Differentiation — Paper 2 · A Level Mathematics 9709

dy 5 Find the value of when x 4 in each of the following cases: dx = (i) y x [4] = ln(x −3), x (ii) y −1 [3] = x 1. +

7 marks

Mark scheme: 1 5 (i) Differentiate ln(x – 3) to obtain B1 x − 3 Attempt to use product rule M1 x Obtain ln ( x − 3 ) + or equivalent A1 x − 3 Substitute 4 to obtain 4 A1 [4] (ii) Use correct quotient or product rule M1 ( x + 1) − ( x − 1) Obtain correct derivative in any form, e.g. A1 ( x + 1)2 2 Substitute 4 to obtain A1 [3] 25 GCE AS/A LEVEL – May/June 2011 9709 21 3

This question in 9709/21 May/June 2011

6 The parametric equations of a curve are x 1 2 sin2θ, y 4 tan θ. = + = dy 1 (i) Show that . [3] dx = sin θ cos3θ (ii) Find the equation of the tangent to the curve at the point where θ 14π, giving your answer in = the form y mx c. [4] = +

7 marks

Mark scheme: dx dy 6 (i) State = 4 sin θ cos θ or equivalent (nothing for = 4 sec 2 θ ) B1 dt dx dy dy dx Use = ÷ M1 dx dθ dθ Obtain given answer correctly A1 [3] π dy (ii) Substitute θ = in and both parametric equations M1 4 dx dy Obtain = 4 and coordinates (2, 4) A1 dx Form equation of tangent at their point M1 State equation of tangent in correct form y = 4x – 4 A1 [4]

This question in 9709/22 Oct/Nov 2011

7 y P Q x O M The diagram shows the curve with parametric equations x 2 t, y 1 3 cos 2t, = −cos = + 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 9709/23 May/June 2016

6 y P x O The diagram shows the curve with parametric equations x 3t y = −6e−2t, = 4t2e−t, for 0 At the point P on the curve, the y-coordinate is 1. ≤t ≤2. 1 1 (i) Show that the value of t at the point P satisfies the equation t 2t. [2] 2e = … … … … … … … 1 1 2tn (ii) Use the iterative formula tn+1 = 2e with t1 = 0.7 to find the value of t at P correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] … … … … … … … … (iii) Find the gradient of the curve at P, giving the answer correct to 2 significant figures. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 6(i) Equate 2 4 e t t − to 1, rearrange to 2 ... t = and hence ... t = M1 Allow M1 for 1 4 t t e− = Confirm 1 2 1 2 e t t = with necessary detail needed as answer is given A1 2 6(ii) Use iterative process correctly at least once M1 Obtain final answer 0.715 t = A1 Show sufficient iterations to 5 sf to justify answer or show a sign change in the interval [0.7145, 0.7155] A1 SC: M1A1 from iterations to 4sf resulting in 0.71 3 Question Answer Marks Guidance 6(iii) Obtain 2 d d 3 12e t x t − = + B1 Use product rule to find d d y t M1 Obtain 2 8 e 4 e t t t t − − − A1 Divide correctly to obtain d d y x M1 Substitute value from part (ii) to obtain 0.31 A1 Allow greater accuracy 5

This question in 9709/22 May/June 2019

6 y P x O The diagram shows the curve with parametric equations x 3t y = −6e−2t, = 4t2e−t, for 0 At the point P on the curve, the y-coordinate is 1. ≤t ≤2. 1 1 (i) Show that the value of t at the point P satisfies the equation t 2t. [2] 2e = … … … … … … … 1 1 2tn (ii) Use the iterative formula tn+1 = 2e with t1 = 0.7 to find the value of t at P correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] … … … … … … … … (iii) Find the gradient of the curve at P, giving the answer correct to 2 significant figures. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 6(i) Equate 2 4 e t t − to 1, rearrange to 2 ... t = and hence ... t = M1 Allow M1 for 1 4 t t e− = Confirm 1 2 1 2 e t t = with necessary detail needed as answer is given A1 2 6(ii) Use iterative process correctly at least once M1 Obtain final answer 0.715 t = A1 Show sufficient iterations to 5 sf to justify answer or show a sign change in the interval [0.7145, 0.7155] A1 SC: M1A1 from iterations to 4sf resulting in 0.71 3 Question Answer Marks Guidance 6(iii) Obtain 2 d d 3 12e t x t − = + B1 Use product rule to find d d y t M1 Obtain 2 8 e 4 e t t t t − − − A1 Divide correctly to obtain d d y x M1 Substitute value from part (ii) to obtain 0.31 A1 Allow greater accuracy 5

This question in 9709/23 May/June 2019

4 A curve has parametric equations x ln 2t 6 t, y t ln t. = + −ln = (a) Find the value of t at the point P on the curve for which x ln 4. [3] = … … … … … … … … … … … … … … … … … … … … … … … (b) Find the exact gradient of the curve at P. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4(a) Equate x to ln4 and use relevant logarithm property M1 Obtain equation with no logarithm present, 2 6 4 + = t t A1 OE Obtain 3 = t A1 3 4(b) Obtain d 2 1 d 2 6 = − + x t t t B1 Use product rule to find d d y t M1 Obtain 1 ln + × t t t A1 Divide to obtain d d y x using their d d y t and d d x t correctly DM1 Must have at least B1 or M1. Do not condone incorrect inverting of terms unless a correct statement is seen initially. Obtain 6(ln3 1) − + A1 or exact equivalents 5

This question in 9709/21 May/June 2021

7 y P x O The diagram shows the curve with parametric equations x 4t e2t, y 6t sin 2t, = + = for 0 The point P on the curve has parameter p and y-coordinate 3. ≤t ≤1. 1 (a) Show that p [1] = 2 sin 2p. … … … (b) Show by calculation that the value of p lies between 0.5 and 0.6. [2] … … … … … (c) Use an iterative formula, based on the equation in part (a), to find the value of p correct to 3 significant figures. Use an initial value of 0.55 and give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … … (d) Find the gradient of the curve at P. [5] … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(a) Equate y to 3 and confirm 1 2sin 2 = p p 1 7(b) Consider sign of 1 2sin 2 − p p or equivalent for 0.5 and 0.6 M1 Obtain 0.09... − and 0.06... or equivalents and justify conclusion A1 AG 2 7(c) Use iteration process correctly at least once M1 Need to see 0.55494… Obtain final answer 0.557 only A1 Allow recovery. Allow if iterations are to 4sf Allow if insufficient iterations seen. Show sufficient iterations to 5 s.f. to justify answer or show sign change in interval [0.5565, 0.5575] A1 If not starting at 0.55 then max marks M1A1A0 3 Question Answer Marks Guidance 7(d) Obtain 2 d 4 2e d = + t x t B1 Use product rule to find d d y t M1 Must be of the form sin 2 cos2 + p t qt t Obtain 6sin 2 12 cos2 + t t t A1 Allow unsimplified. Divide to obtain d d y x using their d d y t and d d x t correctly DM1 Must have either B1 or previous M1. Obtain 0.826 A1 AWRT 5

This question in 9709/22 May/June 2021

7 y P x O The diagram shows the curve with parametric equations x 4t e2t, y 6t sin 2t, = + = for 0 The point P on the curve has parameter p and y-coordinate 3. ≤t ≤1. 1 (a) Show that p [1] = 2 sin 2p. … … … (b) Show by calculation that the value of p lies between 0.5 and 0.6. [2] … … … … … (c) Use an iterative formula, based on the equation in part (a), to find the value of p correct to 3 significant figures. Use an initial value of 0.55 and give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … … (d) Find the gradient of the curve at P. [5] … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(a) Equate y to 3 and confirm 1 2sin 2 = p p 1 7(b) Consider sign of 1 2sin 2 − p p or equivalent for 0.5 and 0.6 M1 Obtain 0.09... − and 0.06... or equivalents and justify conclusion A1 AG 2 7(c) Use iteration process correctly at least once M1 Need to see 0.55494… Obtain final answer 0.557 only A1 Allow recovery. Allow if iterations are to 4sf Allow if insufficient iterations seen. Show sufficient iterations to 5 s.f. to justify answer or show sign change in interval [0.5565, 0.5575] A1 If not starting at 0.55 then max marks M1A1A0 3 Question Answer Marks Guidance 7(d) Obtain 2 d 4 2e d = + t x t B1 Use product rule to find d d y t M1 Must be of the form sin 2 cos2 + p t qt t Obtain 6sin 2 12 cos2 + t t t A1 Allow unsimplified. Divide to obtain d d y x using their d d y t and d d x t correctly DM1 Must have either B1 or previous M1. Obtain 0.826 A1 AWRT 5

This question in 9709/23 May/June 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 9709/23 Oct/Nov 2022

6 y A B x O M The diagram shows the curve with parametric equations x = 1 + t , y = ( ln t + 2)( ln t - 3), for 0 1 t 1 25 . The curve crosses the x-axis at the points A and B and has a minimum point M. dy 4 ln t - 2 (a) Show that = . [4] dx t … … … … … … … … … … … … … … … … (b) Find the exact gradient of the curve at B. [2] … … … … … … … … … … … … … … (c) Find the exact coordinates of M. [3] … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) d y M1 Or chain rule if expression expanded before Use product rule to find differentiation. d t dy 2ln t 1 A1 Obtain = − dt t t d y d x M1 Divide attempt at by attempt at d t d t dy 4ln t − 2 A1 Answer given – necessary detail needed. Confirm = dx t 4 6(b) State or imply t = e 3 at point B B1 − 3 B1 Obtain gradient 10e 2 or exact equivalent 2 6(c) Equate first derivative to zero and obtain value for lnt or t M1 1 A1 Or exact equivalent. Obtain x-coordinate 1 + e 4 Obtain y-coordinate − 25 A1 Or equivalent. 4 3

This question in 9709/22 Feb/March 2024

6 A curve has equation ( x 2 - 3 )ln y + 6 x = 14 . (a) Show that there is no point on the curve at which the y-coordinate is e -1 . [3] … … … … … … … … … … … … … … … (b) Find the equation of the tangent to the curve at the point ( 2, e2 ) . Give your answer in the form y = mx + c , where m and c are exact constants. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) Substitute y = e − 1 in equation and simplify to quadratic equation in x *M1 Evaluate discriminant DM1 OE, such as completion of square or use of formula. Obtain − x 2 + 6 x − 11 = 0, giving discriminant −8 and confirm result A1 OE, such as ( x − 3) 2 + 2 = 0, etc. AG – necessary detail needed. 3 6(b) Attempt use of product rule for differentiation of ( x 2 − 3)ln y *M1 May see in part (a) x 2 − 3 dy A1 OE Obtain 2 x ln y + y dx x 2 − 3 dy A1 FT OE Obtain complete 2 x ln y + + 6 = 0 2 Following their derivative of ( x − 3)ln y. y dx Substitute x = 2 and y = e 2 to find value of gradient of tangent DM1 Need to see attempt at substitution unless correct. −103 implies M1. Obtain gradient −14e 2 A1 Obtain equation y = −14e 2 x + 29e 2 or equivalent of required form A1 6

This question in 9709/22 May/June 2025

6 A curve has equation ( x 2 - 3 )ln y + 6 x = 14 . (a) Show that there is no point on the curve at which the y-coordinate is e -1 . [3] … … … … … … … … … … … … … … … (b) Find the equation of the tangent to the curve at the point ( 2, e2 ) . Give your answer in the form y = mx + c , where m and c are exact constants. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) Substitute y = e − 1 in equation and simplify to quadratic equation in x *M1 Evaluate discriminant DM1 OE, such as completion of square or use of formula. Obtain − x 2 + 6 x − 11 = 0, giving discriminant −8 and confirm result A1 OE, such as ( x − 3) 2 + 2 = 0, etc. AG – necessary detail needed. 3 6(b) Attempt use of product rule for differentiation of ( x 2 − 3)ln y *M1 May see in part (a) x 2 − 3 dy A1 OE Obtain 2 x ln y + y dx x 2 − 3 dy A1 FT OE Obtain complete 2 x ln y + + 6 = 0 2 Following their derivative of ( x − 3)ln y. y dx Substitute x = 2 and y = e 2 to find value of gradient of tangent DM1 Need to see attempt at substitution unless correct. −103 implies M1. Obtain gradient −14e 2 A1 Obtain equation y = −14e 2 x + 29e 2 or equivalent of required form A1 6

This question in 9709/23 May/June 2025

6 A curve has equation ( x 2 - 3 )ln y + 6 x = 14 . (a) Show that there is no point on the curve at which the y-coordinate is e -1 . [3] … … … … … … … … … … … … … … … (b) Find the equation of the tangent to the curve at the point ( 2, e2 ) . Give your answer in the form y = mx + c , where m and c are exact constants. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) Substitute y = e − 1 in equation and simplify to quadratic equation in x *M1 Evaluate discriminant DM1 OE, such as completion of square or use of formula. Obtain − x 2 + 6 x − 11 = 0, giving discriminant −8 and confirm result A1 OE, such as ( x − 3) 2 + 2 = 0, etc. AG – necessary detail needed. 3 6(b) Attempt use of product rule for differentiation of ( x 2 − 3)ln y *M1 May see in part (a) x 2 − 3 dy A1 OE Obtain 2 x ln y + y dx x 2 − 3 dy A1 FT OE Obtain complete 2 x ln y + + 6 = 0 2 Following their derivative of ( x − 3)ln y. y dx Substitute x = 2 and y = e 2 to find value of gradient of tangent DM1 Need to see attempt at substitution unless correct. −103 implies M1. Obtain gradient −14e 2 A1 Obtain equation y = −14e 2 x + 29e 2 or equivalent of required form A1 6

This question in 9709/25 May/June 2025