Hyperbolic functions — Paper 2 · A Level Mathematics - Further 9231

5 The curves C 1 : y = cosh x and C 2 : y = sinh 2x intersect at the point where x = a . (a) Find the exact value of a, giving your answer in logarithmic form. [4] … … … … … … … … … … … … … … … … (b) Sketch C1 and C2 on the same diagram. [2] (c) Find the exact value of the length of the arc of C1 from x = 0 to x = a . [5] … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 5(a) 1 2 cosh 2sinh cosh sinh =  = a a a a M1 A1 ( ) 1 1 1 2 2 1 4 sinh ln 1 − = = + + a M1 ( ) 1 1 2 2 ln 5 = + √ a A1 4 Question Answer Marks 5(b) (B1 for C1 correct, B1 for C2 correct and intersecting C1 in the first quadrant) B1 B1 2 5(c) 2 0 1 sinh d +  a x x M1 2 0 0 cosh d cosh d =   a a x x x x M1 A1 [ ]0 sinh sinh a x a = M1 1 2 A1 5

This question in 9231/21 May/June 2020

5 The curves C 1 : y = cosh x and C 2 : y = sinh 2x intersect at the point where x = a . (a) Find the exact value of a, giving your answer in logarithmic form. [4] … … … … … … … … … … … … … … … … (b) Sketch C1 and C2 on the same diagram. [2] (c) Find the exact value of the length of the arc of C1 from x = 0 to x = a . [5] … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 5(a) 1 2 cosh 2sinh cosh sinh =  = a a a a M1 A1 ( ) 1 1 1 2 2 1 4 sinh ln 1 − = = + + a M1 ( ) 1 1 2 2 ln 5 = + √ a A1 4 Question Answer Marks 5(b) (B1 for C1 correct, B1 for C2 correct and intersecting C1 in the first quadrant) B1 B1 2 5(c) 2 0 1 sinh d +  a x x M1 2 0 0 cosh d cosh d =   a a x x x x M1 A1 [ ]0 sinh sinh a x a = M1 1 2 A1 5

This question in 9231/22 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

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

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

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

2 Find the Maclaurin’s series for lncoshx up to and including the term in x4. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 2 d tanh d = y x x B1 Finds first derivative. 2 2 2 d sech d = y x x B1 Finds second derivative. 2 3 3 3 d sinh 2tanh sech 2 d cosh = − = − y x x x x x M1 A1 Finds third derivative. ( ) ( ) ( ) 4 4 2 2 4 cosh 3sinh 2 co h d s d − − = x x y x x 4 2 2 2sech 4tanh sech = − + x x x B1 Finds fourth derivative. Alternative: 2 2 2 6tanh sech 2sech . − x x x (1) (3) (0) (0) (0) 0 = = = y y y (2)(0) 1 = y (4)(0) 2 = − y M1 Evaluates derivatives at 0. = x 1 1 2 12 2 4 = − y x x A1 CWO 7

This question in 9231/23 May/June 2021

6 (a) Starting from the definitions of sinh and cosh in terms of exponentials, prove that 2 sinh 2 x = cosh 2 x - 1. [3] … … … … … … … … … … … (b) Find the solution to the differential equation d y + y coth x = 4 sinh x d x for which y = 1 when x = ln 3 . [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 6(a) ( ) ( ) 1 1 2 2 sinh e e cosh e e − − = − = + x x x x x x B1 Writes in exponential from. ( ) ( ) ( ) 1 1 1 2 2 2 2 2 2 2 2 e e e 2 e e e 1 − − − − = − + = + − x x x x x x M1 Expands. cosh 2 1 − x A1 AG. 3 6(b) lns oth inh c e e sinh = =  x x x M1 A1 Finds integrating factor. 2 d ( sinh ) 4sinh d = y x x x M1 Correct form on both sides. ( ) 1 2 sinh 2 sinh 2 sinh 2 2 = − + = − + y x x x C x x C M1 A1 Integrates 2 sinh x using identity. 4 40 sinhln3 sinhln9 2ln3 leading to 2ln3 3 9 C C = − + = − + M1 Substitutes initial conditions into their solution. 28 sinh sinh2 2 2ln3 9 = − + − y x x x A1 7

This question in 9231/23 May/June 2021

8 (a) Starting from the definition of cosh in terms of exponentials, prove that 2 cosh 2 A = cosh 2A + 1. [3] … … … … … … The curve C has parametric equations - 4t , for - 2 G t G 2 . x = 2 cosh 2t + 3t, y = 32 cosh 2t 1 1 The area of the surface generated when C is rotated through 2r radians about the y-axis is denoted by A. 1 + 3t) cosh 2t dt . [4] (b) (i) Show that A = 10 r 21 ( 2 cosh 2 t -y 2 … … … … … … … … … … … … … … … … (ii) Hence find A in terms of r and e. [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. … … … … … … … … … … … … … … … … … … … … … …

14 marks

Mark scheme: 8(a) ( ) 1 2 cosh e e− = + A A A B1 Writes in exponential form ( ) ( ) 2 2 2 1 2 2 2 2 1 2cosh e 2 e e e 1 cosh2 1 − − = + + = + + = + A A A A A A M1 A1 Expands, AG. 3 8(b)(i) d d 4sinh 2 3 3sinh 2 4 d d = + = − x y t t t t B1 ( ) ( ) 2 2 2 2 4sinh 2 3 3sinh 2 4 25(sinh 2 1) 25cosh 2 + + − = + = t t t t M1 A1 Expands and applies 2 2 cosh sinh 1 = + A A ( )( ) 1 1 2 2 1 1 2 2 2 2 c 2cosh2 3 2 30 c 2 π 5cosh 2 d π 20 o o h sh s d t t t t t t t t − −   =   +   +   A1 Correct formula for surface area, AG. 4 8(b)(ii) 1 1 2 2 1 1 2 2 2 20 cosh 2 d 10 cosh 4 1d − − = +   t t t t M1 Applies ( ) 2 2 1 cosh cosh2 1 . = + A A ( ) 1 2 1 2 1 1 4 2 10 sinh 4 10 sinh 2 1 − + = +     t t M1 A1 Integrates. [ ] 1 1 1 2 2 2 1 1 1 2 2 2 1 1 2 2 30 cosh2 d 30 sinh 2 sinh2 d − − −   = −       t t t t t t t M1 A1 Integrates by parts. 1 2 1 2 1 2 15 sinh 2 cosh 2 0 − − =     t t t A1 Accept 1 2 1 2 cosh 2 d 0 − = t t t since cosh 2 t t is odd. ( ) ( ) 2 2 1 4 1 e π 0 e 1 − − + A1 64.82457... Question Answer Marks Guidance 8(b)(ii) Alternative method for question 8(b)(ii) 10 ( ) 1 2 1 2 π 2cosh 2 3 cosh 2 d − +  t t t t = ( ) 1 1 2 2 1 1 2 2 3 2 1 ) π cosh 2 sinh 2 ( sinh 2 5π sinh 2 4si h 2 3 d 0 n t t t t t t t − −   + − +    M1 A1 Integrates by parts. 2cosh 2 3 u t t = + , ' 4sinh 2 3 u t = + ' cosh 2 v t = , 1 2 sinh2 v t = 1 1 1 2 2 2 1 1 1 2 2 2 2 1 2 1 d π sinh 4 20π sinh 2 15π sinh 0 2 d t t t t t − − −   − −           = 1 1 1 2 2 2 1 1 1 2 2 2 1 2 d 1 π sinh4 10π cosh4 1 i 0 15π s nh2 d t t t t t − − −   − − −           M1 A1 Applies 2 2sinh 2 cosh 4 1 = − t t ( ) 1 2 1 2 5 10 15 2 4 2 π sinh 4 sinh 4 10 cosh 2 t t t t −   − + −   M1 A1 Integrates ( ) 1 2 1 2 5 15 2 2 π sinh 4 10 cosh 2 t t t −   + −   = ( ) π 5sinh2 10 + = ( ) ( ) 2 2 1 4 1 e π 0 e 1 − − + A1 64.82457... 7

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

8 (a) Starting from the definitions of tanh and sech in terms of exponentials, prove that 1 - tanh 2 x = sec h 2 x . [3] … … … … … … (b) Using the substitution u = tanh x , or otherwise, find sec h 2 x tanh 2 x d x . [2] y … … … … … … … ln 3 = sec h x tanh x d x . It is given that, for n H 0 , I n n 2 y0 (c) Show that, for n H 2 , 3 n - 2 = 5 5 ( n + 1) I n 4 3 + ( n - 2) I n - 2 . [5] b l b l d [You may use the result that ( sec h x) =- tanh x sec h x .] d x … … … … … … … … … … … … … … … … … … … … … … … … … … (d) Find the value of I4. [3] … … … … … … … … … 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) e e 2 tanh sech e e e e − − − − = = + + x x x x x x x x B1 Writes in exponential form. ( ) ( ) ( ) ( ) 2 2 2 2 2 e e e e e e 4 1 e e e e e e − − − − − − + − −   − − = =   +   + + x x x x x x x x x x x x M1 Writes over common denominator. 2 sech x A1 AG 3 8(b) 3 1 3 2 d tanh = + u u x C M1 A1 Uses substitution correctly. Allow by inspection. 2 8(c) l 2 2 n 0 2 3 sech sech tanh d − = n nI x x x x B1 Separates into correct structure. ln3 ln3 1 1 3 3 0 2 3 2 4 0 sech tanh ( 2) sech tanh d n n x x n x x x − −   + −    M1 Uses integration by parts correctly. ( ) ( ) ( ) 3 1 4 1 3 5 5 3 2 3 2 ( 2) n n n n I I − − + − − M1 A1 Uses 2 2 . 1 sech tanh − = x x ( ) ( ) ( ) 2 3 4 5 3 5 2 3 2 ( 2) n n n n I n I − − + − = + − leading to ( ) ( ) 2 3 4 5 3 5 2 ( 1) ( 2) n n n n I n I − − + = + − A1 AG 5 Question Answer Marks Guidance 8(d) ( ) 5 3 7 2 64 1 4 3 5 3 = = I B1 ( ) ( ) 2 3 4 4 2 5 5 3 (4 1) 2 I I + = + leading to 4 4928 0.105 46875 I = = M1 A1 Applies reduction formula with 4. = n 3

This question in 9231/22 Oct/Nov 2021

8 (a) Starting from the definition of cosh in terms of exponentials, prove that 2 cosh 2 A = cosh 2A + 1. [3] … … … … … … The curve C has parametric equations - 4t , for - 2 G t G 2 . x = 2 cosh 2t + 3t, y = 32 cosh 2t 1 1 The area of the surface generated when C is rotated through 2r radians about the y-axis is denoted by A. 1 + 3t) cosh 2t dt . [4] (b) (i) Show that A = 10 r 21 ( 2 cosh 2 t -y 2 … … … … … … … … … … … … … … … … (ii) Hence find A in terms of r and e. [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. … … … … … … … … … … … … … … … … … … … … … …

14 marks

Mark scheme: 8(a) ( ) 1 2 cosh e e− = + A A A B1 Writes in exponential form ( ) ( ) 2 2 2 1 2 2 2 2 1 2cosh e 2 e e e 1 cosh2 1 − − = + + = + + = + A A A A A A M1 A1 Expands, AG. 3 8(b)(i) d d 4sinh 2 3 3sinh 2 4 d d = + = − x y t t t t B1 ( ) ( ) 2 2 2 2 4sinh 2 3 3sinh 2 4 25(sinh 2 1) 25cosh 2 + + − = + = t t t t M1 A1 Expands and applies 2 2 cosh sinh 1 = + A A ( )( ) 1 1 2 2 1 1 2 2 2 2 c 2cosh2 3 2 30 c 2 π 5cosh 2 d π 20 o o h sh s d t t t t t t t t − −   =   +   +   A1 Correct formula for surface area, AG. 4 8(b)(ii) 1 1 2 2 1 1 2 2 2 20 cosh 2 d 10 cosh 4 1d − − = +   t t t t M1 Applies ( ) 2 2 1 cosh cosh2 1 . = + A A ( ) 1 2 1 2 1 1 4 2 10 sinh 4 10 sinh 2 1 − + = +     t t M1 A1 Integrates. [ ] 1 1 1 2 2 2 1 1 1 2 2 2 1 1 2 2 30 cosh2 d 30 sinh 2 sinh2 d − − −   = −       t t t t t t t M1 A1 Integrates by parts. 1 2 1 2 1 2 15 sinh 2 cosh 2 0 − − =     t t t A1 Accept 1 2 1 2 cosh 2 d 0 − = t t t since cosh 2 t t is odd. ( ) ( ) 2 2 1 4 1 e π 0 e 1 − − + A1 64.82457... Question Answer Marks Guidance 8(b)(ii) Alternative method for question 8(b)(ii) 10 ( ) 1 2 1 2 π 2cosh 2 3 cosh 2 d − +  t t t t = ( ) 1 1 2 2 1 1 2 2 3 2 1 ) π cosh 2 sinh 2 ( sinh 2 5π sinh 2 4si h 2 3 d 0 n t t t t t t t − −   + − +    M1 A1 Integrates by parts. 2cosh 2 3 u t t = + , ' 4sinh 2 3 u t = + ' cosh 2 v t = , 1 2 sinh2 v t = 1 1 1 2 2 2 1 1 1 2 2 2 2 1 2 1 d π sinh 4 20π sinh 2 15π sinh 0 2 d t t t t t − − −   − −           = 1 1 1 2 2 2 1 1 1 2 2 2 1 2 d 1 π sinh4 10π cosh4 1 i 0 15π s nh2 d t t t t t − − −   − − −           M1 A1 Applies 2 2sinh 2 cosh 4 1 = − t t ( ) 1 2 1 2 5 10 15 2 4 2 π sinh 4 sinh 4 10 cosh 2 t t t t −   − + −   M1 A1 Integrates ( ) 1 2 1 2 5 15 2 2 π sinh 4 10 cosh 2 t t t −   + −   = ( ) π 5sinh2 10 + = ( ) ( ) 2 2 1 4 1 e π 0 e 1 − − + A1 64.82457... 7

This question in 9231/23 Oct/Nov 2021

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

8 marks

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

This question in 9231/21 May/June 2022

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

8 marks

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

This question in 9231/22 May/June 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

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

5 (a) Starting from the definitions of cosh and sinh in terms of exponentials, prove that 2 cosh 2 x = cosh 2 x + 1. [3] … … … … … … … … … … … … (b) Find the solution of the differential equation dy + 2y tanh x = 1 dx for which y = 1 when x = 0 . Give your answer in the form y = f ( x) . [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 5(a) 1 2 cosh e e x x x    B1     2 2 1 2 2 2 4 e e e e 2 cosh 2 1 x x x x x         M1 A1 Expands, AG. 3 Question Answer Partial Marks Guidance 5(b) d 2lncosh 2 tanh 2 e e cosh x x x x    M1 A1 Finds integrating factor.   2 2 d cosh cosh d y x x x  M1 Correct form on LHS,   d d yI x for their integrating factor I, and attempt to integrate their RHS.   2 2 1 1 1 cosh cosh d cosh2 1d sinh2 2 2 2 y x x x x x x x C         M1 A1 Uses   2 1 cosh cosh 2 1 . 2 x x   M1 A1 is for RHS. 1 C  M1 Finds C.   2 1 1 sech sinh2 1 4 2 y x x x    M1 A1 Divides through by their coefficient of y. 8

This question in 9231/23 May/June 2023

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

6 marks

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

This question in 9231/21 Oct/Nov 2023

6 (a) Starting from the definitions of cosh and sinh in terms of exponentials, prove that sinh 2x = 2 sinh x cosh x . [3] … … … … … … … … … (b) Using the substitution u = sinh x , find sinh 2 2x cosh x dx . [4] y … … … … … … … … … … … … … … … … (c) Find the particular solution of the differential equation d y 2 + y tanh x = sinh 2x , d x given that y = 4 when x = 0 . Give your answer in the form y = f ( x) . [7] … … … … … … … … … … … … … … … … … … … … … … … … …

14 marks

Mark scheme: 6(a) 1 x − x 1 x − x B1 cosh x = e + e sinh x = e − e ( ) ( ) 2 2 1 x − x x − x 1 2 x −2 x M1 A1 Expands, AG. e − e e +e = e − e = sinh2 x ( )( ) ( ) 2 2 3 6(b) u = sinh x  du = cosh x dx B1 2 2 2 2 2 M1 Applies identities to find integral in terms of u.  sinh 2 x cosh x dx = 4  sinh x cosh x du = 4  sinh x ( sinh x + 1) du = 4  u 2 u 2 + 1 du A1 ( )  1 5 1 3   1 5 1 3  A1 = 4 u + u ( + C ) = 4 sinh x + sinh x ( + C )      5 3   5 3  4 tanh xdx lncosh x M1 A1 Finds integrating factor.6(c)  e = e = cosh x d 2 M1 Correct form on LHS and attempt to integrate RHS. ( y cosh x ) = sinh 2 x cosh x dx  1 5 1 3  M1 A1 Integrates RHS using their part (b). y cosh x = 4 sinh x + sinh x + C    5 3  4 = C M1 Substitutes initial conditions.  1 5 1 3  A1 y = 4sech x sinh x + sinh x + 1    5 3  7

This question in 9231/21 Oct/Nov 2023

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. … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

12 marks

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

This question in 9231/22 Oct/Nov 2023

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

6 marks

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

This question in 9231/23 Oct/Nov 2023

6 (a) Starting from the definitions of cosh and sinh in terms of exponentials, prove that sinh 2x = 2 sinh x cosh x . [3] … … … … … … … … … (b) Using the substitution u = sinh x , find sinh 2 2x cosh x dx . [4] y … … … … … … … … … … … … … … … … (c) Find the particular solution of the differential equation d y 2 + y tanh x = sinh 2x , d x given that y = 4 when x = 0 . Give your answer in the form y = f ( x) . [7] … … … … … … … … … … … … … … … … … … … … … … … … …

14 marks

Mark scheme: 6(a) 1 x − x 1 x − x B1 cosh x = e + e sinh x = e − e ( ) ( ) 2 2 1 x − x x − x 1 2 x −2 x M1 A1 Expands, AG. e − e e +e = e − e = sinh2 x ( )( ) ( ) 2 2 3 6(b) u = sinh x  du = cosh x dx B1 2 2 2 2 2 M1 Applies identities to find integral in terms of u.  sinh 2 x cosh x dx = 4  sinh x cosh x du = 4  sinh x ( sinh x + 1) du = 4  u 2 u 2 + 1 du A1 ( )  1 5 1 3   1 5 1 3  A1 = 4 u + u ( + C ) = 4 sinh x + sinh x ( + C )      5 3   5 3  4 tanh xdx lncosh x M1 A1 Finds integrating factor.6(c)  e = e = cosh x d 2 M1 Correct form on LHS and attempt to integrate RHS. ( y cosh x ) = sinh 2 x cosh x dx  1 5 1 3  M1 A1 Integrates RHS using their part (b). y cosh x = 4 sinh x + sinh x + C    5 3  4 = C M1 Substitutes initial conditions.  1 5 1 3  A1 y = 4sech x sinh x + sinh x + 1    5 3  7

This question in 9231/23 Oct/Nov 2023

1 6 (a) Show that cosh x + sinh x 2 = e 2 x . [2] … … … … … … … … … … … … … (b) Find the particular solution of the differential equation 1 d 2 y dy 2 + + 3 y = 5 cosh x + sinh x , 2 dx dx ` j dy 4 given that, when x = 0 , y = 1 and = . [10] dx 3 … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 6(a)       1 1 2 2 1 1 2 2 cosh sinh e e e e x x x x x x         M1 Substitutes sinh and cosh in terms of exponentials.    1 2 1 2 e e x x  A1 AG 2 Question Answer Marks Guidance 6(b) 1 2 2 1 2 3 i 0 11 m m m        M1 Auxiliary equation.   1 2 11 11 2 2 e cos sin x y A x B x    A1 Complimentary function. Allow ‘ y ’ missing. 1 1 1 2 2 2 1 1 2 4 e ' e '' e x x x y k y k y k      B1 Particular integral and its derivatives. 1 1 1 2 2 1 2 2 1 1 1 1 4 2 4 2 e e 3 e 5e 3 5 x x x x k k k k k k        M1 Substitutes and equates coefficients. PI must be correct form ( bx ae ). 4 3 k  A1   2 2 1 1 11 11 4 2 2 3 e cos sin e x x y A x B x     A1 FT Must have ‘ y ’. FT on their CF.     2 1 1 2 2 1 11 11 11 11 11 11 1 2 2 2 2 2 2 2 2 3 d e sin cos e cos sin e d x x x y A x B x A x B x x         B1 4 3 1 A   11 4 1 2 3 2 2 3 B A    1 1 3 11 , A B    M1 A1 Substitutes initial conditions and forms simultaneous equations. For M1, CF must be correct form   2 2 1 1 11 11 1 1 4 3 2 2 3 11 e cos sin e x x y x x      A1 Must have ‘ y ’. 10

This question in 9231/21 May/June 2024

1 6 (a) Show that cosh x + sinh x 2 = e 2 x . [2] … … … … … … … … … … … … … (b) Find the particular solution of the differential equation 1 d 2 y dy 2 + + 3 y = 5 cosh x + sinh x , 2 dx dx ` j dy 4 given that, when x = 0 , y = 1 and = . [10] dx 3 … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 6(a)       1 1 2 2 1 1 2 2 cosh sinh e e e e x x x x x x         M1 Substitutes sinh and cosh in terms of exponentials.    1 2 1 2 e e x x  A1 AG 2 Question Answer Marks Guidance 6(b) 1 2 2 1 2 3 i 0 11 m m m        M1 Auxiliary equation.   1 2 11 11 2 2 e cos sin x y A x B x    A1 Complimentary function. Allow ‘ y ’ missing. 1 1 1 2 2 2 1 1 2 4 e ' e '' e x x x y k y k y k      B1 Particular integral and its derivatives. 1 1 1 2 2 1 2 2 1 1 1 1 4 2 4 2 e e 3 e 5e 3 5 x x x x k k k k k k        M1 Substitutes and equates coefficients. PI must be correct form ( bx ae ). 4 3 k  A1   2 2 1 1 11 11 4 2 2 3 e cos sin e x x y A x B x     A1 FT Must have ‘ y ’. FT on their CF.     2 1 1 2 2 1 11 11 11 11 11 11 1 2 2 2 2 2 2 2 2 3 d e sin cos e cos sin e d x x x y A x B x A x B x x         B1 4 3 1 A   11 4 1 2 3 2 2 3 B A    1 1 3 11 , A B    M1 A1 Substitutes initial conditions and forms simultaneous equations. For M1, CF must be correct form   2 2 1 1 11 11 1 1 4 3 2 2 3 11 e cos sin e x x y x x      A1 Must have ‘ y ’. 10

This question in 9231/22 May/June 2024

7 (a) Show that ( ln ( tanh x)) = 2 cosech 2x . [3] dx … … … … … … … … … … … … … … … (b) Find the solution of the differential equation dy sinh 2x + 2 y = sinh 2 x dx for which y = 5 when x = ln 2 . Give your answer in an exact form. [7] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 7(a) d sech 2 x M1A1 Applies chain rule. ( ln ( tanh x ) ) = dx tanh x 1 2 A1 AG. = = . sinh x cosh x sinh2 x 3 7(b) d y 2 y B1 Divides through by sinh(2 )x . + = 1 d x sinh(2 x ) d M1A1 Correct form on LHS and RHS. ( y tanh x ) = tanh x dx y tanh x = ln(cosh x ) + C M1A1 Integrates RHS. 3 5 ) = ln 5 ( 5 4 +C M1 Substitutes initial conditions. y tanh x = ln(cosh x ) + 3 − ln 54 A1 Accept equivalent exact form. 7

This question in 9231/22 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

2 (a) Starting from the definitions of tanh and sech in terms of exponentials, prove that tanh 2 t + sech 2 t = 1. [3] … … … … … … … … (b) The curve C has parametric equations x = ln ( cosh t) , y = tan -1 ( sinh t) , for 0 G t G 1. Find the length of C. [5] … … … … … … … … … … … … … … …

8 marks

Mark scheme: 2(a) e t − e − t 2 B1 tanh t = sech t = e t + e − t e t + e − t 2 t − t 2 t −2 t M1 Writes in a single fraction over common  e − e  4 e + e + 2  t − t  + 2 = 2 = 1 denominator, AG. e t + e − t e t + e − t  e + e  ( ) ( ) A1 Withold A1 mark if they used mixed variables within a single line of working. 3 2(b) d x B1 = tanh t d t dy cosh t M1 A1 = = sech t dt 1 + sinh 2 t Applies 1 + sinh 2 t = cosh 2 t. 1 1 2 2 M1 Substitutes their derivatives into a correct  0 tanh t + sech t d t =  01d t 2 2  dx   dy  formula for arc length eg.   +   dt  dt   dt  = 1 A1 5

This question in 9231/23 May/June 2025