logarithmic form, the exact value of a. 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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) B1 Correct shape and position, not too truncated. 0, = x 1 = y B1 States equations of asymptotes. 2 8(b) e e 2 coth cosech e e e e − − − + = = − − x x x x x x x x B1 ( ) ( ) 2 2 2 2 2 e e 4 e e 2 1 e e e e e e − − − − −   + + − − = =   −   − − x x x x x x x x x x M1 A1 Writes over common denominator, AG. 3 x Question Answer Marks Guidance 8(c) 2 1 sech d 1 2 1 1 1 d 2tanh 2sinh cosh 2 2 2       = − = −                   x y x x x x Or 2 1 cosech d 1 2 1 1 1 d 2coth 2sinh cosh 2 2 2       = − = −                   x y x x x x M1 A1 Uses chain rule. = 1 cosech sinh( ) − = − x x A1 AG 3 8(d) 2 2 1 cosech d + a a x x M1 Forms correct integral. 2 2 2 coth d coth d =   a a a a x x x x M1 A1 Uses 2 2 coth cosech 1 − = x x . [ ] 2 lnsinh lnsinh2 lnsinh = = − a a x a a M1 Integrates and substitutes limits. ( ) sinh 2 ln ln 2cosh sinh = = a a a M1 Combines logarithms and uses double angle formula. ( ) ln 2cosh ln 4 cosh 2 =  = a a A1 AG ( ) ( ) 2 ln 2 2 1 ln 2 3 = + − = + √ a A1 Must reject ( ) ln 2 . 3 − 7

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