Summation of series — Paper 2 · A Level Mathematics - Further 9231

! 0, 1, - 1. [2]7 (a) Show that z 2 r = -1 , for z z - z r =1 … … … … … … … … … … … … … … … … … … … … … … … … … … (b) By letting z = cos i + i sin i , show that, if sin i ! 0 , n sin ( 2n + 1) i 1 + 2 cos ( 2 ri) = . [5] sin i =/r 1 … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 7(a) ( ) 2 2 2 2 2 2 2 1 2 1 ( ) ( ) 1 − − + + + = −  n n z z z z z z z z M1 Uses sum of geometric series. ( ) 2 2 1 2 1 2 1 1 1 1 − + − − − − − × = − n n z z z z z z z z z A1 Divides numerator and denominator by z. Must see at least 2 2 2 2 , 1 + − − n z z z AG. 2 7(b) 1 2isinθ −= − z z B1 Simplifies denominator. 2 1 1 2 isi cos( 1) 2 n(2 1) cos i i s is n in θ θ θ θ θ + − + + − − − − + = n z z n z n z M1 A1 Applies de Moivre’s theorem to numerator. ( ) 1 2 sin sin(2 1) sin(2 1) 1 cos 2 2sin 2sin θ θ θ θ θ θ = + = − − + =  n r n n r M1 Equates real parts. ( ) 1 sin(2 1) 1 2 cos 2 sin θ θ θ = + + =  n r n r A1 AG 5

This question in 9231/22 Oct/Nov 2020

5 (a) State the sum of the series z + z 2 + z 3 + ... + zn , for z ! 1. [1] … … … (b) Given that z is an nth root of unity and z ! 1, deduce that 1 + z + z 2 + ... + z n - 1 = 0 . [2] … … … … … i + i sin i) , use de Moivre’s theorem to show that (c) Given instead that z = 13 ( cos 3 -m 3 cos i - 1 3 cos m i = . [7] 10 - 6 cos i =/m 1 … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 5(a) 1 1 + − − n z z z or 1 . 1 − n z z z B1 1 5(b) 1 n z = and 1 z ≠ leading to 2 3 ... 0 n z z z z + + + + = leading to 2 1 1 ... 0 n z z z − + + + + = M1 A1 Must see 1. = nz 2 5(c) 1 cos isin 1 3 cos isin m m z z z θ θ θ θ ∞ = + = = − − −  M1 A1 Applies sum to infinity and substitutes for z. ( )( ) ( ) 2 2 i c s isin sin os 3 cos 3 cos in θ θ θ θ θ θ − + + − + M1 A1 Rationalises denominator. ( ) 2 2 2 2 s 3 c cos sin isin cos i in 3 cos s cos 9 6cos in os θ θ θ θ θ θ θ θ θ θ − − + + − + − + M1 Applies 2 2 1 sin cos θ θ + = or i i . e e 2 cos θ θ θ − + = 1 3cos Re 10 6 1 cos θ θ ∞ =  =   − −    m m z M1 A1 Takes the real part, AG. 7

This question in 9231/21 May/June 2021

5 (a) State the sum of the series z + z 2 + z 3 + ... + zn , for z ! 1. [1] … … … (b) Given that z is an nth root of unity and z ! 1, deduce that 1 + z + z 2 + ... + z n - 1 = 0 . [2] … … … … … i + i sin i) , use de Moivre’s theorem to show that (c) Given instead that z = 13 ( cos 3 -m 3 cos i - 1 3 cos m i = . [7] 10 - 6 cos i =/m 1 … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 5(a) 1 1 + − − n z z z or 1 . 1 − n z z z B1 1 5(b) 1 n z = and 1 z ≠ leading to 2 3 ... 0 n z z z z + + + + = leading to 2 1 1 ... 0 n z z z − + + + + = M1 A1 Must see 1. = nz 2 5(c) 1 cos isin 1 3 cos isin m m z z z θ θ θ θ ∞ = + = = − − −  M1 A1 Applies sum to infinity and substitutes for z. ( )( ) ( ) 2 2 i c s isin sin os 3 cos 3 cos in θ θ θ θ θ θ − + + − + M1 A1 Rationalises denominator. ( ) 2 2 2 2 s 3 c cos sin isin cos i in 3 cos s cos 9 6cos in os θ θ θ θ θ θ θ θ θ θ − − + + − + − + M1 Applies 2 2 1 sin cos θ θ + = or i i . e e 2 cos θ θ θ − + = 1 3cos Re 10 6 1 cos θ θ ∞ =  =   − −    m m z M1 A1 Takes the real part, AG. 7

This question in 9231/22 May/June 2021

4 The diagram shows the curve with equation y = 2x for 0 G x G 1, together with a set of N rectangles 1 each of width . N y 2 1 0 x 0 1 2 N - 1 1 N N N 1 (a) By considering the sum of the areas of these rectangles, show that 2x dx 1 U N , where y0 1 N 2 U N = 1 . [4] N 2 N - 1 ` j … … … … … … … … … … … … … … 2x dx . [4](b) Use a similar method to find, in terms of N, a lower bound LN for 1y0 … … … … … … … … … … … … … … … … … (c) Find the least value of N such that U N - L N 1 10 -4 . [2] … … … … … … … … …

10 marks

Mark scheme: 4(a)     1 1 2 1 1 1 1 1 0 2 d 2 2 2 2 N N N N N N x N N N N x         M1 A1 Forms the sum of the areas of the rectangles.     1 1 1 1 1 2 2 2 1 N N N N n n N N      M1 A1 Applies   1 1 , 1 N n N n r r r r      AG. 4 4(b)    2 1 1 1 1 1 1 1 0 2 d 2 2 2 N N N N N x N N N N x          M1 A1 Forms the sum of the areas of appropriate rectangles.     1 1 1 0 1 1 2 2 1 N N N n n N N       M1 A1 Applies 1 0 1. 1 N N n n r r r       4 4(c)     1 1 1 4 4 2 1 1 10 leading to 10 2 1 2 1 N N N N N N N        M1 Simplifies n n U L  to . c n Least value of N is 10001 A1 2

This question in 9231/21 May/June 2022

4 The diagram shows the curve with equation y = 2x for 0 G x G 1, together with a set of N rectangles 1 each of width . N y 2 1 0 x 0 1 2 N - 1 1 N N N 1 (a) By considering the sum of the areas of these rectangles, show that 2x dx 1 U N , where y0 1 N 2 U N = 1 . [4] N 2 N - 1 ` j … … … … … … … … … … … … … … 2x dx . [4](b) Use a similar method to find, in terms of N, a lower bound LN for 1y0 … … … … … … … … … … … … … … … … … (c) Find the least value of N such that U N - L N 1 10 -4 . [2] … … … … … … … … …

10 marks

Mark scheme: 4(a)     1 1 2 1 1 1 1 1 0 2 d 2 2 2 2 N N N N N N x N N N N x         M1 A1 Forms the sum of the areas of the rectangles.     1 1 1 1 1 2 2 2 1 N N N N n n N N      M1 A1 Applies   1 1 , 1 N n N n r r r r      AG. 4 4(b)    2 1 1 1 1 1 1 1 0 2 d 2 2 2 N N N N N x N N N N x          M1 A1 Forms the sum of the areas of appropriate rectangles.     1 1 1 0 1 1 2 2 1 N N N n n N N       M1 A1 Applies 1 0 1. 1 N N n n r r r       4 4(c)     1 1 1 4 4 2 1 1 10 leading to 10 2 1 2 1 N N N N N N N        M1 Simplifies n n U L  to . c n Least value of N is 10001 A1 2

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

7 (a) State the sum of the series 1 + w + w 2 + w 3 + f + w n - 1 , for w ! 1. [1] … … r . [2] (b) Show that ( 1 + i tan i) k = sec k i ( cos ki + i sin ki ) , where i is not an integer multiple of 12 … … … … n 1 (c) By considering ( 1 + i tan i) k , show that -/ k = 0 n 1 sec k i sin ki = cot i ( 1 - sec n i cos ni) , -/ k = 0 r . [5] provided i is not an integer multiple of 12 … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … 6m - 1 r in terms of m. [2](d) Hence find 2k sin 13 k / b l k = 0 … … … … … … … … … … … …

10 marks

Mark scheme: 7(a) w n − 1 B1 w − 1 1 7(b) (1 + i tan) k = seck ( cos+ i sin) k = seck ( cos k+ i sin k) M1 A1 Applies de Moivre’s theorem, AG. 2 7(c) n −1 n M1 A1 Applies part (a). k (1 + itan) − 1 1 + itan) =  ( k = 0 itan − cotsecn ( icos n− sin n) +icot A1 n −1 M1 A1 Takes imaginary part, AG. sec k sin k= cot 1 − sec n cos n  ( ) k = 0 5 6 m −17(d) M1 A1 Sets = 13 π. 1 k 1 1 6 m 1 6 m 1 − 2 2 sin ( 3 kπ ) = 3 3 ( 1 − 2 cos ( 2mπ ) ) = 3 3 ( ) = 3 π  k = 0 2

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

7 (a) State the sum of the series 1 + w + w 2 + w 3 + f + w n - 1 , for w ! 1. [1] … … r . [2] (b) Show that ( 1 + i tan i) k = sec k i ( cos ki + i sin ki ) , where i is not an integer multiple of 12 … … … … n 1 (c) By considering ( 1 + i tan i) k , show that -/ k = 0 n 1 sec k i sin ki = cot i ( 1 - sec n i cos ni) , -/ k = 0 r . [5] provided i is not an integer multiple of 12 … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … 6m - 1 r in terms of m. [2](d) Hence find 2k sin 13 k / b l k = 0 … … … … … … … … … … … …

10 marks

Mark scheme: 7(a) w n − 1 B1 w − 1 1 7(b) (1 + i tan) k = seck ( cos+ i sin) k = seck ( cos k+ i sin k) M1 A1 Applies de Moivre’s theorem, AG. 2 7(c) n −1 n M1 A1 Applies part (a). k (1 + itan) − 1 1 + itan) =  ( k = 0 itan − cotsecn ( icos n− sin n) +icot A1 n −1 M1 A1 Takes imaginary part, AG. sec k sin k= cot 1 − sec n cos n  ( ) k = 0 5 6 m −17(d) M1 A1 Sets = 13 π. 1 k 1 1 6 m 1 6 m 1 − 2 2 sin ( 3 kπ ) = 3 3 ( 1 − 2 cos ( 2mπ ) ) = 3 3 ( ) = 3 π  k = 0 2

This question in 9231/23 Oct/Nov 2022

7 (a) Use the substitution u = x 2 - 1 to find d x . [3] x 2 - 1 … … … … … … … y O 1 2 3 N - 1 N x The diagram shows the curve with equation y = cosh -1 x together with a set of ( N - 1) rectangles of unit width. (b) By considering the sum of the areas of these rectangles, show that N ln r + r 2 - 1 2 N ln N + N 2 - 1 - N 2 - 1 . [5] / ` j ` j r = 2 … … … … … … … … … … … … … … … … … … … … N (c) Use a similar method to find, in terms of N, an upper bound for ln r + r 2 - 1 . [3] / ` j r = 2 … … … … … … … … … … … …

11 marks

Mark scheme: 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

This question in 9231/21 May/June 2023

7 (a) Use the substitution u = x 2 - 1 to find d x . [3] x 2 - 1 … … … … … … … y O 1 2 3 N - 1 N x The diagram shows the curve with equation y = cosh -1 x together with a set of ( N - 1) rectangles of unit width. (b) By considering the sum of the areas of these rectangles, show that N ln r + r 2 - 1 2 N ln N + N 2 - 1 - N 2 - 1 . [5] / ` j ` j r = 2 … … … … … … … … … … … … … … … … … … … … N (c) Use a similar method to find, in terms of N, an upper bound for ln r + r 2 - 1 . [3] / ` j r = 2 … … … … … … … … … … … …

11 marks

Mark scheme: 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

This question in 9231/22 May/June 2023

8 (a) State the sum of the series 1 + z + z 2 + f + z n - 1 , for z ! 1. [1] … (b) By letting z = cos i + i sin i , where cos i ! 1, show that 1 sin ni sin i 1 + cos i + cos 2i + f + cos ( n - 1) i = 1 - cos n i + [7] 2 e 1 - cos i o. … … … … … … … … … … … … … … … … … … … … … … … y 1 O 1 2 n - 1 1 x n n n The diagram shows the curve with equation y = cos x for 0 G x G 1, together with a set of n rectangles of width 1. n (c) By considering the sum of the areas of these rectangles, show that 1 1 1 sin 1 sin n cos xd x 1 1 - cos 1 + . [4] 2n y0 f 1 - cos n1 p … … … … … … (d) Use a similar method to find, in terms of n, a lower bound for cosxd x . [3] 1y0 … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

15 marks

Mark scheme: 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

This question in 9231/21 Oct/Nov 2023

8 (a) State the sum of the series 1 + z + z 2 + f + z n - 1 , for z ! 1. [1] … (b) By letting z = cos i + i sin i , where cos i ! 1, show that 1 sin ni sin i 1 + cos i + cos 2i + f + cos ( n - 1) i = 1 - cos n i + [7] 2 e 1 - cos i o. … … … … … … … … … … … … … … … … … … … … … … … y 1 O 1 2 n - 1 1 x n n n The diagram shows the curve with equation y = cos x for 0 G x G 1, together with a set of n rectangles of width 1. n (c) By considering the sum of the areas of these rectangles, show that 1 1 1 sin 1 sin n cos xd x 1 1 - cos 1 + . [4] 2n y0 f 1 - cos n1 p … … … … … … (d) Use a similar method to find, in terms of n, a lower bound for cosxd x . [3] 1y0 … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … …

15 marks

Mark scheme: 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

This question in 9231/23 Oct/Nov 2023

6 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 of width . N 1 x (a) By considering the sum of the areas of these rectangles, show that b 1 l dx 2 L , where y 2 N 0 1 L = 1 . [4] N 2N `2 N - 1j … … … … … … … … … … … … … … … … 1 x (b) Use a similar method to find, in terms of N, an upper bound UN for y b 12 l dx . [4] 0 … … … … … … … … … … … … … … … … (c) Find the least value of N such that U - L G 10 -3 . [2] N N … … … … … … … … … 1 1 x 1 (d) Given that b l dx = , use the value of N found in part (c) to find upper and lower bounds y 0 2 2 ln 2 for ln2. [4] … … … … … … … … …

14 marks

Mark scheme: 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

This question in 9231/21 Oct/Nov 2024

6 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 of width . N 1 x (a) By considering the sum of the areas of these rectangles, show that b 1 l dx 2 L , where y 2 N 0 1 L = 1 . [4] N 2N `2 N - 1j … … … … … … … … … … … … … … … … 1 x (b) Use a similar method to find, in terms of N, an upper bound UN for y b 12 l dx . [4] 0 … … … … … … … … … … … … … … … … (c) Find the least value of N such that U - L G 10 -3 . [2] N N … … … … … … … … … 1 1 x 1 (d) Given that b l dx = , use the value of N found in part (c) to find upper and lower bounds y 0 2 2 ln 2 for ln2. [4] … … … … … … … … …

14 marks

Mark scheme: 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 Applies r = 1. = 1 ( 2 ) =  N  N N r − 1 1 1 n = 0 n = 0 2 N 2 N1 − 1 1 − N 2 N N ( 2 ) 1 − ( 2 ) ( ) ) = ( 4 6(c) N1 M1 c 2 1 1  10 −3  2 N  103 Simplifies U N − LN to . 1 − 1 = N N 2 N 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

This question in 9231/23 Oct/Nov 2024