Integration — Paper 2 — A Level Mathematics - Further 9231 topical questions

Question 1

y 1 x 0 1 2 n - 1 1 n n n The diagram shows the curve with equation y = x2 for 0 G x G 1 , together with a set of n rectangles of width 1. …

Accepted Answer

4(a) ( )( ) ( )( ) ( )( ) ( )( ) 1 2 2 2 2 2 1 1 1 1 2 1 1 0 d − < + + + +   n n n n n n n n n n x x M1 A1 2 2 3 3 2 1 1 ( 1)(2 1) 2 3 1 6 6 n r n n n n n r n n n = + + + + = =  M1 A1 4 4(b) ( )( ) ( )( ) ( )( ) 1 2 2 2 2 1 1 1 1 2 1 0 d − > + + +   n n n n n n n x x M1 A1 1 2 2 3 3 2 1 1 ( 1)( )(2 2 1) 2 3 1 6 6 − = − − + − + = = =  n r n n n n n r n n n M1 A1 4

Question 2

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…

Accepted Answer

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

Question 3

= 1 - x d x .6 The integral In, where n is an integer, is defined by I n 2 2 2 0 (a) Find the exact value of I1. [2] ......................…

Accepted Answer

6(a) ( ) 1 1 2 2 1 2 2 1 1 0 6 0 1 1 d sin π I x x x − −   = − = =    M1 A1 2 6(b) ( ) ( ) ( ) 1 1 1 2 2 2 2 2 2 2 1 d 1 1 1 d − − − −   − = − + −     n n n x x nx x x x M1 A1 ( ) ( )( ) ( ) 1 1 2 2 1 2 2 2 1 1 1 1 − − − = − − − + − n n n x x x M1 ( ) 1 1 2 2 2 2 0 1 + −   − = − +     n n n n x x nI nI I M1 ( ) 1 2 1 2 3 1 2 2 4 1 2 ( 1) 2 3 ( 1) − + − − + = − −  = + − n n n n n n n nI n I nI n I A1 5 6(c) 1 2 3 3 − = I B1 3 2 5 10 7 2 3 2 5 3 2 3 2 3 − = +  = √ I I I M1 A1 3

Question 4

y 1 x 0 1 2 n - 1 1 n n n The diagram shows the curve with equation y = x2 for 0 G x G 1 , together with a set of n rectangles of width 1. …

Accepted Answer

4(a) ( )( ) ( )( ) ( )( ) ( )( ) 1 2 2 2 2 2 1 1 1 1 2 1 1 0 d − < + + + +   n n n n n n n n n n x x M1 A1 2 2 3 3 2 1 1 ( 1)(2 1) 2 3 1 6 6 n r n n n n n r n n n = + + + + = =  M1 A1 4 4(b) ( )( ) ( )( ) ( )( ) 1 2 2 2 2 1 1 1 1 2 1 0 d − > + + +   n n n n n n n x x M1 A1 1 2 2 3 3 2 1 1 ( 1)( )(2 2 1) 2 3 1 6 6 − = − − + − + = = =  n r n n n n n r n n n M1 A1 4

Question 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…

Accepted Answer

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

Question 6

= 1 - x d x .6 The integral In, where n is an integer, is defined by I n 2 2 2 0 (a) Find the exact value of I1. [2] ......................…

Accepted Answer

6(a) ( ) 1 1 2 2 1 2 2 1 1 0 6 0 1 1 d sin π I x x x − −   = − = =    M1 A1 2 6(b) ( ) ( ) ( ) 1 1 1 2 2 2 2 2 2 2 1 d 1 1 1 d − − − −   − = − + −     n n n x x nx x x x M1 A1 ( ) ( )( ) ( ) 1 1 2 2 1 2 2 2 1 1 1 1 − − − = − − − + − n n n x x x M1 ( ) 1 1 2 2 2 2 0 1 + −   − = − +     n n n n x x nI nI I M1 ( ) 1 2 1 2 3 1 2 2 4 1 2 ( 1) 2 3 ( 1) − + − − + = − −  = + − n n n n n n n nI n I nI n I A1 5 6(c) 1 2 3 3 − = I B1 3 2 5 10 7 2 3 2 5 3 2 3 2 3 − = +  = √ I I I M1 A1 3

Question 7

= 1 + 3 x e d x , where n is an integer.2 Let I n -3 x 0 (a) Show that 3I n n -3 = 1 - 4 e + 3nI n -1 . [3] ...............................…

Accepted Answer

2(a) ( ) ( ) 1 1 1 3 3 1 3 0 0 e 1 3 e 1 3 d n n x x nI x n x x − − −   = − + + +    M1 A1 3 3 1 1 1 1 3 3 e 4 3 1 4 e 3 n n n n n nI I nI − − − − − + +  = − + A1 3 Question Answer Marks 2(b) ( ) 1 1 3 3 3 1 1 0 3 3 0 0e d e 1 e x x I x − − −   = = − = −    . B1 ( ) ( ) 3 3 3 1 1 1 3 3 1 4e 1 e 2 5e I − − − = − + − = − ( ) ( ) 3 3 1 2 3 1 16e 2 2 5e I − − = − + − M1 ( ) 3 1 2 3 5 26e I − = − A1 3

Question 8

y x 0 1 2 3 N - 1 N The diagram shows the curve with equation y = ln x for x H 1, together with a set of ( N - 1) rectangles of unit width.…

Accepted Answer

4(a) + + + M1 1 ln d N x x > M1 [ ]1 1 1 ln d ln 1d ln 1 N N N x x x x x N N N = − = − +   M1 A1 ln ! ln 1 N N N N > − + A1 5 Question Answer Marks 4(b) ln1 ln 2 ln( 1) + + + −  N (or ln 2 ln( 1) + + −  N ) M1 1 ln d N x x < (or 2 ln d N x x < ) M1 ln ! ln 1 ln ( 1) ln 1 N N N N N N N N < − + + = + − + (or ln ! ( 1) ln 2(1 ln 2) N N N N < + − + − ) A1 3

Question 9

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] ................…

Accepted Answer

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

Question 10

i - 6 cos 4i + 15 cos 2i - 10) . [6]8 (a) Use de Moivre’s theorem to show that sin 6 i =- 321 ( cos 6 .....................................…

Accepted Answer

8(a) 1 2i sin z z θ − − = B1 ( ) ( ) ( ) ( ) 6 1 6 6 4 4 2 2 6 15 20 z z z z z z z z − − − − − = + − + + + − M1 A1 ( ) ( ) ( ) 6 2isin 2cos6 6 2cos4 15 2cos2 20 θ θ θ θ = − + − M1 A1 ( ) 6 1 32 sin cos6 6cos4 15cos2 10 θ θ θ θ = − + − + A1 6 8(b) 1 3 6 6 1 1 4 4 0 cos sin d x x x π +  = 1 3 1 8 0 3cos 5d x x π +  M1 A1 [ ] ( ) 1 3 5 3 1 1 8 8 3 2 0 3sin 5 π 3 x x π + = + √ . M1 A1 4 8(c) 2 2 cos 1 sin c c θ θ =  − = B1 ( ) 3 6 2 16 16 1 13 0 6cos4 3 0 c c θ + − − =  −= M1 A1 4θ = 5 7 1 11 3 3 3 3 π, π, π, π M1 ( ) ( ) ( ) ( ) 5 7 1 11 12 12 12 12 cos π , cos π ,cos π ,cos π c = A1 5

Question 11

(a) By differentiating e - x2 , find the Maclaurin’s series for e - x2 up to and including the term in x2. [5] ............................…

Accepted Answer

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

Question 12

y 1 x 0 1 2 n - 1 1 n n n The diagram shows the curve with equation y = 1 - x 3 for 0 G x G 1, together with a set of n rectangles of width…

Accepted Answer

4(a) ( ) 3 1 1 3 3 3 0 1 1 1 1 d 1 ... 1 1   −     − + − + + −          n n x x n n n n  M1 A1 Forms the sum of the areas of the rectangles. ( ) 4 2 2 1 3 4 1 1 1 1 1 4 − = − = − = −  n r n n r n n M1 Applies ( ) 1 2 2 1 4 1 3 1 . − = = −  n r r n n ( ) 2 2 2 2 2 4 1 3 2 1 4 4 − − + − = = n n n n n n A1 AG 4 4(b) ( ) 3 3 3 3 3 3 1 0 1 1 1 1 2 1 1 1 1 . . 1 d .     −     − + − + + −             −  n n n n n x x n n  M1 A1 Forms the sum of the areas of appropriate rectangles. ( ) 4 2 2 1 3 4 1 1 1 1 1 4 − = − − − = − = −  n r n n n n r n n n n M1 Applies ( ) 1 3 1 2 2 1 1 . 4 − = = −  n r r n n 2 2 2 2 4 ( 1) ( 1) 3 2 1 4 4 − − − − − = = n n n n n n n A1 4

Question 13

(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 c…

Accepted Answer

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

Question 14

A curve has equation y = cosh x , for 0 G x G 12 . Find, in terms of r and e, the area of the surface generated when the curve is rotated t…

Accepted Answer

2 2 2 d 1 1 sinh cosh d   + = + =     y x x x M1 A1 Applies 2 2 cosh 1 sinh = + x x 1 2 2 0 2 c π osh d x x  M1 Correct formula, correct limits. 1 0 2 cosh 2 1d π x x = +  M1 Applies 2 2cosh cosh2 1 = + x x or expands ( ) 2 2 e e− + x x 1 2 0 1 π sinh 2 2 x x +   =     A1 Correct integration. ( ) 1 π e 1 2 4 e− − = + A1 6

Question 15

y 0 1 2 n - 1 n x 1 The diagram shows the curve y = for x H 0 , together with a set of n rectangles of unit x 2 + x + 1 width. By consideri…

Accepted Answer

8 Sum ( ) 2 1 1 1 3 7 1 = + + + √ √ √ + +  n n M1A1 Forms sum of the areas of the rectangles, need to see consideration of areas, not just the sum given in the question. ( ) 2 0 1 1 d + < √ +  n x x x M1 Compares with integral. Limits have to be correct. 2 2 1 3 1 2 4   + + = + +     x x x B1 Completes the square 1 1 2 0 0 0 1 d sinh s 1 2 1 2 1 3 2 4 inh 1 3 3 2 − −   +   +           = =            + +        √     √      n n n x x x x M1 A1 Uses formula ( ) ( ) 0 2 0 2 ln l 2 1 2 1 2 1 3 3 2 n 1 3 3                 = =         √ √  + + √ + + + +       +  + n n x x x x x M1 A1 Uses logarithmic form of 1 sinh . − ( ) 2 3 ln l 1 n 3 2 1 2 3       = −     √ √  + + +  √ +   n n n M1 Inserts limits. ( ) 2 1 2 2 ln 3 3 3 1   = + + √     + + n n n A1 AG 10

Question 16

(a) By differentiating e - x2 , find the Maclaurin’s series for e - x2 up to and including the term in x2. [5] ............................…

Accepted Answer

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

Question 17

y 1 x 0 1 2 n - 1 1 n n n The diagram shows the curve with equation y = 1 - x 3 for 0 G x G 1, together with a set of n rectangles of width…

Accepted Answer

4(a) ( ) 3 1 1 3 3 3 0 1 1 1 1 d 1 ... 1 1   −     − + − + + −          n n x x n n n n  M1 A1 Forms the sum of the areas of the rectangles. ( ) 4 2 2 1 3 4 1 1 1 1 1 4 − = − = − = −  n r n n r n n M1 Applies ( ) 1 2 2 1 4 1 3 1 . − = = −  n r r n n ( ) 2 2 2 2 2 4 1 3 2 1 4 4 − − + − = = n n n n n n A1 AG 4 4(b) ( ) 3 3 3 3 3 3 1 0 1 1 1 1 2 1 1 1 1 . . 1 d .     −     − + − + + −             −  n n n n n x x n n  M1 A1 Forms the sum of the areas of appropriate rectangles. ( ) 4 2 2 1 3 4 1 1 1 1 1 4 − = − − − = − = −  n r n n n n r n n n n M1 Applies ( ) 1 3 1 2 2 1 1 . 4 − = = −  n r r n n 2 2 2 2 4 ( 1) ( 1) 3 2 1 4 4 − − − − − = = n n n n n n n A1 4

Question 18

(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 c…

Accepted Answer

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

Question 19

y 0 1 2 n - 1 1 x n n n The diagram shows the curve with equation y = x3 for 0 G x G 1, together with a set of n rectangles of width 1. n 1…

Accepted Answer

3(a) ( )( ) ( )( ) ( )( ) ( )( ) 1 3 0 3 3 3 3 1 1 1 1 2 1 1 d n n n n n n n n n n x x − < + + + +   Need last term for A1. 2 2 3 4 4 1 2 1 ( 1) 1 2 4 n r n n n r n n n = + + = =        M1 A1 Applies 3 1 4 1 2 2 ( 1) , = = +  n r r n n AG. 4 3(b) ( )( ) ( )( ) ( )( ) 1 3 0 3 3 3 1 1 1 1 2 1 d n n n n n n n x x − > + + +   M1 A1 Forms the sum of the areas of appropriate rectangles. Need the last term for A1. 2 2 3 4 4 2 1 1 1 ( 1) 1 2 4 n r n n n r n n n − = − − = =        M1 A1 Applies 3 1 4 1 2 2 ( 1) . = = +  n r r n n Accept 2 1 1. 2 n n n + −       4 3(c) 2 2 3 3 1 1 1 10 leading to 10 2 2 n n n n n n − + − − = < >             M1 Simplifies their − n n U L to . k n Least value of n is 1001. A1 2

Question 20

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 . […

Accepted Answer

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

Question 21

y 0 1 2 n - 1 1 x n n n The diagram shows the curve with equation y = x3 for 0 G x G 1, together with a set of n rectangles of width 1. n 1…

Accepted Answer

3(a) ( )( ) ( )( ) ( )( ) ( )( ) 1 3 0 3 3 3 3 1 1 1 1 2 1 1 d n n n n n n n n n n x x − < + + + +   Need last term for A1. 2 2 3 4 4 1 2 1 ( 1) 1 2 4 n r n n n r n n n = + + = =        M1 A1 Applies 3 1 4 1 2 2 ( 1) , = = +  n r r n n AG. 4 3(b) ( )( ) ( )( ) ( )( ) 1 3 0 3 3 3 1 1 1 1 2 1 d n n n n n n n x x − > + + +   M1 A1 Forms the sum of the areas of appropriate rectangles. Need the last term for A1. 2 2 3 4 4 2 1 1 1 ( 1) 1 2 4 n r n n n r n n n − = − − = =        M1 A1 Applies 3 1 4 1 2 2 ( 1) . = = +  n r r n n Accept 2 1 1. 2 n n n + −       4 3(c) 2 2 3 3 1 1 1 10 leading to 10 2 2 n n n n n n − + − − = < >             M1 Simplifies their − n n U L to . k n Least value of n is 1001. A1 2

Question 22

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 . […

Accepted Answer

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

Question 23

y 0 1 2 3 N – 1 N x x The diagram shows the curve y = 2 for x H 1, together with a set of N - 1 rectangles of unit 2x - 1 width. (a) By con…

Accepted Answer

3(a) 2 1 2 2 1 2 1 2 1 = = = + − −   N N r r r r r r M1 A1 Compares with sum of the areas of the rectangles. < 1 2 1 d 2 1 + −  N x x x M1 Compares with integral. ( ) 2 1 4 2 1 1 d ln 2 1 2 1   = −   −  N N x x x x M1 A1 Finds integral. ( ) 1 4 2 2 1 1 ln 2 1 2 1 = < + − −  N r r N r M1 A1 Inserts limits, AG. 7 3(b) 1 2 2 2 2 1 1 2 1 d 2 1 2 1 2 1 2 1 2 1 − = = = + > + − − − − −    N N N r r r r N x N x r r N x N M1 A1 Compares with integral. ( ) 1 4 2 2 ln 2 1 2 1 N N N − + − A1 CWO. Alternative method for question 3(b) 1 1 1 2 2 d 2 1 2 1 + = > − −   N N r r x x r x M1 A1 Compares with integral. ( ) ( ) 1 4 2 ln 2 1 1 + − N A1 CWO. 3

Question 24

= 4 + x dx .7 The integral In, where n is an integer, is defined by I n 2 2 2 y 0 (a) Find the exact value of I1, expressing your answer in…

Accepted Answer

7(a) ( ) ( ) 3 3 1 2 2 2 2 1 1 2 0 0 1 4 sinh I x x − −     = + =  M1 A1 Recognises integral or uses appropriate substitution. ln 2 A1 Insert limits, must simplify. 3 7(b) ( ) ( ) ( ) 1 1 1 2 2 2 2 2 2 2 1 d 4 4 4 d − − − −   + = − + + +     n n n x x nx x x x M1 A1 Uses the product rule to differentiate. ( )( ) ( ) 1 1 2 2 2 2 2 1 4 4 4 4 n n n x x x − − − − + − + + + M1 Applies 2 2 4 4. = + − x x ( ) 3 1 2 2 2 2 0 4 4 − +   + = − + +     n n n n x x nI nI I M1 Integrates both sides using the limits given. ( ) ( ) 3 3 2 2 2 5 2 5 2 2 (1 ) 4 4 ( 1) + + = − +  = + − n n n n n n n I nI nI n I A1 Substitutes limits and rearranges, AG. Alternative method for question 7(b) ( ) ( ) 3 3 1 2 2 2 1 2 1 2 2 2 0 0 4 4 d − − −   = + + +      n n n I x x n x x x M1 A1 Integrates by parts. ( ) ( )( ) 1 3 3 1 2 2 2 2 1 0 2 2 0 2 4 4 4 4 d − − −   = + + + − +      n n n I x x n x x x M1 Applies 2 2 4 4. = + − x x ( ) ( ) 3 3 2 2 2 5 2 2 5 2 4 4 ( 1) + + = + −  = + − n n n n n n n I nI I nI n I M1 A1 Substitutes limits and rearranges, AG. 5 Question Answer Marks Guidance 7(c) 0 3 3 2 = I B1 Applies reduction formula with 1. = n ( ) 3 8 33 2 125 10 5 00 3 5 12 2 0.033 = +  = = I I I M1 A1 Applies reduction formula with 3. = n 3

Question 25

y 0 1 2 3 4 N – 1 N x ln x The diagram shows the curve with equation y = 2 for x H 2 , together with a set of ( N - 2) rectangles x of unit…

Accepted Answer

4(a) 2 1 3 2 ln ln2 ln 4 = = = +   N N r r r r r r M1 A1 Compares with sum of the areas of the rectangles. M1 for writing out sum, A1 for considering 3 2 ln . = N r r r < 2 2 ln 2 ln d 4 + N x x x M1 Compares with integral. 2 2 2 ln ln 1 d +   = −      N N x x x x x M1 A1 Finds integral. 1 2 ln ln2 ln 1 ln2 1 2 3ln2 1 ln 4 2 4 = + + + +   < + − + = −      N r r N N N N r M1 A1 Inserts limits, AG. A1 requires second M1. 7 4(b) 2 2 2 2 2 1 2 1 2 ln ln ln ln ln d − = = = + > +    N N N r r r r N x N x r r N x N M1 A1 Compares with integral. Lower limit of 1 scores M0. Accept 2 1 2 1 2 ln ln . + = >   N N r r x r x 2 ln 2 1 ln 1 ln 2 + + = − + N N N N A1 Accept ln2 1 ln( 1) 1. 2 1 + + + − + N N 3

Question 26

(a) Starting from the definition of cosh in terms of exponentials, prove that 2 cosh 2 A = cosh 2A + 1. [3] ...............................…

Accepted Answer

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

Question 27

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…

Accepted Answer

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

Question 28

y 1 0 1 2 n - 1 1 x n n n The diagram shows the curve with equation y = 1 - x 2 for 0 G x G 1, together with a set of n rectangles of width…

Accepted Answer

3(a) ( ) ( ) 2 2 2 1 ( 1) 2 1 1 1 1 01 d 1 1 − − < + − + + −   n n n n n n x x M1 A1 Forms the sum of the areas of the rectangles. Three terms, including last for A1. Allow use of sigma. 1 2 3 3 1 1 ( 1) (2 1) 1 1 6 n r n n n r n n − = − − − = −  M1 Applies 1 2 1 6 1 ( 1) (2 1). − = = − −  n r r n n n 2 2 2 2 6 ( 1)(2 1) 4 3 1 6 6 n n n n n n n − − − + − = A1 AG. [Can get this A1 without previous A1.] 4 Question Answer Marks Guidance 3(b) ( ) ( ) ( ) ( ) 2 2 2 2 2 1 1 1 0 2 1 1 1 2 1 1 1 d 1 − − − > − + − + +   n n n n n n n x x M1 A1 Forms the sum of the areas of appropriate rectangles. Three terms, including last for A1. Allow ( ) 2 2 1 1 1 . = −  n r n n r 1 2 3 3 1 1 1 1 ( 1) (2 1) 6 n r n n n n n r n n n n − = − − − − − = −  M1 Applies 1 2 1 6 1 ( 1) (2 1). − = = − −  n r r n n n 2 2 2 6 ( 1) ( 1)(2 1) 4 3 1 6 6 n n n n n n n n − − − − − − = A1 SC: 2 2 4 3 1 1 6 + −− n n n n scores 2/4. 4

Question 29

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)…

Accepted Answer

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

Question 30

(a) Starting from the definitions of tanh and sech in terms of exponentials, prove that 1 - tanh 2 x = sec h 2 x . [3] ....................…

Accepted Answer

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

Question 31

y 0 1 2 3 4 N – 1 N x ln x The diagram shows the curve with equation y = 2 for x H 2 , together with a set of ( N - 2) rectangles x of unit…

Accepted Answer

4(a) 2 1 3 2 ln ln2 ln 4 = = = +   N N r r r r r r M1 A1 Compares with sum of the areas of the rectangles. M1 for writing out sum, A1 for considering 3 2 ln . = N r r r < 2 2 ln 2 ln d 4 + N x x x M1 Compares with integral. 2 2 2 ln ln 1 d +   = −      N N x x x x x M1 A1 Finds integral. 1 2 ln ln2 ln 1 ln2 1 2 3ln2 1 ln 4 2 4 = + + + +   < + − + = −      N r r N N N N r M1 A1 Inserts limits, AG. A1 requires second M1. 7 4(b) 2 2 2 2 2 1 2 1 2 ln ln ln ln ln d − = = = + > +    N N N r r r r N x N x r r N x N M1 A1 Compares with integral. Lower limit of 1 scores M0. Accept 2 1 2 1 2 ln ln . + = >   N N r r x r x 2 ln 2 1 ln 1 ln 2 + + = − + N N N N A1 Accept ln2 1 ln( 1) 1. 2 1 + + + − + N N 3

Question 32

(a) Starting from the definition of cosh in terms of exponentials, prove that 2 cosh 2 A = cosh 2A + 1. [3] ...............................…

Accepted Answer

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

Question 33

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

Accepted Answer

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

Question 34

(a) Use the substitution u = 1 - ( i - 1) 2 to find i - 1 di . [3] y 2 1 - ( i - 1) .......................................................…

Accepted Answer

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

Question 35

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…

Accepted Answer

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

Question 36

y 0 1 2 3 n – 1 n x 1 The diagram shows the curve y = for x 2 0 , together with a set of ( n - 1) rectangles of unit 2 width. x + 2x By con…

Accepted Answer

6 1 1 1 n 1 M1A1 Forms sum of the areas of the rectangles. At least = Sum = + + + 2  2 three terms, including first and last for A1.    n + 2 n ( ) r = 2 r + 2 r SC Areas of these three rectangles written down but sum not clearly formed scores B1. n M1 Compares with integral.  1 x 2 + 2 x  1  ( d x ) 2 2 B1 Completes the square. x + 2 x = ( x + 1) − 1 n M1 A1 Uses formula. Correct limits for A1. n  1 −1 dx =  cosh ( x + 1)  1 ( x + 1) 2 − 1  1  2  n M1 A1 Uses logarithmic form of cosh −1 . ln x + 1 + x + 2 x  ( ) 1 2 M1 Inserts limits. ln n + 1 + n + 2 n − ln 2 + 3 ( ) ( ) n 1 2 1 A1 Adds term from r = 1.  ln n + 1 + n + 2 n − ln 2 + 3 + 3 3 AG ( )  2 ) ( r =1 r + 2 r 10

Question 37

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

Accepted Answer

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

Question 38

(a) Use the substitution u = 1 - ( i - 1) 2 to find i - 1 di . [3] y 2 1 - ( i - 1) .......................................................…

Accepted Answer

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

Question 39

(a) By considering the binomial expansion of ( z + z -1 ) 4 , where z = cos i + i sin i , use de Moivre’s i + 4 cos 2i + 3) . [5] theorem t…

Accepted Answer

3(a) 1 2cos z z     B1 Use of 1 2cos z z     .       4 1 4 4 2 2 4 6 z z z z z z          M1 A1 Expands and groups. M1 A0 for no clear grouping. Correct substitution of n co is s i nz n n     for each term scores M1 A1.     4 2cos 2cos4 4 2cos2 6       M1 Substitutes 2cos . n n z z n    If n co is s i nz n n     used then must cancel sin.   4 1 8 cos cos4 4cos2 3       A1 AG. SC B1 expands   4 i c s os in    and uses trigonometric identities. 5 3(b)   1 1 3 1 2 6 2 6 π π 1 8 0 2 1 4 4 0 0 1 d cos d sin 4 2sin 2 3 x x                M1 A1 Applies substitution (M1) gets to 4 cos d   , changes limits, integration correct (A1).   9 1 16 4 3 π  A1 3

Question 40

= ( 1 + x ) d x .4 The integral In is defined by I n 5 n 0 d 5 n (a) By considering x ( 1 + x ) or otherwise, show that d x ` j, ( 5n + 1) …

Accepted Answer

4(a)       1 5 5 5 5 d 1 5 1 1 d n n n x x nx x x x       M1 A1 Uses the product rule to differentiate.      1 5 5 5 5 1 1 1 1 n n n x x x       M1* Uses 5 5 1 1 x x  .   0 5 1 1 1 5 5 n n n n x x nI nI I            DM1 Integrates both sides using the limits given. Requires previous method mark. 1 1 2 (5 1) 5 (5 1) 2 5 n n n n n n n I nI n I nI          A1 Substitutes limits and rearranges. AG. Alternative method for question 4(a)       1 1 1 5 0 0 1 5 0 5 5 1 d 1 5 1 d n n n nI x x x x n x x x                M1 A1 Integrates by parts.       1 1 5 5 5 1 1 0 0 0 1 5 1 d 5 1 d n n n nI x x n x x n x x                M1* Uses 5 5 1 1 x x  .   1 1 0 5 1 5 5 n n n n I x x nI nI            DM1 Forms recursive formula. Requires previous method mark. 1 (5 1) 2 5 n n n n I nI     A1 Substitutes limits and rearranges. AG. 5 Question Answer Marks Guidance 4(b) 6 1 7 1 1 6 6 0 I x x        or 0 1 I  B1 2 47 2 1 2 33 11 2 10 I I I     M1 A1 Applies reduction formula. 6 2 3 323 3 3 17 16 2 15 I I I     A1 4

Question 41

(a) Use the substitution u = x 2 - 1 to find d x . [3] x 2 - 1 ............................................................................…

Accepted Answer

7(a) 1 2 2 1 d d 1 x x u u x        2 1 x C  A1 Allow with “ C  ” missing. Answer must be in terms of x. 3 7(b)   1 2 cosh ln 1 r r r     B1 1 1 1 cosh 2 cosh 3 cosh N        M1 Forms sum of the areas of the rectangles. 1 1 cosh d N x x   M1 Compares with integral with correct limits. 1 1 2 1 1 1 cosh d cosh d 1 N N N x x x x x x x            A1 Evaluates integral.     2 2 2 2 ln 1 ln 1 1 N r r r N N N N          A1 AG. 5 Question Answer Marks Guidance 7(c) 1 1 1 (cosh 1) cosh 2 cosh ( 1) N         1 1 cosh d N x x   (or 1 2 cosh d N x x   ) M1 A1 Compares with integral with correct limits.     2 2 2 2 ln 1 ( 1)ln 1 1 N r r r N N N N           or       2 2 2 2 ln 1 ( 1)ln 1 1 2ln 2 3 3 N r r r N N N N             A1 Adds   2 ln 1 N N   to both sides. Alternative method for question 7(c) 1 1 1 1 (cosh 1) cosh 2 cosh ( 1) cosh N N             1 1 1 2 1 2 1 cosh d ln 1 1 N N x x x x x x                 (or 1 2 1 cosh d N x x    ) M1 A1 Compares with integral with correct limits.      2 2 2 2 ln 1 1 ln 1 2 2 N r r r N N N N N N           or      2 2 1 ln 1 2 2 2ln 2 3 3 N N N N N N          A1 Second alternative method for question 7(c) 1 1 1 cosh (1 1) cosh (2 1) cosh ( 1 1) N                1 2 1 1 2 cosh 1 d 1 ln 1 2 2 N N x x x x x x x x                 M1 A1 Compares with integral with correct limits.      2 2 1 ln 1 2 2 2ln 2 3 3 N N N N N N          A1 3

Question 42

(a) By considering the binomial expansion of ( z + z -1 ) 4 , where z = cos i + i sin i , use de Moivre’s i + 4 cos 2i + 3) . [5] theorem t…

Accepted Answer

3(a) 1 2cos z z     B1 Use of 1 2cos z z     .       4 1 4 4 2 2 4 6 z z z z z z          M1 A1 Expands and groups. M1 A0 for no clear grouping. Correct substitution of n co is s i nz n n     for each term scores M1 A1.     4 2cos 2cos4 4 2cos2 6       M1 Substitutes 2cos . n n z z n    If n co is s i nz n n     used then must cancel sin.   4 1 8 cos cos4 4cos2 3       A1 AG. SC B1 expands   4 i c s os in    and uses trigonometric identities. 5 3(b)   1 1 3 1 2 6 2 6 π π 1 8 0 2 1 4 4 0 0 1 d cos d sin 4 2sin 2 3 x x                M1 A1 Applies substitution (M1) gets to 4 cos d   , changes limits, integration correct (A1).   9 1 16 4 3 π  A1 3

Question 43

= ( 1 + x ) d x .4 The integral In is defined by I n 5 n 0 d 5 n (a) By considering x ( 1 + x ) or otherwise, show that d x ` j, ( 5n + 1) …

Accepted Answer

4(a)       1 5 5 5 5 d 1 5 1 1 d n n n x x nx x x x       M1 A1 Uses the product rule to differentiate.      1 5 5 5 5 1 1 1 1 n n n x x x       M1* Uses 5 5 1 1 x x  .   0 5 1 1 1 5 5 n n n n x x nI nI I            DM1 Integrates both sides using the limits given. Requires previous method mark. 1 1 2 (5 1) 5 (5 1) 2 5 n n n n n n n I nI n I nI          A1 Substitutes limits and rearranges. AG. Alternative method for question 4(a)       1 1 1 5 0 0 1 5 0 5 5 1 d 1 5 1 d n n n nI x x x x n x x x                M1 A1 Integrates by parts.       1 1 5 5 5 1 1 0 0 0 1 5 1 d 5 1 d n n n nI x x n x x n x x                M1* Uses 5 5 1 1 x x  .   1 1 0 5 1 5 5 n n n n I x x nI nI            DM1 Forms recursive formula. Requires previous method mark. 1 (5 1) 2 5 n n n n I nI     A1 Substitutes limits and rearranges. AG. 5 Question Answer Marks Guidance 4(b) 6 1 7 1 1 6 6 0 I x x        or 0 1 I  B1 2 47 2 1 2 33 11 2 10 I I I     M1 A1 Applies reduction formula. 6 2 3 323 3 3 17 16 2 15 I I I     A1 4

Question 44

(a) Use the substitution u = x 2 - 1 to find d x . [3] x 2 - 1 ............................................................................…

Accepted Answer

7(a) 1 2 2 1 d d 1 x x u u x        2 1 x C  A1 Allow with “ C  ” missing. Answer must be in terms of x. 3 7(b)   1 2 cosh ln 1 r r r     B1 1 1 1 cosh 2 cosh 3 cosh N        M1 Forms sum of the areas of the rectangles. 1 1 cosh d N x x   M1 Compares with integral with correct limits. 1 1 2 1 1 1 cosh d cosh d 1 N N N x x x x x x x            A1 Evaluates integral.     2 2 2 2 ln 1 ln 1 1 N r r r N N N N          A1 AG. 5 Question Answer Marks Guidance 7(c) 1 1 1 (cosh 1) cosh 2 cosh ( 1) N         1 1 cosh d N x x   (or 1 2 cosh d N x x   ) M1 A1 Compares with integral with correct limits.     2 2 2 2 ln 1 ( 1)ln 1 1 N r r r N N N N           or       2 2 2 2 ln 1 ( 1)ln 1 1 2ln 2 3 3 N r r r N N N N             A1 Adds   2 ln 1 N N   to both sides. Alternative method for question 7(c) 1 1 1 1 (cosh 1) cosh 2 cosh ( 1) cosh N N             1 1 1 2 1 2 1 cosh d ln 1 1 N N x x x x x x                 (or 1 2 1 cosh d N x x    ) M1 A1 Compares with integral with correct limits.      2 2 2 2 ln 1 1 ln 1 2 2 N r r r N N N N N N           or      2 2 1 ln 1 2 2 2ln 2 3 3 N N N N N N          A1 Second alternative method for question 7(c) 1 1 1 cosh (1 1) cosh (2 1) cosh ( 1 1) N                1 2 1 1 2 cosh 1 d 1 ln 1 2 2 N N x x x x x x x x                 M1 A1 Compares with integral with correct limits.      2 2 1 ln 1 2 2 2ln 2 3 3 N N N N N N          A1 3

Question 45

(a) Find the Maclaurin series for sin -1 x up to and including the term in x3. [5] ........................................................…

Accepted Answer

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

Question 46

y 1 O 1 2 n - 1 1 x n n n The diagram shows the curve with equation y = ( 1 - x) 2 for 0 G x G 1, together with a set of n rectangles of wi…

Accepted Answer

6(a)   2 2 2 2 1 0 0 1 1 1 1 1 1 d 1 1 1 n x x n n n n n n            M1 A1 Forms the sum of the areas of the rectangles. Must have first two terms and last term for A1. Accept   2 1 1 n n n  for the last term. Accept written in summation form with correct limits.     2 2 2 3 1 3 1 1 1 1 1 1 1 2 n n r r n r n nr r n n n n             3 1 1 2 1 2 1 2 1 1 n n r r r r n n               M1 A1 Setting up correct series so that formulae from MF19 can be applied. Can also recognise sum as 1 2 3 1 n r r n      2 2 2 2 3 3 3 2 2 1 1 1 ( 1)(2 1) 1 2 3 1 2 3 1 6 6 6 n n n n n n n n n n n n n n n n n n               A1 AG. Must have gained previous accuracy mark. 5 Question Answer Partial Marks Guidance 6(b)         2 1 1 1 1 1 1 0 2 2 2 2 1 d 1 1 1 n n n n n n n x x            M1 A1 Forms the sum of the areas of appropriate rectangles. Must have first two and last terms for A1. Accept   1 2 1 n n n  for the last term. Accept written in summation form with correct limits.       1 2 2 2 2 3 3 3 1 3 1 1 1 ( 1)(2 1) 2 6 n r n n n n n n n n nr r n n n n             1 2 2 3 1 2 1 1 n n r r r r n n             M1 Setting up correct series so that formulae from MF19 can be applied. Recognising sum as 1 2 2 3 1 3 1 1 1 1 n n r r r r n n n        without wrong working scores M1 A1 M1. 2 2 2 3 1 6 n n n   A1 4 6(c) 1 n n U L n   M1 Simplifies n n U L  to . c n l 0 n  as n  A1 AG. 2

Question 47

= ( 1 + x ) d x .7 The integral In, where n is an integer, is defined by I n 3 2 21 n 0 (a) Find the exact value of I -1 giving your answer…

Accepted Answer

7(a)   1 4 4 3 3 2 2 1 1 0 0 1 d sinh ln3 I x x x             M1 A1 Recognises integral. 2 7(b)       1 1 1 2 2 2 2 2 1 2 2 d 1 1 1 d n n n x x nx x x x             M1 A1 Uses the product rule to differentiate.      1 1 2 2 2 2 1 2 1 1 1 1 n n n x x x       M1* Uses 2 2 1 1 x x  .   4 1 3 2 2 0 2 1 n n n n x x nI nI I            DM1 Integrates both sides using the limits given. Requires previous method mark.       2 2 5 5 4 4 1 1 3 3 3 3 n n n n n n n I nI n I nI          A1 Substitutes limits and rearranges. AG. 5 7(c) 2 3 2 0 1 4 d s x x    M1 Forms correct integral with correct limits. 4 3 1 0 2 1 1 1 d 2 2 u u I    A1 AG.  1 1 5 4 2 3 3 I I   M1 Applies reduction formula with 1 n . 5 1 ln3 4 9 s   A1 4

Question 48

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] .............…

Accepted Answer

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

Question 49

(a) Starting from the definitions of cosh and sinh in terms of exponentials, prove that sinh 2x = 2 sinh x cosh x . [3] ...................…

Accepted Answer

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

Question 50

(a) State the sum of the series 1 + z + z 2 + f + z n - 1 , for z ! 1. [1] ................................................................…

Accepted Answer

8(a) nz − 1 B1 z − 1 1 8(b) nz − 1 cos n−+1 isi n n B1 = z − 1 cos− 1 + i sin ( cos n−+1 isin n)( cos−−1 isin) M1 Multiplies numerator and denominator by complex ( cos−+1 isin)( cos−−1 isin) conjugate.  nz − 1  cos ncos+ sin nsin− cos n− cos+ 1 M1 Takes real part. Re   = 2 2 cos( n − 1)= cos ncos+ sin nsin  z − 1  ( cos− 1) + sin  cos ncos+ sin nsin− cos n 1 A1 = + 2 (1 − cos) 2 cos n( cos− 1) + sin nsin 1 M1 Factorises. = + 2 (1 − cos) 2 8(b) 1  sin nsin M1 A1 Divides through by denominator. AG. = 1 − cos n+   2  1 − cos  Alternative method for question 8(b) z n − 1 e i n − 1 B1 = z − 1 e i − 1 1 1 1 1 1 ( n − 2 ) e i − e − i 1 2 cos ( n − 2 )+ isin ( n − 2 )− cos 2 + isin ( 2 ) M1 = 2 e i 1 2 − e − i 1 2isin 12   z n − 1  sin( n − 12 )+ sin 12  M1 Takes real part Re   = 1  z − 1  2sin 2  1 sin ( n − 2 ) 1 A1 = + 2sin 12  2 sin ncos 12 − cos nsin 12  1 M1 Uses compound angle identity = + 2 sin 12  2 sin ncos 12  1 1 M1 Divides through by denominator. = − cos n+ 2sin 12  2 2 sin nsin 1 1 s i n nsin 1 1 A1 AG. si n = 2 sin 12 cos 12  and − cos n+ = − cos n+ 1 − cos) 2 2 4s in 2 12  2 2 2 ( 2 sin 2 12 = 1 − cos. 7 8(c) 1 1 1 1 1 2 1 n − 1 M1 A1 Forms sum of areas of rectangles given in the diagram. A0 if comparison with integral missing or unclear. 0 cos x dx  n + n cos n + n cos n +  n cos n  n 1  M1 A1 sin sin Applies result from part (b) with = 1. AG. 1  1 2 n − 1  1  n n n  n =  1 + cos + cos + + cos  =  1 − cos +  n  n n n  2 n  n 1 − cos 1   n  4 8(d) 1 1 1 1 2 1 n M1 A1 Forms sum of areas of rectangles. A0 if comparison with integral missing or unclear. 0 cos x dx  n cos n + n cos n + n cos n  1   1  A1 1  sin1sin n  1 1 1  sin1sin n  =  1 − cos1 +  + cos1 − =  cos1 −+1  2 n 1 n n 2 n 1  1 − cos   1 − cos   n   n  3

Question 51

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 …

Accepted Answer

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

Question 52

(a) Starting from the definitions of cosh and sinh in terms of exponentials, prove that 2 sinh 2 A = cosh 2 A - 1. [3] ....................…

Accepted Answer

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

Question 53

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] .............…

Accepted Answer

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

Question 54

(a) Starting from the definitions of cosh and sinh in terms of exponentials, prove that sinh 2x = 2 sinh x cosh x . [3] ...................…

Accepted Answer

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

Question 55

(a) State the sum of the series 1 + z + z 2 + f + z n - 1 , for z ! 1. [1] ................................................................…

Accepted Answer

8(a) nz − 1 B1 z − 1 1 8(b) nz − 1 cos n−+1 isi n n B1 = z − 1 cos− 1 + i sin ( cos n−+1 isin n)( cos−−1 isin) M1 Multiplies numerator and denominator by complex ( cos−+1 isin)( cos−−1 isin) conjugate.  nz − 1  cos ncos+ sin nsin− cos n− cos+ 1 M1 Takes real part. Re   = 2 2 cos( n − 1)= cos ncos+ sin nsin  z − 1  ( cos− 1) + sin  cos ncos+ sin nsin− cos n 1 A1 = + 2 (1 − cos) 2 cos n( cos− 1) + sin nsin 1 M1 Factorises. = + 2 (1 − cos) 2 8(b) 1  sin nsin M1 A1 Divides through by denominator. AG. = 1 − cos n+   2  1 − cos  Alternative method for question 8(b) z n − 1 e i n − 1 B1 = z − 1 e i − 1 1 1 1 1 1 ( n − 2 ) e i − e − i 1 2 cos ( n − 2 )+ isin ( n − 2 )− cos 2 + isin ( 2 ) M1 = 2 e i 1 2 − e − i 1 2isin 12   z n − 1  sin( n − 12 )+ sin 12  M1 Takes real part Re   = 1  z − 1  2sin 2  1 sin ( n − 2 ) 1 A1 = + 2sin 12  2 sin ncos 12 − cos nsin 12  1 M1 Uses compound angle identity = + 2 sin 12  2 sin ncos 12  1 1 M1 Divides through by denominator. = − cos n+ 2sin 12  2 2 sin nsin 1 1 s i n nsin 1 1 A1 AG. si n = 2 sin 12 cos 12  and − cos n+ = − cos n+ 1 − cos) 2 2 4s in 2 12  2 2 2 ( 2 sin 2 12 = 1 − cos. 7 8(c) 1 1 1 1 1 2 1 n − 1 M1 A1 Forms sum of areas of rectangles given in the diagram. A0 if comparison with integral missing or unclear. 0 cos x dx  n + n cos n + n cos n +  n cos n  n 1  M1 A1 sin sin Applies result from part (b) with = 1. AG. 1  1 2 n − 1  1  n n n  n =  1 + cos + cos + + cos  =  1 − cos +  n  n n n  2 n  n 1 − cos 1   n  4 8(d) 1 1 1 1 2 1 n M1 A1 Forms sum of areas of rectangles. A0 if comparison with integral missing or unclear. 0 cos x dx  n cos n + n cos n + n cos n  1   1  A1 1  sin1sin n  1 1 1  sin1sin n  =  1 − cos1 +  + cos1 − =  cos1 −+1  2 n 1 n n 2 n 1  1 − cos   1 − cos   n   n  3

Question 56

= sech x d x .4 It is given that, for n H 0 , I n n 0 (a) Show that, for n H 2 , 3 n - 2 4 n - 1 I n = + n - 2 I n - 2 . [5] ` j b 5 l b 5 …

Accepted Answer

4(a) l 2 n3 2 0 sech sech d n nI x x x   B1 Separates into correct structure. ln3 ln 0 2 2 2 0 2 2 3 ln3 0 sech sech d sech tanh ( 2) sech tanh d n n n nI x x x x x n x x x              *M1 Uses integration by parts correctly.     3 4 5 5 2 2 ( 2) n n n n I I      DM1 A1 Uses 2 2 . 1 sech tanh x x         2 2 2 2 3 3 4 4 5 5 5 5 1 2 ( 2) ( 1) ( 2) n n n n n n n I n I n I n I               A1 AG 5 Question Answer Marks Guidance 4(b) 4 2 5 I  B1  2 3 4 4 2 5 4 5 236 3 2 0.629 375 I I I      M1 A1 Applies reduction formula with 4. n  3

Question 57

y 0 1 2 n - 1 1 x n n n The diagram shows the curve with equation y = 2x - x 2 for 0 G x G 1, together with a set of n 1 rectangles of widt…

Accepted Answer

5(a)      1 1 1 1 1 2 2 1 0 2 2 2 2 2 d 2 2 2 n n n n n n n n n n n x x x                 M1 A1 Forms the sum of the areas of the rectangles, may be in summation form. M1 is for correct number of rectangles. 2 3 2 3 2 1 1 1 6 2 1 1 ( 1) ( 1)(2 1) n n n n n n r r r r n n n n n            M1 A1 Applies formulae from MF19. Summations must have (correct) limits for A1. Substituted in correctly once for M1.      2 1 1 1 1 1 2 1 2 1 1 6 3 6 3 2 6 1 1 2 1 n n n n n n n           A1 AG. This A1 requires the previous A1. 5 5(b)              1 1 1 1 1 1 2 2 2 2 1 2 2 1 0 2 d 2 2 2 n n n n n n n n n n n x x x             M1 A1 Forms the sum of the areas of appropriate rectangles. M1 is for seeing correct height of last rectangle. 2 3 2 3 1 1 2 2 1 1 1 6 1 1 ( 1)( ) ( 1)( )(2 1) n n n n n n r r r r n n n n n             M1 Applies formulae from MF19, substitutes correct limit. Substituted in correctly once for M1.    2 1 2 1 2 1 1 3 6 3 2 6 1 n n n n      A1    1 2 1 1 3 6 1 n n n    or fully expanded or fully factorised. 4 Question Answer Marks Guidance 5(c)      1 2 1 1 2 1 1 3 6 3 6 1 1 ( ) n n n n n n n U L         M1 Expresses their n n U L  in terms of 1 . n For, M1 they must taking the limit of an expression that tends to a constant. (Must be fully expanded if they do not take the limit at any stage.) 1 0 n  as n  or 2 2 0 3 3   A1 AG, CWO. Their n L must be correct. 2

Question 58

(a) Use the substitution u = 1 + x 2 to find x dx . [2] y 2 1 + x .........................................................................…

Accepted Answer

7(a)     1 2 2 2 1 d d 1 1 x x u u C x C u x             2 7(b) 1 s d d inh y y x x x x    B1 Divides through by .x 1 1 ln d e e x x x x        M1 A1 Finds integrating factor.   1 1 d sinh d x y x x    M1 Correct form on LHS.   d d Iy x for their integrating factor I. 1 1 2 sinh d 1 x y x x x x x       M1 A1 Integrates RHS. RHS must be of the form 1 sinh c x  . 1 1 2 sinh 1 x y x x C x       A1   1 2 sinh ln 1 x x x       1 ln 1 2 2 C     *M1 Substitutes initial conditions.     2 1 2 1 sinh ln 1 2 1 2 y x x x x x x         DM1 A1 Divides through by their integrating factor. Accept   2 1 2 1 1 sinh 1 1 2 . sinh y x x x x x x         10

Question 59

= sech x d x .4 It is given that, for n H 0 , I n n 0 (a) Show that, for n H 2 , 3 n - 2 4 n - 1 I n = + n - 2 I n - 2 . [5] ` j b 5 l b 5 …

Accepted Answer

4(a) l 2 n3 2 0 sech sech d n nI x x x   B1 Separates into correct structure. ln3 ln 0 2 2 2 0 2 2 3 ln3 0 sech sech d sech tanh ( 2) sech tanh d n n n nI x x x x x n x x x              *M1 Uses integration by parts correctly.     3 4 5 5 2 2 ( 2) n n n n I I      DM1 A1 Uses 2 2 . 1 sech tanh x x         2 2 2 2 3 3 4 4 5 5 5 5 1 2 ( 2) ( 1) ( 2) n n n n n n n I n I n I n I               A1 AG 5 Question Answer Marks Guidance 4(b) 4 2 5 I  B1  2 3 4 4 2 5 4 5 236 3 2 0.629 375 I I I      M1 A1 Applies reduction formula with 4. n  3

Question 60

y 0 1 2 n - 1 1 x n n n The diagram shows the curve with equation y = 2x - x 2 for 0 G x G 1, together with a set of n 1 rectangles of widt…

Accepted Answer

5(a)      1 1 1 1 1 2 2 1 0 2 2 2 2 2 d 2 2 2 n n n n n n n n n n n x x x                 M1 A1 Forms the sum of the areas of the rectangles, may be in summation form. M1 is for correct number of rectangles. 2 3 2 3 2 1 1 1 6 2 1 1 ( 1) ( 1)(2 1) n n n n n n r r r r n n n n n            M1 A1 Applies formulae from MF19. Summations must have (correct) limits for A1. Substituted in correctly once for M1.      2 1 1 1 1 1 2 1 2 1 1 6 3 6 3 2 6 1 1 2 1 n n n n n n n           A1 AG. This A1 requires the previous A1. 5 5(b)              1 1 1 1 1 1 2 2 2 2 1 2 2 1 0 2 d 2 2 2 n n n n n n n n n n n x x x             M1 A1 Forms the sum of the areas of appropriate rectangles. M1 is for seeing correct height of last rectangle. 2 3 2 3 1 1 2 2 1 1 1 6 1 1 ( 1)( ) ( 1)( )(2 1) n n n n n n r r r r n n n n n             M1 Applies formulae from MF19, substitutes correct limit. Substituted in correctly once for M1.    2 1 2 1 2 1 1 3 6 3 2 6 1 n n n n      A1    1 2 1 1 3 6 1 n n n    or fully expanded or fully factorised. 4 Question Answer Marks Guidance 5(c)      1 2 1 1 2 1 1 3 6 3 6 1 1 ( ) n n n n n n n U L         M1 Expresses their n n U L  in terms of 1 . n For, M1 they must taking the limit of an expression that tends to a constant. (Must be fully expanded if they do not take the limit at any stage.) 1 0 n  as n  or 2 2 0 3 3   A1 AG, CWO. Their n L must be correct. 2

Question 61

(a) Use the substitution u = 1 + x 2 to find x dx . [2] y 2 1 + x .........................................................................…

Accepted Answer

7(a)     1 2 2 2 1 d d 1 1 x x u u C x C u x             2 7(b) 1 s d d inh y y x x x x    B1 Divides through by .x 1 1 ln d e e x x x x        M1 A1 Finds integrating factor.   1 1 d sinh d x y x x    M1 Correct form on LHS.   d d Iy x for their integrating factor I. 1 1 2 sinh d 1 x y x x x x x       M1 A1 Integrates RHS. RHS must be of the form 1 sinh c x  . 1 1 2 sinh 1 x y x x C x       A1   1 2 sinh ln 1 x x x       1 ln 1 2 2 C     *M1 Substitutes initial conditions.     2 1 2 1 sinh ln 1 2 1 2 y x x x x x x         DM1 A1 Divides through by their integrating factor. Accept   2 1 2 1 1 sinh 1 1 2 . sinh y x x x x x x         10

Question 62

A curve has equation y = e x for ln 4 G x G ln 12 . The area of the surface generated when the curve is 3 5 rotated through 2r radians abou…

Accepted Answer

3(a) ln 125 x 2 x 125 2 1 M1A1 Correct formula and applies substitution, AG. u 1 + u ( u ) du A = 2 π ln 43 e 1 + e dx = 2π 43 2 3(b) u = sinh v  du = cosh v dv B1 ln5 ln5 2 2 M1A1 Applies substitution and cosh 2 x = 1 + sinh 2 x . A = 2 π ln3 1 + sinh v cosh v d v A = 2 π ln3 co sh v d v Need limits for A1. ln5 M1 Applies 2cosh 2 x = cosh2 x + 1. π ln3 ( cosh 2v + 1) d v ln5 A1 Correct integration. 1 = π   2 sinh2v + v  ln3 312 904 π ( 50 + ln5 − 1840 − ln3) = π ( 225 + ln 53 ) A1 AG. 6

Question 63

y 1 0 x 1 2 N - 1 1 N N N x The diagram shows the curve with equation y = b 1 l for 0 G x G 1, together with a set of N rectangles 2 1 each…

Accepted Answer

6(a)  1 1 x  1 1 N1 1 1 N2 1 1 NN−1 1 1 NN M1A1 Forms the sum of the areas of the rectangles. M1 ( N )( 2 ) + ( N )( 2 ) + + ( N )( 2 ) + ( N )( 2 ) for correct number of rectangles.  2 ) dx    0 (  1 N N 1 1 1 1 N M1A1 N n r r N − 1 1 ( 1 ( ( ) 2 ) 2 − 1) ( 2 )( 2 ) n 1 1 N = r = , AG. 1 = ( 2 ) = = N   N N 1 1 r − 1 n =1 n =1 2 N 2 N1 − 1 1 − N N ( N − N ( 2 ) 2 ) ( ) . Applies 4 N − 2 N 1 x N N6(b) 1 1 1 1 1 N 1 1 1 1 0 ( 2 ) d x  ( N ) + ( N )( 2 ) + + ( N )( 2 ) + ( N )( 2 ) −1 M1A1 Formsrectangles.the sum of the areas of appropriate N −1 N n 1 1 M1A1 N −1 n r N − 2 − 1 1 2 1 1 N r = 1. = 1 ( 2 ) =  N  N N 1 r − 1 1 n = 0 n = 0 2 N 2 N1 − 1 1 − N 2 N N ( 2 ) 1 − ( 2 ) ( ) Applies ) = ( 4 6(c) N1 M1 c 2 1 1 .  10 −3  2 N  103 Simplifies U N − LN to 1 − 1 = N N N 2 N 2 N 2 − 1 2 N 2 − 1 ( ) ( ) Least value of N is 500 A1 CWO. 2 N .6(d) LN  2ln21  U N  2U1 N  ln2  2 L1N M1A1FT Forms inequality. FT on their U 0.693 = 2U1 N  ln2  2 L1N = 0.694 A1A1 CWO. Must have used N = 500. 4

Question 64

(a) By considering the binomial expansion of bz + l , where z = cos i + isin i , use de Moivre’s z theorem to show that cos 7i = a cos 7 i …

Accepted Answer

8(a) z + z −1 = 2cos B1 For LHS. −1 7 7 −7 5 −5 3 −3 −1 M1A1 Expands and groups. A0 if not grouped clearly. z + z = z + z + 7 z + z + 21 z + z + 35 z + z ( ) ( ) ( ) ( ) ( ) 27 cos7 = 2cos 7+ 7 ( 2cos 5) + 21( 2cos3) + 35 ( 2cos) M1 Substitutes z n + z − n = 2cos n. cos7 = 641 cos7+ 647 cos5+ 6421 cos3+ 3564 cos A1 5 8(b) 14 π n −1 n −1 14 π 14 π n − 2 2 M1A1 Applies integration by parts. cos cosd =  cos sin cos sin d   0 nI = 0 + ( n − 1) 0 1 n −1 4 π 14 π n − 2 2 M1 Applies sin 2 = 1 − cos 2  cos  1 − cos  d ( ) I n =  cos sin 0 + ( n − 1) 0 − n2 − n2 A1 AG. I n = 2 + (n − 1) I n − 2 − (n − 1) I n  nI n = 2 + ( n − 1) I n − 2 4 8(c) − 92 1 − 12 M1A1 Applies reduction formula from (b). 2 + 8 I 7 9 I 9 = 2 + 8 I 7 = 16 ( ) 14π 1 7 21 35 M1 Applies identity from (a). Allow 64 cos7+ 64 cos5+ 64 cos 3+ 64 cosd missing/incorrect limits. I 7 = 0 14π A1 1 1 7 21 I 7 = 64  7 sin 7+ 5 sin 5+ 3 sin3+ 35sin 0 1 − 12 2 2 − 12 − 12 A1 Check their exact answer when 9I is the subject, 2 −2− 1 −2− 1 2 + 35 2 + 81 9 I 9 = 161 ( ) 7 ( ) + 75 ( ) + 213 ( ) ( ) ) ( like terms collected. ISW. 0.402237 2 2 − 1 = 128670080 2 I 9 = 50286740 ( ) 5

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Questions147 pages

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Questions as text

Q1 · Y 1 x 0 1 2 n - 1 1 n n n The diagram shows the curve with equation y = x2 for 0 G x G 1… 9231/21 May/June 2020

Q2 · The curves C 1 : y = cosh x and C 2 : y = sinh 2x intersect at the point where x = a 9231/21 May/June 2020

Q3 · = 1 - x d x .6 The integral In, where n is an integer, is defined by I n 2 2 2 0 (a) Find… 9231/21 May/June 2020

Q4 · Y 1 x 0 1 2 n - 1 1 n n n The diagram shows the curve with equation y = x2 for 0 G x G 1… 9231/22 May/June 2020

Q5 · The curves C 1 : y = cosh x and C 2 : y = sinh 2x intersect at the point where x = a 9231/22 May/June 2020

Q6 · = 1 - x d x .6 The integral In, where n is an integer, is defined by I n 2 2 2 0 (a) Find… 9231/22 May/June 2020

Q7 · = 1 + 3 x e d x , where n is an integer.2 Let I n -3 x 0 (a) Show that 3I n n -3 = 1 - 4… 9231/23 May/June 2020

Q8 · Y x 0 1 2 3 N - 1 N The diagram shows the curve with equation y = ln x for x H 1… 9231/23 May/June 2020

Q9 · The curve C has parametric equations x = 12 t 2 1 - ln t , y = t2 + 1, for 2 G t G 2 9231/23 May/June 2020

Q10 · I - 6 cos 4i + 15 cos 2i - 10) 9231/23 May/June 2020

Q11 · By differentiating e - x2 , find the Maclaurin’s series for e - x2 up to and including… 9231/21 Oct/Nov 2020

Q12 · Y 1 x 0 1 2 n - 1 1 n n n The diagram shows the curve with equation y = 1 - x 3 for 0 G x… 9231/21 Oct/Nov 2020

Q13 · Sketch the graph of y = coth x for x 2 0 and state the equations of the asymptotes 9231/21 Oct/Nov 2020

Q14 · A curve has equation y = cosh x , for 0 G x G 12 9231/22 Oct/Nov 2020

Q15 · Y 0 1 2 n - 1 n x 1 The diagram shows the curve y = for x H 0 , together with a set of n… 9231/22 Oct/Nov 2020

Q16 · By differentiating e - x2 , find the Maclaurin’s series for e - x2 up to and including… 9231/23 Oct/Nov 2020

Q17 · Y 1 x 0 1 2 n - 1 1 n n n The diagram shows the curve with equation y = 1 - x 3 for 0 G x… 9231/23 Oct/Nov 2020

Q18 · Sketch the graph of y = coth x for x 2 0 and state the equations of the asymptotes 9231/23 Oct/Nov 2020

Q19 · Y 0 1 2 n - 1 1 x n n n The diagram shows the curve with equation y = x3 for 0 G x G 1… 9231/21 May/June 2021

Q20 · The curve C has parametric equations - x = 2 cosh t , y = 32 t 1 4 sinh t2 , for 0 G t G 1 9231/21 May/June 2021

Q21 · Y 0 1 2 n - 1 1 x n n n The diagram shows the curve with equation y = x3 for 0 G x G 1… 9231/22 May/June 2021

Q22 · The curve C has parametric equations - x = 2 cosh t , y = 32 t 1 4 sinh t2 , for 0 G t G 1 9231/22 May/June 2021

Q23 · Y 0 1 2 3 N – 1 N x x The diagram shows the curve y = 2 for x H 1, together with a set of… 9231/23 May/June 2021

Q24 · = 4 + x dx .7 The integral In, where n is an integer, is defined by I n 2 2 2 y 0 (a)… 9231/23 May/June 2021

Q25 · Y 0 1 2 3 4 N – 1 N x ln x The diagram shows the curve with equation y = 2 for x H 2… 9231/21 Oct/Nov 2021

Q26 · Starting from the definition of cosh in terms of exponentials, prove that 2 cosh 2 A =… 9231/21 Oct/Nov 2021

Q27 · It is given that y = sinh ( x 2 ) + cosh ( x 2 ) 9231/22 Oct/Nov 2021

Q28 · Y 1 0 1 2 n - 1 1 x n n n The diagram shows the curve with equation y = 1 - x 2 for 0 G x… 9231/22 Oct/Nov 2021

Q29 · The curve C has parametric equations + bt , for 1 G t G 2 , x = 3 t + 2t -1 + at 3 , y =… 9231/22 Oct/Nov 2021

Q30 · Starting from the definitions of tanh and sech in terms of exponentials, prove that 1… 9231/22 Oct/Nov 2021

Q31 · Y 0 1 2 3 4 N – 1 N x ln x The diagram shows the curve with equation y = 2 for x H 2… 9231/23 Oct/Nov 2021

Q32 · Starting from the definition of cosh in terms of exponentials, prove that 2 cosh 2 A =… 9231/23 Oct/Nov 2021

Q33 · The curve C has parametric equations = 4e ( t - 2 ) , for 0 G t G 2 9231/21 Oct/Nov 2022

Q34 · Use the substitution u = 1 - ( i - 1) 2 to find i - 1 di 9231/21 Oct/Nov 2022

Q35 · G x G 1 9231/22 Oct/Nov 2022

Q36 · Y 0 1 2 3 n – 1 n x 1 The diagram shows the curve y = for x 2 0 , together with a set of… 9231/22 Oct/Nov 2022

Q37 · The curve C has parametric equations = 4e ( t - 2 ) , for 0 G t G 2 9231/23 Oct/Nov 2022

Q38 · Use the substitution u = 1 - ( i - 1) 2 to find i - 1 di 9231/23 Oct/Nov 2022

Q39 · By considering the binomial expansion of ( z + z -1 ) 4 , where z = cos i + i sin i , use… 9231/21 May/June 2023

Q40 · = ( 1 + x ) d x .4 The integral In is defined by I n 5 n 0 d 5 n (a) By considering x ( 1… 9231/21 May/June 2023

Q41 · Use the substitution u = x 2 - 1 to find d x 9231/21 May/June 2023

Q42 · By considering the binomial expansion of ( z + z -1 ) 4 , where z = cos i + i sin i , use… 9231/22 May/June 2023

Q43 · = ( 1 + x ) d x .4 The integral In is defined by I n 5 n 0 d 5 n (a) By considering x ( 1… 9231/22 May/June 2023

Q44 · Use the substitution u = x 2 - 1 to find d x 9231/22 May/June 2023

Q45 · Find the Maclaurin series for sin -1 x up to and including the term in x3 9231/23 May/June 2023

Q46 · Y 1 O 1 2 n - 1 1 x n n n The diagram shows the curve with equation y = ( 1 - x) 2 for 0… 9231/23 May/June 2023

Q47 · = ( 1 + x ) d x .7 The integral In, where n is an integer, is defined by I n 3 2 21 n 0… 9231/23 May/June 2023

Q48 · The curve C has parametric equations 3 2 2 - t2 1 , y = t2 + 5 , for 0 1 t G 3 9231/21 Oct/Nov 2023

Q49 · Starting from the definitions of cosh and sinh in terms of exponentials, prove that sinh… 9231/21 Oct/Nov 2023

Q50 · State the sum of the series 1 + z + z 2 + f + z n - 1 , for z ! 9231/21 Oct/Nov 2023

Q51 · Y M O x The diagram shows part of the curve y = x sech 2 x and its maximum point M 9231/22 Oct/Nov 2023

Q52 · Starting from the definitions of cosh and sinh in terms of exponentials, prove that 2… 9231/22 Oct/Nov 2023

Q53 · The curve C has parametric equations 3 2 2 - t2 1 , y = t2 + 5 , for 0 1 t G 3 9231/23 Oct/Nov 2023

Q54 · Starting from the definitions of cosh and sinh in terms of exponentials, prove that sinh… 9231/23 Oct/Nov 2023

Q55 · State the sum of the series 1 + z + z 2 + f + z n - 1 , for z ! 9231/23 Oct/Nov 2023

Q56 · = sech x d x .4 It is given that, for n H 0 , I n n 0 (a) Show that, for n H 2 , 3 n - 2… 9231/21 May/June 2024

Q57 · Y 0 1 2 n - 1 1 x n n n The diagram shows the curve with equation y = 2x - x 2 for 0 G x… 9231/21 May/June 2024

Q58 · Use the substitution u = 1 + x 2 to find x dx 9231/21 May/June 2024

Q59 · = sech x d x .4 It is given that, for n H 0 , I n n 0 (a) Show that, for n H 2 , 3 n - 2… 9231/22 May/June 2024

Q60 · Y 0 1 2 n - 1 1 x n n n The diagram shows the curve with equation y = 2x - x 2 for 0 G x… 9231/22 May/June 2024

Q61 · Use the substitution u = 1 + x 2 to find x dx 9231/22 May/June 2024

Q62 · A curve has equation y = e x for ln 4 G x G ln 12 9231/21 Oct/Nov 2024

Q63 · Y 1 0 x 1 2 N - 1 1 N N N x The diagram shows the curve with equation y = b 1 l for 0 G x… 9231/21 Oct/Nov 2024

Q64 · By considering the binomial expansion of bz + l , where z = cos i + isin i , use de… 9231/21 Oct/Nov 2024

Q65 · Find the value of dx , giving your answer in the form ln ( a + b) , where a and b are 2… 9231/22 Oct/Nov 2024

Q66 · The curve C has parametric equations 1 t2 1 3 1 x = e - t - , y = 2et ( t - 1 ) , for 0 G… 9231/22 Oct/Nov 2024

Q67 · 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… 9231/22 Oct/Nov 2024

Q68 · A curve has equation y = e x for ln 4 G x G ln 12 9231/23 Oct/Nov 2024

Q69 · Y 1 0 x 1 2 N - 1 1 N N N x The diagram shows the curve with equation y = b 1 l for 0 G x… 9231/23 Oct/Nov 2024

Q70 · By considering the binomial expansion of bz + l , where z = cos i + isin i , use de… 9231/23 Oct/Nov 2024

Q71 · Let I = `1 - xj sinh x dx , where n is a non-negative integer 9231/21 May/June 2025

Q72 · Y x 0 1 2 3 n-1 n 1 x The diagram shows the curve with equation y = e for x H 1, together… 9231/21 May/June 2025

Q73 · Let I = `1 - xj sinh x dx , where n is a non-negative integer 9231/22 May/June 2025

Q74 · Y x 0 1 2 3 n-1 n 1 x The diagram shows the curve with equation y = e for x H 1, together… 9231/22 May/June 2025

Q75 · Starting from the definitions of tanh and sech in terms of exponentials, prove that tanh… 9231/23 May/June 2025

Q76 · Y 0 x 1 2 n - 1 1 n n n 1 The diagram shows the curve with equation y = for 0 G x G 1… 9231/23 May/June 2025