Differentiation — Paper 2 · A Level Mathematics - Further 9231

2 It is given that y = 2x . d y x (a) By differentiating lny with respect to x, show that = 2 ln 2 . [3] d x … … … … … … … … … … d2 y (b) Write down 2 . [1] d x … … … (c) Hence find the first three terms in the Maclaurin’s series for 2x. [3] … … … … … … … … … … …

7 marks

Mark scheme: 2(a) 1 d ln ln 2 ln 2 d =  = y y x y x M1 A1 d ln2 2 ln2 d = = x y y x AG A1 3 2(b) ( ) 2 2 2 d 2 ln2 d = x y x B1 1 Question Answer Marks 2(c) ( ) 2 (0) 1, '(0) ln2, ''(0) ln 2 = = = y y y B1 ( ) ( ) 2 2 2 ln 2 ''(0) 2 (0) '(0) 1 ln 2 2! 2 = + + + = + + +   x y y y x x x x M1 A1 3

This question in 9231/21 May/June 2020

2 It is given that y = 2x . d y x (a) By differentiating lny with respect to x, show that = 2 ln 2 . [3] d x … … … … … … … … … … d2 y (b) Write down 2 . [1] d x … … … (c) Hence find the first three terms in the Maclaurin’s series for 2x. [3] … … … … … … … … … … …

7 marks

Mark scheme: 2(a) 1 d ln ln 2 ln 2 d =  = y y x y x M1 A1 d ln2 2 ln2 d = = x y y x AG A1 3 2(b) ( ) 2 2 2 d 2 ln2 d = x y x B1 1 Question Answer Marks 2(c) ( ) 2 (0) 1, '(0) ln2, ''(0) ln 2 = = = y y y B1 ( ) ( ) 2 2 2 ln 2 ''(0) 2 (0) '(0) 1 ln 2 2! 2 = + + + = + + +   x y y y x x x x M1 A1 3

This question in 9231/22 May/June 2020

5 The curve C has parametric equations x = 12 t 2 1 - ln t , y = t2 + 1, for 2 G t G 2 . (a) Find the exact length of C. [5] … … … … … … … … … … … … … … … … … … … … … … … … … d 2 y (b) Find 2 in terms of t, simplifying your answer. [4] d x … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 5(a) 1, 2 x t t y − = − =   B1 ( ) 2 2 2 2 2 1 2 4 x y t t t t − − + = −+ + = +   M1 A1 1 1 2 2 2 2 1 2 15 1 2 8 d ln 2ln 2 t t t t t −   + = + = +    M1 A1 5 Question Answer Marks 5(b) 1 d 2 d y y x x t t − = = −   B1 ( ) ( ) 2 2 1 1 2 1 d 2 d t t t t t t − − − − +  =   −   − B1 ( ) ( ) 2 2 3 2 1 1 2 1 d d 2 d d d d t y t x t t t x t t − − − − +   = × =   −   − M1 A1 4

This question in 9231/23 May/June 2020

6 (a) Starting from the definitions of tanh and sech in terms of exponentials, prove that 1 - tanh 2 i = sech 2 i . [3] … … … … … … … … r . r for - 14 r 1 x 1 34 The variables x and y are such that tanh y = cos x + 14 b l, r with respect to x, show that (b) By differentiating the equation tanh y = cos x + 14 b l d y 1 r [4] =- cosec x + 4 d x b l. … … … … … … … … … … … … … … r) in the form(c) Hence find the first three terms in the Maclaurin’s series for tanh -1 cos ( x + 14 b l 1 2 2 lna + bx + cx , giving the exact values of the constants a, b and c. [5] … … … … … … … … … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 6(a) e e 2 tanh sech e e e e θ θ θ θ θ θ θ θ − − − − = = + + B1 ( ) ( ) ( ) ( ) 2 2 2 2 2 e e e e e e 4 1 e e e e e e θ θ θ θ θ θ θ θ θ θ θ θ − − − − − − + − −   − − = =   +   + + M1 2 sech θ = A1 3 Question Answer Marks 6(b) 2 1 1 4 4 d tanh cos( π) sech sin( π) d y y x y x x = +  = − + M1 A1 ( ) 2 1 1 4 4 d 1 cos ( π) sin( π) d y x x x − + = − + M1 1 4 1 4 2 1 4 sin( π) d cosec( π) d sin ( π) x y x x x + = − = − + + A1 4 6(c) 2 1 1 4 4 2 d cot( π)cosec( π) d y x x x = + + B1 1 4 '(0) cosec( π) ''(0) y y =− = 1 1 4 4 cot( π)cosec( π) M1 ( ) 1 1 1 2 2 2 2 (0) tanh 2 ln 2 2 y −   + √ = √ =   −√   M1 2 1 1 2 2 ln(3 2 2) 2 2 y x x = + √ −√+ √ M1 A1 5

This question in 9231/23 May/June 2020

1 (a) By differentiating e - x2 , find the Maclaurin’s series for e - x2 up to and including the term in x2. [5] … … … … … … … … … … … … … … … … … … 1 5 e - x 2 d x , giving your answer as a rational fraction in its lowest (b) Deduce an approximation to y terms. 0 [2] … … … … … … …

7 marks

Mark scheme: 1(a) 2 f '( ) 2 e− = − x x x B1 Finds first derivative. 2 2 2 f ''( ) 4 e 2e − − = − x x x x B1 Finds second derivative. f(0) 1 f '(0) 0 f ''(0) 2 = = = − M1 Evaluates derivatives at zero. 2 2 e 1 − = − x x M1 A1 5 1(b) 3 1 1 5 2 5 0 0 1 74 1 d 3 375   − = − =      x x x x M1 A1 Substitutes 2 1−x or better. 2

This question in 9231/21 Oct/Nov 2020

5 It is given that x = sinh -1 t , y = cos -1 t , where - 1 1 t 1 1. dy 1 (a) By differentiating cosy with respect to t, show that =- . [4] dt 1 - t 2 … … … … … … … … … … … … … … … … … … … … … … … … d 2 y (b) Find 2 in terms of t, simplifying your answer. [5] dx … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 5(a) d sin 1 dt − = y y M1 A1 Differentiates both sides with respect to t. 2 d 0 1 cos 1 n dt si 0 π < < − −  >  = y y y y M1 Applies 2 2 sin cos 1. + = y y 2 d 1 dt 1 = − − y t A1 AG, justifies taking positive square root. 4 5(b) 2 d 1 dt 1 = + x t B1 2 2 d 1 d 1 + = − − y t x t B1 Finds first derivative. ( ) ( ) ( ) 2 1 1 1 1 2 2 2 2 2 2 2 2 2 2 1 1 (1 ) 1 d 1 dt 1 1 − −   − + + + − +   − = −   − −   t t t t t t t t t M1 Differentiates d d y x with respect to t. 2 2 2 2 d d 1 d dt d d 1   +   = − ×   −   y t t x x t M1 Applies chain rule. ( ) ( ) ( ) 1 1 2 2 3 2 2 2 2 2 2 1 (1 ) 1 2 1 1 −     − + + −         = − = −   −     −   t t t t t t t A1 OE (simplified). 5

This question in 9231/21 Oct/Nov 2020

8 (a) Sketch the graph of y = coth x for x 2 0 and state the equations of the asymptotes. [2] (b) Starting from the definitions of coth and cosech in terms of exponentials, prove that coth 2 x - cosech 2 x = 1. [3] … … … … … … … … … … … … … … … … … The curve C has equation y = ln coth 12 x for x 2 0 . a k dy (c) Show that =- cosechx . [3] dx … … … … … … (d) It is given that the arc length of C from x = a to x = 2a is ln4, where a is a positive constant. Show that cosha = 2 and find, in logarithmic form, the exact value of a. [7] … … … … … … … … … … … … … … … … … … Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

15 marks

Mark scheme: 8(a) B1 Correct shape and position, not too truncated. 0, = x 1 = y B1 States equations of asymptotes. 2 8(b) e e 2 coth cosech e e e e − − − + = = − − x x x x x x x x B1 ( ) ( ) 2 2 2 2 2 e e 4 e e 2 1 e e e e e e − − − − −   + + − − = =   −   − − x x x x x x x x x x M1 A1 Writes over common denominator, AG. 3 x Question Answer Marks Guidance 8(c) 2 1 sech d 1 2 1 1 1 d 2tanh 2sinh cosh 2 2 2       = − = −                   x y x x x x Or 2 1 cosech d 1 2 1 1 1 d 2coth 2sinh cosh 2 2 2       = − = −                   x y x x x x M1 A1 Uses chain rule. = 1 cosech sinh( ) − = − x x A1 AG 3 8(d) 2 2 1 cosech d + a a x x M1 Forms correct integral. 2 2 2 coth d coth d =   a a a a x x x x M1 A1 Uses 2 2 coth cosech 1 − = x x . [ ] 2 lnsinh lnsinh2 lnsinh = = − a a x a a M1 Integrates and substitutes limits. ( ) sinh 2 ln ln 2cosh sinh = = a a a M1 Combines logarithms and uses double angle formula. ( ) ln 2cosh ln 4 cosh 2 =  = a a A1 AG ( ) ( ) 2 ln 2 2 1 ln 2 3 = + − = + √ a A1 Must reject ( ) ln 2 . 3 − 7

This question in 9231/21 Oct/Nov 2020

r up to and including the term in x2. [5]1 Find the Maclaurin’s series for tan x + 14 … … … … … … … … … … … …

5 marks

Mark scheme: 1 2 1 '( ) s π ec 4 f x x   = +     B1 Finds first derivative. 2 1 1 ''( ) 2sec t n 4 π 4 π a f x x x     = + +         B1 Finds second derivative. (0) 1 '(0) 2 ''(0) 4 = = = f f f M1 Evaluates derivatives at zero. 2 1 tan 1 2 2 4 π x x x   + = + +     M1 A1 5

This question in 9231/22 Oct/Nov 2020

5 The curve C has equation y 2 + ( xy + 1) 2 = 5 . dy 2 (a) Show that, at the point ( 1, 1) on C, =- 3 . [3] dx … … … … … … … … d 2 y (b) Find the value of 2 at the point ( 1, 1) . [5] dx … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(a) ( ) 2 d 2 ' d = y yy x B1 Differentiates 2 y correctly. ( ) ( ) ( ) 2 d 1 2( 1) ' d + = + + xy xy xy y x B1 Differentiates ( ) 2 1 + xy correctly. ( ) 2 (1) ' ((1) 1) ' 1 0 ' 3 + + + =  = − y y y B1 Substitutes (1,1), AG. 2 2 ' + = − + + xy y y y x y x 3 5(b) ( ) 2 '' ' + yy y B1 Differentiates '. yy ( ) ( )( ) ( 1) '' ' ' ' ' 0 + + + + + + + = xy xy y y xy y xy y B1 B1 Differentiates ( ) ( 1) ' . + + xy xy y 2 2 4 2 1 '' 2 '' 1 0 3 3 3 3        + − + − + − + =               y y M1 Substitutes (1,1) and 2 ' 3 = − y 19 '' 27 = y A1 5 Question Answer Marks Guidance 5(b) Alternative ( ) 2 2 2 2 2 2 ( )(2 ' ') ( )( ' 2 ' 1) '' + + + + − + + + + = − + + y x y x xyy y y xy y y xy x y y y x y x M1 A1 A1 Differentiate 2 2 ' + = − + + xy y y y x y x using quotient rule. A1 for differentiating 2 xy correctly. A1 for everything correct. 19 '' 27 = y M1 A1 Substitutes (1,1) and 2 ' . 3 = − y 5

This question in 9231/22 Oct/Nov 2020

1 (a) By differentiating e - x2 , find the Maclaurin’s series for e - x2 up to and including the term in x2. [5] … … … … … … … … … … … … … … … … … … 1 5 e - x 2 d x , giving your answer as a rational fraction in its lowest (b) Deduce an approximation to y terms. 0 [2] … … … … … … …

7 marks

Mark scheme: 1(a) 2 f '( ) 2 e− = − x x x B1 Finds first derivative. 2 2 2 f ''( ) 4 e 2e − − = − x x x x B1 Finds second derivative. f (0) 1 f '(0) 0 f ''(0) 2 = = = − M1 Evaluates derivatives at zero. 2 2 e 1 − = − x x M1 A1 5 1(b) 3 1 1 5 2 5 0 0 1 74 1 d 3 375   − = − =      x x x x M1 A1 Substitutes 2 1−x or better. 2

This question in 9231/23 Oct/Nov 2020

5 It is given that x = sinh -1 t , y = cos -1 t , where - 1 1 t 1 1. dy 1 (a) By differentiating cosy with respect to t, show that =- . [4] dt 1 - t 2 … … … … … … … … … … … … … … … … … … … … … … … … d 2 y (b) Find 2 in terms of t, simplifying your answer. [5] dx … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 5(a) d sin 1 dt − = y y M1 A1 Differentiates both sides with respect to t. 2 d 0 1 cos 1 n dt si 0 π < < − −  >  = y y y y M1 Applies 2 2 sin cos 1. + = y y 2 d 1 dt 1 = − − y t A1 AG, justifies taking positive square root. 4 5(b) 2 d 1 dt 1 = + x t B1 2 2 d 1 d 1 + = − − y t x t B1 Finds first derivative. ( ) ( ) ( ) 2 1 1 1 1 2 2 2 2 2 2 2 2 2 2 1 1 (1 ) 1 d 1 dt 1 1 − −   − + + + − +   − = −   − −   t t t t t t t t t M1 Differentiates d d y x with respect to t. 2 2 2 2 d d 1 d dt d d 1   +   = − ×   −   y t t x x t M1 Applies chain rule. ( ) ( ) ( ) 1 1 2 2 3 2 2 2 2 2 2 1 (1 ) 1 2 1 1 −     − + + −         = − = −   −     −   t t t t t t t A1 OE (simplified). 5

This question in 9231/23 Oct/Nov 2020

8 (a) Sketch the graph of y = coth x for x 2 0 and state the equations of the asymptotes. [2] (b) Starting from the definitions of coth and cosech in terms of exponentials, prove that coth 2 x - cosech 2 x = 1. [3] … … … … … … … … … … … … … … … … … The curve C has equation y = ln coth 12 x for x 2 0 . a k dy (c) Show that =- cosechx . [3] dx … … … … … … (d) It is given that the arc length of C from x = a to x = 2a is ln4, where a is a positive constant. Show that cosha = 2 and find, in logarithmic form, the exact value of a. [7] … … … … … … … … … … … … … … … … … … Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

15 marks

Mark scheme: 8(a) B1 Correct shape and position, not too truncated. 0, = x 1 = y B1 States equations of asymptotes. 2 8(b) e e 2 coth cosech e e e e − − − + = = − − x x x x x x x x B1 ( ) ( ) 2 2 2 2 2 e e 4 e e 2 1 e e e e e e − − − − −   + + − − = =   −   − − x x x x x x x x x x M1 A1 Writes over common denominator, AG. 3 x Question Answer Marks Guidance 8(c) 2 1 sech d 1 2 1 1 1 d 2tanh 2sinh cosh 2 2 2       = − = −                   x y x x x x Or 2 1 cosech d 1 2 1 1 1 d 2coth 2sinh cosh 2 2 2       = − = −                   x y x x x x M1 A1 Uses chain rule. = 1 cosech sinh( ) − = − x x A1 AG 3 8(d) 2 2 1 cosech d + a a x x M1 Forms correct integral. 2 2 2 coth d coth d =   a a a a x x x x M1 A1 Uses 2 2 coth cosech 1 − = x x . [ ] 2 lnsinh lnsinh2 lnsinh = = − a a x a a M1 Integrates and substitutes limits. ( ) sinh 2 ln ln 2cosh sinh = = a a a M1 Combines logarithms and uses double angle formula. ( ) ln 2cosh ln 4 cosh 2 =  = a a A1 AG ( ) ( ) 2 ln 2 2 1 ln 2 3 = + − = + √ a A1 Must reject ( ) ln 2 . 3 − 7

This question in 9231/23 Oct/Nov 2020

7 (a) It is given that y = sec h -1 x + 12 . dy 1 Express cosh y in terms of x and hence show that sinh y = - 2 . [3] d x 1 x + 2 b l … … … … … … … … … … … … (b) Find the first three terms in the Maclaurin’s series for sec h –1 x + 12 in the form b l lna + bx + cx2 , where a, b and c are constants to be determined. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 7(a) ( ) 1 1 1 2 2 sech cosh cosh y x y x y − = = +  = + B1 Relates to cosh ,y d sinh d y y x B1 Differentiates LHS. ( ) 1 2 2 x − − + B1 Differentiates RHS. AG. 3 7(b) 2 3 2 2 d d 1 sinh cosh 2 d 2 d y y y x x x −     + = +         M1 A1 M1 A1 for LHS. B1 for RHS. B1 ( ) ( ) 1 1 1 2 (0) sech cosh (2) ln 2 3 − − = = = + y M1 A1 Relates to 1 cosh− and uses logarithmic form. 4 '(0) 3 = − y 16 ''(0) 3 3 = y M1 Evaluates derivatives at 0. = x ( ) 2 4 8 ln 2 3 3 3 3 = + − + y x x A1 Alternative method for question 7(b) ( ) ( ) ( ) 2 2 2 3 1 1 1 2 4 2 2 d 1 1 d 1 y x x x x x x − = − = − + − − + + − B1 ( ) ( ) ( ) ( ) ( ) 3 1 2 2 2 1 2 2 3 3 1 1 1 2 2 4 2 2 4 2 d 1 2 d y x x x x x x x x − − − − = + − − −− + − − + M1 A1 ( ) ( ) 1 1 1 2 (0) sech cosh (2) ln 2 3 − − = = = + y M1 A1 Relates to 1 cosh− and uses logarithmic form. Question Answer Marks Guidance 7(b) 4 '(0) 3 = − y 16 ''(0) 3 3 = y M1 Evaluates derivatives at 0. = x ( ) 2 4 8 ln 2 3 3 3 3 = + − + y x x A1 7

This question in 9231/21 May/June 2021

8 The curve C has parametric equations - x = 2 cosh t , y = 32 t 1 4 sinh t2 , for 0 G t G 1. dx dy 2 (a) Find and show that = 1 - sinh t . [3] dt dt … … … … … … The area of the surface generated when C is rotated through 2r radians about the x-axis is denoted by A. 1 t - 4 sinh 2t ( 1 + cosh 2t)dt . [4] (b) (i) Show that A = r y 32 1 0 b l … … … … … … … … … … … … … … … … … (ii) Hence find A in terms of r, sinh2 and cosh2. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

13 marks

Mark scheme: 8(a) d 2sinh d = x t t B1 ( ) 3 1 1 1 2 2 2 2 2 d cosh 2 1 cosh 2 1 sinh d = − = − − = − y t t t t M1 A1 Applies 2 2 sinh cosh 2 1, = − t t AG. 3 8(b)(i) 2 2 2 2 2 2 4 2 2 2 d d 4 ( 1) 4 2 1 ( 1) d d     + = + − = + − + = +         x y s s s s s s t t M1 Factorises 2 2 d d . d d     +         x y t t 4 cosh t A1 ( ) ( )( ) 1 1 2 3 3 1 1 2 4 2 4 0 0 2π sinh 2 cosh d π sinh 2 cosh 2 1 d t t t t t t t t − = − +   M1 A1 Correct formula for surface area, AG. A0 if limits missing. 2 2 d d d d     +         x y t t does not need to be simplified for M1. 4 Question Answer Marks Guidance 8(b)(ii) ( ) ( ) 1 1 1 1 1 1 4 16 16 4 2 0 2 0 sinh2 cosh2 1 d (1 cosh2 ) 1 cosh2   + = + = + −    t t t t M1 A1 Integrates. ( ) ( ) 1 1 1 3 3 3 1 1 2 2 2 2 2 0 0 0 cosh2 1 d sinh2 sinh2 d   + = + − +     t t t t t t t t t M1 A1 Integrates by parts. ( ) ( ) 1 3 3 3 1 1 1 1 1 2 2 4 2 2 2 4 4 2 0 sinh 2 cosh 2 sinh 2 cosh 2 t t t t t   + − − = − +   A1 ( ) 8 2 3 3 1 6 11 4 8 1 π sinh 2 cosh 2 (1 cosh 2) − + − + A1 OE. Must be exact. (Decimal answer is 3.980131435…) 6

This question in 9231/21 May/June 2021

7 (a) It is given that y = sec h -1 x + 12 . dy 1 Express cosh y in terms of x and hence show that sinh y = - 2 . [3] d x 1 x + 2 b l … … … … … … … … … … … … (b) Find the first three terms in the Maclaurin’s series for sec h –1 x + 12 in the form b l lna + bx + cx2 , where a, b and c are constants to be determined. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 7(a) ( ) 1 1 1 2 2 sech cosh cosh y x y x y − = = +  = + B1 Relates to cosh ,y d sinh d y y x B1 Differentiates LHS. ( ) 1 2 2 x − − + B1 Differentiates RHS. AG. 3 7(b) 2 3 2 2 d d 1 sinh cosh 2 d 2 d y y y x x x −     + = +         M1 A1 M1 A1 for LHS. B1 for RHS. B1 ( ) ( ) 1 1 1 2 (0) sech cosh (2) ln 2 3 − − = = = + y M1 A1 Relates to 1 cosh− and uses logarithmic form. 4 '(0) 3 = − y 16 ''(0) 3 3 = y M1 Evaluates derivatives at 0. = x ( ) 2 4 8 ln 2 3 3 3 3 = + − + y x x A1 Alternative method for question 7(b) ( ) ( ) ( ) 2 2 2 3 1 1 1 2 4 2 2 d 1 1 d 1 y x x x x x x − = − = − + − − + + − B1 ( ) ( ) ( ) ( ) ( ) 3 1 2 2 2 1 2 2 3 3 1 1 1 2 2 4 2 2 4 2 d 1 2 d y x x x x x x x x − − − − = + − − −− + − − + M1 A1 ( ) ( ) 1 1 1 2 (0) sech cosh (2) ln 2 3 − − = = = + y M1 A1 Relates to 1 cosh− and uses logarithmic form. Question Answer Marks Guidance 7(b) 4 '(0) 3 = − y 16 ''(0) 3 3 = y M1 Evaluates derivatives at 0. = x ( ) 2 4 8 ln 2 3 3 3 3 = + − + y x x A1 7

This question in 9231/22 May/June 2021

8 The curve C has parametric equations - x = 2 cosh t , y = 32 t 1 4 sinh t2 , for 0 G t G 1. dx dy 2 (a) Find and show that = 1 - sinh t . [3] dt dt … … … … … … The area of the surface generated when C is rotated through 2r radians about the x-axis is denoted by A. 1 t - 4 sinh 2t ( 1 + cosh 2t)dt . [4] (b) (i) Show that A = r y 32 1 0 b l … … … … … … … … … … … … … … … … … (ii) Hence find A in terms of r, sinh2 and cosh2. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

13 marks

Mark scheme: 8(a) d 2sinh d = x t t B1 ( ) 3 1 1 1 2 2 2 2 2 d cosh 2 1 cosh 2 1 sinh d = − = − − = − y t t t t M1 A1 Applies 2 2 sinh cosh 2 1, = − t t AG. 3 8(b)(i) 2 2 2 2 2 2 4 2 2 2 d d 4 ( 1) 4 2 1 ( 1) d d     + = + − = + − + = +         x y s s s s s s t t M1 Factorises 2 2 d d . d d     +         x y t t 4 cosh t A1 ( ) ( )( ) 1 1 2 3 3 1 1 2 4 2 4 0 0 2π sinh 2 cosh d π sinh 2 cosh 2 1 d t t t t t t t t − = − +   M1 A1 Correct formula for surface area, AG. A0 if limits missing. 2 2 d d d d     +         x y t t does not need to be simplified for M1. 4 Question Answer Marks Guidance 8(b)(ii) ( ) ( ) 1 1 1 1 1 1 4 16 16 4 2 0 2 0 sinh2 cosh2 1 d (1 cosh2 ) 1 cosh2   + = + = + −    t t t t M1 A1 Integrates. ( ) ( ) 1 1 1 3 3 3 1 1 2 2 2 2 2 0 0 0 cosh2 1 d sinh2 sinh2 d   + = + − +     t t t t t t t t t M1 A1 Integrates by parts. ( ) ( ) 1 3 3 3 1 1 1 1 1 2 2 4 2 2 2 4 4 2 0 sinh 2 cosh 2 sinh 2 cosh 2 t t t t t   + − − = − +   A1 ( ) 8 2 3 3 1 6 11 4 8 1 π sinh 2 cosh 2 (1 cosh 2) − + − + A1 OE. Must be exact. (Decimal answer is 3.980131435…) 6

This question in 9231/22 May/June 2021

1 Find the Maclaurin’s series for ex tanx from first principles up to and including the term in x2. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 1 ( ) 2 e tan sec + x x x B1 Finds first derivative. ( ) 2 2 e tan 2sec 2sec tan + + x x x x x M1 A1 Finds second derivative. (0) 0, = y '(0) 1, = y ''(0) 2 = y M1 Evaluates derivatives at 0. = x 2 = + y x x A1 5

This question in 9231/21 Oct/Nov 2021

3 The curve C has equation xy 3 - 4x 3 y = 3 . d y (a) Show that, at the point ( - 1, 1) on C, = 11. [3] d x … … … … … … … … … … … … … … … … … … … … … … … … … d 2 y (b) Find the value of 2 at the point ( - 1, 1) . [5] dx … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 3(a) 2 3 3 '+ xy y y B1 Differentiates 3. xy ( ) 3 2 4 ' 3 0 − + = x y x y B1 Differentiates 3 . x y ( ) 3 3 2 3( 1)(1) ' 1 4 ( 1) ' 3( 1) (1) 0 y y − + − − + − = leading to ' 11 y = B1 Substitutes ( 1,1), − AG. ( ) 2 3 2 3 ' 3 4 12 − = − y xy x x y y 3 Question Answer Marks Guidance 3(b) ( ) ( ) 2 3 2 2 3 4 '' ' 6 ' 3 12 − + + − xy x y y xyy y x B1 B1 Differentiates ( ) 2 3 3 4 '. − xy x y ( ) 2 2 3 ' 12 ' 2 0 + − + = y y x y xy B1 Differentiates 3 2 12 . − y x y ( ) 2 2 3 2 2 3 '' 4 '' 6 ' 6 ' 24 ' 24 0 − + + − − = xy y x y xy y y y x y xy ( )( ) ( )( ) ( ) 2 3 2 2 2 3 2 2 2 2 3 3 4 12 ' 24 3 ' 12 6 ' 3 12 '' 3 4 − + − − − + − = − xy x x y xy y y x y y xyy y x y xy x '' 11( 75) 3(11) 12(9) 0 + − + − = y M1 Substitutes (–1, 1) '' 900 = y A1 5

This question in 9231/21 Oct/Nov 2021

1 It is given that y = sinh ( x 2 ) + cosh ( x 2 ) . (a) Use standard results from the list of formulae (MF19) to find the Maclaurin’s series for y in terms of x up to and including the term in x4. [2] … … … … … … … … d 4 y (b) Deduce the value of 4 when x = 0 . [1] dx … … … (c) Use your answer to part (a) to find an approximation to 2 y dx , giving your answer as a rational 1y0 fraction in its lowest terms. [2] … … … … … … … … … … … …

5 marks

Mark scheme: 1(a) ( ) ( ) 2 2 sinh cosh + x x attempt at all four derivatives of y. 4 1 2 2 1+ + x x A1 2 1(b) 1 2 4! 12 × = B1 1 1(c) 1 2 1 2 0 2 4 1 d + +  x x x M1 Substitutes their power series, must be at least 2. + a bx 1 2 523 1 1 3 10 960 0 3 5   = + + =   x x x A1 2

This question in 9231/22 Oct/Nov 2021

5 The curve C has parametric equations + bt , for 1 G t G 2 , x = 3 t + 2t -1 + at 3 , y = 4t - 32 t -1 3 where a and b are constants. =- (a) It is given that a = 23 and b 1 2 . 2 dy 2 25 2 -2 2 dx Show that + = 4 ( t + t ) and find the exact length of C. [6] e dt o e dt o … … … … … … … … … … … … … … … … … … … … … … (b) It is given instead that a = b = 0 . d 2 y Find the value of 2 when t = 1. [4] dx … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 5(a) 2 2 3 2 2 − = − + x t t 2 2 3 3 2 2 4 − = + − y t t B1 Differentiates x and y with respect to t. 4 4 2 2 4 2 2 4 4 12 23 9 12 9 4 12 1 12 2 4 4 + + − + + + + + − t t t t t t t t M1 Expands 2 2. +   x y Accept ( ) 2 25 2 2 4 25 . − + − t t ( ) 2 25 25 25 25 4 2 4 4 2 2 4 4 t t t t − − + + = + M1 A1 Factorises 2 2. +   x y AG. 2 2 3 1 5 5 85 1 2 2 3 2 12 1 1 2 d − −   + = − =   t t t t t M1 A1 Applies correct formula for arc length. 6 Question Answer Marks Guidance 5(b) 2 3 2 2 4 d d 3 2 − − + = = −   t y y x x t B1 Finds d . d y x ( )( ) ( )( ) ( ) 2 2 3 2 3 2 2 2 3 3 2 2 3 2 3 4 4 4 d d 3 2 3 2 − − − − − − − − − − +   + =     − −   t t t t t t t t B1 Differentiates d d y x with respect to t. ( ) 2 2 3 2 3 2 2 3 2 4 d 5 5 d d d 3 2 d 3 2 d 2 2 − − − −   + =     − × −   − = = − t t y t t t x t x when 1. = t M1 A1 Applies chain rule. 4

This question in 9231/22 Oct/Nov 2021

1 Find the Maclaurin’s series for ex tanx from first principles up to and including the term in x2. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 1 ( ) 2 e tan sec + x x x B1 Finds first derivative. ( ) 2 2 e tan 2sec 2sec tan + + x x x x x M1 A1 Finds second derivative. (0) 0, = y '(0) 1, = y ''(0) 2 = y M1 Evaluates derivatives at 0. = x 2 = + y x x A1 5

This question in 9231/23 Oct/Nov 2021

3 The curve C has equation xy 3 - 4x 3 y = 3 . d y (a) Show that, at the point ( - 1, 1) on C, = 11. [3] d x … … … … … … … … … … … … … … … … … … … … … … … … … d 2 y (b) Find the value of 2 at the point ( - 1, 1) . [5] dx … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 3(a) 2 3 3 '+ xy y y B1 Differentiates 3. xy ( ) 3 2 4 ' 3 0 − + = x y x y B1 Differentiates 3 . x y ( ) 3 3 2 3( 1)(1) ' 1 4 ( 1) ' 3( 1) (1) 0 y y − + − − + − = leading to ' 11 y = B1 Substitutes ( 1,1), − AG. ( ) 2 3 2 3 ' 3 4 12 − = − y xy x x y y 3 Question Answer Marks Guidance 3(b) ( ) ( ) 2 3 2 2 3 4 '' ' 6 ' 3 12 − + + − xy x y y xyy y x B1 B1 Differentiates ( ) 2 3 3 4 '. − xy x y ( ) 2 2 3 ' 12 ' 2 0 + − + = y y x y xy B1 Differentiates 3 2 12 . − y x y ( ) 2 2 3 2 2 3 '' 4 '' 6 ' 6 ' 24 ' 24 0 − + + − − = xy y x y xy y y y x y xy ( )( ) ( )( ) ( ) 2 3 2 2 2 3 2 2 2 2 3 3 4 12 ' 24 3 ' 12 6 ' 3 12 '' 3 4 − + − − − + − = − xy x x y xy y y x y y xyy y x y xy x '' 11( 75) 3(11) 12(9) 0 + − + − = y M1 Substitutes (–1, 1) '' 900 = y A1 5

This question in 9231/23 Oct/Nov 2021

5 The variables x and y are such that y = 0 when x = 0 and ( x + 1) y + ( x + y + 1 ) 3 = 1. d y 3 (a) Show that = - when x = 0 . [3] d x 4 … … … … … … … … … … … … … (b) Find the Maclaurin’s series for y up to and including the term in x2. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 5(a)   d ( 1) ( 1) ' d x y x y y x     correctly.        3 2 d 1 3 1 ' 1 d x y x y y x       B1 Differentiates   3 1 x y   correctly.   3 4 (1) ' 3 ' 1 0 ' y y y      B1 Substitutes (0,0), AG. 3 5(b)  1 '' 2 ' x y y   B1 Differentiates ( 1) ' . x y y        2 2 3 1 '' 6 1 ' 1 0 x y y y x y         B1 B1 Differentiates   2 3( 1) 1 ' . x y y    Expect       6 1 1 6 1 1           x y y x y y y instead of    2 6 1 1 .     y x y    3 1 4 4 2 '' 2 3 '' 6 0 y y      M1 Substitutes (0,0). 9 32 '' y  A1 3 9 64 2 4 2 ''(0) (0) '(0) 2! y y y y x x x x      M1 A1 Finds Maclaurin’s series. 7

This question in 9231/21 May/June 2022

5 The variables x and y are such that y = 0 when x = 0 and ( x + 1) y + ( x + y + 1 ) 3 = 1. d y 3 (a) Show that = - when x = 0 . [3] d x 4 … … … … … … … … … … … … … (b) Find the Maclaurin’s series for y up to and including the term in x2. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 5(a)   d ( 1) ( 1) ' d x y x y y x     correctly.        3 2 d 1 3 1 ' 1 d x y x y y x       B1 Differentiates   3 1 x y   correctly.   3 4 (1) ' 3 ' 1 0 ' y y y      B1 Substitutes (0,0), AG. 3 5(b)  1 '' 2 ' x y y   B1 Differentiates ( 1) ' . x y y        2 2 3 1 '' 6 1 ' 1 0 x y y y x y         B1 B1 Differentiates   2 3( 1) 1 ' . x y y    Expect       6 1 1 6 1 1           x y y x y y y instead of    2 6 1 1 .     y x y    3 1 4 4 2 '' 2 3 '' 6 0 y y      M1 Substitutes (0,0). 9 32 '' y  A1 3 9 64 2 4 2 ''(0) (0) '(0) 2! y y y y x x x x      M1 A1 Finds Maclaurin’s series. 7

This question in 9231/22 May/June 2022

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

5 marks

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

This question in 9231/21 Oct/Nov 2022

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 … …

12 marks

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

This question in 9231/21 Oct/Nov 2022

2 A curve has equation ( x + 1 ) y + y 2 = 2 . dy 2 (a) Show that = - at the point ( 0, - 2) . [3] dx 3 … … … … … … … … d 2 y (b) Find the value of 2 at the point ( 0, - 2) . [4] dx … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 2(a) d B1 Differentiates ( x + 1) y correctly. ( ( x + 1) y ) = ( x + 1) y '+ y dx d 2 B1 Differentiates y 2 correctly. y = 2 yy ' ( ) dx (1) y '− 2 + 2( −2) y ' = 0 leading to 'y = − 23 B1 Substitutes (0, −2), AG. 3 2(b) ( x + 1) y ''+ 2 y ' B1 Differentiates ( x + 1) y '+ y. +2 yy ''+ 2 ( y ' ) 2 = 0 B1 Differentiates 2 yy '. 2 2 2 M1 Substitutes (0, −2) into their expression with y ''. y ''+ 2 ( − 3 ) + 2( −2) y ''+ 2 ( − 3 ) = 0 y '' = − 274 A1 Alternative method for question 2(b) − y ( x + 1 + 2 y )( − y ' ) + y (1 + 2 y ' ) B1 B1 B1 for ( x + 1 + 2 y )( − y ' ) in numerator. y ' = y '' = x + 1 + 2 y ( x + 1 + 2 y ) 2 B1 for y (1 + 2 y ' ) in numerator. 2 2 M1 Substitutes (0, −2) and y ' = − 23 into their ( 0 + 1 + 2 ( −2 ) )( 1 + 2 ( − 3 3 ) − 2 ( ) ) y '' = ( 0 + 1 + 2 ( −2 ) ) 2 expression. y '' = − 274 A1 4

This question in 9231/22 Oct/Nov 2022

G x G 1. Find, in terms of r and e, the area of the3 (a) A curve has equation y = e x + 14 e -x , for 0 surface generated when the curve is rotated through 2r radians about the x-axis. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Using standard results from the list of formulae (MF19), or otherwise, find the Maclaurin’s series for e x + 14 e -x up to and including the term in x2. [2] … … … … … … … … … … … … …

8 marks

Mark scheme: 2 M1 23(a) x 1 − x 2 x 1 1 −2 x 2 x 1 1 −2 x 1 + ( e − 4 e ) = 1 + e − 2 + 16 e = e + 2 + 16 e Finds 1 +  dy   dx  x 1 − x 2 x 1 − x = e + 4 e ( A1 e + 4 e ) 1 x 1 − x x 1 − x M1 Correct formula for surface area with limits. 2 π d x e + 4 e )( e + 4 e ) 0 ( 1 2 x 1 1 − 2 x A1 Expanded integrand. 2π 0 e + 2 + 1 6 e d x 1 1 2 x 1 1 −2 x M1 Integrates and substitutes limits. Integrand must be 2  2 e + 2 x − 3 2 e  0 2 x −2 x of the form ae + be + c , with a and b nonzero. 1 2 1 −2 1 2 1 −2 1 A1 2π = π ( 2 e − 32 e + 32 ) ( e − 16 e + 16 ) 6 3(b)  x 2  1  ( − x ) 2  M1 Uses expansion of ex or finds first and second  1 + x + +  +  1 − x + +  x 1 x x 1 − x  2!  4  2!  derivatives e − 4 e− and e + 4 e . 5 4 + 43 x + 85 x 2 A1 2

This question in 9231/22 Oct/Nov 2022

8 It is given that y = cosh u , where u 2 0 , and 2 2 2 d u du d u -x cosh u - 1 2 + + cosh u - 2 cosh u = 4e . e dx dx o e d x o (a) Show that d 2 y dy -x + - 2y = 4 e . [4] 2 dx dx … … … … … … … … … du (b) Find u in terms of x, given that, when x = 0 , u = ln 3 and = 3 . [10] dx … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

14 marks

Mark scheme: 8(a) dy du B1 = sinh u dx dx 2 2 2 B1 d y d u  du  = sinh u + cosh u 2 2   dx dx  dx  2 2 2 M1 Uses substitution to find y-x equation, AG. d y dy d u  du  du + − 2 y = sinh u + cosh u + sinh u − 2cosh u 2 2   dx dx dx  dx  dx 2  d 2 u du   du  2 − x A1 AG cosh u − 1  2 +  + cosh u   − 2cosh u = 4e  dx dx   dx  8(a) Alternative for 8(a) du −1 dy B1 = ( sinh u ) dx dx d 2 u  1  d 2 y d y  cosh u  d u B1 2 =   2 −  2  d x  sinh u  d x d x  sinh u  d x d 2 u  1  d 2 y  dy 2 cosh u  = − 2   2   3  dx  sinh u  dx  dx  sinh u    1  d 2 y  dy 2 cosh u  −1 dy  cosh u  dy  2 M1 Uses substitution to find y-x equation, AG. 2cosh u sinh u    2 −   3  + ( sinh u )  + 2   −   sinh u  dx  dx  sinh u  dx  sinh u  dx    d 2 y  dy 2 cosh u  dy cosh u  dy  2 − + + − 2cosh u 2   2  2   dx  dx  sinh u  dx sinh u  dx  d 2 y dy − x A1 AG + − 2 y = 4e dx 2 d x 4 8(b) m 2 + m − 2 = 0  m = 1, − 2 M1 Auxiliary equation. y = A e x + B e− 2 x A1 Complimentary function. Allow ‘ y = ’ missing. y = ke − x y ' = − ke − x y '' = ke − x B1 Particular integral and its derivatives. ke − x − ke − x − 2 ke − x = 4e − x M1 Substitutes and equates coefficients. k = −2 A1 WWW y = cosh u = Ae x + Be −2 x − 2e − x A1 Must have ‘ y = ’ or ‘cosh u = ’ . du x −2 x − x B1 y ' = sinh u = Ae − 2 Be + 2e dx A + B − 2 = 53 A − 2B + 2 = 4 A = 289 , B = 95 M1 A1 Substitutes initial conditions and forms simultaneous equations. For M1, allow substitution into their y and 'y if two linear equations in two unknowns derived. −1 28 x 5 −2 x − x A1 Substitutes for y and finds u in terms of x. u = cosh ( 9 e + 9 e − 2e ) 10

This question in 9231/22 Oct/Nov 2022

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

5 marks

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

This question in 9231/23 Oct/Nov 2022

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 … …

12 marks

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

This question in 9231/23 Oct/Nov 2022

8 (a) Starting from the definitions of sech and tanh in terms of exponentials, prove that 1 - sech2 t = tanh 2 t . [3] … … … … … … … … The curve C has parametric equations + ln sech t, y = 1 + tanh 4 t, for t 2 0 . x = 12 tanh 2 t d y (b) Show that = - 4 sech2 t . [5] d x … … … … … … … … … … … … … … d 2 y 9(c) Find the coordinates of the point on C with 2 = - , giving your answer in the form ( a + ln b, c) dx 2 where a, b and c are rational numbers. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

14 marks

Mark scheme: 8(a) 2 sech e e t t t    e e tanh e e t t t t t                2 2 2 2 2 2 2 2 e e 4 e e 2 e 2 e 4 1 e e e e e e e e t t t t t t t t t t t t t t                            M1 A1 Expands, gets to   2 2 2 e 2 e 4 e e t t t t       for M1, AG. 3 8(b) 3 2 d 4tanh sech d y t t t  B1     2 2 3 d tanh sech tanh tanh sech 1 tanh d x t t t t t t t      M1 A1 M1 sensible attempt at derivative of x. 3 2 2 3 d d sech d d d 4tanh 4s t ech d tanh y y t t t t x t x      M1 A1 Applies chain rule, must substitute their d d y t and d d x t for M1, AG. 5 8(c) 2 2 2 2 3 2 h d a d d d h d d 8sec t n sech 8 d tanh ta h d n t t t x t y y t t x x t             M1 A1 Finds 2 2 d . d y x For M1 2 . d s d e d d ch tanh y c t t t x       AEF. Accept 2 8cosech .t    2 16 25 2 2 2 9 9 2 2 2 sech 8 8 1 tanh tanh tanh tanh t t t t t         M1 A1 Sets equal to 9 2 , uses identity from (a) or equivalent. Accept 16 9 2 sinh t  or 5 2 2 9 cosh . t     8 3 881 25 5 625 , ln , x y   A1 A1 A1 for each correct coordinate. ln3 t  6

This question in 9231/21 May/June 2023

8 (a) Starting from the definitions of sech and tanh in terms of exponentials, prove that 1 - sech2 t = tanh 2 t . [3] … … … … … … … … The curve C has parametric equations + ln sech t, y = 1 + tanh 4 t, for t 2 0 . x = 12 tanh 2 t d y (b) Show that = - 4 sech2 t . [5] d x … … … … … … … … … … … … … … d 2 y 9(c) Find the coordinates of the point on C with 2 = - , giving your answer in the form ( a + ln b, c) dx 2 where a, b and c are rational numbers. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

14 marks

Mark scheme: 8(a) 2 sech e e t t t    e e tanh e e t t t t t                2 2 2 2 2 2 2 2 e e 4 e e 2 e 2 e 4 1 e e e e e e e e t t t t t t t t t t t t t t                            M1 A1 Expands, gets to   2 2 2 e 2 e 4 e e t t t t       for M1, AG. 3 8(b) 3 2 d 4tanh sech d y t t t  B1     2 2 3 d tanh sech tanh tanh sech 1 tanh d x t t t t t t t      M1 A1 M1 sensible attempt at derivative of x. 3 2 2 3 d d sech d d d 4tanh 4s t ech d tanh y y t t t t x t x      M1 A1 Applies chain rule, must substitute their d d y t and d d x t for M1, AG. 5 8(c) 2 2 2 2 3 2 h d a d d d h d d 8sec t n sech 8 d tanh ta h d n t t t x t y y t t x x t             M1 A1 Finds 2 2 d . d y x For M1 2 . d s d e d d ch tanh y c t t t x       AEF. Accept 2 8cosech .t    2 16 25 2 2 2 9 9 2 2 2 sech 8 8 1 tanh tanh tanh tanh t t t t t         M1 A1 Sets equal to 9 2 , uses identity from (a) or equivalent. Accept 16 9 2 sinh t  or 5 2 2 9 cosh . t     8 3 881 25 5 625 , ln , x y   A1 A1 A1 for each correct coordinate. ln3 t  6

This question in 9231/22 May/June 2023

1 (a) Find the Maclaurin series for sin -1 x up to and including the term in x3. [5] … … … … … … … … … … … … … … … … … … … 1 5 1 (b) Deduce an approximation to d u , giving your answer as a fraction. [1] 2 y0 1 - u … … … … … … …

6 marks

Mark scheme: 1(a) 1 2 2 '( ) 1 f x x    B1 Correct first derivative.   3 2 2 ''( ) 1 f x x x    B1 Correct second derivative.     5 3 2 2 2 2 2 '''( ) 3 1 1 f x x x x       M1 Differentiates their ''( ) f x using product rule. (0) 0 '(0) 1 ''(0) 0 '''(0) 1 f f f f     M1 Evaluates their derivatives at 0 x  . Must have attempted all three derivatives. 3 1 6 1 sin x x x    A1 CWO. Alternative method for question 1(a) sin '( ) sec y x f x y    (B1) Finds first derivative.   2 ''( ) tan sec f x y y  (B1) Finds second derivative.   2 2 3 4 5 '''( ) 2tan sec sec sec 3sec 2sec f x y y y y y y     (M1) Differentiates   2 tan sec y y using product and chain rule. (0) 0 '(0) 1 ''(0) 0 '''(0) 1 f f f f     (M1) Evaluates their derivatives at 0 x  . Must have attempted all three derivatives. 3 1 6 1 sin x x x    (A1) CWO. 5 Question Answer Partial Marks Guidance 1(b) 151 750 B1 1

This question in 9231/23 May/June 2023

4 The curve C has equation 4y 3 + ( x + y) 6 = 109 . dy 1 (a) Show that, at the point ( - 4 , 3) on C, = . [3] dx 17 … … … … … … … … … … … … … … … … … … … … … … … … … d 2 y (b) Find the value of 2 at the point ( - 4 , 3). [5] dx … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4(a)   3 2 d 4 12 ' d y y y x  3 y correctly.     6 5 d 6( ) (1 ') d x y x y y x     B1 Differentiates   6 x y  correctly.      2 1 12 3 ' 6 1 1 ' 0 ' 17 y y y       B1 Substitutes ( 4,3),  AG. 3 Question Answer Partial Marks Guidance 4(b)   2 2 2 '' 4 y y y y  B1 Differentiates 2 . y y      5 4 2 5 1 0 x y y x y y         B1 B1 Differentiates   5 ( ) 1 . x y y       2 2 4 0 1 1 1 7 12 5 1 18 96 ( 1) 0 17 7 17 7 y y          M1 Substitutes ( 4,3).  96 289 y A1 5

This question in 9231/23 May/June 2023

5 The curve C has parametric equations 3 2 2 - t2 1 , y = t2 + 5 , for 0 1 t G 3 . x = 23 t (a) Find the exact length of C. [5] … … … … … … … … … … … … … … … … … … … … … … … … … d 2 y (b) Find the set of values of t for which 2 2 0 . [5] dx … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

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

This question in 9231/21 Oct/Nov 2023

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

5 marks

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

This question in 9231/22 Oct/Nov 2023

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 … … … … … … … … … … … … … … … …

7 marks

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

This question in 9231/22 Oct/Nov 2023

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 … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

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

This question in 9231/22 Oct/Nov 2023

5 The curve C has parametric equations 3 2 2 - t2 1 , y = t2 + 5 , for 0 1 t G 3 . x = 23 t (a) Find the exact length of C. [5] … … … … … … … … … … … … … … … … … … … … … … … … … d 2 y (b) Find the set of values of t for which 2 2 0 . [5] dx … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

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

This question in 9231/23 Oct/Nov 2023

2 Find the Maclaurin’s series for e 1 + x 2 + e 1 - x up to and including the term in x2 . [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 2 2 1 2 e e e e 1 x x x       B1 Changes 2 1e x  or 1e x  so that the formula given in the list of formulae (MF19) can be applied.   2 1 2 1 e e 1 x x x     B1     2 1 2 2 e 1 e 1 x x x        M1 Sums power series. 2 3 1 2 2 1 e e 2e e e x x x x         A1 Alternative method for question 2 2 1 1 f '( ) 2 e e x x x x     B1 Finds first derivative. 2 2 2 1 1 1 f ''( ) 4 e 2e e x x x x x       B1 Finds second derivative. f (0) 2e f '(0) e f ''(0) 3e    M1 Evaluates their derivatives at zero. 2 3 1 2 2 1 e e 2e e e x x x x         A1 4

This question in 9231/21 May/June 2024

3 It is given that x = sin -1 t and y = t cos -1 t , for 0 G t 1 1. dy 2 -1 (a) Show that = - t + 1 - t cos t . [3] dx … … … … … … … … … … … … … … … … … … … … … … … … … d 2 y (b) Find 2 in terms of t. [4] dx … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 3(a) 1 2 d cos 1 d y t t t t     , 2 d 1 d 1 x t t   2 1 1 d d dt d dt os d c y y t x t x t       M1 A1 Uses chain rule, AG. 3 3(b) 2 1 1 2 2 2 d d 1 cos cos 1 2 d d 1 1 1 y t t t t t t x t t t               M1 A1 Applies product rule. 2 2 1 2 d 2 1 cos d y t t t x     M1 A1 Uses chain rule. 4

This question in 9231/21 May/June 2024

2 Find the Maclaurin’s series for e 1 + x 2 + e 1 - x up to and including the term in x2 . [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 2 2 1 2 e e e e 1 x x x       B1 Changes 2 1e x  or 1e x  so that the formula given in the list of formulae (MF19) can be applied.   2 1 2 1 e e 1 x x x     B1     2 1 2 2 e 1 e 1 x x x        M1 Sums power series. 2 3 1 2 2 1 e e 2e e e x x x x         A1 Alternative method for question 2 2 1 1 f '( ) 2 e e x x x x     B1 Finds first derivative. 2 2 2 1 1 1 f ''( ) 4 e 2e e x x x x x       B1 Finds second derivative. f (0) 2e f '(0) e f ''(0) 3e    M1 Evaluates their derivatives at zero. 2 3 1 2 2 1 e e 2e e e x x x x         A1 4

This question in 9231/22 May/June 2024

3 It is given that x = sin -1 t and y = t cos -1 t , for 0 G t 1 1. dy 2 -1 (a) Show that = - t + 1 - t cos t . [3] dx … … … … … … … … … … … … … … … … … … … … … … … … … d 2 y (b) Find 2 in terms of t. [4] dx … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 3(a) 1 2 d cos 1 d y t t t t     , 2 d 1 d 1 x t t   2 1 1 d d dt d dt os d c y y t x t x t       M1 A1 Uses chain rule, AG. 3 3(b) 2 1 1 2 2 2 d d 1 cos cos 1 2 d d 1 1 1 y t t t t t t x t t t               M1 A1 Applies product rule. 2 2 1 2 d 2 1 cos d y t t t x     M1 A1 Uses chain rule. 4

This question in 9231/22 May/June 2024

2 It is given that 1 -1 x = 1 + and y = cos t for 0 1 t 1 1. t dy t 2 (a) Show that = . [2] d x 2 1 - t … … … … … … … … … … … … … … … … … … … … … … … … b d 2 y a 2(b) Show that = - t `1 - t j `2 - t 2j, where a and b are constants to be determined. [4] dx 2 … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 2(a) dy dy dt 1 2 t 2 M1A1 Uses chain rule, AG. =  = − −t = 2 ( ) 2 dx dt dx 1 − t 1 − t 2 2(b) 2 3 2 − 12 M1A1 Applies Quotient rule. 1 − t ) d  dy 2t 1 − t + t (  = 2 dt  dx  1 − t − 12 M1A1 Uses chain rule. 2 3 2 1 − t d 2 y 2t 1 − t + t ( ) 2 3 2 − 12 5 2 − 32 = −t = −2t 1 − t − t 1 − t 2 2 ( ) ( ) ( ) dx 1 − t 3 2 − 32 2 2 3 2 − 32 2 = −t 1 − t 2 1 − t + t = −t 1 − t 2 − t ( ) ( ( ) ) ( ) ( ) 4

This question in 9231/21 Oct/Nov 2024

2 The curve C has equation 4 y 2 + 4 ln ( xy) = 1. 1 d y 1 (a) Show that, at the point b,2 l on C, = - . [3] 2 d x 6 … … … … … … … … … … … … … … … … … … … … … … … … … d 2 y 1 (b) Find the value of at the point b,2 l. [4] dx 2 2 … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 2(a) d 2 B1 Differentiates y 2 correctly. 4 y = 8 yy ' ( ) dx d xy '+ y −1 −1 B1 Differentiates ln( xy ) correctly. ( 4ln( xy ) ) = 4 = 4 y ' y + 4 x dx xy 1 1 1 B1 Substitutes (2, 12 ) into 8 ( 2 ) y '+ 4 ( 2 ) y '+ 4 ( 2 ) = 0  y ' = − 6 8 y ' y + 4 y ' y −1 + 4 x −1 = 0, AG. 3 2(b) EITHER Solution 1 (B1B1 Differentiates 8 y ' y + 4 y ' y −1 + 4 x −1 = 0. 2 + 8 y '' y and B1 for other terms. 8 ( y ' )2 + 8 y '' y − 4 ( y ' ) 2 y −2 + 4 y −1 y ''− 4 x −2 = 0 B1 for 8 ( y ' ) 2 1 1 1 2 M1) 1 8 ( − 6 ) + 8 y '' ( 2 ) − 16 ( − 6 ) + 8 y ''−=1 0 Substitutes ( 2, 2 ) . OR Solution 2 (B1B1 −1 −1 −2 −2 2 y + y 2 y '− y y ' and B1 −1 B1 for x ( ) ( ) −1 −1 y ' = − x 2 y + y ( ) −1 2 y + y −1 −2 −1 for x −2 ( ) . −1 −1 −2 −2 −1 y '' = x 2 y + y 2 y '− y y ' + x 2 y + y ( ) ( ) ( ) 1 −2 1 1 −1 M1) 1 1 and y ' = − 6 . y '' = 2 (1 + 2 ) ( − 6 )( 2 − 4 ) + 4 (1 + 2 ) Substitutes ( 2, 2 ) 12 y ''− 119 = 0  y '' = 10811 A1 4

This question in 9231/22 Oct/Nov 2024

6 y 0 x 1 2 n - 1 1 n n n The diagram shows the curve with equation y = e 1 - x for 0 G x G 1, together with a set of n rectangles of width 1. n 1 (a) By considering the sum of the areas of these rectangles, show that e 1 - x dx 1 U , where n 0y e - 1 U = . [4] n - 1n n (1 - e ) … … … … … … … … … … … … … 1 e 1 - x dx . [4](b) Use a similar method to find, in terms of n, a lower bound Ln for y 0 … … … … … … … … … … … … … … … … … … (c) Show that lim ( U - L ) = 0 . [2] n " 3 n n … … … … … … … … (d) Use the Maclaurin’s series for ex given in the list of formulae (MF19) to find the first three - 1z terms of the series expansion of z ( 1 - e ) , in ascending powers of 1, and deduce the value of z lim ( Un ). [3] n " 3 … … … … … … … … … … …

16 marks

Mark scheme: 0 M1A1 Forms the sum of the areas of the rectangles.6(a)  1 1− x  1 1− n 1 1− 1n 1 1− nn−1 e dx  ( n ) e + ( n ) e + + ( n ) e M1 for correct number of rectangles.    0  r e  1 − e −1  e − 1 M1A1 Applies sum of geometric progression, AG. e n −1 − n = e = =    n n n n r = 0 n 1 − e − 1  1 − e − 1  ( ) 4 16(b) 1 1− x 1 1− n 1 1− n2 1 1− nn−1 1 1− nn M1A1 Forms the sum of the areas of appropriate 0 e d x  ( n ) e + ( n ) e + ( n ) e + ( n ) e rectangles. M1 for correct number rectangles. e − 1 e 1 e − 1 M1A1 Applies sum of geometric progression. = + = −1 n n n e1− 1 1 − e n n e 1 − 1n ( e − ) ( ) n 1 − e − 1 n e 1 − 1 ( ) − ( ) OE. E.g. or . 1 1 − − n n n 1 − e n 1 − e ( ) ( ) 4 6(c) e − 1 M1 c U n − Ln = Simplifies U n − Ln to , where c is a n n nonzero constant. e −→1 0 as n → A1 AG. CWO. n 2 6(d) 1 z M1A1 Substitutes power series for e −z1 . A1 for z 1 − e − 1 = z 1 − + 2 − 2 + ( 1 − 1z + 21z 6 z 3 = 1 − 21z + 6 1z ( ) ) ( ) enough terms. lim (U n ) = e − 1 B1 n → 3

This question in 9231/22 Oct/Nov 2024

2 It is given that 1 -1 x = 1 + and y = cos t for 0 1 t 1 1. t dy t 2 (a) Show that = . [2] d x 2 1 - t … … … … … … … … … … … … … … … … … … … … … … … … b d 2 y a 2(b) Show that = - t `1 - t j `2 - t 2j, where a and b are constants to be determined. [4] dx 2 … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 2(a) dy dy dt 1 2 t 2 M1A1 Uses chain rule, AG. =  = − −t = 2 ( ) 2 dx dt dx 1 − t 1 − t 2 2(b) 2 3 2 − 12 M1A1 Applies Quotient rule. 1 − t ) d  dy 2t 1 − t + t (  = 2 dt  dx  1 − t − 12 M1A1 Uses chain rule. 2 3 2 1 − t d 2 y 2t 1 − t + t ( ) 2 3 2 − 12 5 2 − 32 = −t = −2t 1 − t − t 1 − t 2 2 ( ) ( ) ( ) dx 1 − t 3 2 − 32 2 2 3 2 − 32 2 = −t 1 − t 2 1 − t + t = −t 1 − t 2 − t ( ) ( ( ) ) ( ) ( ) 4

This question in 9231/23 Oct/Nov 2024

6 (a) Starting from the definitions of tanh and sech in terms of exponentials, prove that 1 - tanh 2 u = sech 2 u . [3] … … … … … … … … … … … … d -1 1 (b) Show that ( sech t) =- . [4] d t 2 t 1 - t … … … … … … … … … … … … It is given that x = tanh -1 t and y = t sech -1 t , for 0 1 t 1 1. d y 2 2 -1(c) Show that =- 1 - t + `1 - t j sech t . [4] d x … … … … … … … … d 2 y (d) Find in terms of t. [4] dx 2 … … … … … … … … … … … … … …

15 marks

Mark scheme: 6(a) e u − e − u 2 B1 Writes in exponential from. tanh u = sech u = e u + e − u e u + e − u 2 2 e u + e − u − e u − e − u  e u − e − u  2 ( ) ( ) 4 M1 Writes over common denominator. 1 −  u − u  = 2 = 2 e u + e − u e u + e − u  e + e  ( ) ( ) = sech 2 u A1 Withold this mark if working with LHS and RHS simultaneously. 3 6(b) −1 du M1 A1 Differentiates implicitly. u = sech t  sech u = t − tanh u sech u = 1 dt du 1 1 M1 A1 Uses identity from (a), AG. = − = − dt tanh u sech u t 1 − t 2 First alternative method for question 6(b) −1 1 du 1 M1 A1 Differentiates implicitly. u = sech t  cosh u =  sinh u = − t dt t 2 du 1 1 1 M1 A1 Uses cosh 2 u − sinh 2 u = 1 , AG. = − = − = − d t t 2 cosh 2 u − 1 t 2 t −2 − 1 t 1 − t 2 6(b) Second alternative method for question 6(b) −1 1 −1 1 M1 A1 −1 1 u = sech t  cosh u =  u = cosh  Differentiates cosh  . t t t du 1 −2 1 M1 A1 Applies formula with chain rule and simplifies, AG. =  −t = − 2 ( ) 2 1 dt ( t ) − 1 t 1 − t Third alternative method for question 6(b) 2  M1 A1 Uses logarithmic form of sech −1 t 1  1 1   1 + 1 − t  sech u = t  cosh u = + = ln   u = ln  − 1     t t  t t 2      d u −−1 1 − t 2  t  1 M1 A1 Differentiates logarithmic form using chain and =     = − quotient rules, AG.    d t   1 + 1 − t 2  t 1 − t 2  t 2 1 − t 2  4 6(c) dy −1 1 dx 1 B1 B1 = sech t − , = d t 1 − t 2 dt 1 − t 2 dy dy dt 2 2 −1 M1 A1 Uses chain rule, AG. =  = − 1 − t + 1 − t sech t ( ) dx dt dx 4 6(d) d  dy t 1 − t 2 −1 M1 A1 Applies product rule. Might see − − 2t sech t  = 2t 2 − 1 dt  dx  1 − t 2 t 1 − t 2 − 2t sech −1 t . t 1 − t 2 3 2 2 M1 A1 Uses chain rule. 2 1 − t d y 2 ( ) 2 −1 = t 1 − t − − 2t 1 − t sech t 2 ( ) dx t 4

This question in 9231/21 May/June 2025

6 (a) Starting from the definitions of tanh and sech in terms of exponentials, prove that 1 - tanh 2 u = sech 2 u . [3] … … … … … … … … … … … … d -1 1 (b) Show that ( sech t) =- . [4] d t 2 t 1 - t … … … … … … … … … … … … It is given that x = tanh -1 t and y = t sech -1 t , for 0 1 t 1 1. d y 2 2 -1(c) Show that =- 1 - t + `1 - t j sech t . [4] d x … … … … … … … … d 2 y (d) Find in terms of t. [4] dx 2 … … … … … … … … … … … … … …

15 marks

Mark scheme: 6(a) e u − e − u 2 B1 Writes in exponential from. tanh u = sech u = e u + e − u e u + e − u 2 2 e u + e − u − e u − e − u  e u − e − u  2 ( ) ( ) 4 M1 Writes over common denominator. 1 −  u − u  = 2 = 2 e u + e − u e u + e − u  e + e  ( ) ( ) = sech 2 u A1 Withold this mark if working with LHS and RHS simultaneously. 3 6(b) −1 du M1 A1 Differentiates implicitly. u = sech t  sech u = t − tanh u sech u = 1 dt du 1 1 M1 A1 Uses identity from (a), AG. = − = − dt tanh u sech u t 1 − t 2 First alternative method for question 6(b) −1 1 du 1 M1 A1 Differentiates implicitly. u = sech t  cosh u =  sinh u = − t dt t 2 du 1 1 1 M1 A1 Uses cosh 2 u − sinh 2 u = 1 , AG. = − = − = − d t t 2 cosh 2 u − 1 t 2 t −2 − 1 t 1 − t 2 6(b) Second alternative method for question 6(b) −1 1 −1 1 M1 A1 −1 1 u = sech t  cosh u =  u = cosh  Differentiates cosh  . t t t du 1 −2 1 M1 A1 Applies formula with chain rule and simplifies, AG. =  −t = − 2 ( ) 2 1 dt ( t ) − 1 t 1 − t Third alternative method for question 6(b) 2  M1 A1 Uses logarithmic form of sech −1 t 1  1 1   1 + 1 − t  sech u = t  cosh u = + = ln   u = ln  − 1     t t  t t 2      d u −−1 1 − t 2  t  1 M1 A1 Differentiates logarithmic form using chain and =     = − quotient rules, AG.    d t   1 + 1 − t 2  t 1 − t 2  t 2 1 − t 2  4 6(c) dy −1 1 dx 1 B1 B1 = sech t − , = d t 1 − t 2 dt 1 − t 2 dy dy dt 2 2 −1 M1 A1 Uses chain rule, AG. =  = − 1 − t + 1 − t sech t ( ) dx dt dx 4 6(d) d  dy t 1 − t 2 −1 M1 A1 Applies product rule. Might see − − 2t sech t  = 2t 2 − 1 dt  dx  1 − t 2 t 1 − t 2 − 2t sech −1 t . t 1 − t 2 3 2 2 M1 A1 Uses chain rule. 2 1 − t d y 2 ( ) 2 −1 = t 1 − t − − 2t 1 − t sech t 2 ( ) dx t 4

This question in 9231/22 May/June 2025

1 Find the Maclaurin’s series for e b x + 2 l up to and including the term in x2. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: Question Answer Marks Guidance 1 −2 ( x + 2 ) −1 B1 Finds first derivative. f '( x ) = −( x + 2) e −1 B1 Finds second derivative. −4 ( x + 2) -1 −3 ( x + 2) -1 2 x + 5 ( x + 2 ) e f ''( x ) = ( x + 2) e + 2 ( x + 2 ) e = 4 ( x + 2 ) 2 *M1 Evaluates derivatives at zero. Must have f (0) = e 1 2 f '(0) = − 14 e 1 2 f ''(0) = 165 e 1 attempted second derivative using the product rule. ( x + 2 ) −1 12 1 12 5 12 2 DM1 A1 e = e − 4 e x + 32 e x 5

This question in 9231/23 May/June 2025

3 The curve C has equation 9y 2 - 3 sinh -1 ( xy) = 1 - 3 ln 3 . 1 dy 1 (a) Show that, at the point ( 4 , ) on C, =- . [4] 3 dx 2 … … … … … … … … … … … … … … … … … … … … … … … … d 2 y 1 (b) Find the value of at the point ( 4 , ) . [5] dx 2 3 … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 3(a) d 2 B1 Differentiates y 2 correctly. 9 y = 18 yy ' ( ) d x d −1 xy '+ y M1 A1 Differentiates sinh −xy1 ( ) correctly. −3sinh ( xy ) = −3 ( ) 2 2 dx 1 + x y 1 18 A1 Substitutes( 4, 3 ) and shows at least one 3 y '− 3 4 y5' + 1 3 = 0  y ' = − 12 3 intermediate step, AG. 4 3(b) 2 B1 Differentiates 18 yy ' 18 yy ''+ 18 ( y ' )  2 2 2 2 2 − 12  B1 B1 xy '+ y 1 + x y . Differentiates −3 xy ) 1 + x y ( xy ''+ 2 y ' ) − ( xy '+ y ) ( ) (   2 2 −3 2 2  1 + x y 1 + x y     2 5 5 3 4 1 M1 1 ( ) y = − 3 ( 4 y ''− 1) − 3 ) ( 5 )( 3 18 Substitutes ( 4, 3 ) . and = 0 2 3 y ''+ 184 − 3 25 9 45 ( 4 y ''− 1) − 60 18 18 3 y ''+ 4 − = 0 25 − 65 y ''+ 1087 = 0  y '' = 294 A1 5

This question in 9231/23 May/June 2025