Cambridge A Level Mathematics - Further 9231 — 2022 Oct/Nov Paper 2 · Variant 1

1 Find the Maclaurin’s series for ln ( 1 + ex) up to and including the term in x2. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 e x B1 Finds first derivative. y ' = 1 + e x e x B1 Finds second derivative. y '' = 2 1 + e x ( ) y (0) = ln 2, y '(0) = 12 , y ''(0) = 14 B1 Evaluates at x = 0. y = y (0) + y '(0) x + 21! y ''(0) x 2 + M1 Allow 2! missing. ln2 + 12 x + 18 x 2 A1 Decimal used for ln2 scores A0. 5

More questions on Differentiation

2 (a) Show that the system of equations x - y + 2z = 4, x - y - 3z = a, x - y + 7z = 13, where a is a constant, does not have a unique solution. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Given that a = - 5 , show that the system of equations in part (a) is consistent. Interpret this situation geometrically. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Given instead that a ! - 5 , show that the system of equations in part (a) is inconsistent. Interpret this situation geometrically. [2] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) 1 −1 2 M1 A1 Shows that determinant is zero. 1 −1 −3 = −−+7 3 10 − 2(0) = 0 1 −1 7 2 2(b) x − y + 2 z = 4, M1 A1 M1 for row operations or eliminating one x − y − 3 z = −5,  z = 95 , x − y = 52 variable. A1 for deriving one correct equation with two x − y + 7 z = 13, unknowns. The three planes form a sheaf. B1 All three planes have the same common line. 3 2(c) x − y + 2 z = 4, B1 Derives contradiction. x − y − 3 z = a ,  z = 95 , 3 z + a = 4 − 2 z  a = −5 x − y + 7 z = 13, The three planes form a triangular prism. B1 2

More questions on Matrices

3 The curve C has parametric equations = 4e ( t - 2 ) , for 0 G t G 2 . x = et - 13 t 3 , y 21 t Find, in terms of e, the length of C. [6] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 3 x = et − t 2 B1 1 2 t B1 y = 2te 2 t 2 2 t 2 t 2 t 4 t 2 M1 A1 Finds x 2 + y 2. e − t + 4t e = e + 2t e + t = e + t ( ) 2 2 t 2 t 1 3 e + t dt =   e + 3 t  0 = e + 3 0 2 5 M1 A1 6

More questions on Integration

4 (a) Starting from the definitions of cosh and sinh in terms of exponentials, prove that cosh 2 x - sinh 2 x = 1. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ d -1 (b) Show that tan ( sinh x) = sec h x . [3] dx ` j ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (c) Sketch the graph of y = sec h x , stating the equation of the asymptote. [2] ............................................................................................................................................................ (d) By considering a suitable set of n rectangles of unit width, use your sketch to show that n sec h r 1 tan -1 (sinh n) . [3] =/r 1 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 3 (e) Hence state an upper bound, in terms of r, for sec h r . [1] =/r 1 ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 4(a) 1 x − x 1 x − x B1 e + e e − e cosh x = 2 ( ) sinh x = 2 ( ) 1 2 x −2 x 2 x −2 x M1 A1 Expands, AG. Clear LHS to RHS for A1. e + 2 + e − e + 2 − e = 1 4 ( ) 3 4(b) d −1 cosh x M1 A1 d −1 1 tan sinh x = Applies tan u = . ( ) 2 ( ) 2 dx sinh x + 1 du u + 1 cosh x A1 AG. = = sech x cosh 2 x 3 4(c) y B1 Correct shape, symmetrical about x = 0. y = 0 B1 Accept labels on their sketch. 2 4(d) n n M1 A1 Compares sum with integral. Limits correct for sech x dx A1.  sech r  0 r =1 −1 n −1 A1 AG. =  tan sinh x  = tan sinh n   0 3 4(e) 1 B1 π 2 1

More questions on Summation of series

5 Find the particular solution of the differential equation d 2 y dy 2 2 2 + 2 + y = 4x + 3x + 3 , d x dx dy given that, when x = 0 , y = = 0 . [10] dx .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 5 2m2 + 2m + 1 = 0  m = − 12  12 i M1 Correct auxiliary equation. − 12 x 1 1 A1 Complementary function. Accept “ y = ” y = e ( A cos 2 x + B sin 2 x ) missing. y = px 2 + qx + r leading to y ' = 2 px + q leading to y '' = 2 p B1 Particular integral and its derivatives. p = 4 4 p + q = 3 4 p + 2q + r = 3 M1 Substitutes and equates coefficients. q = −13 r = 13 A1 − 12 x 1 1 2 A1 General solution. y = e ( A cos 2 x + B sin 2 x ) + 4 x − 13 x + 13 − 12 x 1 1 1 1 1 − 12 x 1 1 M1 Full use of product rule. y must be of form y ' = e ( − 2 A sin 2 x + 2 B cos 2 x ) − 2 e ( A cos 2 x + B sin 2 x ) + 8 x − 13 shown in the general solution above with p  0. A+ 13 = 0 12 B − 12 A − 13 = 0 A = −13, B = 13 M1 A1 Uses initial conditions to derive two linear equations in two unknowns. − 12 x 1 1 2 A1 y = e ( −13cos 2 x + 13sin 2 x ) + 4 x − 13 x + 13 10

More questions on Differential equations

6 The matrix A is given by 2 - 3 - 7 A = 0 5 7 f0 0 - 2p. (a) Find a matrix P and a diagonal matrix D such that A 5 = PDP -1 . [7] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Use the characteristic equation of A to show that A 4 = aA 2 + bI , where a and b are integers to be determined. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a) Eigenvalues of A are 2, 5 and −2. B1 Lower diagonal matrix or characteristic equation. i j k  −12  1 M1 A1 Uses vector product (or equations) to find    corresponding eigenvectors. = 2: 0 3 7 = 0 ~ 0       0 0 −4  0  0 i j k  −21   1  A1     = 5: −3 −3 −7 = 21 −1         0 0 7  0   0  i j k  28   1  A1     = −2: 4 −3 −7 = −28 −1         0 7 7  28   1   1 1 1   32 0 0  M1 A1 Or correctly matched permutations of columns.     Their eigenvectors must be non-zero and Thus P = 0 −1 −1 and D = 0 3125 0     correctly matched to their eigenvalues raised to      0 0 1   0 0 −32  the fifth power for M1. 7 6(b) (− 2 )(− 5)(+ 2 ) = 3 − 52 − 4+ 20 = 0 B1 Characteristic equation. A 3 = 5 A 2 + 4 A − 20I M1 Substitutes for A and makes A 3 the subject. Tolerate I missing. A 4 = 5A3 + 4 A 2 − 20 A = 5 5A 2 + 4 A − 20I + 4 A 2 − 20 A M1 Multiplies by A and substitutes for A 3 . ( ) A 4 = 29 A 2 − 100I A1 Has I in answer or states the values of a and b. 4

More questions on Matrices

7 (a) State the sum of the series 1 + w + w 2 + w 3 + f + w n - 1 , for w ! 1. [1] ............................................................................................................................................................ ............................................................................................................................................................ r . [2] (b) Show that ( 1 + i tan i) k = sec k i ( cos ki + i sin ki ) , where i is not an integer multiple of 12 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ n 1 (c) By considering ( 1 + i tan i) k , show that -/ k = 0 n 1 sec k i sin ki = cot i ( 1 - sec n i cos ni) , -/ k = 0 r . [5] provided i is not an integer multiple of 12 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 6m - 1 r in terms of m. [2](d) Hence find 2k sin 13 k / b l k = 0 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 7(a) w n − 1 B1 w − 1 1 7(b) (1 + i tan) k = seck ( cos+ i sin) k = seck ( cos k+ i sin k) M1 A1 Applies de Moivre’s theorem, AG. 2 7(c) n −1 n M1 A1 Applies part (a). k (1 + itan) − 1 1 + itan) =  ( k = 0 itan − cotsecn ( icos n− sin n) +icot A1 n −1 M1 A1 Takes imaginary part, AG. sec k sin k= cot 1 − sec n cos n  ( ) k = 0 5 6 m −17(d) M1 A1 Sets = 13 π. 1 k 1 1 6 m 1 6 m 1 − 2 2 sin ( 3 kπ ) = 3 3 ( 1 − 2 cos ( 2mπ ) ) = 3 3 ( ) = 3 π  k = 0 2

More questions on Summation of series

8 (a) Use the substitution u = 1 - ( i - 1) 2 to find i - 1 di . [3] y 2 1 - ( i - 1) ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Find the solution of the differential equation d y 2 -1 i - y = i sin ( i - 1 ) , di where 0 1 i 1 2 , given that y = 1 when i = 1. Give your answer in the form y = f ( i) . [11] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 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) du 1 B1 −2(− 1) leading to (− 1) d = − 2 du d=  − 1 1 1 2 M1 A1 Applies substitution. For M1, integrand must be du = − u + C = − 1 − (− 1) + C d= − 2   of the form k , where k  0 is constant. For u  u  1 − (− 1) 2 A1, allow “ +C ” missing. 3 8(b) dy y −1 B1 Divides through by . − = sin (− 1) d  − −1 d − ln −1 M1 A1 Finds integrating factor. e = e =  d −1 −1 M1 Correct form on LHS using their integrating  y = sin (− 1) ( ) factor. d −1 −1   M1 A1 Integrates s in −1 (− 1) .  y = sin (− 1) − d 2  1 − (− 1)    − 1  1 M1 Applies part (a) and uses  2 d=  2 d+  2 d 1 dx = sin −1 x + C . 2  1 − (− 1)  1 − (− 1)  1 − (− 1)  1 − x −1 −1 2 −1 A1  y = sin (− 1) + 1 − (− 1) − sin (− 1) + C 8(b) 1 = 1 + C M1 Substitutes initial conditions into their expression in y and . −1 2 M1 A1 Divides through by their integrating factor. y = ( − 1) sin (− 1) +  1 − (− 1) 11

More questions on Differential equations