Hyperbolic functions

7(a) sech y = 1 = x+ 1  cosh y = ( x+ 1 )−1 B1 Relates to cosh y, cosh y 2 2 dy B1 Differentiates LHS. sinhy dx − ( x+ 1 )−2 B1 Differentiates RHS. AG. 2 3 7(b) d2y dy 2  1 −3 M1 A1 M1 A1 for LHS. sinh +coshy   =2  x+  B1 for RHS. dx2 dx  2 B1 y(0)=sech −1 ( 1 ) =cosh −1(2)=ln ( 2+ 3 ) M1 A1 Relates to cosh−1 and uses logarithmic form. 2 4 16 M1 Evaluates derivatives at x=0. y'(0)=− y''(0)= 3 3 3 ( ) 4 8 A1 y =ln 2+ 3 − x+ x2 3 3 3 Alternative method for question 7(b) dy 1 1 B1 = − = − dx ( x+ 1 )2 ( x+ 1 )−2 −1 ( x+ 1 ) 3 − x− x2 2 2 2 4 d2y = 1 ( x+ 1 )−1 ( 3 − x− x2 )−3 2 (−1−2x )+ ( 3 − x− x2 )−1 2 ( x+ 1 )−2 M1 A1 dx2 2 2 4 4 2 y(0)=sech −1 ( 1 ) =cosh −1(2)=ln ( 2+ 3 ) M1 A1 Relates to cosh−1 and uses logarithmic form. 2 © UCLES 2021 Page 11 of 13 7(b) 4 16 M1 Evaluates derivatives at x=0. y'(0)=− y''(0)= 3 3 3 ( ) 4 8 A1 y =ln 2+ 3 − x+ x2 3 3 3 7 Question Answer Marks Guidance

7(a) sech y = 1 = x+ 1  cosh y = ( x+ 1 )−1 B1 Relates to cosh y, cosh y 2 2 dy B1 Differentiates LHS. sinhy dx − ( x+ 1 )−2 B1 Differentiates RHS. AG. 2 3 7(b) d2y dy 2  1 −3 M1 A1 M1 A1 for LHS. sinh +coshy   =2  x+  B1 for RHS. dx2 dx  2 B1 y(0)=sech −1 ( 1 ) =cosh −1(2)=ln ( 2+ 3 ) M1 A1 Relates to cosh−1 and uses logarithmic form. 2 4 16 M1 Evaluates derivatives at x=0. y'(0)=− y''(0)= 3 3 3 ( ) 4 8 A1 y =ln 2+ 3 − x+ x2 3 3 3 Alternative method for question 7(b) dy 1 1 B1 = − = − dx ( x+ 1 )2 ( x+ 1 )−2 −1 ( x+ 1 ) 3 − x− x2 2 2 2 4 d2y = 1 ( x+ 1 )−1 ( 3 − x− x2 )−3 2 (−1−2x )+ ( 3 − x− x2 )−1 2 ( x+ 1 )−2 M1 A1 dx2 2 2 4 4 2 y(0)=sech −1 ( 1 ) =cosh −1(2)=ln ( 2+ 3 ) M1 A1 Relates to cosh−1 and uses logarithmic form. 2 © UCLES 2021 Page 11 of 13 7(b) 4 16 M1 Evaluates derivatives at x=0. y'(0)=− y''(0)= 3 3 3 ( ) 4 8 A1 y =ln 2+ 3 − x+ x2 3 3 3 7 Question Answer Marks Guidance

2 dy B1 Finds first derivative. =tanhx dx d2y B1 Finds second derivative. =sech2x dx2 d3y sinhx M1 A1 Finds third derivative. =−2tanhxsech2x=−2 dx3 cosh3x d4y cosh2 ( x )−3sinh2 ( x ) B1 Finds fourth derivative. Alternative: dx4 =−2 cosh4 ( x ) =−2sech4 x+4tanh2 xsech2 x 6tanh2 xsech2 x−2sech2 x. y(0)= y(1)(0)= y(3)(0)=0 y(2)(0)=1 y(4)(0)=−2 M1 Evaluates derivatives at x=0. y= 1 x2 − 1 x4 A1 CWO 2 12 7 © UCLES 2021 Page 7 of 15

1(a) sinh x2 +cosh x2 M1 Combines using correct power series. Or sensible attempt at all four derivatives of y. 1+x2 +1x4 A1 2 2 1(b) 1×4!=12 B1 2 1 1(c)  1 21+x2 + 1x4 dx M1 Substitutes their power series, must be at least a+bx2. 2 0 1 A1 =x+1x3 + 1 x52 = 523  3 10  0 960 2 © UCLES 2021 Page 6 of 14

8(a) ex −e −x 2 B1 Writes in exponential form. tanhx= sechx= ex +e −x ex +e −x ex −e −x  2 ( ex +e −x )2 − ( ex −e −x )2 4 M1 Writes over common denominator. 1−  ex +e −x   = ( ex +e −x )2 = ( ex +e −x )2 sech2 x A1 AG 3 8(b) u2 du= 1tanh3x+C M1 A1 Uses substitution correctly. Allow by inspection. 3 2 8(c) I = ln3 sechn−2 xsech2 xtanh2 xdx B1 Separates into correct structure. n 0 1sechn−2 xtanh3x ln3 +1(n−2) ln3 sechn−2 xtanh4 xdx M1 Uses integration by parts correctly. 3  0 3 0 1 ( 3 )n−2( 4 )3 + 1(n−2) ( I −I ) M1 A1 Uses 1−sech2x=tanh2x. 3 5 5 3 n−2 n ( 3+n−2 ) I = ( 3 )n−2( 4 )3 +(n−2)I leading to A1 AG n 5 5 n−2 (n+1)I = ( 3 )n−2( 4 )3 +(n−2)I n 5 5 n−2 5 © UCLES 2021 Page 13 of 14 8(d) I = 1 ( 4 )3 = 64 B1 2 3 5 375 (4+1)I = ( 3 )2( 4 )3 +2I leading to I = 4928 =0.105 M1 A1 Applies reduction formula with n=4. 4 5 5 2 4 46875 3 © UCLES 2021 Page 14 of 14

2(a) coshx 1  ex ex sinhx 1  ex ex B1 2 2 1  ex ex2 1 1  e2x e2x cosh2x M1 A1 Expands, AG. 2 2 3 2(b) 2sinh2 xksinhx10 M1 A1 Applies identity. k2 80 M1 A1 Sets discriminant positive. A1 k  8, k  8 5 © UCLES 2022 Page 6 of 13

2(a) coshx 1  ex ex sinhx 1  ex ex B1 2 2 1  ex ex2 1 1  e2x e2x cosh2x M1 A1 Expands, AG. 2 2 3 2(b) 2sinh2 xksinhx10 M1 A1 Applies identity. k2 80 M1 A1 Sets discriminant positive. A1 k  8, k  8 5 © UCLES 2022 Page 6 of 13

4(a) coshx= 1 ( ex +e−x) sinhx= 1 ( ex −e−x) B1 2 2 1 ( e2x +2+e−2x −e2x +2−e−2x) =1 M1 A1 Expands, AG. Clear LHS to RHS for A1. 4 3 © UCLES 2022 Page 7 of 13 4(b) d ( tan−1sinhx ) = coshx M1 A1 Applies d ( tan−1u ) = 1 . dx sinh2x+1 du u2 +1 coshx A1 AG. = =sechx cosh2x 3 y 4(c) B1 Correct shape, symmetrical about x=0. y=0 B1 Accept labels on their sketch. 2 4(d) n M1 A1 Compares sum with integral. Limits correct for n sechr sechxdx A1. 0 r=1 n A1 AG. =tan−1sinhx =tan−1sinhn   0 3 4(e) 1 B1 π 2 1 © UCLES 2022 Page 8 of 13

4(a) coshx= 1 ( ex +e−x) sinhx= 1 ( ex −e−x) B1 2 2 1 ( e2x +2+e−2x −e2x +2−e−2x) =1 M1 A1 Expands, AG. Clear LHS to RHS for A1. 4 3 © UCLES 2022 Page 7 of 13 4(b) d ( tan−1sinhx ) = coshx M1 A1 Applies d ( tan−1u ) = 1 . dx sinh2x+1 du u2 +1 coshx A1 AG. = =sechx cosh2x 3 y 4(c) B1 Correct shape, symmetrical about x=0. y=0 B1 Accept labels on their sketch. 2 4(d) n M1 A1 Compares sum with integral. Limits correct for n sechr sechxdx A1. 0 r=1 n A1 AG. =tan−1sinhx =tan−1sinhn   0 3 4(e) 1 B1 π 2 1 © UCLES 2022 Page 8 of 13

3 1 B1 ex dy 2 = dx 1 1− e2x 4  1 1  1  1  B1 1− e2x  ex − ex − e2x  d2y  4 2  2  2  = dx2  1  2 1− e2x   4  2 10 M1 Evaluates derivatives at x=0. f '(0)= f ''(0)= 3 9 1 1 3  M1 Uses logarithmic form of tanh−1. f(0)=tanh−1   = ln  2  2 2 2  1 2 5 M1 A1 1 ln3+ x+ x2 Applies f(x)= f(0)+ f '(0)x+ f ''(0)x2 2 3 9 2! 6 Question Answer Marks Guidance

5(a) dy M1 A1 Differentiating using product rule =−2xsech2xtanhx+sech2x and chain rule for M1. dx =1−tanh2x−2xtanhx+2xtanh3x   sech2x02xtanhx−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 M1 A1 Compares sum with integral. n rsech2r xsech2xdx Consistent limits for M1. 1 r=2  n xsech2xdx=xtanhxn − n tanhxdx M1 A1 Integrates by parts. 1 1 1 =xtanhx+lnsechxn A1 1 =ntanhn+lnsechn−(tanh1+lnsech1) A1 AG. Must have gained all previous marks. 6 Question Answer Marks Guidance

3 1 B1 ex dy 2 = dx 1 1− e2x 4  1 1  1  1  B1 1− e2x  ex − ex − e2x  d2y  4 2  2  2  = dx2  1  2 1− e2x   4  2 10 M1 Evaluates derivatives at x=0. f '(0)= f ''(0)= 3 9 1 1 3  M1 Uses logarithmic form of tanh−1. f(0)=tanh−1   = ln  2  2 2 2  1 2 5 M1 A1 1 ln3+ x+ x2 Applies f(x)= f(0)+ f '(0)x+ f ''(0)x2 2 3 9 2! 6 Question Answer Marks Guidance

2(a) xsinht y cosht B1 Differentiates x and y with respect to t. x2  y2 sinh2tcosh2t cosh2t M1 A1 Applies cosh2tsinh2tcosh2t. May use exponential form or  dy 2 1 2  cosh2t 1 2 1   dx1  sinhtdt.   dx    sinh2t  3 A1 AG s5 cosh2tdt 0 4 © Cambridge University Press & Assessment 2024 Page 6 of 15 2(b) d  cosh2t  sinh2t sech2t sinh2t(cosh2t)1 2 B1 Finds first derivative. dt d2  cosh2t  2 cosh2t tanh22t cosh2t 2 cosh2t sinh22tcosh2t3 2 B1 Finds second derivative dt2 3 3 M1 A1 Applies Maclaurin’s series. s51t2dt t1t35 0  3  0  84 0.672 A1 CWO. 125 5 Question Answer Marks Guidance

6(a) eu −e−u 2 B1 Writes in exponential from. tanhu= sechu= eu +e−u eu +e−u eu −e−u  2 ( eu +e−u)2 − ( eu −e−u)2 4 M1 Writes over common denominator. 1−  = = eu +e−u  ( eu +e−u)2 ( eu +e−u)2 =sech2u A1 Withold this mark if working with LHS and RHS simultaneously. 3 6(b) du M1 A1 Differentiates implicitly. u=sech−1tsechu=t−tanhusechu =1 dt du 1 1 M1 A1 Uses identity from (a), AG. =− =− dt tanhusechu t 1−t2 First alternative method for question 6(b) 1 du 1 M1 A1 Differentiates implicitly. u=sech−1tcoshu= sinhu =− t dt t2 du 1 1 1 M1 A1 Uses cosh2u−sinh2u=1, AG. =− =− =− dt t2 cosh2u−1 t2 t−2 −1 t 1−t2 © Cambridge University Press & Assessment 2025 Page 13 of 18 6(b) Second alternative method for question 6(b) 1 1 M1 A1 1 u=sech−1tcoshu= u=cosh−1   Differentiatescosh−1   . t t t du = 1  ( −t−2) =− 1 M1 A1 Applies formula with chain rule and simplifies, AG. dt (1)2 −1 t 1−t2 t Third alternative method for question 6(b) 1 1 1   1+ 1−t2  M1 A1 Uses logarithmic form of sech−1t sechu=tcoshu= u=ln + −1=ln  t  t t2   t      du  −1− 1−t2  t  1 M1 A1 Differentiates logarithmic form using chain and =  =− quotient rules, AG. dt   t2 1−t2    1+ 1−t2   t 1−t2 4 6(c) dy 1 dx 1 B1 B1 =sech−1t− , = dt 1−t2 dt 1−t2 dy = dy  dt =− 1−t2 + ( 1−t2) sech−1t M1 A1 Uses chain rule, AG. dx dt dx 4 © Cambridge University Press & Assessment 2025 Page 14 of 18 6(d) d dy t 1−t2 M1 A1 Applies product rule. Might see dt  dx   = 1−t2 − t 1−t2 −2tsech−1t 2t2 −1 −2tsech−1t. t 1−t2 d2y =t 1−t2 − ( 1−t2)3 2 −2t ( 1−t2) sech−1t M1 A1 Uses chain rule. dx2 t 4 © Cambridge University Press & Assessment 2025 Page 15 of 18

6(a) eu −e−u 2 B1 Writes in exponential from. tanhu= sechu= eu +e−u eu +e−u eu −e−u  2 ( eu +e−u)2 − ( eu −e−u)2 4 M1 Writes over common denominator. 1−  = = eu +e−u  ( eu +e−u)2 ( eu +e−u)2 =sech2u A1 Withold this mark if working with LHS and RHS simultaneously. 3 6(b) du M1 A1 Differentiates implicitly. u=sech−1tsechu=t−tanhusechu =1 dt du 1 1 M1 A1 Uses identity from (a), AG. =− =− dt tanhusechu t 1−t2 First alternative method for question 6(b) 1 du 1 M1 A1 Differentiates implicitly. u=sech−1tcoshu= sinhu =− t dt t2 du 1 1 1 M1 A1 Uses cosh2u−sinh2u=1, AG. =− =− =− dt t2 cosh2u−1 t2 t−2 −1 t 1−t2 © Cambridge University Press & Assessment 2025 Page 13 of 18 6(b) Second alternative method for question 6(b) 1 1 M1 A1 1 u=sech−1tcoshu= u=cosh−1   Differentiatescosh−1   . t t t du = 1  ( −t−2) =− 1 M1 A1 Applies formula with chain rule and simplifies, AG. dt (1)2 −1 t 1−t2 t Third alternative method for question 6(b) 1 1 1   1+ 1−t2  M1 A1 Uses logarithmic form of sech−1t sechu=tcoshu= u=ln + −1=ln  t  t t2   t      du  −1− 1−t2  t  1 M1 A1 Differentiates logarithmic form using chain and =  =− quotient rules, AG. dt   t2 1−t2    1+ 1−t2   t 1−t2 4 6(c) dy 1 dx 1 B1 B1 =sech−1t− , = dt 1−t2 dt 1−t2 dy = dy  dt =− 1−t2 + ( 1−t2) sech−1t M1 A1 Uses chain rule, AG. dx dt dx 4 © Cambridge University Press & Assessment 2025 Page 14 of 18 6(d) d dy t 1−t2 M1 A1 Applies product rule. Might see dt  dx   = 1−t2 − t 1−t2 −2tsech−1t 2t2 −1 −2tsech−1t. t 1−t2 d2y =t 1−t2 − ( 1−t2)3 2 −2t ( 1−t2) sech−1t M1 A1 Uses chain rule. dx2 t 4 © Cambridge University Press & Assessment 2025 Page 15 of 18

2(a) et −e−t 2 B1 tanht= secht= et +e−t et +e−t 2 M1 Writes in a single fraction over common et −e−t  4 e2t +e−2t +2   + = =1 denominator, AG. et +e−t  ( et +e−t)2 ( et +e−t)2 A1 Withold A1 mark if they used mixed variables within a single line of working. 3 © Cambridge University Press & Assessment 2025 Page 9 of 17 2(b) dx B1 =tanht dt dy cosht M1 A1 = =secht dt 1+sinh2t Applies 1+sinh2t=cosh2t. 1 1 M1 Substitutes their derivatives into a correct  tanh2t+sech2t dt = 1dt 0 0 2 2 dx dy formula for arc length eg. + dt     dt  dt  =1 A1 5 Question Answer Marks Guidance

2(a) 1+2sinh2 x=6sinh2 xsinhx=1 M1 A1 Sets equal and uses 2 cosh2x=1+2sinh2 x. Acceptcoshx= 1 5 or tanhx= 1 2 5 or substituting exponential form to reach e4x −3e2x +1=0 ore2x −3+e−2x =0 x=sinh−1(1)=ln ( 1 + 1 +1 ) M1 Uses logarithmic form or 2 2 4 ( ) e2x = 1 3 5 . 2 x=ln ( 1 + 15 ) A1 Must be exact and in the form lna, 2 2 where a is simplified. 4 © Cambridge University Press & Assessment 2025 Page 10 of 19 2(b) B1 C or C correct. 1 2 B1 C andintersections with C correct. 2 1 2 © Cambridge University Press & Assessment 2025 Page 11 of 19

1 © Cambridge University Press & Assessment 2026 Page 9 of 18 f '( x ) = ( 1 + ( x + 1 ) 2 ) − 12 B1 Finds first derivative. Accept x + 1 + 1 ( x + 1 ) 2 + 1  1 + 2 ( 2 x x + + 1 2 2 ) + 1  or implicit y 'c o s h y = 1 . f ''( x ) = − ( 1 + ( x + 1 ) 2 ) − 32 ( x + 1 ) B1 Finds second derivative. Allow not simplified or implicit y ''c o s h y + ( y )' 2 s i n h y = 0 . sinh−11=ln ( 1+ 2 ) B1 1 f '(0)= 2 f ''( 0 ) = − 2 1 2 M1 Evaluates their derivatives at zero only. Must have attempted both derivatives. s i n h − 1 ( x + 1 ) = ln ( 1 + 2 ) + 1 2 x − 4 1 2 x 2 A1 CWO. 5 Trace ID: PC-80D328BA4A Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/21 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

1 © Cambridge University Press & Assessment 2026 Page 9 of 18 f '( x ) = ( 1 + ( x + 1 ) 2 ) − 12 B1 Finds first derivative. Accept x + 1 + 1 ( x + 1 ) 2 + 1  1 + 2 ( 2 x x + + 1 2 2 ) + 1  or implicit y 'c o s h y = 1 . f ''( x ) = − ( 1 + ( x + 1 ) 2 ) − 32 ( x + 1 ) B1 Finds second derivative. Allow not simplified or implicit y ''c o s h y + ( y )' 2 s i n h y = 0 . sinh−11=ln ( 1+ 2 ) B1 1 f '(0)= 2 f ''( 0 ) = − 2 1 2 M1 Evaluates their derivatives at zero only. Must have attempted both derivatives. s i n h − 1 ( x + 1 ) = ln ( 1 + 2 ) + 1 2 x − 4 1 2 x 2 A1 CWO. 5 Trace ID: PC-45F835F92B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/22 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

5(a) © Cambridge University Press & Assessment 2026 Page 12 of 19 s e c h t = e t 2 + e − t t a n h t = e e t t − + e e − − t t B1 1 −  e t 2 + e − t  2 = ( e t + ( t e e + − t e ) − 2 t − 2 ) 4 = e 2 t + ( e 2 t + + − 2 t e 2 ) − t e − 4 = ( ( e e t t − + e e − − t t ) ) 2 2 M1 A1 Expands, gets to e 2 t + ( e 2 t + + − 2 t e 2 ) − t e − 4 for M1, AG. Variable used must be consistent throughout. A0 if a mixture of variables is used. 3 5(b) d d y t = t s i n h t + c o s h t , d d x t = t c o s h t + s i n h t B1 CAO d d y x = d d y t  d d t x = t t s c i n o h s h t t + + c s o i s n h h t t M1 Uses chain rule. t t s c i n o h s h t t + + c s o i s n h h t t  s s e e c c h h t t = t t a t + n h t a t n + h 1 t A1 Divides numerator and denominator by c o s h t , AG. 3 Trace ID: PC-D4137E1CBF Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/23 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 5(c) © Cambridge University Press & Assessment 2026 Page 13 of 19 d d t  d d y x  = ( t + t a n h t ) ( t s e c h 2 t + ( t a t n + h t ) t a n − h ( t 2 ) t t a n h t + 1 ) ( 1 + s e c h 2 t ) *M1 A1 Applies quotient rule.  t 2 s e c h 2 t + t t a n h t + t s e c h 2 t t a n h t + ( t t a + n t h a 2 t n h − t ) t 2 t a n h t − t t a n h t s e c h 2 t − 1 − s e c h 2 t   t 2 s e c h 2 t + ( t t a + n t h a 2 n t h − t ) s 2 e c h 2 t − 1  t 2 s e c h 2 t + ( 1 ( t − + s e c h t a n h ) 2 t 2 ) t − s e c h 2 t − 1 M1 Simplifies numerator to the form ( t2 −2 ) sechnt by applying t a n h 2 t = 1 − s e c h 2 t . = ( 2 t ( t − + 2 t ) s a n e h c t h ) 2 2 t A1 d d 2 x y 2 = ( 2 t ( t − + 2 t ) s a n e h c t h ) 2 2 t  t c o s h 1 t + s i n h t = ( 2 t ( t − + 2 t ) s a n e h c t h ) 3 3 t dM1 A1 dt Uses chain rule, multiplies by their for M1. dx Convincing argument to final result for A1 6 Trace ID: PC-D4137E1CBF Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/23 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

4(a) © Cambridge University Press & Assessment 2026 Page 13 of 19 x = 4 s i n h 2 t + 3 , y = 3 s i n h 2 t − 4 B1 Differentiates x and y with respect to t. x 2 + y 2 = ( 4 s i n h 2 t + 3 ) 2 + ( 3 s i n h 2 t − 4 ) 2 = 2 5 ( s i n h 2 2 t + 1 ) M1 Expands and simplifies x 2 + y 2 . Must be in terms of t. =25cosh22t A1 5  1 0 c o s h 2 t d t = 52  s i n h 2 t 1 0 = 54 ( e 2 − e − 2 ) M1A1 Applies correct formula for arc length. M1 for integrating the correct form a c o s h 2 t . 5 4(b) y x = 3 4 s s i n i n h h 2 2 t t − + 4 3 B1 Finds first derivative. d 3sinh2t−4 (4sinh2t+3)(6cosh2t)−(3sinh2t−4)(8cosh2t) *M1  = Differentiates dt4sinh2t+3 (4sinh2t+3)2 d d y x with respect to t. d d y x must have the correct form a c s s i n i n h h 2 2 t t + + b d . 24sinh2tcosh2t+18cosh2t−24sinh2tcosh2t+32cosh2t A1 Correct numerator expanded showing that (4sinh2t+3)2 24sinh2tcosh2twill cancel.   50cosh2t =   (4sinh2t+3)2    d d 2 x y 2 = d d t  3 4 s s i n i n h h 2 2 t t − + 4 3   d d t x = ( 4 5 s 0 i n c h o s 2 h t 2 + t 3 ) 3 DM1A1 Applies chain rule. 5 Trace ID: PC-F3F6C8368E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/2a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/24 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/