Differentiation — Paper 2 · A Level Mathematics 9709

6 ln x The diagram shows the part of the curve y = for 0 < x ≤4. The curve cuts the x-axis at A and its x maximum point is M. (i) Write down the coordinates of A. [1] (ii) Show that the x-coordinate of M is e, and write down the y-coordinate of M in terms of e. [5] (iii) Use the trapezium rule with three intervals to estimate the value of 4 ln x dx, x 1 correct to 2 decimal places. [3] (iv) State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (iii). [1]

10 marks

Mark scheme: 6 (i) State coordinates (1, 0) B1 1 (ii) Use quotient or product rule M1 − ln x 1 Obtain correct derivative, e.g. + A1 x 2 x 2 Equate derivative to zero and solve for x M1 Obtain x = e A1 1 Obtain y = A1 5 e (iii) Show or imply correct coordinates 0, 0.34657..., 0.36620..., 0.34657,,, B1 Use correct formula, or equivalent, with h = 1 and four ordinates A1 Obtain answer 0.89 with no errors seen A1 3 (iv) Justify statement that the rule gives an under-estimate B1 1

This question in 9709/21 May/June 2005

3 The parametric equations of a curve are x = 3t + ln(t −1), y = t2 + 1, for t > 1. dy (i) Express in terms of t. [3] dx (ii) Find the coordinates of the only point on the curve at which the gradient of the curve is equal to 1. [4]

7 marks

Mark scheme: 3 (i) dx 1 dy State = 3 + or = 2t dt t − 1 dt B1 dy dy dx Use = ÷ dx dt dt M1 dy 2t (t − 1) Obtain in any correct form, e.g. dx 3t − 2 A1 [3] (ii) Equate derivative to 1 and solve for t M1 1 Obtain roots 2 and A1 2 State or imply that only t = 2 is admissible c.w.o. A1 Obtain coordinates (6, 5) A1 [4] GCE A/AS LEVEL – May/June 2007 9709 02

This question in 9709/21 May/June 2007

7 The diagram shows the part of the curve y = ex cos x for 0 ≤x ≤12π. The curve meets the y-axis at the point A. The point M is a maximum point. (i) Write down the coordinates of A. [1] (ii) Find the x-coordinate of M. [4] (iii) Use the trapezium rule with three intervals to estimate the value of 12π ex cos x dx, 0 giving your answer correct to 2 decimal places. [3] (iv) State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (iii). [1]

9 marks

Mark scheme: 7 (i) State coordinates (0, 1) for A B1 [1] (ii) Differentiate using the product rule M1* Obtain derivative in any correct form A1 Equate derivative to zero and solve for x M1* 1 Obtain x = π or 0.785 (allow 45°) A1 [4] 4 (ii) Show or imply correct ordinates 1, 1.4619…, 1.4248…, 0 B1 1 Use correct formula or equivalent with h = π and four ordinates M1 6 Obtain correct answer 1.77 with no errors seen A1 [3] (iv) Justify statement that the trapezium rule gives and underestimate B1 [1]

This question in 9709/21 May/June 2007

4 The equation of a curve is y = 2x −tan x, where x is in radians. Find the coordinates of the stationary points of the curve for which −12π < x < 12π. [5]

5 marks

Mark scheme: 4 State derivative 2 – sec2 x, or equivalent B1 Equate derivative to zero and solve for x M1 1 Obtain x = π , or 0.785 (± 45° gains A1) A1 4 1 Obtain x = – π , (allow negative of first solution) A1√ 4 1 1 Obtain corresponding y-values π – 1 and – π + 1, ± 0.571 A1 [5] 2 2

This question in 9709/21 Oct/Nov 2007

8 The diagram shows the curve y = x2e−x and its maximum point M. (i) Find the x-coordinate of M. [4] (ii) Show that the tangent to the curve at the point where x = 1 passes through the origin. [3] (iii) Use the trapezium rule, with two intervals, to estimate the value of 3 x2e−x dx, 1 giving your answer correct to 2 decimal places. [3]

10 marks

Mark scheme: 8 (i) Differentiate using product or quotient rule M1 Obtain derivative in any correct form A1 Equate derivative to zero and solve for x M1 Obtain answer x = 2 correctly, with no other solution A1 [4] (ii) Find the gradient of the curve when x = 1, must be simplified, allow 0.368 B1 Form the equation of the tangent when x = 1 M1 Show that it passes through the origin A1 [3] (iii) State or imply correct ordinates 0.36787…, 0.54134…, 0.44808… B1 Use correct formula, or equivalent, correctly with h = 1 and three ordinates M1 Obtain answer 0.95 with no errors seen A1 [3]

This question in 9709/21 Oct/Nov 2007

8 (i) (a) Prove the identity + sin x sec2x + sec x tan x ≡1 . cos2x (b) Hence prove that 1 sec2x + sec x tan x ≡ 1 −sin x. [3] 1 dy (ii) By differentiating cos x, show that if y = sec x then dx = secx tan x. [3] (iii) Using the results of parts (i) and (ii), find the exact value of 14π 1 dx. [3] 1 −sin x 0

9 marks

Mark scheme: 8 (i) (a) Use trig formulae and justify given result B1 (b) Use 1 – sin2 x = cos2 x M1 Obtain given result correctly A1 [3] (ii) Use quotient or chain rule M1 Obtain correct derivative in any form A1 Obtain given result correctly A1 [3] (iii) Obtain integral tan x + sec x B1 Substitute limits correctly M1 Obtain exact answer 2 , or equivalent A1 [3]

This question in 9709/21 Oct/Nov 2008

4 The parametric equations of a curve are x 4 sin θ, y 3 cos 2θ, = = −2 1 dy where 2π θ 2π. Express in terms of θ, simplifying your answer as far as possible. [5] dx −1 < <

5 marks

Mark scheme: dx 4 State = 4 cos θ B1 dθ dy State = 4 sin 2θ , or equivalent B1 dθ dy dy dx Use = ÷ M1 dx dθ dθ dy sin 2θ Obtain in any correct form, e.g. A1 dx cos θ Simplify and obtain answer 2 sinθ A1√ [5] [The f.t. is on gradients of the form k sin 2θ / cos θ, or equivalent.] 2 2 2 2

This question in 9709/21 May/June 2009

7 y x O M The diagram shows the curve y xe2x and its minimum point M. = (i) Find the exact coordinates of M. [5] (ii) Show that the curve intersects the line y 20 at the point whose x-coordinate is the root of the equation = 1 20 x ln . 2 x = [1] (iii) Use the iterative formula 1 20 2 xn xn+1 = ln , with initial value x1 1.3, to calculate the root correct to 2 decimal places, giving the result of = each iteration to 4 decimal places. [3]

9 marks

Mark scheme: 7 (i) Use product rule M1* Obtain derivative in any correct form A1 Equate derivative to zero and solve for x M1(dep*) Obtain answer x = − 1 correctly A1 2 Obtain y = –1/(2e) or exact equivalent A1 [5] (ii) Show that 20 = xe2x is equivalent to x = 1 ln(20 / x) or vice versa B1 [1] 2 (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.35 A1 Show sufficient iterations to justify its accuracy to 2 d.p. A1 [3] 1

This question in 9709/21 May/June 2009

8 (a) Find the equation of the tangent to the curve y at the point where x 1. [4] = ln(3x −2) = (b) (i) Find the value of the constant A such that 6x A 3x 3x ≡2 + [2] −2 −2. 6 6x 8 (ii) Hence show that dx 8 ln 2. [5] 3 3x ä = + 2 −2

11 marks

Mark scheme: 8 (a) State derivative is k/(3x –2) where k = 3.1, or 1 M1 3 State correct derivative 3/(3x –2) A1 Form the equation of the tangent at the point where x = 1 M1 Obtain answer y = 3x –3, or equivalent A1 [4] (b) (i) Carry out a complete method for finding A M1 Obtain A = 4 A1 [2] (ii) Integrate and obtain term 2x B1 Obtain second term of the form aln(3x –2) M1 Obtain second term 4 ln(3x – 2) A1√ 3 Substitute limits correctly M1 Obtain given answer following full and correct working A1 [5]

This question in 9709/21 May/June 2009

8 The equation of a curve is y2 2xy 2. + −x2 = (i) Find the coordinates of the two points on the curve where x 1. [2] = (ii) Show by differentiation that at one of these points the tangent to the curve is parallel to the x-axis. Find the equation of the tangent to the curve at the other point, giving your answer in the form ax by c 0. [7] + + =

9 marks

Mark scheme: 8 (i) EITHER: Substitute x = 1 and attempt to solve 3-term quadratic in y M1 Obtain answers (1, 1) and (1, –3) A1 OR: State answers (1, 1) and (1, –3) B1 + B1 [2] dy (ii) State 2y as derivative of y2 B1 dx dy State 2y + 2x as derivative of 2xy B1 dx dy Substitute for x and y, and solve for M1 dx d y Obtain = 0 when x = 1 and y = 1 A1 dx d y Obtain = –2 when x = 1 and y = –3 A1√ dx Form the equation of the tangent at (1, –3) M1 Obtain answer 2x + y + 1 = 0 A1 [7]

This question in 9709/21 Oct/Nov 2009

7 y M 1 x O 2p The diagram shows the curve y x2 cos x, for 0 2π, and its maximum point M. = ≤x ≤1 (i) Show by differentiation that the x-coordinate of M satisfies the equation 2 tan x x. = [4] (ii) Verify by calculation that this equation has a root (in radians) between 1 and 1.2. [2] 2 (iii) Use the iterative formula tan−1 to determine this root correct to 2 decimal places. xn+1 = xn Give the result of each iteration to 4 decimal places. [3]

9 marks

Mark scheme: 7 (i) Use product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and express tan x in terms of x M1 Obtain given answer A1 [4] 2 (ii) Consider sign of tan x – at x = 1 and x = 1.2, or equivalent M1 x Complete the argument with correct calcuations A1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.08 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (1.075, 1.085) A1 [3] 1

This question in 9709/22 Oct/Nov 2009

cos x dy 8 (i) By differentiating , show that if y cot x then [3] sin x dx = = −cosec2x. (ii) By expressing cot2x in terms of cosec2x and using the result of part (i), show that 12π cot2x dx 1 4π. 1 ã 4π = −1 [4] 1 1 (iii) Express cos 2x in terms of sin2x and hence show that can be expressed as 2 cosec2x. 1 2x Hence, using the result of part (i), find −cos 1 dx. ä 1 2x [3] −cos

10 marks

Mark scheme: 8 (i) Use quotient rule M1 Obtain correct derivative in any form A1 Obtain given result correctly A1 [3] (ii) State cot2 x ≡ –1 + cos ec2x , or equivalent B1 Obtain integral –x – cotx (f.t. on signs in the identity) B1√ Substitute correct limits correctly M1 Obtain given answer A1 [4] 1 (iii) Use trig formulae to convert integrand to 2 where k = ±2, or ±1 M1 k sin x Obtain given answer 1 cos ec2x correctly A1 2 Obtain answer – 1 cot x + c, or equivalent B1 [3] 2

This question in 9709/21 May/June 2010

5 The equation of a curve is y = x3e−x. (i) Show that the curve has a stationary point where x 3. [3] = (ii) Find the equation of the tangent to the curve at the point where x 1. [4] =

7 marks

Mark scheme: 5 (i) Use product rule M1 Obtain correct derivative in any form A1 Show that derivative is equal to zero when x = 3 A1 [3] (ii) Substitute x = 1 into gradient function, obtaining 2e–l or equivalent M1 State or imply required y-coordinate is e–1 B1 Form equation of line through (l, e–1) with gradient found (NOT the normal) M1 Obtain equation in any correct form A1 [4] 2

This question in 9709/22 May/June 2010

5 The equation of a curve is y = x3e−x. (i) Show that the curve has a stationary point where x 3. [3] = (ii) Find the equation of the tangent to the curve at the point where x 1. [4] =

7 marks

Mark scheme: 5 (i) Use product rule M1 Obtain correct derivative in any form A1 Show that derivative is equal to zero when x = 3 A1 [3] (ii) Substitute x = 1 into gradient function, obtaining 2e–l or equivalent M1 State or imply required y-coordinate is e–1 B1 Form equation of line through (l, e–1) with gradient found (NOT the normal) M1 Obtain equation in any correct form A1 [4] 2

This question in 9709/23 May/June 2010

4 The parametric equations of a curve are 9 x 1 y t , for t 2. = + ln(t −2), = + t > dy (i) Show that (t2 −9)(t −2) . [3] dx = t2 (ii) Find the coordinates of the only point on the curve at which the gradient is equal to 0. [3]

6 marks

Mark scheme: dx 1 dy − 24 (i) State = or = 1 − 9t B1 dt t − 2 dt dy dy dx Use = ÷ M1 dx dt dt Obtain given answer correctly A1 [3] (ii) Equate derivative to zero and solve for t M1 State or imply that t = 3 is admissible c.w.o., and note t = –3, 2 cases A1 Obtain coordinates (1, 6) and no others A1 [3]

This question in 9709/21 Oct/Nov 2010

8 y Q p x O The diagram shows the curve y x sin x, for 0 The point Q 12π, 12π lies on the curve. = ≤x ≤π. (i) Show that the normal to the curve at Q passes through the point [5] (π, 0). d (ii) Find x cos [2] dx(sin −x x). 12π (iii) Hence evaluate x sin x dx. [3] ã 0

10 marks

Mark scheme: 8 (i) Use product rule M1 Obtain correct derivative in any form A1 1 Substitute x = π , and obtain gradient of –1 for normal A1√ 2 from y ′ = sin x − x cos x ONLY  1 1  Show that line through  π , π  with gradient –1 passes through (π , 0 ) M1  2 2  A1 [5] (ii) Differentiate sin x and use product rule to differentiate xcos x M1 Obtain xsin x , or equivalent A1 [2] (iii) State that integral is sin x − x cos x (+ c ) B1 π Substitute limits 0 and correctly M1 2 Obtain answer 1 A1 [3] S. R. Feeding limits into original integrand, 0/3

This question in 9709/21 Oct/Nov 2010

4 The parametric equations of a curve are 9 x 1 y t , for t 2. = + ln(t −2), = + t > dy (i) Show that (t2 −9)(t −2) . [3] dx = t2 (ii) Find the coordinates of the only point on the curve at which the gradient is equal to 0. [3]

6 marks

Mark scheme: dx 1 dy − 24 (i) State = or = 1 − 9t B1 dt t − 2 dt dy dy dx Use = ÷ M1 dx dt dt Obtain given answer correctly A1 [3] (ii) Equate derivative to zero and solve for t M1 State or imply that t = 3 is admissible c.w.o., and note t = –3, 2 cases A1 Obtain coordinates (1, 6) and no others A1 [3]

This question in 9709/22 Oct/Nov 2010

8 y Q p x O The diagram shows the curve y x sin x, for 0 The point Q 12π, 12π lies on the curve. = ≤x ≤π. (i) Show that the normal to the curve at Q passes through the point [5] (π, 0). d (ii) Find x cos [2] dx(sin −x x). 12π (iii) Hence evaluate x sin x dx. [3] ã 0

10 marks

Mark scheme: 8 (i) Use product rule M1 Obtain correct derivative in any form A1 1 Substitute x = π , and obtain gradient of –1 for normal A1√ 2 from y ′ = sin x − x cos x ONLY  1 1  Show that line through  π , π  with gradient –1 passes through (π , 0 ) M1  2 2  A1 [5] (ii) Differentiate sin x and use product rule to differentiate xcos x M1 Obtain xsin x , or equivalent A1 [2] (iii) State that integral is sin x − x cos x (+ c ) B1 π Substitute limits 0 and correctly M1 2 Obtain answer 1 A1 [3] S. R. Feeding limits into original integrand, 0/3

This question in 9709/22 Oct/Nov 2010

7 y M x O 1 ln x The diagram shows the curve y and its maximum point M. = x2 (i) Find the exact coordinates of M. [5] (ii) Use the trapezium rule with three intervals to estimate the value of 4 ln x dx, x2 ä 1 giving your answer correct to 2 decimal places. [3]

8 marks

Mark scheme: 7 (i) Use product or quotient rule M1* Obtain correct derivative in any form A1 Equate derivative to zero and solve for x M1*(dep) Obtain x = e0.5 or e A1 1 Obtain , or equivalent A1 [5] 2e (ii) State or imply correct ordinates 0, 0.17328..., 0.12206..., 0.08664... B1 Use correct formula, or equivalent, correctly with h = 1 and four ordinates M1 Obtain answer 0.34 with no errors seen A1 [3] dy 2

This question in 9709/23 Oct/Nov 2010

2 A curve has parametric equations x 3t sin 2t, y 4 2 cos 2t. = + = + Find the exact gradient of the curve at the point for which t 16π. [4] =

4 marks

Mark scheme: dx dy 2 State = 3 + 2 cos 2t or = −4 sin 2t (or both) B1 dt dt d y d y d x Use = ÷ M1 d x d t d t − 4 sin 2t Obtain or imply A1 3 + 2 cos 2t 1 1 Substitute π to obtain − 3 or exact equivalent A1 [4] 6 2

This question in 9709/21 May/June 2011

5 A curve has equation x2 2y2 5x 6y 10. Find the equation of the tangent to the curve at the + + + = point Give your answer in the form ax by c 0, where a, b and c are integers. [6] (2, −1). + + =

6 marks

Mark scheme: dy 5 Obtain 4 y as derivative of 2y2 B1 dx dy Differentiate LHS term by term to obtain expression including at least one M1 dx dy dy Obtain 2 x + 4 y + 5 + 6 A1 dx dx dy Substitute 2 and –1 to attempt value of M1 dx 9 Obtain − A1 2 Obtain equation 9x + 2y –16 = 0 or equivalent of required form A1 [6] GCE AS/A LEVEL – May/June 2011 9709 22

This question in 9709/22 May/June 2011

6 The curve y 4x2 ln x has one stationary point. = (i) Find the coordinates of this stationary point, giving your answers correct to 3 decimal places. [5] (ii) Determine whether this point is a maximum or a minimum point. [2]

7 marks

Mark scheme: 6 (i) Attempt differentiation using product rule M1 Obtain 8 x ln x + 4 x (a.c.f.) A1 Equate first derivative to zero and attempt solution M1 Obtain 0.607 A1 Obtain –0.736 following their x-coordinate A1√ [5] (ii) Use an appropriate method for determining nature of stationary point M1 Conclude point is a minimum (with no errors seen, second derivative = 8) A1 [2]

This question in 9709/22 May/June 2011

5 A curve has equation x2 2y2 5x 6y 10. Find the equation of the tangent to the curve at the + + + = point Give your answer in the form ax by c 0, where a, b and c are integers. [6] (2, −1). + + =

6 marks

Mark scheme: dy 5 Obtain 4 y as derivative of 2y2 B1 dx dy Differentiate LHS term by term to obtain expression including at least one M1 dx dy dy Obtain 2 x + 4 y + 5 + 6 A1 dx dx dy Substitute 2 and –1 to attempt value of M1 dx 9 Obtain − A1 2 Obtain equation 9x + 2y –16 = 0 or equivalent of required form A1 [6] GCE AS/A LEVEL – May/June 2011 9709 23

This question in 9709/23 May/June 2011

6 The curve y 4x2 ln x has one stationary point. = (i) Find the coordinates of this stationary point, giving your answers correct to 3 decimal places. [5] (ii) Determine whether this point is a maximum or a minimum point. [2]

7 marks

Mark scheme: 6 (i) Attempt differentiation using product rule M1 Obtain 8 x ln x + 4 x (a.c.f.) A1 Equate first derivative to zero and attempt solution M1 Obtain 0.607 A1 Obtain –0.736 following their x-coordinate A1√ [5] (ii) Use an appropriate method for determining nature of stationary point M1 Conclude point is a minimum (with no errors seen, second derivative = 8) A1 [2]

This question in 9709/23 May/June 2011

3 y 1 1 x O 4 p 2 p The diagram shows the part of the curve y 1 tan 2x for 0 Find the x-coordinates of the 2 2π. = ≤x ≤1 points on this part of the curve at which the gradient is 4. [5]

5 marks

Mark scheme: 1 3 Obtain derivative of the form k sec2 2x, where k = 1 or k = M1 2 Obtain correct derivative sec2 2x A1 Use correct method for solving sec2 2x = 4 M1 1 Obtain answer x = π (or 0.524 radians) A1 6 1 Obtain answer x = π (or 1.05 radians) and no others in range A1 [5] 3

This question in 9709/21 Oct/Nov 2011

3 y M x O The diagram shows the curve y x ln x and its minimum point M. = −2 (i) Find the x-coordinate of M. [2] (ii) Use the trapezium rule with three intervals to estimate the value of 5 ln dx, ã 2 (x −2 x) giving your answer correct to 2 decimal places. [3] (iii) State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the integral in part (ii). [1]

6 marks

Mark scheme: 3 (i) Obtain correct derivative B1 Obtain x = 2 only B1 [2] (ii) State or imply correct ordinates 0.61370..., 0.80277..., 1.22741..., 1.78112... B1 Use correct formula, or equivalent, correctly with h = 1 and four ordinates Ml Obtain answer 3.23 with no errors seen Al [3] (iii) Justify statement that the trapezium rule gives an over-estimate B1 [1]

This question in 9709/22 Oct/Nov 2011

1 Find the gradient of the curve y at the point where x 4. [3] = ln(5x + 1) =

3 marks

Mark scheme: k 1 1 Obtain derivative of the form , where k = 1, 5 or M1 5 x + 1 5 5 Obtain correct derivative A1 5 x + 1 5 Substitute x = 4 into expression for derivative and obtain A1√ [3] 21 2 2

This question in 9709/23 Oct/Nov 2011

7 y x O P 1 The diagram shows the curve y The curve has a gradient of 3 at the point P. = (x −4)e 2x. (i) Show that the x-coordinate of P satisfies the equation 2 x = + 6e −12x. [4] (ii) Verify that the equation in part (i) has a root between x 3.1 and x 3.3. [2] = = 2xn 2 6e−1 to determine this root correct to 2 decimal places. + (iii) Use the iterative formula xn+1 = Give the result of each iteration to 4 decimal places. [3]

9 marks

Mark scheme: 7 (i) At any stage, state the correct derivative of e 2 B1 Use product rule M1 Obtain correct derivative in any form A1 Equate derivative to 3 and obtain given equation correctly A1 [4] 1 − x (ii) Consider sign of 2 + 6e 2 – x, or equivalent M1 Complete the argument correctly with appropriate calculations A1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 3.21 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (3.205, 3.215) B1 [3] dy 2

This question in 9709/23 Oct/Nov 2011

5 y M x O 12x The diagram shows the curve y 4e 3 and its minimum point M. = −6x + (i) Show that the x-coordinate of M can be written in the form ln a, where the value of a is to be stated. [5] (ii) Find the exact value of the area of the region enclosed by the curve and the lines x 0, x 2 = = and y 0. [4] =

9 marks

Mark scheme: 2 x5 (i) Differentiate to obtain expression of form ke + m M1 1 2 x Obtain correct 2e − 6 A1 Equate attempt at first derivative to zero and attempt solution DM1 Obtain 12 x = ln 3 or equivalent A1 Conclude x = ln 9 or a = 9 A1 [5] 1 2 x 2 (ii) Integrate to obtain expression of form ae + bx + cx M1 1 2 x 2 Obtain correct 8e − 3 x + 3 x A1 Substitute correct limits and attempt simplification DM1 Obtain 8e – 14 A1 [4] GCE AS/A LEVEL – May/June 2012 9709 21 3

This question in 9709/21 May/June 2012

6 A curve has parametric equations 1 x , y = = √(t + 2). (2t + 1)2 The point P on the curve has parameter p and it is given that the gradient of the curve at P is −1. 1 (i) Show that p 6 2. [6] = (p + 2) −1 (ii) Use an iterative process based on the equation in part (i) to find the value of p correct to 3 decimal places. Use a starting value of 0.7 and show the result of each iteration to 5 decimal places. [3]

9 marks

Mark scheme: 6 (i) Obtain derivative of form k(2t + 1)–3 M1 Obtain –4(2t + 1)–3 or equivalent as derivative of x A1 1 − 12 Obtain 2 (t + 2) or equivalent as derivative of y B1 dy Equate attempt at to –1 M1 dx 3 12 Obtain ( 2 p + )1 = 8( p + 2) or equivalent A1 1 Confirm given answer p = ( p + 2) 6 − 12 A1 [6] (ii) Use iteration process correctly at least once M1 Obtain final answer 0.678 A1 Show sufficient iterations to 5 decimal places to justify answer or show a sign change in the interval (0.6775, 0.6785) A1 [3] [0.7 → 0.68003 → 0.67857 → 0.67847 → 0.67846] 2 2

This question in 9709/21 May/June 2012

5 The parametric equations of a curve are x y e2t 2t. = ln(t + 1), = + dy (i) Find an expression for in terms of t. [4] dx (ii) Find the equation of the normal to the curve at the point for which t 0. Give your answer in = the form ax by c 0, where a, b and c are integers. [4] + + =

8 marks

Mark scheme: d x 1 5 (i) State = B1 d t t + 1 dy 2 t State = 2e + 2 B1 dt dy Attempt expression for M1 dx dy 2 t Obtain = ( 2e + 2)(t + )1 or equivalent A1 [4] dx (ii) Substitute t = 0 and attempt gradient of normal M1 dy 1 Obtain − 4 following their expression for A1√ dx Attempt to find equation of normal through point (0, 1) M1 Obtain x + 4y – 4 = 0 A1 [4] GCE AS/A LEVEL – May/June 2012 9709 22

This question in 9709/22 May/June 2012

6 y M 1 x O a p 2 sin 2x for 0 The diagram shows the curve y The x-coordinate of the maximum point M ≤x ≤12π. = x 2 is denoted by α. + dy (i) Find and show that α satisfies the equation tan 2x 2x 4. [4] dx = + (ii) Show by calculation that α lies between 0.6 and 0.7. [2] 2 tan−1(2xn + 4) (iii) Use the iterative formula xn+1 = 1 to find the value of α correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3]

9 marks

Mark scheme: 6 (i) Attempt use of quotient rule or equivalent M1 2( x + 2) cos 2 x − sin 2 x Obtain or equivalent A1 ( x + 22) Equate numerator to zero and attempt rearrangement M1 Confirm given result tan 2x = 2x + 4 A1 [4] (ii) Consider sign of tan 2x – 2x – 4 for 0.6 and 0.7 or equivalent M1 Obtain –2.63 and 0.40 or equivalents and justify conclusion A1 [2] (iii) Use iteration process correctly at least once M1 Obtain final answer 0.694 A1 Show sufficient iterations to 5 decimal places to justify answer or show a sign change in the interval (0.6935, 0.6945) A1 [3] [0.6 → 0.69040 → 0.69352 → 0.69363 0.65 → 0.69215 → 0.69358 → 0.69363 0.7 → 0.69384 → 0.69364 → 0.69363] 2 2

This question in 9709/22 May/June 2012

5 y M x O 12x The diagram shows the curve y 4e 3 and its minimum point M. = −6x + (i) Show that the x-coordinate of M can be written in the form ln a, where the value of a is to be stated. [5] (ii) Find the exact value of the area of the region enclosed by the curve and the lines x 0, x 2 = = and y 0. [4] =

9 marks

Mark scheme: 2 x5 (i) Differentiate to obtain expression of form ke + m M1 1 2 x Obtain correct 2e − 6 A1 Equate attempt at first derivative to zero and attempt solution DM1 Obtain 12 x = ln 3 or equivalent A1 Conclude x = ln 9 or a = 9 A1 [5] 1 2 x 2 (ii) Integrate to obtain expression of form ae + bx + cx M1 1 2 x 2 Obtain correct 8e − 3 x + 3 x A1 Substitute correct limits and attempt simplification DM1 Obtain 8e – 14 A1 [4] GCE AS/A LEVEL – May/June 2012 9709 23 3

This question in 9709/23 May/June 2012

6 A curve has parametric equations 1 x , y = = √(t + 2). (2t + 1)2 The point P on the curve has parameter p and it is given that the gradient of the curve at P is −1. 1 (i) Show that p 6 2. [6] = (p + 2) −1 (ii) Use an iterative process based on the equation in part (i) to find the value of p correct to 3 decimal places. Use a starting value of 0.7 and show the result of each iteration to 5 decimal places. [3]

9 marks

Mark scheme: 6 (i) Obtain derivative of form k(2t + 1)–3 M1 Obtain –4(2t + 1)–3 or equivalent as derivative of x A1 1 − 12 Obtain 2 (t + 2) or equivalent as derivative of y B1 dy Equate attempt at to –1 M1 dx 3 12 Obtain ( 2 p + )1 = 8( p + 2) or equivalent A1 1 Confirm given answer p = ( p + 2) 6 − 12 A1 [6] (ii) Use iteration process correctly at least once M1 Obtain final answer 0.678 A1 Show sufficient iterations to 5 decimal places to justify answer or show a sign change in the interval (0.6775, 0.6785) A1 [3] [0.7 → 0.68003 → 0.67857 → 0.67847 → 0.67846] 2 2

This question in 9709/23 May/June 2012

4 The parametric equations of a curve are 2 x y , for t 0. t = ln(1 −2t), = < dy 1 (i) Show that . [3] −2t dx = t2 (ii) Find the exact coordinates of the only point on the curve at which the gradient is 3. [3]

6 marks

Mark scheme: dx 2 dy − 24 (i) State = or = −2t B1 dt 1 − 2t dt dy dy dx Use = ÷ M1 dx dt dt Obtain given answer correctly A1 [3] (ii) Equate derivative to 3 and solve for t M1 State or imply that t = –1 c.w.o. A1 Obtain coordinates (ln 3, –2) A1 [3] GCE AS LEVEL – October/November 2012 9709 21

This question in 9709/21 Oct/Nov 2012

1 dy 8 (i) By differentiating , show that if y sec θ then tan θ sec θ. [3] cos θ dθ = = (ii) Hence show that d2y a sec3θ bsec θ, dθ2 = + giving the values of a and b. [4] (iii) Find the exact value of 14π tan2θ sec θ tan dθ. ã 0 (1 + −3 θ) [5]

12 marks

Mark scheme: 8 (i) Differentiate using chain or quotient rule M1 Obtain derivative in any correct form A1 Obtain given answer correctly A1 [3] (ii) Differentiate using product rule M1 State derivative of tan θ = sec2 θ B1 Use trig identity 1 + tan2 θ = sec2 θ correctly M1 Obtain 2sec3 θ – sec θ A1 [4] (iii) Use tan 2 x = sec 2 θ − 1 to integrate tan 2 x M1 Obtain 3sec θ from integration of 3sec θ tan θ B1 Obtain tan θ − 3sec θ A1 Attempt to substitute limits, using exact values M1 Obtain answer 4 − 3 2 A1 [5]

This question in 9709/21 Oct/Nov 2012

4 The parametric equations of a curve are 2 x y , for t 0. t = ln(1 −2t), = < dy 1 (i) Show that . [3] −2t dx = t2 (ii) Find the exact coordinates of the only point on the curve at which the gradient is 3. [3]

6 marks

Mark scheme: dx 2 dy − 24 (i) State = or = −2t B1 dt 1 − 2t dt dy dy dx Use = ÷ M1 dx dt dt Obtain given answer correctly A1 [3] (ii) Equate derivative to 3 and solve for t M1 State or imply that t = –1 c.w.o. A1 Obtain coordinates (ln 3, –2) A1 [3] GCE AS LEVEL – October/November 2012 9709 23

This question in 9709/23 Oct/Nov 2012

1 dy 8 (i) By differentiating , show that if y sec θ then tan θ sec θ. [3] cos θ dθ = = (ii) Hence show that d2y a sec3θ bsec θ, dθ2 = + giving the values of a and b. [4] (iii) Find the exact value of 14π tan2θ sec θ tan dθ. ã 0 (1 + −3 θ) [5]

12 marks

Mark scheme: 8 (i) Differentiate using chain or quotient rule M1 Obtain derivative in any correct form A1 Obtain given answer correctly A1 [3] (ii) Differentiate using product rule M1 State derivative of tan θ = sec2 θ B1 Use trig identity 1 + tan2 θ = sec2 θ correctly M1 Obtain 2sec3 θ – sec θ A1 [4] (iii) Use tan 2 x = sec 2 θ − 1 to integrate tan 2 x M1 Obtain 3sec θ from integration of 3sec θ tan θ B1 Obtain tan θ − 3sec θ A1 Attempt to substitute limits, using exact values M1 Obtain answer 4 − 3 2 A1 [5]

This question in 9709/23 Oct/Nov 2012

2 Find the gradient of each of the following curves at the point for which x 0. = (i) y 3 sin x tan 2x [3] = + 6 (ii) y [3] = 1 e2x +

6 marks

Mark scheme: 2 (i) Differentiate to obtain form k1 cos x + k 2 sec 2 2 x M1 Obtain correct second term 2 sec 2 2 x A1 Obtain 3 cos x + 2 sec 2 2 x and hence answer 5 A1 [3] 2 x 2 x −2 (ii) Differentiate to obtain form ke ( 1 + e ) M1 2 x 2 x −2 Obtain correct − 12e ( 1 + e ) or equivalent (may be implied) A1 Obtain –3 A1 [3]

This question in 9709/21 May/June 2014

5 y M x O The diagram shows part of the curve y 2 cos x 2x = −cos and its maximum point M. The shaded region is bounded by the curve, the axes and the line through M parallel to the y-axis. (i) Find the exact value of the x-coordinate of M. [4] (ii) Find the exact value of the area of the shaded region. [4]

8 marks

Mark scheme: 5 (i) Differentiate to obtain − 2 sin x + 2 sin 2 x or equivalent B1 Use sin 2 x = 2 sin x cos x or equivalent B1 Equate first derivative to zero and solve for x M1 Obtain 13 π A1 [4] (ii) Integrate to obtain form k1 sin x + k 2 sin 2 x M1 Obtain correct 2 sin x − 12 sin 2 x A1 Apply limits 0 and their answer from part (i) M1 Obtain 34 3 or exact equivalent A1 [4]

This question in 9709/22 Oct/Nov 2014

3 The equation of a curve is y 6 sin x cos 2x. = −2 Find the equation of the tangent to the curve at the point 1 2 . Give the answer in the form 60, y mx c, where the values of m and c are correct to 3 significant figures. [5] = +

5 marks

Mark scheme: 3 Differentiate to obtain form p cos x + q sin 2 x or equivalent M1 Obtain correct 6cos x + 4sin 2 x or equivalent A1 Substitute 16π to obtain derivative equal to 5 3 or 8.66 A1 Form equation of tangent (not normal) using numerical value of gradient obtained by differentiation M1 Obtain y = 8.66 x − 2.53 cao A1 [5]

This question in 9709/21 May/June 2015

4 y M x O The diagram shows the curve y = ex + 4e−2x and its minimum point M. (i) Show that the x-coordinate of M is ln 2. [3] (ii) The region shaded in the diagram is enclosed by the curve and the lines x = 0, x = ln 2 and y = 0. Use integration to show that the area of the shaded region is 2.5 [4]

7 marks

Mark scheme: 4 (i) Differentiate to obtain e x − 8e −2 x B1 Use correct process to solve equation of form a e x + b e −2 x = 0 M1 Confirm given answer ln 2 correctly A1 [3] (ii) Integrate to obtain expression of form p e x + q e − 2 x M1 Obtain correct e x − 2e −2 x A1 Apply both limits correctly M1 depM 5 Confirm given answer A1 [4] 2 4

This question in 9709/22 May/June 2015

4 y M x O The diagram shows the curve y = ex + 4e−2x and its minimum point M. (i) Show that the x-coordinate of M is ln 2. [3] (ii) The region shaded in the diagram is enclosed by the curve and the lines x = 0, x = ln 2 and y = 0. Use integration to show that the area of the shaded region is 2.5 [4]

7 marks

Mark scheme: 4 (i) Differentiate to obtain e x − 8e −2 x B1 Use correct process to solve equation of form a e x + b e −2 x = 0 M1 Confirm given answer ln 2 correctly A1 [3] (ii) Integrate to obtain expression of form p e x + q e − 2 x M1 Obtain correct e x − 2e −2 x A1 Apply both limits correctly M1 depM 5 Confirm given answer A1 [4] 2 4

This question in 9709/23 May/June 2015

2 A curve has equation 3x 1 y + . = x −5 Find the coordinates of the points on the curve at which the gradient is [5] −4.

5 marks

Mark scheme: 2 Use quotient rule or, after adjustment, product rule M1* 3 x − 15 − 3 x − 1 Obtain or equivalent A1 ( x − 52) Equate first derivative to –4 and solve for x M1 dep Obtain x-coordinates 3 and 7 or one correct pair of coordinates A1 Obtain y-coordinates –5 and 11 respectively or other correct pair of coordinates A1 [5]

This question in 9709/21 Oct/Nov 2015

7 y A D B x O C The parametric equations of a curve are x 6 sin2t, y 2 sin 2t 3 cos 2t, = = + for 0 The curve crosses the x-axis at points B and D and the stationary points are A and C, as ≤t < 0. shown in the diagram. dy 2 (i) Show that cot 2t [5] 3 −1. dx = (ii) Find the values of t at A and C, giving each answer correct to 3 decimal places. [3] (iii) Find the value of the gradient of the curve at B. [3]

11 marks

Mark scheme: dx 7 (i) Obtain 12 sin t cos t or equivalent for B1 dt dy Obtain 4 cos 2t − 6 sin 2t or equivalent for B1 dt dy Obtain expression for in terms of t M1 dx Use 2 sin t cos t = sin 2t A1 dy 2 Confirm given answer = cot 2t − 1 with no errors seen A1 [5] dx 3 2 (ii) State or imply tan 2t = B1 3 Obtain t = .0 294 B1 Obtain t = .1 865 B1 [3] (iii) Attempt solution of 2 sin 2t + 3 cos 2t = 0 at least as far as tan t2 = K M1 3 Obtain tan 2t = − or equivalent A1 2 13 Substitute to obtain − A1 [3] 9

This question in 9709/21 Oct/Nov 2015

6 y M x O 120 The diagram shows the part of the curve y sin 2x for 0 and the stationary point M. = 3e−x ≤x ≤120, (i) Find the equation of the tangent to the curve at the origin. [4] (ii) Find the coordinates of M, giving each coordinate correct to 3 decimal places. [4]

8 marks

Mark scheme: 6 (i) Use product rule to obtain expression of form k1e − x sin 2 x + k 2 e − x cos2 x M1 Obtain correct −3e − x sin2 x + 6e − x cos2 x A1 Substitute x = 0 in first derivative to obtain equation of form y = mx M1 Obtain y = 6 x or equivalent with no errors in solution A1 [4] (ii) Equate first derivative to zero and obtain tan2 x = k M1* Carry out correct process to find value of x dep M1* Obtain x = 0.554 A1 Obtain y = 1.543 A1 [4] dy

This question in 9709/22 Feb/March 2016

3x2 6 The equation of a curve is y At the point on the curve with positive x-coordinate p, the = x2 4. gradient of the curve is 1 + 2. _P48p Q (i) Show that p . [5] −16 = p2 8 + (ii) Show by calculation that 2 p 3. [2] < < (iii) Use an iterative formula based on the equation in part (i) to find the value of p correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3]

10 marks

Mark scheme: 6 (i) Use quotient rule or equivalent *M1 6 x ( x 2 + 4) − 6 x 3 Obtain 2 2 or equivalent A1 ( x + 4) Equate first derivative to 12 and remove algebraic denominators dep on *M1 DM1 Obtain 48 p = p 4 + 8 p 2 + 16 or 48 x = x 4 + 8 x 2 + 16 or equivalent A1 48 p − 16 Confirm given result p = 2 A1 [5] p + 8 48 p − 16 (ii) Consider sign of p − 2 at 2 and 3 or equivalent M1 p + 8 Complete argument correctly with appropriate calculations A1 [2] (iii) Carry out iteration process correctly at least once M1 Obtain final answer 2.728 A1 Show sufficient iterations to justify accuracy to 4 sf or show sign change in interval (2.7275, 2.7285) B1 [3] 2 2

This question in 9709/21 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 9709/22 May/June 2016

3 A curve has equation y 2 sin 2x cos 2x 6 and is defined for 0 Find the x-coordinates of the stationary points of= the curve,−5giving your+ answers correct to 3≤xsignificant≤0. figures. [6]

6 marks

Mark scheme: 3 Differentiate to obtain 4cos2 x + 10sin 2 x B1 Equate first derivative to zero and arrange to tan 2 x = ... *M1 Obtain tan 2 x = − 0.4 A1 Carry out correct method for finding at least one value of x, dependent *M DM1 Obtain x = 1.38 A1 Obtain x = 2.95 and no others between 0 and π A1 [6] 2

This question in 9709/21 Oct/Nov 2016

7 The parametric equations of a curve are x t3 6t 1, y t4 4t2 5. = + + = −2t3 + −12t + dy dy (i) Find and use division to show that can be written in the form at b, where a and b are dx dx + constants to be found. [5] … … … … … … … … … … … … … … … … … … … … … … (ii) The straight line x 9 0 is the normal to the curve at the point P. Find the coordinates −2y + = of P. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 7(i) Differentiate x and y and form d d y x M1 Obtain 3 2 2 4 6 8 12 3 6 t t t t − + − + A1 First 2 marks may be implied by an attempt at division Carry out division at least as far as kt or equivalent M1 For M1, it must be division by a quadratic factor. Allow attempt at factorisation with same conditions as for division Obtain 4 3 t A1 Obtain 4 3 2 t − with complete division shown and no errors seen A1 Total: 5 Question Answer Marks Guidance 7(ii) State or imply gradient of straight line is 1 2 B1 Allow B1 if 1 9 2 2 y x = + is seen Attempt value of t from their d d y x = their negative reciprocal of gradient of line M1 Obtain 0 t = and hence ( ) 1,5 A1 Total: 3

This question in 9709/21 May/June 2017

e4x 4 Find the equation of the tangent to the curve y at the point on the curve for which x 0. 2x 3 = = Give your answer in the form ax by c 0 where a,+b and c are integers. [5] + + = … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 4 Use quotient rule (or product rule) to find first derivative M1 Obtain ( ) 4 4 2 8 e 10e 2 3 x x x x + + or equivalent A1 Substitute 0 x = to obtain gradient 10 9 A1 Form equation of tangent through ( ) 1 3 0, with numerical gradient M1 Obtain 10 9 3 0 x y − + = A1 Total: 5

This question in 9709/22 May/June 2017

e4x 4 Find the equation of the tangent to the curve y at the point on the curve for which x 0. 2x 3 = = Give your answer in the form ax by c 0 where a,+b and c are integers. [5] + + = … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 4 Use quotient rule (or product rule) to find first derivative M1 Obtain ( ) 4 4 2 8 e 10e 2 3 x x x x + + or equivalent A1 Substitute 0 x = to obtain gradient 10 9 A1 Form equation of tangent through ( ) 1 3 0, with numerical gradient M1 Obtain 10 9 3 0 x y − + = A1 Total: 5

This question in 9709/23 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 9709/23 May/June 2017

7 y P Q x O The diagram shows the curve y x2 3x 1 5 cos 12x. = + + + The curve crosses the y-axis at the point P and the gradient of the curve at P is m. The point Q on the curve has x-coordinate q and the gradient of the curve at Q is −m. (i) Find the value of m and hence show that q satisfies the equation x a sin 12x b, = + where the values of the constants a and b are to be determined. [4] … … … … … … … … … … … … … … (ii) Show by calculation that q [2] −4.5 < < −4.0. … … … … … … … … … (iii) Use an iterative formula based on the equation in part (i) to find the value of q correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(i) Differentiate to obtain form k1 x + k 2 + k 3 sin 12 x *M1 Obtain correct 2 x + 3 − 52 sin 12 x and deduce or A1 imply gradient at P is 3 Equate first derivative to their − 3 and rearrange DM1 Obtain x = 54 sin 12 x − 3 A1 4 7(ii) Consider sign of their 2 x + 6 − 52 sin 12 x at − 4.5 M1 and − 4.0 or equivalent Complete argument correctly for correct A1 expression with appropriate calculations 2 7(iii) Use iteration formula correctly at least once M1 Obtain final answer − 4.11 A1 Show sufficient iterations to justify accuracy to A1 3 sf or show sign change in interval ( −4.115, − 4.105) 3

This question in 9709/21 Oct/Nov 2017

3 The equation of a curve is y tan 12x 3 sin 12x. The curve has a stationary point M in the interval = + x Find the coordinates of M, giving each coordinate correct to 3 significant figures. [6] 0 < < 20. … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 3 Differentiate to obtain form M1 If a factor of 0.5 is missed, can still k1 sec 2 12 x + k 2 cos 12 x get 5/6, penalise at first A1 Obtain 12 sec 2 12 x + 23 cos 12 x A1 Equate first derivative to zero and produce *M1 cos 3 12 x = k 3 Use correct process to find one value of x DM1 Dep on *M, allow for obtaining 1.609….. , 92.2 o or 268 o Obtain x = 4.67 A1 Allow x = 4.67 or better for A1 Obtain y = 1.12 A1 Allow y = 1.12 from x = 4.66 but nothing else 6

This question in 9709/22 Oct/Nov 2017

5 y 0, 9 P Q x O The diagram shows the curve y and a straight line. The curve crosses the y-axis at the point P. The straight line crosses the y-axis= 4e−2xat the point 0, 9 and its gradient is equal to the gradient of the curve at P. The straight line meets the curve at two points, one of which is Q as shown. (i) Show that the x-coordinate of Q satisfies the equation x 9 [6] 8 = −12e−2x. … … … … … … … … … … … … … … … (ii) Use an iterative formula based on the equation in part (i) to find the x-coordinate of Q correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 5(i) Obtain derivative of the form ke − 2 x *M1 Condone k = 4 for M1 State or imply gradient of curve at P is –8 A1 Form equation of straight line through (0, 9) *DM1 dep on *M with negative gradient Obtain y = − 8 x + 9 or equivalent A1 Equate equation of curve and equation of DM1 dep on both *M straight line Rearrange to confirm x = 89 − 12 e − 2 x A1 6 5(ii) Use iterative process correctly at least once M1 Obtain final answer 1.07 A1 Show sufficient iterations to 5 sf to justify A1 answer or show sign change in interval (1.065,1.075) 6

This question in 9709/22 Oct/Nov 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 9709/23 Oct/Nov 2017

7 y P Q x O The diagram shows the curve y x2 3x 1 5 cos 12x. = + + + The curve crosses the y-axis at the point P and the gradient of the curve at P is m. The point Q on the curve has x-coordinate q and the gradient of the curve at Q is −m. (i) Find the value of m and hence show that q satisfies the equation x a sin 2x1 b, = + where the values of the constants a and b are to be determined. [4] … … … … … … … … … … … … … … (ii) Show by calculation that q [2] −4.5 < < −4.0. … … … … … … … … … (iii) Use an iterative formula based on the equation in part (i) to find the value of q correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(i) Differentiate to obtain form k1 x + k 2 + k 3 sin 12 x *M1 Obtain correct 2 x + 3 − 52 sin 12 x and deduce or A1 imply gradient at P is 3 Equate first derivative to their − 3 and rearrange DM1 Obtain x = 54 sin 12 x − 3 A1 4 7(ii) Consider sign of their 2 x + 6 − 52 sin 12 x at − 4.5 M1 and − 4.0 or equivalent Complete argument correctly for correct A1 expression with appropriate calculations 2 7(iii) Use iteration formula correctly at least once M1 Obtain final answer − 4.11 A1 Show sufficient iterations to justify accuracy to A1 3 sf or show sign change in interval ( −4.115, − 4.105) 3

This question in 9709/23 Oct/Nov 2017

2 A curve has equation y 4x sin 2x.1 Find the equation of the tangent to the curve at the point for which = x [4] = 0. … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 Differentiate using product rule *M1 Obtaining form 1 1 1 2 2 2 sin cos + k x k x x Obtain correct 1 1 2 2 4sin 2 cos + x x x or unsimplified equivalent A1 Attempt equation of tangent with numerical value for gradient DM1 Dependent on first M1 Obtain 4 = y x A1 4

This question in 9709/22 Feb/March 2018

4 y M x O P 5 ln x The diagram shows the curve with equation y 2x 1. The curve crosses the x-axis at the point P = and has a maximum point M. + (i) Find the gradient of the curve at the point P. [3] … … … … … … … … … … … … … … … … … … … x 0.5 (ii) Show that the x-coordinate of the point M satisfies the equation x . [2] + ln x = … … … … … … … … … … … … (iii) Use an iterative formula based on the equation in part (ii) to find the x-coordinate of M correct to 4 significant figures. Show the result of each iteration to 6 significant figures. [3] … … … … … … … … … … …

8 marks

Mark scheme: 4(i) Use quotient rule or equivalent M1 Obtaining two terms in numerator and 2 (2 1) + x in denominator for a quotient Obtain correct 5 2 (2 1) 10ln (2 1) + − + x x x x or equivalent, or ( ) ( ) 1 2 5 2 1 10ln 2 1 − − + − + x x x x or equivalent A1 Obtaining one term with ( ) 1 2 1 − + x oe and a second term with ( ) 2 2 1 − + x oe for a product Condone poor use of brackets if recovered later Substitute 1 = x to obtain 15 9 or 5 3 or equivalent, www A1 3 4(ii) Equate numerator to zero and attempt relevant arrangement M1 For M1, need to see at least one line of working after either 5 10 10ln 0 + − = x x or their numerator (which must have at least 2 terms, one involving ln x ) = 0 Confirm 0.5 ln + = x x x A1 AG; necessary detail needed 2 4(iii) Use iteration process correctly at least once M1 Obtain final answer 3.181 A1 Show sufficient iterations to 6 sf to justify answer or show sign change in interval (3.1805, 3.1815) A1 3

This question in 9709/21 May/June 2018

5 The parametric equations of a curve are x 2 cos 3 sin y 3 cos = 21 + 1, = 1 for 0 1 < 1 < 20. (i) Find the gradient of the curve at the point for which 1 radian. [4] 1 = … … … … … … … … … … … … … … … … … … … … … … (ii) Find the value of sin at the point on the curve where the tangent is parallel to the y-axis. [3] 1 … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(i) Obtain d d 4sin2 3cos θ θ θ = − + x B1 Use d d d d d d / θ θ = y y x x in terms of θ or with 1 already substituted M1 Obtain or imply d 3sin d 4sin 2 3cos θ θ θ − = − + y x A1 Substitute 1 to obtain 1.25 A1 Or greater accuracy 1.252013… 4 5(ii) Equate denominator of first derivative to zero M1 Use sin2 2sin cos θ θ θ = A1 Obtain 3 8 sinθ = A1 3

This question in 9709/21 May/June 2018

2 A curve has equation y 3 ln 2x 9 ln x. = + −2 (i) Find the x-coordinate of the stationary point. [4] … … … … … … … … … … … … … … … … (ii) Determine whether the stationary point is a maximum or minimum point. [2] … … … … … … …

6 marks

Mark scheme: 2(i) Differentiate to obtain form 1 2 2 9 − + k k x x M1 Obtain correct 6 2 2 9 − + x x A1 Equate first derivative to zero and attempt solution to ... = x M1 Dependent on previous M1 Obtain 9 = x A1 4 2(ii) Use appropriate method for determining nature of stationary point M1 Second derivative or gradient or value of y Conclude minimum with no errors seen A1 2

This question in 9709/22 May/June 2018

5 A curve has equation y3 sin 2x 4y 8. + = Find the equation of the tangent to the curve at the point where it crosses the y-axis. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 5 Use product rule to differentiate first term obtaining form 2 3 1 2 d sin 2 cos2 d + y k y x k y x x M1 Obtain correct 2 3 d 3 sin 2 2 cos2 d + y y x y x x A1 State 2 3 d d 3 sin 2 2 cos2 4 0 d d + + = y y y x y x x x A1 Identify 0, 2 = = x y as relevant point B1 Find equation of tangent through (0, 2) with numerical gradient M1 Dependent on previous M1 Obtain 4 2 = − + y x or equivalent A1 6

This question in 9709/22 May/June 2018

2 A curve has equation y 3 ln 2x 9 ln x. = + −2 (i) Find the x-coordinate of the stationary point. [4] … … … … … … … … … … … … … … … … (ii) Determine whether the stationary point is a maximum or minimum point. [2] … … … … … … …

6 marks

Mark scheme: 2(i) Differentiate to obtain form 1 2 2 9 − + k k x x M1 Obtain correct 6 2 2 9 − + x x A1 Equate first derivative to zero and attempt solution to ... = x M1 Dependent on previous M1 Obtain 9 = x A1 4 2(ii) Use appropriate method for determining nature of stationary point M1 Second derivative or gradient or value of y Conclude minimum with no errors seen A1 2

This question in 9709/23 May/June 2018

5 A curve has equation y3 sin 2x 4y 8. + = Find the equation of the tangent to the curve at the point where it crosses the y-axis. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 5 Use product rule to differentiate first term obtaining form 2 3 1 2 d sin 2 cos2 d + y k y x k y x x M1 Obtain correct 2 3 d 3 sin 2 2 cos2 d + y y x y x x A1 State 2 3 d d 3 sin 2 2 cos2 4 0 d d + + = y y y x y x x x A1 Identify 0, 2 = = x y as relevant point B1 Find equation of tangent through (0, 2) with numerical gradient M1 Dependent on previous M1 Obtain 4 2 = − + y x or equivalent A1 6

This question in 9709/23 May/June 2018

3 y M x O The diagram shows the curve with equation y 5 sin 2x tan 2x = −3 for values of x such that 0 1 Find the x-coordinate of the stationary point M, giving your answer correct to 3 significant≤xfigures.< 40. [5] … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 B1 Differentiate to obtain 2 6sec 2 − x B1 Equate first derivative to zero and find value for 3 cos 2x M1 Use correct process for finding x from 3 cos 2 = x k M1 Obtain 0.284 nfww A1 Or greater accuracy 5

This question in 9709/22 Oct/Nov 2018

5 A curve has parametric equations x t ln t 1 , y 3te2t. = + + = (i) Find the equation of the tangent to the curve at the origin. [5] … … … … … … … … … … … … … … … … … … … … … … … (ii) Find the coordinates of the stationary point, giving each coordinate correct to 2 decimal places. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 5(i) Use product rule to differentiate y obtaining 2 2 1 2 e e t t k k t + M1 Obtain correct 2 2 3e 6 e t t t + A1 State derivative of x is 1 1 1 t + + B1 Use d d d / d d d y y x x t t = with 0 t = to find gradient M1 Obtain 3 2 y x = or equivalent A1 5 Question Answer Marks Guidance 5(ii) Equate d d y x or d dt y to zero and solve for t M1 Allow full marks if correct solution is obtained but d d x t is incorrect Obtain 1 2 t = − A1 Obtain 1.19 x = − A1 Obtain 0.55 y = − A1 4

This question in 9709/23 Oct/Nov 2018

7 The parametric equations of a curve are x 3 sin y 1 2 tan = 21, = + 21, for 0 1 ≤1 < 40. (i) Find the exact gradient of the curve at the point for which 1 [4] 1 = 60. … … … … … … … … … … … … … … … … … … … … … … (ii) Find the value of at the point where the gradient of the curve is 2, giving the value correct to 3 significant figures.1 [4] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 7(i) Obtain d d 6cos2 x θ θ = Obtain d 2 d 4sec 2 y θ θ = B1 Divide d d y θ by d d x θ with θ equated to 1 6 π M1 Obtain 16 3 or exact equivalent A1 Allow FT on A1 if d d 3cos2 x θ θ = and d 2 d 2sec 2 y θ θ = 4 7(ii) Equate expression for d d y x to 2 with only one trigonometry ratio used *M1 Either cos2 or sec2 θ θ Obtain 3 1 3 cos 2θ = or 3 sec 3 = A1 Attempt correct steps to find a value of θ from 3 cos 2 m θ = , 0 1 m < < DM1 Obtain 0.402 θ = and no others within the range A1 AWRT SC: Allow FT if d d 3cos2 x θ θ = and d 2 d 2sec 2 y θ θ = 4

This question in 9709/22 Oct/Nov 2019

6 A curve has equation y x3e0.2x where x At the point P on the curve, the gradient of the curve is 15. = ≥0. _ 75e−0.2x (a) Show that the x-coordinate of P satisfies the equation x . [4] 15 x = + … … … … … … … … … … … … … … … … … … … … … … … (b) Use the equation in part (a) to show by calculation that the x-coordinate of P lies between 1.7 and 1.8. [2] … … … … … … … … … (c) Use an iterative formula, based on the equation in part (a), to find the x-coordinate of P correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3] … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) Differentiate using the product rule *M1 Obtain 2 0.2 3 0.2 3 e 0.2 e x x x x + A1 OE Equate first derivative to 15 and rearrange to ... x = DM1 Confirm 0.2 75e 15 x x x − = + A1 AG – necessary detail needed 4 6(b) Consider sign of 0.2 75e 15 x x x − − + or equivalent for 1.7 and 1.8 M1 Obtain 0.08... − and 0.03... or equivalents and justify conclusion A1 2 6(c) Use iterative process correctly at least once M1 Answer required to exactly 4 sf Obtain final answer 1.771 A1 Show sufficient iterations to 6 sf to justify answer or show a sign change in the interval [1.7705, 1.7715] A1 3

This question in 9709/22 Feb/March 2020

7 y A x O The diagram shows part of the curve with equation y 4 sin2x 8 sin x 3, = + + where x is measured in radians. The curve crosses the x-axis at the point A and the shaded region is bounded by the curve and the lines x 0 and y 0. = = (a) Find the exact x-coordinate of A. [2] … … … … … … … (b) Find the exact gradient of the curve at A. [3] … … … … … … … … … … … … … (c) Find the exact area of the shaded region. [5] … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 7(a) Solve equation 0 y = to find value of x M1 Obtain 7 6 π A1 2 7(b) Attempt first derivative using chain rule M1 OE Obtain d 8sin cos 8cos d y x x x x = + A1 OE Substitute value from part (a) to find gradient 2 3 − A1 Or exact equivalent 3 7(c) Express integrand in the form 1 2 3 cos2 sin k k x k x + + *M1 Obtain correct 5 2cos2 8sin x x − + A1 OE. Allow unsimplified Integrate to obtain 5 sin2 8cos x x x − − A1 Apply limits 0 and their value from part (a) correctly DM1 Obtain 35 7 3 8 6 2 π + + or exact equivalent A1 5

This question in 9709/22 Feb/March 2020

3 A curve has parametric equations x et y 3e2t 1. = −2e−t, = + Find the equation of the tangent to the curve at the point for which t 0. [5] = … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 State 2 d d e 2e , 6e d d − = + = t t t x y t t B1 Use d d d / d d d = y y x x t t either in terms of t or after substitution of 0 = t *M1 Obtain gradient of tangent is 2 A1 Attempt equation of tangent with numerical gradient and coordinates DM1 Obtain 2 6 = + y x or equivalent A1 5

This question in 9709/21 May/June 2020

5 y M x O The diagram shows part of the curve with equation y x3 cos 2x. The curve has a maximum at the point M. = 3 (a) Show that the x-coordinate of M satisfies the equation x 1.5x2 cot 2x. [3] = … … … … … … … … … … … … … … … … … … (b) Use the equation in part (a) to show by calculation that the x-coordinate of M lies between 0.59 and 0.60. [2] … … … … … … … … … (c) Use an iterative formula, based on the equation in part (a), to find the x-coordinate of M correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(a) Differentiate using the product rule to obtain 2 3 cos2 sin 2 − ax x bx x M1 Obtain 2 3 3 cos2 2 sin 2 − x x x x A1 Equate first derivative to zero and confirm 3 2 1.5 cot 2 = x x x AG A1 3 5(b) Consider sign of 3 2 1.5 cot 2 − x x x or equivalent for 0.59 and 0.60 M1 Obtain 0.009... − and 0.005... or equivalents and justify conclusion A1 2 5(c) Use iteration correctly at least once M1 Obtain final answer 0.596 A1 Show sufficient iterations to 5 sf to justify answer or show sign change in interval [0.5955, 0.5965] A1 3

This question in 9709/21 May/June 2020

1 2 Find the exact coordinates of the stationary point on the curve with equation y 5xe 2x. [5] = … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 Differentiate using product rule to obtain 1 1 2 2 e e + x x a bx *M1 Obtain correct 1 1 2 2 5 5e e 2 + x x x OE A1 Equate first derivative to zero and solve for x DM1 Obtain x-coordinate –2 A1 Obtain y-coordinate 1 10e− − A1 5

This question in 9709/22 May/June 2020

3 The equation of a curve is cos 3x 5 sin y 3. + = Find the gradient of the curve at the point 1 1 . [5] 9π, 6π … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 Differentiate cos3x to obtain 3sin3 − x Differentiate 5sin y to obtain d 5cos d y y x B1 Obtain d 3sin3 5cos 0 d − + = y x y x OE B1 Substitute x and y values to find value of first derivative M1 Obtain 3 5 A1 5

This question in 9709/22 May/June 2020

1 2 Find the exact coordinates of the stationary point on the curve with equation y 5xe 2x. [5] = … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 Differentiate using product rule to obtain 1 1 2 2 e e + x x a bx *M1 Obtain correct 1 1 2 2 5 5e e 2 + x x x OE A1 Equate first derivative to zero and solve for x DM1 Obtain x-coordinate –2 A1 Obtain y-coordinate 1 10e− − A1 5

This question in 9709/23 May/June 2020

3 The equation of a curve is cos 3x 5 sin y 3. + = Find the gradient of the curve at the point 1 1 . [5] 9π, 6π … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 Differentiate cos3x to obtain 3sin3 − x Differentiate 5sin y to obtain d 5cos d y y x B1 Obtain d 3sin3 5cos 0 d − + = y x y x OE B1 Substitute x and y values to find value of first derivative M1 Obtain 3 5 A1 5

This question in 9709/23 May/June 2020

3 The parametric equations of a curve are x e2t cos 4t, y 3 sin 2t. = = Find the gradient of the curve at the point for which t 0. [5] = … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 Attempt use of product rule to find d d x t M1 Obtain 2 2 2e cos4 4e sin 4 t t t t − A1 Obtain d 6cos2 d y t t = B1 Divide to find d d y x with 0 t = substituted M1 Obtain 3 A1 CWO 5

This question in 9709/22 Feb/March 2021

4 y x O 1 3 5x The shaded region is bounded by the The diagram shows part of the curve with equation y = 4x3 1. curve and the lines x 1, x 3 and y 0. + = = = dy (a) Find and hence find the x-coordinate of the maximum point. [4] dx … … … … … … … … … … … … … … … … … (b) Use the trapezium rule with two intervals to find an approximation to the area of the shaded region. Give your answer correct to 2 significant figures. [3] … … … … … … … … … … … … … … … … … (c) State, with a reason, whether your answer to part (b) is an over-estimate or under-estimate of the exact area of the shaded region. [1] … … … … … …

8 marks

Mark scheme: 4(a) Differentiate using quotient rule (or product rule) M1* Obtain 3 3 3 2 5(4 1) 60 (4 1) x x x + − + A1 OE Equate first derivative to zero and attempt solution DM1 Obtain 1 2 x = A1 4 Question Answer Marks Guidance 4(b) Use y values 5 10 15 , , 5 33 109 or decimal equivalents B1 Use correct formula, or equivalent, with 1 h = M1 Obtain 1 20 15 1 2 33 109   + +     or equivalent and hence 0.87 A1 3 4(c) State over-estimate with reference to top of each trapezium above curve B1 Or clear equivalent. 1

This question in 9709/22 Feb/March 2021

5 y M x O 3x 2 The diagram shows the curve with equation y . The curve has a minimum point M. + ln x = dy 3x 2 (a) Find an expression for and show that the x-coordinate of M satisfies the equation x . + dx 3 ln x = [3] … … … … … … … … … … … … … … … … (b) Use the equation in part (a) to show by calculation that the x-coordinate of M lies between 3 and 4. [2] … … … … … … … … … (c) Use an iterative formula, based on the equation in part (a), to find the x-coordinate of M correct to 5 significant figures. Give the result of each iteration to 7 significant figures. [3] … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(a) Use quotient rule (or equivalent) to find first derivative M1 Obtain 2 1 3ln (3 2) d d (ln ) − + = x x y x x x A1 OE Equate first derivative to zero and confirm 3 2 3ln + = x x x A1 AG 3 5(b) Consider 3 2 3ln + −x x x or equivalent for values 3 and 4 M1 M0 if using d d y x Obtain 0.33... − and 0.63... or equivalents and justify conclusion A1 AG 2 5(c) Use iteration process correctly at least once M1 Obtain final answer 3.3223 A1 Answer required to exactly 5 s.f. Show sufficient iterations to 7 s.f. to justify answer or show sign change in the interval [3.32225, 3.32235] A1 3

This question in 9709/21 May/June 2021

6 y A B x O M The diagram shows the curve with equation y ln x 2 ln x. = −2 The curve crosses the x-axis at the points A and B, and has a minimum point M. (a) Find the exact value of the gradient of the curve at each of the points A and B. [6] … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the exact x-coordinate of M. [2] … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) Obtain 1 = x 0e . Must come from correct work, e.g. ln 0 = x . Obtain 2 e = x B1 Differentiate to obtain at least one correct term *M1 Obtain correct first derivative 2ln 2 − x x x A1 Allow ln ln 2 + − x x x x x Allow 2 2 −x x Substitute at least one of their x-values corresponding to 0 = y to find gradient DM1 Allow unsimplified. Obtain gradient 2 − [at A] and gradient 2 2e− at [B] A1 Must be simplified. 6 6(b) Equate first derivative to zero M1 Their derivative must have at least 2 terms. Obtain e = x A1 Allow 1e 2

This question in 9709/22 May/June 2021

6 y A B x O M The diagram shows the curve with equation y ln x 2 ln x. = −2 The curve crosses the x-axis at the points A and B, and has a minimum point M. (a) Find the exact value of the gradient of the curve at each of the points A and B. [6] … … … … … … … … … … … … … … … … … … … … … … … … (b) Find the exact x-coordinate of M. [2] … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) Obtain 1 = x 0 e . Must come from correct work, e.g. ln 0 = x . Obtain 2 e = x B1 Differentiate to obtain at least one correct term *M1 Obtain correct first derivative 2ln 2 − x x x A1 Allow ln ln 2 + − x x x x x Allow 2 2 −x x Substitute at least one of their x-values corresponding to 0 = y to find gradient DM1 Allow unsimplified. Obtain gradient 2 − [at A] and gradient 2 2e− at [B] A1 Must be simplified. 6 6(b) Equate first derivative to zero M1 Their derivative must have at least 2 terms. Obtain e = x A1 Allow 1e 2

This question in 9709/23 May/June 2021

4 The curve with equation y xe2x has a minimum point M. = + 5e−x (a) Show that the x-coordinate of M satisfies the equation x 1 ln 5 ln 1 2x . [5] = 3 −13 + … … … … … … … … … … … … … … … … … … … … … … … … (b) Use an iterative formula, based on the equation in part (a), to find the x-coordinate of M correct to 3 significant figures. Use an initial value of 0.35 and give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4(a) Attempt use of product rule to differentiate 2 e x x Obtain 2 2 e 2 e 5e− + − x x x x A1 Equate first derivative to zero and multiply by ex to obtain an equation involving 3e x M1 Obtain 3e (1 2 ) 5 + = x x or equivalent A1 Confirm given result 1 1 3 3 ln5 ln(1 2 ) = − + x x with sufficient detail A1 AG 5 4(b) Use iteration process correctly at least once M1 Need 0.35 and 2 correct values. Obtain final answer 0.357 A1 Answer required to exactly 3sf. Allow recovery. Show sufficient iterations to 5sf to justify answer or show sign change in interval [0.3565, 0.3575] A1 3

This question in 9709/21 Oct/Nov 2021

3 The curve with equation y 5x tan 2x = −2 has exactly one stationary point in the interval 0 1 ≤x < 4π. Find the coordinates of this stationary point, giving each coordinate correct to 3 significant figures. [6] … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 3 Differentiate to obtain first derivative of form 2 1 2 sec 2 + k k x 1 2 0 ≠ k k . Obtain correct 2 5 4sec 2 − x A1 Equate first derivative of form 2 1 2 sec 2 + k k x to zero and obtain value for 2 cos 2x M1 OE Or value for 2 tan 2xwhich must be > 0. –1 ⩽ cos 2x ⩽ 1. Obtain 2 cos2 5 = x or (decimal) equivalent A1 Or 1 2 tan2 = x Obtain 0.232 = x A1 AWRT Obtain 0.159 = y A1 AWRT 6

This question in 9709/22 Oct/Nov 2021

5 A curve has equation x2 4x cos 3y 6. + = Find the exact value of the gradient of the normal to the curve at the point 2, 1 . [6] 12π … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 5 to obtain 1 2 d cos3 sin3 d + y k y k x y x *M1 1 2 0 ≠ k k Obtain d 4cos3 12 sin3 d − y y x y x A1 Allow unsimplified. Obtain d 2 4cos3 12 sin3 0 d + − = y x y x y x A1 Allow unsimplified. Substitute x- and y-values to find value of first derivative DM1 Apply d 1/ d − y x to find gradient of normal M1 Obtain 3 2 − or 3 2 2 − or exact equivalent A1 6

This question in 9709/22 Oct/Nov 2021

4 The curve with equation y xe2x 5e x has a minimum point M. (a) Show that the x-coordinate of M satisfies the equation x 1 ln 5 ln 1 2x . [5] = 3 −13 + … … … … … … … … … … … … … … … … … … … … … … … … (b) Use an iterative formula, based on the equation in part (a), to find the x-coordinate of M correct to 3 significant figures. Use an initial value of 0.35 and give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4(a) Attempt use of product rule to differentiate 2 e x x M1 Obtain 2 2 e 2 e 5e− + − x x x x A1 Equate first derivative to zero and multiply by e x to obtain an equation involving 3e x M1 Obtain 3e (1 2 ) 5 + = x x or equivalent A1 Confirm given result 1 1 3 3 ln5 ln(1 2 ) = − + x x with sufficient detail A1 AG 5 4(b) Use iteration process correctly at least once M1 Need 0.35 and 2 correct values. Obtain final answer 0.357 A1 Answer required to exactly 3sf. Allow recovery. Show sufficient iterations to 5sf to justify answer or show sign change in interval [0.3565, 0.3575] A1 3

This question in 9709/23 Oct/Nov 2021

6 y P M x O 4e2x 9 The diagram shows the curve with equation y . The curve has a minimum point M and + ex 2 = crosses the y-axis at the point P. + (a) Find the exact value of the gradient of the curve at P. [4] … … … … … … … … … … … … … … (b) Find the exact coordinates of M. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) Differentiate using quotient rule M1 OE x 2 x 2 x x A1 OE (e + 2) 8e − (4e + 9) e Obtain (e x + 2) 2 Substitute x = 0 in first derivative and attempt evaluation M1 Obtain 119 A1 4 6(b) Equate first derivative to zero and attempt factorisation or M1 equivalent Solve a three-term quadratic equation in ex to obtain e x = ... M1 (2e x + 9)(2e x − 1) = 0 . Obtain x-coordinate ln 12 or − ln2 A1 Obtain y-coordinate 4 A1 4

This question in 9709/22 Feb/March 2023

7 y P x O The diagram shows the curve with parametric equations x k tan t, y 3 sin 2t sin t, = = −4 for 0 t 1 It is given that k is a positive constant. The curve crosses the x-axis at the point P. < < 2π. (a) Find the value of cos t at P, giving your answer as an exact fraction. [3] … … … … … … … … … … … … … … … … … dy (b) Express in terms of k and cos t. [4] dx … … … … … … … … … … … … (c) Given that the normal to the curve at P has gradient 10,9 find the value of k, giving your answer as an exact fraction. [3] … … … … … … … … … … …

10 marks

Mark scheme: 7(a) Use identity sin2t = 2sin t cos t B1 Attempt solution of y = 0 for cost M1 Obtain cost = 23 A1 Or exact equivalent. 3 7(b) d x d y *M1 Differentiate to obtain at least one of and correct d t d t dy 6cos2t − 4cos t A1 Obtain = dx k sec 2 t d y DM1 Using correct identities. Attempt to express in terms of cost d x dy 6(2cos 2 t − 1)cos 2 t − 4cos 3 t A1 OE Obtain = dx k 4 7(c) d y *M1 Substitute value from part (a) in expression for involving k d x and cost Equate to − 109 and solve for k DM1 Obtain k = 43 A1 3

This question in 9709/22 Feb/March 2023

2 3 ln x 2 A curve has equation y . + 1 2x = + Find the equation of the tangent to the curve at the point 1, 2 . Give your answer in the form 3 ax by c 0, where a, b and c are integers. [5] + + = … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 Use quotient rule (or equivalent) to find first derivative M1 Obtain 3 2 (1 2 ) (2 3ln )2 (1 2 )     x x x x A1 OE Substitute 1  x and obtain 5 9 A1 Attempt equation of tangent through 2 3 (1, ) with their numerical gradient M1 Must have made an attempt at differentiation. Obtain 5 9 1 0   x y or equivalent of required form A1 5

This question in 9709/21 May/June 2023

5 y C A B x O 2x The diagram shows the curve with equation y x2 4 . The curve crosses the x-axis at the = e−1 −5x + points A and B, and has a maximum at the point C. (a) Find the exact gradient of the curve at B. [5] … … … … … … … … … … … … … … … … … (b) Find the exact coordinates of C. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 5(a) Attempt use of product rule to find first derivative *M1 Obtain 1 1 2 2 2 1 2 e ( 5 4) e (2 5)        x x x x x A1 OE Obtain 4  x for point B B1 Substitute 4  x to find the value of the derivative DM1 Obtain 2 3e A1 or exact equivalent. 5 5(b) Equate their first derivative to zero and simplify as far as quadratic equation *M1 allow if it appears in part (a). Obtain at least 2 9 14 0    x x A1 OE Solve to find relevant x value and substitute to find the value of y DM1 Obtain 7  x and 7 2 18e   y A1 or exact equivalent. 4

This question in 9709/22 May/June 2023

5 y C A B x O 2x The diagram shows the curve with equation y x2 4 . The curve crosses the x-axis at the = e−1 −5x + points A and B, and has a maximum at the point C. (a) Find the exact gradient of the curve at B. [5] … … … … … … … … … … … … … … … … … (b) Find the exact coordinates of C. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 5(a) Attempt use of product rule to find first derivative *M1 Obtain 1 1 2 2 2 1 2e ( 5 4) e (2 5)        x x x x x A1 OE Obtain 4  x for point B B1 Substitute 4  x to find the value of the derivative DM1 Obtain 2 3e  A1 or exact equivalent. 5 5(b) Equate their first derivative to zero and simplify as far as quadratic equation *M1 allow if it appears in part (a). Obtain at least 2 9 14 0    x x A1 OE Solve to find relevant x value and substitute to find the value of y DM1 Obtain 7  x and 7 2 18e   y A1 or exact equivalent. 4

This question in 9709/23 May/June 2023

7 The curve with equation e2x y3 y 11 has a stationary point at p, q . −18x + + = (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] … … … … … … … … …

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 9709/21 Oct/Nov 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 9709/23 Oct/Nov 2023

5 y P O x x 3 The diagram shows part of the curve with equation y = . At the point P, the gradient of the curve x + 2 is 6. (a) Show that the x-coordinate of P satisfies the equation x = 3 12 x + 12 . [4] … … … … … … … … … … … … … … … … … … (b) Show by calculation that the x-coordinate of P lies between 3.8 and 4.0 . [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. Show the result of each iteration to 5 significant figures. [3] … … … … … … … … … … … … … … …

9 marks

Mark scheme: 5(a) Differentiate using quotient rule *M1 OE 2 3 A1 OE ( x + 2)3 x − x Obtain ( x + 2) 2 Equate first derivative to 6 and simplify at least as far as x 3 = ... DM1 Confirm x = 3 12 x + 12 A1 Answer given – necessary detail needed. 4 5(b) Consider sign of x − 3 12 x + 12 , or equivalent, for 3.8 and 4.0 M1 Obtain −0.06... and 0.08..., or equivalents, and justify conclusion A1 Answer given – necessary detail needed. 2 5(c) Use iterative process correctly at least once M1 Obtain final answer 3.88 A1 Answer required to exactly 3 sf. Show sufficient iterations to 5 sf to justify answer or show sign change in the A1 interval [3.875, 3.885] 3

This question in 9709/22 Feb/March 2024

1 A curve has equation y = 2 tan x - 5 sin x for 0 G x 1 12 r. Find the x-coordinate of the stationary point of the curve. Give your answer correct to 3 significant figures. [3] … … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 1 Differentiate to obtain 2 2sec 5cos x x  B1 Equate first derivative to zero and solve correctly at least as far as cos ... x  M1 B1 M1 for o 42.5 x  . Obtain 3 cos 0.4 x  and hence 0.742 x  A1 Or greater accuracy (0.74261…). 3

This question in 9709/21 May/June 2024

2 A curve has equation x 2 ln y + y 2 + 4x = 9 . Find the gradient of the curve at the point (2, 1). [5] … … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 Attempt use of product rule for differentiating x y Allow if d d y x missing. Obtain 2 d 2 ln d x y x y y x  A1 Obtain d 2 d y y x  B1 Substitute 2, 1 x y   in equation equal to 0 involving at least one d d y x and solve for d d y x M1 Correct equation: 2 d 2 ln d x y x y y x  d 2 d y y x  4 0   Obtain 2 3  A1 Or exact equivalent. Not from wrong working. 5

This question in 9709/21 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 9709/22 May/June 2024

6 y M O x ln ( 2x + 1) The diagram shows the curve with equation y = . The curve has a maximum point M. x + 3 dy (a) Find an expression for . [2] dx … … … … … … … … … … x + 3 (b) Show that the x-coordinate of M satisfies the equation x = - 0. 5 . [2] ln ( 2x + 1) … … … … … … … … … … (c) Show by calculation that the x-coordinate of M lies between 2.5 and 3.0 . [2] … … … … … … … … … … … … … (d) Use an iterative formula based on the equation in part (b) to find the x-coordinate of M correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3] … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) Attempt use of quotient rule M1 Or equivalent method. Obtain 2 2 1 2 ( 3) ln(2 1) ( 3)      x x x x A1 OE 2 6(b) Equate first derivative to zero and arrange as far as 2 1 ...  x M1 Confirm 3 0.5 ln(2 1)     x x x A1 Answer given – necessary detail needed. 2 Question Answer Marks Guidance 6(c) Consider sign of 3 0.5 ln(2 1)     x x x or equivalent for 2.5 and 3.0 M1 Obtain   0.07 0.0696...  and   0.4 0.4166... and justify conclusion A1 Answer given – necessary detail needed. 2 Alternative Method for Question 6(c) Consider the values of    3 f 0.5 ln 2 1     x x x and obtain   f 2.5 2.57 (2.5696…) and  f 3 2.58 (2.58339…) (M1) Conclude   f 2.5 3  and  f 3 2.5  so root lies in given interval (A1) Answer given – necessary detail needed. 2 6(d) Use iterative process correctly at least once M1 Obtain final answer 2.569 A1 Answer required to exactly 4sf. Show sufficient iterations to 6 sf to justify answer or show sign change in interval [2.5685, 2.5695] A1 3

This question in 9709/22 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 1 r. 2 Find the equation of the normal to the curve at the point for which t = 1 r . Give your answer in the 6 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 **M1 FT 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 9709/23 May/June 2024

3 The function f is defined by f ( x) = tan 2 a 1 x k for 0 G x 1 r. 2 (a) Find the exact value of f la 2 r k. [3] 3 … … … … … … … … … … … … … … … … … … … … … … … … … … 1 r 2 (b) Find the exact value of `f ( x) + sin xj d x . [4] y 0 … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 3(a) Differentiate to obtain form k tan 12 x sec 2 12 x M1 OE. May use identities before differentiation. Obtain correct tan 12 x sec 2 12 x A1 OE. Allow unsimplified. Substitute 23 π to obtain 4 3 A1 3 3(b) Express integrand as sec 2 12 x −+1 sin x B1 Integrate to obtain k1 tan 12 x − x + k 2 cos x M1 Where k1k 2  0. Obtain correct 2tan 12 x − x − cos x A1 Apply limits correctly to obtain 3 − 12 π or exact equivalent A1 4

This question in 9709/21 Oct/Nov 2024

3 The function f is defined by f ( x) = tan 2 a 1 x k for 0 G x 1 r. 2 (a) Find the exact value of f la 2 r k. [3] 3 … … … … … … … … … … … … … … … … … … … … … … … … … … 1 r 2 (b) Find the exact value of `f ( x) + sin xj d x . [4] y 0 … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 3(a) Differentiate to obtain form k tan 12 x sec 2 12 x M1 OE. May use identities before differentiation. Obtain correct tan 12 x sec 2 12 x A1 OE. Allow unsimplified. Substitute 23 π to obtain 4 3 A1 3 3(b) Express integrand as sec 2 12 x −+1 sin x B1 Integrate to obtain k1 tan 12 x − x + k 2 cos x M1 Where k1k 2  0. Obtain correct 2tan 12 x − x − cos x A1 Apply limits correctly to obtain 3 − 12 π or exact equivalent A1 4

This question in 9709/23 Oct/Nov 2024

4 y x O The diagram shows the curve with equation y = 6e 2 x - e 3 x . The shaded region is bounded by the axes and the curve. (a) Find the exact x-coordinate of the maximum point. [3] … … … … … … … … … … … … … … … … … … … … (b) Find the area of the shaded region. Give your answer in the form p, where p and q are integers. q [4] … … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(a) 2 x 3 x B1 Differentiate to obtain 12e − 3e Attempt solution for x of equation of form k1e 2 x + k 2 e 3 x = 0 (but not for y = 0 ), to M1 SOI obtain e =x k Obtain x = ln4 or x = 2ln2 A1 x = 1.386 scores A0. 3 4(b) Identify x = ln6 as point where curve meets x-axis B1 May see in part (a) but must be used here. Integrate to obtain the form k 3 e 2 x + k 4 e 3 x M1 Where k3  6 and k4 −1 Obtain 3e 2 x − 13 e 3 x A1 May see inclusion of + c. Apply limits to obtain answer 1003 only A1 4

This question in 9709/21 May/June 2025

8x 3 Find the coordinates of the stationary points of the curve with equation y = - 6 x + 5 . [5] 2x + 3 … … … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 8 x −12 x 2 + 15 *M1 Attempt use of quotient rule (or equivalent) to differentiate or 2 x + 3 ( 2 x + 3 ) d y 24 A1 OE Obtain = − 6 2 Allow unsimplified. d x (2 x + 3) Equate first derivative to zero, and attempt solution of a quadratic equation to find DM1 4 x 2 + 12 x + 5 = 0 two values of x Obtain at least one of the stationary points ( − 5 , 30) and ( − 1 , 6), A1 2 2 or both x-values, − 5 and − 1 2 2 Obtain both stationary points A1 5

This question in 9709/22 May/June 2025

8x 3 Find the coordinates of the stationary points of the curve with equation y = - 6 x + 5 . [5] 2x + 3 … … … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 8 x −12 x 2 + 15 *M1 Attempt use of quotient rule (or equivalent) to differentiate or 2 x + 3 ( 2 x + 3 ) d y 24 A1 OE Obtain = − 6 d x (2 x + 3) 2 Allow unsimplified. Equate first derivative to zero, and attempt solution of a quadratic equation to find DM1 4 x 2 + 12 x + 5 = 0 two values of x Obtain at least one of the stationary points ( − 5 , 30) and ( − 1 , 6), A1 2 2 or both x-values, − 5 and − 1 2 2 Obtain both stationary points A1 5

This question in 9709/23 May/June 2025

8x 3 Find the coordinates of the stationary points of the curve with equation y = - 6 x + 5 . [5] 2x + 3 … … … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 8 x −12 x 2 + 15 *M1 Attempt use of quotient rule (or equivalent) to differentiate or 2 x + 3 ( 2 x + 3 ) d y 24 A1 OE Obtain = − 6 2 Allow unsimplified. d x (2 x + 3) Equate first derivative to zero, and attempt solution of a quadratic equation to find DM1 4 x 2 + 12 x + 5 = 0 two values of x Obtain at least one of the stationary points ( − 5 , 30) and ( − 1 , 6), A1 2 2 or both x-values, − 5 and − 1 2 2 Obtain both stationary points A1 5

This question in 9709/25 May/June 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] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …

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 2 2 k1e1− 2 x (3x − 1) 1 + k2 e1− 2 x (3x − 1)− 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 9709/21 Oct/Nov 2025

5 y A B O x The diagram shows the curve with equation y = 4 cos 2 x + 8 sin x for 0 G x G r . The maximum points on the curve are denoted by A and B, and the shaded region is bounded by the line segment AB and the curve. (a) Find the coordinates of A and B. [5] … … … … … … … … … … … … … … … … … … … (b) Find the exact area of the shaded region. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 5(a) Differentiate to obtain −8sin2 x + 8cos x B1 Attempt to solve equation of form k1 sin2 x + k 2 cos x = 0 using correct identity *M1 Must have attempted differentiation. Obtain at least sin x = 12 A1 Attempt to find both coordinates of at least one maximum point DM1 5 1 and no others between 0 and π A1 Angles must be in radians. Obtain ( and ( 6 π, 6 ) 6 π, 6 ) 5 5(b) Integrate to obtain expression of form k1 sin2 x + k 2 cos x *M1 Obtain correct 2sin2 x − 8cos x A1 Condone poor notation for this mark. Apply their x-limits correctly to evaluate area under curve between A and B *DM1 Condone use of degrees for this mark. Obtain 6 3 or exact unsimplified equivalent A1 Carry out correct process to find area of shaded region DDM1 The y-coordinates of both points must be the same. Rectangle – their 6 3. 5 and hence 4π − 6 3 A1 Or exact equivalent. Obtain 6 ( 6 π − 16 π ) − 6 3, 6

This question in 9709/22 Oct/Nov 2025