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

r up to and including the term in x2. [5]1 Find the Maclaurin’s series for tan x + 14 .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 2 1 '( ) s π ec 4 f x x   = +     B1 Finds first derivative. 2 1 1 ''( ) 2sec t n 4 π 4 π a f x x x     = + +         B1 Finds second derivative. (0) 1 '(0) 2 ''(0) 4 = = = f f f M1 Evaluates derivatives at zero. 2 1 tan 1 2 2 4 π x x x   + = + +     M1 A1 5

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2 A curve has equation y = cosh x , for 0 G x G 12 . Find, in terms of r and e, the area of the surface generated when the curve is rotated through 2r radians about the x-axis. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 2 2 d 1 1 sinh cosh d   + = + =     y x x x M1 A1 Applies 2 2 cosh 1 sinh = + x x 1 2 2 0 2 c π osh d x x  M1 Correct formula, correct limits. 1 0 2 cosh 2 1d π x x = +  M1 Applies 2 2cosh cosh2 1 = + x x or expands ( ) 2 2 e e− + x x 1 2 0 1 π sinh 2 2 x x +   =     A1 Correct integration. ( ) 1 π e 1 2 4 e− − = + A1 6

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3 Find all the roots of the equation ( w + 1 ) 6 = 1, giving your answers in the form x + yi where x and y are real and exact. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 3 2 π isin π, 0, ,5 2 1 cos 6 6 n w n n = = + +  M1 A1 Uses the 6th roots of unity. 1 3 0, i 3, i 3, 2 1 1 2 2 2 2 = − √ ± √ ± − − w A1 A1 A1 for 3 correct roots. A1 Gives all six roots of the equation in exact form. 4

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4 Find the solution of the differential equation dy x + 2y = ex dx for which y = 3 when x = 1. Give your answer in the form y = f ( x) . [8] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 4 d 2 1 e d + = x y y x x x B1 Divides through by x. 1 2 2 d −  = x x e x M1 A1 Finds integrating factor, must be integrating 1. − x ( ) 2 d e d = x yx x x M1 Correct form on LHS and attempt to integrate 1 ex x multiplied by their integrating factor. 2 ( 1)e = − + x yx x C A1 3 = C M1 Finds C. 2 ( 1)e 3 − + = x x y x M1 A1 Divides through by coefficient of y. 8

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5 The curve C has equation y 2 + ( xy + 1) 2 = 5 . dy 2 (a) Show that, at the point ( 1, 1) on C, =- 3 . [3] dx ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ d 2 y (b) Find the value of 2 at the point ( 1, 1) . [5] dx ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a) ( ) 2 d 2 ' d = y yy x B1 Differentiates 2 y correctly. ( ) ( ) ( ) 2 d 1 2( 1) ' d + = + + xy xy xy y x B1 Differentiates ( ) 2 1 + xy correctly. ( ) 2 (1) ' ((1) 1) ' 1 0 ' 3 + + + =  = − y y y B1 Substitutes (1,1), AG. 2 2 ' + = − + + xy y y y x y x 3 5(b) ( ) 2 '' ' + yy y B1 Differentiates '. yy ( ) ( )( ) ( 1) '' ' ' ' ' 0 + + + + + + + = xy xy y y xy y xy y B1 B1 Differentiates ( ) ( 1) ' . + + xy xy y 2 2 4 2 1 '' 2 '' 1 0 3 3 3 3        + − + − + − + =               y y M1 Substitutes (1,1) and 2 ' 3 = − y 19 '' 27 = y A1 5 Question Answer Marks Guidance 5(b) Alternative ( ) 2 2 2 2 2 2 ( )(2 ' ') ( )( ' 2 ' 1) '' + + + + − + + + + = − + + y x y x xyy y y xy y y xy x y y y x y x M1 A1 A1 Differentiate 2 2 ' + = − + + xy y y y x y x using quotient rule. A1 for differentiating 2 xy correctly. A1 for everything correct. 19 '' 27 = y M1 A1 Substitutes (1,1) and 2 ' . 3 = − y 5

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6 Find the particular solution of the differential equation d 2 x dx + 8 + 15 x = 102 cos 3 t , 2 dt dt dx given that, when t = 0 , x = 1 and = 0 . [11] dt .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... 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Mark scheme: 6 2 8 15 0 3, 5 + + =  = −− m m m M1 Auxiliary equation. 5 3 e e − − = + t t x A B A1 Complementary function. sin3 cos3 3 cos3 3 sin3 9 sin3 9 cos3 = +  = −  = − −   x p t q t x p t q t x p t q t M1 A1 Particular integral and its derivatives. 9 24 15 0 6 24 0 − − + =  − = p q p p q 9 24 15 102 4 17 − + + =  + = q p q p q M1 Substitutes and equates coefficients. 4, 1 = = p q A1 5 3 e e 4sin3 cos3 − − = + + + t t x A B t t A1 5 3 5 e 3 e 12cos3 3sin3 − − = − − + −  t t x A B t t M1 Differentiating the correct form. 1 1 = + + A B 0 5 3 12 6, 6 = − − +  = = − A B A B M1 A1 Forms simultaneous equations using initial conditions. 5 3 6e 6e 4sin3 cos3 − − = − + + t t x t t A1 11

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! 0, 1, - 1. [2]7 (a) Show that z 2 r = -1 , for z z - z r =1 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) By letting z = cos i + i sin i , show that, if sin i ! 0 , n sin ( 2n + 1) i 1 + 2 cos ( 2 ri) = . [5] sin i =/r 1 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 7(a) ( ) 2 2 2 2 2 2 2 1 2 1 ( ) ( ) 1 − − + + + = −  n n z z z z z z z z M1 Uses sum of geometric series. ( ) 2 2 1 2 1 2 1 1 1 1 − + − − − − − × = − n n z z z z z z z z z A1 Divides numerator and denominator by z. Must see at least 2 2 2 2 , 1 + − − n z z z AG. 2 7(b) 1 2isinθ −= − z z B1 Simplifies denominator. 2 1 1 2 isi cos( 1) 2 n(2 1) cos i i s is n in θ θ θ θ θ + − + + − − − − + = n z z n z n z M1 A1 Applies de Moivre’s theorem to numerator. ( ) 1 2 sin sin(2 1) sin(2 1) 1 cos 2 2sin 2sin θ θ θ θ θ θ = + = − − + =  n r n n r M1 Equates real parts. ( ) 1 sin(2 1) 1 2 cos 2 sin θ θ θ = + + =  n r n r A1 AG 5

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8 y 0 1 2 n - 1 n x 1 The diagram shows the curve y = for x H 0 , together with a set of n rectangles of unit x 2 + x + 1 width. By considering the sum of the areas of these rectangles, show that n + . [10] 1 ln 3 3 n + 3 n + n + 1 / 1 1 2 2 2 b l + r + 1 r =1 r 2 .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... 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Mark scheme: 8 Sum ( ) 2 1 1 1 3 7 1 = + + + √ √ √ + +  n n M1A1 Forms sum of the areas of the rectangles, need to see consideration of areas, not just the sum given in the question. ( ) 2 0 1 1 d + < √ +  n x x x M1 Compares with integral. Limits have to be correct. 2 2 1 3 1 2 4   + + = + +     x x x B1 Completes the square 1 1 2 0 0 0 1 d sinh s 1 2 1 2 1 3 2 4 inh 1 3 3 2 − −   +   +           = =            + +        √     √      n n n x x x x M1 A1 Uses formula ( ) ( ) 0 2 0 2 ln l 2 1 2 1 2 1 3 3 2 n 1 3 3                 = =         √ √  + + √ + + + +       +  + n n x x x x x M1 A1 Uses logarithmic form of 1 sinh . − ( ) 2 3 ln l 1 n 3 2 1 2 3       = −     √ √  + + +  √ +   n n n M1 Inserts limits. ( ) 2 1 2 2 ln 3 3 3 1   = + + √     + + n n n A1 AG 10

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9 It is given that a is a positive constant. (a) Show that the system of equations ax + ( 2a + 5) y + ( a + 1) z = 1, - 4y = 2, 3y - z = 3, has a unique solution and interpret this situation geometrically. 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The matrix A is given by a 2a + 5 a + 1 A = 0 - 4 0 f0 3 - 1 p. (b) Show that the eigenvalues of A are a, - 1 and - 4 . 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(c) Find a matrix P such that a 0 0 A = P 0 - 1 0 P -1 . 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Mark scheme: 9(a) 0 2 5 1 0 4 0 4 0 3 1 + + ≠ − = − a a a a or 11 8 2 1 2 9 2 = + = − = − x a y z solves equations and finds at least two correct values. The three planes intersect at a single point. B1 3 9(b) 2 5 1 0 4 0 0 0 3 1 λ λ λ − + + −− = −− a a a M1 Equates determinant to zero. ( ) )( 1 ) 0 , 4 1 4 ( , λ λ λ λ −− =  = −− − −− a a A1 AG 2 Question Answer Marks Guidance 9(c) ( )( ) 4 1 1 : 0 4 0 0 0 0 3 1 0 0 λ −− −−      = −− =       −−     a a a a a i j k M1 A1 Uses vector product (or equations) to find corresponding eigenvectors. 3( 1) 1 1: 1 2 5 1 0 0 0 3 0 3( 1) 1 a a a a a λ +         = − + + + =         − − + −     i j k  A1 3 12 1 4 : 4 2 5 1 (3 12) 1 0 3 3 3 12 1 a a a a a a λ +         = − + + + = − + −         +     i j k  A1 1 1 1 0 0 1 0 1 1     = −     −   P A1 OE 5 Question Answer Marks Guidance 9(d) 3 2 ( (5 ) ( )( 1)( 4) 4 0 4 5 ) a a a a λ λ λ λ λ λ + − + − − + + = − = B1 Finds characteristic equation. 1 2 4 (5 ) (4 5 ) a a a −= + − + − A A A I M1 A1 Multiplies through by 1 − A . 2 2 2 2 2 17 1 0 16 0 0 15 1 a a a   − −     −  =   A B1 2 2 2 1 2 5 1 1 0 0 0 4 0 (4 5 ) 0 1 0 0 3 2 17 1 4 0 16 0 (5 ) 0 15 1 1 0 0 1 a a a a a a a a a −   − −  + −    + +         = − + −         −   −     A M1 Substitutes for and A in correct equation. 1 4 5 8 4 4 1 0 0 4 0 3 4 a a a a a a − + +     = −     − −   A A1 6

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