Cambridge A Level Mathematics - Further 9231 — 2022 May/June Paper 2 · Variant 1

41 The curve C has polar equation r = e 3 i for 0 G i G a. Given that the length of C is s, find a in terms of s. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 1 3 3 3 2 2 4 9 16 0 5 4 0 e e d e d s              3 3 4 4 5 5 3 3 0 e e 1 s           M1 A1 Integrates.   3 4 3 5 ln 1 s   A1 5

More questions on Polar coordinates

2 (a) Starting from the definitions of cosh and sinh in terms of exponentials, prove that cosh 2x = 2 sinh 2 x + 1. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the set of values of k for which cosh 2x = k sinh x has two distinct real roots. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) 1 1 2 2 cosh e e sinh e e x x x x x x           2 2 2 1 1 2 2 e e 1 e e cosh 2 x x x x x       M1 A1 Expands, AG. 3 2(b) 2 2sinh sinh 1 0 x k x   M1 A1 Applies identity. 2 8 0 k   M1 A1 Sets discriminant positive. 8, k  8 k  A1 5

More questions on Hyperbolic functions

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

Mark scheme: 3(a) 2 3 2 1 2 1 0 leadin i g to z z m     M1 Auxiliary equation.   1 2 3 3 2 2 e sin cos t x A t B t    A1 Complimentary function. 2 leading to leading to 2 2         x pt q x p x pt qt r B1 Particular integral and its derivatives. 2 2 2 2 1 p pt q pt qt r t        M1 Substitutes and equates coefficients. PI must have the correct form. 1, 2, 1 r q p    A1   1 2 3 2 3 2 2 e sin cos 2 1 t x A t B t t t       A1 6 3(b) 1 2 e 0 t   as t  M1 2 x   A1 FT 2

More questions on Differential equations

4 The diagram shows the curve with equation y = 2x for 0 G x G 1, together with a set of N rectangles 1 each of width . N y 2 1 0 x 0 1 2 N - 1 1 N N N 1 (a) By considering the sum of the areas of these rectangles, show that 2x dx 1 U N , where y0 1 N 2 U N = 1 . [4] N 2 N - 1 ` j ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 2x dx . [4](b) Use a similar method to find, in terms of N, a lower bound LN for 1y0 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Find the least value of N such that U N - L N 1 10 -4 . 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Mark scheme: 4(a)     1 1 2 1 1 1 1 1 0 2 d 2 2 2 2 N N N N N N x N N N N x         M1 A1 Forms the sum of the areas of the rectangles.     1 1 1 1 1 2 2 2 1 N N N N n n N N      M1 A1 Applies   1 1 , 1 N n N n r r r r      AG. 4 4(b)    2 1 1 1 1 1 1 1 0 2 d 2 2 2 N N N N N x N N N N x          M1 A1 Forms the sum of the areas of appropriate rectangles.     1 1 1 0 1 1 2 2 1 N N N n n N N       M1 A1 Applies 1 0 1. 1 N N n n r r r       4 4(c)     1 1 1 4 4 2 1 1 10 leading to 10 2 1 2 1 N N N N N N N        M1 Simplifies n n U L  to . c n Least value of N is 10001 A1 2

More questions on Summation of series

5 The variables x and y are such that y = 0 when x = 0 and ( x + 1) y + ( x + y + 1 ) 3 = 1. d y 3 (a) Show that = - when x = 0 . 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(b) Find the Maclaurin’s series for y up to and including the term in x2. 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Mark scheme: 5(a)   d ( 1) ( 1) ' d x y x y y x     correctly.        3 2 d 1 3 1 ' 1 d x y x y y x       B1 Differentiates   3 1 x y   correctly.   3 4 (1) ' 3 ' 1 0 ' y y y      B1 Substitutes (0,0), AG. 3 5(b)  1 '' 2 ' x y y   B1 Differentiates ( 1) ' . x y y        2 2 3 1 '' 6 1 ' 1 0 x y y y x y         B1 B1 Differentiates   2 3( 1) 1 ' . x y y    Expect       6 1 1 6 1 1           x y y x y y y instead of    2 6 1 1 .     y x y    3 1 4 4 2 '' 2 3 '' 6 0 y y      M1 Substitutes (0,0). 9 32 '' y  A1 3 9 64 2 4 2 ''(0) (0) '(0) 2! y y y y x x x x      M1 A1 Finds Maclaurin’s series. 7

More questions on Differentiation

6 Use the substitution y = vx to find the solution of the differential equation d y 2 2 x = y + 9x + y d x for which y = 0 when x = 1. Give your answer in the form y = f ( x) , where f ( x) is a polynomial in x. 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Mark scheme: 6 d d d d y v v x x x   2 2 2 d 9 d v x v x vx x x v x           M1 Substitutes and derives first order separable equation. 2 d 9 d v x v x   A1 2 1 1 d d 9 v x x v     M1 Separates variables and integrates both sides. 1 sinh ln 3 v x C        A1 0 C  B1 Substitutes initial conditions.   2 3 9 ln 1 ln v v x    M1 Converts to logarithmic form. 2 2 2 9 3 y y x x    A1 4 2 2 2 2 9 6 9 0 x x y y y x      M1 Squares to eliminate radical. 2 2 3 ( 1) y x   A1 10

More questions on Differential equations

7 (a) Use de Moivre’s theorem to show that cosec 7 i cosec 7i = 6 4 2 . 7 cosec i - 56 cosec i + 112 cosec i - 64 [6] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Hence obtain the roots of the equation x 7 - 14x 6 + 112x 4 - 224x 2 + 128 = 0 in the form cosecqr, where q is rational. 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Mark scheme: 7(a) 7 Im i 7 i s n c s   7 2 5 4 3 6 sin7 21 35 7 s c s c s c s       A1       2 3 7 2 5 2 3 2 sin7 21 1 35 1 7 1 s s s s s s s              7 2 5 2 4 3 2 4 6 sin7 21 1 35 1 2 7 1 3 3 s s s s s s s s s s           M1 Uses 2 2 1 . c s  7 5 3 1 s 5 in7 64 12 6 7 s s s s     A1 7 5 3 1 64 112 5 cose 6 c7 7 s s s s     M1 Divides numerator and denominator by 7. s 6 4 7 2 cosec cosec7 cosec cosec cosec 7 56 112 64          A1 AG CWO 6 7(b)   7 7 6 4 2 6 4 2 2 7 56 112 64 leading to 2 7 56 112 64 x x x x x x x x         M1 A1 Relates with equation in part (a). 1 2 2 leadi c ng to osec7 si n7     M1 Solves 1 2 sin7 .    1 42 π cosec x  A1 Gives one correct solution.   19 7 5 13 17 11 42 42 42 42 42 42 cosec , , , , π , , x q q     A1 Gives six other solutions. Allow different values of q as long as all seven solutions are found. 5

More questions on Complex numbers

8 (a) Find the value of a for which the system of equations 3x + ay = 0, 5x - y = 0, x + 3y + 2z = 0, does not have a unique solution. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The matrix A is given by 3 0 0 A = 5 - 1 0 f1 3 2p. (b) Find a matrix P and a diagonal matrix D such that A 2 = PDP -1 . 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(c) Use the characteristic equation of A to show that ( A + 6I) 2 = A 4 ( A + bI) 2 , where b is an integer to be determined. 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Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................ ............................................................................................................................................................................

Mark scheme: 8(a) 3 0 5 1 0 0 10 6 0 1 3 2 a a      3 5 a  A1 2 8(b) Eigenvalues of A are 3, 1  and 2. B1 Lower diagonal matrix or characteristic equation. 4 3: 5 4 0 5 1 3 1 19               i j k M1 A1 Uses vector product (or equations) to find corresponding eigenvectors. 0 0 1: 4 0 0 12 1 1 3 3 12 1                         i j k  A1 0 0 2: 1 0 0 0 0 1 3 0 3 1             i j k  A1 Thus 4 0 0 5 1 0 19 1 1             P and 9 0 0 0 1 0 0 0 4           D M1 A1 Or correctly matched permutations of columns. 7 Question Answer Marks Guidance 8(c)     3 2 3 1 2 6 0 4               B1 Characteristic equation. 3 2 6 4    A I A A M1 Substitutes for A and makes 6  A I the subject.       2 4 2 2 3 2 6 4 4      A I A A A A I M1 A1 Squares and factorises. 4

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