Integration — Paper 3 · A Level Mathematics 9709

2 1 The diagram shows a sketch of the curve y = for values of x from −0.6 to 0.6. 1 + x3 (i) Use the trapezium rule, with two intervals, to estimate the value of 0.6 1 dx, 1 + x3 −0.6 giving your answer correct to 2 decimal places. [3] (ii) Explain, with reference to the diagram, why the trapezium rule may be expected to give a good approximation to the true value of the integral in this case. [1]

4 marks

Mark scheme: 2 (i) Show or imply correct decimal ordinates 1.2755…, 1, 0.8223… B1 Use correct formula, or equivalent, with h = 0.6 and three ordinates M1 Obtain correct answer 1.23 with no errors seen A1 3 [SR: if the area is calculated with one interval, or three or more, give D1 for a correct answer.] (ii) Give an adequate justification, e.g. one trapezium over-estimates area and the other under-estimates, or errors cancel out B1 1

This question in 9709/31 May/June 2005

4 (i) Use the substitution x = tan θ to show that 1 −x2 dx = cos 2θ dθ. [4] (1 + x2)2 (ii) Hence find the value of 1 1 −x2 dx. [3] (1 + x2)2 0

7 marks

Mark scheme: d x 2 24 (i) State or imply d x = sec θ d θ or = sec θ B1 d θ Substitute for x and dx throughout the integral M1 Obtain integral in terms of θ in any correct form A1 Reduce to the given form correctly A1 4 (ii) State integral 1 sin 2θ B1 2 Use limits θ = 0 and θ = 41 π correctly in integral of the form k sin 2θ M1 Obtain answer ½ or 0.5 A1 3

This question in 9709/31 May/June 2005

8 (i) Using partial fractions, find 1 dy. [4] y(4 −y) (ii) Given that y = 1 when x = 0, solve the differential equation dy = y(4 −y), dx obtaining an expression for y in terms of x. [4] (iii) State what happens to the value of y if x becomes very large and positive. [1]

9 marks

Mark scheme: 8 (i) Attempt to express integrand in partial fractions, A B e.g. obtain A or B in + M1 y 4 − y 1 1 1 Obtain ( + ) , or equivalent A1 4 y 4 − y Integrate and obtain 41 ln y − 41 ln ( 4 − y ) , or equivalent A1√ + A1√ 4 A B (ii) Separate variables correctly, integrate + and obtain further y 4 − y term x, or equivalent M1* Use y = 1 and x = 0 to evaluate a constant, or as limits M1(dep*) Obtain answer in any correct form A1 Obtain final answer y = 4 /( 3 e −x4 + )1 , or equivalent A1 4 (iii) State that y approaches 4 as x becomes very large B1 1

This question in 9709/31 May/June 2005

9 x The diagram shows part of the curve y = and its maximum point M. The shaded region R is x2 + 1 bounded by the curve and by the lines y = 0 and x = p. (i) Calculate the x-coordinate of M. [4] (ii) Find the area of R in terms of p. [3] (iii) Hence calculate the value of p for which the area of R is 1, giving your answer correct to 3 significant figures. [2]

9 marks

Mark scheme: 9 (i) Use quotient or product rule M1 Obtain derivative in any correct form A1 Equate derivative to zero and solve for x or x2 M1 Obtain x = 1 correctly A1 4 [Differentiating ( x 2 + )1 y = x using the product rule can also earn the first M1A1.] [SR: if the quotient rule is misused, with a ‘reversed’ numerator or v instead of v² in the denominator, award M0A0 but allow the following M1A1.] (ii) Obtain indefinite integral of the form k ln ( x 2 + )1 , where k = ½, 1 or 2 M1* Use limits x = 0 and x = p correctly, or equivalent M1(dep*) Obtain answer ½ ln(p2 +1) A1 3 [Also accept –ln cos θ or ln cos θ , where x = tan θ , for the first M1*.] (iii) Equate to 1 and convert equation to the form p 2 + 1 = exp(1/ k ) M1 Obtain answer p = 2.53 A1 2 A AND AS LEVEL – JUNE 2005 9709/8719 3

This question in 9709/31 May/June 2005

6 (i) Use the substitution x = sin2 θ to show that x dx = 2 sin2 θ dθ. [4] 1 −x (ii) Hence find the exact value of 1 4 x dx. [4] 1 −x 0

8 marks

Mark scheme: dx 6 (i) State = 2sin θ cos θ , or dx = 2sin θ cos θ dθ B1 dθ Substitute for x and dx throughout M1 Obtain any correct form in terms of θ A1 Reduce to the given form correctly A1 [4] (ii) Use cos 2A formula, replacing integrand by a + bcos 2θ , where ab ≠ 0 M1* Integrate and obtain θ − 1 sin 2θ A1√ 2 Use limits θ = 0 and θ = 1 π M1(dep*) 6 Obtain exact answer 1 π − 1 3 , or equivalent A1 [4] 6 4

This question in 9709/31 Oct/Nov 2005

3 Use integration by parts to show that 4 ln x dx = 6 ln 2 −2. [4] 2

4 marks

Mark scheme: 1 3 Using 1 and ln x as parts reach xln x ± ∫ x. x dx M1* Obtain indefinite integral xln x –x A1 Substitute correct limits correctly M1(dep*) Obtain given answer A1 [4]

This question in 9709/31 Oct/Nov 2007

9 y M R x O The diagram shows the curve y = e−12x√(1 + 2x) and its maximum point M. The shaded region between the curve and the axes is denoted by R. (i) Find the x-coordinate of M. [4] (ii) Find by integration the volume of the solid obtained when R is rotated completely about the x-axis. Give your answer in terms of π and e. [6]

10 marks

Mark scheme: 9 (i) Either use correct product or quotient rule, or square both sides, use correct product rule and make a reasonable attempt at applying the chain rule M1 Obtain correct result of differentiation in any form A1 Set derivative equal to zero and solve for x M1 Obtain x = 1 only, correctly A1 [4] 2 + 2 x )d x B1 (ii) State or imply the indefinite integral for the volume is π ∫ e − x (1 x M1 Integrate by parts and reach ± e − x (1 + 2 x ) ± ∫ 2e − x d x , or equivalent A1 Obtain − e − x (1 + 2 x ) + ∫ 2e − x d Complete integration correctly, obtaining − e − x (1 + 2 x ) − 2e − x , or equivalent A1 Use limits x = − 1 and x = 0 correctly, having integrated twice M1 2 Obtain exact answer π (2 e − 3) , or equivalent A1 [6] [If π omitted initially or 2π or π/2 used, give B0 and then follow through.] GCE A/AS LEVEL – May/June 2008 9709 03

This question in 9709/31 May/June 2008

a 1 2x9 The constant a is such that xe dx = 6. 0 (i) Show that a satisfies the equation −1 x = 2 + e 2x. [5] (ii) By sketching a suitable pair of graphs, show that this equation has only one root. [2] (iii) Verify by calculation that this root lies between 2 and 2.5. [2] (iv) Use an iterative formula based on the equation in part (i) to calculate the value of a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

12 marks

Mark scheme: 2 x9 (i) Integrate by parts and reach kxe 2 x d x M1 − k ∫ e 2 x − 2 e 1 Obtain 2 xe 1 ∫ 2 x d x A1 1 2 x 12 x Complete the integration, obtaining 2 xe − 4e , or equivalent A1 Substitute limits correctly and equate result to 6, having integrated twice M1 −a12 Rearrange and obtain a = e + 2 A1 [5] −x12 (ii) Make recognizable sketch of a relevant exponential graph, e.g. y = e + 2 B1 Sketch a second relevant straight line graph, e.g. y = x, or curve, and indicate the root B1 [2] −x12 (iii) Consider sign of x − e − 2 at x = 2 and x = 2.5, or equivalent M1 Justify the given statement with correct calculations and argument A1 [2] − 12 x n (iv) Use the iterative formula x n +1 = 2 + e correctly at least once, with 2 ≤ x n ≤ 5.2 M1 Obtain final answer 2.31 A1 Show sufficient iterations to at least 4 d.p. to justify its accuracy to 2 d.p., or show there is a sign change in the interval (2.305, 2.315) A1 [3]

This question in 9709/31 Oct/Nov 2008

2 y 1 x O 4p 2 for 0 The diagram shows the curve y = √(1 + tan2x) ≤x ≤14π. (i) Use the trapezium rule with three intervals to estimate the value of 14π 2 dx, ã 0 √(1 + tan2x) giving your answer correct to 2 decimal places. [3] (ii) The estimate found in part (i) is denoted by E. Explain, without further calculation, whether another estimate found using the trapezium rule with six intervals would be greater than E or less than E. [1]

4 marks

Mark scheme: 2 (i) State or imply 3 of the 4 ordinates 1, 1.069389..., 1.290994..., 1.732050... B1 Use correct formula, or equivalent, with h = 121 π and four ordinates M1 Obtain answer 0.98 with no errors seen A1 3 [Accept h = 0.26 but not h = 15 when awarding the M1] [SR: if only 5 and/or 3 are given, and decimals are not seen, the B1 is available] 3 [SR: solutions with 2 or 4 intervals can score only the M1 for a correct expression] (ii) Justify statement that the second estimate would be less than E B1 1

This question in 9709/31 May/June 2009

10 y M x O The diagram shows the curve y for x and its maximum point M. = x2√(1 −x2) ≥0 (i) Find the exact value of the x-coordinate of M. [4] (ii) Show, by means of the substitution x sin θ, that the area A of the shaded region between the = curve and the x-axis is given by 12π A 1 sin2 2θ dθ. = 4 ã 0 [3] (iii) Hence obtain the exact value of A. [4]

11 marks

Mark scheme: 10 (i) EITHER Use product and chain rule M1 Obtain correct derivative in any form A1 OR Square and differentiate LHS by chain rule and RHS by product rule or as powers M1 Obtain correct result in any form A1 dy Set equal to zero and make reasonable attempt to solve for x ≠ 0 M1 dx Obtain answer x = 2 , or exact equivalent, correctly A1 4 3 dx (ii) State or imply dx = cos θ dθ or = cos θ B1 dθ Substitute for x and dx throughout the integral ∫ ydx M1 Obtain the given form correctly with no errors seen A1 3 (iii) Attempt integration and reach indefinite integral of the form aθ + bsin 4θ , where ab ≠ 0 M1* Obtain indefinite integral 18 θ − 321 sin 4θ , or equivalent A1 Substitute limits correctly M1(dep*) Obtain exact answer 161 π A1 4 [Working to carry out the change of limits is needed for the A mark in (ii) but, if omitted, can be earned retrospectively if it is seen in part (iii)]

This question in 9709/31 May/June 2009

5 (i) Prove the identity cos 4θ cos 2θ 3 sin4 θ. [4] −4 + ≡8 (ii) Using this result find, in simplified form, the exact value of 13π sin4 θ dθ. ã 1 [4] 6π

8 marks

Mark scheme: 5 (i) EITHER: Use double angle formulae correctly to express LHS in terms of trig functions of 2θ M1 Use trig formulae correctly to express LHS in terms of sin θ, converting at least two terms M1 Obtain expression in any correct form in terms of sin θ A1 Obtain given answer correctly A1 OR: Use double angle formulae correctly to express RHS in terms of trig functions of 2θ M1 Use trig formulae correctly to express RHS in terms of cos 4θ and cos 2θ M1 Obtain expression in any correct form in terms of cos 4θ and cos 2θ A1 Obtain given answer correctly A1 [4] (ii) State indefinite integral 1 sin 4θ – 4 sin 2θ + 3θ, or equivalent B2 4 2 (award B1 if there is just one incorrect term) Use limits correctly, having attempted to use the identity M1 Obtain answer 1 (2π – 3 ), or any simplified exact equivalent A1 [4] 32

This question in 9709/31 Oct/Nov 2009

9 y M A x O 4 ln x The diagram shows the curve y and its maximum point M. The curve cuts the x-axis at the = √x point A. (i) State the coordinates of A. [1] (ii) Find the exact value of the x-coordinate of M. [4] (iii) Using integration by parts, show that the area of the shaded region bounded by the curve, the x-axis and the line x 4 is equal to 8 ln 2 [5] = −4.

10 marks

Mark scheme: 9 (i) State coordinates (1, 0) B1 [1] (ii) Use correct quotient or product rule M1 Obtain derivative in any correct form A1 Equate derivative to zero and solve for x M1 Obtain x = e2 correctly A1 [4] GCE A/AS LEVEL – October/November 2009 9709 31 1 (iii) Attempt integration by parts reaching a x ln x ± a ∫ x x dx M1* 1 Obtain 2 x ln x − 2 ∫ dx A1 x Integrate and obtain 2 x ln x − 4 x A1 Use limits x = 1 and x = 4 correctly, having integrated twice M1(dep*) Justify the given answer A1 [5] dA

This question in 9709/31 Oct/Nov 2009

4 (i) Using the expansions of and prove that cos(3x −x) cos(3x + x), 1 2x 3x sin x. 2(cos −cos 4x) ≡sin [3] (ii) Hence show that 13π sin 3x sin x dx 18 √3. 1 ã 6π = [3]

6 marks

Mark scheme: 4 (i) State correct expansion of cos(3x – x) or cos(3x + x) B1 Substitute expansions in 12 (cos 2 x − cos 4 x ) , or equivalent M1 Simplify and obtain the given identity correctly A1 [3] (ii) Obtain integral 14 sin 2 x − 18 sin 4 x B1 Substitute limits correctly in an integral of the form a sin 2 x + b sin 4 x M1 Obtain given answer following full, correct and exact working A1 [3] GCE AS/A LEVEL – May/June 2010 9709 31

This question in 9709/31 May/June 2010

2 8 (i) Express (x + 1)(x + 3) in partial fractions. [2] (ii) Using your answer to part (i), show that 2 2 1 1 1 1 x 1 x 3 ≡ − + + [2] (x + 1)(x + 3) (x + 1)2 + + (x + 3)2. 1 4 7 3 (iii) Hence show that dx 12 2. [5] ä 0 = −ln (x + 1)2(x + 3)2

9 marks

Mark scheme: A B 8 (i) State or imply the form + and use a relevant method to find A or B M1 x + 1 x + 3 Obtain A = 1, B = −1 A1 [2] (ii) Square the result of part (i) and substitute the fractions of part (i) M1 Obtain the given answer correctly A1 [2] 1 1 (iii) Integrate and obtain − − ln ( x + 1) + ln( x + 3) − B3 x + 1 x + 3 Substitute limits correctly in an integral containing at least two terms of the correct form M1 Obtain given answer following full and exact working A1 [5] GCE AS/A LEVEL – May/June 2010 9709 31

This question in 9709/31 May/June 2010

π 2 Show that x2 sin x dx π2 [5] ã 0 = −4.

5 marks

Mark scheme: 2 Integrate by parts and reach ± x 2 cos x ± ∫ 2 x cos x d x M1 Obtain − x 2 cos x + ∫ 2 x cos x dx , or equivalent A1 Complete the integration, obtaining − x 2 cos x + 2 x sin x + 2 cos x , or equivalent A1 Substitute limits correctly, having integrated twice M1 Obtain the given answer correctly A1 [5]

This question in 9709/32 May/June 2010

5 y M x O p The diagram shows the curve y and its maximum point M. The x-coordinate of M is = e−x −e−2x denoted by p. (i) Find the exact value of p. [4] (ii) Show that the area of the shaded region bounded by the curve, the x-axis and the line x p is 1 = equal to 8. [4]

8 marks

Mark scheme: 5 (i) State derivative − e − x − ( −2e) −2 x , or equivalent B1 + B1 Equate derivative to zero and solve for x M1 Obtain p = ln 2, or exact equivalent A1 [4] (ii) State indefinite integral − e − x − ( − 12 e) −2 x , or equivalent B1 + B1 Substitute limits x = 0 and x = p correctly M1 Obtain given answer following full and correct working A1 [4] GCE AS/A LEVEL – May/June 2010 9709 33

This question in 9709/33 May/June 2010

7 (i) Prove the identity cos 3θ cos3θ cos θ. [4] ≡4 −3 (ii) Using this result, find the exact value of 12π cos3θ dθ. ã 1 [4] 3π

8 marks

Mark scheme: 7 (i) Use correct cos(A + B) formula to express cos 3θ in terms of trig functions of 2θ and θ M1 Use correct trig formulae and Pythagoras to express cos 3θ in terms of cosθ M1 Obtain a correct expression in terms of cosθ in any form A1 Obtain the given identity correctly A1 [4] [SR: Give M1 for using correct formulae to express RHS in terms of cos θ and cos 2θ , then M1A1 for expressing in terms of either only cos 3θ and cos θ , or only cos 2θ , sin 2θ , cos θ , and sin θ , and A1 for obtaining the given identity correctly.] 1 1 1 (ii) Use identity and integrate, obtaining terms ( sin 3θ ) and (3 sin θ ) , or equivalent B1 + B1 4 3 4 Use limits correctly in an integral of the form ksin 3θ + lsin θ M1 2 3 Obtain answer − 3 , or any exact equivalent A1 [4] 3 8 3 2

This question in 9709/33 May/June 2010

1 x2 5 Let I dx. = ä 0 √(4 −x2) (i) Using the substitution x 2 sin θ, show that = 16π I 4 sin2θ dθ. = ã 0 [3] (ii) Hence find the exact value of I. [4]

7 marks

Mark scheme: dx 5 (i) State or imply dx = 2 cos θ dθ, or = 2 cos θ, or equivalent B1 dθ Substitute for x and dx throughout the integral M1 Obtain the given answer correctly, having changed limits and shown sufficient working A1 [3] (ii) Replace integrand by 2 – 2 cos 2θ, or equivalent B1 Obtain integral 2θ – sin 2θ, or equivalent B1√ Substitute limits correctly in an integral of the form aθ ± b sin 2θ, where ab Þ 0 M1 1 3 Obtain answer π − or exact equivalent A1 [4] 3 2 [The f.t. is on integrands of the form a + c cos 2θ, where ac Þ 0.]

This question in 9709/31 Oct/Nov 2010

9 y x O 2 M The diagram shows the curve y x3 ln x and its minimum point M. = (i) Find the exact coordinates of M. [5] (ii) Find the exact area of the shaded region bounded by the curve, the x-axis and the line x 2. [5] =

10 marks

Mark scheme: 9 (i) Use correct product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and find non-zero x M1 1 Obtain x = exp (− 3 ) , or equivalent A1 Obtain y = –l/(3e), or any ln-free equivalent A1 [5] 1 (ii) Integrate and reach kx 4 ln x + l ∫ x 4 . x dx M1 Obtain 14 x 4 ln x − 14 ∫ x 3 dx A1 Obtain integral 14 x 4 ln x − 161 x 4 , or equivalent A1 Use limits x = 1 and x = 2 correctly, having integrated twice M1 15 Obtain answer 4 ln 2 − , or exact equivalent A1 [5] 16 GCE A/AS LEVEL – October/November 2010 9709 31 dx ( )

This question in 9709/31 Oct/Nov 2010

1 x2 5 Let I dx. = ä 0 √(4 −x2) (i) Using the substitution x 2 sin θ, show that = 16π I 4 sin2θ dθ. = ã 0 [3] (ii) Hence find the exact value of I. [4]

7 marks

Mark scheme: dx 5 (i) State or imply dx = 2 cos θ dθ, or = 2 cos θ, or equivalent B1 dθ Substitute for x and dx throughout the integral M1 Obtain the given answer correctly, having changed limits and shown sufficient working A1 [3] (ii) Replace integrand by 2 – 2 cos 2θ, or equivalent B1 Obtain integral 2θ – sin 2θ, or equivalent B1√ Substitute limits correctly in an integral of the form aθ ± b sin 2θ, where ab Þ 0 M1 1 3 Obtain answer π − or exact equivalent A1 [4] 3 2 [The f.t. is on integrands of the form a + c cos 2θ, where ac Þ 0.]

This question in 9709/32 Oct/Nov 2010

9 y x O 2 M The diagram shows the curve y x3 ln x and its minimum point M. = (i) Find the exact coordinates of M. [5] (ii) Find the exact area of the shaded region bounded by the curve, the x-axis and the line x 2. [5] =

10 marks

Mark scheme: 9 (i) Use correct product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and find non-zero x M1 1 Obtain x = exp (− 3 ) , or equivalent A1 Obtain y = –l/(3e), or any ln-free equivalent A1 [5] 1 (ii) Integrate and reach kx 4 ln x + l ∫ x 4 . x dx M1 Obtain 14 x 4 ln x − 14 ∫ x 3 dx A1 Obtain integral 14 x 4 ln x − 161 x 4 , or equivalent A1 Use limits x = 1 and x = 2 correctly, having integrated twice M1 15 Obtain answer 4 ln 2 − , or exact equivalent A1 [5] 16 GCE A/AS LEVEL – October/November 2010 9709 32 dx ( )

This question in 9709/32 Oct/Nov 2010

9 y x e O M The diagram shows the curve y x2 ln x and its minimum point M. = (i) Find the exact values of the coordinates of M. [5] (ii) Find the exact value of the area of the shaded region bounded by the curve, the x-axis and the line x e. [5] =

10 marks

Mark scheme: 9 (i) Use product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero1 and solve for x M1 Obtain answer x = e– 2 , or equivalent A1 Obtain answer y = – 1 e–1, or equivalent A1 [5] 2 1 (ii) Attempt integration by parts reaching kx3 ln x ± k ∫ x 3 . x dx M1* Obtain 1 x 3 ln x − 1 x 2 d x , or equivalent A1 3 3 ∫ Integrate again and obtain 1 x 3 ln x − 1 x 3 , or equivalent A1 3 9 Use limits x = 1 and x = e, having integrated twice M1(dep*) Obtain answer 1 (2e3 + 1), or exact equivalent A1 [5] 9 [SR: An attempt reaching ax2 (x ln x – x) + b ∫ 2 x ( x ln x −)x dx scores M1. Then give the first A1 for I = x2 (x ln x – x) – 2I + ∫ 2x 2 dx, or equivalent.]

This question in 9709/31 Oct/Nov 2011

9 y x e O M The diagram shows the curve y x2 ln x and its minimum point M. = (i) Find the exact values of the coordinates of M. [5] (ii) Find the exact value of the area of the shaded region bounded by the curve, the x-axis and the line x e. [5] =

10 marks

Mark scheme: 9 (i) Use product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero1 and solve for x M1 Obtain answer x = e– 2 , or equivalent A1 Obtain answer y = – 1 e–1, or equivalent A1 [5] 2 1 (ii) Attempt integration by parts reaching kx3 ln x ± k ∫ x 3 . x dx M1* Obtain 1 x 3 ln x − 1 x 2 d x , or equivalent A1 3 3 ∫ Integrate again and obtain 1 x 3 ln x − 1 x 3 , or equivalent A1 3 9 Use limits x = 1 and x = e, having integrated twice M1(dep*) Obtain answer 1 (2e3 + 1), or exact equivalent A1 [5] 9 [SR: An attempt reaching ax2 (x ln x – x) + b ∫ 2 x ( x ln x −)x dx scores M1. Then give the first A1 for I = x2 (x ln x – x) – 2I + ∫ 2x 2 dx, or equivalent.]

This question in 9709/32 Oct/Nov 2011

a 5 It is given that x ln x dx 22, where a is a constant greater than 1. ã 1 = r 87 (i) Show that a . [5] = 2 ln a −1 (ii) Use an iterative formula based on the equation in part (i) to find the value of a correct to 2 decimal places. Use an initial value of 6 and give the result of each iteration to 4 decimal places. [3]

8 marks

Mark scheme: 5 (i) Either Use integration by parts and reach an expression kx2 lnx ± n ∫ x2. 1x dx M1 Obtain 12 x2 ln x – ∫ 12 x dx or equivalent A1 Obtain 12 x2 ln x – 14 x2 A1 Or Use Integration by parts and reach an expression kx(xlnx – x) ± m ∫ xlnx – xdx M1 Obtain I = (x2 lnx – x2) – I + ∫ xdx A1 Obtain 12 x2 ln x – 14 x2 A1 Substitute limits correctly and equate to 22, having integrated twice DM1* 87 Rearrange and confirm given equation a = A1 [5] 2ln a − 1 (ii) Use iterative process correctly at least once M1 Obtain final answer 5.86 A1 Show sufficient iterations to 4 d.p. to justify 5.86 or show a sign change in the A1 interval (5.855, 5.865) (6 → 5.8030 → 5.8795 → 5.8491 → 5.8611 → 5.8564) [3] GCE AS/A LEVEL – October/November 2011 9709 33 2 3

This question in 9709/33 Oct/Nov 2011

5 y a x O The diagram shows the curve y 8 sin 2x1 2x1 = −tan for 0 π. The x-coordinate of the maximum point is α and the shaded region is enclosed by the curve≤xand<the lines x α and y 0. = = (i) Show that α 23π. [3] = (ii) Find the exact value of the area of the shaded region. [4]

7 marks

Mark scheme: 1 1 2 1 5 (i) Differentiate to obtain 4 cos x − sec x B1 2 2 2 1 Equate to zero and find value of cos x M1 2 1 1 2 Obtain cos x = and confirm α = π A1 [3] 2 2 3 1 (ii) Integrate to obtain − 16 cos x … B1 2 1 … + 2 ln cos x or equivalent B1 2 2 1 1 Using limits 0 and π in a cos x + b ln cos x M1 3 2 2 1 Obtain 8+ 2 ln or exact equivalent A1 [4] 2 dy 2

This question in 9709/31 May/June 2012

4x2 5x 3 9 By first expressing in partial fractions, show that + + 2x2 5x 2 + + 4 4x2 5x 3 dx 8 9. + + 2x2 5x 2 = −ln [10] ä 0 + +

10 marks

Mark scheme: B C 9 State or imply form A + + B1 2 x + 1 x + 2 State or obtain A = 2 B1 Use correct method for finding B or C M1 Obtain B = 1 A1 Obtain C = –3 A1 1 Obtain 2 x + ln( 2 x + )1 − 3 ln( x + 2) [Deduct B1 for each error or omission] B3 2 Substitute limits in expression containing aln(2x + 1) + bln(x + 2) M1 1 Show full and exact working to confirm that 8 + ln 9 − 3 ln 6 + 3 ln 2 , or an equivalent 2 expression, simplifies to given result 8 – ln 9 A1 [10] [SR: If A omitted from the form of fractions, give B0B0M1A0A0 in (i); B0 B1 B1 M1A0 in (ii).] M Nx Px Q [SR: For a solution starting with + or + , give B0B0M1A0A0 in (i); 2 x + 1 x + 2 2 x + 1 x + 2 B1 B1 B1 , if recover correct form, M1A0 in (ii).] B Dx + E [SR: For a solution starting with + , give M1A1 for one of B = 1, D = 2, E = 1 2 x + 1 x + 2 and A1 for the other two constants; then give B1B1 for A = 2, C = −3.] Fx + G C [SR: For a solution starting with + , give M1A1 for one of C = −3, F = 4, G = 3 2 x + 1 x + 2 and A1 for the other constants or constant; then give B1B1 for A = 2, B = 1.] GCE AS/A LEVEL – May/June 2012 9709 31 4 3 2

This question in 9709/31 May/June 2012

9 y R e x O 1 The diagram shows the curve y x 2 ln x. The shaded region between the curve, the x-axis and the line x e is denoted by R. = = (i) Find the equation of the tangent to the curve at the point where x 1, giving your answer in the form y mx c. = [4] = + (ii) Find by integration the volume of the solid obtained when the region R is rotated completely about the x-axis. Give your answer in terms of π and e. [7] [Question 10 is printed on the next page.]

11 marks

Mark scheme: 9 (i) Use correct product rule M1 ln x x Obtain derivative in any correct form, e.g. + A1 2 x x Carry out a complete method to form an equation of the tangent at x = 1 M1 Obtain answer y = x – 1 A1 [4] (ii) State or imply that the indefinite integral for the volume is π ∫ x (ln x ) 2 d x B1 ln x 2 2 2 Integrate by parts and reach ax (ln x ) + b ∫ x . x dx M1* 1 Obtain x 2 (ln x ) 2 − ∫ x ln x dx , or unsimplified equivalent A1 2 1 Attempt second integration by parts reaching cx 2 ln x + d ∫ x 2 . x d x M1(dep*) 1 2 2 1 2 1 2 Complete the integration correctly, obtaining x (ln x ) − x ln x + x A1 2 2 4 Substitute limits x = 1 and x = e, having integrated twice M1(dep*) 1 2 Obtain answer π e( − )1 , or exact equivalent A1 [7] 4 [If π omitted, or 2π or π/2 used, give B0 and then follow through.] [Integration using parts x ln x and ln x is also viable.] GCE AS/A LEVEL – May/June 2012 9709 32

This question in 9709/32 May/June 2012

7 y R x O p2 The diagram shows part of the curve y for x where x is in radians. The shaded region = cos(√x) ≥0, between the curve, the axes and the line x p2, where p 0, is denoted by R. The area of R is equal = > to 1. p2 3 cos p (i) Use the substitution x u2 to find dx. Hence show that sin p −2 . [6] = ã 0 cos(√x) = 2p 3 cos pn (ii) Use the iterative formula sin−1 −2 , with initial value p1 1, to find the value of pn+1 = 2pn = p correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

9 marks

Mark scheme: 7 (i) Substitute for x and dx throughout the integral M1 Obtain ∫ 2u cos u d u A1 Integrate by parts and obtain answer of the form au sin u + b cos u , where ab ≠ 0 M1 Obtain 2u sin u + 2 cos u A1 Use limits u = 0, u = p correctly and equate result to 1 M1 Obtain the given answer A1 [6] (ii) Use the iterative formula correctly at least once M1 Obtain final answer p = 1.25 A1 Show sufficient iterations to 4 d.p. to justify its accuracy to 2 d.p., or show there is a sign change in the interval (1.245, 1.255) A1 [3] GCE AS/A LEVEL – May/June 2012 9709 33 B C

This question in 9709/33 May/June 2012

1 dy 5 (i) By differentiating show that if y sec x then sec x tan x. [2] cos x, dx = = 1 (ii) Show that x tan x. [1] secx x ≡sec + −tan 1 (iii) Deduce that sec2x 2 sec x tan x. [2] x ≡2 −1 + (sec −tan x)2 14π 1 1 (iv) Hence show that dx [3] ä 0 = 4(8√2 −π). (secx −tan x)2

8 marks

Mark scheme: 5 (i) Use correct quotient or chain rule M1 Obtain the given answer correctly having shown sufficient working A1 [2] (ii) Use a valid method, e.g. multiply numerator and denominator by sec x + tan x, and a version of Pythagoras to justify the given identity B1 [1] (iii) Substitute, expand (sec x + tan x)2 and use Pythagoras once M1 Obtain given identity A1 [2] (iv) Obtain integral 2 tan x – x + 2 sec x B1 Use correct limits correctly in an expression of the form a tan x + bx + c sec x, or equivalent, where abc 0 M1 Obtain the given answer correctly A1 [3]

This question in 9709/31 Oct/Nov 2012

1 dy 5 (i) By differentiating show that if y sec x then sec x tan x. [2] cos x, dx = = 1 (ii) Show that x tan x. [1] secx x ≡sec + −tan 1 (iii) Deduce that sec2x 2 sec x tan x. [2] x ≡2 −1 + (sec −tan x)2 14π 1 1 (iv) Hence show that dx [3] ä 0 = 4(8√2 −π). (secx −tan x)2

8 marks

Mark scheme: 5 (i) Use correct quotient or chain rule M1 Obtain the given answer correctly having shown sufficient working A1 [2] (ii) Use a valid method, e.g. multiply numerator and denominator by sec x + tan x, and a version of Pythagoras to justify the given identity B1 [1] (iii) Substitute, expand (sec x + tan x)2 and use Pythagoras once M1 Obtain given identity A1 [2] (iv) Obtain integral 2 tan x – x + 2 sec x B1 Use correct limits correctly in an expression of the form a tan x + bx + c sec x, or equivalent, where abc 0 M1 Obtain the given answer correctly A1 [3]

This question in 9709/32 Oct/Nov 2012

4 8 (a) Show that 4x ln x dx 56 ln 2 [5] Ó2 = −12. 1 240 (b) Use the substitution u sin 4x to find the exact value of cos34x dx. [5] = Ó 0

10 marks

Mark scheme: 8 (a) Carry out integration by parts and reach ax 2 ln x + b ∫ 12 x 2 d x M1* Obtain 2 x 2 ln x −∫ 1x . 2 x 2 d x A1 Obtain 2 x 2 ln x − x 2 A1 Use limits, having integrated twice M1 (dep*) Confirm given result 56ln 2 − 12 A1 [5] GCE AS/A LEVEL – May/June 2013 9709 31 (b) State or imply ddux = 4cos4 x B1 Carry out complete substitution except limits M1 Obtain ∫ ( 14 − 14 u 2 ) d u or equivalent A1 Integrate to obtain form k1u + k 2 u 3 with non-zero constants k1 , k 2 M1 Use appropriate limits to obtain 1196 A1 [5]

This question in 9709/31 May/June 2013

9 y M x O 12 p The diagram shows the curve y sin22x cos x for 0 and its maximum point M. = ≤x ≤120, (i) Find the x-coordinate of M. [6] (ii) Using the substitution u sin x, find by integration the area of the shaded region bounded by the curve and the x-axis. = [4]

10 marks

Mark scheme: 9 (i) Use product rule M1 Obtain correct derivative in any form, e.g. 4sin2x cos2x cos x – sin2 2x sin x A1 Equate derivative to zero and use a double angle formula M1* Reduce equation to one in a single trig function M1(dep*) Obtain a correct equation in any form, e.g. 10 cos3 x = 6 cos x, 4 = 6 tan2 x or 4 = 10 sin2 x A1 Solve and obtain x = 0.685 A1 [6] (ii) Using du = ± cos x dx, or equivalent, express integral in terms of u and du M1 Obtain ∫ 4u 2 1( − u 2 ) du , or equivalent A1 Use limits u = 0 and u = 1 in an integral of the form au3 + bu5 M1 8 Obtain answer (or 0.533) A1 [4] 15

This question in 9709/33 May/June 2013

4 ln x 3 Find the exact value of dx. [5] x 1

5 marks

Mark scheme: 1 3 EITHER: Integrate by parts and reach kx 2 ln x − m ∫ x 2 . x dx M1* 1 1 Obtain 2 x 2 ln x − 2 ∫ 1 d x , or equivalent A1 x 2 1 1 Integrate again and obtain 2 x 2 ln x − 4 x 2 , or equivalent A1 Substitute limits x = 1 and x = 4, having integrated twice M1(dep*) Obtain answer 4(ln 4 − )1 , or exact equivalent A1 1 1 u u OR1: Using u = ln x, or equivalent, integrate by parts and reach kue 2 − m ∫ e 2 d u M1* 1 1 u u Obtain 2u e 2 − 2 ∫ e 2 d u , or equivalent A1 1 1 u u Integrate again and obtain 2ue 2 − 4e 2 , or equivalent A1 Substitute limits u = 0 and u = ln4, having integrated twice M1(dep*) Obtain answer 4 ln 4 −,4 or exact equivalent A1 1 OR2: Using u = x , or equivalent, integrate and obtain ku ln u − m ∫ u . u d u M1* Obtain 4u ln u − 4 ∫ 1du , or equivalent A1 Integrate again and obtain 4u ln u − 4u , or equivalent A1 Substitute limits u = 1 and u = 2, having integrated twice or quoted ∫ ln u du as u ln u ± u M1(dep*) Obtain answer 8 ln 2 − 4 , or exact equivalent A1 x ln x ± x x ln x ± x OR3: Integrate by parts and reach I = + k ∫ dx M1* x x x x ln x − x 1 1 1 I − Obtain I = + 2 2 ∫ dx A1 x x Integrate and obtain I = 2 x ln x − 4 x , or equivalent A1 Substitute limits x = 1 and x = 4, having integrated twice M1(dep*) Obtain answer 4 ln 4 −,4 or exact equivalent A1 [5] GCE A LEVEL – October/November 2013 9709 32

This question in 9709/32 Oct/Nov 2013

p 2x5 It is given that Ó 0 4xe−1 dx = 9, where p is a positive constant. @8p 16 A (i) Show that p 2 ln . [5] + 7 = (ii) Use an iterative process based on the equation in part (i) to find the value of p correct to 3 significant figures. Use a starting value of 3.5 and give the result of each iteration to 5 significant figures. [3]

8 marks

Mark scheme: 5 (i) Use integration by parts to obtain axe 2 + ∫ b e 2 dx M1* 1 1 − x − x Obtain − 8 xe 2 + ∫ 8e 2 d x or unsimplified equivalent A1 1 1 − x − x Obtain − 8 xe 2 − 16 e 2 A1 Use limits correctly and equate to 9 M1 (d*M)  8 p + 16  Obtain given answer p = 2 ln   correctly A1 [5]  7  GCE A LEVEL – October/November 2013 9709 33 (ii) Use correct iteration formula correctly at least once M1 Obtain final answer 3.77 A1 Show sufficient iterations to 5sf or better to justify accuracy 3.77 or show sign change in interval (3.765, 3.775) A1 [3] [ 5.3 → .3 6766 → .3 7398 → .3 7619 → .3 7696 → .3 7723 ]

This question in 9709/33 Oct/Nov 2013

9 y M x O 120 The diagram shows the curve y e2 sinx cosx for 0 and its maximum point M. = ≤x ≤120, (i) Using the substitution u sin x, find the exact value of the area of the shaded region bounded by = the curve and the axes. [5] (ii) Find the x-coordinate of M, giving your answer correct to 3 decimal places. [6]

11 marks

Mark scheme: 9 (i) Substitute for x and dx throughout using u = sinx and du = cos x dx, or equivalent M1 Obtain integrand e 2 u A1 Obtain indefinite integral 12 e 2 u A1 Use limits u = 0, u = 1 correctly, or equivalent M1 Obtain answer 12 e( 2 − )1 , or exact equivalent A1 5 GCE A LEVEL – May/June 2014 9709 33 (ii) Use chain rule or product rule M1 Obtain correct terms of the derivative in any form, e.g. 2cosx e 2 sin x cos x − e 2 sin x sin x A1 + A1 Equate derivative to zero and obtain a quadratic equation in sin x M1 Solve a 3-term quadratic and obtain a value of x M1 Obtain answer 0.896 A1 6

This question in 9709/33 May/June 2014

2 (i) Use the trapezium rule with 3 intervals to estimate the value of 230 cosecx dx, Ó 1 60 giving your answer correct to 2 decimal places. [3] (ii) Using a sketch of the graph of y cosec x, explain whether the trapezium rule gives an = overestimate or an underestimate of the true value of the integral in part (i). [2]

5 marks

Mark scheme: 2 (i) State or imply ordinates 2, 1.1547…, 1, 1.1547… B1 1 Use correct formula, or equivalent, with h = π and four ordinates M1 6 Obtain answer 1.95 A1 [3] (ii) Make recognisable sketch of y = cosec x for the given interval B1 Justify a statement that the estimate will be an overestimate B1 [2] 1

This question in 9709/31 Oct/Nov 2014

2 (i) Use the trapezium rule with 3 intervals to estimate the value of 230 cosecx dx, Ó 1 60 giving your answer correct to 2 decimal places. [3] (ii) Using a sketch of the graph of y cosec x, explain whether the trapezium rule gives an = overestimate or an underestimate of the true value of the integral in part (i). [2]

5 marks

Mark scheme: 2 (i) State or imply ordinates 2, 1.1547…, 1, 1.1547… B1 1 Use correct formula, or equivalent, with h = π and four ordinates M1 6 Obtain answer 1.95 A1 [3] (ii) Make recognisable sketch of y = cosec x for the given interval B1 Justify a statement that the estimate will be an overestimate B1 [2] 1

This question in 9709/32 Oct/Nov 2014

5 (a) Find 4 tan22x dx. [3] Ó + 1 1 20 sin x (b) Find the exact value of + 60 dx. [5] Ô 1 sin x 40

8 marks

Mark scheme: 5 (a) Use identity tan 2 2 x = sec 2 2 x − 1 B1 Obtain integral of form ax + b tan 2 x M1 1 Obtain correct 3x + tan 2x , condoning absence of + c A1 [3] 2 1 1 (b) State sin x cos π + cos x sin π B1 2 6 1 cos x sin 16 π Simplify integrand to cos π + or equivalent B1 6 sin x Integrate to obtain at least term of form a In(sin x ) *M1 Apply limits and simplify to obtain two terms M1 dep *M 1 1 Obtain π 3 −1 ln( ) or equivalent A1 [5] 8 2 2 2 

This question in 9709/31 May/June 2015

9 y M x O The diagram shows the curve y and its maximum point M. = x2e2−x (i) Show that the x-coordinate of M is 2. [3] 2 (ii) Find the exact value of Ó 0 x2e2−x dx. [6]

9 marks

Mark scheme: 9 (i) Use product rule to find first derivative M1 Obtain 2 xe 2 −−x x 2 e 2 − x A1 Confirm x = 2 at M A1 [3] (ii) Attempt integration by parts and reach ± x 2 e 2 −±x ∫ 2 xe 2 − x dx *M1 Obtain − x 2 e 2 −+x ∫ 2 xe 2 − x dx A1 Attempt integration by parts and reach ± x 2 e 2 − x ± 2 xe 2 − x ± 2e 2 − x *M1 Obtain − x 2 e 2 − x − 2 xe 2 − x − 2e 2 − x A1 Use limits 0 and 2 having integrated twice M1 dep *M Obtain 2e 2 − 10 A1 [6] dx 2 dy 2

This question in 9709/31 May/June 2015

a 6 It is given that x cosx dx 0.5, where 0 a 1 Ó 0 = < < 20. 1.5 (i) Show that a satisfies the equation sin a −cosa . [4] = a (ii) Verify by calculation that a is greater than 1. [2] (iii) Use the iterative formula P1.5 −cosan Q sin−1 an+1 = an to determine the value of a correct to 4 decimal places, giving the result of each iteration to 6 decimal places. [3]

9 marks

Mark scheme: 6 (i) Integrate and reach ± x sin x m ∫ sin x dx M1* Obtain integral x sin x + cos x A1 Substitute limits correctly, must be seen since AG, and equate result to 0.5 M1(dep*) Obtain the given form of the equation A1 4 (ii) EITHER: Consider the sign of a relevant expression at a = 1 and at another relevant value, π e.g. a = 1.5 Y M1 2 OR: Using limits correctly, consider the sign of [x sin x + cos x ]0a − 5.0 , or compare the value of [x sin x + cos x ]a0 with 0.5, for a =1 AND for another relevant value, π e.g a = 1.5 Y . M1 2 Complete the argument, so change of sign, or above and below stated, both with correct calculated values A1 2 (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.2461 A1 Show sufficient iterations to 6 d.p. to justify 1.2461 to 4 d.p., or show there is a sign change in the interval (1.24605, 1.24615) A1 3

This question in 9709/33 May/June 2015

10 y M R x O 1 p x2 The diagram shows the curve y = for x ≥0, and its maximum point M. The shaded region R 1 + x3 is enclosed by the curve, the x-axis and the lines x = 1 and x = p. (i) Find the exact value of the x-coordinate of M. [4] (ii) Calculate the value of p for which the area of R is equal to 1. Give your answer correct to 3 significant figures. [6]

10 marks

Mark scheme: 10 (i) Use the quotient rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and solve for x M1 Obtain answer x = 3 2 , or exact equivalent A1 [4] (ii) State or imply indefinite integral is of the form k ln(1 + 3x ) M1 1 ln(1 + x 3 ) A1 State indefinite integral 3 3 ) M1 Substitute limits correctly in an integral of the form k ln(1 + x 1 State or imply that the area of R is equal to ln(1 + p 3 ) − 1 ln 2 , or equivalent A1 3 3 Use a correct method for finding p from an equation of the form ln(1 + p 3 ) = a or ln((l + p 3 /) 2) = b M1 Obtain answer p = 3.40 A1 [2]

This question in 9709/31 Oct/Nov 2015

10 y M R x O 1 p x2 The diagram shows the curve y = for x ≥0, and its maximum point M. The shaded region R 1 + x3 is enclosed by the curve, the x-axis and the lines x = 1 and x = p. (i) Find the exact value of the x-coordinate of M. [4] (ii) Calculate the value of p for which the area of R is equal to 1. Give your answer correct to 3 significant figures. [6]

10 marks

Mark scheme: 10 (i) Use the quotient rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and solve for x M1 Obtain answer x = 3 2 , or exact equivalent A1 [4] (ii) State or imply indefinite integral is of the form k ln(1 + 3x ) M1 1 ln(1 + x 3 ) A1 State indefinite integral 3 3 ) M1 Substitute limits correctly in an integral of the form k ln(1 + x 1 State or imply that the area of R is equal to ln(1 + p 3 ) − 1 ln 2 , or equivalent A1 3 3 Use a correct method for finding p from an equation of the form ln(1 + p 3 ) = a or ln((l + p 3 /) 2) = b M1 Obtain answer p = 3.40 A1 [2]

This question in 9709/32 Oct/Nov 2015

7 y M x O 2 12x The diagram shows part of the curve y 2x e and its maximum point M. = −x2 (i) Find the exact x-coordinate of M. [4] (ii) Find the exact value of the area of the shaded region bounded by the curve and the positive x-axis. [5]

9 marks

Mark scheme: 7 (i) Use the correct product rule M1 1 2 x 1 2 12 x Obtain correct derivative in any form, e.g. (2 − 2 x )e + 2 (2 x − x )e A1 Equate derivative to zero and solve for x M1 Obtain x = 5 − 1 only A1 [4] (2 − 2 x )e 1 (ii) Integrate by parts and reach a (2 x − x 2 )e 1 2 x dx M1* 2 x + b ∫ 1 (2 − 2 x )e d x , or equivalent A1 Obtain 2 e 2 x (2 x − x 2 ) − 2 ∫ 12 x 2 x , or equivalent A1 Complete the integration correctly, obtaining (12 x − 2 x 2 − 24)e 1 Use limits x = 0, x = 2 correctly having integrated by parts twice DM1 Obtain answer 24 – 8e, or exact simplified equivalent A1 [5]

This question in 9709/31 Oct/Nov 2016

5 (i) Prove the identity tan tan sec [4] 21 −tan 1  1 21. 160 1 3 (ii) Hence show that tan sec ln [4] 2 2. Ó 0 1 21 d1 =

8 marks

Mark scheme: 5 (i) EITHER: Use tan 2A formula to express LHS in terms of tanθ M1 Express as a single fraction in any correct form A1 Use Pythagoras or cos 2A formula M1 Obtain the given result correctly A1 OR: Express LHS in terms of sin 2θ, cos 2θ, sin θand cos θ M1 Express as a single fraction in any correct form A1 Use Pythagoras or cos 2A formula or sin(A – B) formula M1 Obtain the given result correctly A1 [4] (ii) Integrate and obtain a term of the form a ln(cos2θ) or b ln(cosθ) (or secant equivalents) M1* Obtain integral − 12 ln(cos 2θ) + ln(cosθ) , or equivalent A1 Substitute limits correctly (expect to see use of both limits) DM1 Obtain the given answer following full and correct working A1 [4]

This question in 9709/32 Oct/Nov 2016

7 y M x O 2 12x The diagram shows part of the curve y 2x e and its maximum point M. = −x2 (i) Find the exact x-coordinate of M. [4] (ii) Find the exact value of the area of the shaded region bounded by the curve and the positive x-axis. [5]

9 marks

Mark scheme: 7 (i) Use the correct product rule M1 1 2 x 1 2 12 x Obtain correct derivative in any form, e.g. (2 − 2 x )e + 2 (2 x − x )e A1 Equate derivative to zero and solve for x M1 Obtain x = 5 − 1 only A1 [4] (2 − 2 x )e 1 (ii) Integrate by parts and reach a (2 x − x 2 )e 1 2 x dx M1* 2 x + b ∫ 1 (2 − 2 x )e d x , or equivalent A1 Obtain 2 e 2 x (2 x − x 2 ) − 2 ∫ 12 x 2 x , or equivalent A1 Complete the integration correctly, obtaining (12 x − 2 x 2 − 24)e 1 Use limits x = 0, x = 2 correctly having integrated by parts twice DM1 Obtain answer 24 – 8e, or exact simplified equivalent A1 [5]

This question in 9709/32 Oct/Nov 2016

7 h m A water tank has vertical sides and a horizontal rectangular base, as shown in the diagram. The area of the base is 2 m2. At time t 0 the tank is empty and water begins to flow into it at a rate of 1 m3 per hour. At the same time water= begins to flow out from the base at a rate of 0.2 h m3 per hour, where h m is the depth of water in the tank at time t hours. (i) Form a differential equation satisfied by h and t, and show that the time T hours taken for the depth of water to reach 4 m is given by 4 10 T dh. [3] 5 h = Ô0 − … … … … … … … … … … … … … … … (ii) Using the substitution u 5 h, find the value of T. [6] = − … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(i) dV dh B1 State or imply = 2 dt dt dV B1 State or imply = 1 − 0.2 h dt Obtain the given answer correctly B1 Total: 3 7(ii) 1 B1 State or imply d u = − d h , or equivalent 2 h Substitute for h and dh throughout M1 5 A1 20(5 − u ) Obtain T d u , or equivalent =∫ u 3 Integrate and obtain terms 100ln u − 20u , or equivalent A1 Substitute limits u = 3 and u = 5 correctly M1 Obtain answer 11.1, with no errors seen A1 Total: 6

This question in 9709/32 Feb/March 2017

10 y P Q x e O The diagram shows the curve y ln x 2. The x-coordinate of the point P is equal to e, and the normal to the curve at P meets the x-axis= at Q. (i) Find the x-coordinate of Q. [4] … … … … … … … … … … … … (ii) Show that ln x dx x ln x c, where c is a constant. [1] Ó = −x + … … … … … (iii) Using integration by parts, or otherwise, find the exact value of the area of the shaded region between the curve, the x-axis and the normal PQ. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(i) ln x B1 State or imply derivative is 2 x State or imply gradient of the normal at x = e is − 12 e , or equivalent B1 Carry out a complete method for finding the x-coordinate of Q M1 2 A1 Obtain answer x = e + , or exact equivalent e Total: 4 10(ii) Justify the given statement by integration or by differentiation B1 Total: 1 10(iii) 2 ln x M1* x. dx Integrate by parts and reach ax (ln x ) + b ∫ x Complete the integration and obtain x (ln x ) 2 − 2 x ln x + 2 x , or equivalent A1 Use limits x = 1 and x = e correctly, having integrated twice DM1 Obtain exact value e – 2 A1 1 B1 Use x- coordinate of Q found in part (i) and obtain final answer e − 2 + e Total: 5

This question in 9709/32 Feb/March 2017

3 It is given that x ln 1 y, where 0 y 1. = −y −ln < < e−x (i) Show that y . [2] = 1 e−x + … … … … … … … … … … … … … … … … … … … … … … … … 1 @ A 2e (ii) Hence show that y dx ln . [4] Ó 0 = e 1 + … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: Question Answer Marks 3(i) 1 − y B1 Remove logarithms correctly and obtain ex = y e − x B1 Obtain the given answer y = − x following full working 1 + e Total: 2 3(ii) State integral k ln(1 + e − x ) where k = ± 1 *M1 State correct integral − ln(1 + e − x ) A1 Use limits correctly DM1  2e  A1 Obtain the given answer ln   following full working  e + 1  Total: 4

This question in 9709/31 May/June 2017

10 y M x O 140 The diagram shows the curve y sin x cos22x for 0 and its maximum point M. = ≤x ≤140 (i) Using the substitution u cosx, find by integration the exact area of the shaded region bounded = by the curve and the x-axis. [6] … … … … … … … … … … … … … … … … … … … (ii) Find the x-coordinate of M. Give your answer correct to 2 decimal places. [6] … … … … … … … … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 10(i) State or imply d u = − sin x d x B1 Using correct double angle formula, express the integral in terms of u and du M1 Obtain integrand ± (2u 2 − 1) 2 A1 1 A1 Change limits and obtain correct integral (2u 2 − 1) 2 d u with no errors seen ∫ 1 2 Substitute limits in an integral of the form au 5 + bu 3 + cu M1 Obtain answer 151 (7 − 4 2) , or exact simplified equivalent A1 Total: 6 10(ii) Use product rule and chain rule at least once M1 Obtain correct derivative in any form A1 Equate derivative to zero and use trig formulae to obtain an equation in M1 cos x and sin x Use correct methods to obtain an equation in cos x or sin x only M1 Obtain 10cos 2 x = 9 or 10sin 2 x = 1 , or equivalent A1 Obtain answer 0.32 A1 Total: 6

This question in 9709/31 May/June 2017

1 dy 7 (i) Prove that if y then sec tan [2] cos = = 1 1. 1 d1 … … … … … … … … … 1 sin (ii) Prove the identity 2 2 sec tan [3] + 1 1  sec21 + 1 1 −1. −sin 1 … … … … … … … … … … … … … … … … … … … … … … … … 1 40 1 sin (iii) Hence find the exact value of [4] + 1 1 d1. Ô0 −sin 1 … … … … … … … … … … … … …

9 marks

Mark scheme: 7(i) Use quotient or chain rule M1 Obtain given answer correctly A1 Total: 2 7(ii) EITHER: (M1 Multiply numerator and denominator of LHS by 1 + sinθ Use Pythagoras and express LHS in terms of sec θand tanθ M1 Complete the proof A1) OR1: (M1 Express RHS in terms of cos θand sin θ Use Pythagoras and express RHS in terms of sin θ M1 Complete the proof A1) OR2: (M1 Express LHS in terms of secθ and tanθ Multiply numerator and denominator by secθ + tanθ and use Pythagoras M1 Complete the proof A1) Total: 3 7(iii) Use the identity and obtain integral 2tanθ+ 2secθ− θ B2 Use correct limits correctly in an integral containing terms a tanθand b secθ M1 1 A1 Obtain answer 2 2 − 4π Total: 4

This question in 9709/32 May/June 2017

10 y M x O p 140 The diagram shows the curve y x2 cos 2x for 0 The curve has a maximum point at M where x p. = ≤x ≤140. = 1 @1 A (i) Show that p satisfies the equation p . [3] 2 tan−1 p = … … … … … … … … @ A 1 1 (ii) Use the iterative formula to determine the value of p correct to 2 decimal 2 tan−1 pn+1 = pn places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … (iii) Find, showing all necessary working, the exact area of the region bounded by the curve and the x-axis. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 10(i) Use correct product rule M1 2 A1 Obtain correct derivative in any form y ′ = 2 x cos2 x − 2 x sin 2 x ( ) Equate to zero and derive the given equation A1 Total: 3 10(ii) Use the iterative formula correctly at least once e.g. M1 0.5 → 0.55357 → 0.53261 → 0.54070 → 0.53755 Obtain final answer 0.54 A1 Show sufficient iterations to 4 d.p. to justify 0.54 to 2 d.p., or show there is a sign change in A1 the interval (0.535, 0.545) Total: 3 10(iii) 2 *M1 Integrate by parts and reach ax sin 2 x + b ∫ x sin 2 x dx 1 2 1 A1 Obtain x sin 2 x −∫ 2 x. sin 2 x dx 2 2 1 2 1 1 A1 Complete integration and obtain x sin 2 x + x cos2 x − sin 2 x , or equivalent 2 2 4 1 DM1 Substitute limits x = 0, x = 4π , having integrated twice 1 2 A1 Obtain answer (π − 8) , or exact equivalent 32 Total: 5

This question in 9709/32 May/June 2017

1 20 14 Find the exact value of sin [4] Ó 0 1 21 d1. … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 4 1 1 *M1 cos 2 θ dθ Integrate by parts and reach aθcos 2 θ+ b ∫ Complete integration and obtain indefinite integral −2θcos 12 θ+ 4sin 12 θ A1 Substitute limits correctly, having integrated twice DM1 Obtain final answer ( 4 − π) / 2 , or exact equivalent A1 Total: 4

This question in 9709/33 May/June 2017

9 y R x O 2 2x 1 The diagram shows the curve y The shaded region R is enclosed by the curve, = + x2 e−1 for x ≥0. the x-axis and the lines x 0 and x 2. = = (i) Find the exact values of the x-coordinates of the stationary points of the curve. [4] … … … … … … … … … … … … … … … … … … (ii) Show that the exact value of the area of R is 18 . [5] −42e … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 9(i) Use correct product or quotient rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and obtain a 3 term quadratic equation in x M1 Obtain answers x = 2 ± 3 A1 4 9(ii) 2 − 12 x − 12 x *M1 xe d x Integrate by parts and reach k (1 + x )e + l ∫ 2 − 12 x − 12 x A1 xe dx , or equivalent Obtain −2(1 + x )e + 4 ∫ 2 − 12 x A1 Complete the integration and obtain ( −18 − 8 x − 2 x )e , or equivalent Use limits x = 0 and x = 2 correctly, having fully integrated twice by parts DM1 Obtain the given answer A1 5

This question in 9709/31 Oct/Nov 2017

a 1 9 It is given that x 2 ln x dx 2, where a 1. Ó 1 = > 3 3 7 2a 2 (i) Show that a 2 . [5] + 3 ln a = … … … … … … … … … … … … … … … … … … … … … … … (ii) Show by calculation that a lies between 2 and 4. [2] … … … … … … … … … … … (iii) Use the iterative formula ` 3 a23 7 2a2n + 3 ln an an+1 = to determine a correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] … … … … … … … … … …

10 marks

Mark scheme: 3 9(i) 3 2 2 1 *M1 x . dx Integrate by parts and reach ax ln x + b ∫ x 2 32 2 12 A1 Obtain 3 x ln x 3 x d x −∫ 2 32 4 32 A1 Obtain integral 3 x ln x − 9 x , or equivalent Substitute limits correctly and equate to 2 DM1 Obtain the given answer correctly AG A1 5 9(ii) Evaluate a relevant expression or pair of expressions at x = 2 and x = 4 M1 Complete the argument correctly with correct calculated values A1 2 9(iii) Use the iterative formula correctly at least once M1 Obtain final answer 3.031 A1 Show sufficient iterations to 5 d.p. to justify 3.031 to 3 d.p., or show there is a sign A1 change in the interval (3.0305, 3.0315) 3

This question in 9709/32 Oct/Nov 2017

2 sin x 2x sin x 4 (i) Show that [4] −sin 1 2x 1 cosx.  −cos + … … … … … … … … … … … … … … … … … … … … … … … … … 1 20 2 sin x 2x (ii) Hence, showing all necessary working, find dx, giving your answer in the −sin Ô 1 1 2x 30 −cos form ln k. [4] … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4(i) Use correct double angle formulae and express LHS in terms of cos x and sin x M1 ( ) 2 2sin 2sin cos 1 2cos 1 − − x x x x Obtain a correct expression A1 Complete method to get correct denominator e.g. by factorising to remove a factor of 1 cos − x M1 Obtain the given RHS correctly OR (working R to L): A1 2 sin 1 cos sin sin cos 1 cos 1 cos 1 cos − − × = + − − x x x x x x x x M1A1 2 2sin 2sin cos 2 2cos − = − x x x x Given answer so check working carefully 2sin sin2 1 cos2 − = − x x x M1A1 4 4(ii) State integral of the form ln(1 cos ) + a x M1* If they use the substitution 1 cos = + u x allow M1A1 for ln − u Obtain integral ln(1 cos ) − + x A1 Substitute correct limits in correct order M1(dep)* Obtain answer ( ) 3 ln 2 , or equivalent A1 4

This question in 9709/32 May/June 2018

8 y M x O 3x The diagram shows the curve y x 1 e−1 and its maximum point M. = + (i) Find the x-coordinate of M. [4] … … … … … … … … … … … … … … … … … … … (ii) Find the area of the shaded region enclosed by the curve and the axes, giving your answer in terms of e. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 8(i) Use correct product or quotient rule M1 ( ) 1 1 1 3 3 3 d 1 e e d − − − = + + x x y x x or ( ) 1 1 3 3 2 3 1 e 1 e d 3 d e − + = x x x x y x Obtain complete correct derivative in any form A1 Equate derivative to zero and solve for x M1 Obtain answer x = 2 with no errors seen A1 4 8(ii) Integrate by parts and reach ( ) 1 1 3 3 1 e e d − − + + ∫ x x a x b x M1* Obtain ( ) 1 1 3 3 3 1 e 3 e d − − − + + ∫ x x x x , or equivalent A1 1 1 1 3 3 3 3 3e d 3e − − − − + − ∫ x x x xe x Complete integration and obtain ( ) 1 1 3 3 3 1 e 9e − − − + − x x x , or equivalent A1 Use correct limits x = – 1 and x = 0 in the correct order, having integrated twice M1(dep*) Obtain answer 1 3 9e 12 − , or equivalent A1 5

This question in 9709/32 May/June 2018

7 y M R x O 120 The diagram shows the curve y 5 sin2x cos3x for 0 and its maximum point M. The shaded = ≤x ≤120, region R is bounded by the curve and the x-axis. (i) Find the x-coordinate of M, giving your answer correct to 3 decimal places. [5] … … … … … … … … … … … … … … … … … … … (ii) Using the substitution u sin x and showing all necessary working, find the exact area of R. [4] = … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(i) Use product rule M1* Obtain correct derivative in any form A1 Equate derivative to zero and obtain an equation in a single trig function depM1* Obtain a correct equation, e.g. 2 3 tan 2 x = A1 Obtain answer x = 0.685 A1 5 Question Answer Marks Guidance 7(ii) Use the given substitution and reach ( ) 2 4 d a u u u ∫ − M1 Obtain correct integral with a = 5 and limits 0 and 1 A1 Use correct limits in an integral of the form 3 5 1 1 3 5 a u u   −     M1 Obtain answer 2 3 A1 4

This question in 9709/31 Oct/Nov 2018

10 y M x O 120 The diagram shows the curve y sin3x cosx for 0 and its maximum point M. = ≤x ≤120, (i) Using the substitution u cosx, find by integration the exact area of the shaded region bounded = by the curve and the x-axis. [6] … … … … … … … … … … … … … … … … … … … (ii) Showing all your working, find the x-coordinate of M, giving your answer correct to 3 decimal places. [6] … … … … … … … … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 10(i) State or imply du = – sin x dx B1 Using Pythagoras express the integral in terms of u M1 Obtain integrand ( ) 2 1 ± − u u A1 Integrate and obtain 3 7 2 2 2 2 3 7 − + u u , or equivalent A1 Change limits correctly and substitute correctly in an integral of the form 3 7 2 2 + au bu M1 Or substitute original limits correctly in an integral of the form 3 2 (cos ) a x + 7 2 (cos ) b x Obtain answer 8 21 A1 6 10(ii) Use product rule and chain rule at least once M1 Obtain correct derivative in any form A1 + A1 Equate derivative to zero and obtain a horizontal equation in integral powers of sin x and cos x M1 Use correct methods to obtain an equation in one trig function M1 Obtain 2 tan 6 = x , 2 7cos 1 = x or 2 7sin 6 = x , or equivalent, and obtain answer 1.183 A1 6

This question in 9709/32 Feb/March 2019

1 Use the trapezium rule with 3 intervals to estimate the value of 3 2x dx. [3] Ó0 −4 … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 1 State or imply ordinates 3, 2, 0, 4 B1 These and no more Accept in unsimplified form 0 2 4 − etc. Use correct formula, or equivalent, with h = 1 and four ordinates M1 Obtain answer 5.5 A1 3

This question in 9709/31 May/June 2019

a 39 It is given that 3, where the constant a is such that 0 a x cos 3x1 dx = < < 20. Ó 0 (i) Show that a satisfies the equation 4 cos 13a −3 . [5] a = sin 13a … … … … … … … … … … … … … … … … … … … … … … (ii) Verify by calculation that a lies between 2.5 and 3. [2] … … … … … … … … … … … (iii) Use an iterative formula based on the equation in part (i) to calculate a correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] … … … … … … … … … … … …

10 marks

Mark scheme: 9(i) Commence integration by parts, reaching 1 1 sin sin d 3 3 ax x b x x −∫ *M1 Obtain 1 1 3 sin 3 sin d 3 3 x x x x −∫ A1 Complete integration and obtain 1 1 3 sin 9cos 3 3 x x x + A1 Substitute limits correctly and equate result to 3 in an integral of the form 1 1 sin cos 3 3 px x q x + DM1 ( ) 3 3 sin 9cos 0 9 3 3 a a a = + − − Obtain 4 3cos 3 sin 3 a a a − = correctly A1 With sufficient evidence to show how they reach the given equation 5 9(ii) Calculate values at a = 2.5 and a = 3 of a relevant expression or pair of expressions. M1 2.5 2.679 and 3 2.827 < > If using 2.679 and 2.827 must be linked explicitly to 2.5 and 3. Solving f(a) = 0, f(2.5) = 0.179. and f(3) = –0.173 or if 1 1 f ( ) sin 3cos 4 f (2.5) 0.13..,f (3) 0.145... 3 3 a a a a = + − ⇒ = − = Complete the argument correctly with correct calculated values A1 Accept values to 1 sf. or better 2 Question Answer Marks Guidance 9(iii) Use the iterative process 1 na + = 1 3 1 1 3 4 3cos sin n n n a a a + − correctly at least once M1 Show sufficient iterations to at least 5 d.p. to justify 2.736 to 3d.p., or show a sign change in the interval (2.7355, 2.7365) A1 Obtain final answer 2.736 A1 3

This question in 9709/32 Oct/Nov 2019

8 y x O 120 The diagram shows the graph of y sec x for 0 1 = ≤x < 20. 1.2 (i) Use the trapezium rule with 2 intervals to estimate the value of sec x dx, giving your answer Ó 0 correct to 2 decimal places. [3] … … … … … … … (ii) Explain, with reference to the diagram, whether the trapezium rule gives an overestimate or an underestimate of the true value of the integral in part (i). [1] … … … … … … … (iii) P is the point on the part of the curve y sec x for 0 1 at which the gradient is 2. By first = ≤x < 20 1 differentiating cosx, find the x-coordinate of P, giving your answer correct to 3 decimal places. [6] … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 8(i) State or imply ordinates 1, 1.2116…, 2.7597... B1 Use correct formula, or equivalent, with h = 0.6 M1 Obtain answer 1.85 A1 3 8(ii) Explain why the rule gives an overestimate B1 1 8(iii) Differentiate using quotient or chain rule M1 Obtain correct derivative in terms of sin x and cos x A1 Equate derivative to 2, use Pythagoras and obtain an equation in sin x M1 Obtain 2 2sin sin 2 0 + − = x x A1 OE Solve a 3-term quadratic for x M1 Obtain answer x = 0.896 only A1 6

This question in 9709/33 Oct/Nov 2019

10 y M R x 0 O The diagram shows the graph of y ecos x sin3x for 0 and its maximum point M. The shaded = ≤x ≤0, region R is bounded by the curve and the x-axis. (i) Find the x-coordinate of M. Show all necessary working and give your answer correct to 2 decimal places. [5] … … … … … … … … … … … … … … … … … (ii) By first using the substitution u cosx, find the exact value of the area of R. [7] = … … … … … … … … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 10(i) Use product rule and chain rule at least once M1 Obtain correct derivative in any form A1 Equate derivative to zero, use Pythagoras and obtain an equation in cos x M1 Obtain 2 cos 3cos 1 0 + −= x x , or 3-term equivalent A1 Obtain answer x = 1.26 A1 5 10(ii) Using du = ± sin x dx express integrand in terms of u and du M1 Obtain integrand ( ) 2 e 1 − u u A1 OE Commence integration by parts and reach ( ) 2 e 1 e d − + ∫ u u a u b u u *M1 Obtain ( ) 2 e 1 2 e d − − ∫ u u u u u A1 OE Complete integration, obtaining ( ) 2 e 2 1 − + u u u A1 OE Substitute limits u = 1 and u = – 1 (or x = 0 and x = π ), having integrated completely DM1 Obtain answer 4 e , or exact equivalent A1 7

This question in 9709/33 Oct/Nov 2019

5 (a) Find the quotient and remainder when 2x3 6x 3 is divided by x2 3. [3] −x2 + + + … … … … … … … … … … … … … … … … … … … … … … … … … 3 2x3 6x 3 (b) Using your answer to part (a), find the exact value of dx. [5] −x2 + + x2 3 Ô1 + … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(a) M1 Obtain quotient 2x – 1 A1 Obtain remainder 6 A1 3 5(b) Obtain terms x2 – x (FT on quotient of the form 2x + k) B1FT Obtain term of the form 1 tan 3 −      x a M1 Obtain term 1 6 tan 3 3 −      x (FT on a constant remainder) A1FT Use x = 1 and x = 3 as limits in a solution containing a term of the form ( ) 1 tan− a bx M1 Obtain final answer 1 6 3 π + , or exact equivalent A1 5

This question in 9709/31 May/June 2020

10 r h A A tank containing water is in the form of a hemisphere. The axis is vertical, the lowest point is A and the radius is r, as shown in the diagram. The depth of water at time t is h. At time t 0 the tank is = full and the depth of the water is r. At this instant a tap at A is opened and water begins to flow out at a rate proportional to h. The tank becomes empty at time t 14. = The volume of water in the tank is V when the depth is h. It is given that V 1 3rh2 . = 3π −h3 (a) Show that h and t satisfy a differential equation of the form dh B 1 3 , dt = − 2 2 2rh −h where B is a positive constant. [4] … … … … … … … … … … … … … … (b) Solve the differential equation and obtain an expression for t in terms of h and r. [8] … … … … … … … … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 10(a) State or imply d d V k h t = − B1 State or imply 2 d 2π π d V rh h h = − , or equivalent B1 Use d d d . d d d V V h t h t = M1 Obtain the given answer correctly A1 4 10(b) Separate variables and attempt integration of at least one side M1 Obtain terms 3 5 2 2 4 2 3 5 rh h − and –Bt A3, 2, 1, 0 Use t = 0, h = r to find a constant of integration c M1 Use t = 14, h = 0 to find B M1 Obtain correct c and B, e.g. 5 5 2 2 14 1 , 15 15 c r B r = = A1 Obtain final answer 3 5 2 2 14 20 6 h h t r r     = − +         , or equivalent A1 8

This question in 9709/33 May/June 2020

10 y x O M The diagram shows the curve y = 2 −x e−12x, and its minimum point M. (a) Find the exact coordinates of M. [5] … … … … … … … … … … … … … … … … … … … (b) Find the area of the shaded region bounded by the curve and the axes. Give your answer in terms of e. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(a) Use correct product or quotient rule *M1 ( ) 1 1 2 2 d 1 2 e e d 2 − −   = − − −     x x y x x M1 requires at least one of derivatives correct Obtain correct derivative in any form A1 Equate derivative to zero and solve for x DM1 Obtain x = 4 A1 ISW Obtain y = –2e–2, or exact equivalent A1 5 Question Answer Marks Guidance 10(b) Commence integration and reach ( ) 1 1 2 2 2 e e d − − − +  x x a x b x *M1 Condone omission of dx ( ) 1 1 2 2 2 2 e 4e − − − − + x x x or 1 2 2 e −x x Obtain ( ) 1 1 2 2 2 2 e 2 e d − − − − − x x x x A1 OE Complete integration and obtain 1 2 2 e −x x A1 OE Use correct limits, x = 0 and x = 2, correctly, having integrated twice DM1 Ignore omission of zeros and allow max of 1 error Obtain answer 4e–1, or exact equivalent A1 ISW Alternative method for question 10(b) 1 2 1 1 2 2 d 2 e 2e e d − − −        = − x x x x x x *M1 A1 1 2 2 e − ∴ x x A1 Use correct limits, x = 0 and x = 2, correctly, having integrated twice DM1 Ignore omission of zeros and allow max of 1 error Obtain answer 4e–1, or exact equivalent A1 ISW 5

This question in 9709/31 Oct/Nov 2020

10 y R 3 x O a 2π M The diagram shows the curve y x cos x, for 0 and its minimum point M, where x a. = ≤x ≤32π, = The shaded region between the curve and the x-axis is denoted by R. 1 (a) Show that a satisfies the equation tan a [3] = 2a. … … … … … … @ A 1 , with initial value (b) The sequence of values given by the iterative formula an+1 = π + tan−1 2an x1 3, converges to a. = Use this formula to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … (c) Find the volume of the solid obtained when the region R is rotated completely about the x-axis. Give your answer in terms of [6] π. … … … … … … … … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 10(a) Use correct product rule M1 Obtain correct derivative in any form A1 e.g. d 1 cos sin d 2 y x x x x x = − . Accept in a or in x Equate derivative to zero and obtain 1 tan 2 a a = A1 Obtain given answer from correct working. The question says ‘show that ..’ so there should be an intermediate step e.g. cos 2 sin x x x = . Allow 1 tan 2 x x = 3 10(b) Use the iterative process correctly at least once (get one value and go on to use it in a second use of the formula) M1 Must be working in radians Degrees gives 1, 12.6039, 5.4133, ... M0 Obtain final answer 3.29 A1 Clear conclusion Show sufficient iterations to at least 4 d.p.to justify 3.29, or show there is a sign change in the interval (3.285, 3.295) A1 3, 3.3067, 3.2917, 3.2923 Allow more than 4d.p. Condone truncation. 3 Question Answer Marks Guidance 10(c) State or imply the indefinite integral for the volume is ( ) 2 π cos d x x x  B1 [If π omitted, or 2π or 1 π 2 used, give B0 and follow through. 4/6 available] Use correct cos 2A formula, commence integration by parts and reach ( sin2 ) sin2 d x ax b x ax b x x + ±  + *M1 Alternative: 2 1 sin 2 sin 2 d 4 4 4 x x x x x + − Obtain 1 1 1 1 ( sin2 ) sin2 d 2 4 2 4 x x x x x x + − + , or equivalent A1 Complete integration and obtain 2 1 1 1 sin2 cos2 4 4 8 x x x x + + A1 OE Substitute limits x = 0 and x = 1 π 2 , having integrated twice DM1 2 π π 1 1 0 0 0 2 8 4 4   + − − − −     Obtain answer ( ) 2 1 π π 4 16 − , or exact equivalent A1 CAO 6

This question in 9709/32 Oct/Nov 2020

10 y x O M The diagram shows the curve y = 2 −x e−12x, and its minimum point M. (a) Find the exact coordinates of M. [5] … … … … … … … … … … … … … … … … … … … (b) Find the area of the shaded region bounded by the curve and the axes. Give your answer in terms of e. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(a) Use correct product or quotient rule *M1 ( ) 1 1 2 2 d 1 2 e e d 2 − −   = − − −     x x y x x M1 requires at least one of derivatives correct Obtain correct derivative in any form A1 Equate derivative to zero and solve for x DM1 Obtain x = 4 A1 ISW Obtain y = –2e–2, or exact equivalent A1 5 Question Answer Marks Guidance 10(b) Commence integration and reach ( ) 1 1 2 2 2 e e d − − − +  x x a x b x *M1 Condone omission of dx ( ) 1 1 2 2 2 2 e 4e − − − − + x x x or 1 2 2 e −x x Obtain ( ) 1 1 2 2 2 2 e 2 e d − − − − − x x x x A1 OE Complete integration and obtain 1 2 2 e −x x A1 OE Use correct limits, x = 0 and x = 2, correctly, having integrated twice DM1 Ignore omission of zeros and allow max of 1 error Obtain answer 4e–1, or exact equivalent A1 ISW Alternative method for question 10(b) 1 2 1 1 2 2 d 2 e 2e e d − − −        = − x x x x x x *M1 A1 1 2 2 e − ∴ x x A1 Use correct limits, x = 0 and x = 2, correctly, having integrated twice DM1 Ignore omission of zeros and allow max of 1 error Obtain answer 4e–1, or exact equivalent A1 ISW 5

This question in 9709/33 Oct/Nov 2020

10 y M x O 1 2π The diagram shows the curve y sin 2x cos2x for 0 and its maximum point M. = ≤x ≤12π, (a) Using the substitution u sin x, find the exact area of the region bounded by the curve and the = x-axis. [5] … … … … … … … … … … … … … … … … … … … (b) Find the exact x-coordinate of M. [6] … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 10(a) State or imply du = cos x dx B1 Using double angle formula for sin2x and Pythagoras, express integral in terms of u and du. M1 Obtain integral ( ) 3 2 d  − u u u A1 OE Use limits u = 0 and u = 1 in an integral of the form 2 4 + au bu , where 0 ≠ ab M1 a + b or a + b − 0 1 1 and 2 a b   = = −     Obtain answer 1 2 A1 5 10(b) Use product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and use a double angle formula *M1 Obtain an equation in one trig variable DM1 Obtain 2 4 sin 1 = x , 2 4 cos 3 = x or 2 3 tan 1 = x A1 Obtain answer 1 π 6 x = A1 6

This question in 9709/32 Feb/March 2021

1 4 (a) Prove that [2] −cos 21 1 cos  tan21. + 21 … … … … … … … … 1 3π 1 (b) Hence find the exact value of [4] −cos 21 Ô 1 1 cos d1. 6π + 21 … … … … … … … … … … … … … …

6 marks

Mark scheme: 4(a) Use correct double angle formula or t-substitution twice M1 Obtain 2 1 cos2 tan 1 cos2 θ θ θ − = + from correct working A1 AG 2 Question Answer Marks Guidance 4(b) Express 2 tan θ in terms of 2 sec θ M1 ( ) 2 π 3 π 6 sec 1 d θ θ ±          Integrate and obtain terms tanθ θ − A1 Accept with a mixture of x and θ Substitute limits correctly in an integral of the form tanθ θ + a b , where 0 ≠ ab M1 3 π 1 π 3 6 3 − − +       Allow if trig. not substituted Obtain answer 2 1 3 π 3 6 − A1 or equivalent exact 2-term expression 4

This question in 9709/31 May/June 2021

0. The curve has one stationary point.9 The equation of a curve is y 3 ln x for x = x−2 > (a) Find the exact coordinates of the stationary point. [5] … … … … … … … … … … … … … … … … … … … … … … … … 8 (b) Show that y dx 18 ln 2 [5] Ó 1 = −9. … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 9(a) Use correct product rule or correct quotient rule M1 Obtain correct derivative in any form A1 2 5 3 3 2 ln 3 ' x x x x y − − − = Equate 2 term derivative to zero and solve for x M1 Obtain answer 3 2 e = x A1 Or exact equivalent Obtain answer = y 3 2e A1 Or exact equivalent 5 Question Answer Marks Guidance 9(b) Commence integration and reach 1 1 3 3 1 ln . d ax x b x x x +  *M1 Obtain 1 1 3 3 1 3 ln 3 . d − x x x x x A1 Complete the integration and obtain 1 1 3 3 3 ln 9 − x x x A1 OE Use limits correctly in an expression of the form 1 1 3 3 ln + px x qx ( ) 0 pq ≠ DM1 6ln8 9 2 0 9 −× − + Obtain 18ln 2 9 − from full and correct working A1 AG need to see ln8 3ln 2 = 5

This question in 9709/31 May/June 2021

8 y M x O ln x The diagram shows the curve y and its maximum point M. = x4 (a) Find the exact coordinates of M. [4] … … … … … … … … … … … … … … … … … … a ln x 1 (b) By using integration by parts, show that for all a 1, dx 9. [6] x4 < > Ô 1 … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 8(a) Use quotient or product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and solve for x M1 Obtain x = 4 e and y = 1 4e , or exact equivalents A1 4 8(b) Commence integration and reach 3 3 ln . − − +  ax x b x 1 x dx *M1 Obtain 3 3 1 1 ln . 3 3 − − − +  x x x 1 x dx A1 OE Complete integration and obtain 3 3 1 1 ln 3 9 − − − − x x x A1 Substitute limits correctly, having integrated twice DM1 Obtain answer 3 3 1 1 1 ln 9 3 9 − − − − a a a A1 OE Justify the given statement A1 6

This question in 9709/33 May/June 2021

π 14 Find the exact value of x sin 2x dx. [5] Ó 1 3π … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 4 Commence integration and reach 1 1 cos cos d 2 2 ax x b x x +  *M1 Obtain 1 1 2 cos 2 cos d 2 2 x x x x − +  A1 OE Complete integration obtaining 1 1 2 cos 4sin 2 2 x x x − + A1 OE Use limits correctly, having integrated twice DM1 Obtain answer 2 + 3 3 π, or exact equivalent A1 5

This question in 9709/33 Oct/Nov 2021

8 (a) Find the quotient and remainder when 8x3 4x2 2x 7 is divided by 4x2 1. [3] + + + + … … … … … … … … … … … … … … … … … … … … … … … … … 1 2 8x3 4x2 2x 7 (b) Hence find the exact value of dx. [5] + + + 4x2 1 Ô0 + … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 8(a) Commence division and reach quotient of the form 2x ± 1 + r Obtain (quotient) 2x + 1 A1 Obtain (remainder) 6 A1 3 Question Answer Marks Guidance 8(b) Obtain terms 2 + x x B1 OE Obtain term of the form 1 tan 2 − a x M1 Obtain term 1 3tan 2 − x A1 OE Use x = 0 and 1 2 = x as limits in a solution containing a term of the form 1 tan 2 − a x M1 2 1 1 π 2 2 4 a æ ö÷ ç + + ÷ ç ÷ çè ø , need π 4 seen or implied Obtain final answer 3 (1 π 4 + ), or exact equivalent A1 ISW, Answers in degrees score A0. 5

This question in 9709/32 Feb/March 2022

11 y M x O 1 2π The diagram shows the curve y sin x cos 2x for 0 and its maximum point M. = ≤x ≤12π, (a) Find the x-coordinate of M, giving your answer correct to 3 significant figures. [6] … … … … … … … … … … … … … … … … … (b) Using the substitution u cosx, find the area of the shaded region enclosed by the curve and the x-axis in the first quadrant,= giving your answer in a simplified exact form. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 11(a) Use correct product rule or chain rule M1 Obtain correct derivative in any form A1 cos x.cos 2x − sin x.2sin 2x Equate derivative to zero and use a correct double angle formula *M1 If chain rule used then derivative set to 0 gains M1 since correct double angle formula has already been used. Obtain an equation in one trigonometric variable DM1 Allow following from coefficient errors in differentiation only Obtain 2 6sin 1 = x , 2 6cos 5 = x or 2 5tan 1 = x A1 One of these 3 expressions Obtain final answer x = 0.421 A1 Must be 3s.f. 6 Question Answer Marks Guidance 11(b) State or imply du = sin − x dx B1 Using double angle formula, express integral in terms of u and du M1 Use cos2x = 2cos2x − 1 Integrate and obtain ± 3 2 3   −     u u A1 Use limits u = 1, u = 1 2 in an integral of the form 3 + au bu , where ab ≠ 0 M1 Require both limits substituted twice in 3 + au bu for M1. Do not condone decimals. Obtain ( ) 1 2 1 3 − or 1 1 2 1 1 2 or or 3 3 3 3 2       simplified equivalent A1 ISW 5

This question in 9709/32 Feb/March 2022

3 27 6 Let I dx. 2 = Ô0 9 x2 + 1 4π (a) Using the substitution x 3 tan show that I [4] = 1, = Ó0 cos21 d1. … … … … … … … … … … … … … … … … … … … … … … … (b) Hence find the exact value of I. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) State or imply 2 d 3sec d x    Substitute throughout for x and dx M1 Obtain any correct form in terms of  A1 e.g.   2 2 2 81sec d 9 9tan      Justify change of limits and obtain π 4 2 0 cos d   correctly A1 AG 4 6(b) Obtain indefinite integral of the form cos 2 d a b     , where 0 ab *M1 Obtain 1 1 sin 2 2 4    A1 Use correct limits correctly in an expression containing and sin 2 p q   where 0 pq  DM1   π 1 8 4 0   Obtain answer   1 π 2 8  A1 Or exact equivalent e.g. 1 1 π 8 4  . 4

This question in 9709/31 May/June 2022

a 10 The constant a is such that x2 ln x dx 4. Ó 1 = @ A1 35 3 (a) Show that a . [5] = 3 ln a −1 … … … … … … … … … … … … … … … … … … … … … … … (b) Verify by calculation that a lies between 2.4 and 2.8. [2] … … … … … … … … … … … … (c) Use an iterative formula based on the equation in part (a) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … …

10 marks

Mark scheme: 10(a) Commence integration and reach 3 3 ln   ax x b x . 1 x dx Obtain 3 3 1 1 ln . 3 3   x x x 1 x dx A1 OE Allow omission of dx. Complete integration and obtain 3 3 1 1 ln 3 9  x x x A1 Allow  1 3 3 1 3       x . Use limits correctly and equate to 4, having integrated twice DM1 3 3 1 1 ln 3 9  a a a  (0  1 9 ) = 4 allow one sign error OR one numerical error, but 0 may be absent or expressed as 3 ln1 3 a . Allow  1 3 3 1 3       ax and  1 3 1 3       . Obtain given result correctly A1 1 3 35 3ln 1        a a AG After substitution, any errors even if corrected A0. Need to see at least one line of working between substitution and the given answer. 5 10(b) Calculate the values of a relevant expression or pair of expressions at a = 2.4 and a = 2.8 All values must be correct for M1 (numerical question) M1 Justify the given statement with correct calculated values A1 2.4 < 2.7(8) and 2.8 > 2.5(6) sign change here insufficient OR –0.3(8) and 0.2(4) < 0, > 0 or change of sign. 2 Question Answer Marks Guidance 10(c) Use the iterative process 1  n a 1 3 35 3ln 1        n a correctly at least twice M1 Obtain final answer a = 2.64 A1 Must be 2 dp. Show sufficient iterations to 4 dp to justify 2.64 to 2 dp, or show there is a sign change in (2.635, 2.645) A1 2.635 (35/(3lna  1))1/3 a = 0.0029(4) > 0 2.645 (35/(3lna  1))1/3 a =  0.012 < 0 3

This question in 9709/33 May/June 2022

9 y x O M The diagram shows part of the curve y = 3 −x e−13x for x ≥0, and its minimum point M. (a) Find the exact coordinates of M. [5] … … … … … … … … … … … … … … … … … (b) Find the area of the shaded region bounded by the curve and the axes, giving your answer in terms of e. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 9(a) Use correct product or quotient rule *M1 Obtain correct derivative in any form A1 dy − 3x 1 − 3x = −e − 3 − x ) e e.g. dx 3 ( Equate their derivative to zero and solve for x DM1 Obtain x = 6 A1 Obtain y = − 3e−2 A1 Or exact equivalent. 5 9(b) − 1 x − 1 x *M1 3 + b  e 3 dx, where ab  0 Commence integration and reach a ( 3 − x ) e 1 1 − x − x A1 3 −3 e 3 dx, or equivalent Obtain −3 ( 3 − x ) e 1 A1 − 3x − 3x −x 3 −3e ( 3 − x ) + 9e Complete integration and obtain 3 xe , or equivalent Substitute limits x = 0 and x = 3, having integrated twice DM1 9 A1 Obtain answer , or exact equivalent e 5

This question in 9709/31 Oct/Nov 2022

8 y x O a The diagram shows part of the curve y sin x. This part of the curve intersects the x-axis at the point = where x a. = (a) State the exact value of a. [1] … … … (b) Using the substitution u x, find the exact area of the shaded region in the first quadrant = bounded by this part of the curve and the x-axis. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 2 .8(a) State (a =) π 2 B1 Allow 32400, 180 2 . Accept ( x = ) 1 8(b) State or imply dx =2u du or equivalent B1 e.g. ddux = 2 1 x Incorrect statements e.g. du = 1 is B0. 2 x Substitute for x and dx throughout the integral M1 Obtain u2 sin u du A1 Allow with missing du. Commence integration of ku sin u du by parts and reach *M1 ku cos u  k cos u du Obtain integral −ku cos u + k sin u A1 Substitute limits u = 0 and u = their a , a ≠0 , a in radians DM1 −2πcosπ + 2sin π ( +0 − 2sin0 ) or x = 0 and their a Need limits stated but condone if zeros not shown in in the complete integral substitution. Obtain answer 2π A1 7

This question in 9709/32 Oct/Nov 2022

14π 3 Find the exact value of x sec2x dx. [5] Ó 0 … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 Commence integration by parts and reach x tan x  tan x.1 dx *M1 Use a correct method to integrate tan x M1 Obtain integral x tan x − lnsec x , or equivalent A1 Use limits correctly, having integrated twice DM1 1 1 A1 Obtain answer π − ln 2 , or exact equivalent 4 2 5

This question in 9709/33 Oct/Nov 2022

8 y 1 2 x O M The diagram shows the curve y x3 ln x, for x 0, and its minimum point M. = > (a) Find the exact coordinates of M. [4] … … … … … … … … … … … … … … … … … … (b) Find the exact area of the shaded region bounded by the curve, the x-axis and the line x 12. [5] = … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 8(a) Use the product rule correctly *M1 x3 d/dx(lnx) + d/dx(x3) lnx. Obtain the correct derivative in any form A1 x 3 2 e.g. + 3 x ln x . x Equate derivative to zero and solve exactly for x DM1 Reaching x = ea.  1 1  A1 ISW Obtain answer  3 , −  or exact equivalent  e 3e  4 8(b) Integrate by parts and reach ax 4 ln x + b  ( x 4 / x )dx *M1 x 4 1 4 A1 OE Obtain ln x −  ( x / x )dx 4 4 x 4 x 4 A1 OE Complete integration and obtain ln x − 4 16 1 DM1 Correct substitution [(1/4)ln1 or 0 − 1/16] – [(1/64)ln(1/2) – Use limits of x = and x = 1 in the correct order, having integrated twice (1/16)2] or minus this value CWO. 2 Allow omission of (1/4)ln1 or 0. 15 1 A1 Obtain answer − ln2 or exact equivalent final answer 256 64 5

This question in 9709/32 Feb/March 2023

a 19 The constant a is such that xe−2x dx 8. Ó 0 = (a) Show that a 1 ln 4a 2 . [5] = 2 + … … … … … … … … … … … … … … … … … … … … … … … … (b) Verify by calculation that a lies between 0.5 and 1. [2] … … … … … … … … … … … (c) Use an iterative formula based on the equation in (a) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … … …

10 marks

Mark scheme: 9(a) Commence integration and reach 2 2 e e d x x px q x    Obtain 2 2 1 1 2 2 e e d x x x x     A1 OE Complete integration and obtain 2 2 1 1 e e 2 4 x x x     A1 Use limits correctly and equate to 1 , 8 having integrated twice DM1 2 2 1 1 1 1 e e 2 4 4 8 a a a       . Obtain   1 ln 4 2 2 a a   correctly A1 AG 5 9(b) Calculate the values of a relevant expression or pair of expressions at a = 0.5 and a = 1 M1 Justify the given statement with correct calculated values A1 e.g. 0.5 < 0.69…, 1 > 0.89… 0.193 > 0, –1.105 < 0 0.066 < 0.125, 0.148 > 0.125 if put limits in the integral. Condone if they use calculator for the definite integral. 2 9(c) Use the iterative process   1 1 2 ln 4 2 n n a a   correctly at least once. M1 Obtain final answer 0.84 A1 Show sufficient iterations to at least 4 d.p. to justify 0.84 to 2 d.p. or show that there is a sign change in (0.835, 0.845) A1 e.g. 0.75, 0.8047, 0.8261, 0.8343, 0.8373, 0.8385 1, 0.8959, 0.8599, 0.8469, 0.8420, 0.8402 . 3

This question in 9709/31 May/June 2023

9 y M x O 3 The diagram shows the curve y = xe−14x2 , for x ≥0, and its maximum point M. (a) Find the exact coordinates of M. [4] … … … … … … … … … … … … … … … … … … … (b) Using the substitution x = u, or otherwise, find by integration the exact area of the shaded region bounded by the curve, the x-axis and the line x = 3. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 9(a) Use the correct product rule *M1 Condone error in chain rule. 2 x Obtain correct derivative in any form A1 − d y x 2 − x42 4 e.g. = − e + e . d x 2 Equate derivative to zero and solve for x DM1  − 1  A1 Or exact equivalent. Obtain answer  2, 2e 2  Can state the components separately.     4 9(b) 1 − 12 B1 Or equivalent e.g. du = 2 xdx . State or imply dx = u du 1 2 2 Alternative substitution: u = − x . 4 Substitute for x and dx M1 u A1 OE 1 − 14 Obtain correct integral  e d u 2 1 1 − u − x 2 M1 u = 9 and u = 0 or x = 3 and x = 0. Use correct limits in an integral of the form a e 4 or ae 4 9 A1 Or exact equivalent. − Obtain answer 2 − 2e 4 Alternative Method for Question 9(b) 1 1 − x 2 − x 2 M1 Recognition used. xe 4 dx = ae 4  a negative A1 a = − 2 A1 1 − x 2 M1 x = 3 and x = 0. Use correct limits in an integral of the form ae 4 9 A1 Or exact equivalent. − Obtain answer 2 − 2e 4 5

This question in 9709/31 Oct/Nov 2023

9 y M x O a π R The diagram shows the curve y sin x cos 2x, for 0 and a maximum point M, where x a. The shaded region between the curve= and the x-axis is≤xdenoted≤π, by R. = (a) Find the value of a correct to 2 decimal places. [5] … … … … … … … … … … … … … … … … … … (b) Find the exact area of the region R, giving your answer in simplified form. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 9(a) Use correct product rule *M1 As far as p cos x cos2 x + q sin x sin2 x or full working (u, v, du/dx, dv/dx) shown. dy A1 OE Obtain = cos x cos2 x − 2sin x sin2 x dx Equate derivative to zero and use correct double angle formulae DM1 Allow if only have one double angle in their derivative. 2 A1 2 2 Obtain cos x 1 − 6sin x = 0 or equivalent e.g. cos x 6cos x − 5 = 0 , 5tan x = 1 . ( ) ( ) Simplified but not necessarily factorised - like terms must be collected. Obtain a = 0.42 A1 Only. Accept x = 0.42 . Alternative method for question 9(a) Use correct double angle formula *M1 Obtain sin x − 2sin 3 x or equivalent A1 Use correct chain rule or product rule to differentiate and equate the DM1 derivative to zero 2 A1 OE Obtain cos x 1 − 6sin x = 0 ( ) Obtain a = 0.42 A1 Only. Accept x = 0.42 . 5 2cos 2 x sin x − sin x dx .9(b) Use double angle formula and obtain p cos 3 x + q cos x correctly *M1 e.g. from   2 3  A1 Obtain − cos x + cos x  Correct for their integral.  3  Correct use of limits 14 π and 34 π (or use double the integral from 14 π to 12 π ) DM1 OE   3 3   2  −1   1  1 1 −    −    − −  .   3   2   2   2 2  2 2 A1 Or simplified exact equivalent. Obtain 3 Final answer must be positive. Alternative method 1 for question 9(b) Use integration by parts twice and obtain r cos x cos2 x + s sin x sin2 x *M1 Seen, not just implied. 1 2 A1 Obtain cos x cos2 x + sin x sin2 x Accept  (correct for their integral). 3 3 Correct use of limits 14 π and 34 π (or use double the integral from 14 π to 12 π ) DM1 OE 1  1 1   0 + 2  −−1 0 − 2   1 .   3  2 2  2 2 A1 Or simplified exact equivalent. Obtain 3 Final answer must be positive.

This question in 9709/32 Oct/Nov 2023

10 y x O The diagram shows the curve y = x cos 2x, for x ≥0. (a) Find the equation of the tangent to the curve at the point where x = 12π. [4] … … … … … … … … … … … … … … … … (b) Find the exact area of the shaded region shown in the diagram, bounded by the curve and the x-axis. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 10(a) Use the product rule correctly on y = x cos 2x M1 dx/dx cos 2x + x d/dx(cos 2x) attempted. Obtain the correct derivative in any form A1 e.g. cos 2x – 2x sin 2x. If cos 2x + x–2sin 2x, not recovered, max M1A0A1FTA0 but can recover for full marks by seeing correct substitution. π dy π A1FT d y π Obtain y = − and = −1 when x = FT their with x = substituted. 2 dx 2 d x 2 Obtain answer x + y = 0 A1 π OE CWO Need to see y and dy/dx at x = . 2 4 10(b) Integrate by parts and reach ax sin2 x + b  sin2 xdx *M1 1 1 A1 OE Obtain x sin2 x −  sin2 xdx 2 2 1 1 A1 OE Complete integration and obtain x sin2 x + cos2 x 2 4 π DM1 1  π  2π 1 2π 1 Use limits of x = 0 and x = in the correct order, having integrated twice If correct,   sin + cos − cos0 4 2  4  4 4 4 4 to obtain ax sin 2x + ccos 2x 1  π  2π 1 or   sin − cos0 . 2  4  4 4 Max one substitution error. π 1 A1 π − 2 Obtain answer − or exact simplified two term equivalent ISW Accept . 8 4 8 1 1 Accept x sin2 x + cos2 x then final answer. 2 4 5

This question in 9709/33 Oct/Nov 2023

8 Use the substitution u = 1 - sin x to find the exact value of 2 3 r sin 2x dx . yr 1 - sin x Give your answer in the form a + b 2 where a and b are rational numbers to be determined. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 8 State or imply du = – cosx dx B1 Use sin 2x = 2sinxcosx and write the integral in terms of u *M1 Obtain (1 ) 2 d   u u u or equivalent A1 Integrate correctly to obtain 1 3 2 2  au bu DM1 Obtain correct 1 3 2 2 4 4 3   u u A1 Correctly use limits u = 2 and 0 in an expression of the form 1 3 2 2  au bu OR limits 3 2 π x  and 1 2 π in an expression of the form 1 3 2 2 (1 sin ) (1 sin )    a x b x DM1 Obtain 8 4 3 3 2  A1 7

This question in 9709/31 May/June 2024

6 y M O x The diagram shows the curve y = xe -ax , where a is a positive constant, and its maximum point M. (a) Find the exact coordinates of M. [4] … … … … … … … … … … … … … … … … … … … … … … 2 (b) Find the exact value of xe -ax d x . [5] y0a … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) Use correct product rule *M1 Or equivalent. Condone incorrect chain rule. M0 if a value is used for a (not equivalent work). Obtain correct derivative A1 E.g. d e e d ax ax y ax x     Equate derivative to zero and solve for x DM1 Obtain 1 1 e , a a x y   A1 ISW Or exact equivalent. 4 6(b) Use integration by parts to obtain e e d ax ax px q x    *M1 Condone sign error in parts formula and omission of dx. M0 if a value is used for a (not equivalent work). Obtain 1 1 e e d ax ax a a x x     A1 OE Complete integration to obtain 2 1 1 e e ax ax a a x     A1 OE Correct use of limits 0 and 2 a in an expression of the form e e ax ax rx s    DM1   2 2 2 2 2 2 1 1 e e 0 a a a       Obtain   2 2 1 1 3e a   A1 ISW Or simplified 2-term equivalent, e.g. 2 2 2 e 3. e a  5

This question in 9709/32 May/June 2024

8 (a) Express 3 cos 2x - 3 sin 2 x in the form R cos ( 2x + a) , where R 2 0 and 0 1 a 1 1 r . Give the 2 exact values of R and a. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … 1 r 12 3 (b) Hence find the exact value of dx , simplifying your answer. [5] y 2 0 `3 cos 2x - 3 sin 2xj … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 8(a) State R = 12 or exact equivalent B1 ISW Use trig formula to find α M1 Allow 1 3 30 or tan 3             or cos−1 3 2          or sin−1 1 2        Allow M1 if – 1 3 tan 3         etc. NB: If cos = 3 and sin = 3 seen then M0 A0. Obtain α = 1 π 6 A1 CWO, so A0 if from 1 3 tan . 3          3 Question Answer Marks Guidance 8(b) Express integral in the form A   2 sec 2 ... d x x   or A   2 sec 2 ... d x x   B1FT FT α from (a). Integrate and reach B   tan 2 ... x  or B   tan 2 ... x  B1FT FT α from (a). Where B = A or 2A or 0.5A. Obtain   1 tan 2 ... 8 x  B1FT OE FT α from (a). Allow 1 8 as 1 1. 4 2  Coefficient must be correct. Use limits of 0 x  and 1 12 π x  in the correct order in expression of form B   tan 2 ... x  so B tan ... 6         −B  tan ... or B tan ... 6         −B   tan ...  M1 Allow with tan still present. FT α from (a). SC: B1 3 12 OE after 1 1 tan 2 π 8 6 x        with no working. Obtain answer 1 12 3 or 1 4 3 or 1 48 or single term exact equivalent A1 1 8 ( 3 – 1 3 ) = 1 8 3 1 3        needs simplifying. 5 Note: allow all marks in (b) even if α = 1 π 6 found by an incorrect method in (a).

This question in 9709/33 May/June 2024

6 y R 3 x O r M 4 The diagram shows the curve y = sin 2 x ( 1 + sin 2x) , for 0 G x G 3 r , and its minimum point M. The 4 shaded region bounded by the curve that lies above the x-axis and the x-axis itself is denoted by R. (a) Given that the x-coordinate of M lies in the interval 1 r 1 x 1 3 r , find the exact coordinates 2 4 of M. [4] … … … … … … … … … … … … … … … … … … … (b) Find the exact area of the region R. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) Correct use of product rule to differentiate *M1 2cos2 x (1 + sin2 x ) + sin2 x  2cos2 x , or 2cos 2 x − 2sin 2 x + 8sin x cos 3 x − 8cos x sin 3 x. All terms needed but could have errors in the coefficients. Obtain 2cos2 x + 4cos2 x sin2 x A1 OE 1 1 Equate derivative to zero and solve for 2x or x DM1 2cos2 x (1 + 2sin2 x ) = 0  x = 2 sin −1 ( − 2 ) Condone if they only consider cos 2x = 0. 7 1 A1 Mark degrees as a misread. Obtain x = π, y = − The Q asks for an exact answer. 12 4 4 6(b) Use correct double angle formula to integrate M1 1 − cos4 x sin2 x + dx  2 Or use integration by parts and correct double angle formula 1  1 + cos4 x Or − cos2 x (1 + sin2 x ) + dx. 2  2 1 x 1 A1 1 1 x 1 Obtain − cos2 x + − sin4 x ( +C ) Or − cos2 x − cos2 x sin2 x + + sin4 x ( + C ) 2 2 8 2 2 2 8 1 M1 1 1 1 + π − 0 + − 0 + 0 Use limits 0 and π correctly in a solution containing p cos2 x and q sin4 x 2 4 2 2 1 A1 Obtain π + 1 4 4

This question in 9709/31 Oct/Nov 2024

9 (a) Find the quotient and remainder when x 4 + 16 is divided by x 2 + 4 . [3] … … … … … … … … … … … … … … … … … … … … … … … … … … … c 2 3 4 x + 16 4 (b) Hence show that dd 2 dx = ( r + 4 ) . [5] x + 4 3 e2 … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 9(a) Divide to obtain quotient x 2 + k M1 k is a constant. Obtain quotient x 2 − 4 A1 If quotient stated separately, mark at this stage. Obtain remainder 32 A1 If remainder stated separately, mark at this stage. Need not state which is quotient and remainder, but if stated wrongly, max 2/3. After a correct division, still allow the marks if 2 32 then written as x − 4 + . x 2 + 4 Alternative Method for Question 9(a) Expands brackets to get B = 0 M1 ( x 2 + 4 )( x 2 + Bx + C ) + D = 2 + 4 Bx + 4C + D x 4 + Bx 3 + ( C + 4 ) x C = – 4 A1 D = 32 A1 Need not state which is quotient and remainder, but if stated wrongly, max 2/3. 3 9(b) 1 3 B1 FT Follow their quotient of form Ax2 + B. x − 4 x 3 1 1 M1 Obtain p tan− qx where q = 2 or q = 2 −1 1 A1 FT Follow their constant remainder, Obtain 16tan x 2  their constant remainder  −1 1 i.e.   tan x.  2  2 1 1 M1 Terms need not be evaluated, e.g. Use limits correctly in an expression containing p tan− qx where q = 2 or q = 8   2  8 3 − 8 3  + 16tan −1 3 − − 8 + 16tan −1 1   3   and rx + sx  3  8 16 −1 16π or − 8 can be − , 16tan 3 can be , 3 3 3 16tan −1 1 can be 4π. 4 A1 AG Obtain ( π + 4 ) from full and correct working 3 5

This question in 9709/33 Oct/Nov 2024

11 y R a O r 2r x M The diagram shows the curve y = 2 sin x 2 + cos x , for 0 G x G 2 r , and its minimum point M, where x = a . (a) Find the value of a correct to 2 decimal places. [5] … … … … … … … … … … … … … … … … … … (b) Use the substitution u = 2 + cos x to find the exact area of the shaded region R. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 11(a) Use of correct product rule and correct chain rule M1 dy Bsin xsin x = A cos x 2 + cos x + dx 2 + cos x d y 2sin 2 x A1 OE Obtain = 2cos x 2 + cos x − d x 2 2 + cos x Equate the derivative to zero and obtain a horizontal 3 term quadratic equation or 4 *M1 Accept in cos .x term quartic equation in cosa E.g. 3cos2x + 4cosx – 1 = 0. If M0 earlier then needs that expression to be such that arrive at 3 term quadratic or E.g. 3cos4x + 16cos3x + 18cos2x – 1 = 0. 4 term quartic equation in cos x without further trig errors. 1 − 2 to be The only error in the form of the differential allowed is for ( 2 + cos x ) 1 3 2 ( 2 + cos x ) + − 2 or ( 2 + cos x ) Solve for cos a DM1  −+2 7   cos a = or 0.215     3  Allow presence of other solution(s). Obtain a = 4.93 A1 Allow more accurate, e.g. 4.929… even though question states 2 dp. If x = 1.35 leads to x = 4.93 award A1 BOD. If x = 1.35 and x = 4.93 award A0. 5 11(b) State or imply du = − sin x dx B1 OE If B0, max M1M1M1. Substitute throughout for u and du M1 Obtain − 2 udu A1 OE. Ignore limits if − 2 udu , but if + 2 udu , 3 u d u . then must have correct limits  12 (See final M1) 3 2 M1 Constant of integration not required Integrate to obtain ku ( +C ) 3 3 M1 1 and 3 for u, or 0 and π for x. Use correct limits correctly in an expression of the form ku 2 or k (2 + cos x ) 2 4 4 4 4 A1 3 Obtain 3 3 − 1 or 4 3 − or 27 − OE. Allow, e.g., 3 for 27. ( ) 3 3 3 3 ISW but don’t ignore e.g. multiplying throughout by 3. If the answer is changed from negative to positive value at end, then A0. Last M1A1 can use modulus, providing no errors seen. 6

This question in 9709/33 Oct/Nov 2024

r 11 Find the exact value of x 2 cos 1 x dx . [6] 3 0y … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 11 2 1 1 *M1 OE Commence integration by parts and reach Ax sin x  Bx sin x dx BOD on ± otherwise scores 0/6. 3 3 2 1 1 A1 OE Obtain 3 x sin x − 6 x sin x dx Allow 3×2 for 6. 3 3 2 1 1 1 *DM1 OE Complete integration by parts and reach Ax sin x + Bx cos x + C sin x 3 3 3 2 1 1 1 A1 Allow 6 × 3 for 18, 9 × 6 for 54 OE. Obtain 3 x sin x + 18 x cos x − 54sin x oe 3 3 3 Substitute limits correctly in an expression of the form DM1 Dependent on both previous M1 marks. 2 1 1 1 π 3 π 1 Ax sin x + Bx cos x + C sin x , where ABC ≠ 0 Need to use sin = , and cos = to obtain 3 3 3 3 2 3 2 2 3 Bπ C 3 Aπ + + . 2 2 2 3 3 2 A1 27 3 3 18 54 Obtain answer π + 9π − 27 3 or exact equivalent ISW Allow for , for 9, for 27, 2 2 2 2 2 2187 for 27 3 etc. Alternative Method for first 4 marks: 2 1 1 *M1 Commence integration by parts and reach Ax sin x + Bx cos x 3 3 2 1 1 A1 OE Obtain 3 x sin x + 18 x cos x Allow 6 × 3 for 18. 3 3 2 1 1 1 *DM1 Complete integration by parts and reach Ax sin x + Bx cos x + C sin x 3 3 3 11 2 1 1 1 A1 OE Obtain 3 x sin x + 18 x cos x − 54sin x Allow 6 × 3 for 18, 9 × 6 for 54, OE. 3 3 3 6

This question in 9709/32 Feb/March 2025

11 y M x O a 1 r 2 The diagram shows the curve y = cos x sin 2x for 0 G x G 1 r. The curve has a maximum point at M, 2 where x = a . (a) Find the exact value of a. [6] … … … … … … … … … … … … … … … … … … … … … (b) The region enclosed between the x-axis and the curve is rotated through 2r radians about the x-axis. Find the exact volume of the solid generated. [5] … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 11(a) d 2cos2 x cos2 x B1 SOI sin 2 x = Accept k dx 2 sin 2 x sin 2 x Use correct product rule M1* Obtain correct derivative in any form A1 dy 2cos2 x E.g. = − sin x sin2 x + cos x  . dx 2 sin2 x Equate the derivative to zero and obtain a horizontal equation DM1 E.g. − sin a sin2a + cos a cos2a = 0 Use correct trig formulae to obtain an equation in one trig function DM1 E.g. cos3a = 0 or sin 2 a = 14 . Obtain a = 16 π A1 Exact answer in radians only. 6 11(b) 2 M1* y dx Use π  Use double angle formula to obtain a form that can be integrated directly, DM1 1 1 Or 4 sin4 x + 2 sin2 x dx,  2 3 e.g.  cos x sin2 x dx =  2cos x sin x dx 3 or  2 ( u − u ) du by substituting u = sin x. 1 Obtain − ( π )  2 cos 4 x A1 Or − (π 16 1 cos4 x + 14 cos2 x ) , OE. Use correct limits correctly DM1 1 4 π2 1 1 1  π , ( 2 cos ( 2 π ) − 2 )  − π  2 cos x  0 = −  or − π 1 cos2π + 1 cosπ − 1 − 1 OE. ( 16 4 16 4 ) Obtain final answer 12 π A1 If the factor of π is missing throughout, allow the first 4 marks and A0 here. 5

This question in 9709/31 May/June 2025

10 (a) Find the quotient and remainder when x2 is divided by 1 + 4x 2 . [2] … … … … … … … … … … … … … … … … … … 0 .5 (b) Find the exact value of ; x tan -1 ( 2 x) d x . [6] 0 … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 10(a) 1 B1 Could be found by using long division or by writing Obtain quotient 2 2 4 x = q 1 + 4 x + r and comparing coefficients: ( ) 1 = 4q, 0 = q + r. 1 B1 Allow B1B1 if implied by correct division and no Obtain remainder − further working, but do not ISW. 4 Allow for a correct statement of the identity, but not for an incorrect statement of the remainder. 2 10(b) 2 −1 2 B *M1 OE 2 Commence integration by parts and reach Ax tan 2 x   x C + Dx dx 1 2 −1 x 2 A1 OE dx 2 Obtain 2 x tan 2 x −  1 + 4 x x 2 −1 DM1 2 Integrate  1 + 4 x dx to obtain an expression of the form p tan 2 x + qx Complete integration and obtain 12 x 2 tan −1 2 x + 18 tan −1 2 x − 14 x A1 FT OE FT their constant quotient and remainder from (a), 2 k − 1 k kx and 8 tan 2 x − 4 x from their 2 . 1 + 4 x Substitute limits correctly in an expression of the form DM1 Need some evidence that they have considered the Fx 2 tan −1 2 x + G tan −1 2 x + Hx lower limit, e.g. sight of 0 in the working. No need to evaluate trigonometry. If in stages, then the 0 needs to be seen for each part. Obtain answer 161 π − 81 or exact one- or two-term equivalent with trigonometry A1 evaluated 6

This question in 9709/32 May/June 2025

11 y M O 1 r x 2 The diagram shows the graph of y = 5 sin 2x cos 2 x for 0 G x G 1 r and its maximum point M. 2 (a) Find the exact x-coordinate of M. [6] … … … … … … … … … … … … … … … … … … … (b) By using the substitution u = cos x , find the area of the region bounded by the curve, the x-axis between x = 0 and x = 1 r , and the line x = 1 r . [5] 4 4 … … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 11(a) 2 B1 OE Differentiate cos x to obtain −2sin x cos x Could be stated as − sin2 .x Use correct product rule *M1 With a ‘+’ in the middle. 2 A1 OE Obtain derivative −10sin2 x sin x cos x + 10cos2 x cos x Equate derivative to zero and obtain an equation in one trig function DM1 Trigonometry formulas used need to be correct. Obtain 3 tan2 x = 1, 4 sin2 x = 1 or 4 cos2 x = 3 or cos3x = 0 A1 OE Obtain x = 16 π only A1 Alternative Method for the Question 11(a) Use double angle formula to obtain y = 10sin x cos 3 x B1 Use correct product rule *M1 4 2 2 A1 OE Obtain derivative 10cos x − 30sin x cos x Equate derivative to zero and obtain an equation in one trig function DM1 Trigonometry formulas used need to be correct. Obtain 3 tan2 x = 1, 4 sin2 x = 1 or 4 cos2 x = 3 or cos3x = 0 A1 OE Obtain x = 16 π only A1 11(a) Alternative Method 2 for the Question 11(a) Use double angle formula to obtain y = 52 sin2 x ( cos2 x + 1) B1 Use double angle formula to obtain y = 54 sin4 x + 52 sin2 x *M1 Obtain derivative 5cos4 x + 5cos2 x A1 Equate derivative to zero and obtain an equation in cos2x DM1 2cos 2 2 x + cos2 x −=1 0 Obtain cos2x = 12 only A1 Obtain x = 16 π only A1 6 11(b) d u B1 SOI = − sin x d x 3 *M1 OE Reach an integral of the form Au du  The question requires use of the substitution method. 3 A1 OE Obtain 10 u du − Ignore limits, but check order of limits if no minus sign. Substitute correct limits correctly in an expression of the form Cu 4 or C cos 4 x DM1 2 u = 1 and u = 2 1 x = 0 and x = π 4 1 1 10 2 u 3 du or 10 u 3 du 1 −1  2 Allow the correct answer from the correct integration and relevant limits to imply M1. 15 A1 WWW Obtain answer or 1.875 ISW 8 5

This question in 9709/32 May/June 2025

1 4 Find the exact value of x tan -1 x dx . [6] y 0 … … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 4 2 −1 x 2 M1* Condone sign error in formula for integration by parts. Begin integration by parts and obtain px tan x + q dx 2  1 + x x 2 −1 1 x 2 A1 OE Obtain tan x − dx 2 Allow with arctanx or tan-1 x. 2 2  1 + x DM1 Split the fraction and integrate. x 2  Use dx and integrate to obtain x + tan −1 x  2 dx =  1 − 1 2  1 + x  1 + x A1 OE x 2 −1 1 −1 Obtain tan x − x − tan x ( ) 2 2 Substitute correct limits correctly in an expression of the form DM1 π 1 π 2 −1 −1 − 0 − + + 0 ( −0 ) x tan x + x + tan x 8 2 8 Dependent on both previous M marks. Need to see evidence of the use of the lower limit at least once. Must evaluate the trigonometry using radians. π 1 A1 ISW Obtain − Or exact two term equivalent 4 2 6

This question in 9709/32 Oct/Nov 2025

11 y M O x 1 1 - r r 4 4 The diagram shows the graph of y = sec 2 x 3 + 2 tan x for - 1 r G x G 1 r , and its minimum point M. 4 4 (a) Find the x-coordinate of M. [6] … … … … … … … … … … … … … … … … … … … … … … … … … (b) Using the substitution u = 3 + 2 tan x , find the exact value of the area of the region bounded by the curve, the x-axis and the lines x =- 1 r and x = 1 r . [6] 4 4 … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 11(a) 2 − 1 B1 OE (can be unsimplified). Differentiate 3 + 2tan x to obtain sec x ( 3 + 2tan x ) 2 Not dependent on product rule, but must be convincing or seen in isolation, not as a derivative of the whole expression. B0B0 for e.g. 1 2 . 2sec x sec x tan x sec 2 x ( 3 + 2 tan x ) − Differentiate sec2x to obtain 2secx secx tanx B1 OE Not dependent on product rule but must be convincing or seen in isolation, not as a derivative of the whole expression. B0B0 for e.g. 1 2 OE, e.g. 2sec x sec x tan x sec 2 x ( 3 + 2 tan x ) − 2sin x 3 . cos x Use correct product (or quotient) rule M1 d d 2 sec2x (√……..) + √……… (sec x) dx dx [Obtain derivative, if correct M1 1 − 1 2 here. 1 1 Must include (…) 2 and (…) 2 4 − 2 ] 2 + sec x ( 3 + 2tan x ) 2sec x tan x ( 3 + 2tan x ) Arithmetical errors only for this M1. and equate derivative to zero and obtain an equation in one trig function May work in terms of sin x and cos x. Obtain 5tan2 x + 6 tan x + 1 = 0 A1 OE, e.g. 5tan4 x + 6tan3 x + 6tan2 x + 6tan x + 1 = 0 52sin4 x – 28sin2 x + 1 = 0 52cos4 x – 76cos2 x + 25 = 0 cot2 x + 6 cot x + 5 = 0 Obtain AWRT x = – 0.197 only A1 ISW May be more accurate. 6 11(b) du 2 B1 SOI = 2sec x dx *M1 OE Reach an integral of the form ∫ A u du 1 12 A1 OE Obtain ∫ u du 2 1 32 A1FT OE FT their coefficient. Obtain 3u 3 DM1 OE Substitute correct limits correctly in an expression of the form Bu 2 u = 1 and u = 5 3 2 1 1 or B (3 + 2tan x ) x = − π and x = π 4 4  3 3  2 2 Do not allow only decimals. and obtain c  5 − 1    A1 5 5 1 125 − Obtain answer − Or exact equivalent, e.g. 1. 3 3 3 ISW 6

This question in 9709/33 Oct/Nov 2025

a 1 10 The constant a is such that xe 2 x dx = 6. y 0 - 21 a (a) Show that a = 2 + e . [5] … … … … … … … … … … … … … … … … … … … … … … … … … (b) Verify by calculation that a lies between 2.2 and 2.4. [2] … … … … … … … … … (c) Use an iterative formula based on the equation in part (a) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(a) 12 x 12 x *M1 e dx Use integration by parts to obtain pxe + q  1 2 x 12 x A1 Obtain 2 xe 2 e dx − 1 2 x 12 x A1 Complete the integration to obtain 2 xe − 4 e 1 Use limits 0 and a correctly and equate the answer to 6 DM1 2ae 2 a − 4e 12 a ( −0 ) + 4 = 6 − 12 a A1 AG Obtain a = 2 + e from full and correct working 5 10(b) Calculate the value of a relevant expression or values of a relevant pair of M1 Example 1 2.2 < 2.33 and 2.4 > 2.30 expressions at 2.2 and 2.4. Values to at least 2 sf. − 1 a 2 OE Need all relevant values but only one (pair) needs to be correct to score M1. Example 2 f ( a ) = a −−2 e Complete set of values for their expression. If not comparing with 0, then the f ( 2.2 ) = −0.132... comparison must be clear. f ( 2.4 ) = 0.0988... 1 1 a a 2 − 4e 2 + 4 Example 3 f ( a ) = 2ae f ( 2.2 ) = 5.20... f ( 2.4 ) = 6.65... 1 1 a a 2 − 4e 2 − 2 Example 4 f ( a ) = 2ae f ( 2.2 ) = −0.80... f ( 2.4 ) = 0.65... Complete the argument correctly with correct calculated values. Accept truncated A1 Example 1 A clear explanation is needed. values. Allow work on a smaller interval. Example 2 f ( 2.2 ) = −0.132...  0 f ( 2.4 ) = 0.0988...  0 OE Example 3 f ( 2.2 ) = 5.20...  6 f ( 2.4 ) = 6.65...  6 Example 4 f ( 2.2 ) = −0.80...  0 f ( 2.4 ) = 0.65...  0 2 10(c) Use iterative process correctly at least once M1 Obtain final answer 2.31 A1 Show sufficient iterations to at least 4 dp to justify 2.31 to 2 dp A1 Allow recovery. Allow truncation. E.g. 2.3, 2.3166, 2.3140, 2.3144. 3

This question in 9709/35 Oct/Nov 2025