Cambridge A Level Mathematics - Further 9231 — 2023 Oct/Nov Paper 2 · Variant 1

1 Show that the system of equations 14x - 4y + 6z = 5 , x + y + kz = 3, - 21x + 6 y - 9z = 14, where k is a constant, does not have a unique solution and interpret this situation geometrically. [4] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 14 −4 6 M1 A1 Evaluates determinant. Can expand along any row e.g. 1 k 1 k 1 1 36 − 36 ) + ( −126 + 126 ) + k ( 84 − 84 ) . 1 1 k = 14 + 4 + 6 − ( 6 −9 −21 −9 −21 6 −21 6 −9 If using row operations, they must show an inconsistent = 14 ( −−9 6 k ) + 4 ( −+9 21k ) + 6 ( 6 + 21) = 0 system for M1. All their row operations must be correct for A1. Two parallel planes, not identical. B1 Other plane not parallel. B1 4

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2 Find the roots of the equation ( z + 5i) 3 = 4 + 4 3 i , giving your answers in the form r cos i + i ( r sin i - 5) , where r 2 0 and 0 1 i 1 2 r . [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 12 π B1 Finds modulus and argument of 4 + 4i 3. 3 3 ( z + 5i) = 4 + 4i 3 = 8e  1 1  1  1  M1 A1 Finds one root. z1 = 2  cos π + isin π − 5i = 2cos π + i 2sin π − 5   9 9  9  9  7  7  13  13  A1 FT Finds other two roots. FT on their modulus. z 2 = 2cos π + i  2 sin π − 5  , z3 = 2cos π + i  2 sin π − 5  9  9  9  9  A1 FT 5

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+ bx + cx , giving3 Find the first three terms in the Maclaurin’s series for tanh -1 12 ex in the form 12 lna 2 the exact values of the constants a, b and c. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 3 1 x B1 e d y 2 = d x 1 2 x 1 − e 4  1 2 x  1 x  1 x  1 2 x  B1 2  1 − e  e  − e  − e  d y  4  2  2  2  = d x 2  1 2 x  2  1 − e   4  2 10 M1 Evaluates derivatives at x = 0. f '(0) = f ''(0) = 3 9 −1 1  1  3  M1 Uses logarithmic form of tanh −1 . f (0) = tanh = ln  2      2  2  2  1 2 5 2 M1 A1 1 2 ln3 + x + x Applies f ( x ) = f (0) + f '(0) x + f ''(0) x 2 3 9 2! 6

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4 Find the particular solution of the differential equation d 2 y dy 2 + 2 + 3 y = 27 x , 2 d x dx dy given that, when x = 0 , y = 2 and = - 8 . 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Mark scheme: 4 m 2 + 2 m + 3 = 0 M1 Auxiliary equation. A1 Complementary function. Allow with “ y = ” missing. [ y =]e − x A cos 2 x + B sin 2 x ( ) y = px 2 + qx + r  y ' = 2 px + q  y '' = 2 p B1 Particular integral and its derivatives. 3 p = 27 3q + 4 p = 0 2 p + 2q + 3r = 0 M1 Substitutes and equates coefficients. 4 p = 9 q = −12 r = 2 A1 y = e − x A cos 2 x + B sin 2 x + 9 x 2 − 12 x + 2 A1 General solution. Must have “ y = ”. ( ) − x − x M1* Differentiates. Must use product rule. y ' = e − 2 A sin 2 x + 2 B cos 2 x − e A cos 2 x + B sin 2 x + 18 x − 12 ( ) ( ) A + 2 = 2 2 B − A − 12 = −8 DM1 Uses initial conditions A1  A = 0, B = 2 2 y = 2 2e − x sin 2 x + 9 x 2 − 12 x + 2 A1 Must have “ y = ”. 10

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5 The curve C has parametric equations 3 2 2 - t2 1 , y = t2 + 5 , for 0 1 t G 3 . x = 23 t (a) Find the exact length of C. 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Mark scheme: 5(a) 1 − 1 B1 Differentiates x and y with respect to t. x = t 2 − t 2 , y = 2 1 −  1 − 1  2  1  2 M1 A1 Factorises x 2 + y 2 . x 2 + y 2 =  t 2 − t 2  + 4 = t + 2 + t −1 =  t 2 + t 2          3 1 1 3 1 M1 A1 Applies correct formula for arc length. 3 − 2 2  2 2 2  3 t0 + t d t =  t + 2t  = 4 3 2 2 x + y d t .  3  0 M1 for their 0 Answer must be simplified to 4 3 for A1. 5 5(b) dy y 2 B1 Finds first derivative. = = dx x 1 − 1 t 2 − t 2 d y − 3  B1 Differentiates with respect to t.  1 − 12 1 2 −2  t + t   d x   2 2  d    2 =  − dt  1 1 2  1 − 1  2 2  t − t  2 2  t − t      3 −  M1 A1 Applies chain rule. OE. Does not have to be simplified  1 − 12 1 2 −2  t + t  for A1.    2 2  d 2 y d   t + 1  2  d t =  = = − 1 − d x 2 dt  1 1 3 3 −  dx  1  ( t − 1) 2 2  t − t  2 2  t − t      0 t 1 A1 Accept −1 t 1 . CWO. 5

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6 (a) Starting from the definitions of cosh and sinh in terms of exponentials, prove that sinh 2x = 2 sinh x cosh x . 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(b) Using the substitution u = sinh x , find sinh 2 2x cosh x dx . 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(c) Find the particular solution of the differential equation d y 2 + y tanh x = sinh 2x , d x given that y = 4 when x = 0 . Give your answer in the form y = f ( x) . 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Mark scheme: 6(a) 1 x − x 1 x − x B1 cosh x = e + e sinh x = e − e ( ) ( ) 2 2 1 x − x x − x 1 2 x −2 x M1 A1 Expands, AG. e − e e +e = e − e = sinh2 x ( )( ) ( ) 2 2 3 6(b) u = sinh x  du = cosh x dx B1 2 2 2 2 2 M1 Applies identities to find integral in terms of u.  sinh 2 x cosh x dx = 4  sinh x cosh x du = 4  sinh x ( sinh x + 1) du = 4  u 2 u 2 + 1 du A1 ( )  1 5 1 3   1 5 1 3  A1 = 4 u + u ( + C ) = 4 sinh x + sinh x ( + C )      5 3   5 3  4 tanh xdx lncosh x M1 A1 Finds integrating factor.6(c)  e = e = cosh x d 2 M1 Correct form on LHS and attempt to integrate RHS. ( y cosh x ) = sinh 2 x cosh x dx  1 5 1 3  M1 A1 Integrates RHS using their part (b). y cosh x = 4 sinh x + sinh x + C    5 3  4 = C M1 Substitutes initial conditions.  1 5 1 3  A1 y = 4sech x sinh x + sinh x + 1    5 3  7

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7 The matrix A is given by - 6 2 13 A = 0 - 2 5 f 0 0 8p. (a) Find a matrix P and a diagonal matrix D such that A -1 = PDP -1 . 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(b) Use the characteristic equation of A to find A -1 . 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Mark scheme: 7(a) = −6, = −2, = 8 B1 i j k 8 1 M1 A1 Uses vector product (or equations) to find corresponding   eigenvectors. = −6: 0 4 5 = 0 0     0 0 2 0 0 i j k  10  1 A1    = −2: −4 2 13 = 20 2       0 0 5  0  0 i j k  140  2 A1    = 8: −14 2 13 = 70 1       0 −10 5  140  2  1 1 2   − 16 0 0  M1 A1 Or correctly matched permutations of columns.    1  M1 for their (non-zero) eigenvectors matched to their and D = Thus P = 0 2 1    0 − 2 0  eigenvalues.    1   0 0 2   0 0 8  7 7(b) A3 − 52 A − 96I = 0 M1 Substitutes A into characteristic equation. 96 A −=1 A 2 − 52I M1 Multiples through by A −1.  36 −16 36  B1 2   A = 0 4 30      0 0 64  7(b)  − 16 − 16 83  A1 −1  1 5  A =  0 − 2 16   1   0 0 8  4

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8 (a) State the sum of the series 1 + z + z 2 + f + z n - 1 , for z ! 1. [1] ............................................................................................................................................................ (b) By letting z = cos i + i sin i , where cos i ! 1, show that 1 sin ni sin i 1 + cos i + cos 2i + f + cos ( n - 1) i = 1 - cos n i + [7] 2 e 1 - cos i o. ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ y 1 O 1 2 n - 1 1 x n n n The diagram shows the curve with equation y = cos x for 0 G x G 1, together with a set of n rectangles of width 1. n (c) By considering the sum of the areas of these rectangles, show that 1 1 1 sin 1 sin n cos xd x 1 1 - cos 1 + . [4] 2n y0 f 1 - cos n1 p ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (d) Use a similar method to find, in terms of n, a lower bound for cosxd x . 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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) nz − 1 B1 z − 1 1 8(b) nz − 1 cos n−+1 isi n n B1 = z − 1 cos− 1 + i sin ( cos n−+1 isin n)( cos−−1 isin) M1 Multiplies numerator and denominator by complex ( cos−+1 isin)( cos−−1 isin) conjugate.  nz − 1  cos ncos+ sin nsin− cos n− cos+ 1 M1 Takes real part. Re   = 2 2 cos( n − 1)= cos ncos+ sin nsin  z − 1  ( cos− 1) + sin  cos ncos+ sin nsin− cos n 1 A1 = + 2 (1 − cos) 2 cos n( cos− 1) + sin nsin 1 M1 Factorises. = + 2 (1 − cos) 2 8(b) 1  sin nsin M1 A1 Divides through by denominator. AG. = 1 − cos n+   2  1 − cos  Alternative method for question 8(b) z n − 1 e i n − 1 B1 = z − 1 e i − 1 1 1 1 1 1 ( n − 2 ) e i − e − i 1 2 cos ( n − 2 )+ isin ( n − 2 )− cos 2 + isin ( 2 ) M1 = 2 e i 1 2 − e − i 1 2isin 12   z n − 1  sin( n − 12 )+ sin 12  M1 Takes real part Re   = 1  z − 1  2sin 2  1 sin ( n − 2 ) 1 A1 = + 2sin 12  2 sin ncos 12 − cos nsin 12  1 M1 Uses compound angle identity = + 2 sin 12  2 sin ncos 12  1 1 M1 Divides through by denominator. = − cos n+ 2sin 12  2 2 sin nsin 1 1 s i n nsin 1 1 A1 AG. si n = 2 sin 12 cos 12  and − cos n+ = − cos n+ 1 − cos) 2 2 4s in 2 12  2 2 2 ( 2 sin 2 12 = 1 − cos. 7 8(c) 1 1 1 1 1 2 1 n − 1 M1 A1 Forms sum of areas of rectangles given in the diagram. A0 if comparison with integral missing or unclear. 0 cos x dx  n + n cos n + n cos n +  n cos n  n 1  M1 A1 sin sin Applies result from part (b) with = 1. AG. 1  1 2 n − 1  1  n n n  n =  1 + cos + cos + + cos  =  1 − cos +  n  n n n  2 n  n 1 − cos 1   n  4 8(d) 1 1 1 1 2 1 n M1 A1 Forms sum of areas of rectangles. A0 if comparison with integral missing or unclear. 0 cos x dx  n cos n + n cos n + n cos n  1   1  A1 1  sin1sin n  1 1 1  sin1sin n  =  1 − cos1 +  + cos1 − =  cos1 −+1  2 n 1 n n 2 n 1  1 − cos   1 − cos   n   n  3

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