Differentiation — Paper 3 · A Level Mathematics 9709

7 The diagram shows the curve y = x2e−12x. (i) Find the x-coordinate of M, the maximum point of the curve. [4] (ii) Find the area of the shaded region enclosed by the curve, the x-axis and the line x = 1, giving your answer in terms of e. [5]

9 marks

Mark scheme: 7 (i) Use product or quotient rule M1* 1 1 − 2 x 1 2 − 2 x Obtain first derivative 2 xe − x e or equivalent A1 2 Equate derivative to zero and solve for non-zero x M1(dep*) Obtain answer x = 4 A1 4 1 1 − x − x (ii) Integrate by parts once, obtaining kx 2 e 2 + l ∫ x e 2 d x , where kl ≠ 0 M1 1 1 − x − x Obtain integral − 2 x 2 e 2 + 4 ∫ xe 2 d x , or any unsimplified equivalent A1 1 − x 2 or equivalent A1 Complete the integration, obtaining − 2(x 2 + 4 x + 8 )e Having integrated by parts twice, use limits x = 0 and x = 1 in the complete integral M1 1 − Obtain simplified answer 16 − 26 e 2 or equivalent A1 5 A Bx + C

This question in 9709/31 Oct/Nov 2004

4 The curve with equation y = e−x sin x has one stationary point for which 0 ≤x ≤π. (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 product or quotient rule M1 Obtain derivative in any correct form A1 Equate derivative to zero and solve for x M1 Obtain answer x = 1 π or 0.785 with no errors seen A1 [4] 4 (ii) Use an appropriate method for determining the nature of a stationary point M1 Show the point is a maximum point with no errors seen A1 [2] [SR: for the answer 45° deduct final A1 in part (i), and deduct A1 in part (ii) if this value in degrees is used in the exponential.]

This question in 9709/31 Oct/Nov 2007

6 The equation of a curve is xy(x + y) = 2a3, where a is a non-zero constant. Show that there is only one point on the curve at which the tangent is parallel to the x-axis, and find the coordinates of this point. [8]

8 marks

Mark scheme: dy 26 EITHER State x + 2 xy , or equivalent, as derivative of x2y B1 dx dy 2 State y + 2 xy , or equivalent, as derivative of xy2 B1 dx dy OR State xy 1( + ) , or equivalent, as a term in an attempt to apply the product dx rule B1 dy State ( y + x )( x + y ) , or equivalent, in an attempt to apply the product rule dx B1 dy Equate attempted derivative of LHS to zero and set equal to zero dx M1 Obtain a horizontal equation, e.g. y2 = –2xy, or y = −2x, or equivalent A1√ Explicitly reject y = 0 as a possibility A1 Obtain an equation in x (or in y) M1 Obtain x = a A1 Obtain y = −2a only A1 [8] [The first M1 is dependent on at least one B mark having been earned.] [SR: for an attempt using (x + y) = 2a3 / xy, the B marks are given for the correct derivatives of the two sides of the equation, and the M1 for setting dy equal to zero.] dx [SR: for an attempt which begins by expressing y in terms of x, give M1A1 dy for a reasonable attempt at differentiation, M1A1√ for setting equal to dx zero and obtaining an equation free of surds, A1 for solving and obtaining x = a; then M1 for obtaining an equation for y, A1 for y = −2a and A1 for finding and rejecting y = a as a possibility.] GCE A/AS LEVEL – May/June 2008 9709 03 B C

This question in 9709/31 May/June 2008

ex 13 The curve y = cos x, for −12π < x < 2π, has one stationary point. Find the x-coordinate of this point. [5]

5 marks

Mark scheme: 3 Use correct quotient or product rule M1 e x cos x + e x sin x Obtain correctly the derivative in any form, e.g. A1 cos 2 x Equate derivative to zero and reach tan x = k M1* Solve for x M1(dep*) Obtain x = − 14 π (or −0.785) only (accept x in [−0.79, −0.78] but not in degrees) A1 [5] [The last three marks are independent. Fallacious log work forfeits the M1*. For the M1(dep*) the solution can lie outside the given range and be in degrees, but the mark is not available if k = 0. The final A1 is only given for an entirely correct answer to the whole question.] dx dy

This question in 9709/31 Oct/Nov 2008

8 An underground storage tank is being filled with liquid as shown in the diagram. Initially the tank is empty. At time t hours after filling begins, the volume of liquid is V m3 and the depth of liquid is h m. It is given that V = 43h3. The liquid is poured in at a rate of 20 m3 per hour, but owing to leakage, liquid is lost at a rate dh proportional to h2. When h = 1, = 4.95. dt (i) Show that h satisfies the differential equation dh 5 = −1 [4] dt h2 20. 20h2 2000 (ii) Verify that ≡−20 + [1] 100 −h2 (10 −h)(10 + h). (iii) Hence solve the differential equation in part (i), obtaining an expression for t in terms of h. [5]

10 marks

Mark scheme: dV 2 dh dV 28 (i) State or obtain = 4 h , or = 4 h , or equivalent B1 dt dt d h dV 2 State or imply = 20 − kh B1 dt Use the given values to evaluate k M1 Show that k = 0.2, or equivalent, and obtain the given equation A1 [4] [The M1 is dependent on at least one B mark having been earned.] (ii) Fully justify the given identity B1 [1] (iii) Separate variables correctly and attempt integration of both sides M1 Obtain terms –20h and t, or equivalent A1  10 + h  Obtain terms aln(10 + h) + bln(10 – h), where ab ≠ 0, or k ln   M1  10 − h  Obtain correct terms, i.e. with a = 100 and b = −100, or k = 2000/20, or equivalent A1 Evaluate a constant and obtain a correct expression for t in terms of h A1 [5] 1 x 1 x ∫

This question in 9709/31 Oct/Nov 2008

4 A curve has equation y tan x. Find the x-coordinates of the stationary points on the curve in the = e−3x 1 interval Give your answers correct to 3 decimal places. [6] x −12π < < 2π.

6 marks

Mark scheme: 4 Use product or quotient rule M1 Obtain derivative in any correct form A1 Equate derivative to zero and obtain an equation of the form a sin 2x = b, or a quadratic in tan x, sin2 x, or cos2 x M1* Carry out correct method for finding one angle M1(dep*) Obtain answer, e.g. 0.365 A1 Obtain second answer 1.206 and no others in the range (allow 1.21) A1 [6] [Ignore answers outside the given range.] [Treat answers in degrees, 20.9° and 69.1°, as a misread.]

This question in 9709/31 Oct/Nov 2009

10 In a model of the expansion of a sphere of radius r cm, it is assumed that, at time t seconds after the start, the rate of increase of the surface area of the sphere is proportional to its volume. When t 0, = dr r 5 and 2. = dt = (i) Show that r satisfies the differential equation dr 0.08r2. dt = [4] [The surface area A and volume V of a sphere of radius r are given by the formulae A 4πr2, 4 = V = 3πr3.] (ii) Solve this differential equation, obtaining an expression for r in terms of t. [5] (iii) Deduce from your answer to part (ii) the set of values that t can take, according to this model. [1]

10 marks

Mark scheme: dA 10 (i) State or imply = kV M1* dt dr dr 4 Obtain equation in r and , e.g. 8πr = k πr3 A1 dt dt 3 dr Use = 2, r = 5 to evaluate k M1(dep*) dt Obtain given answer A1 [4] (ii) Separate variables correctly and integrate both sides M1 1 Obtain terms – and 0.08t, or equivalent A1 + A1 r Evaluate a constant or use limits t = 0, r = 5 with a solution containing terms of the form a and bt M1 r 5 Obtain solution r = , or equivalent A1 [5] 1( − 4.0t ) (iii) State the set of values 0 Y t < 2.5, or equivalent B1 [1] [Allow t < 2.5 and 0 < t < 2.5 to earn B1.]

This question in 9709/31 Oct/Nov 2009

3 The equation of a curve is x3 3. −x2y −y3 = dy (i) Find in terms of x and y. [4] dx (ii) Find the equation of the tangent to the curve at the point giving your answer in the form ax by c 0. (2, 1), [2] + + =

6 marks

Mark scheme: dy 3 (i) State 2xy + x2 as derivative of x2y B1 dx dy State 3y2 as derivative of y3 B1 dx dy Equate derivative of LHS to zero and solve for M1 dx 3 x 2 − 2 xy Obtain answer 2 2 , or equivalent A1 [4] x + 3 y (ii) Find gradient of tangent at (2, 1) and form equation of tangent M1 Obtain answer 8x – 7y – 9 = 0, or equivalent A1√ [2]

This question in 9709/32 Oct/Nov 2009

9 y P x –1 O 1 r 1 The diagram shows the curve y . −x 1 x = + 1 dy (i) By first differentiating obtain an expression for in terms of x. Hence show that the −x 1 dx [5] gradient of the normal to +thex,curve at the point is (x, y) (1 + x) √(1 −x2). (ii) The gradient of the normal to the curve has its maximum value at the point P shown in the diagram. Find, by differentiation, the x-coordinate of P. [4]

9 marks

Mark scheme: 9 (i) Use quotient or product rule to differentiate (1 – x)/(1 + x) M1 Obtain correct derivative in any form A1 dy Use chain rule to find M1 dx Obtain a correct expression in any form A1 Obtain the gradient of the normal in the given form correctly A1 [5] (ii) Use product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and solve for x M1 Obtain x = 12 A1 [4]

This question in 9709/31 May/June 2010

6 The equation of a curve is x ln y 2x 1. = + dy (i) Show that [4] dx = −yx2. (ii) Find the equation of the tangent to the curve at the point where y 1, giving your answer in the form ax by c 0. = [4] + + =

8 marks

Mark scheme: 1 dy 6 (i) EITHER: State or imply as derivative of ln y B1 y dx x dy State correct derivative of LHS, e.g. ln y + B1 y dx dy Differentiate RHS and obtain an expression for M1 dx Obtain given answer A1 2 x + 1 OR 1: State ln y = , or equivalent, and differentiate both sides M1 x 1 dy State correct derivative of LHS, e.g. B1 y dx State correct derivative of RHS, e.g. − 1 / x 2 B1 Rearrange and obtain given answer A1 OR 2: State y = exp( 2 + 1 / x ) , or equivalent, and attempt differentiation by chain rule M1 State correct derivative of RHS, e.g. − exp(2 + 1 / x /) x 2 B1 + B1 Obtain given answer A1 [4] [The B marks are for the exponential term and its multiplier.] (ii) State or imply x = − 12 when y = 1 B1 Substitute and obtain gradient of −4 B1√ Correctly form equation of tangent M1 Obtain final answer y + 4x + 1 = 0, or equivalent A1 [4]

This question in 9709/32 May/June 2010

5 The curve with equation 6e2x key e2y c, + + = where k and c are constants, passes through the point P with coordinates 3, ln (ln 2). (i) Show that 58 2k c. [2] + = (ii) Given also that the gradient of the curve at P is find the values of k and c. [5] −6,

7 marks

Mark scheme: 5 (i) Use at least one of e2x = 9, e y = 2 and e2y = 4 B1 Obtain given result 58 + 2k = c AG B1 [2] dy dy y (ii) Differentiate left-hand side term by term, reaching ae2x + be + ce2y M1 dx dx dy dy y Obtain 12e2x + ke + 2e2y A1 dx dx Substitute (ln 3, ln 2) in an attempt involving implicit differentiation at least once, where RHS = 0 M1 Obtain 108 – 12k – 48 = 0 or equivalent A1 Obtain k = 5 and c = 68 A1 [5] 2 2

This question in 9709/31 May/June 2011

6 A certain curve is such that its gradient at a point (x, y) is proportional to xy. At the point (1, 2) the gradient is 4. (i) By setting up and solving a differential equation, show that the equation of the curve is y = 2ex2−1. [7] (ii) State the gradient of the curve at the point (−1, 2) and sketch the curve. [2]

9 marks

Mark scheme: dy 6 (i) Show that the differential equation is = 2 xy B1 dx Separate variables correctly and attempt integration of both sides M1 Obtain term ln y, or equivalent A1 Obtain term x2, or equivalent A1 Evaluate a constant, or use limits x = 1, y = 2, in a solution containing terms aln y and bx2 M1 Obtain correct solution in any form A1 Obtain the given answer correctly A1 [7] (ii) State that the gradient at (–1, 2) is –4 B1 Show the sketch of curve with correct concavity, positive y-intercept and axis of symmetry x = 0 B1 [2] [SR: A solution with k≠ 2, or not evaluated, can earn B0M1A1A1M1A1A0 in part (i).] dy [SR: If given answer is assumed valid, give B1 if is shown correctly to be equal to dx 2xy, is stated to be proportional to xy, and shown to be equal to 4 at (1, 2).] GCE AS/A LEVEL – May/June 2011 9709 32

This question in 9709/32 May/June 2011

10 y M P x O 3 The diagram shows the curve y = x2e−x. (i) Show that the area of the shaded region bounded by the curve, the x-axis and the line x = 3 is equal to 2 −17 . [5] e3 (ii) Find the x-coordinate of the maximum point M on the curve. [4] (iii) Find the x-coordinate of the point P at which the tangent to the curve passes through the origin. [2]

11 marks

Mark scheme: 10 (i) Attempt integration by parts and reach ± x 2 e −±x ∫ 2 xe − x dx M1* Obtain − x 2 e −+x ∫ 2 xe − x d x , or equivalent A1 Integrate and obtain –x2e–x – 2xe–x – 2e–x, or equivalent A1 Use limits x = 0 and x = 3, having integrated by parts twice M1(dep*) Obtain the given answer correctly A1 [5] (ii) Use correct product or quotient rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and solve for non-zero x M1 Obtain x = 2 with no errors send A1 [4] (iii) Carry out a complete method for finding the x-coordinate of P M1 Obtain answer x =1 A1 [2]

This question in 9709/32 May/June 2011

ln x 2 The curve y has one stationary point. Find the x-coordinate of this point. [4] = x3

4 marks

Mark scheme: 2 Use correct quotient or product rule M1 3 ln x 1 Obtain correct derivative in any form, e.g. − + A1 x 4 x 4 Equate derivative to zero and solve for x an equation of the form ln x = a , where a > 0 M1 1 Obtain answer exp( ), or 1.40, from correct work A1 [4] 3 1 1 ( ) − x − x ∫

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

e2x 2 The equation of a curve is y . Show that the gradient of the curve at the point for which = 1 e2x 9 + x ln 3 is 50. [4] =

4 marks

Mark scheme: 2 Use correct quotient or product rule or equivalent M1 2 x 2 x 2 x 2 x (1 + e ).2e − e .2e Obtain 2 x 2 or equivalent A1 (1 + e ) Substitute x = ln 3 into attempt at first derivative and show use of relevant logarithm M1 property at least once in a correct context Confirm given answer 509 legitimately A1 [4]

This question in 9709/33 Oct/Nov 2011

7 The equation of a curve is 1. ln(xy) −y3 = dy y (i) Show that [4] dx = x(3y3 −1). (ii) Find the coordinates of the point where the tangent to the curve is parallel to the y-axis, giving each coordinate correct to 3 significant figures. [4]

8 marks

Mark scheme: 1 1 dy 7 (i) EITHER: State or imply as derivative of ln xy, or equivalent B1 x dx dy State or imply 3y2 as derivative of y3, or equivalent B1 dx dy Equate derivative of LHS to zero and solve for M1 dx Obtain the given answer A1 dy OR Obtain xy = exp (1 + y3) and state or imply y + x dxas derivative of xy B1 d State or imply 3y2 exp (1 + y3) as derivative of (1 + y3) B1 dx dy Equate derivatives and solve for M1 dx Obtain the given answer A1 [4] [The M1 is dependent on at least one of the B marks being earned] (ii) Equate denominator to zero and solve for y M1* Obtain y = 0.693 only A1 Substitute found value in the equation and solve for x M1(dep*) Obtain x = 5.47 only A1 [4] GCE AS/A LEVEL – October/November 2012 9709 31

This question in 9709/31 Oct/Nov 2012

7 The equation of a curve is 1. ln(xy) −y3 = dy y (i) Show that [4] dx = x(3y3 −1). (ii) Find the coordinates of the point where the tangent to the curve is parallel to the y-axis, giving each coordinate correct to 3 significant figures. [4]

8 marks

Mark scheme: 1 1 dy 7 (i) EITHER: State or imply as derivative of ln xy, or equivalent B1 x dx dy State or imply 3y2 as derivative of y3, or equivalent B1 dx dy Equate derivative of LHS to zero and solve for M1 dx Obtain the given answer A1 dy OR Obtain xy = exp (1 + y3) and state or imply y + x dxas derivative of xy B1 d State or imply 3y2 exp (1 + y3) as derivative of (1 + y3) B1 dx dy Equate derivatives and solve for M1 dx Obtain the given answer A1 [4] [The M1 is dependent on at least one of the B marks being earned] (ii) Equate denominator to zero and solve for y M1* Obtain y = 0.693 only A1 Substitute found value in the equation and solve for x M1(dep*) Obtain x = 5.47 only A1 [4] GCE AS/A LEVEL – October/November 2012 9709 32

This question in 9709/32 Oct/Nov 2012

5 For each of the following curves, find the gradient at the point where the curve crosses the y-axis: 1 x2 (i) y ; [3] + = 1 e2x + (ii) 2x3 5xy y3 8. [4] + + =

7 marks

Mark scheme: 5 (i) Use correct quotient rule or equivalent M1 (1 + e 2 x )2 x − (1 + x 2 )2e 2 x Obtain 2 x 2 or equivalent A1 (1 + e ) 1 Substitute x = 0 and obtain − or equivalent A1 [3] 2 d y 3 2 (ii) Differentiate y and obtain 3 y B1 d x d y Differentiate 5xy and obtain 5 y + 5 x B1 d x 2 d y 2 d y Obtain 6 x + 5 y + 5 x + 3 y = 0 B1 d x d x GCE AS/A LEVEL – May/June 2013 9709 31 5 Substitute x = 0, y = 2 to obtain − or equivalent following correct work B1 [4] 6

This question in 9709/31 May/June 2013

4 The parametric equations of a curve are x = e−t cost, y = e−t sin t. dy Show that = tan t −1 . [6] dx 4

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 d x 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 9709/32 Oct/Nov 2013

8 y x 0 O The diagram shows the curve y = x cos 12x for 0 ≤x ≤0. dy 1 (i) Find and show that 4d2y + y + 4 sin = 0. [5] 2x dx dx2 (ii) Find the exact value of the area of the region enclosed by this part of the curve and the x-axis. [5]

10 marks

Mark scheme: 8 (i) Use product rule M1 Obtain derivative in any correct form A1 Differentiate first derivative using the product rule M1 Obtain second derivative in any correct form, e.g. − 12 sin 12 x − 14 x cos 12 x − 12 sin 12 x A1 Verify the given statement A1 5 2 x d x M1* (ii) Integrate and reach kx sin 12 x + l ∫ sin 1 2 x d x , or equivalent A1 Obtain 2 x sin 12 x − 2 ∫ sin 1 Obtain indefinite integral 2 x sin 12 x + 4 cos 12 x A1 Use correct limits x = 0, x = π correctly M1(dep*) Obtain answer 2π − 4 , or exact equivalent A1 5 dN

This question in 9709/32 May/June 2014

4 The equation of a curve is y 3 cos 2x 7 sin x 2. = + + Find the x-coordinates of the stationary points in the interval 0 Give each answer correct to 3 significant figures. ≤x ≤0. [7]

7 marks

Mark scheme: 4 Differentiate to obtain form a sin 2 x + b cos x M1 Obtain correct − 6 sin 2 x + 7 cos x A1 Use identity sin 2 x = 2 sin x cos x B1 Solve equation of form c sin x cos x + d cos x = 0 to find at least one value of x M1 Obtain 0.623 A1 Obtain 2.52 A1 1 Obtain 1.57 or π from equation of form c sin x cos x + d cos x = 0 A1 2 Treat answers in degrees as MR – 1 situation [7] 2 2

This question in 9709/31 May/June 2015

e2x 4 The curve with equation y has one stationary point. Find the exact values of the coordinates = 4 e3x of this point. + [6]

6 marks

Mark scheme: 4 Use correct quotient or product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and obtain a horizontal equation M1 Carry out complete method for solving an equation of the form a3e x = b , or ae 5 x = be 2 x M1 Obtain x = ln 2 , or exact equivalent A1 1 Obtain y = , or exact equivalent A1 6 3 d x 3 d y 3

This question in 9709/33 May/June 2015

6 A curve has equation sin y ln x x sin y, = −2 for −120 ≤y ≤120. dy (i) Find in terms of x and y. [5] dx (ii) Hence find the exact x-coordinate of the point on the curve at which the tangent is parallel to the x-axis. [3]

8 marks

Mark scheme: 6 (i) EITHER: State correct derivative of sin y with respect to x B1 Use product rule to differentiate the LHS M1 Obtain correct derivative of the LHS A1 Obtain a complete and correct derived equation in any form A1 dy Obtain a correct expression for in any form A1 dx OR: State correct derivative of sin y with respect to x B1 Rearrange the given equation as sin y = x / (ln x + 2) and attempt to differentiate both sides B1 Use quotient or product rule to differentiate the RHS M1 Obtain correct derivative of the RHS A1 dy Obtain a correct expression for in any form A1 [5] dx dy (ii) Equate to zero and obtain a horizontal equation in ln x or sin y M1 dx Solve for ln x M1 Obtain final answer x = 1/ e , or exact equivalent A1 [3]

This question in 9709/32 Feb/March 2016

7 The equation of a curve is x3 y3 3. −3x2y + = dy x2 (i) Show that −2xy . [4] dx = x2 −y2 (ii) Find the coordinates of the points on the curve where the tangent is parallel to the x-axis. [5]

9 marks

Mark scheme: 2 dy 27 (i) State or imply 6 xy + 3 x as derivative of 3x y B1 dx 2 dy 3 State 3 y as derivative of y B1 dx dy Equate attempted derivative of the LHS to zero and solve for M1 dx Obtain the given answer A1 [4] (ii) Equate numerator to zero M1* Obtain x = 2y, or equivalent A1 Obtain an equation in x or y DM1* Obtain the point (−2, −1) A1 State the point (0, 1.44) B1 [5] A B C

This question in 9709/31 May/June 2016

8 y x O a 0 The diagram shows the curve y cosecx for 0 x and part of the curve y When x a, the = < = e−x. = < 0 tangents to the curves are parallel. 1 dy (i) By differentiating show that if y cosecx then cotx. [3] sin x, = dx = −cosecx (ii) By equating the gradients of the curves at x a, show that = @ A ea . [2] a = tan−1 sin a (iii) Verify by calculation that a lies between 1 and 1.5. [2] (iv) Use an iterative formula based on the equation in part (ii) to determine a correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3]

10 marks

Mark scheme: 8 (i) Use correct quotient or chain rule M1 Obtain correct derivative in any form A1 Obtain the given answer correctly A1 [3] (ii) State a correct equation, e.g. − e − a = − cosec a cot a B1 Rearrange it correctly in the given form B1 [2] (iii) Calculate values of a relevant expression or pair of expressions at x = 1 and x = 1.5 M1 Complete the argument correctly with correct calculated values A1 [2] (iv) Use the iterative formula correctly at least once M1 Obtain final answer 1.317 A1 Show sufficient iterations to 5 d.p. to justify 1.317 to 3 d.p., or show there is a sign change in the interval (1.3165, 1,3175) A1 [3]

This question in 9709/32 May/June 2016

4 The parametric equations of a curve are x t cost, y ln 1 sin t , = + = + where t 1 −120 < < 20. dy (i) Show that sec t. [5] dx = (ii) Hence find the x-coordinates of the points on the curve at which the gradient is equal to 3. Give your answers correct to 3 significant figures. [3]

8 marks

Mark scheme: dx 4 (i) State = 1 − sin t B1 dt Use chain rule to find the derivative of y M1 dy cos t Obtain = , or equivalent A1 dt 1 + sin t d y dy d x Use = ÷ M1 d x dt dt Obtain the given answer correctly A1 [5] (ii) State or imply t = cos −1 ( 13 ) B1 Obtain answers x = 1.56 and x = − 0.898 B1 + B1 [3]

This question in 9709/33 May/June 2016

5 The curve with equation y tan x, where a is a positive constant, has only one point in the interval e−ax 0 x 1 at which the tangent= is parallel to the x-axis. Find the value of a and state the exact value of<the <x-coordinate20 of this point. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5 Use product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero, use Pythagoras and obtain a quadratic equation in tan x M1 Obtain tan 2 x − a tan x + 1 = 0 , or equivalent A1 Use the condition for a quadratic to have only one root M1 Obtain answer a = 2 A1 Obtain answer x = 14π A1 Total: 7

This question in 9709/32 Feb/March 2017

7 y M x O 12x e The diagram shows a sketch of the curve y for x 0, and its minimum point M. x = > (i) Find the x-coordinate of M. [4] … … … … … … … … … … … … … … … … … … … (ii) Use the trapezium rule with two intervals to estimate the value of 3 12x e dx, x Ô1 giving your answer correct to 2 decimal places. [3] … … … … … … … … … … … … … … (iii) The estimate found in part (ii) is denoted by E. Explain, without further calculation, whether another estimate found using the trapezium rule with four intervals would be greater than E or less than E. [1] … … … … … …

8 marks

Mark scheme: 7(i) Use correct quotient rule or product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and solve for x M1 Obtain x = 2 A1 Total: 4 7(ii) State or imply ordinates 1.6487…, 1.3591…, 1.4938… B1 Use correct formula, or equivalent, with h = 1 and three ordinates M1 Obtain answer 2.93 only A1 Total: 3 7(iii) Explain why the estimate would be less than E B1 Total: 1

This question in 9709/33 May/June 2017

5 The equation of a curve is 2x4 xy3 y4 10. + + = dy y3 (i) Show that + . [4] dx = −8x33xy2 4y3 + … … … … … … … … … … … … … … … … … … … … … … … … (ii) Hence show that there are two points on the curve at which the tangent is parallel to the x-axis and find the coordinates of these points. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(i) 3 2 dy 3 B1 State or imply y + 3 xy as derivative of xy dx 3 dy 4 B1 State or imply 4 y as derivative of y dx d y M1 Equate derivative of the LHS to zero and solve for d x Obtain the given answer A1 4 5(ii) Equate numerator to zero *M1 Obtain y = −2x, or equivalent A1 Obtain an equation in x or y DM1 Obtain final answer x = −1, y = 2 and x = 1, y = −2 A1 4

This question in 9709/31 Oct/Nov 2017

6 The equation of a curve is x3y 2a4, where a is a non-zero constant. −3xy3 = dy 3x2y (i) Show that . [4] −3y3 dx = 9xy2 −x3 … … … … … … … … … … … … … … … … … … … … … … … … (ii) Hence show that there are only two points on the curve at which the tangent is parallel to the x-axis and find the coordinates of these points. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(i) 2 3 dy B1 State or imply 3 x y + x as derivative of 3x y dx 2 dy 3 3 B1 State or imply 9 xy + 3 y as derivative of 3xy dx d y M1 Equate derivative of the LHS to zero and solve for d x Obtain the given answer AG A1 4 6(ii) Equate numerator to zero and use x = – y to obtain an equation in x or in y M1 Obtain answer x = a and y = – a A1 Obtain answer x = – a and y = a A1 Consider and reject y = 0 and x = y as possibilities B1 4

This question in 9709/32 Oct/Nov 2017

5 The equation of a curve is x2 x 3y 3. + −y3 = dy x2 2xy (i) Show that . [4] + dx = y2 −x2 … … … … … … … … … … … … … … … … … … … … … … … … (ii) Hence find the exact coordinates of the two points on the curve at which the gradient of the normal is 1. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(i) State or imply 3 2 d d y y x as derivative of 3 y B1 State or imply 2 d 6 3 d + y xy x x as derivative of 2 3x y OR State or imply ( ) 2 d 2 3 1 3d   + + +     y x x y x x as derivative of ( ) 2 3 + x x y B1 2 2 2 d d 3 6 3 3 0 d d + + − = y y x xy x y x x Equate derivative of the LHS to zero and solve for d d y x M1 Given answer so check working carefully Obtain the given answer A1 4 5(ii) Equate derivative to – 1 and solve for y M1* Use their y = – 2x or equivalent to obtain an equation in x or y M1(dep*) Obtain answer (1, – 2) A1 Obtain answer ( 3 3 , 0) B1 Must be exact e.g. 1 ln3 3 e but ISW if decimals after exact value seen 4

This question in 9709/32 May/June 2018

4 The parametric equations of a curve are x 2 sin sin y 2 cos cos = 1 + 21, = 1 + 21, where 0 < 1 < 0. dy (i) Obtain an expression for in terms of [3] dx 1. … … … … … … … … … … … … … … … … … … … … … … (ii) Hence find the exact coordinates of the point on the curve at which the tangent is parallel to the y-axis. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(i) Obtain d d x θ 2cos 2cos2 θ θ = + or d d y θ = 2sin 2sin 2 θ θ − − B1 Use dy/dx = dy/dθ ÷ dx/dθ M1 Obtain correct d d y x in any form, e.g. – 2sin 2sin2 2cos 2cos2 θ θ θ θ + + A1 3 4(ii) Equate denominator to zero and use any correct double angle formula M1* Obtain correct 3-term quadratic in cos θ in any form A1 Solve for θ depM1* Obtain x = 3 3 / 2 and y = 1 2 , or exact equivalents A1 4

This question in 9709/31 Oct/Nov 2018

4 The parametric equations of a curve are x 2 sin sin y 2 cos cos = 1 + 21, = 1 + 21, where 0 < 1 < 0. dy (i) Obtain an expression for in terms of [3] dx 1. … … … … … … … … … … … … … … … … … … … … … … (ii) Hence find the exact coordinates of the point on the curve at which the tangent is parallel to the y-axis. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(i) Obtain d d x θ 2cos 2cos2 θ θ = + or d d y θ = 2sin 2sin 2 θ θ − − B1 Use dy/dx = dy/dθ ÷ dx/dθ M1 Obtain correct d d y x in any form, e.g. – 2sin 2sin2 2cos 2cos2 θ θ θ θ + + A1 3 4(ii) Equate denominator to zero and use any correct double angle formula M1* Obtain correct 3-term quadratic in cos θ in any form A1 Solve for θ depM1* Obtain x = 3 3 / 2 and y = 1 2 , or exact equivalents A1 4

This question in 9709/33 Oct/Nov 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 9709/33 Oct/Nov 2018

1 4 The equation of a curve is y + e−x , for x 0. = 1 > −e−x dy (i) Show that is always negative. [3] dx … … … … … … … … … … … … … … … … … … … … … … … (ii) The gradient of the curve is equal to when x a. Show that a satisfies the equation −1 = e2a 1 0. Hence find the exact value of a. [4] −4ea + = … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(i) Use the quotient or product rule M1 Obtain correct derivative in any form A1 Reduce to – ( ) 2 2e 1 e − − − x x , or equivalent, and explain why this is always negative A1 3 Question Answer Marks Guidance 4(ii) Equate derivative to – 1 and obtain the given equation B1 State or imply 2 4 1 0 − + = u u , or equivalent in e a B1 Solve for a M1 Obtain answer a = ( ) ln 2 3 + and no other A1 4

This question in 9709/33 May/June 2019

7 The equation of a curve is x3 3xy2 5. + −y3 = dy x2 y2 (a) Show that + [4] dx = y2 −2xy. … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the coordinates of the points on the curve where the tangent is parallel to the y-axis. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(a) State or imply 2 3 6 + y xy d d y x as derivative of 2 3xy B1 State or imply 2 3y d d y x as derivative of 3y B1 Equate attempted derivative of LHS to zero and solve for d d y x M1 Need to see d d y x factorised out prior to AG Obtain the given answer correctly A1 AG 4 7(b) Equate denominator to zero *M1 Obtain y = 2x, or equivalent A1 Obtain an equation in x or y DM1 Obtain the point (1, 2) A1 State the point ( ) 3 5, 0 B1 Alternatively (1.71, 0). 5

This question in 9709/32 Feb/March 2020

6 The parametric equations of a curve are t x ln 2 3t , y . 2 3t = + = + (a) Show that the gradient of the curve is always positive. [5] … … … … … … … … … … … … … … … … … … … … … … … (b) Find the equation of the tangent to the curve at the point where it intersects the y-axis. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) Use correct chain rule or correct quotient rule to differentiate x or y M1 Obtain d d x t = 3 2 3 + t or d d y t = ( ) 2 2 2 3 + t A1 OE Use d d y x = d d y t ÷ d d x t M1 Obtain answer ( ) 2 3 2 3 + t A1 OE. Express as a simple fraction but not necessarily fully cancelled. Explain why this is always positive A1 For correct gradient. e.g. x is only defined for 2 3 0 + > t hence gradient > 0 Alternative method for Question 6(a) Form equation in x and y only M1 Obtain e 2 1 2 e 3 3 3e − −   = = −     x x x y A1 OE Differentiate M1 Obtain 2 e 3 x y − ′ = A1 OE Explain why this is always positive A1 5 Question Answer Marks Guidance 6(b) Obtain 1 3 y = − when x = 0 B1 Use a correct method to form the given tangent M1 2 3 1 3 y x + =           Obtain answer 3 2 1 = − y x A1 OE 3

This question in 9709/31 May/June 2021

7 y P x O M N For the curve shown in the diagram, the normal to the curve at the point P with coordinates x, y meets the x-axis at N. The point M is the foot of the perpendicular from P to the x-axis. The curve is such that for all values of x in the interval 0 1 the area of triangle PMN is equal to tan x. ≤x < 2π, MN dy (a) (i) Show that [1] y dx. = … … … … … 1 dy (ii) Hence show that x and y satisfy the differential equation 2y2 tan x. [2] dx = … … … … … … … (b) Given that y 1 when x 0, solve this differential equation to find the equation of the curve, expressing y in= terms of x.= [6] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(a)(i) Justify the given statement MN y = d d y x B1 1 7(a)(ii) Express the area of PMN in terms of y and d d y x and equate to tan x M1 Obtain the given equation correctly A1 2 7(b) Separate variables and integrate at least one side M1 Obtain term 3 1 6 y A1 Obtain term of the form lncos ± x M1 Evaluate a constant or use x = 0 and y = 1 in a solution containing terms 3 ay and lncos ± x , or equivalent M1 Obtain correct answer in any form, e.g. 3 1 1 lncos 6 6 = − + y x A1 Obtain final answer 3 (1 6ln cos ) = − y x A1 OE 6

This question in 9709/33 May/June 2021

7 (a) Given that y ln ln x , show that = dy 1 [1] dx x ln x. = … … … … … The variables x and t satisfy the differential equation dx x ln x t 0. dt + = It is given that x e when t 2. = = (b) Solve the differential equation obtaining an expression for x in terms of t, simplifying your answer. [7] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (c) Hence state what happens to the value of x as t tends to infinity. [1] … … … …

9 marks

Mark scheme: 7(a) Show sufficient working to justify the given answer B1 1 7(b) Correct separation of variables B1 e.g. 1 1 ln d d t x x t x − =   Obtain term ( ) ln ln x B1 Obtain term ln − t B1 Evaluate a constant or use x = e and t = 2 as limits in an expression involving ( ) ln ln x M1 Obtain correct solution in any form, e.g. ln(ln ) ln ln 2 = − + x t A1 Use log laws to enable removal of logarithms M1 Obtain answer 2 e = t x , or simplified equivalent A1 7 7(c) State that x tends to 1 coming from e = k t x B1 1

This question in 9709/31 Oct/Nov 2021

9 The equation of a curve is ye2x 2. −y2ex = dy 2yex (a) Show that −y2 . [4] dx = 2y −ex … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the exact coordinates of the point on the curve where the tangent is parallel to the y-axis. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 9(a) State correct derivative of 2 e x y with respect to x B1 2 2 d 2 e e d x x y y x + State correct derivative of 2ex y with respect to x B1 2 d 2 e e d x x y y y x + Equate attempted derivative of the LHS to zero and solve for d d y x M1 Obtain 2 d 2 e d 2 e x x y y y x y − = − A1 Obtain the given answer correctly. Condone multiplication by 1 1 − − and cancelling of ex without comment. Alternative method for Question 9(a) Rearrange as ( ) 2 2 2 2 d d e e 2e e e d d e e x x x x x x x y y y y x x y =  − = − − − B1 Other rearrangements are possible e.g. ( ) 2 2 2 2 d d 2e e e 2 e e d d x x x x x y y y y y y x x − − − − − = + = − ( ) 2 2 2 d 2 d 2e e e d d e e x x x x x y y y x x y   = − × − −     − B1 2 d d 4 2 e e d d x x x y y e y y x x − − −  = − + − Solve for d d y x M1 Obtain 2 d 2 e d 2 e x x y y y x y − = − A1 Obtain the given answer correctly. 4 Question Answer Marks Guidance 9(b) Equate denominator to zero and substitute for y or for ex in the equation of the curve *M1 Obtain equation of the form 3e x a b = or 3 cy d = DM1 ( ) 3 3 e 8, 1 x y = = SOI Obtain x = ln 2 A1 Accept 1 ln8 3 ISW Obtain y = 1 A1 4

This question in 9709/32 Oct/Nov 2021

8 The equation of a curve is x3 y3 2xy 8 0. + + + = dy (a) Express in terms of x and y. [4] dx … … … … … … … … … … … … … … … … … … … … … … … … The tangent to the curve at the point where x 0 and the tangent at the point where y 0 intersect atthe acute angle = = !. (b) Find the exact value of tan [5] !. … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 8(a) State or imply 2 3y d d y x as derivative of 3y State or imply 2 2 y x  d d y x as derivative of 2xy B1 Complete differentiation and equate attempted derivative to zero and solve for d d y x M1 Obtain answer  2 2 3 2 3 2 x y y x   A1 4 8(b) Find gradient at either (0, – 2) or (– 2, 0) M1 Obtain answers 1 3 and 3 A1 A1 Use   tan A B  formula to find tan M1 Obtain answer 4 tan 3  A1 5

This question in 9709/31 May/June 2022

7 The equation of a curve is x3 3x2y 3. + −y3 = dy x2 2xy (a) Show that + . [4] dx = y2 −x2 … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the coordinates of the points on the curve where the tangent is parallel to the x-axis. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(a) State or imply 2 6 3 xy x  d d y x as derivative of 2 3x y State or imply 2 3y d d y x as derivative of 3y B1 Allow B1 B1 for     2 2 2 3 d 6 d 3 d 3 d 0 x x xy x x y y y     Equate attempted derivative of left-hand side to zero and solve to obtain an equation with d d y x as subject M1 Allow if zero implied by subsequent working. Allow if recover from an extra d d y x  at the beginning of the left-hand side. Obtain 2 2 2 d 2 d y x xy x y x    correctly A1 AG Accept d d for y x y . 4 7(b) Equate numerator to zero *M1 Must be using the given derivative. Obtain x = – 2y, or equivalent A1 An equation with x or y as the subject SOI. Use 3 2 3 3 3 x x y y    to obtain an equation in x or y DM1 3 3 3 8 12 3 y y y     or 3 3 3 3 1 2 8 3 x x x    or any equivalent form (do not need to evaluate powers). Obtain the point (– 2, 1) and no others from solving their cubic equation A1 Allow if each component stated separately. ISW. State the point (0, 3 3  ), or equivalent from correct work B1 Accept (0, 3 3 ), or (0, – 1.44) (-1.44225). Allow if each component stated separately. ISW. 5

This question in 9709/32 May/June 2022

x 1 7 The equation of a curve is y = cos2x, for 0 ≤x < 2π. At the point where x = a, the tangent to the curve has gradient equal to 12. ` _ a 3 cos a + 2a sin a (a) Show that a = cos−1 . [3] 12 … … … … … … … … … … … … … … … … … … … … … … … (b) Verify by calculation that a lies between 0.9 and 1. [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 or quotient rule M1 Obtain correct derivative in any form A1 2 dy cos x + 2 x sin x cos x e.g. = 4 or dx cos x dy 2 2 = sec x + 2 x sec x tan x dx A1 AG  cos a + 2 a sin a  Equate derivative at x = a to 12 and obtain a = cos −1  3   12  3 7(b) Evaluate a relevant expression or pair of expressions at a = 0.9 and a = 1 M1 Must be calculated in radians. Complete the argument correctly with correct calculated values A1 cos 0.9 = 0.622  0.553 0.9  0.985 0.0846  0 e.g. or or cos1 = 0.540  0.570 1  0.964 −0.0358  0 or could be looking at values of the gradient 8.46 & 14.1 2 7(c) M1 Must be working in radians.  cosan + 2 an sinan   3  correctly at least once Use the process a n +1 = cos−1    12  Obtain final answer 0.97 A1 Show sufficient iterations to 4 d.p. to justify 0.97 to 2 d.p., or show there A1 e.g. 0.95, 0.9743, 0.9694, 0.9704 is a sign change in the interval (0.965, 0.975) 3

This question in 9709/31 Oct/Nov 2022

7 The variables x and satisfy the differential equation 1 dx x cot sin21 = tan21 −2 1, d1 for 0 1 and x 0. It is given that x 2 when 1 < 1 < 2π > = 1 = 4π. d cot (a) Show that 1 . cot21 = −2 d1 sin21 (You may assume without proof that the derivative of cot with respect to is [1] 1 1 −cosec21.) … … … … … … … … (b) Solve the differential equation and find the value of x when 1 [7] 1 = 6π. … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 7(a) Show sufficient working to justify the given statement B1 e.g. see 2cot − cosec 2 in the working or express in terms of sinand cos and use quotient rule to obtain the given result. Solution must have θ present throughout and must reach the given answer. 1 7(b) Separate variables correctly B1 tan 2  2cot  xdx = 2 − 2 d  sin  sin  Check for relevant working in (a) Condone incorrect notation e.g. missing dx. Need either the integral sign or the dx, dθ. 1 2 B1 Obtain term x 2 Obtain terms tan+ cot 2 B1 + B1  2cot   2cos 1 d = − C ) Alternative:  2 d =  3 2 ( +  sin   sin  sin  1 M1 Need to have 3 terms. Constant of correct form. Form an equation for the constant of integration, or use limits x = 2, = π , in 4 a solution with at least two correctly obtained terms of the form ax 2 , btan and ccot 2, where abc ≠ 0 A1 1 or 12 x 2 = tan+ cosec 2− 1 x 2 = tan+ cot 2 State correct solution in any form, e.g. 2 If everything else is correct, allow a correct final answer to imply this A1. 1 A1 18 + 2 3 Substitute = π and obtain answer x = 2.67 2.6748… 3 If see a correctly rounded value ISW. 6 7

This question in 9709/32 Oct/Nov 2022

x3 8 The curve with equation y has a stationary point at x p, where p 0. = ex = > −1 (a) Show that p 3 1 . [3] = −e−p … … … … … … … … … … … … … … … … … … … … … … … … (b) Verify by calculation that p lies between 2.5 and 3. [2] … … … … … … … … … … … … (c) Use an iterative formula based on the equation in part (a) to determine p correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … …

8 marks

Mark scheme: 8(a) Use quotient or product rule M1 Obtain correct derivative in any form A1 Equate derivative at x = p to zero and obtain the given equation A1 3 8(b) Evaluate a relevant expression or pair of relevant pair of expressions at p = 2.5 M1 and p = 3 Complete the argument with correct calculated values A1 2 8(c) pn M1 correctly at least once 1 − e− Use the iterative formula np +1 = 3 ( ) Obtain final answer p = 2.82 A1 Show sufficient iterations to 4 d.p.to justify 2.82 to 2 d.p., or show there is a sign A1 change in the interval (2.815, 2.825) 3

This question in 9709/33 Oct/Nov 2022

5 The parametric equations of a curve are x te2t, y t2 t 3. = = + + dy (a) Show that [3] dx = e−2t. … … … … … … … … … … … … … … … … … … … … … … … @ A (b) Hence show that the normal to the curve, where t passes through the point 0, 3 . = −1, −1e4 [3] … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 5(a) dx 2 t 2 t B1 OE Obtain = e + 2te dt dy dy dt M1 d y 2t + 1 Use =  = 2 t . dx dt dx d x e (1 + 2t ) d y −2t A1 AG Need to see e2t (1 + 2t) in denominator. Obtain the given answer = e d x 3 5(b) 2 1 B1 Obtain x = − e− or − and y = 3 at t = − 1 e 2 2 1 B1 Obtain gradient of normal = − e− or − e 2 x = 0 substituted into equation of normal or use of gradients to give B1 −2 −2 Equation of normal y −=3 −e x −−e . ( ) 1 y = 3 − with no errors AG 4 e SC Decimals B0 B1 B0 − 0.135 . 3

This question in 9709/32 Feb/March 2023

5 The equation of a curve is x2y 4a3, where a is a non-zero constant. −ay2 = dy 2xy (a) Show that . [4] dx = 2ay −x2 … … … … … … … … … … … … … … … … … … … … … … … … (b) Hence find the coordinates of the points where the tangent to the curve is parallel to the y-axis. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(a) State or imply 2 d 2 d y xy x x  as derivative of 2 x y 2xy x  . State or imply d 2 d y ay x as derivative of 2 ay B1 Accept partial: 2 2 x ay y    . Equate attempted derivative to zero and solve for d d y x M1 Obtain answer 2 d 2 d 2 y xy x ay x   from correct working A1 AG 4 5(b) State or imply 2 2 0 ay x   *M1 Substitute into equation of curve to obtain equation in x and a or in y and a DM1 e.g. 2 2 3 2 4 ay ay a   or 4 4 3 4 2 4 x x a a a   . Obtain one correct point A1 e.g.   2 , 2 a a . Obtain second correct point and no others A1 e.g.   2 , 2 a a  . 4 SC: Allow A1 A0 for 2 x a  or for y = 2a.

This question in 9709/31 May/June 2023

7 The equation of a curve is 3x2 4xy 3y2 5. + + = dy 2y (a) Show that + [4] dx = −3x2x 3y. + … … … … … … … … … … … … … … … … … … … … … … … … (b) Hence find the exact coordinates of the two points on the curve at which the tangent is parallel to y 2x 0. [5] + = … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(a) State or imply d 6 d y y x as the derivative of 3y2 B1 Allow y for d d y x throughout. Accept 6 4 f x y x   . State or imply d 4 4 d y x y x  as the derivative of 4xy B1 Accept 4 6 f x y y    . Equate derivative of LHS to zero and solve for d d y x M1 Allow an extra d d y x in front of their differentiated equation. Allow if ‘= 0’ is implied but not seen. Allow d d f y x f x y      Obtain d 3 2 d 2 3 y x y x x y    A1 AG – must come from correct working. The position of the negative must be clear. 4 Question Answer Marks Guidance 7(b) Equate d d y x to –2 and solve for x in terms of y or for y in terms of x *M1 Must be using the given derivative. Obtain 4 x y  or 4 x y  A1 Seen or implied by correct later work. Substitute 4 their x y  or 4 x their y  in curve equation DM1 Allow unsimplified. Obtain 1 7 y  or 4 7 x  A1 Or exact equivalent. Or 4 1 and 7 7 x y   or exact equivalent. Obtain both pairs of values A1 Or 4 1 and 7 7 x y   or exact equivalent. A1 A0 for incorrect final pairing. 5

This question in 9709/32 May/June 2023

e3x2−1 5 Find the exact coordinates of the stationary points of the curve y = . [6] 1 −x2 … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 5 Use correct product or quotient rule M1 Need attempt at both derivatives condone errors in chain rule. In quotient rule allow BOD in formula if ± 2x seen unless clear that incorrect formula has been used. If omit denominator or forget to square or complete reversal of signs then M0 A0 M1 A1 A1 A1. 2 3 x 2 −1 3 x 2 −1 A1 2 3 x 2 −1 3 x 2 −1 6 x (1 − x )e + 2 xe If 6 x 1 − x e + 2 xe = 0 from the start, with no ( ) Obtain correct derivative in any form, e.g. 2 1 − x 2 wrong formula seen, award M1A1. ( ) Equate derivative (or its numerator) to zero and solve for x M1 6x – 6x3 + 2x = 0 and solve. Allow for just one x value. Allow if from solution of 3 term quadratic equation, but if they get x = 0 the x must factorise out Obtain the point (0,e −1 ) or exact equivalent A1 2 3 Or for all three x coordinates found 0,  oe and no 3 extras but if this is the case then one pair of correct coordinates A1 and both other pairs of correct coordinates A1. Accept, e.g. x = 0, y = e–1 ISW for last 3 marks.  2 3 3  A1 Allow √(4/3). Obtain the point  , −3e  or exact equivalent    3   2 3 3  A1 Obtain the point  − , −3e  or exact equivalent    3  6

This question in 9709/33 Oct/Nov 2023

5 The equation of a curve is y = 2 for 0 G x G 2 r. cos x dy Find and hence find the x-coordinates of the stationary points of the curve. [7] dx … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5 Use correct quotient (or product) rule *M1 Obtain correct derivative   sin 3 sin 2 2 e cos ( 2e sin cos ) cos  x x x x x x or equivalent A1 Equate numerator to zero DM1 Obtain equation in one unknown DM1 E.g. sin2x – 2sinx – 1 = 0. Solve a 3 term quadratic in sin x to find a value for x M1 Obtain a correct solution to the quadratic equation, e.g. 3.57° A1 At least 3sf. Obtain a further correct solution, e.g. x = 5.86° and no others in the interval A1FT At least 3sf. FT 3π – their 3.57. Alternative Method for the first 3 marks: Take logarithms of both sides and simplify (*M1) ln sin 2lncos   y x x or equivalent. Obtain 1 d sin cos 2 d cos   y x x y x x (A1) Or equivalent. Equate d d y x to zero (DM1) Continue as for the original 7

This question in 9709/31 May/June 2024

10 (a) Given that 2x = tan y , show that = 2 . [3] dx 1 + 4x … … … … … … … … … … … … … 3 (b) Hence find the exact value of 2 x tan -1 ( 2x) dx . [7] y1 2 … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(a) Obtain 2 = sec2y d d y x or equivalent d x y = sec2y by differentiation with respect to y. Use sec2y = 1 + tan2 y M1 Replace tan y with 2x and rearrange to obtain given answer 2 d 2 d 1 4   y x x A1 3 10(b) Integrate by parts and reach 2 2 1 2 tan 2 d 1 4     x ax x b x x *M1 Obtain 2 2 1 1 2 2 tan 2 d 1 4 x x x x x     A1 OE Reduce integral to expression of the form 2 d 1 4    n m x x M1 Complete integration and reach 2 1 1 tan 2 tan 2     px x qx r x M1 Obtain 2 1 1 1 1 1 2 4 8 tan 2 tan 2 x x x x     A1 OE Use limits of 1 2 x  and 1 2 3 x  in the correct order, having integrated twice DM1 Obtain answer 5 1 1 48 8 8 π 3   or exact equivalent A1 7

This question in 9709/31 May/June 2024

4 The equation of a curve is y e 2 x + y 2 e x = 6 . Find the gradient of the curve at the point where y = 1. [6] … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 4 Substitute 1 y  and obtain ex a  , where 0 M1 Must come from a quadratic in e .   2e e 6 0 x x    Ignore any negative solution. Obtain e 2 x  only A1 Or equivalent e.g. ln2. x  Condone x = 0.693….. State or imply   2 2 2 d d e 2 e e d d x x x y y y x x   B1 Accept y for d . d y x State or imply   2 2 d d e e 2 e d d x x x y y y y x x   B1 Accept y for d . d y x Differentiate RHS of given equation to obtain zero (could be implied by subsequent work), substitute for x and y and obtain d ... d y x  M1 Independent. d d 2 4 4 2 4 0 d d y y x x            NB: Could also rearrange to obtain d d y x then substitute. Both steps are needed for M1. Obtain 5 4  or 1.25  A1 Correct answer from correct working only Accept 10 8 .  -1.2499 is A0. Question Answer Marks Guidance 4 Alternative method for Question 4: Dividing through by x e Substitute 1 y  and obtain ex a  , where 0 a  (M1) Must come from a quadratic in e . x   2e e 6 0 x x    Ignore any negative solution. Obtain e 2 x  only (A1) Or equivalent e.g. ln2. x  Condone x = 0.693….. State or imply   2 d d 2 d d y y y x x  (B1) Accept y for d . d y x State or imply   d d e e d d x x x y ye y x x   (B1) Accept y for d . d y x Differentiate RHS of given equation to obtain 6e x   , substitute for x and y and obtain d … d y x  (M1) Independent. d d 6 1 2 2 2 d d 2 y y x x           NB: Could also rearrange to obtain d d y x then substitute. Both steps are needed for M1. Obtain 5 4  or 1.25  (A1) Correct answer from correct working only Accept 10 8 .  -1.2499 is A0. 6

This question in 9709/32 May/June 2024

10 (a) By writing y = sec 3i as cos3i, show that di = 3 sin i sec 4 i . [2] … … … … … … … … … … … … (b) The variables x and i satisfy the differential equation 2 d i 4 x + 9 sin i = ( x + 3) cos i . ` j d x r . It is given that x = 3 when i = 13 Solve the differential equation to find the value of cosi when x = 0 . Give your answer correct to 3 significant figures. [8] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(a) Use of correct chain rule (and correct quotient rule) and 3 cos   4 cos sin k      or equivalent.   4 4 d 3 sin cos 3sin sec d y         Must be expressed in the given form A1 Obtain given answer from full and correct working (signs must be shown), but condone   3 d d sec …   and '( ). y  2 Question Answer Marks Guidance 10(b) Separate variables:     4 2 3 sin d d cos 9 x x x            B1 Or 4 2 d d . 3sin 3 9 cos 9 x x x             Condone missing integral signs or missing d or d , x  but not both. Obtain   3 sec p A  B1 Correct form, p any constant but not 0. Use 2 2 2 d d 3 9 3 9 9 9 9 x x x x x x x                  and obtain   2 ln 9 q x  or   1 3 tan . x r C   *M1 Might have one third of both sides. Alt: substitute 3tan x   to obtain 1 tan d ; q    condone if have  in place of  in this method. Obtain   2 ln 9 q x  and   1 3 tan x r C   DM1 Obtain     ln cos q    OE. Obtain     3 2 1 3 2 3 sec ln 9 3tan x x C       or equivalent A1 Or might see a third of both sides. Must have 2 different variables. Use 1 3 , 3 x     in an equation including 3 sec , p    2 ln 9 q x  and 1 3 tan x r  to evaluate the constant of integration M1 Or as limits in a definite integral. Limits for  are 0 and 1 4 .  Obtain constant = 3 3 2 4 8 ln18    A1 OE, e.g. 1.308… to at least 3sf. Obtain cos 0.601  A1 Accept AWRT 0.601. 8

This question in 9709/32 May/June 2024

9 10- x x x A container in the shape of a cuboid has a square base of side x and a height of ( 10- )x . It is given that x varies with time, t, where t 2 0 . The container decreases in volume at a rate which is inversely proportional to t. When t = 1 , x = 1 and the rate of decrease of x is 20. 10 2 37 (a) Show that x and t satisfy the differential equation dx - 1 = . [5] td 2t `20 x - 3x 2j … … … … … … … … … … … … … … … … … (b) Solve the differential equation, obtaining an expression for t in terms of x. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 9(a) Obtain  d d V k t t  or  d 1 d V t kt  B1 Obtain 2 d 20 3 d V x x x   B1 Correct use of chain rule involving k M1 Use d d V t = d d . d d V x x t  Expressions for d d V t and d d V x must be seen to get M1. Obtain   2 d d 20 3 x k t t x x   or equivalent, A1 If this expression is first seen with numerical values, allow A1 when their value of k is substituted back into the general expression. Use 1 10 t  , 1 2 x  and d 20 d 37 x t  to obtain given answer which must be stated d 20 d 37 x t  needed to score final A1 A1   2 d 1 d 2 20 3 x t t x x    AG Need to at least see 20 37  = 1 3 10 10 4 k        if k t or 20 37  = 1 3 10 10 4 k         if k t  in working for correct k. d 20 d 37 x t  seen anywhere, then A0. 5 Question Answer Marks Guidance 9(b) Separate variables correctly & integrate at least one side correctly B1 Obtain terms 10x2 – x3 B1 May see –10x2 + x3 if negative sign moved across or e.g. 20x2 – 2x3 if 2 moved across. Allow 2 3 20 3 . 2 3 x x  Obtain term ln t with ‘correct’ coefficient from their separation of variables, for example a ln t for a t . B1FT FT sign and position of 2 from their separation but B0 if error from later manipulation. Use 1 10 t  , 1 2 x  to evaluate a constant or as limits in a solution containing terms of the form x2, x3 and ln t (or ln 2t) M1 Allow numerical and sign errors and decimals. Allow if exponentiate before substitution, even if exponentiation done incorrectly, allow for c or ec. Obtain correct answer in any form, for example 2 3 ln 19 ln0.1 10 2 8 2 t x x     A1 2 3 ln2 19 ln0.2 10 2 8 2 t x x     or 2 3 ln 10 2.5 0.125 1.15 2 t x x       Allow 1.14 to 1.16 for 1.15 and allow 2.44 to 2.46 for 2.45 Obtain answer 3 2 19 2 20 1 4 10 e x x t    or equivalent A1 ISW Need t =……… E.g. 3 3 2 2 2 3 19 2 19 4 2 20 4 19 20 20 2 4 0 . .1 e 1 , , e e 10 10e e x x x x x x     Allow decimals, allow 2.44 to 2.46 for 2.45, e.g. 3 2 2 20 2.45 e . x x   A0 if 1 ln10 e present in final answer. 6

This question in 9709/33 May/June 2024

3 The equation of a curve is ln ( x + y) = 3x 2 y . Find the gradient of the curve at the point ( 1 , 0) . [4] … … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 3 1+ ddyx B1 State or imply as the derivative of ln ( x + y ) x + y 2 dy 2 B1 State or imply 6 xy + 3 x as the derivative of 3x y dx d y M1 Having the correct form for at least one of the above. Substitute (1, 0 ) and solve for d x dy 1 A1 Obtain = or 0.5 dx 2 Alternative Method for Question 3: x 2 y dy B1 Rewrite as x + y = 3e and state or imply 1 + as the derivative of the LHS dx  2 dy  3 x 2 y x 2 y B1 State or imply  6 xy + 3 x  e as the derivative of 3e  dx  d y M1 Having the correct form for at least one of the above. Substitute (1, 0 ) and solve for d x dy 1 A1 Obtain = or 0.5 dx 2 4

This question in 9709/31 Oct/Nov 2024

10 h m 4 m A large cylindrical tank is used to store water. The base of the tank is a circle of radius 4 metres. At time t minutes, the depth of the water in the tank is h metres. There is a tap at the bottom of the tank. When the tap is open, water flows out of the tank at a rate proportional to the square root of the volume of water in the tank. d h (a) Show that =- m h , where m is a positive constant. [4] d t … … … … … … … … … … … … … … … … … … (b) At time t = 0 the tap is opened. It is given that h = 4 when t = 0 and that h = 2.25 when t = 20. Solve the differential equation to obtain an expression for t in terms of h, and hence find the time taken to empty the tank. [6] … … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(a) dV dV dh B1 SOI =  k V or = 16π dt dt dt Correct use of chain rule and V = 16πh M1 dV dh OE, e.g. = 16π . dt dt dh dV dV − k 16πh A1 Any equivalent form in terms of h. Obtain =  = dt dt dh 16π  k  k A1 Obtain given answer from full and correct working. = −   h = − h since is constant  4 π  4 π 4 10(b) Separate variables correctly and commence integration *M1 1 dt OE  dh =  − h Obtain −t = 2 h ( +C ) A1 Use the boundary conditions in an equation containing pt and q h to form DM1 OE, e.g. 0 = 4 + C or − 20= 2  1.5 + C. an equation in and/or C Use the boundary conditions in an equation containing pt and q h to form a DM1 1 C = −4, = second equation in and/or C and solve 20 t OE, e.g. − = 2 h − 4. 20 Hence t = 80 − 40 h A1 Must be seen. Time to empty the tank is 80 minutes A1 6

This question in 9709/31 Oct/Nov 2024

2 e 2 x 11 Let f ( )x = . e 2 x - 3 e x + 2 (a) Find f l ( x) and hence find the exact coordinates of the stationary point of the curve with equation y = f ( )x . [5] … … … … … … … … … … … … … … … … … … … … … … … … … ln 5 (b) Use the substitution u = ex and partial fractions to find the exact value of f ( )x d x . y ln 3 Give your answer in the form lna, where a is a rational number in its simplest form. [9] … … … … … … … … … … … … … … … … … … … … … … … … … …

14 marks

Mark scheme: 11(a) Use correct quotient rule NB the question asks for f x( ) so need complete form M1 Or correct product rule. 4e 2 x (e 2 x − 3e x + 2) − 2e 2 x (2e 2 x − 3e x ) A1 Obtain correct derivative in any form, e.g. 2 e 2 x − 3e x + 2 ( ) Equate their derivative to zero *M1 Can be implied by numerator equated to zero for quotient rule. 8 = 6ex ( ) Solve for x to obtain x = lna DM1 a positive. 4 A1 No errors seen. Obtain x = ln and y = –16 3 8 Accept equivalent exact forms, e.g. x = ln . 6 11(a) Alternative Method for Question 11(a) Complete method to express f ( x ) in partial fractions M1 As far as p + q + r with values for p, e x − 2 e x − 1  8 2  q and r  2 + x − x  .  e − 2 e − 1  u = e x Allow in u ( ) . s e x t e x *M1 Note: the question requires f’(x) so if they have Differentiate to obtain f  ( x ) = 2 + 2 substituted for ex, they will also need chain rule. e x − 2 e x − 1 ( ) ( ) − 8e x 2e x A1 From correct work. Obtain f  ( x ) = 2 + 2 e x − 2 e x − 1 ( ) ( ) Equate derivative to zero and solve for x to obtain x = lna DM1 Must follow correctly to give a positive value of a. 4 A1 No errors seen. Obtain x = ln and y = –16 3 8 Accept x = ln , or equivalent. 6 5 11(b) du x B1 State or imply = e dx 2u B1 Correct expression in u. Obtain  u du or equivalent 2 − 3u + 2 Condone missing du or missing integral but not both.  2  8 2 + Or1   − du Allow FT if using their partial fractions from   u  u ( u − 2 ) u ( u − 1) (a). A B B1 FT Complete reduction to partial fractions. State or imply partial fractions of the form + Correct form for their integrand. u − 1 u − 2 C D E 2u − 3 3 2u − 3 F G Or1 + + Or2 + = + + u u − 2 u − 1 u 2 − 3u + 2 u 2 − 3u + 2 u 2 − 3u + 2 u − 2 u − 1 Use a correct method for finding a constant M1 Available if they have incorrect form. −2 4 A1 Obtain correct + u − 1 u − 2 2u − 3 3 3 Or2 + − u 2 − 3u + 2 u − 2 u − 1 Integrate to obtain a ln (u – 1) + b ln (u – 2) or equivalent *M1 M0 if they have additional terms that do not cancel out. Obtain correct –2 ln (u – 1) + 4 ln (u – 2) or equivalent A1FT FT values of their partial fraction coefficients. Correctly use limits u = 5 and 3 in an expression of the form a ln (u – 1) + b ln (u – 2) DM1 or x = ln 5 and ln 3 in an expression of the form a ln (ex – 1) + b ln (ex – 2) 81 A1 Accept ln 20.25. Obtain ln 4 9

This question in 9709/32 Oct/Nov 2024

4 The parametric equations of a curve are x = etant , y = 3 tan 2 t . Find the equation of the tangent to the curve at the point (e, 3). Give your answer in the form y = mx + c , where m and c are exact. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: π B1 SOI4 At ( e, 3 ) , tan t = 1 or t = 4 d x 2 tan t B1 = sec t  e d t dy 2 B1 = 6tan t  sec t dt dy 6tan t  sec 2 t − tan t *M1 Correct use of their derivatives. = = 6tan t  e 2 tan t ( ) dx sec t  e y − 3 = 6 x − e ) DM1 Substitute for t, and use correct method for the e ( equation of the line. y = 6e x − 3 A1 Or exact equivalent. Alternative Method for Question 4 y y 2 B1 SOI 3 Correct cartesian form, e.g. x = e or ln x = 3 or y = 3 ( ln x ) Differentiate function of a function *M1 Complete method. y A1 dx e 3 d y 1 Obtain = k y e or = k  ln x dy d x x y A1 dx 1 3 3 d y 1 Obtain = 6 y e or = 6  ln x dy d x x y − 3 = 6 x − e ) DM1 Use correct method for the equation of the line. e ( 4 y = 6e x − 3 A1 Or exact equivalent. 6

This question in 9709/31 May/June 2025

5 The equation of a curve is xy + y 2 e -x = 4 . dy y 2 - ye x (a) Show that = x . [4] dx xe + 2y … … … … … … … … … … … … … … … … (b) Find the gradients of the tangents to the curve when x = 0 . [2] … … … … … … … … …

6 marks

Mark scheme: 5(a) dy B1 State or imply y + x as the derivative of xy dx − x d y 2 − x − x B1 State or imply 2 ye − y e as the derivative of y 2e d x d y M1 Need implicit differentiation and attempt at a Equate attempted derivative to zero and solve for product. d x dy y 2 − ye x A1 AG Obtain = from correct working Need to see sufficient correct detail. x dx xe + 2 y Alternative Method for Question 5(a) x dy x x B1 Using xye x + y 2 − 4e x = 0. State or imply xe + y xe + e as the derivative of xy ex ( ) dx dy x 2 B1 Using xye x + y 2 − 4e x = 0. State or imply 2 y − 4e as the derivative of y − 4ex dx d y M1 Must make use of 4 − xy = y 2 e − x Equate attempted derivative to zero and solve for d x dy y 2 − ye x A1 AG Obtain = from correct working Need to see sufficient correct detail. x dx xe + 2 y 4 5(b) Obtain one correct gradient B1 1 E.g. at ( 0, 2 ) . 2 Obtain second correct gradient B1 − 3 E.g. at ( 0, − 2 ) . 2 2

This question in 9709/33 May/June 2025

4 Find the exact coordinates of the stationary point of the curve with equation y = 3x 3 ln x 4 , for x 2 0 . [5] … … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 4 Use the correct product rule *M1 4 d 3 3 d 4 Attempt ln x 3 x + 3 x ln x with their derivatives. (Note: may start with 12 x3 ln x) ( ) ( ) dx dx Obtain the correct derivative in any form e.g. 9 x 2 ln x 4 + 12 x 2 A1 2 4 3 4 x 3 2 2 E.g. 9 x ln x + 3 x  4 . (If starting with 12 x3 ln x, should get e.g. 36 x ln x + 12 x ) x d 4 4 May see ln x = . ( ) d x x 4 − DM1 E.g. ax4 = eb, or cx3 = ed or x = e f. Equate to zero and eliminate ln, e.g. x4 = e 3 Allow this mark even if in decimals. Allow SCB1 if incorrect sign in product rule resulting in 4 x4 = e 3 , OE. 1 − A1 ISW Obtain x = e 3 only or exact simplified equivalent 1 4  − 4  x =  e 3  scores A0.   1 − x =  e 3 scores A0. Answers with no working score no marks. 4 A1 ISW Obtain y = − only or exact simplified equivalent 4 e −1 − 3 y = 3e lne scores A0. Answers with no working score no marks. 5

This question in 9709/35 May/June 2025

7 The equation of a curve is 2y 3 - 3 x 2 y - x 3 = 16 . dy x 2 + 2 xy (a) Show that = . [4] dx 2y 2 - x 2 … … … … … … … … … … … … … … … … … … … … … … … … … (b) Hence find the coordinates of the points on the curve at which the normal is parallel to the y-axis. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 7(a) State or imply 6 y 2 dy as the derivative of 2 y 3 B1 dx State or imply 3 x 2 dy + 6 xy as the derivative of 3x 2 y B1 dx Complete the differentiation and equate the derivative of the LHS to zero and M1 d y Needs to be clear how the value of was obtained, d y d x solve for (the = 0 can be implied) d x e.g. collecting like terms and using brackets. 2 A1 AG dy x + 2 xy Obtain = from correct working 2 dx 2 y 2 − x 2 dy 2 xy + x Accept = . dx 2 y 2 − x 2 No incorrect statements, e.g. ‘cancel by 3’ or ‘divide top and bottom by 3’ are acceptable, but to have dy 3 x 2 + 6 xy = and say “divide by 3” at the final step dx 6 y 2 − 3 x 2 scores A0. 4 7(b) Equate derivative to 0 and solve for x or for x in terms of y. *M1 E.g., x 2 + 2 xy = 0  x = 0 or x = −2 y. Must be using the numerator. Obtain ( 0, 2 ) B1 Allow if the values are stated separately. Do not ISW. Use their x = −2 y to form an equation in one unknown dM1 E.g., 2 y 3 − 12 y 3 + 8 y 3 = 16 or Or equivalent e.g. substitute y = −x2 2x  1 3   1 3  3 −2  x  + 3  x  − x = 16.  8   2  Obtain ( 4, − 2 ) A1 Allow if the values are stated separately. Do not ISW 4

This question in 9709/32 Oct/Nov 2025

4 The equation of a curve is x 2 ln 2y - y ln `2 + x 2j = ln 6. Find the exact value of the gradient of the curve at the point (2, 3). Give your answer in simplified form. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 4 d 2 2 1 d y B1 SOI x ln 2 y = 2 x ln 2 y + x  d x y dx d 2 dy 2 2 xy B1 y ln ( 2 + x ) =  ln ( 2 + x ) + 2 dx dx 2 + x Equate derivative of right-hand side to zero and substitute x = 2, y = 3 M1 Must be using implicit differentiation and have attempted differentiation of products where appropriate. Need to see evidence of substitution. 4 dy dy A1 OE Obtain 4ln 6 + . − ln 6. − 2 = 0 3 dx dx 6 − 12ln 6 10 A1 OE Simplify to obtain or 4 − Do not ISW. 4 − 3ln 6 4 − 3ln 6 No fractions within fractions. 5

This question in 9709/35 Oct/Nov 2025