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

1 Find the Maclaurin’s series for ln ( x + 2) + ln ( x 2 + 5) up to and including the term in x2. [5] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: Question Answer Marks Guidance 1 EITHER Solution 1 M1 2 x + 5 Changes ln ( x + 2 ) or ln ( ) 1 1 2 (1 + = ln2 + ln (1 + ln ( x + 2 ) = ln ( 2 x ) ) 2 x ) so that the formula given in the list of formulae (MF19) can be applied. 1 ln ( x + 2 ) = ln2 + 2 x − 81 x 2 + A1 2 1 2 1 2 A1 ln x + 5 = ln 5 = ln5 + 5 x + ( ) ( ( 1 + 5 x ) ) 1 1 2 1 2 M1 Sums power series + ( ln2 + 2 x − 8 x + ) ( ln5 + 5 x + ) OR Solution 2 (M1 A1) Finds first derivative. −1 2 −1 x + 5 f '( x ) = ( x + 2 ) + 2 x ( ) −2 2 2 −2 2 −1 (A1) Finds second derivative. x + 5 + 2 x + 5 f ''( x ) = − ( x + 2 ) − ( 2 x ) ( ) ( ) f (0) = ln10 f '(0) = 12 f ''(0) = 203 (M1) Evaluates derivatives at zero. 2 1 3 2 A1 Accept ln10 written as ln2 + ln5 x + 5 = ln10 + 2 x + 40 x ln ( x + 2 ) + ln ( ) but do not accept decimals. WWW. 5

More questions on Differentiation

2 It is given that 1 x = 1 + and y = tet . t dy 3 2 (a) Show that = - et ( t + t ) . [3] dx ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ d 2 y (b) Find 2 in terms of t. [4] dx ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 2(a) dy t dx −2 B1 = e ( t + 1) , = −t dt dt dy dy dt 2 t t 3 2 M1 A1 Uses chain rule, AG. t + t =  = −t e ( t + 1) = −e ( ) dx dt dx 3 M1 A1 Applies product rule.2(b) d  dy = − e t 3t 2 + 2t − e t t 3 + t 2 = − e t t 3 + 4t 2 + 2t   ( ) ( ) ( ) dt  dx  d 2 y 3 t 2 M1 A1 Uses chain rule, multiplies by their = t e t + 4t + 2 2 ( ) d t dx for M1. d x Alternative method for question 2(b) d 2 y t  2 d t d t  t 3 2 d t (M1 A1) Differentiates −et t 3 + t 2 ( ) t + t 2 = − e  3t + 2t  − e ( ) dx  dx dx  dx implicitly. d 2 y t 2 3 dt 3 t 2 (M1 A1) dt 2 = e −4t − 2t − t = t e t + 4t + 2 Substitutes = −t . 2 ( ) ( ) dx dx dx 4

More questions on Differentiation

3 (a) Use de Moivre’s theorem to show that cos 5i = 16 cos 5 i - 20 cos 3 i + 5 cos i . [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) Hence obtain the roots of the equation 32x 5 - 40x 3 + 10x - 2 = 0 in the form cos ( qr ) , where q is a rational number. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 3(a) 5 5 2 3 4 M1 A1 Expands and takes real part. Accept cos 5= Re ( cos+ isin) = cos − 10sin cos + 5 sin cos 1 RHS to LHS using 2 cos= z + . z = cos 5 − 10cos 3  1 − cos 2  + 5cos 1 − 2cos 2 + cos 4  M1 Applies sin 2 = 1 − cos 2 . ( ) ( ) = 16 cos 5 − 20cos 3 + 5cos A1 AG 4 3(b) x = cos, cos5= 12 2 M1 Applies identify given in (b). 5 =  14 + 2k M1 Solves cos5= 12 2 1 π ) A1 Gives one correct solution. Accept cos ( 20 q = 20.1 cos( 209 π), cos( 1720 π),cos( 2250 π),cos( 3230 π) A1 Gives other solutions. OE. A0 for repeated roots. 4

More questions on Complex numbers

4 Find the solution of the differential equation dy + 3 y = sin x dx for which y = 1 when x = 0 . Give your answer in the form y = f ( x) . [9] .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... .................................................................................................................................................................... ....................................................................................................................................................................

Mark scheme: 4  3 dx 3 x M1 A1 Finds integrating factor. e = e d 3 x 3 x M1 Correct form on LHS and attempt ye = e sin x ( ) to integrate RHS. dx EITHER M1 A1 Integrates by parts once or uses  e3 x sin x dx = −e 3 x cos x + 3 e 3 x cos x dx sin x = e i x − e -i x . 2i OR e 3 x cos x dx  e3 x sin x dx = 13 e 3 x sin x − 13  EITHER M1 Integrates by parts again or  e 3 x sin x dx = − e 3 x cos x + 3 ( e 3 x sin x − 3  e 3 x sin x dx ) substitutes e ix − e -ix = sin x and 2i OR ix -ix e + e 3 x 1 3 x 1 1 3 x 1 3 x = cos x. e sin x dx e sin x dx = 3 e sin x − 3 ( 3  e cos x + 3  ) 2 ye3 x = 110 e3 x ( 3sin x − cos x ) + C A1 Must not see i. 1 = − 101 + C M1 Finds C. Substitutes into their expression (must be integrated). y = 103 sin x − 101 cos x + 1011 e−3 x A1 Divides through by coefficient of y. 9

More questions on Differential equations

5 y M O x The diagram shows part of the curve y = x sech 2 x and its maximum point M. (a) Show that, at M, 2x tanh x - 1 = 0 and verify that this equation has a root between 0.7 and 0.8. [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) By considering a suitable set of rectangles, use the diagram to show that n r sech 2 r 1 n tanh n + lnsech n - tanh 1 - lnsech 1 . [6] =/r 2 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 5(a) dy 2 2 M1 A1 Differentiating using product rule = −2 x sech x tanh x + sech x and chain rule for M1. dx  = 1 − tanh 2 x − 2 x tanh x + 2 x tanh 3 x    sech 2 x  0  2 x tanh x −=1 0 A1 AG. 2(0.7)tanh(0.7) −=1 −0.154  0 B1 Shows sign change. Must write 2(0.8)tanh(0.8) −=1 0.062  0 down values correct to at least 1dp for B1. 4 5(b) n n M1 A1 Compares sum with integral. x sech 2 x dx Consistent limits for M1.  r sech 2 r  1 r = 2 n 2 n n M1 A1 Integrates by parts. tanh x dx x sech x dx =  x tanh x 1 1 − 1 =  x tanh x + lnsech x 1n A1 = n tanh n + lnsech n − ( tanh1 + lnsech1) A1 AG. Must have gained all previous marks. 6

More questions on Integration

6 The matrix P is given by 1 - 1 1 P = 0 2 1 f 0 0 - 1 p. (a) State the eigenvalues of P. [1] ............................................................................................................................................................ ............................................................................................................................................................ (b) Use the characteristic equation of P to find P -1 . [4] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ The 3 # 3 matrix A has distinct non-zero eigenvalues a, 1,2 2 with corresponding eigenvectors 1 - 1 1 0 , 2 , 1 , f 0 p f 0 p f - 1 p respectively. (c) Find A -1 in terms of a. [5] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 6(a) 1, 2, − 1 B1 1 6(b) P 3 − 2 P 2 − P + 2I = 0 B1 States that P satisfies its characteristic equation. 2P −=1 I + 2P − P 2 M1 Multiplies through by P −1 .  1 −3 −1   1 12 23  M1 A1 2   −1  1 1  P = 0 4 1  P =  0 2 2        0 0 1 0 0 −1     4 6(c)  1a 0 0  B1   D = 0 2 0    1   0 0 2   1a 0 0  M1 Applies A −1 = PDP −1. − 1   −1 A = P 0 2 0 P    1   0 0 2   1a −2 12   1 12 23  M1 A1 Multiplies two adjacent matrices.  1   1 1  =  0 4 2   0 2 2   1    0 0 − 2  0 0 −1   1a 1−22aa 3 −2 3a a  A1  3   0 2 2   1   0 0 2  5

More questions on Matrices

7 (a) Starting from the definitions of cosh and sinh in terms of exponentials, prove that 2 sinh 2 A = cosh 2 A - 1. [3] ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ (b) A curve has equation y = x2 , for 0 G x G 23 . The area of the surface generated when the curve is rotated through 2r radians about the x-axis is denoted by S. = r - ln 3 [9] Use the substitution x = 12 sinh u to show that S 1 820 32 81 b l. ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................

Mark scheme: 7(a) 1 2 A −2 A 1 A − A B1 e + e e − e cosh2 A = 2 ( ) sinh A = 2 ( ) 2 2 1 A − A 1 2 A −2 A M1 A1 Expands, AG. A0 for mixing e − e e − 2 + e = cosh2 A − 1 2sinh A = 2 ( ) = 2 ( ) e − e . variables e.g. sinh A = 12 ( x − x ) 3 7(b) 23 2 2 M1 A1 Correct formula with correct limits.   dy  3 2 2 S = 2y 1 + dx = 2 x 1 + 4 x dx Correct limits for M1. (Limits may 0  dx  be recovered.) 0   sinh −1 43 1 2 2 M1 Applies given substitution to their 4 sinh u 1 + sinh u cosh u du expression with correct limits. S = π 0 (Limits may be recovered.) sinh −1 43 2 2 A1 Must be simplified. 1 = 4 π 0 sinh u cosh u du sinh −1 43 2 M1 Applies sinh2u = 2sinh u cosh u . 1 sinh 2u du May use double angle formulae for = 16 π 0 cosh and sinh instead. sinh −1 43 M1 Applies sinh 2 A = 12 ( cosh2 A − 1) cosh4u − 1du = 312 π 0 . S must have the form a sinh 2 u cosh 2 u du.  sinh −1 43 A1 1 1 = 32 π  4 sinh 4u − u  0 sinh −1 43 = ln3 B1 1 1 ln81 − ln81 1 1 1 1 820 A1 AG. e − e − ln3 = 32 π ( 81 − ln3 ) = 32 π ( 8 ( 81 − 81 ) − ln3 ) ) S = 32 π ( 8 ( ) 9

More questions on Integration

8 It is given that v = y4 and 3 d 2 y 2 dy 2 3 dy 4 -2 x y 2 + 3 y + y + y = e . dx e dx o dx (a) Show that d 2 v dv -2 x + + 4v = 4 e . [4] 2 dx dx ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ dy 3(b) Find y in terms of x, given that, when x = 0 , y = 1 and = - . [10] dx 8 ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ ............................................................................................................................................................ 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) dv 3 dy B1 v = y 4 = 4 y dx dx 2 2 2 B1 d v 3 d y 2  dy  = 4 y + 12 y 2 2   dx dx  dx  2 2 2 M1 Uses substitution to find v-x d v dv 3 d y 2  dy  3 dy 4 + + 4v = 4 y + 12 y + 4 y + 4 y equation, AG.   2 2 dx dx dx  dx  dx  3 d 2 y 2  dy  2 3 dy 4  −2 x A1 AG. = 4  y 2 + 3 y   + y + y  = 4e  dx  dx  dx    Alternative method for question 8(a) dy 1 dv 1 dv (B1) y = v 1 = 4 = 3 3 4 dx dx 4 y dx 4v 2 2 2 (B1) d y 1 d v 3  dv  2 = 3 2 − 7   4  dx  dx 4v 4 dx 16v (M1) Uses substitution to find v-x 3 1  1 d 2 v 3  dv  2  3  1 dv  −2 x 4 2  1 dv  2 4 v  3 2 − 7    + 3v  3  + v  3  + v = e equation, AG. 4    dx  4 dx   4v 4 dx   4v 4 dx 16v   4v 2 2 2 (A1) 1 d v 3  dv  3  dv  1 dv −2 x − + + + v = e 2     4 dx 16v  dx  16v  dx  4 dx 4 8(b) m 2 + m + 4 = 0 M1 Auxiliary equation.  v =  e − 12 x ( 15 15 A1 Complimentary function. Allow A cos 2 x + B sin 2 x ) ‘v =’ missing. v = ke −2 x  v ' = −2ke −2 x  v '' = 4ke −2 x B1 Particular integral and its derivatives. 4 ke −2 x − 2 ke −2 x + 4 ke −2 x = 4e −2 x M1 Substitutes and equates coefficients. k = 23 A1 4 − 12 x 15 15 2 −2 x A1 Substitutes for y and find v in terms v = y = e + 3 e ( A cos 2 x + B sin 2 x ) of x. Must not see i. 1 B1 3 dy − 2 x 15 15 15 15 1 − 12 x 15 15 4 −2 x v ' = 4 y = e − 3 e ( − 2 A sin 2 x + 2 B cos 2 x ) − 2 e ( A cos 2 x + B sin 2 x ) dx 2 3 15 1 4 1 M1 A1 Substitutes initial conditions and 1 = A + 3 −=2 2 B − 2 A − 3  A = 3 , B = 0 forms simultaneous equations. Must have used the product rule when differentiating their v for M1. 1 A1 Accept 1 − 12 x 15 2 −2 x 4 1 y = 1 ( 3 e cos 2 x + 3 e ) 1 − 2 x 15 2 −2 x y =  4 ( 3 e cos 2 x + 3 e ) 10

More questions on Differential equations