Cambridge A Level Mathematics - Further 9231 — 2021 May/June Paper 2 · Variant 3

1 (a) Find a and b such that z 8 - iz 5 - z 3 + i = ( z 5 - a)( z 3 - b) . [1] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Hence find the roots of z 8 - iz 5 - z 3 + i = 0 , giving your answers in the form re ii, where r 2 0 and 0 G i 1 2r . [6] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 1(a) = B1 1 1(b) 2 5 π ie k z = M1 Finds one fifth root of unity. 2 5 π ie , 0,1,2,3,4 k z k = = A1 Gives all fifth roots of unity. Accept 1 not in exponential form. A0 if 1 = r not seen or implied. Argument of i is 1 2 π. B1 1 6i π e z = M1 A1 Finds one root of 3 i. = z A0 if first root not given in exponential form. 5 3 6 2 i π i π e , e z = A1 Finds other two roots of 3 i. = z A0 if 1 = r not seen or implied. Withhold final A1 if a and b reversed in part (b) but all other work correct. 6 SC If ( ) 2 1 10 5 π π ie , 0,1,2,3,4 k z k + = = award M1 A1 A1 only. SC If ( ) 2 3 π ie , 0,1,2 k z k = = award M1 A1 only.

More questions on Complex numbers

2 Find the Maclaurin’s series for lncoshx up to and including the term in x4. [7] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 2 d tanh d = y x x B1 Finds first derivative. 2 2 2 d sech d = y x x B1 Finds second derivative. 2 3 3 3 d sinh 2tanh sech 2 d cosh = − = − y x x x x x M1 A1 Finds third derivative. ( ) ( ) ( ) 4 4 2 2 4 cosh 3sinh 2 co h d s d − − = x x y x x 4 2 2 2sech 4tanh sech = − + x x x B1 Finds fourth derivative. Alternative: 2 2 2 6tanh sech 2sech . − x x x (1) (3) (0) (0) (0) 0 = = = y y y (2)(0) 1 = y (4)(0) 2 = − y M1 Evaluates derivatives at 0. = x 1 1 2 12 2 4 = − y x x A1 CWO 7

More questions on Hyperbolic functions

3 y 0 1 2 3 N – 1 N x x The diagram shows the curve y = 2 for x H 1, together with a set of N - 1 rectangles of unit 2x - 1 width. (a) By considering the sum of the areas of these rectangles, show that N r 1 2 1 4 ln ( 2 N - 1) + 1. [7] 2 2r - 1 =/r 1 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ N r(b) Use a similar method to find, in terms of N, a lower bound for 2 . [3] 2r - 1 =/r 1 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) 2 1 2 2 1 2 1 2 1 = = = + − −   N N r r r r r r M1 A1 Compares with sum of the areas of the rectangles. < 1 2 1 d 2 1 + −  N x x x M1 Compares with integral. ( ) 2 1 4 2 1 1 d ln 2 1 2 1   = −   −  N N x x x x M1 A1 Finds integral. ( ) 1 4 2 2 1 1 ln 2 1 2 1 = < + − −  N r r N r M1 A1 Inserts limits, AG. 7 3(b) 1 2 2 2 2 1 1 2 1 d 2 1 2 1 2 1 2 1 2 1 − = = = + > + − − − − −    N N N r r r r N x N x r r N x N M1 A1 Compares with integral. ( ) 1 4 2 2 ln 2 1 2 1 N N N − + − A1 CWO. Alternative method for question 3(b) 1 1 1 2 2 d 2 1 2 1 + = > − −   N N r r x x r x M1 A1 Compares with integral. ( ) ( ) 1 4 2 ln 2 1 1 + − N A1 CWO. 3

More questions on Integration

4 By considering the binomial expansions of z + and z - , where z = cos i + i sin i , use de z z Moivre’s theorem to show that sin 5i - a sin 3i + b sin i tan 5i = , cos 5i + a cos 3i + b cos i where a and b are integers to be determined. [7] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 4 1 2isinθ − − = z z and 1 2cosθ − + = z z B1 Use of 1 2isinθ − − = z z and 1 2cosθ − + = z z ( ) ( ) ( ) ( ) 5 3 3 1 5 1 5 5 10 − − − − − = − − − + − z z z z z z z z M1 A1 Expands and groups. A0 if grouping not shown. ( ) ( ) ( ) ( ) 5 3 3 1 5 1 5 5 10 − − − − + = + + + + + z z z z z z z z M1 A1 Expands and groups. A0 if grouping not shown. ( ) ( ) 5 5 5 5 ) 2 0 isin 2isin5 2cos5 2 co i 5(2isin3 1 (2 sin ) 5 2cos3 10 s 2cos θ θ θ θ θ θ θ θ = − + + + M1 Substitutes 2cos θ − + = n n z z n and 2isin . θ − − = n n z z n 5 sin5 5sin3 10sin . cos5 5cos3 10c t os an θ θ θ θ θ θ θ − + = + + A1 Justifies cancellation of constants. 7

More questions on Complex numbers

5 The variables x and y are related by the differential equation d 2 y dy -x - 2 - 3y = 4 e . 2 dx dx (a) Find the value of the constant k such that y = kxe -x is a particular integral of the differential equation. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ dy 1 (b) Find the solution of the differential equation for which y = = 2 when x = 0 . [6] dx ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...................................................................................................................................................

Mark scheme: 5(a) ( ) ( ) e leading to ' e 1 leading to '' e 2 x x x y kx y k x y k x − − − = = − = − B1 B1 Differentiates particular integral. ( ) ( ) e 2 2 e 1 3 e 4e leading to 4 4 x x x x k x k x kx k − − − − − − − − = − = M1 Substitutes. 1 = − k A1 4 5(b) 2 2 3 0 leading to 1,3 m m m − − = = − M1 Auxiliary equation. 3 e e − = + x x y A B A1 Complementary function. (Can be awarded if seen in the general solution.) 3 3 e e e ( )e e − − − = + − = − + x x x x x y A B x A x B A1 FT General solution. FT on their k. Must have “ = y ”. 3 ' ( )e e 3 e − − = − − − + x x x y A x B B1 Derivative of general solution. 1 1 1 2 2 2 , 1 3 leading to 0, A B A B A B + = − −+ = = = M1 Substitutes initial conditions into general solution. 3 1 2 e e− = − x x y x A1 Must have “ = y ”. 6

More questions on Differential equations

6 (a) Starting from the definitions of sinh and cosh in terms of exponentials, prove that 2 sinh 2 x = cosh 2 x - 1. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the solution to the differential equation d y + y coth x = 4 sinh x d x for which y = 1 when x = ln 3 . [7] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ...................................................................................................................................................

Mark scheme: 6(a) ( ) ( ) 1 1 2 2 sinh e e cosh e e − − = − = + x x x x x x B1 Writes in exponential from. ( ) ( ) ( ) 1 1 1 2 2 2 2 2 2 2 2 e e e 2 e e e 1 − − − − = − + = + − x x x x x x M1 Expands. cosh 2 1 − x A1 AG. 3 6(b) lns oth inh c e e sinh = =  x x x M1 A1 Finds integrating factor. 2 d ( sinh ) 4sinh d = y x x x M1 Correct form on both sides. ( ) 1 2 sinh 2 sinh 2 sinh 2 2 = − + = − + y x x x C x x C M1 A1 Integrates 2 sinh x using identity. 4 40 sinhln3 sinhln9 2ln3 leading to 2ln3 3 9 C C = − + = − + M1 Substitutes initial conditions into their solution. 28 sinh sinh2 2 2ln3 9 = − + − y x x x A1 7

More questions on Differential equations

= 4 + x dx .7 The integral In, where n is an integer, is defined by I n 2 2 2 y 0 (a) Find the exact value of I1, expressing your answer in logarithmic form. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ d 2 - 21 n (b) By considering x 4 + x or otherwise, show that dx b ` j l, n = 2 5 4nI n + 2 3 2 + ( n - 1) I n . [5] b l ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Find the value of I5. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 7(a) ( ) ( ) 3 3 1 2 2 2 2 1 1 2 0 0 1 4 sinh I x x − −     = + =  M1 A1 Recognises integral or uses appropriate substitution. ln 2 A1 Insert limits, must simplify. 3 7(b) ( ) ( ) ( ) 1 1 1 2 2 2 2 2 2 2 1 d 4 4 4 d − − − −   + = − + + +     n n n x x nx x x x M1 A1 Uses the product rule to differentiate. ( )( ) ( ) 1 1 2 2 2 2 2 1 4 4 4 4 n n n x x x − − − − + − + + + M1 Applies 2 2 4 4. = + − x x ( ) 3 1 2 2 2 2 0 4 4 − +   + = − + +     n n n n x x nI nI I M1 Integrates both sides using the limits given. ( ) ( ) 3 3 2 2 2 5 2 5 2 2 (1 ) 4 4 ( 1) + + = − +  = + − n n n n n n n I nI nI n I A1 Substitutes limits and rearranges, AG. Alternative method for question 7(b) ( ) ( ) 3 3 1 2 2 2 1 2 1 2 2 2 0 0 4 4 d − − −   = + + +      n n n I x x n x x x M1 A1 Integrates by parts. ( ) ( )( ) 1 3 3 1 2 2 2 2 1 0 2 2 0 2 4 4 4 4 d − − −   = + + + − +      n n n I x x n x x x M1 Applies 2 2 4 4. = + − x x ( ) ( ) 3 3 2 2 2 5 2 2 5 2 4 4 ( 1) + + = + −  = + − n n n n n n n I nI I nI n I M1 A1 Substitutes limits and rearranges, AG. 5 Question Answer Marks Guidance 7(c) 0 3 3 2 = I B1 Applies reduction formula with 1. = n ( ) 3 8 33 2 125 10 5 00 3 5 12 2 0.033 = +  = = I I I M1 A1 Applies reduction formula with 3. = n 3

More questions on Integration

8 (a) Find the value of a for which the system of equations 13 x + 18 y - 28 z = 0, - 4x - ay + 8z = 0, 2x + 6y - 5z = 0, does not have a unique solution. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The matrix A is given by 13 18 - 28 A = - 4 - 1 8 f 2 6 - 5p. 2 (b) Find the eigenvalue of A corresponding to the eigenvector 0 [1] f1p. ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Find a matrix P and a diagonal matrix D such that A = PDP -1 . [8] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (d) Use the characteristic equation of A to find A -1 in terms of A. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 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) 13 18 28 4 8 9 24 2 6 5 − − − = − − a a M1 Finds determinant. 8 3 a = A1 Accept 24. 9 2 8(b) 13 18 28 2 4 1 8 0 1. 0 2 6 1 2 1 5 λ  =   − −         − −  = −         − −      B1 1 Question Answer Marks Guidance 8(c) 13 18 28 4 1 8 0 2 6 5 λ λ λ − − − −− = −− B1 Sets ( ) det 0. λ − = I A ( )( )( ) 3 2 7 7 15 0 1 3 5 0 λ λ λ λ λ λ − + − − =  + − − = M1 Expands determinant and factorises. 1, 3, 5 λ λ λ = − = = A1 16 1 3: 4 4 8 16 1 2 6 8 16 1 λ −       = − − = −       − −     i j k 12 1 5: 4 6 8 24 2 2 6 10 12 1 λ −         = − − = −         − −      i j k M1 Uses vector product (or equations) to find corresponding eigenvectors. A1 A1 A1 for each correct Eigenvector. Thus 2 1 1 0 1 2 1 1 1 −     =       P and 1 0 0 0 3 0 0 0 5 −     =       D M1 A1 Or correctly matched permutations of columns. Accept scalar multiples of eigenvectors. At least two (non-zero) Eigenvectors for M1. 8 Question Answer Marks Guidance 8(d) 3 2 7 7 15 − + − − = A A A I 0 M1 Substitutes A into characteristic equation and multiplies through by 1. − A Allow missing I. ( ) 1 2 1 7 7 15 −= − + − A A A I A1 Accept with 13 18 28 4 1 8 2 6 5 −     = − −     −   A and 2 41 48 80 32 23 64 8 0 17 −     − −     −   = A substituted to give 1 116 8 . 43 78 1 4 9 15 22 42 59 − −     −     −   = A 2

More questions on Matrices