Differentiation — Paper 2 · A Level Mathematics 9709

1 dy 5 (i) By differentiating , show that if y = sec θ then = sec θ tan θ. [3] cos θ dθ (ii) The parametric equations of a curve are x = 1 + tan θ, y = sec θ, 1 dy for −12π < θ < 2π. Show that = sin θ. [3] dx (iii) Find the coordinates of the point on the curve at which the gradient of the curve is 12. [3]

9 marks

Mark scheme: 5 (i) Differentiate using chain or quotient rule M1 Obtain derivative in any correct form A1 Obtain given answer correctly A1 3 (ii) State dx = sec2 θ, or equivalent B1 dθ d y d y d x Use = ÷ M1 d x d θ d θ Obtain given answer correctly A1 3 GCE AS LEVEL – JUNE 2005 9709 2 π (iii) State that θ = B1 6 1 Obtain x-coordinate 1 + , or equivalent B1 3 2 Obtain y-coordinate , or equivalent B1 3 3

This question in 9709/21 May/June 2005

6 It is given that the curve y = (x −2)ex has one stationary point. (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: 6 (i) Use product rule M1* Obtain correct derivative in any form, e.g. (x – 1)ex A1 Equate derivative to zero and solve for x M1* (dep) Obtain x = 1 A1 Obtain y = –e A1 [5] (ii) Carry out a method for determining the nature of a stationary point M1 Show that the point is a minimum point, with no errors seen A1 [2] GCE A/AS LEVEL – May/June 2008 9709 02 dy 2

This question in 9709/21 May/June 2008

7 The equation of a curve is x2 + y2 −4xy + 3 = 0. dy 2y −x (i) Show that = [4] dx y −2x. (ii) Find the coordinates of each of the points on the curve where the tangent is parallel to the x-axis. [5]

9 marks

Mark scheme: dy 7 (i) State 2y as derivative of y2, or equivalent B1 dx dy State 4y + 4x as derivative of 4xy, or equivalent B1 dx dy Equate derivative of LHS to zero and solve for M1 dx Obtain given answer correctly A1 [4] [The M1 is dependent on at least one of the B marks being obtained.] (ii) State or imply that the coordinates satisfy 2y – x = 0 B1 Obtain an equation in x2 (or y2) M1 Solve and obtain x2 = 4 (or y2 = 1) A1 State answer (2, 1) A1 State answer (–2, –1) A1 [5] 1 2

This question in 9709/21 May/June 2008

d2y 2x6 Find the exact coordinates of the point on the curve y = xe−1 at which = 0. [7] dx2

7 marks

Mark scheme: x x 6 At any stage, state the correct derivative of e 2 or 2e B1 Use product or quotient rule M1 Obtain correct first derivative in any form A1 Obtain correct second derivative in any form B1√ Equate second derivative to zero and solve for x M1 Obtain x = 4 A1 Obtain y = 4e– 2, or equivalent A1 [7] GCE A/AS LEVEL – October/November 2008 9709 02

This question in 9709/21 Oct/Nov 2008

6 The curve with equation y x ln x has one stationary point. = (i) Find the exact coordinates of this point, giving your answers in terms of e. [5] (ii) Determine whether this point is a maximum or a minimum point. [2]

7 marks

Mark scheme: 6 (i) Use product rule M1* Obtain derivative in any correct form A1 Equate derivative to zero and solve for x M1(dep*) Obtain x = 1/e, or exact equivalent A1 Obtain y = –1/e, or exact equivalent A1 [5] (ii) Carry out complete method for determining the nature of a stationary point M1 Show that at x = 1/e there is a minimum point, with no errors seen A1 [2]

This question in 9709/21 Oct/Nov 2009

4 The parametric equations of a curve are 1 et x y = −e−t, = + e−t. dy (i) Show that e2t [3] dx = −1. (ii) Hence find the exact value of t at the point on the curve at which the gradient is 2. [2]

5 marks

Mark scheme: dx d y 4 (i) State = e–t or = et – e–t B1 dt d t dy d y dx Use = ÷ M1 dx d t dt Obtain given answer correctly A1 [3] dy (ii) Substitute = 2 and use correct method for solving an equation of the form e2t = a, dx where a > 0 M1 Obtain answer t = 1 ln 3, or equivalent A1 [2] 2

This question in 9709/22 Oct/Nov 2009

6 The equation of a curve is x2y y2 6x. + = dy 6 (i) Show that . [4] −2xy dx = x2 2y + (ii) Find the equation of the tangent to the curve at the point with coordinates giving your answer in the form ax by c 0. (1, 2), [3] + + =

7 marks

Mark scheme: dy 6 (i) State 2xy + x2 as derivative of x2y B1 dx dy State 2y as derivative of y2 B1 dx dy Equate derivatives of LHS and RHS, and solve for M1 dx Obtain given answer A1 [4] (ii) Substitute and obtain gradient 2 , or equivalent B1 5 Form equation of tangent at the given point (1, 2) M1 Obtain answer 2x – 5y + 8 = 0, or equivalent A1 [3] [The M1 is dependent on at least one of the B marks being obtained.]

This question in 9709/21 May/June 2010

8 The equation of a curve is x2 2xy 8 0. + −y2 + = (i) Show that the tangent to the curve at the point is parallel to the x-axis. [4] (−2, 2) (ii) Find the equation of the tangent to the curve at the other point on the curve for which x giving your answer in the form y mx c. = −2,[5] = +

9 marks

Mark scheme: dy 8 (i) State 2y as derivative of y2, or equivalent B1 dx dy State 2y + 2x as derivative of 2xy, or equivalent B1 dx dy Substitute x = –2 and y = 2 and evaluate M1 dx Obtain zero correctly and make correct conclusion A1 [4] (ii) Substitute x = –2 into given equation and solve M1 Obtain y = –6 correctly A1 dy Obtain = 2 correctly B1 d x Form the equation of the tangent at (–2, –6) M1 Obtain answer y = 2x – 2 Al [5]

This question in 9709/23 Oct/Nov 2010

7 The parametric equations of a curve are x e3t, y t2et 3. = = + dy (i) Show that . [4] t(t + 2) dx = 3e2t (ii) Show that the tangent to the curve at the point is parallel to the x-axis. [2] (1, 3) (iii) Find the exact coordinates of the other point on the curve at which the tangent is parallel to the x-axis. [2]

8 marks

Mark scheme: 7 (i) Use product rule to differentiate y M1 Obtain correct derivative in any form in t for y A1 dy dy dx Use = ÷ M1 dx dt dt Obtain given answer correctly A1 [4] (ii) State t = 0 M1 dy State that = 0 and make correct conclusion A1 [2] dx (iii) Substitute t = –2 into equation for x or y M1 Obtain (e–6, 4e–2 + 3) A1 [2]

This question in 9709/21 Oct/Nov 2011

8 The equation of a curve is 2x2 y2 6. −3x −3y + = dy 4x (i) Show that −3 [3] dx = 3 −2y. (ii) Find the coordinates of the two points on the curve at which the gradient is [6] −1.

9 marks

Mark scheme: dy 8 (i) State 2 y as derivative of y2, or equivalent B1 dx dy Equate derivative of LHS to zero and solve for M1 dx Obtain given answer correctly A1 [3] (ii) Equate gradient expression to –1 and rearrange M1 Obtain y = 2x A1 Substitute into original equation to obtain an equation in x2 (or y2) M1 Obtain 2x2 – 3x – 2 = 0 (or y2 – 3y – 4 = 0) A1 Correct method to solve their quadratic equation M1 State answers (– 1 2 , –1) and (2, 4) A1 [6]

This question in 9709/23 Oct/Nov 2011

sin 2x 2 The curve with equation y has one stationary point in the interval 0 2π. Find the exact = e2x ≤x ≤1 x-coordinate of this point. [4]

4 marks

Mark scheme: 2 Use quotient rule or product rule, correctly M1 Obtain correct derivative in any form A1 Equate derivative to zero and solve for x M1 π Obtain x = A1 [4] 8 2 2

This question in 9709/22 Oct/Nov 2012

7 The equation of a curve is 3x2 2y2 0. −4xy + −6 = dy 3x (i) Show that [4] −2y dx 2x = −2y. (ii) Find the coordinates of each of the points on the curve where the tangent is parallel to the x-axis. [5]

9 marks

Mark scheme: dy 27 (i) State 4 y as derivative of 2y , or equivalent B1 dx dy State 4 y + 4 x as derivative of 4xy, or equivalent B1 dx dy Equate derivative of LHS to zero and solve for M1 dx Obtain given answer correctly A1 [4] (ii) State or imply that the coordinates satisfy 3x – 2y = 0 B1 Obtain an equation in x2 (or y2) M1 Solve and obtain x2 = 4 (or y2 = 9) A1 State answer (2 , 3) A1 State answer (−2, −3) A1 [5]

This question in 9709/22 Oct/Nov 2012

5 The parametric equations of a curve are x e2t, y 4tet. = = dy (i) Show that 2(t + 1) . [4] dx = et (ii) Find the equation of the normal to the curve at the point where t 0. [4] =

8 marks

Mark scheme: 5 (i) Use product rule to differentiate y M1 Obtain correct derivative in any form A1 d y d y d x Use = ÷ M1 d x d t d t Obtain given answer correctly A1 [4] d y (ii) Substitute t = 0 in and both parametric equations B1 d x dy Obtain = 2 and coordinates (1, 0) B1 dx dy Form equation of the normal at their point, using negative reciprocal of their M1 dx 1 1 State correct equation of normal y = − x + or equivalent A1 [4] 2 2

This question in 9709/21 May/June 2013

5 The equation of a curve is x2 −2x2y + 3y = 9. dy 2x −4xy (i) Show that = . [4] dx 2x2 −3 (ii) Find the equation of the normal to the curve at the point where x = 2, giving your answer in the form ax + by + c = 0. [4]

8 marks

Mark scheme: d y 5 (i) State 3 as derivative of 3y, or equivalent B1 d x d y State 4xy + 2x2 as a derivative of 2x2y, or equivalent B1 d x d y Equate derivative of LHS to zero and solve for M1 d x Obtain given answer correctly A1 [4] (ii) Substitute x = 2 into given equation and solve for y M1 12 Obtain gradient = correctly A1 5 d y Form equation of the normal at their point, using negative recip of their M1 d x State correct equation of normal 5x + 12y + 2 = 0 or equivalent A1 [4]

This question in 9709/22 May/June 2013

7 (a) Find the exact area of the region bounded by the curve y = 1 + e2x−1, the x-axis and the lines x = 1 and x = 2. [4] 2 (b) y M 1 x O 2p e2x The diagram shows the curve y = for 0 < x < 120, and its minimum point M. Find the sin 2x exact x-coordinate of M. [5]

9 marks

Mark scheme: 7 (a) Obtain one term of form ke2x–1 with any non-zero k M1 1 Obtain correct integral x + e2x–1 A1 2 Substitute limits, giving exact values M1 1 Correct answer e3 + 1 A1 [4] 2 (b) Use product or quotient rule M1* Obtain correct derivative in any form A1 Equate derivative to zero and solve for x M1* dep Obtain tan 2x = 1 A1 π Obtain x = A1 [5] 8

This question in 9709/22 May/June 2013

5 The parametric equations of a curve are x e2t, y 4tet. = = dy (i) Show that 2(t + 1) . [4] dx = et (ii) Find the equation of the normal to the curve at the point where t 0. [4] =

8 marks

Mark scheme: 5 (i) Use product rule to differentiate y M1 Obtain correct derivative in any form A1 dy dy dx Use = ÷ M1 dx dt dt Obtain given answer correctly A1 [4] d y (ii) Substitute t = 0 in and both parametric equations B1 d x d y Obtain = 2 and coordinates (1, 0) B1 d x dy Form equation of the normal at their point, using negative reciprocal of their M1 dx 1 1 State correct equation of normal y = − x + or equivalent A1 [4] 2 2

This question in 9709/23 May/June 2013

3 The equation of a curve is y 12e2x 4x. Find the exact x-coordinate of each of the stationary points of the curve and determine= the−5exnature+ of each stationary point. [6]

6 marks

Mark scheme: 3 Obtain derivative e2x – 5ex + 4 B1 Equate derivative to zero and carry out recognisable solution method for a quadratic in ex M1 Obtain ex = 1 or ex = 4 A1 Obtain x = 0 and x = ln 4 A1 Use an appropriate method for determining nature of at least one stationary point M1  d 2 y 2 x x d 2 y d 2 y  = 2e − 5e , when x = ,0 = − (3), x = ln ,4 = + (12 )  dx 2 dx 2 dx 2   Conclude maximum at x = 0 and minimum at x = ln 4 (no errors seen) A1 [6] ( )

This question in 9709/21 Oct/Nov 2013

5 The parametric equations of a curve are x cos y 4 = 21 −cos 1, = sin21, for 0 ≤1 ≤0. dy 8 cos (i) Show that [4] 1 dx 1 cos = −4 1. (ii) Find the coordinates of the point on the curve at which the gradient is [4] −4.

8 marks

Mark scheme: dx dy 5 (i) State = −2 sin 2θ + sin θ or = 8 sin θ cos θ B1 d θ d θ dy d y d x Use = ÷ M1 d x d θ dθ Use sin 2θ = 2sinθ cosθ M1 Obtain given answer correctly A1 [4] (ii) Equate derivative to −4 and solve for cos θ M1 Obtain cos θ = ½ A1 Obtain x = −1 A1 Obtain y = 3 A1 [4] 2

This question in 9709/21 Oct/Nov 2013

e3x−12 The curve y has one stationary point. Find the coordinates of this stationary point. [5] 2x =

5 marks

Mark scheme: 2 Use quotient or product rule M1 Obtain correct derivative in any form A1 Equate (numerator) of derivative to zero and solve for x DM1 Obtain x = 13 A1 Obtain y = 32 A1 [5]

This question in 9709/22 Oct/Nov 2013

3 The equation of a curve is y 12e2x 4x. Find the exact x-coordinate of each of the stationary points of the curve and determine= the−5exnature+ of each stationary point. [6]

6 marks

Mark scheme: 3 Obtain derivative e2x – 5ex + 4 B1 Equate derivative to zero and carry out recognisable solution method for a quadratic in ex M1 Obtain ex = 1 or ex = 4 A1 Obtain x = 0 and x = ln 4 A1 Use an appropriate method for determining nature of at least one stationary point M1  d 2 y 2 x x d 2 y d 2 y  = 2e − 5e , when x = ,0 = − (3), x = ln ,4 = + (12 )  dx 2 dx 2 dx 2   Conclude maximum at x = 0 and minimum at x = ln 4 (no errors seen) A1 [6] ( )

This question in 9709/23 Oct/Nov 2013

5 The parametric equations of a curve are x cos y 4 = 21 −cos 1, = sin21, for 0 ≤1 ≤0. dy 8 cos (i) Show that [4] 1 dx 1 cos = −4 1. (ii) Find the coordinates of the point on the curve at which the gradient is [4] −4.

8 marks

Mark scheme: dx dy 5 (i) State = −2 sin 2θ + sin θ or = 8 sin θ cos θ B1 d θ d θ dy d y d x Use = ÷ M1 d x d θ dθ Use sin 2θ = 2sinθ cosθ M1 Obtain given answer correctly A1 [4] (ii) Equate derivative to −4 and solve for cos θ M1 Obtain cos θ = ½ A1 Obtain x = −1 A1 Obtain y = 3 A1 [4] 2

This question in 9709/23 Oct/Nov 2013

8 y M x O The diagram shows the curve y tan x cos2x, for 0 1 = ≤x < 20, and its maximum point M. dy (i) Show that 4 cos2x [5] dx = −sec2x −2. (ii) Hence find the x-coordinate of M, giving your answer correct to 2 decimal places. [4]

9 marks

Mark scheme: 8 (i) Differentiate using product rule M1 Obtain sec2 x cos 2 x − 2 tan x sin 2 x A1 Use cos 2 x = 2 cos 2 x − 1 or sin 2 x = 2 sin x cos x or both B1 Express derivative in terms of sec x and cos x only M1 Obtain 4 cos 2 x − sec 2 x − 2 with no errors seen (AG) A1 [5] (ii) State 4 cos 4 x − 2 cos 2 x − 1 = 0 B1 Apply quadratic formula to a 3 term quadratic equation in terms of cos 2 x to find the least positive value of cos 2 x M1 1 + 5 Obtain or imply cos2 x = or 0.809… A1 4 Obtain 0.45 A1 [4]

This question in 9709/22 May/June 2014

8 y M x O The diagram shows the curve y tan x cos2x, for 0 1 = ≤x < 20, and its maximum point M. dy (i) Show that 4 cos2x [5] dx = −sec2x −2. (ii) Hence find the x-coordinate of M, giving your answer correct to 2 decimal places. [4]

9 marks

Mark scheme: 8 (i) Differentiate using product rule M1 Obtain sec2 x cos 2 x − 2 tan x sin 2 x A1 Use cos 2 x = 2 cos 2 x − 1 or sin 2 x = 2 sin x cos x or both B1 Express derivative in terms of sec x and cos x only M1 Obtain 4 cos 2 x − sec 2 x − 2 with no errors seen (AG) A1 [5] (ii) State 4 cos 4 x − 2 cos 2 x − 1 = 0 B1 Apply quadratic formula to a 3 term quadratic equation in terms of cos 2 x to find the least positive value of cos 2 x M1 1 + 5 Obtain or imply cos2 x = or 0.809… A1 4 Obtain 0.45 A1 [4]

This question in 9709/23 May/June 2014

7 The equation of a curve is y3 4xy 16. + = dy 4y (i) Show that [4] dx = − 3y2 4x. + (ii) Show that the curve has no stationary points. [2] (iii) Find the coordinates of the point on the curve where the tangent is parallel to the y-axis. [4]

10 marks

Mark scheme: 2 dy 37 (i) Obtain 3 y as derivative of y B1 dx dy Obtain 4 y + 4 x as derivative of 4xy B1 dx dy Equate derivative of left-hand side to zero and solve for , must be from dx implicit differentiation M1 dy 4 y Confirm given answer = − correctly A1 [4] dx 3 y 2 + 4 x (ii) State or imply y = 0 B1 Substitute in equation of curve and show contradiction B1 [2] (iii) State or imply 3 y 2 + 4 x = 0 B1 Eliminate one variable from equation of curve using 3 y 2 + 4 x = 0 M1 Obtain y = − 2 A1 Obtain x = − 3 A1 [4]

This question in 9709/21 May/June 2015

5 Find the x-coordinates of the stationary points of the following curves: [3] (i) y = 4xe−3x; 4x2 (ii) y x 1. [5] = +

8 marks

Mark scheme: 5 (i) Use product rule to obtain form k1e −3 x + k 2 xe −3 x M1 Obtain correct 4 e − 3 x − 12 x e − 3 x A1 Obtain x = 13 or 0.333 or better and no other A1 [3] (ii) Use quotient rule or equivalent M1* Obtain correct numerator 8 x ( x + )1 − 4 x 2 or equivalent A1 Equate numerator to zero and solve to find at least one value M1 dep Obtain x = − 2 A1 Obtain x = 0 A1 [5] dx

This question in 9709/22 Oct/Nov 2015

sin 2x 7 The equation of a curve is y cosx 1. = + dy 2 cos2x cosx (i) Show that . [7] + −1 dx cosx 1 = + (ii) Find the x-coordinate of each stationary point of the curve in the interval x Give each answer correct to 3 significant figures. −0 < < 0. [3]

10 marks

Mark scheme: 7 (i) Use quotient rule or equivalent to find first derivative M1 2 cos 2 x (cos x + )1 + sin 2 x sin x Obtain or equivalent A1 (cos x + 2)1 Use at least one of cos 2 x = 2 cos 2 x − 1 and 2 x = 2 sin x cos x B1 Express first derivative in terms of cos x only M1 2 cos 3 x + 4 cos 2 x − 2 Obtain 2 or equivalent A1 (cos x + )1 Factorise numerator or divide numerator by (cos x + )1 or equivalent M1 2 (cos 2 x + cos x − )1 Confirm given answer correctly A1 [7] cos x + 1 (ii) Use quadratic formula or equivalent to find value of cos x M1 Obtain x-coordinate 0.905 A1 Obtain x-coordinate –0.905 and no others in range A1 [3]

This question in 9709/23 Oct/Nov 2015

7 The equation of a curve is 2x3 y3 24. + = dy (i) Express in terms of x and y, and show that the gradient of the curve is never positive. [4] dx (ii) Find the coordinates of the two points on the curve at which the gradient is [5] −2.

9 marks

Mark scheme: dy 27 (i) State 3 y as derivative of 3y B1 dx d y Equate derivative of left-hand side to zero and solve for M1 d x d y 6 x 2 Obtain = − 2 or equivalent A1 d x 3 y Observe x 2 and y 2 never negative and conclude appropriately A1 [4] (ii) Equate first derivative to − 2 and rearrange to y 2 = x 2 or equivalent B1 Substitute in original equation to obtain at least one equation in 3x or 3y M1 Obtain 3 x 3 = 24 or x 3 = 24 or 3 y 3 = 24 or − y 3 = 24 A1 Obtain (2, 2) A1 Obtain ( 3 24, − 3 24) or (2.88, − 2.88) and no others A1 [5] cos x

This question in 9709/22 Feb/March 2016

7 y P Q x O M The diagram shows the curve with parametric equations x 2 y 1 3 cos 2t, = −cost, = + for 0 t The minimum point is M and the curve crosses the x-axis at points P and Q. < < 0. dy (i) Show that cos t. [4] dx = −12 (ii) Find the coordinates of M. [2] (iii) Find the gradient of the curve at P and at Q. [4]

10 marks

Mark scheme: dx dy 7 (i) State dt = sin t and dt = −6sin2t B1 Use sin2t = 2sin t cos t B1 d y Form expression for d x in terms of t M1 Confirm −12cost A1 [4] (ii) Identify 12π as value of t B1 Obtain (2, − 2) B1 [2] (iii) Identify cos2t = − 13 B1 Attempt to find value of t (or of cost ) for at least one of the two points M1 Obtain 0.955 (or 1 ) or 2.186 (or − 1 ) A1 3 3 Obtain − 12 or − 4 3 or −6.93 and 12 or 4 3 or 6.93 A1 [4] 3 3

This question in 9709/22 May/June 2016

1 5 The equation of a curve is y 6xe 3x. At the point on the curve with x-coordinate p, the gradient of the curve is 40. = @ A 20 (i) Show that p 3 ln . [4] p 3 = + (ii) Show by calculation that 3.3 p 3.5. [2] < < (iii) Use an iterative formula based on the equation in part (i) to find the value of p correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3]

9 marks

Mark scheme: 1 3 x 13 x5 (i) Use product rule to obtain form k1e + k 2 xe *M1 1 3 x 13 x Obtain correct 6e + 2 xe A1 Equate first derivative to 40 and obtain equation without e present, dep *M DM1 Confirm p = 3ln p20+ 3 or x = 3ln x20+ 3 A1 [4] (ii) Consider sign of p − 3ln p20+ 3 at 3.3 and 3.5 or equivalent M1 Complete argument correctly with appropriate calculations A1 [2] (iii) Carry out iterative process correctly at least once M1 Obtain final answer 3.412 A1 Show sufficient iterations to justify accuracy to 3 dp or show sign change in interval (3.4115, 3.4125) B1 [3] 2

This question in 9709/23 May/June 2016

6 A curve has parametric equations x ln t 1 , y t2 ln t. = + = dy (i) Find an expression for in terms of t. [5] dx (ii) Find the exact value of t at the stationary point. [2] (iii) Find the gradient of the curve at the point where it crosses the x-axis. [2]

9 marks

Mark scheme: 6 (i) State ddxt = t+11 B1 Use product rule for derivative of y M1 Obtain 2t ln t + t or equivalent A1 Use ddyx = ddyt ÷ ddxt M1 Obtain (t + 1)(2t ln t + t ) A1 [5] (ii) Solve 2ln t + 1 = 0 M1 − 12 Obtain t = e A1 [2] (iii) Identify t = 1 only B1 Obtain 2 B1 [2] 3 4

This question in 9709/23 Oct/Nov 2016

8 y P x O M The diagram shows the curve with equation y 3x2 ln 16x . = The curve crosses the x-axis at the point P and has a minimum point M. (i) Find the gradient of the curve at the point P. [5] … … … … … … … … … … … … … … … (ii) Find the exact coordinates of the point M. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 8(i) Apply product rule to find first derivative *M1 Obtain ( ) 1 6 6 1n 3 x x x + or equivalent A1 Allow unsimplified for A1 Identify 6 x = at P B1 Substitute their value of x at P into attempt at first derivative DM1 dep *M Obtain 18 A1 Total: 5 Question Answer Marks Guidance 8(ii) Equate their first derivative to zero and attempt solution of equation of form ( ) 1 6 1n 0 k x m + = *M1 Obtain x–coordinate of form 2 1 a a e DM1 dep *M Obtain 1 2 6e x − = or exact equivalent A1 Substitute exact x–value in the form 2 1 a a e and attempt simplification to remove ln M1 Obtain 1 54e− − or exact equivalent A1 Total: 5

This question in 9709/21 May/June 2017

8 y 1, 4 3, 3 x O The diagram shows the curve with parametric equations x 2 2t, y 2 sin3t 3 cos3t 1 = −cos = + + for 0 The end-points of the curve are 1, 4 and 3, 3 . ≤t ≤120. dy 3 (i) Show that sin t cos t. [5] dx 2 4 = −9 … … … … … … … … … … … … … … … (ii) Find the coordinates of the minimum point, giving each coordinate correct to 3 significant figures. [3] … … … … … … … … … … … … (iii) Find the exact gradient of the normal to the curve at the point for which x 2. [3] = … … … … … … … … … … …

11 marks

Mark scheme: 8(i) Obtain d d 2sin2 x t t = B1 Obtain d 2 2 d 6sin cos 9cos sin y t t t t t = − B1 Use d d d / d d d y y x x t t = for their first derivatives M1 Use identity sin 2 2sin cos t t t = B1 Simplify to obtain 3 9 2 4 sin cos t t − with necessary detail present A1 Total: 5 Question Answer Marks Guidance 8(ii) Equate d d y x to zero and obtain tant k = M1 Obtain 3 2 tant = or equivalent A1 Substitute value of t to obtain coordinates ( ) 2.38, 2.66 A1 Total: 3 8(iii) Identify 1 4 t π = B1 Substitute to obtain exact value for gradient of the normal M1 Obtain gradient 4 3 2 , 8 3 2 or similarly simplified exact equivalent A1 Total: 3

This question in 9709/22 May/June 2017

6 The parametric equations of a curve are x 2e2t 4et, y 5te2t. = + = dy (i) Find in terms of t and hence find the coordinates of the stationary point, giving each coordinate dx correct to 2 decimal places. [6] … … … … … … … … … … … … … … … … … … … … … … (ii) Find the gradient of the normal to the curve at the point where the curve crosses the x-axis. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(i) Obtain ddxt = 4e 2 t + 4e t B1 Use product rule to find ddyt M1 dy 5e 2 t + 10te 2 t A1 Obtain = or equivalent dx 4e 2 t + 4e t ae 2 t + bte 2 t M1 Equate first derivative of the form ce 2 t + de t to zero and solve to find t Obtain t = − 12 from completely correct work A1 Obtain (3.16, − 0.92) A1 6 6(ii) Identify t = 0 B1 Substitute t = 0 in expression for first derivative M1 and find negative reciprocal Obtain − 85 or equivalent A1 3

This question in 9709/21 Oct/Nov 2017

3x 3 Find the exact coordinates of the stationary point of the curve with equation y [5] ln x. = … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 Use quotient rule to find first derivative or equivalent *M1 Obtain 1 2 3ln 3 d d (ln ) x x x y x x − × = or equivalent A1 Condone lack of brackets in denominator unless specifically simplified to 2ln x Equate first derivative to zero and attempt value of x from ln x k = oe DM1 Must get as far as x = Obtain e x = A1 Allow 1e Obtain 3e y = A1 Allow 3 1e SC1: If 1 3ln 3 0 x x x − × = seen with no reference to d d y x , then allow M1 A1 then following marks SC2: If denominator incorrect and numerator correct/reversed/added then max marks M0A0M1A1A1 SC3: If numerator reversed then max marks M1A0M1A1A1 5

This question in 9709/22 May/June 2019

3x 3 Find the exact coordinates of the stationary point of the curve with equation y [5] ln x. = … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 Use quotient rule to find first derivative or equivalent *M1 Obtain 1 2 3ln 3 d d (ln ) x x x y x x − × = or equivalent A1 Condone lack of brackets in denominator unless specifically simplified to 2ln x Equate first derivative to zero and attempt value of x from ln x k = oe DM1 Must get as far as x = Obtain e x = A1 Allow 1e Obtain 3e y = A1 Allow 3 1e SC1: If 1 3ln 3 0 x x x − × = seen with no reference to d d y x , then allow M1 A1 then following marks SC2: If denominator incorrect and numerator correct/reversed/added then max marks M0A0M1A1A1 SC3: If numerator reversed then max marks M1A0M1A1A1 5

This question in 9709/23 May/June 2019

4 y x O 14 x The diagram shows the curve with equation y The shaded region is bounded by the curve −2 = x2 8. and the lines x 14 and y 0. + = = dy (a) Find and hence determine the exact x-coordinates of the stationary points. [4] dx … … … … … … … … … … … … … … … … … (b) Use the trapezium rule with three intervals to find an approximation to the area of the shaded region. Give the answer correct to 2 significant figures. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(a) Differentiate using quotient rule (or product rule) *M1 Obtain 2 2 2 ( 8) 2 ( 2) ( 8) x x x x + − − + A1 OE Equate first derivative to zero and attempt solution to get x = … DM1 Obtain 2 12 ± or exact equivalents A1 4 4(b) Use y values 4 8 12 (0), , , 44 108 204 or decimal equivalents B1 Decimal equivalents need to be to at least 2 decimal places Use correct formula, or equivalent, with 4 h = M1 Obtain 4 8 12 2 0 2 2 44 108 204   + × + × +     or equivalent and hence 0.78 A1 3

This question in 9709/22 Oct/Nov 2020

5 y x O B A The diagram shows the curve with parametric equations 2t x ln 2t 3 , y −3 = + = 2t 3. + The curve crosses the y-axis at the point A and the x-axis at the point B. dy 6 (a) Show that [4] dx = 2t 3. + … … … … … … … … … … … … … … (b) Find the gradient of the curve at A. [2] … … … … … … … … … … … … (c) Find the gradient of the curve at B. [2] … … … … … … … … … … … …

8 marks

Mark scheme: 5(a) Obtain d 2 d 2 3 = + x t t Use quotient rule, or equivalent, to find d d y t M1 Obtain 2 d 2(2 3) 2(2 3) d (2 3) + − − = + y t t t t A1 OE Divide to confirm d 6 d 2 3 = + y x t A1 AG 4 5(b) Attempt to find value of t corresponding to 0 = x M1 Obtain 1 = − t and hence gradient is 6 A1 2 5(c) Attempt to find value of t corresponding to 0 = y M1 Obtain 3 2 = t and hence gradient is 1 A1 2

This question in 9709/21 Oct/Nov 2021

5 y x O B A The diagram shows the curve with parametric equations 2t x ln 2t 3 , y −3 = + = 2t 3. + The curve crosses the y-axis at the point A and the x-axis at the point B. dy 6 (a) Show that [4] dx = 2t 3. + … … … … … … … … … … … … … … (b) Find the gradient of the curve at A. [2] … … … … … … … … … … … … (c) Find the gradient of the curve at B. [2] … … … … … … … … … … … …

8 marks

Mark scheme: 5(a) Obtain d 2 d 2 3 = + x t t Use quotient rule, or equivalent, to find d d y t M1 Obtain 2 d 2(2 3) 2(2 3) d (2 3) + − − = + y t t t t A1 OE Divide to confirm d 6 d 2 3 = + y x t A1 AG 4 5(b) Attempt to find value of t corresponding to 0 = x M1 Obtain 1 = − t and hence gradient is 6 A1 2 5(c) Attempt to find value of t corresponding to 0 = y M1 Obtain 3 2 = t and hence gradient is 1 A1 2

This question in 9709/23 Oct/Nov 2021

7 y B A x O 3 2 ln x The diagram shows the curve with equation y 3x 1. The curve crosses the x-axis at the point A = and has a maximum point B. The shaded region is bounded+ by the curve and the lines x 3 and y 0. = = (a) Find the gradient of the curve at A. [3] … … … … … … … … … … … … … … … … … (b) Show by calculation that the x-coordinate of B lies between 3.0 and 3.1. [3] … … … … … … … … … … … … (c) Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 decimal places. [3] … … … … … … … … … … …

9 marks

Mark scheme: 7(a) Differentiate using quotient rule (or product rule) M1 2 A1 OE (3 x + 1) − 6ln x x Obtain (3 x + 1) 2 Substitute x = 1 to obtain 1 A1 OE 2 3 7(b) Equate numerator of first derivative to zero M1 May be implied. 2 M1 OE Consider sign of (3 x + 1) − 6ln x for 3.0 and 3.1 x Obtain 0.074… and –0.14… or equivalents and justify conclusion A1 AG – necessary detail needed. 0.00075 and – 0.001275 . 3 7(c) Use y-values [0], 2 ln2 or 0.1980 and 2 ln3 or 0.2197 B1 7 10 Use correct formula, or equivalent, with h = 1 M1 Obtain 0.31 A1 3

This question in 9709/21 Oct/Nov 2022

5 y M x O The diagram shows the curve with parametric equations x 4e2t, y cos 2t, = = 5e−t for The curve has a maximum point M. −14π ≤t ≤14π. dy (a) Find an expression for in terms of t. [3] dx … … … … … … … … … … … … … … … … … (b) Find the coordinates of M, giving each coordinate correct to 3 significant figures. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(a) Use product rule to find d d y t Obtain d 5e cos2 10e sin2 d     t t y t t t A1 or (unsimplified) equivalent (do not condone poor use of brackets. Obtain 2 d 5e cos2 10e sin 2 d 8e      t t t y t t x A1 OE following their expression for d d y t . 3 5(b) Equate d d y x to zero and simplify at least as far as tan 2 ...  t M1* now condoning any error with d d x t . Obtain 1 2 tan 2  t A1 Obtain 0.231...  t A1 allow 0.232 t  . Substitute negative value of t in expressions for x and y DM1 Obtain 2.52  x and 5.64  y A1 or greater accuracy. 5

This question in 9709/21 May/June 2023

2 A curve has equation y 3 tan 2x1 cos 2x. = Find the gradient of the curve at the point for which x 1 [5] = 3π. … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 1 1 2 1 B1 State or imply derivative of tan x is sec x or derivative of 2 2 2 cos2x is −2sin2x Attempt use of product rule to find first derivative *M1 3 2 1 1 A1 or (unsimplified) equivalent. Obtain correct sec x cos2 x − 6tan x sin2 x 2 2 2 1 DM1 Substitute π into attempt at first derivative and attempt evaluation to 3 find the gradient Obtain −4 A1 5

This question in 9709/21 Oct/Nov 2023

3 y A B x O 2 The diagram shows the curve with 2x. The points on the curve with equation y x-coordinates 0 and 2 are denoted and =The6e−1shaded region is enclosed by the curve, the line by A B respectively. through A parallel to the x-axis and the line through B parallel to the y-axis. (a) Find the exact gradient of the curve at B. [2] … … … … … … … … … … … … … … … (b) Find the exact area of the shaded region. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3(a) − 12 x M1 For any non-zero k except 6. Differentiate to obtain form ke 1 3 A1 Substitute x = 2 to obtain − 3e− or − e 2 3(b) 2 x B1 OE Integrate to obtain −12e − 1 x M1 For any non-zero k except 6 1 Use limits 0 and 2 correctly to an integral of the form ke −, 2 retaining − or equivalent perhaps involving integration of 6 − 6e 2 x . exactness 1 12 A1 Subtract from 12 to obtain final answer 12e− or e 3

This question in 9709/22 Oct/Nov 2023

6 y B A x O The diagram shows the curve with parametric equations x 3 ln 2t , y 4t ln t. = −3 = The curve crosses the y-axis at the point A. At the point B, the gradient of the curve is 12. (a) Find the exact gradient of the curve at A. [5] … … … … … … … … … … … … … (b) Show that the value of the parameter t at B satisfies the equation 9 3 t [2] 1 ln t 2. = + + … … … … … … … … … … … … … (c) Use an iterative formula, based on the equation in (b), to find the value of t at B, giving your answer correct to 3 significant figures. Use an initial value of 5 and give the result of each iteration to 5 significant figures. [3] … … … … … … … …

10 marks

Mark scheme: 6(a) dx 6 B1 Obtain = dt 2t − 3 d y M1 Allow unsimplified. Use product rule to find d t dy (4ln t + 4)(2t − 3) A1 OE Obtain = dx 6 Attempt to find t corresponding to point A using a complete and correct M1 method Obtain t = 2 and hence gradient is 23 ln 2 + 23 A1 Or exact equivalent. 5 6(b) d y k M1 Equate to 12 and attempt rearrangement to 2t −=3 d x 4ln t + 4 9 3 A1 AG Confirm given result t = + with sufficient detail 1 + ln t 2 2 6(c) Use iteration process correctly at least once M1 Need to see 4.9626 . Obtain final answer 4.96 A1 Answer required to exactly 3 s.f. Show sufficient iterations to 5 sf to justify answer or show sign change in A1 interval [4.955, 4.965] 3

This question in 9709/22 Oct/Nov 2023

7 The curve with equation e2x 18x y3 y 11 has a stationary point at p, q . (a) Find the exact value of p. [4] … … … … … … … … … … … … … … … … … … … … … … … … (b) Show that q 3 2 18 ln 3 [2] = + −q. … … … … … … … … (c) Show by calculation that the value of q lies between 2.5 and 3.0. [2] … … … … … (d) Use an iterative formula, based on the equation in (b), to find the value of q correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3] … … … … … … … … … Additional Page If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown. … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(a) 3 2 d y B1 Differentiate y to obtain 3 y d x Differentiate complete equation to produce at least one term involving M1 d y using implicit differentiation. d x 2 x 2 dy dy A1 Obtain 2e − 18 + 3 y + = 0 dx dx dy 1 A1 Substitute = 0 to obtain either p = 2 ln9 or p = ln3 dx 4 7(b) Substitute value of p in original equation and rearrange as far as y 3 = ... M1 Allow in terms of ln9 . or q3 = … Obtain given result q = 3 2 + 18ln3 − q or y = 3 2 + 18ln3 − y with A1 AG sufficient detail 2 7(c) Consider sign of q − 3 2 + 18ln3 − q or equivalent for 2.5 and 3.0 M1 Obtain −0.18... and 0.34... with sufficient detail and justify A1 OE conclusion 2 7(d) Use iteration process correctly at least once M1 Obtain final answer q = 2.673 A1 Answer required to exactly 4 s.f. Show sufficient iterations to 6 sf to justify answer or show sign change A1 in the interval [2.6725, 2.6735] 3

This question in 9709/23 Oct/Nov 2023

5 A curve has equation y = . The curve has exactly one stationary point P. 1 + 3x dy 1 1 -2 x (a) Find and hence show that the x-coordinate of P satisfies the equation x = 6 + 2 e . [4] dx … … … … … … … … … … … … … … … … … … … … … … … … … (b) Show by calculation that the x-coordinate of P lies between 0.35 and 0.45 . [2] … … … … … … … … … … … … … (c) Use an iterative formula based on the equation in part (a) to find the x-coordinate of P correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … … … …

9 marks

Mark scheme: 5(a) Differentiate using quotient rule *M1 OE Obtain 2 2 2 (1 3 ) 2e (1 e ) 3 (1 3 ) x x x x     A1 OE Equate first derivative to zero and arrange as far as ... x  DM1 Confirm 2 1 1 6 2 e x x    A1 Answer given – necessary detail needed. Must have exact terms. 4 5(b) Consider sign of 2 1 1 6 2 e x x    M1 OE Obtain 0.06...  (– 0.064959...) and 0.08... (0.0800…) or equivalents and justify conclusion A1 Answer given – necessary detail needed. Alternative Method 1 for Question 5(b) Consider  2 1 1 6 2 f e x x    and obtain   f 0.35 0.42 (0.4149…) and   f 0.45 0.37 (0.36995….) (M1) Conclude   f 0.35 0.45  and   f 0.45 0.35  so root lies in given interval. (A1) Alternative Method 2 for Question 5(b) Consider the sign of their d d y x from part (a) (M1) Obtain 0.19...  (– 0.187...) and 0.21... (0.2139…) or equivalents and justify conclusion (A1) 2 Question Answer Marks Guidance 5(c) Use iterative process correctly at least once M1 Obtain final answer 0.394 A1 Answer required to exactly 3sf. Show sufficient iterations to 5 sf to justify answer or show sign change in interval [0.3935, 0.3945] A1 3

This question in 9709/21 May/June 2024

4 A curve is defined by the parametric equations x = 4 cos 2 t , y = 3 sin 2t , for values of t such that 0 1 t 1 12 r. r . Give your answer in the Find the equation of the normal to the curve at the point for which t = 16 form ax + by + c = 0 where a, b and c are integers. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4 Obtain forms 1 d cos sin d  x k t t t or 2 d cos2 d  y k t t *M1 Obtain correct 8cos sin  t t or 4sin2  t and 2 3cos2t A1 Attempt value of d d y x when 1 6 π  t *DM1 Need to see attempt at substitution. Obtain d 1 d 2 y x  A1 State or imply gradient of normal is 2 **M1FT Following their value of the first derivative. Attempt equation of normal **DM1 Not tangent and with attempt to find coordinates 3 3, 2       . Obtain 4 2 9 0    x y A1 Or equivalent of requested form. 7

This question in 9709/22 May/June 2024

3 y A B x O The diagram shows the curve with equation y = 8e -x - e 2 x . The curve crosses the y-axis at the point A and the x-axis at the point B. The shaded region is bounded by the curve and the two axes. (a) Find the gradient of the curve at A. [3] … … … … … … … … … … … … … … … … (b) Show that the x-coordinate of B is ln2 and hence find the area of the shaded region. [5] … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 3(a) Differentiate to obtain form 2 1 2 e e  x x k k M1 Where 1 2 0 k k , 1 8 k and 2 1 k . Obtain 2 8e 2e    x x A1 Substitute 0  x to obtain –10 A1 3 3(b) Attempt to find x-coordinate of B M1 2 8e e 0   x x . Obtain 3e 8  x and hence ln2  x A1 AG so necessary detail needed. A0 if decimals used. Integrate to obtain 2 1 2 8e e    x x B1 Use limits 0 and ln2 correctly to find area M1 For integral of form 2 3 4 e e  x x k k where 3 4 0  k k . 1 8  k and 2 1  k . Obtain 5 2 A1 OE 5

This question in 9709/23 May/June 2024

6 A curve has parametric equations e 2 t - 2 3 t x = , y = e + 1. e 2 t + 1 d y (a) Find an expression for in terms of t. [4] d x … … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the exact gradient of the curve at the point where the curve crosses the y-axis. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 6(a) Differentiate x using quotient rule or correct equivalent *M1 (e 2 t + 1)2e 2 t − (e 2 t − 2)2e 2 t A1 Obtain or equivalent (e 2 t + 1) 2 d y DM1 Attempt expression for in terms of t d x Obtain 1 2 et (e 2 t + 1) 2 or (unsimplified) equivalent A1 No fractions within fractions. Attempt to simplify (e 2 t + 1)2e 2 t − (e 2 t − 2)2e 2 t must be seen. 4 6(b) Identify t = 12 ln 2 at point where curve crosses y-axis B1 d y M1 Substitute non-zero value of t in their expression for and attempt simplification d x Obtain 9 2 2 or exact equivalent A1 3

This question in 9709/21 Oct/Nov 2024

3 A curve has equation 6e -x y 2 + e 2 x - 12y + 7 = 0 . Find the gradient of the curve at the point ( ln3, 2) . [6] … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 3 Use product rule to differentiate 6e−x y 2 M1 Allow correct use of quotient rule. d y Must have from implicit differentiation. d x − x 2 − x dy A1 SOI Obtain −6e y + 12e y dx 2 x dy B1 May see separately. Obtain +2e − 12 dy dx 2 x +2e − 12 + 7 scores B0. dx dy M1 (Correct equation: Rearrange correctly to obtain = ... dy 2 x dy dx − x 2 − x −6e y + 12e y + 2e − 12 = 0) dx dx d y Must be at least one from implicit differentiation d x present. May already have used substitution dy 6 y 2 e − x − 2e 2 x = . dx 12e − x y − 12 d y M0 for inclusion of an incorrect . d x Substitute for x and y to obtain an exact final answer M1 d y Must be at least one from implicit differentiation d x present. Obtain 52 or 2.5 A1 6

This question in 9709/22 Oct/Nov 2024

6 A curve has parametric equations e 2 t - 2 3 t x = , y = e + 1. e 2 t + 1 d y (a) Find an expression for in terms of t. [4] d x … … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the exact gradient of the curve at the point where the curve crosses the y-axis. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 6(a) Differentiate x using quotient rule or correct equivalent *M1 (e 2 t + 1)2e 2 t − (e 2 t − 2)2e 2 t A1 Obtain or equivalent (e 2 t + 1) 2 d y DM1 Attempt expression for in terms of t d x Obtain 1 2 et (e 2 t + 1) 2 or (unsimplified) equivalent A1 No fractions within fractions. Attempt to simplify (e 2 t + 1)2e 2 t − (e 2 t − 2)2e 2 t must be seen. 4 6(b) Identify t = 12 ln 2 at point where curve crosses y-axis B1 d y M1 Substitute non-zero value of t in their expression for and attempt simplification d x Obtain 9 2 2 or exact equivalent A1 3

This question in 9709/23 Oct/Nov 2024

4 sin x 4 A curve has equation y = for values of x such that 0 G x G 2 r . 3 + cos 2x dy (a) Find . [2] dx … … … … … (b) Hence find the coordinates of the stationary points of the curve. [4] … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 4(a) Attempt use of quotient rule or equivalent *M1 4cos x (3 + cos2 x ) + 8sin x sin2 x A1 Obtain (3 + cos2 x ) 2 2 4(b) Equate first derivative to zero and attempt to express numerator in DM1 terms of cos x Obtain 24cos x − 8cos 3 x = 0 A1 OE Attempt solution to find coordinates of at least one stationary point M1 From equation involving two powers of cos .x Obtain 2(1 π, 2) and 2(3 π, − 2) and no others A1 4

This question in 9709/22 Feb/March 2025

8 A curve has equation 3 e 2 x y + 4 e 3 x + y 3 = 18 . dy - 2e 2 x y - 4e 3 x (a) Show that = . [4] dx e 2 x + y 2 … … … … … … … … … … … … … … … … … … … … … … … … … (b) Show that the curve has no stationary points. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 8(a) Attempt use of product rule for differentiating 3e 2 x y M1 2 x 2 x dy A1 Obtain 6e y + 3e dx 3 x 2 dy B1 Obtain remaining terms 12e + 3 y dx Equate sum of correct terms to zero and rearrange to confirm A1 AG given result Necessary detail needed. 4 8(b) Equate derivative to zero and obtain y = −2e x or equivalent B1 Substitute for y in original equation and attempt appropriate M1 simplification Obtain −10e3 x = 18 or equivalent and conclude no possible value A1 AG Necessary detail needed. of x 3

This question in 9709/22 Feb/March 2025

2 dy1 Given that y = 6x cos `x + 1j, find an expression for . [2] dx … … … … … … … … … … … … … … … … … … … … … … … … … … …

2 marks

Mark scheme: Question Answer Marks Guidance 1 Attempt use of product rule M1 Allow for 6cos x 2 + 1 − 12 x sin x 2 + 1 ( ) ( ) 6cos x 2 + 1 − 12sin x 2 + 1 ( ) ( ) 6cos x 2 + 1 − 6 x sin x 2 + 1 ( ) ( ) 6cos x 2 + 1 − 6 x 2 sin x 2 + 1 ( ) ( ) Obtain 6cos( x 2 + 1) − 12 x 2 sin( x 2 + 1) A1 Allow unsimplified. 2

This question in 9709/21 May/June 2025

7 A curve has equation 5x 2 y + 4e 2 y - 7x + 10 = 0 . dy (a) Find an expression in terms of x and y for and hence find the gradient of the curve at the point dx for which y = 0 . [6] … … … … … … … … … … … … … … … (b) Show that there is no point on the curve at which the tangent is parallel to the y-axis. [2] … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 7(a) Attempt use of product rule for differentiating 5x 2 y *M1 Must have evidence of implicit differentiation. 2 dy A1 Obtain 10 xy + 5 x dx 2 y 2 y dy B1 Differentiate 4e and obtain 8e dx 2 dy 2 y dy dy 7 − 10 xy A1 OE Obtain 10 xy + 5 x + 8e − 7 = 0 and hence = dx dx dx 5 x 2 + 8e 2 y Attempt to find relevant value of x and substitute that and y = 0 to find gradient DM1 x = 2 Obtain x = 2 and gradient 14 A1 OE 6 7(b) Attempt to show that denominator of derivative cannot be zero M1 d y Must not be using when y = 0. d x State that both terms are non-negative or equivalent and conclude appropriately A1 FT Following their denominator of form k1 x 2 + k 2e 2 y , provided that k1k 2  0. 2

This question in 9709/22 Oct/Nov 2025

8 The equation of a curve is y = 4e 1 - 2x 3x - 1 . Find the exact coordinates of the stationary point of the curve. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 28 2 *M1 Where k1k 2  0. Differentiate to obtain the form k1e1− 2 x (3x − 1) 1 + k2 e1− 2 x (3x − 1)− 1 1− 2 x 12 A1 OE. Allow unsimplified Obtain correct first term −8e (3x − 1) 1− 2 x 12 A1 OE. Allow unsimplified Obtain correct second term + 6e (3x − 1)− Equate first derivative to zero and attempt solution as far as 3x −=1 k DM1 OE. Must be in the form 1− 2 x 12 1− 2 x − 12 k1e (3x − 1) + k2 e (3x − 1) . Obtain 3 x −=1 34 OE, and hence x = 127 A1 Or exact equivalent. − 16 A1 Or exact equivalent. Obtain y = 2 3e 6

This question in 9709/23 Oct/Nov 2025

8 The equation of a curve is y = 4e 1 - 2 x 3x - 1 . Find the exact coordinates of the stationary point of the curve. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 28 2 *M1 Where k1k 2  0. Differentiate to obtain the form k1e1− 2 x (3x − 1) 1 + k2 e1− 2 x (3x − 1)− 1 1− 2 x 12 A1 OE. Allow unsimplified Obtain correct first term −8e (3x − 1) 1− 2 x 12 A1 OE. Allow unsimplified Obtain correct second term + 6e (3x − 1)− Equate first derivative to zero and attempt solution as far as 3x −=1 k DM1 OE. Must be in the form 1− 2 x 12 1− 2 x − 12 k1e (3x − 1) + k2 e (3x − 1) . Obtain 3 x −=1 34 OE, and hence x = 127 A1 Or exact equivalent. − 16 A1 Or exact equivalent. Obtain y = 2 3e 6

This question in 9709/25 Oct/Nov 2025