Differentiation — Paper 3 · A Level Mathematics 9709

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

3 The equation of a curve is y = x + cos 2x. Find the x-coordinates of the stationary points of the curve for which 0 ≤x ≤π, and determine the nature of each of these stationary points. [7]

7 marks

Mark scheme: 3 State correct derivative 1 –2sin 2x B1 Equate derivative to zero and solve for x M1 Obtain answer x = 1 π A1 12 Carry out an appropriate method for determining the nature of a stationary point M1 Show that x = 1 π is a maximum with no errors seen A1 12 Obtain second answer x = 5 π in range A1√ 12 Show this is a minimum point A1 [7] 3

This question in 9709/31 Oct/Nov 2005

8 y P x O T N 12 In the diagram the tangent to a curve at a general point P with coordinates (x, y) meets the x-axis at T. The point N on the x-axis is such that PN is perpendicular to the x-axis. The curve is such that, for all values of x in the interval 0 < x < 12π, the area of triangle PTN is equal to tan x, where x is in radians. PN (i) Using the fact that the gradient of the curve at P is , show that TN dy 1 = 2y2 cot x. [3] dx (ii) Given that y = 2 when x = 16π, solve this differential equation to find the equation of the curve, expressing y in terms of x. [6]

9 marks

Mark scheme: y d y 8 (i) State = , or equivalent B1 TN d x dy Express area of PTN in terms of y and , and equate to tan x M1 dx Obtain given relation correctly A1 [3] (ii) Separate variables correctly B1 2 Integrate and obtain term − , or equivalent B1 y Integrate and obtain term ln(sin x), or equivalent B1 Evaluate a constant or use limits y = 2, x = 1 π in a solution containing a term of the 6 form a/y or bln(sin x) M1 2 Obtain correct solution in any form, e.g. − = ln (2 sin x ) − 1 A1 y Rearrange as y = 2 / (1 − ln (2 sin x )) , or equivalent A1 [6] [Allow decimals, e.g. as in a solution y = 2 / (3.0 − ln (sin x )) .]

This question in 9709/31 May/June 2008

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

4 The parametric equations of a curve are x = a(2θ −sin 2θ), y = a(1 −cos 2θ). dy Show that = cot θ. [5] dx

5 marks

Mark scheme: dx dy 4 State or imply = a ( 2 − 2 cos 2θ ) or = 2 a sin 2θ B1 dθ dθ dy dy dx Use = ÷ M1 dx d θ dθ dy sin 2θ Obtain = , or equivalent A1 dx 1( − cos 2θ ) Make use of correct sin 2A and cos 2A formulae M1 Obtain the given result following sufficient working A1 [5] [SR: An attempt which assumes a is the parameter and θ a constant can only earn the two M marks. One that assumes θ is the parameter and a is a function of θ can earn B1M1A0M1A0.] [SR: For an attempt that gives a a value, e.g. 1, or ignores a, give B0 but allow the remaining marks.] GCE A/AS LEVEL – October/November 2008 9709 03 2

This question in 9709/31 Oct/Nov 2008

6 The parametric equations of a curve are x a sin3t, = a cos3t, y = where a is a positive constant and 0 t 12π. < < dy (i) Express in terms of t. [3] dx (ii) Show that the equation of the tangent to the curve at the point with parameter t is x sin t y cos t a sin t cos t. + = [3] (iii) Hence show that, if this tangent meets the x-axis at X and the y-axis at Y, then the length of XY is always equal to a. [2]

8 marks

Mark scheme: dx 2 dy 2 6 (i) EITHER State = −3a cos t sin t or = 3a sin t cos t , or equivalent B1 d t d t dy dy dx Use = ÷ M1 dx dt dt 2 − 13 2 − 13 23 23 OR State 3 x d x or 3 y d y as differentials of x or y respectively, or equivalent B1 dy Obtain in terms of t, having taken the differential of a constant to be zero M1 dx dy Obtain in any correct form A1 3 dx (ii) Form the equation of the tangent M1 Obtain the equation in any correct form A1 Obtain the given answer A1 3 (iii) State the x-coordinate of X or the y-coordinate of Y in any correct form B1 Obtain the given answer with no errors seen B1 2 GCE A/AS LEVEL – May/June 2009 9709 03

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

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 y 1 x O p 2p M sin x The diagram shows the curve y for 0 x and its minimum point M. x = < ≤2π, (i) Show that the x-coordinate of M satisfies the equation x tan x. = [4] (ii) The iterative formula π xn+1 = tan−1(xn) + can be used to determine the x-coordinate of M. Use this formula to determine the x-coordinate of M correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

7 marks

Mark scheme: 4 (i) Use correct quotient or product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and solve for x M1 Obtain the given answer correctly A1 [4] (ii) Use the iterative formula correctly at least once M1 Obtain final answer 4.49 A1 Show sufficient iterations to at least 4 d.p. to justify its accuracy to 2 d.p., or show that there is a sign change in the interval (4.485, 4.495) A1 [3]

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

ln x 6 The curve y has one stationary point. = x 1 + (i) Show that the x-coordinate of this point satisfies the equation x 1 x + , = ln x and that this x-coordinate lies between 3 and 4. [5] (ii) Use the iterative formula xn 1 + xn+1 = ln xn to determine the x-coordinate correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

8 marks

Mark scheme: 6 (i) Use correct quotient or product rule M1 1 ln x Obtain correct derivative in any form, e.g. − A1 x ( x + )1 ( x + 2)1 Equate derivative to zero and obtain the given equation correctly A1 ( x + )1 Consider the sign of x − at x = 3 and x = 4, or equivalent M1 ln x Complete the argument with correct calculated values A1 [5] (ii) Use the iterative formula correctly at least once, using or reaching a value in the interval (3, 4) M1 Obtain final answer 3.59 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 (3.585, 3.595) A1 [3]

This question in 9709/33 May/June 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

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

2 The parametric equations of a curve are t x y e−2t. 2t 3, = = + Find the gradient of the curve at the point for which t 0. [5] =

5 marks

Mark scheme: 2 Use of correct quotient or product rule to differentiate x or t M1 3 Obtain correct or unsimplified equivalent A1 (2t + 3)2 Obtain –2e–2t for derivative of y B1 dy ddyt Use = dx or equivalent M1 dx dt Obtain –6 cwo A1 [5] Alternative:  1−−62 xx   y = e  B1 Eliminate parameter and attempt differentiation     Use correct quotient or product rule M1 Use chain rule M1 dy − 6 1−−62 xx Obtain = 2 e A1 dx (1 − 2 x ) Obtain –6 cwo A1 2

This question in 9709/33 Oct/Nov 2010

5 The parametric equations of a curve are x = ln(tan t), y = sin2t, where 0 < t < 12π. dy (i) Express in terms of t. [4] dx (ii) Find the equation of the tangent to the curve at the point where x = 0. [3]

7 marks

Mark scheme: dx 25 (i) EITHER: State = sec t / tan t , or equivalent B1 dt dy State = 2 sin t cos t , or equivalent B1 d t dy dy dx Use = ÷ M1 dx dt dt Obtain correct answer in any form, e.g. 2 sin 2 t cos 2 t A1 OR: Obtain y = e2x / (1 + e2x), or equivalent B1 Use correct quotient or product rule M1 Obtain correct derivative in any form, e.g. 2e2x / (1 + e2x)2 A1 Obtain correct derivative in terms of t in any form, e.g. (2tan2t) / (1 + tan2t)2 A1 [4] 1 (ii) State or imply t = π when x = 0 B1 4 Form the equation of the tangent at x = 0 M1 1 1 Obtain correct answer in any horizontal form, e.g. y = x + A1 [3] 2 2 1 [SR: If the OR method is used in part (i), give B1 for stating or implying y = or 2 d y 1 = when x = 0.] d x 2 dy

This question in 9709/32 May/June 2011

8 y M 1 x O 2p The diagram shows the curve y 5 sin3x cos2x for 0 2π, and its maximum point M. = ≤x ≤1 (i) Find the x-coordinate of M. [5] (ii) Using the substitution u cos x, find by integration the area of the shaded region bounded by the = curve and the x-axis. [5]

10 marks

Mark scheme: 8 (i) Use product and chain rule M1 Obtain correct derivative in any form, e.g. 15 sin 2 x cos 3 x − 10 sin 4 x cos x A1 Equate derivative to zero and obtain a relevant equation in one trigonometric function M1 Obtain 2 tan 2 x = 3 , 5 cos 2 x = 2 , or 5 sin 2 x = 3 A1 Obtain answer x = 0.886 radians A1 [5] du (ii) State or imply d u = − sin x d x , or = − sin x , or equivalent B1 dx Express integral in terms of u and du M1 Obtain ± 5(u 2 − u 4 ) ∫ d u , or equivalent A1 1 Integrate and use limits u = 1 and u = 0 (or x = 0 and x = π ) M1 2 2 Obtain answer , or equivalent, with no errors seen A1 [5] 3 dx ( )( )

This question in 9709/33 May/June 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 9709/31 Oct/Nov 2011

2 The parametric equations of a curve are x y 2 cos3t. = 3(1 + sin2t), = dy Find in terms of t, simplifying your answer as far as possible. [5] dx

5 marks

Mark scheme: 2 EITHER: Use chain rule M1 dx obtain = 6 sin t cos t , or equivalent A1 dt d y 2 obtain = −6 cos t sin t , or equivalent A1 d t dy dy dx Use = ÷ M1 dx dt dt d y Obtain final answer = − cos t A1 d x OR: Express y in terms of x and use chain rule M1 1 dy x ) = k ( 2 − Obtain 2 , or equivalent A1 dx 3 1 dy x ) = − ( 2 − Obtain 2 , or equivalent A1 dx 3 Express derivative in terms of t M1 d y Obtain final answer = − cos t A1 [5] d x 2 2

This question in 9709/32 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

8 y x O The diagram shows the curve with parametric equations x sin t cos t, y sin3t cos3t, = + = + t 54π. for 14π < < dy (i) Show that sin t cos t. [3] dx = −3 (ii) Find the gradient of the curve at the origin. [2] (iii) Find the values of t for which the gradient of the curve is 1, giving your answers correct to 2 significant figures. [4]

9 marks

Mark scheme: 8 (i) Differentiate y to obtain 3sin2 t cos t – 3cos2 t sin t o.e. B1 dy dy dt Use = / M1 dx dt dx Obtain given result –3sin t cos t A1cwo [3] 3 (ii) Identify parameter at origin as t = 4π B1 3 3 Use t = 4π to obtain 2 B1 [2] (iii) Rewrite equation as equation in one trig variable B1 e.g. sin2t = − 23 , 9 sin4 x – 9 sin2 x + 1 = 0, tan2 x + 3 tan x + 1 = 0 Find at least one value of t from equation of form sin 2t = k o.e. M1 Obtain 1.9 A1 Obtain 2.8 and no others A1 [4] GCE AS/A LEVEL – October/November 2011 9709 33

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

6 The equation of a curve is 3x2 y2 45. −4xy + = (i) Find the gradient of the curve at the point [4] (2, −3). (ii) Show that there are no points on the curve at which the gradient is 1. [3]

7 marks

Mark scheme: dy 6 (i) Obtain 2 y as derivative of y2 B1 dx d y Obtain − 4 y − 4 x as derivative of –4xy B1 d x dy Substitute x = 2 and y = –3 and find value of dx d ( 45) (dependent on at least one B1 being earned and = 0 ) M1 dx 12 Obtain or equivalent A1 [4] 7 dy dy (ii) Substitute = 1 in an expression involving , x and y and obtain ay = bx M1 dx dx Obtain y = x or equivalent A1 Uses y = x in original equation and demonstrate contradiction A1 [3]

This question in 9709/31 May/June 2012

6 The equation of a curve is y 3 sin x 4 cos3x. = + (i) Find the x-coordinates of the stationary points of the curve in the interval 0 x π. [6] < < (ii) Determine the nature of the stationary point in this interval for which x is least. [2]

8 marks

Mark scheme: 6 (i) State derivative in any correct form, e.g. 3 cos x − 12 cos 2 x sin x B1 + B1 Equate derivative to zero and solve for sin 2x, or sin x or cos x M1 1 Obtain answer x = π A1 12 5 Obtain answer x = π A1 12 1 Obtain answer x = π and no others in the given interval A1 [6] 2 (ii) Carry out a method for determining the nature of the relevant stationary point M1 1 Obtain a maximum at π correctly A1 [2] 12 [Treat answers in degrees as a misread and deduct A1 from the marks for the angles.]

This question in 9709/32 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

3 The parametric equations of a curve are x sin 2θ y cos 2θ 2 sin θ. = −θ, = + dy 2 cos θ Show that . [5] dx = 1 2 sin θ +

5 marks

Mark scheme: d x dy 3 Obtain = 2 cos 2θ − 1 or = −2 sin 2θ + 2 cos θ , or equivalent B1 d θ dθ dy dy dx Use = ÷ M1 d x dθ dθ dy − 2 sin 2θ + 2 cos θ Obtain = , or equivalent A1 dx 2 cos 2θ − 1 At any stage use correct double angle formulae throughout M1 Obtain the given answer following full and correct working A1 [5]

This question in 9709/33 May/June 2012

e2x 4 The curve with equation y has one stationary point. = x3 (i) Find the x-coordinate of this point. [4] (ii) Determine whether this point is a maximum or a minimum point. [2]

6 marks

Mark scheme: 4 (i) Use correct quotient or product rule M1 2 e 2 x 3e 2 x Obtain correct derivative in any form, e.g. − A1 x 3 x 4 Equate derivative to zero and solve a 2-term equation for non-zero x M1 3 Obtain x = correctly A1 [4] 2 (ii) Carry out a method for determining the nature of a stationary point, e.g. test derivative either side M1 Show point is a minimum with no errors seen A1 [2] GCE AS/A LEVEL – May/June 2012 9709 33

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

3 The parametric equations of a curve are 4t x y 2 = 2t 3, = ln(2t + 3). + dy (i) Express in terms of t, simplifying your answer. [4] dx (ii) Find the gradient of the curve at the point for which x 1. [2] =

6 marks

Mark scheme: 3 (i) Either Use correct quotient rule or equivalent to obtain dx 4( 2t + )3 − 8t = or equivalent B1 dt ( 2t + 2)3 dy 4 Obtain = or equivalent B1 dt 2t + 3 dy dy dt = or equivalent M1 Use dx dx dt 1 Obtain (2t + 3 ) or similarly simplified equivalent A1 3 3 x Or Express t in terms of x or y e.g. t = B1 4 − 2 x  6  Obtain Cartesian equation e.g. y = 21n   B1  2 − x  dy 2 Differentiate and obtain = M1 dx 2 − x 1 Obtain (2t + 3 ) or similarly simplified equivalent A1 [4] 3 3 (ii) Obtain 2t = 3 or t = B1 2 dy Substitute in expression for and obtain 2 B1 [2] dx GCE A LEVEL – October/November 2012 9709 33

This question in 9709/33 Oct/Nov 2012

5 y M x –a 3a The diagram shows the curve with equation x3 xy2 ay2 0, + + −3ax2 = where a is a positive constant. The maximum point on the curve is M. Find the x-coordinate of M in terms of a. [6]

6 marks

Mark scheme: dy 5 EITHER: State 2 ay as derivative of ay2 B1 dx dy 2 State y + 2 xy as derivative of xy2 B1 dx d y Equate derivative of LHS to zero and set equal to zero M1 d x Obtain 3 x 2 + y 2 − 6 ax = 0 , or horizontal equivalent A1 Eliminate y and obtain an equation in x M1 Solve for x and obtain answer x = 3a A1 2 3ax 2 − x 3 OR1: Rearrange equation in the form y = and attempt differentiation of one x + a side B1 Use correct quotient or product rule to differentiate RHS M1 Obtain correct derivative of RHS in any form A1 d y Set equal to zero and obtain an equation in x M1 d x Obtain a correct horizontal equation free of surds A1 Solve for x and obtain answer x = 3a A1 1  3ax 2 − x 3  2 OR2: Rearrange equation in the form y =   and differentiation of RHS B1 x + a   Use correct quotient or product rule and chain rule M1 Obtain correct derivative in any form A1 GCE AS/A LEVEL – May/June 2013 9709 32 Equate derivative to zero and obtain an equation in x M1 Obtain a correct horizontal equation free of surds A1 Solve for x and obtain answer x = 3a A1 [6]

This question in 9709/32 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

1 x 1 The equation of a curve is y for x 2. Show that the gradient of the curve is always + 1 2x = > −1 negative. + [3]

3 marks

Mark scheme: 1 Use correct quotient or product rule M1 Obtain correct derivative in any form A1 Justify the given statement A1 [3] 2 ( ) 2 ( ) 2

This question in 9709/31 Oct/Nov 2013

4 The parametric equations of a curve are x cos t, y sin t. = e−t = e−t dy Show that tan t . [6] dx = −140

6 marks

Mark scheme: 4 Use correct product or quotient rule at least once M1* d x − t − t d y − t − t Obtain = e sin t − e cos t or = e cos t − e sin t , or equivalent A1 d t d t d y d y d x Use = ÷ M1 d x d t d t d y sin t − cos t Obtain = , or equivalent A1 dx sin t + cos t d y EITHER: Express in terms of tan t only M1(dep*) d x 1  Show expression is identical to tan − t π  A1  4  1  t M1 OR: Express tan − t π  in terms of tan  4  d y Show expression is identical to A1 [6] d x

This question in 9709/31 Oct/Nov 2013

1 + x 1 The equation of a curve is y = for x > −1 Show that the gradient of the curve is always 2. 1 + 2x negative. [3]

3 marks

Mark scheme: 1 Use correct quotient or product rule M1 Obtain correct derivative in any form A1 Justify the given statement A1 [3] 2 ( ) 2 ( ) 2

This question in 9709/32 Oct/Nov 2013

10 y T1 T3 x O T4 T2 The diagram shows the curve y 2x sin 4x for x The stationary points are labelled T1, T2, = 10e−1 ≥0. … as shown. T3, (i) Find the x-coordinates of T1 and T2, giving each x-coordinate correct to 3 decimal places. [6] (ii) It is given that the x-coordinate of Tn is greater than 25. Find the least possible value of n. [4]

10 marks

Mark scheme: 10 (i) Use of product or quotient rule M1 1 1 − x − x Obtain − 5e 2 sin 4 x + 40e 2 cos 4 x A1 dy Equate to zero and obtain tan4z = k or R cos(4x ± α) M1 dx  −1 1  Obtain tan 4x = 8 or 65 cos 4 x ± tan  A1  8  Obtain 0.362 or 20.7° A1 Obtain 1.147 or 65.7° A1 [6] 1 (ii) State or imply that x-coordinates of Tn are increasing by π or 45° B1 4 Attempt solution of inequality (or equation) of form x1 + (n – 1)kπ . 25 M1 4 Obtain n > (25 − .0362 ) + 1 , following through on their value of x1 A1 π n = 33 A1 [4]

This question in 9709/31 May/June 2014

6 y M x O 2 The diagram shows the curve x2 y2 2 x2 and one of its maximum points M. Find the + = −y2 coordinates of M. [7]

7 marks

Mark scheme: 6 Obtain correct derivative of RHS in any form B1 Obtain correct derivative of LHS in any form B1 d y Set equal to zero and obtain a horizontal equation M1 d x Obtain a correct equation, e.g. x 2 + y 2 = 1 , from correct work A1 By substitution in the curve equation, or otherwise, obtain an equation in x 2 or y 2 M1 Obtain x = 12 3 A1 Obtain y = 12 A1 7 2

This question in 9709/33 May/June 2014

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

4 The parametric equations of a curve are 1 x , y tan3t, = cos3t = where 0 1 ≤t < 20. dy (i) Show that sin t. [4] dx = (ii) Hence show that the equation of the tangent to the curve at the point with parameter t is y x sint t. [3] = −tan

7 marks

Mark scheme: 4 (i) Use chain rule correctly at least once M1 dx 3sint dy 2 2 Obtain either = 4 or = 3tan t sec t , or equivalent A1 d t cos t dt dy dy dx Use = ÷ M1 dx dt dt Obtain the given answer A1 [4] (ii) State a correct equation for the tangent in any form B1 Use Pythagoras M1 Obtain the given answer A1 [3] 1 + 2i

This question in 9709/31 Oct/Nov 2014

4 The parametric equations of a curve are 1 x , y tan3t, = cos3t = where 0 1 ≤t < 20. dy (i) Show that sin t. [4] dx = (ii) Hence show that the equation of the tangent to the curve at the point with parameter t is y x sint t. [3] = −tan

7 marks

Mark scheme: 4 (i) Use chain rule correctly at least once M1 dx 3sint dy 2 2 Obtain either = 4 or = 3tan t sec t , or equivalent A1 d t cos t dt dy dy dx Use = ÷ M1 dx dt dt Obtain the given answer A1 [4] (ii) State a correct equation for the tangent in any form B1 Use Pythagoras M1 Obtain the given answer A1 [3] 1 + 2i

This question in 9709/32 Oct/Nov 2014

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

10 y O x P The diagram shows part of the curve with parametric equations x 2 ln t 2 , y t3 2t 3. = + = + + (i) Find the gradient of the curve at the origin. [5] (ii) At the point P on the curve, the value of the parameter is p. It is given that the gradient of the curve at P is 12. 1 (a) Show that p [1] = 3p2 2 −2. + (b) By first using an iterative formula based on the equation in part (a), determine the coordinates of the point P. Give the result of each iteration to 5 decimal places and each coordinate of P correct to 2 decimal places. [4]

10 marks

Mark scheme: dx 2 dy 210 (i) Obtain = and = 3t + 2 B1 dt t + 2 dt dy dy dx Use = ÷ M1 dx dt dt dy 1 2 Obtain = (3t + 2)(t + 2) A1 dx 2 Identify value of t at the origin as –1 B1 5 Substitute to obtain as gradient at the origin A1 [5] 2 1 1 (ii) (a) Equate derivative to and confirm p = − 2 B1 [1] 2 2 3 p + 2 (b) Use the iterative formula correctly at least once M1 Obtain value p = − .1924 or better (–1.92367…) A1 Show sufficient iterations to justify accuracy or show a sign change in appropriate interval A1 Obtain coordinates (–5.15, –7.97) A1 [4]

This question in 9709/31 May/June 2015

3 A curve has equation y cos x cos 2x. Find the x-coordinate of the stationary point on the curve in the interval 0 x 1 giving= your answer correct to 3 significant figures. [6] < < 20,

6 marks

Mark scheme: 3 EITHER: Use correct product rule M1 Obtain correct derivative in any form, e.g. − sin x cos 2 x − 2 cos x sin 2 x A1 Use the correct double angle formulae to express derivative in cos x and sin x, or cos 2x and sin x M1 OR1: Use correct double angle formula to express y in terms of cos x and attempt differentiation M1 Use chain rule correctly M1 Obtain correct derivative in any form, e.g. − 6 cos 2 x sin x + sin x A1 OR2: Use correct factor formula and attempt differentiation M1 3 1 Obtain correct derivative in any form, e.g. − sin3x − sin x A1 2 2 Use correct trig formulae to express derivative in terms of cos x and sin x, or sin x M1 Equate derivative to zero and obtain an equation in one trig function M1 Obtain 6 cos 2 x = 1 , 6 sin 2 x = 5 , tan 2 x = 5 or 3 cos 2 x = − 2 A1 Obtain answer x = 1.15 (or 65.9°) and no other in the given interval A1 [6] [Ignore answers outside the given interval.] [SR: Solution attempts following the EITHER scheme for the first two marks can earn the second and third method marks as follows: Equate derivative to zero and obtain an equation in tan 2x and tan x M1 Use correct double angle formula to obtain an equation in tan x M1]

This question in 9709/32 May/June 2015

5 The parametric equations of a curve are x a cos4t, y a sin4t, = = where a is a positive constant. dy (i) Express in terms of t. [3] dx (ii) Show that the equation of the tangent to the curve at the point with parameter t is x sin2t y cos2t a sin2t cos2t. + = (iii) Hence show that if the tangent meets the x-axis at P and the y-axis at Q, then OP OQ a, + = where O is the origin. [2]

5 marks

Mark scheme: d x 3 d y 35 (i) State = − 4 a cos t sin t , or = 4 a sin t cos t B1 d t d t d y d y d x Use = ÷ M1 d x d t d t d y Obtain correct expression for in a simplified form A1 3 d x (ii) Form the equation of the tangent M1 Obtain a correct equation in any form A1 Obtain the given answer A1 3 (iii) State the x-coordinate of P or the y-coordinate of Q in any form B1 Obtain the given result correctly B1 2 ∫

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

3 A curve has equation 2 x y −tan 1 tan x. = + 1 Find the equation of the tangent to the curve at the point for which x giving the answer in the form y mx c where c is correct to 3 significant figures. = 40, [6] = +

6 marks

Mark scheme: 3 Use correct quotient rule or equivalent to find first derivative M1* − (1 + tan x ) sec 2 x − sec 2 x ( 2 − tan x ) Obtain or equivalent A1 1( + tan x ) 2 1 Substitute x = π to find gradient dep M1* 4 3 Obtain − A1 2 1 Form equation of tangent at x = π M1 4 3 Obtain y = − x + .1 68 or equivalent A1 [6] 2 d y y& dy

This question in 9709/33 Oct/Nov 2015

5 The curve with equation y sin x cos 2x has one stationary point in the interval 0 x 1 Find the = < < 20. x-coordinate of this point, giving your answer correct to 3 significant figures. [6]

6 marks

Mark scheme: 5 Use product rule M1 Obtain correct derivative in any form, e.g. cos x cos2 x − 2sin x sin 2 x A1 Equate derivative to zero and use double angle formulae M1 Remove factor of cos x and reduce equation to one in a single trig function M1 Obtain 6sin 2 x = 1 , 6cos 2 x = 5 or 5tan 2 x = 1 A1 Solve and obtain x = 0.421 A1 [6] [Alternative: Use double angle formula M1.Use chain rule to differentiate M1. Obtain correct derivative e.g. cos θ − 6sin 2 θ cos θ A1, then as above.]

This question in 9709/31 May/June 2016

6 The curve with equation y x2 cos 2x1 has a stationary point at x p in the interval 0 x = = < < 0. 1 4 (i) Show that p satisfies the equation tan 2p p. [3] = (ii) Verify by calculation that p lies between 2 and 2.5. [2] @ A 4 (iii) Use the iterative formula 2 to determine the value of p correct to 2 decimal tan−1 pn+1 = pn places. Give the result of each iteration to 4 decimal places. [3]

8 marks

Mark scheme: 6 (i) Use the product rule M1 Obtain correct derivative in any form A1 Equate 2-term derivative to zero and obtain the given answer correctly A1 [3] (ii) Use calculations to consider the sign of a relevant expression at p = 2 and p = 2.5, or compare values of relevant expressions at p= 2 and p = 2.5 M1 Complete the argument correctly with correct calculated values A1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 2.15 A1 Show sufficient iterations to 4 d.p. to justify 2.15 to 2 d.p., or show there is a sign change in the interval (2.145,2.155) A1 [3]

This question in 9709/33 May/June 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 9709/31 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

sin x 2 The equation of a curve is y for x Show that the gradient of the curve is positive 1 cosx, = −0 < < 0. for all x in the given interval. + [4]

4 marks

Mark scheme: 2 Use correct quotient or product rule M1 Obtain correct derivative in any form A1 1 Use Pythagoras to simplify the derivative to , or equivalent A1 1 + cos x Justify the given statement, − 1 < cosx < 1 statement, or equivalent A1 [4]

This question in 9709/33 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

4 The parametric equations of a curve are x ln cos y = 1, = 31 −tan 1, where 0 1 ≤1 < 20. dy (i) Express in terms of tan [5] dx 1. … … … … … … … … … … … … … … … … … … … … … … (ii) Find the exact y-coordinate of the point on the curve at which the gradient of the normal is equal to 1. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4(i)  d x sin θ  M1 Use chain rule to differentiate x  = −   dθ cos θ  dy 2 B1 State = 3 − sec θ dθ dy dy dx M1 Use = ÷ dx dθ dθ d y 3 − sec 2 θ A1 Obtain correct in any form e.g. d x − tan θ dy tan 2 θ − 2 A1 Obtain = , or equivalent dx tan θ Total: 5 4(ii) Equate gradient to −1 and obtain an equation in tanθ M1 Solve a 3 term quadratic ( tan 2 θ + tan θ − 2 = 0 ) in tanθ M1 π 3π A1 Obtain θ = and y = − 1 only 4 4 Total: 3

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

4 The parametric equations of a curve are x t2 1, y 4t ln 2t . = + = + −1 dy (i) Express in terms of t. [3] dx … … … … … … … … … … … … … … … … … … … … … … … (ii) Find the equation of the normal to the curve at the point where t 1. Give your answer in the form ax by c 0. = [3] + + = … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 4(i) dy 2 B1 State = 4 + dt 2t − 1 dy dy dx M1 Use = ÷ dx dt dt dy 8t − 2 2 2 A1 Obtain answer = , or equivalent e.g. + dx 2t (2t − 1) t 2 4t − 2t Total: 3 4(ii) Use correct method to find the gradient of the normal at t = 1 M1 Use a correct method to form an equation for the normal at t = 1 M1 Obtain final answer x + 3 y − 14 = 0 , or horizontal equivalent A1 Total: 3

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

5 A curve has equation y 2 ln 1 3 cos2x for 0 3 = + ≤x ≤120. dy (i) Express in terms of tan x. [4] dx … … … … … … … … … … … … … … … … … … … … … … … … (ii) Hence find the x-coordinate of the point on the curve where the gradient is Give your answer correct to 3 significant figures. −1. [2] … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 5(i) Use the chain rule M1 Obtain correct derivative in any form A1 Use correct trigonometry to express derivative in terms of tan x M1 d y 4tan x A1 Obtain = − , or equivalent d x 4 + tan 2 x Total: 4 5(ii) Equate derivative to –1 and solve a 3–term quadratic for tan x M1 Obtain answer x=1.11 and no other in the given interval A1 Total: 2

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

2 x 14 The curve with equation y has one stationary point in the interval x −sin cosx = −120 < < 20. (i) Find the exact coordinates of this point. [5] … … … … … … … … … … … … … … … … … … … … … … … … (ii) Determine whether this point is a maximum or a minimum point. [2] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(i) Use correct product or quotient rule or rewrite as 2sec x − tan x and differentiate M1 Obtain correct derivative in any form A1 Equate the derivative to zero and solve for x M1 Obtain x = 16 π A1 Obtain y = 3 A1 5 4(ii) Carry out an appropriate method for determining the nature of a stationary point M1 Show the point is a minimum point with no errors seen A1 2

This question in 9709/32 Oct/Nov 2017

5 The parametric equations of a curve are x 2t sin 2t, y 1 cos 2t, = + = −2 for t 1 −120 < < 20. dy (i) Show that 2 tan t. [5] dx = … … … … … … … … … … … … … … … … … … … … … … (ii) Hence find the x-coordinate of the point on the curve at which the gradient of the normal is 2. Give your answer correct to 3 significant figures. [2] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(i) State correct derivative of x or y with respect to t B1 d y d y d x M1 Use = ÷ d x d t d t d y 4sin 2t A1 Obtain = , or equivalent d x 2 + 2cos2t Use double angle formulae throughout M1 Obtain the given answer correctly AG A1 5 5(ii) −1 1  B1 State or imply t = tan  −   4  Obtain answer x = – 0.961 B1 2

This question in 9709/32 Feb/March 2018

e3x 3 A curve has equation y Find the x-coordinates of the stationary points of the curve in the 1 = tan 2x. interval 0 x Give your answers correct to 3 decimal places. [6] < < 0. … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 3 Use quotient or product rule M1 Obtain correct derivative in any form A1 1 M1* Equate derivative to zero and obtain a quadratic in tan x or an equation of the 2 form a sin x = b Solve for x M1(dep*) Obtain answer 0.340 A1 Obtain second answer 2.802 and no other in the given interval A1 6

This question in 9709/31 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

ln x 4 The curve with equation y has a stationary point at x p. 3 x = = + 3 (i) Show that p satisfies the equation ln x 1 . [3] x = + … … … … … … … … … … … … … … … … … … … … … … … (ii) By sketching suitable graphs, show that the equation in part (i) has only one root. [2] 3 x (iii) It is given that the equation in part (i) can be written in the form x . Use an iterative + ln x = formula based on this rearrangement to determine the value of p correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … … …

8 marks

Mark scheme: 4(i) Use the quotient or product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and obtain the given equation A1 Total: 3 4(ii) Sketch a relevant graph, e.g. y = ln x B1 3 B1 Sketch a second relevant graph, e.g. y = 1 + , and justify the given statement x Total: 2 4(iii) 3 + x M1 Use iterative formula nx +1 = correctly at least once ln x n Obtain final answer 4.97 A1 Show sufficient iterations to 4 d.p.to justify 4.97 to 2 d.p. or show there is a sign A1 change in the interval (4.965, 4.975) Total: 3

This question in 9709/33 May/June 2018

8 The equation of a curve is 2x3 2a3, where a is a non-zero constant. −y3 −3xy2 = dy 2x2 (i) Show that [4] −y2 dx = y2 2xy. + … … … … … … … … … … … … … … … … … … … … … … … … (ii) Find the coordinates of the two points on the curve at which the tangent is parallel to the y-axis. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 8(i) 2 d y B1 State or imply 3y as derivative of 3y d x 2 d y 2 B1 State or imply 3 y + 6 xy as derivative of 3xy d x d y M1 Equate derivative of LHS to zero and solve for d x Obtain the given answer A1 Total: 4 8(ii) Equate denominator to zero and solve for y M1* Obtain y = 0 and x = a A1 Obtain y = αx and substitute in curve equation to find x or M1(dep*) to find y Obtain x = – a A1 Obtain y = 2a A1 Total: 5

This question in 9709/33 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

3 cos x 7 A curve has equation y for 2 sin x, = −120 ≤x ≤120. + (i) Find the exact coordinates of the stationary point of the curve. [6] … … … … … … … … … … … … … … … … … … … … … … … … a 3 cosx (ii) The constant a is such that dx 1. Find the value of a, giving your answer correct 2 sin x = Ô0 + to 3 significant figures. [4] … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 7(i) Use correct quotient or product rule M1 Obtain correct derivative in any form A1 ( ) ( ) 2 d 3sin 2 sin 3cos cos d 2 sin − + − = + y x x x x x x Condone invisible brackets if recovery implied later. Equate numerator to zero M1 Use 2 2 cos sin 1 + = x x and solve for sin x M1 6sin 3 0 − − = x ⇒ sinx =…. Obtain coordinates / 6 π = − x and 3 = y ISW A1 + A1 From correct working. No others in range SR: A candidate who only states the numerator of the derivative, but justifies this, can have full marks. Otherwise they score M0A0M1M1A0A0 6 7(ii) State indefinite integral of the form k ln (2 + sin x) M1* Substitute limits correctly, equate result to 1 and obtain 3 ln (2 + sin a) – 3 ln 2 = 1 A1 or equivalent Use correct method to solve for a M1(dep*) Allow for a correct method to solve an incorrect equation, so long as that equation has a solution. 1 3 1 2 1 sin e + = a ( ) 1 3 1 sin 2 e 1 − ⇒ = −     a Can be implied by 52.3° Obtain answer a = 0.913 or better A1 Ignore additional solutions. Must be in radians. 4

This question in 9709/32 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 5 (i) Differentiate with respect to [2] 1. sin21 … … … … … … … … … … … … (ii) The variables x and satisfy the differential equation 1 dx x tan 0, 1 + cosec21 = d1 for 0 1 and x 0. It is given that x 4 when 1 Solve the differential equation, < 1 < 20 > = 1 = 60. obtaining an expression for x in terms of [6] 1. … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(i) Use chain rule M1 3 2 cos sin cosec cot θ θ θ θ − = − k k Allow M1 for 1 2cos sin θ θ − − Obtain correct answer in any form A1 e.g. 2 2cosec cot θ θ − , 3 2cos sin θ θ − Accept 4 2sin cos sin θ θ θ − 2 5(ii) Separate variables correctly and integrate at least one side B1 2 d cosec cot d θ θ θ = − ∫ ∫ x x Obtain term 2 1 2 x B1 Obtain term of the form 2 sin k θ M1* or equivalent Obtain term 2 1 2sin θ A1 or equivalent Use x = 4, 1 6 θ π = to evaluate a constant, or as limits, in a solution with terms 2 ax and 2 sin b θ , where ab ≠ 0 DM1 Dependent on the preceding M1 Obtain solution ( ) 2 cosec 12 x θ = + A1 or equivalent 6

This question in 9709/31 May/June 2019

x 4 Find the exact coordinates of the point on the curve y at which the gradient of the tangent = 1 ln x 1 + is equal to 4. [7] … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4 Use correct quotient rule M1 Allow use of correct product rule on 1 1 ln − × + x x Obtain correct derivative in any form A1 ( ) ( ) ( ) 2 2 1 1 ln d 1 1 d 1 ln 1 ln 1 ln + −×     = = −   + + +   x x y x x x x x Equate derivative to 1 4 and obtain a quadratic in ln x or (1+ ln x) M1 Horizontal form. Accept ( ) 2 1 ln 1 ln 4 = + x x Reduce to 2 (ln ) 2ln 1 0 − + = x x A1 or 3-term equivalent. Condone 2 lnx if later used correctly Solve a 3-term quadratic in ln x for x M1 Must see working if solving incorrect quadratic Obtain answer x = e A1 Accept 1e Obtain answer y = 1 2 e A1 Exact only with no decimals seen before the exact value. Accept 1e 2 but not e 1 ln e + 7

This question in 9709/32 May/June 2019

e−2x dy 2 The curve with equation y has a stationary point in the interval x 1. Find and = 1 −1 < < dx −x2 hence find the x-coordinate of this stationary point, giving the answer correct to 3 decimal places. [5] … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 Use correct quotient rule or correct product rule M1 Obtain correct derivative in any form A1 ( ) ( ) 2 2 2 2 2 2e 1 2 e d d 1 x x x x y x x − − − − + = − Equate derivative to zero and obtain a 3 term quadratic in x M1 Obtain a correct 3-term equation e.g. 2 2 2 2 0 x x + − = or 2 1 x x + = A1 From correct work only Solve and obtain x = 0.618 only A1 From correct work only 5

This question in 9709/32 Oct/Nov 2019

5 The equation of a curve is 2x2y a3, where a is a positive constant. Show that there is only one −xy2 = point on the curve at which the tangent is parallel to the x-axis and find the y-coordinate of this point. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5 State 2 4 2 xy x + d d y x , or equivalent, as derivative of 2 2x y B1 State 2 2 y xy + d d y x , or equivalent, as derivative of 2 xy B1 Equate attempted derivative of LHS to zero and set d d y x equal to zero (or set numerator equal to zero) *M1 2 2 d 4 d 2 2 y y xy x x xy − = − Reject y = 0 B1 Allow from 2 0 y kxy − = Obtain y = 4x A1 OE from correct numerator. ISW Obtain an equation in y (or in x) and solve for y (or for x) in terms of a DM1 3 3 3 3 3 3 8 16 or 8 4 y y x x a a − = − = Obtain y = – 2a A1 With no errors seen 7 Question Answer Marks Guidance 5 Alternative method for question 5 Rewrite as 3 2 2 a y x xy = − and differentiate M1 Correct use of function of a function and implicit differentiation Obtain correct derivative (in any form) A1 ( ) 3 2 2 d 4 d d d 2 y a x y x y x x x xy   − − −     = − set d d y x equal to zero (or set numerator equal to zero) *M1 Obtain 4x – y = 0 A1 Confirm 2 2 0 x xy − ≠ B1 0 x = and 2x y = both give 0 a = Obtain an equation in y (or in x) and solve for y (or for x) DM1 3 3 3 3 3 3 8 16 or 8 4 y y x x a a − = − = Obtain y = – 2a A1 With no errors seen 7

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

4 The curve with equation y e2x sin x 3 cos x has a stationary point in the interval 0 = + ≤x ≤π. (a) Find the x-coordinate of this point, giving your answer correct to 2 decimal places. [4] … … … … … … … … … … … … … … … (b) Determine whether the stationary point is a maximum or a minimum. [2] … … … … … … … …

6 marks

Mark scheme: 4(a) Use product rule M1 Obtain derivative in any correct form e.g. ( ) ( ) 2 2 2e sin 3cos e cos 3sin x x x x x x + + − A1 Equate derivative to zero and obtain an equation in one trigonometric ratio M1 Obtain x = 1.43 only A1 4 4(b) Use a correct method to determine the nature of the stationary point e.g. 2.84 2.88 1.42, 0.06e 0 1.44, 0.07e 0 x y x y ′ = = > ′ = = − < M1 Show that it is a maximum point A1 2

This question in 9709/31 May/June 2020

4 A curve has equation y cos x sin 2x. = Find the x-coordinate of the stationary point in the interval 0 x 1 giving your answer correct to < < 2π, 3 significant figures. [6] … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 4 Use correct product rule M1 Obtain correct derivative in any form, e.g. –sin x sin 2x + 2cos x cos 2x A1 Use double angle formula to express derivative in terms of sin x and cos x M1 Equate derivative to zero and obtain an equation in one trig function M1 Obtain 3 sin 2x = 1, or 3 cos 2x = 2 or 2 tan 2x = 1 A1 Solve and obtain x = 0.615 A1 6

This question in 9709/32 May/June 2020

6 y M x O 1 x The diagram shows the curve y , for x and its maximum point M. = 1 3x4 ≥0, + (a) Find the x-coordinate of M, giving your answer correct to 3 decimal places. [4] … … … … … … … … … … … … … … … … … … … … (b) Using the substitution u 3x2, find by integration the exact area of the shaded region bounded = by the curve, the x-axis and the line x 1. [5] = … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) Use quotient or product rule M1 Obtain correct derivative in any form e.g. ( ) ( ) 4 3 2 4 1 3 12 1 3 x x x x + − × + A1 Equate derivative to zero and solve for x M1 Obtain answer 0.577 A1 4 Question Answer Marks 6(b) State or imply d 2 3 d = u x x, or equivalent B1 Substitute for x and dx M1 Obtain integrand ( ) 2 1 2 3 1+ u , or equivalent A1 State integral of the form ܽtanିଵݑ and use limits u = 0 and u = 3 (or x = 0 and x = 1) correctly M1 Obtain answer 3 π 18 , or exact equivalent A1 5

This question in 9709/32 May/June 2020

3 The parametric equations of a curve are x = 3 −cos 21, y = 21 + sin 21, for 0 < 1 < 12π. dy Show that = cot 1. [5] dx … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 State or imply d dθ x = 2sin 2θ or d dθ y = 2 2cos2θ + B1 Use d d y x = d dθ y ÷ d dθ x M1 Obtain correct answer d d y x = 2 2cos2 2sin2 θ θ + A1 OE Use correct double angle formulae M1 Obtain the given answer correctly d cot d θ = y x A1 AG. Must have simplified numerator in terms of cosθ. Alternative method for question 3 Start by using both correct double angle formulae e.g. x = 3 – (2cos2θ − 1), y = 2θ + 2sinθcosθ M1 d dθ x or d dθ y B1 d d y x = ( ) ( ) 2 2 n 2 2 cos 4cos sin si + − θ θ θ θ M1 A1 Simplify to given answer correctly d cot d θ = y x A1 AG

This question in 9709/31 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 9709/31 Oct/Nov 2020

5 y P x O The diagram shows the curve with parametric equations x tan y = 1, = cos21, for 1 −12π < 1 < 2π. (a) Show that the gradient of the curve at the point with parameter is sin [3] 1 −2 1 cos31. … … … … … … … … … … … … … … … … … … The gradient of the curve has its maximum value at the point P. (b) Find the exact value of the x-coordinate of P. [4] … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(a) State d d x θ = 2 sec θ or d d y θ = 2sin cos θ θ − B1 CWO, AEF. Use d d y x = d d y θ ÷ d d x θ M1 Obtain 3 d 2sin cos d y x θ θ = − from correct working A1 AG Alternative method for question 5(a) Convert to Cartesian form and differentiate M1 1 2 1 y x = + ( ) 2 2 d 2 d 1 y x x x − = + A1 OE Obtain 3 d 2sin cos d y x θ θ = − from correct working A1 AG 3 Question Answer Marks Guidance 5(b) Use correct product rule to obtain ( ) 3 d 2cos sin d θ θ θ ± M1 Condone incorrect naming of the derivative For work done in correct context Obtain correct derivative in any form A1 e.g. ( ) 4 2 2 2cos 6sin cos θ θ θ ± − + Equate derivative to zero and obtain an equation in one trig ratio A1 e.g. 2 3 tan 1 θ = , or 2 4 sin 1 θ = or 2 4 cos 3 θ = Obtain answer x = – 1 3 A1 Or 3 3 − Alternative method for question 5(b) Use correct quotient rule to obtain 2 2 d d y x M1 Obtain correct derivative in any form A1 ( ) ( ) ( ) 2 2 2 4 2 2 1 2 2 2 1 1 x x x x x − + + × × + + Equate derivative to zero and obtain an equation in x2 A1 e.g. 6x2 = 2 Obtain answer x = – 1 3 A1 4

This question in 9709/32 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

3 The parametric equations of a curve are x = 3 −cos 21, y = 21 + sin 21, for 0 < 1 < 12π. dy Show that = cot 1. [5] dx … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 State or imply d dθ x = 2sin 2θ or d dθ y = 2 2cos2θ + B1 Use d d y x = d dθ y ÷ d dθ x M1 Obtain correct answer d d y x = 2 2cos2 2sin2 θ θ + A1 OE Use correct double angle formulae M1 Obtain the given answer correctly d cot d θ = y x A1 AG. Must have simplified numerator in terms of cosθ. Alternative method for question 3 Start by using both correct double angle formulae e.g. x = 3 – (2cos2θ − 1), y = 2θ + 2sinθcosθ M1 d dθ x or d dθ y B1 d d y x = ( ) ( ) 2 2 n 2 2 cos 4cos sin si + − θ θ θ θ M1 A1 Simplify to given answer correctly d cot d θ = y x A1 AG

This question in 9709/33 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

7 y M x O a tan−1x The diagram shows the curve y and its maximum point M where x a. x = = (a) Show that a satisfies the equation @ A 2a a tan . [4] = 1 a2 + … … … … … … … … … … … … … … … … … (b) Verify by calculation that a lies between l.3 and 1.5. [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] … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(a) Use correct quotient rule or correct product rule M1 e.g. 1 2 1 1 . tan . d 1 2 d x x y x x x x − − + = Obtain correct derivative in any form A1 Equate derivative to zero and remove inverse tangent M1 Obtain 2 2 tan 1 a a a   =   +   from correct working A1 AG. Accept with x in place of a. 4 Question Answer Marks Guidance 7(b) Calculate the value of a relevant expression or pair of expressions at a = 1.3 and a = 1.5 M1 Must be using radians Complete the argument correctly with correct calculated values A1 e.g.1.3 1.448, 1.5 1.322 < > ( ) 0.148, 0.178 − 2 7(c) Use the iterative process 1 + = na tan 2 2 1     +   n n a a correctly at least twice M1 Obtain final answer 1.39 A1 Show sufficient iterations to at least 4 d.p. to justify 1.39 to 2 d.p. or show there is a sign change in the interval (1.385, 1.395) A1 Allow recovery 3

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

18 The equation of a curve is y tan2x for x = e−5x −12π < < 2π. Find the x-coordinates of the stationary points of the curve. Give your answers correct to 3 decimal places where appropriate. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 8 Use correct product (or quotient) rule M1 At least 3 of 4 terms correct Obtain 5 2 5 2 d 5e tan 2e tan sec d − − = − + x x y x x x x A1 OE. Equate their derivative to zero, use 2 2 sec 1 tan = + x x and obtain an equation in tan x M1 Obtain 2 2tan 5tan 2 0 − + = x x A1 Allow 3 2 2tan 5tan 2tan 0 − + = x x x State answer x = 0 B1 From correct derivative. Solve a 3 term quadratic in tan x and obtain a value of x M1 Must be in radians Obtain answer, e.g. 0.464 A1 Must be 3 d.p. as specified in the question. Allow A1A0 if both values given to 2 d.p. or > 3 d.p. Obtain second non-zero answer, e.g. 1.107 and no other in the given interval A1 Alternative method for Question 8 Use correct product (or quotient) rule M1 At least 3 of 4 terms correct Obtain 5 2 5 2 d 5e tan 2e tan sec d − − = − + x x y x x x x A1 OE Equate their derivative to zero and obtain an equation in sin x and cos x M1 Obtain 5cos sin 2 = x x A1 Or simplified equivalent (i.e. cancelled) State answer x = 0 B1 From correct derivative. Use double angle formula or square both sides and solve for x M1 Or equivalent method. Must be in radians. Obtain answer, e.g. 0.464 A1 Must be 3 d.p. as specified in the question. Allow A1A0 if both values given to 2 d.p. or > 3 d.p. Obtain second non-zero answer, e.g. 1.107 and no other in the given interval A1 8

This question in 9709/32 May/June 2021

3 The parametric equations of a curve are x t ln t 2 , y t = + + = −1 e−2t, where t > −2. dy (a) Express in terms of t, simplifying your answer. [5] dx … … … … … … … … … … … … … … (b) Find the exact y-coordinate of the stationary point of the curve. [2] … … … … … … …

7 marks

Mark scheme: 3(a) State d d x t = 1 + 1 2 + t B1 Use product rule M1 Obtain d d y t = ( ) 2 2 e 2 1 e − − − − t t t A1 OE Use d d y x = d d y t ÷ d d x t M1 Obtain correct answer in any simplified form, e.g. ( )( ) 3 2 2 3 − + + t t t 2 e−t A1 5 3(b) Equate derivative to zero and solve for t M1 Obtain t = 3 2 and obtain answer y = 3 1 e 2 −, or exact equivalent A1 2

This question in 9709/33 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

3 The curve with equation y has one stationary point. = xe1−2x (a) Find the coordinates of this point. [4] … … … … … … … … … … … … … … (b) Determine whether the stationary point is a maximum or a minimum. [2] … … … … … … … … …

6 marks

Mark scheme: 3(a) Use correct product rule M1 Obtain correct derivative in any form A1 1 2 1 2 d e 2 e d x x y x x − − = − Equate derivative to zero and solve for x M1 Obtain x = 1 2 and y = 1 2 A1 4 3(b) Use a correct method for determining the nature of a stationary point M1 e.g. ( ) 2 1 2 1 2 2 d 2e 2 1 2 e d x x y x x − − = − − − Show that it is a maximum point A1 2

This question in 9709/31 Oct/Nov 2021

4 The parametric equations of a curve are x 1 y cos cos 4 = −cos 1, = 1 −1 21. dy 1 Show that sin2 . [5] dx = −2 21 … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 4 State d dθ dθ y = 1 sin sin 2 2 θ θ − + Use d d y x = d dθ y ÷ d dθ x M1 Obtain correct answer in any form A1 e.g. 1 sin sin 2 2 sin θ θ θ − + Use double angle correctly to obtain d d y x in terms of θ M1 sin 2θ = 2sin θ cos θ Obtain the given answer with no errors seen − 2sin2 1 2θ       A1 AG. Requires correct cancellation of ALL sin θ terms and cos θ = 1 − 2sin2 1 2θ       seen SC For incorrect signs, consistent throughout max. B0, M1, A0, M1, A1 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

10 y x O a π The curve y x sin x has one stationary point in the interval 0 x where x a (see diagram). = < < π, = (a) Show that tan a 2a. [4] = −1 … … … … … … … … … … … … … … … … … … (b) Verify by calculation that a lies between 2 and 2.5. [2] … … … … … … … (c) Show that if a sequence of values in the interval 0 x given by the iterative formula 1 converges, then it converges to a,<the root< π of the equation in part (a). [2] xn+1 = π −tan−1 2xn … … … … … … … (d) Use the iterative formula given in part (c) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … …

11 marks

Mark scheme: 10(a) Use correct product rule M1 Condone incorrect / missing chain rule Obtain correct derivative in any form A1 e.g. d cos sin d 2 sin y x x x x x   or 2 d 2 2 sin cos d y y x x x x x   Equate derivative to zero and obtain an equation in tan x or tan a M1 Obtain 1 2 tana a  correctly A1 AG 4 10(b) Calculate the value of a relevant expression or pair of expressions at a = 2 and a = 2.5 M1 Must be working in radians At least one correct Complete the argument correctly with correct calculated values A1 e.g. 1 2.18 and 1.25 0.747    2 10(c) State a suitable equation, e.g. 1 1 π tan 2 x x         B1 A correct equation without subscripts or quote   tan tan      Using   tan A B  formula, or otherwise, rearrange this as 1 tan 2 x x  B1 Complete argument correctly 2 Question Answer Marks Guidance 10(d) Use the iterative process correctly at least once M1 Must be working in radians Obtain answer a = 2.29 A1 Show sufficient iterations to 4 dp to justify 2.29 to 2 dp or show there is a sign change in the interval (2.285, 2.295) A1 e.g. 2.25, 2.2974, 2.2871, 2.2893, 2.2888, … 3

This question in 9709/31 May/June 2022

4 The equation of a curve is y cos3x sin x. It is given that the curve has one stationary point in the 1 = interval 0 x < < 2π. Find the x-coordinate of this stationary point, giving your answer correct to 3 significant figures. [6] … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 4 Use the correct product rule and then the chain rule to differentiate either 3 cos or sin x x M1 e.g. two terms with one part of 3 2 d cos cos cos sin sin d sin y x x p x x x q x x   . Obtain correct derivative in any form e.g. 3 2 d cos cos 3cos sin sin d 2 sin y x x x x x x x   A1 A1 A1 for each correct term substituted in the complete derivative. Equate their derivative to zero and obtain a horizontal equation with positive integer powers of sin x and/or cos x from an equation including sin x or sin 1 x using sensible algebra. M1 e.g. 2 2 4 1 2 3cos sin cos 0 x x x    Use correct formula(s) to express their equation/derivative in terms of one trigonometric function M1 Can be awarded before the previous M1. May involve more than one trigonometric term. Obtain 2 7cos 6 x  , 2 7sin 1 x , or 2 6tan 1 x , or equivalent, and obtain answer x = 0.388 A1 CAO. The question asks for 3 sf. Ignore additional answers outside   π 2 0, . 22.2 is A0. 6

This question in 9709/32 May/June 2022

14 The curve y tan x has two stationary points in the interval 0 = e−4x ≤x < 2π. dy (a) Obtain an expression for and show it can be written in the form sec2x a b sin 2x e−4x, where + dx a and b are constants. [4] … … … … … … … … … … … … … … … … … … … … … … … (b) Hence find the exact x-coordinates of the two stationary points. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(a) Use correct product rule or quotient rule, and attempt at chain rule M1 ke4x tan x + e4x sec2 x or 4 2 4 4 2 e sec tan ( e ) (e )  x x x x x k Need to see d(tan x)/dx = sec2 x (formula sheet) and attempt at ke4x, where k ≠ 1. Obtain correct derivative in any form A1  4e4x tan x + e4x sec2 x or 4 2 4 4 2 e sec tan (4e ) (e )  x x x x x Use trigonometric formulae to express derivative in the form ke4x sin x cos x sec2 x + ae4x sec2 x or ke4x sin cos cos cos x x x x + ae4x sec2 x or sec2 x(ke4x sin x cos x + ae4x) Allow 2 1 cos x instead of sec2 x M1 Need to use 2 tan sin cos sec  x x x x or sin cos tan . cos cos  x x x x x OE. M1 is independent of previous M1, but expression must be of appropriate form. Obtain correct answer with a = 1 and b = – 2 A1 At least one line of trigonometric working is required from 4e4x tan x + e4x sec2 x to given answer sec2 x(1 – 2 sin 2x) e4x with elements in any order. If only error: 4 sin x cos x = 4 sin 2x M1 A1 M1 A0. 4 Question Answer Marks Guidance 4(b) Equate derivative to zero and use correct method to solve for x M1 sin 2x = 1 2 , hence x = 1 2 sin–1 1 2 or x = tan–1(2 ± 3 ) Allow M1 for correct method for non-exact value. Obtain answer, e.g. x = 1 π 12 A1 [0.262 M1 A0] Obtain second answer, e.g. 5 π 12 and no other in the given interval A1 FT FT – 2 2  their x if exact values; x must be 2   . Ignore answers outside the given interval. Treat answers in degrees as a misread. 15°, 75°. SC No values found for a and b in 4(a) but chooses values in 4(b): max M1 for x . 3

This question in 9709/33 May/June 2022

1 16 The parametric equations of a curve are x y ln tan t, where 0 t = cos t, = < < 2π. dy cos t (a) Show that [5] dx = sin2t. … … … … … … … … … … … … … … … … … … … … … … … (b) Find the equation of the tangent to the curve at the point where y 0. [3] = … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) Use chain rule at least once M1 Needs d 1 d (tan ) d tan d  y t t t t or d d x t = (  1) 2 d (cos ) (cos ) d t t t  . BOD if + and ( 1) (  1) not seen. d d x t = sec t tan t (from List of Formulae MF19) M1 A1. If d d x t =  sec t tan t M1 A0. Obtain d d x t = sec tan t t A1 OE e.g. sin t (cos t)–2 . If e.g. d d x t = sec tan x x or sec tan   or sec tan t x , condone recovery on next line. Obtain d d y t = 2 sec tan t t A1 OE e.g. 1 sin cos t t . If e.g. d d y t = 2 sec tan x x or 2 sec tan  , condone recovery on next line. Only penalise notation errors once in d d x t and d d y t if no recovery. Use d d y x = d d y t ÷ d d x t M1 Allow even if previous M0 scored, but must be using derivatives. Obtain given answer 2 cos sin t t A1 AG After d d y x = d d y t ÷ d d x t used, any notation error A0. Must cancel cos t correctly. 5 Question Answer Marks Guidance 6(b) State or imply t = 1 π 4 when y = 0 B1 Form the equation of the tangent at y = 0 or find c M1 x = 2 , d d y x = 2 and y = 0, their coordinates and gradient used in y = mx + c. Obtain answer 2 2   y x A1 OE e.g. 2 ( 2   y x ) ISW. Allow y = 1.41x  2[.00] or 1.41(x  1.41). 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

3 The equation of a curve is y sin x sin 2x. The curve has a stationary point in the interval 0 x 1 = < < 2π. Find the x-coordinate of this point, giving your answer correct to 3 significant figures. [6] … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 3 Use correct product rule on given expression *M1 Obtain correct derivative in any form A1 e.g.cos x sin2 x + 2sin x cos2 x Use correct double angle formulae to express derivative in terms of sin x and *M1 cos x Equate derivative to zero and obtain an equation in one trig variable DM1 dependent on the 2 previous M Marks. Obtain 3sin 2 x = 2 , 3cos 2 x = 1 or tan 2 x = 2 A1 OE Solve and obtain x = 0.955 A1 3 sf only. Final answer in degrees is A0. Ignore any attempt to find the corresponding value of y. Alternative method for the first three marks Use correct double angle formula to obtain y = 2cos x − 2cos 3 x *M1 or y = 2sin 2 x cos x Use chain rule and / or product rule *M1 Obtain derivative y = −2sin x + 6sin x cos 2 x A1 y = −2sin 3 x + 4sin x cos 2 x Alternative method for the second and third M marks Equate derivative to zero and obtain an equation in tan x and tan2x *M1 Use correct double angle formula to obtain an equation in tan x DM1 6

This question in 9709/32 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

10 y M x O The diagram shows the curve y x 5 3 and its maximum point M. = + −2x (a) Find the exact coordinates of M. [5] … … … … … … … … … … … … … … … … … … (b) Using the substitution u 3 find by integration the area of the shaded region bounded by = −2x, the curve and the x-axis. Give your answer in the form a 13, where a is a rational number. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(a) Use the product rule correctly to obtain 1 2 ( 5)(3 2 ) (3 2 ) n p x x q x     BOD over sign errors unless an incorrect rule is quoted. Obtain correct derivative in any form A1 e.g. 1 1 2 2 ( 5)(3 2 ) (3 2 ) x x x       . Equate derivative to zero and obtain a linear equation DM1 Allow with surd factor e.g.        1 2 3 2 5 3 2 0 x x x        . Obtain a correct linear equation. A1 e.g. –(x + 5) + 3 – 2x = 0. Obtain answer 2 13 39 , 3 9          . A1 Or exact equivalent e.g. 2 13 13 , 3 3 3          or 2 2197 , 3 27          . Accept with x, y stated separately. ISW Alternative Method for Question 10(a) Obtain y2 and differentiate *M1 Ignore their left hand side i.e. their 2 d d y x . Obtain correct derivative in any form A1 e.g. 2 6 34 20 x x    . Equate derivative to zero and solve for x DM1 Obtain 2 3  A1 Ignore –5 if seen. Obtain answer 2 13 39 , 3 9          only A1 Or exact equivalent e.g. 2 13 13 , 3 3 3          or 2 2197 , 3 27          . ISW 5 Question Answer Marks Guidance 10(b) Use the given substitution and reach 1 2 13 d 2 2 u a u u         *M1 OE Need to see -2 or -½ used. Condone if du missing or the integral sign is missing. Allow M1A0 for complete substitution into 3 2 d x x x   to obtain first term of the line below. Obtain correct integral 1 2 1 13 d 2 2 2 u u u          A1 OE e.g. 1 3 d 5 d 2 2 u u u u u            . Ignore limits at this stage. Condone if du missing. x = –5 and 3 2 B1 SOI e.g. by u = 13 and 0. In any order and at any stage. Use correct limits the right way round in an integral of the form 3 5 2 2 26 2 3 5 a u u          DM1 Obtain answer 169 13 15 or a = 169 15 A1 or exact equivalents. 5

This question in 9709/32 May/June 2023

5 y M x O a 6π1 The diagram shows the part of the curve y x2 cos 3x for 0 and its maximum point M, where = ≤x ≤16π, x a. = @ A 1 2 (a) Show that a satisfies the equation a = 3 tan−1 3a . [3] … … … … … … … … … … … … … … … … … … (b) 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] … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 5(a) Use correct product rule M1 d dx (x2)cos(3x) + x2 d dx (cos 3x). Obtain correct derivative in any form A1 e.g. 2 2 cos3 3 sin3  x x x x . Equate derivative to zero and obtain 1 1 2 tan . 3 3        a a A1 AG Condone 1 1 2 tan 3 3   a a . Must at least reach expression 2x = 3x2 tan(3x) or better before final answer to gain A1. Final answer must be in terms of a. Can work with x and switch to a at very end. Look for 2 3 a or 2 3 x in working not immediately corrected or as penultimate line A0. 3 5(b) Use the iterative process 1 1 1 2 tan 3 3          n n a a correctly at least twice during successive iterations in the numerous iterations M1 Degrees 0/3. Obtain final answer 0.36 A1 Must be 2d.p. Show sufficient iterations to 4 or more d.p. to justify 0.36 to 2 d.p. or show there is a sign change in the interval   0.355, 0.365 A1 Allow small errors in 4th d.p. Allow errors at start if self corrects later. 0.5 0.4 0.3 0.2 0.1 /6 /12 0.3091 0.3435 0.3826 0.4264 0.4740 0.3017 0.3989 0.3789 0.3650 0.3499 0.3339 0.3176 0.3820 0.3439 0.3513 0.3566 0.3625 0.3688 0.3754 0.3502 0.3649 0.3619 0.3599 0.3576 0.3552 0.3526 0.3624 0.3567 0.3578 0.3604 0.3614 0.3576 0.3580 3

This question in 9709/33 May/June 2023

x2 1 Find the exact coordinates of the points on the curve y = at which the gradient of the tangent 1 −3x is equal to 8. [5] … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: Question Answer Marks Guidance 1 Use correct quotient or product rule *M1 Obtain correct derivative in any form A1 2  2  (1 − 3 x )2 x − x ( −3) 2 x − 3 x e.g. 2  = 2  or (1 − 3 x )   ( 1 − 3 x )  2 −2 −1 3x (1 − 3x ) + 2 x (1 − 3x ) . Equate derivative to 8 and solve for x DM1 75 x 2 − 50 x + 8 = (15 x − 4 )( 5 x − 2 ) . 2 4 A1 Exact values required. Obtain answers x = and 5 15 4 16 A1 Allow A1 for one correct point. Obtain answers y = − and 5 45 5

This question in 9709/31 Oct/Nov 2023

6 The parametric equations of a curve are x = t + 3, y = ln t, for t > 0. dy (a) Obtain a simplified expression for in terms of t. [3] dx … … … … … … … … … … … … (b) Hence find the exact coordinates of the point on the curve at which the gradient of the normal is −2. [3] … … … … … … … …

6 marks

Mark scheme: 6(a) State correct derivative of x or y with respect to t B1 dx 1 − 12 dy 1 = t , = . dt 2 dt t dy dy dt M1 Use correct chain rule. Use =  dx dt dx − dy 2 A1 2 t 2 Obtain answer = . Or simplified equivalent e.g 2t 1 or dx t t 3 6(b) dy 1 M1 State or imply their = dx 2 Obtain t = 4 A1 Or equivalent. Obtain answer (7, ln 16) A1 Or exact equivalent. Can state the two components separately. 3

This question in 9709/31 Oct/Nov 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

7 The equation of a curve is x3 + y2 + 3x2 + 3y = 4. dy + 6x (a) Show that = −3x2 . [3] dx 2y + 3 … … … … … … … … … … … … … … … … … … … … … … … … (b) Hence find the coordinates of the points on the curve at which the tangent is parallel to the x-axis. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 7(a) d y B1 Allow for 3x2dx + 2ydy or Fx = 3 x 2 + 6 x and Fy = 2 y + 3 . as the derivative of y2 State or imply 2 y dx d y M1 dy d y Equate derivative of LHS to zero and solve for 3x2 + 2 y + 6x + 3 = 0 d x dx d x dy Fx or 3x2dx + 2 ydy + 6xdx + 3dy = 0 or = − need dx Fy evidence from B1 mark or formula must be seen. Allow errors. Obtain the given answer A1 dy 3 x 2 + 6 x −3 x 2 − 6 x AG = − not . dx 2 y + 3 2 y + 3 d y dy Must factorise with e.g. 3x2 + 6x + (2y + 3) = 0 d x dx or 3x2dx + 6xdx + d y ( 2 y + 3 ) = 0. 3 7(b) Equate numerator to zero and solve for x *M1 Allow for just one x value. Obtain x = 0 and x = –2 only A1 Substitute their x, [x = 0 or x = –2] in curve equation to obtain quadratic DM1 y2 + 3y – 4 = 0 or y2 + 3y = 0. equation in y equal to 0 Obtain y = 1 and y = –4 [when x = 0] A1 Obtain y = 0 and y = –3 [when x = –2] A1 ISW If forget x = 0 then max 3/5. 5

This question in 9709/33 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

6 The equation of a curve is 2y 2 + 3xy + x = x 2 . d y 2x - 3y - 1 (a) Show that = . [4] d x 4y + 3x … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Hence show that the curve does not have a tangent that is parallel to the x-axis. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: y introduced instead of d then allow B1 dx6(a) State or imply 4 y ddyx as the derivative of 2y 2 B1 SC If dd x for both, followed by correct method M1 Max 2. y = correct expression to collect all x State or imply 3 y + 3 x ddyx as the derivative of 3xy B1 Allow extra dd marks if correct. Complete the differentiation, all 4 terms, isolate 2 d y terms on LHS or bracket d y M1 d x d x terms and solve for dy dx dy 2 x − 3 y − 1 A1 Answer Given – need to have seen 4y d y + 3x dy Obtain = dx dx dx 4 y + 3 x dy = 2x −3y −1 or (4y + 3x) − 2x +3y = −1. dx Need to see = 2x or = 0 consistently throughout otherwise M1 A0. No recovery allowed. When all terms are included then must be an equation. 4 Allow all marks if using dx and dy. 6(b) Equate numerator to zero, obtaining 2x = 3y + 1 or 3y = 2x −1 and form equation in M1* 2 2 2 e.g. 9 ( 2 x − 1) + x ( 2 x − 1) + x = x x only or y only from 2y2 +3xy + x = x2 or 2 y 2 + 32 (1 + 3 y ) y + 12 (1 + 3 y ) = 14 (1 + 3 y ) 2 . Allow errors. Obtain 2 ( 2 x − 1) 2 = − x 2 or a 3 term quadratic in one unknown and try to solve. DM1 e.g. 17 x 2 − 8 x + 2 = 0 ( b 2 − 4 ac = −72 ) 9 If errors in quadratic formulation allow solution, applying usual rules for solution of 2 2 or 17 y + 6 y + 1 = 0 ( b − 4 ac = −32 ) . quadratic equation, and allow M1 x = 4/17 ± (3√2/17)i, y = − 3//17 ± (2√2/17)i . Conclude that the equation has no [real] roots A1 Given Answer. CWO 3

This question in 9709/32 Feb/March 2024

7 y a x O M The diagram shows the curve y = xe 2 x - 5x and its minimum point M, where x = a . 1 5 (a) Show that a satisfies the equation a = ln [3] 2 b 1 + 2a l. … … … … … … … … … … … … … … … … (b) Verify by calculation that a lies between 0.4 and 0.5 . [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] … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 7(a) Use correct product rule M1 y Obtain correct derivative in any form A1 e.g. d = e 2 x + 2 xe 2 x − 5 d x 1  5  A1 Given answer – need to see e2x= 5/(1 + 2x) Equate derivative to zero and obtain α= ln   or ln e2x = ln (5/(1 + 2x)) in working. 2  1 + 2α Must be in terms of α not x. Allow α to be used before equating to 0. 3 7(b) Calculate the value of a relevant expression or values of a pair of expressions at M1 Need to attempt BOTH values and have one x = 0.4 and x = 0.5 correct. Complete the argument correctly with correct calculated values A1 e.g. 0.4 < 0.51[ 08 ] and 0.5 > 0.458 or 0.46 or 0.45 or – 0.11[08] < 0 and 0.042 > 0 If use original derivative −0.994 (0.4) and 0.437 (0.5). 2 7(c) 1  5  M1 Obtain one value and then substitute it into the Use the iterative process αn +1 = ln   correctly at least twice anywhere in formula to obtain a second value. 2  1 + 2αn  iteration process Obtain final answer 0.47 A1 Show sufficient iterations to 4 d.p. to justify 0.47 to 2 d.p. or show there is a sign A1 0.4,0.5108,0.4528,0.4823,0.4670,0.4749 change in the interval ( 0.465, 0.475 ) 0.45,0.4838,0.4663,0.4753,0.4707,0.4730 0.5,0.4581,0.4795,0.4685,0.4742 Allow self correction. 3 SC B1 No working 0.47

This question in 9709/32 Feb/March 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

2 Find the exact coordinates of the stationary point of the curve y = e 2 x sin 2x for 0 G x G 1 r . [5] 2 … … … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 Use correct product rule cos 2x may be 1 – 2 sin2x or … Allow M1 if only error is ex instead of e2x in one of terms, then maximum 1/5. Obtain correct derivative 2 2 2e sin2 2e cos2 x x x x  A1 OE, e.g.   2 2 2 2 4e sin cos 2e cos sin . x x x x x x   Equate derivative of the form ae2xsin2x + e2xbcos2x to 0 and solve for 2x or x using a correct method Note may have substituted for sin2x and/or cos2x M1 Obtain 2x = tan−1(− their b/their a) OE. Allow one slip in rearranging. Allow degrees. Variety of other methods available, such as solving quadratic equation in sin x or tan x e.g. tan² x – 2tan x – 1 = 0 leading to x = tan-1(1 + √2). Obtain x = 3 8 π only or exact equivalent A1 CWO 67.5° gets A0. Ignore any answers outside interval 0 ⩽ x ⩽ π . 2 Obtain y = 3π 4 1 2e 2 only or exact simplified equivalent A1 CWO, ISW. Not 3π 4 3 sin πe 4         . Ignore any answers using x outside interval 0 ⩽ x ⩽ π . 2 5

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

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

2 The equation of a curve is xy 2 + ln `x + 2yj = 1. Find the gradient of the curve at the point where x = 0 . [5] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 dy B1 1 + 2 dx State or imply as the derivative of ln (x + 2y) x + 2 y d y B1 2 dy 2 State derivative of xy2 is x2y + y2 x y + y d x dx dy B1 OE 1 + 2 2 dy dx May be implied by correct final answer. Obtain y + 2 xy + = 0 dx x + 2 y 1 B1 OE Obtain y = e when x = 0 1 Allow 2  2.718 or 1.36 or better e.g. 1.359… 2 May be implied by correct final answer. dy 1 3 B1 OE Obtain = − e + 4 ( ) Accept AWRT –3.01. ISW. dx 8 5

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

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

11 y M O a 1 x r 2 The diagram shows the curve y = x sin 2x for 0 G x G 1 r. The curve has a maximum point at M, 2 where x = a . (a) Show that tan 2a =-4a [4] … … … … … … … … … … … … … (b) Show by calculation that 0.9 1 a 1 0. 95 . [2] … … … … … … (c) Show that if a sequence of values given by the iterative formula = x 1 -1 n + 1 2 br - tan `4x njl converges, then it converges to a. [2] … … … … … … … … … … … (d) Use the iterative formula in part (c) to calculate a correct to 4 decimal places. Give the result of each iteration to 6 decimal places. [3] … … … … … … … … … … … … …

11 marks

Mark scheme: 11(a) Use the correct product rule to differentiate *M1 Could be working in terms of a. p Obtain the form sin2 x + q x cos2 x. x dy 1 A1 Obtain = sin2 x + 2 x cos2 x dx 2 x Equate the derivative to zero and form an equation without surds DM1 E.g. sin2a + 4a cos2a = 0. Could be working in terms of x. Obtain tan2a = −4a from correct work A1 AG 4 11(b) Calculate the values of a relevant expression or pair of expressions at x = 0.9 and M1 Allow smaller interval, provided it contains the x = 0.95 root. Must be working in radians. Complete the argument correctly with correct calculated values A1 E.g.  tan1.8 + 3.6 = −0.686...  0   tan1.9 + 3.8 = 0.873...  0 2 11(c) 1 −1 −1 B1 Or work from right to left. π − tan 4 a and rearrange to π − 2a = tan 4a State a = 2 ( ) Allow working in x or a. State tan ( π − 2 a ) = 4 a and rearrange to tan2a = −4a B1 Allow working in x or a. 2 11(d) Use the iterative process correctly at least once M1 M0 if working in degrees. Obtain final answer 0.9183 A1 Show sufficient iterations to at least 6 d.p. to justify 0.9183 to 4 d.p. A1 E.g. or show there is a sign change in the interval ( 0.91825, 0.91835 ) 0.9,0.920872,0.917944,0.918347,0.918292... 3

This question in 9709/33 May/June 2025

6 The parametric equations of a curve are 2 x = and y = tan t3 , cos t3 for 0 G t G 2r . dy (a) Show that can be written as A cosec t3 , where A is a constant to be found. [5] dx … … … … … … … … … … … … … … … … … … … … … … … (b) Find an equation of the normal to the curve at the point where t = 1 r . Give your answer in the 12 form y = mx + c , where the constants m and c are exact. [4] … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) dy 2 B1 OE Obtain = 3sec 3t dt −1 d x M1 dx −2 d Attempt chain rule on 2 ( cos3t ) for or attempt to differentiate 2sec 3t Attempt = −2 ( cos3t ) ( cos3t ) with their derivative. d t dt dt d x A1 −2 cos3t ) sin3t . Obtain = 6sec 3t tan 3t OE, e.g. 6 ( d t Allow unsimplified, e.g. (−1)  2  (−1)  3 for 6. dy dy dt dy dt M1 2 1 Use =  with their and their Expect, for example, 3sec 3t  dx dt dx dt dx 6sec 3t tan 3t 2 1 or 3sec 3t  −2 if correct. 6 ( cos3t ) sin 3t dy 1 A1 WWW Obtain = cosec3t Must be in this form of answer given in the question. dx 2 Not required to state A = 12 . d y d x d y Allow slips in notation for , and . d t d t d x 6(a) Alternative Method for Question 6(a) 2 B1 2 x Convert to Cartesian form e.g. 1 + y = 4 d 2 d y B1 Correct use of implicit differentiation e.g. y = 2 y d y d x dy x B1 OE Obtain 2 y = dx 2 dy 2sec3t M1 Convert to parametric form e.g. 2tan3t = dx 2 dy 1 A1 WWW Obtain = cosec3t Must be in this form of answer given in the question. dx 2 5 6(b) 1 B1 Must be exact. Obtain x = 2 2 and y = 1 when t = 12 π 1 dy M1 Expect gradient of normal = − 2 if correct. Substitute t = 12 π into −1 their dx Allow if in decimals. Allow a small slip, but not with their coefficient A If their coefficient A is dealt with incorrectly M0, but allow second M1. 2 M1 E.g. y – 1 = − 2 x − 2 2 if correct or find c in equation Form equation of the normal with their (x, y), found using x = and ( ) cos3t of line. dy y = tan 3t , and −1 their Allow M1 even if decimals. dx M0 if using gradient of tangent. A1 CAO Obtain equation of normal y = − 2 x + 5 Require y = mx + c and exact m and c 2 Accept y = − x + 5. 2 4

This question in 9709/35 May/June 2025

4 y M O 1 x r 4 The diagram shows the graph of y = e sin 2 x cos 4x for 0 G x G 1 r , and its maximum point M. 4 Find the x-coordinate of M. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 4 Use correct product rule: p sin 4 xe sin2 x + q cos2 x cos4 xe sin2 x M1 Condone sign error. Condone if no attempt at chain rule seen. Obtain derivative −4sin4 xesin2 x + 2cos2 x cos4 xe sin2 x A1 OE Equate derivative to zero and obtain an equation in one trig function M1 Allow arithmetical errors. Obtain 2sin 2 2 x + 4sin2 x −=1 0 A1 OE Obtain x = 0.113 and no other root A1 May be more accurate. 5

This question in 9709/31 Oct/Nov 2025

7 The parametric equations of a curve are 2 t x = t - ln ( 2 t + 1), y = . 2t + 1 dy Obtain a simplified expression for in terms of t. [5] dx … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 7 dx 2 B1 State = 2t − dt 2t + 1 Use correct quotient rule or correct product rule M1 Might split first as 1 2 − 4 t+1 2. dy ( 2t + 1) .1 − t ( 2 ) A1 1 Obtain = oe Simplifies to 2 2 dt ( 2t + 1) ( 2t + 1) dy dy dx M1 dx Use =  Allow for . dx dt dt dy 1 A1 OE. Must have 1 in numerator. Obtain answer 2 ( 2t + 1)( 2t − 1)( t + 1) 1 May have, e.g., . ISW. 4t 2 + 2t − 2 ( 2t + 1)( ) 2 B1 2  y   2 y   y   1  Obtain x =   − ln  + 1  OE, e.g. x =   − ln   .  1 − 2 y   1 − 2 y   1 − 2 y   1 − 2 y  Use correct quotient rule to differentiate M1 dy (1 − 2 y ) 3 (1 − 2 y ) 3 d x  2 y  (1 − 2 y ) + 2 y  2 A1 = = OE,   − Obtain =  dx −2 (1 − 4 y )(1 − y )   2 y − 2 (1 − 2 y ) 2 d y 1 − 2 y  1 − 2 y 1 − 2 y ) 2  (  2 t 3 M1 Express in terms of t. dy (1 − 2 t +1 ) 1 = = 4 t t dx −2 (1 − 2 t + 1 )(1 − 2 t +1 ) −2 ( 2t + 1)(1 − 2 t )( t + 1 ) 1 A1 OE Obtain answer 2 ( 2t + 1)( 2t − 1)( t + 1) 5

This question in 9709/31 Oct/Nov 2025

5 (a) It is given that f ( x) = ( x - a ) 2 g ( x) , where f ( x) and g ( x) are polynomials. Show that ( x - a ) is a factor of fl( )x . [2] … … … … … … … … (b) It is given that ( x - 3 ) 2 is a factor of 2x 3 - 4 x 2 + px + q , where p and q are constants. Find the values of p and q. [5] … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(a) Use correct product rule M1 2 f  ( x ) = ( x − a ) g ( x ) + 2 ( x − a ) g ( x ) Allow incorrect chain rule. Obtain correct derivative and state a clear conclusion A1 AG E.g. take out a factor of (x – a) and show the factorised form as far as ( x − a )h( x ) correctly (no further comment needed), or show that f  ( a ) = 0 and state that ‘(x – a) is a factor’. 2 5(b) Use f ( 3 ) = 0 M1 2  33 − 4  32 + 3 p + q = 0 Obtain 3 p + q = −18 A1 OE Powers should be evaluated but do not need to be simplified. Use f  ( 3 ) = 0 M1 54 − 24 + p = 0 Obtain p = −30 A1 OE Obtain p = −30, q = 72 A1 Correct only. Alternative Method for Question 5(b) 2 M1 Q  0 Attempt division of f(x) by( x − 3) as far as ( 2 x + Q ) Obtain quotient ( 2 x + 8 ) A1 Equate linear remainder to zero and compare coefficients M1 2 Or expand ( 2 x + 8 )( x − 3) and compare coefficients. Method to obtain an equation in p or q. obtain one of p = −30, q = 72 A1 Obtain p = −30, q = 72 A1 Correct answers only. 5(b) Alternative Method 2 for Question 5(b) Use f ( 3 ) = 0 M1 54 − 36 + 3 p + q = 0 Obtain 3 p + q = −18 A1 OE Powers evaluated. 2 M1 Using this as part of a hybrid method they need to be Divide f(x) by( x − 3) as far as ( 2 x + Q ) and form an equation in p only or q working towards a second equation by considering only or in p and q coefficients in the remainder or use f ( −4 ) = 0 or equivalent for their ( 2 x + 8 ) . Obtain correct equation in p or q A1 Obtain p = −30, q = 72 A1 Correct answer only. Alternative Method 3 for Question 5(b) f  ( x ) = 6 x 2 − 8 x + p B1 = ( x − 3 )( 6 x + 10 ) M1 Use the factor x − 3. Obtain p = −30 A1 2 3 2 M1 2 = 2 x − 4 x − 30 x + q Use the factor ( x − 3) . f ( x ) = ( x − 3) ( 2 x + Q ) ( )  Q = 8, q = 72 A1 5(b) Alternative Method 4 for Question 5(b) 2 M1 A = 2 can be found by inspection. Expand f ( x ) = ( x − 3) ( Ax + B ) and compare at least one coefficient other than for x3 Obtain q = 9 B and p = 9 A − 6 B A1 or obtain B = 8 Use their A and B to solve for p or q M1 A = 2 B = 8 Obtain one of p = −30, q = 72 A1 Obtain p = −30, q = 72 A1 Correct answer only. 5

This question in 9709/32 Oct/Nov 2025

8 The curve with equation y = e -5 x ln 5x has a stationary point at x = p. 1 (a) Show that p satisfies the equation ln 5p = . [3] 5p … … … … … … … … … … … (b) By sketching a suitable pair of graphs, show that the equation in part (a) has only one root. [2] (c) Show by calculation that 0.2 1 p 1 0. 6 . [2] … … … … … … … … … … … … 1 1 (d) It is given that the equation in part (a) can be written in the form p = exp e o, where exp (x) 5 5p denotes ex. Use an iterative formula based on this rearrangement to calculate p correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … … …

10 marks

Mark scheme: 8(a) d d M1 M0 if y = e−5p ln 5p seen prior to differentiation. Use the correct product or quotient rule, e.g. e−5x (ln 5x) + ln 5x (e−5x) Accept if only seen in actual derivative = 0. dx dx 1 −5 x −5 x A1 Obtain the correct derivative in any form e.g. e − 5e ln5 x x 1 A1 AG Obtain the given answer ln5 p = after full and correct working 1 −5 x −5 x 5 p May go from e − 5e ln5 x = 0 , or x 1 −5 x −5 x e = 5e ln5 x to the given answer without x intermediate working. 3 8(b) 1 M1 For both marks: Sketch an acceptable graph, e.g. y = ln 5x or y = Note: Allow without scale on either axis, but if 5x y = ln5x = 0 identified to be not x = 0.2, then 0 marks for y = ln 5x. Allow graphs not labelled, or labelled with p instead of x. Allow ln 5x starting at the x-axis. If either graph shown in other quadrants, must be correct. 1 For y = , asymptotic behaviour needed for at 5x least one axis. Must not touch axes. 1 A1 Sketch a second acceptable graph, e.g. y = or ln 5x, and justify the given 5x statement by dot, cross or statement only one intersection. 2 8(c) Calculate the values of a relevant expression or pair of expressions at p = 0.2 M1 1 f(p) = ln5 p − and p = 0.6 5 p f(0.2) = –1 < 0, f(0.6) = 0.765 > 0 Note can use, e.g., p = 0.3 and p = 0.5, or any smaller interval which works At least one correct value to at least 2sf. 1 Or comparing ln5 p and . 5 p At least 3 correct values to at least 2sf. Complete the argument correctly with correct calculated values A1 2 8(d) Use the iterative formula correctly at least twice M1 M0 for 0.3526, 0.3526, 0.3526… Obtain final answer p = 0.35, Answer = 0.35, or just 0.35 stated A1 Allow, e.g., a1, a2, a3 … or x1, x2, x3 … or answer1 , answer2, answer3 … for M1 and second A1. For first A1, must be p = 0.35 or answer = 0.35 unless just 0.35 is stated, e.g. not x = 0.35, p7 =…, p∞ = … etc. Show sufficient iterations to 4 dp to justify 0.35 to 2 dp or show there is a sign A1 E.g. 0.4, 0.3297, 0.3668, 0.3450, 0.3571, 0.3502, change in the interval (0.345, 0.355) 0.3541. Allow M1(A1 or A0) A1 if more values are to at 0.2, 0.5437, 0.2889, 0.3996, 0.3299, 0.3667, 0.3451, 0.3571, 0.3502, 0.3541 least 4dp than to 3dp. 0.25, 0.4451, 0.3135, 0.3786, 0.3392, 0.3607, 0.3482, 0.3552, 0.3512, 0.3535 SC B1 for starting from either 0.3526 or 0.3527 and 0.3, 0.3895, 0.3342,0.3639, 0.3465, 0.3562, 0.3507, 0.3538 0.3520, 0.3530 using iterative formula correctly at least twice if the 0.45, 0.3119, 0.3797, 0.3387, 0.3610, 0.3480, 0.3553, 0.3512, 0.3535 sequence shows a correct change in the 4th decimal 0.5, 0.2984,0.3910, 0.3336, 0.3643, 0.3463, 0.3563, 0.3506, 0.3538 place (and SC DB1 for getting p = 0.35), but 0 0.55, 0.2877, 0.4008, 0.3294, 0.3670, 0.3449, 0.3572, 0.3501, 0.3541 marks otherwise. 0.6, 0.2791, 0.4095, 0.3260, 0.3694, 0.3437, 0.3579, 0.3497, 0.3543 3

This question in 9709/33 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

tan x 6 A curve has equation y = . 5 + 2 sin x d y (a) Express as a simplified fraction in terms of sinx. [4] d x … … … … … … … … … … … … … … … (b) Show that the curve has no stationary points. [2] … … … … … … … … … …

6 marks

Mark scheme: 6(a) Use correct quotient rule (or correct product rule) *M1 Allow one error. 2 x − tan x  2cos x A1 dy ( 5 + 2sin x ) sec Obtain = 2 dx ( 5 + 2sin x ) Express derivative in terms of sin x and cos ,x and simplify to an expression in DM1 1 sin x ( 5 + 2sin x ) 2 −  2cos x sin x only dy cos x cos x = 2 dx ( 5 + 2sin x ) 3 A1 OE without fractions within a fraction. dy 5 + 2sin x = 2 2 Denominator must be in factorised form. dx (1 − sin x )( 5 + 2sin x ) ISW once correct form is seen. Alternative Method for Question 6(a) Use correct quotient rule (or correct product rule) *M1 Allow one error. dy cos x ( 5 + 2sin x ) cos x − sin x ( cos x ( 2cos x ) + ( 5 + 2sin x )( − sin x ) ) A1 Obtain = dx cos 2 x ( 5 + 2sin x ) 2 Simplify to an expression in sin x only DM1 3 A1 OE without fractions within a fraction. dy 5 + 2sin x = 2 2 Denominator must be in factorised form. dx (1 − sin x )( 5 + 2sin x ) ISW once correct form is seen. 6(a) Alternative Method 2 for Question 6(a) Use correct quotient rule (or correct product rule) *M1 Allow one error. dy ( 5cos x + sin 2 x ) cos x − sin x ( −5sin x + 2cos2 x ) A1 = dx cos 2 x ( 5 + 2sin x ) 2 Simplify to an expression in sin x only DM1 3 A1 OE without fractions within a fraction. dy 5 + 2sin x = 2 2 Denominator must be in factorised form. dx (1 − sin x )( 5 + 2sin x ) ISW once correct form is seen. 4 6(b) Set numerator of their derivative equal to zero and attempt to solve for sin x M1 Must be a polynomial in sin .x Show that there are no solutions and hence no stationary points. A1 AG From correct working only using a correct numerator from part (a). Reference to −1 sin x  1 OE must be made. Alternative Method for Question 6(b) Use of −1 sin x 1 leading to 3 5 + 2sin 3 x 7 M1  5 + 2sin 3 x  0 A1 OE 2

This question in 9709/35 Oct/Nov 2025