Integration — 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

7 (i) By expanding sin(2x + x) and using double-angle formulae, show that sin 3x = 3 sin x −4 sin3x. [5] (ii) Hence show that 13π sin3x dx = 24.5 [5] 0

10 marks

Mark scheme: 7 (i) Make relevant use of the sin(A + B) formula B1 Make relevant use of sin2A and cos2A formulae M1 Obtain a correct expression in terms of sin x and cos x A1 Use cos2 x = 1 – sin2 x to obtain an expression in terms of sin x M1(dep*) Obtain given answer correctly A1 5 3 1 (ii) Replace integrand by sin x – sin 3x, or equivalent B1 4 4 3 1 Integrate, obtaining − cos x + cos 3 x, or equivalent B1√ + B1√ 4 12 Use limits correctly M1 Obtain given answer correctly A1 5

This question in 9709/21 May/June 2005

6 (i) Express cos2x in terms of cos 2x. [1] (ii) Hence show that 13π cos2x dx = 1 + 1 √3. [4] 6π 8 0 (iii) By using an appropriate trigonometrical identity, deduce the exact value of 3π! sin2x dx. 0 [3]

8 marks

Mark scheme: 1 1 6 (i) State correct expression + cos 2 x , or equivalent B1 [1] 2 2 (ii) Integrate an expression of the form a + b cos 2 x , where ab ≠ 0 , correctly M1 1 1 State correct integral x + sin 2 x , or equivalent A1 2 4 Use correct limits correctly M1 Obtain given answer correctly A1 [4] (iii) Use identity sin 2 x = 1 − cos 2 x and attempt indefinite integration M1  1 1  Obtain integral x −  x − sin 2 x  , or equivalent A1  2 4  1 3 Use limits and obtain answer π − A1 [3] 6 8 [Solutions that use the result of part (ii), score M1A1 for integrating 1 and A1 for the final answer.] 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

1 Show that 4 1 1 dx = ln 3. [4] 2x + 1 2 1

4 marks

Mark scheme: 1 1 State indefinite integral of the form k In(2x + 1), where k = , 1 or 2 M1 2 1 State correct integral In(2x + 1) A1 2 Use limits correctly, allow use of limits x = 4 and x = 1 in an incorrect form M1 Obtain given answer A1 [4]

This question in 9709/21 Oct/Nov 2007

7 (i) Prove the identity (cos x + 3 sin x)2 ≡5 −4 cos 2x + 3 sin 2x. [4] (ii) Using the identity, or otherwise, find the exact value of 14π (cos x + 3 sin x)2 dx. [4] 0

8 marks

Mark scheme: 7 (i) Expand and use sin 2A formula M1 Use cos 2A formula at least once M1 Obtain any correct expression in terms of cos 2x and sin 2x only – can be implied A1 Obtain given answer correctly A1 [4] 3 (ii) State indefinite integral 5x – 2sin 2x – cos 2x B2 2 [Award B1 if one error in one term] Substitute limits correctly – must be correct limits M1 1 Obtain answer (5π – 2), or exact simplified equivalent A1 [4] 4

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

3 Find the exact value of (cos 2x + sin x) dx. [5] 0

5 marks

Mark scheme: 3 Obtain integral 12 sin2x – cos x B1 + B1 Substitute limits correctly in an integral of the form a sin 2x + b cos x M1 1 π ) M1 Use correct exact values, e.g. of cos (6 1 Obtain answer 1 – 3 , or equivalent A1 [5] 4

This question in 9709/21 May/June 2008

1 8 The constant a, where a > 1, is such that x + dx = 6. x 1 (i) Find an equation satisfied by a, and show that it can be written in the form a = √(13 −2 ln a). [5] (ii) Verify, by calculation, that the equation a = √(13 −2 ln a) has a root between 3 and 3.5. [2] (iii) Use the iterative formula = √(13 −2 ln an+1 an), with a1 = 3.2, to calculate the value of a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

10 marks

Mark scheme: 8 (i) Obtain terms 12 x2 and ln x B1 + B1 Substitute limits correctly M1 a2 + ln a – 1 Obtain correct equation in any form, e.g. 1 = 6 A1 2 2 Obtain given answer correctly A1 [5] (ii) Consider sign of a – (13 − 2ln a ) at a = 3 and a = 3.5, or equivalent M1 Complete the argument correctly with correct calculations A1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 3.26 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (3.255, 3.265) B1 [3]

This question in 9709/21 May/June 2008

3 y 1 x O 1 2 1 The diagram shows the curve y for values of x from 0 to 2. 1 = √x + (i) Use the trapezium rule with two intervals to estimate the value of 2 1 dx, 1 √x ä 0 + giving your answer correct to 2 decimal places. [3] (ii) 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 (i). [1]

4 marks

Mark scheme: 3 (i) Show or imply correct ordinates 1, 0.5, 0.414213 ... B1 Use correct formula, or equivalent, with h = 1 and three ordinates M1 Obtain answer 1.21 with no errors seen A1 [3] (ii) Justify the statement that the rule gives an over-estimate B1 [1] dx

This question in 9709/21 May/June 2009

5 (i) Express cos2 2x in terms of cos 4x. [2] 18π (ii) Hence find the exact value of cos2 2x dx. [4] ã 0

6 marks

Mark scheme: 5 (i) Use double angle formulae and obtain a + bcos 4x M1 Obtain answer 1 + 1 cos 4x, or equivalent A1 [2] 2 2 (ii) Integrate and obtain 1 x + 1 sin 4x A1√ + A1√ 2 8 Substitute limits correctly M1 Obtain answer 1 π + 1 , or exact equivalent A1 [4] 16 8 GCE A/AS LEVEL – October/November 2009 9709 21

This question in 9709/21 Oct/Nov 2009

7 y 1 R x O p 1 The diagram shows the curve y The shaded region R is bounded by the curve and the lines y = e−x. = and x p, where p is a constant. = (i) Find the area of R in terms of p. [4] (ii) Show that if the area of R is equal to 1 then p 2 = −e−p. [1] (iii) Use the iterative formula pn+1 = 2 −e−pn, 2, to calculate the value of p correct to 2 decimal places. Give the result with initial value p1 = of each iteration to 4 decimal places. [3]

8 marks

Mark scheme: 7 (i) EITHER: Integrate 1 – e–x obtaining x ± e–x M1 Obtain indefinite integral x – e–x A1 Substitute limits x = 0, x = p correctly M1 Obtain answer p + e–p – 1, or equivalent A1 OR: Integrate e–x obtaining ± e–x M1 Substitute limits x = 0, x = p correctly M1 Obtain area below curve is 1 – e–p A1 Obtain answer p + e–p – 1, or equivalent A1 [4] (ii) Show that p + e–p – 1 = 1 is equivalent to p = 2 – e–p or vice versa B1 [1] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.84 A1 Show sufficient iterations to justify its accuracy to 2 d.p. A1 [3]

This question in 9709/21 Oct/Nov 2009

3 (i) Use the trapezium rule with two intervals to estimate the value of 13π sec x dx, ã 0 giving your answer correct to 2 decimal places. [3] (ii) Using a sketch of the graph of y sec x for 0 3π, explain whether the trapezium rule gives an under-estimate or an over-estimate= of the true≤x value≤1 of the integral in part (i). [2]

5 marks

Mark scheme: 3 (i) Show or imply correct ordinates 1, 1.15470…, 2 B1 Use correct formula, or equivalent, with h = 1 π and three ordinates M1 6 Obtain answer 1.39 with no errors seen A1 [3] (ii) Make recognisable sketch of y = sec x for 0 @ x @ 1 π B1 3 Using a correct graph, explain that the rule gives an over-estimate B1 [2] dx t d y t t

This question in 9709/22 Oct/Nov 2009

13π 8 (a) Find the exact value of 2x dx. [5] ã 0 (sin + sec2x) 4 1 1 (b) Show that dx ln 5. [4] 2x x 1 + = ä1 +

9 marks

Mark scheme: 8 (a) Integrate and obtain term k cos 2x, where k = ± 1 or ±1 M1 2 Obtain term – 1 cos 2x A1 2 Obtain term tan x B1 Substitute correct limits correctly M1 3 Obtain exact answer + 3 A1 [5] 4 (b) Integrate and obtain 1 ln x + ln(x + 1) or 1 ln 2x + ln(x + 1) B1 + B1 2 2 Substitute correct limits correctly M1 Obtain given answer following full and correct working A1 [4]

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

2 y R x O 1 2 3 The diagram shows part of the curve y The shaded region R is bounded by the curve and by = xe−x. the lines x 2, x 3 and y 0. = = = (i) Use the trapezium rule with two intervals to estimate the area of R, giving your answer correct to 2 decimal places. [3] (ii) State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the area of R. [1]

4 marks

Mark scheme: 2 (i) State or imply correct ordinates 0.27067..., 0.20521..., 0.14936... B1 Use correct formula, or equivalent, correctly with h = 0.5 and three ordinates M1 Obtain answer 0.21 with no errors seen A1 [3] (ii) Justify statement that the trapezium rule gives an over-estimate B1 [1] 2 2

This question in 9709/22 May/June 2010

4π 4 (a) Show that cos 2x dx 12. [2] ã 0 = (b) By using an appropriate trigonometrical identity, find the exact value of 13π 3 tan2x dx. ã 1 [4] 6π

6 marks

Mark scheme:  1  4 (a) Obtain integral a sin 2x with a = ±  ,1 2 or  M1  2  1 Use limits and obtain (AG) A1 [2] 2 (b) Use tan2 x = sec2 x – 1 and attempt to integrate both terms M1 Obtain 3tan x – 3x A1 Attempt to substitute limits, using exact values M1 π Obtain answer 2 3 − A1 [4] 2

This question in 9709/22 May/June 2010

2 y R x O 1 2 3 The diagram shows part of the curve y The shaded region R is bounded by the curve and by = xe−x. the lines x 2, x 3 and y 0. = = = (i) Use the trapezium rule with two intervals to estimate the area of R, giving your answer correct to 2 decimal places. [3] (ii) State, with a reason, whether the trapezium rule gives an under-estimate or an over-estimate of the true value of the area of R. [1]

4 marks

Mark scheme: 2 (i) State or imply correct ordinates 0.27067..., 0.20521..., 0.14936... B1 Use correct formula, or equivalent, correctly with h = 0.5 and three ordinates M1 Obtain answer 0.21 with no errors seen A1 [3] (ii) Justify statement that the trapezium rule gives an over-estimate B1 [1] 2 2

This question in 9709/23 May/June 2010

4π 4 (a) Show that cos 2x dx 12. [2] ã 0 = (b) By using an appropriate trigonometrical identity, find the exact value of 13π 3 tan2x dx. ã 1 [4] 6π

6 marks

Mark scheme:  1  4 (a) Obtain integral a sin 2x with a = ±  ,1 2 or  M1  2  1 Use limits and obtain (AG) A1 [2] 2 (b) Use tan2 x = sec2 x – 1 and attempt to integrate both terms M1 Obtain 3tan x – 3x A1 Attempt to substitute limits, using exact values M1 π Obtain answer 2 3 − A1 [4] 2

This question in 9709/23 May/June 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

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

4 (a) Find dx. [2] ã e1−2x (b) Express sin23x in terms of cos 6x and hence find sin23x dx. [4] ã

6 marks

Mark scheme: 4 (a) Obtain integral of the form ke1 – 2x with any non-zero k M1 Correct integral A1 [2] (b) Attempt to use double angle formula to expand cos (3x + 3x) M1 1 1 State correct expression − cos6x or equivalent A1 2 2 Integrate an expression of the form a + b cos6x, where ab ≠ 0, correctly M1 1 1 State correct integral x – sin6x, or equivalent A1 [4] 2 12

This question in 9709/23 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

6 (a) Find dx. [4] ã 4ex(3 + e2x) 14π (b) Show that 2 dθ 1 [4] ã (3 + tan2θ) = 2(8 + π). −14π

8 marks

Mark scheme: 6 (a) Rewrite integrand as 12ex + 4e3x B1 Integrate to obtain 12ex … B1 4 3 x Integrate to obtain … + e B1 3 Include … + c B1 [4] (b) Use identity tan2θ = sec2θ – 1 B1 Integrate to obtain 2tanθ + θ or equivalent B1 Use limits correctly for integral of form αtanθ + bθ M1 1 Confirm given answer (8 + π ) A1 [4] 2 2

This question in 9709/21 May/June 2011

2 y 3 B A x O 2 The diagram shows the curve y Region A is bounded by the curve and the lines x 0, = √(1 + x3). = x 2 and y 0. Region B is bounded by the curve and the lines x 0 and y 3. = = = = (i) Use the trapezium rule with two intervals to find an approximation to the area of region A. Give your answer correct to 2 decimal places. [3] (ii) Deduce an approximation to the area of region B and explain why this approximation under- estimates the true area of region B. [2]

5 marks

Mark scheme: 2 (i) Show or imply correct ordinates 1, 2 or 1.414, 3 B1 Use correct formula, or equivalent, with h = 1 M1 Obtain 3.41 A1 [3] (ii) Obtain 6 – 3.41 and hence 2.59, following their answer to (i) provided less than 6 B1√ Refer, in some form, to two line segments replacing curve and conclude with clear justification of given result that answer is an under-estimate. B1 [2]

This question in 9709/22 May/June 2011

2 y 3 B A x O 2 The diagram shows the curve y Region A is bounded by the curve and the lines x 0, = √(1 + x3). = x 2 and y 0. Region B is bounded by the curve and the lines x 0 and y 3. = = = = (i) Use the trapezium rule with two intervals to find an approximation to the area of region A. Give your answer correct to 2 decimal places. [3] (ii) Deduce an approximation to the area of region B and explain why this approximation under- estimates the true area of region B. [2]

5 marks

Mark scheme: 2 (i) Show or imply correct ordinates 1, 2 or 1.414, 3 B1 Use correct formula, or equivalent, with h = 1 M1 Obtain 3.41 A1 [3] (ii) Obtain 6 – 3.41 and hence 2.59, following their answer to (i) provided less than 6 B1√ Refer, in some form, to two line segments replacing curve and conclude with clear justification of given result that answer is an under-estimate. B1 [2]

This question in 9709/23 May/June 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

4 (i) Express cos2x in terms of cos 2x. [1] (ii) Hence show that 16π 1 1 sin dx 1 √3 8 12π 4. + + [5] ã 0 (cos2x + 2x) =

6 marks

Mark scheme: 1 1 4 (i) State correct expression + cos 2 x , or equivalent B1 [1] 2 2 (ii) Integrate an expression of the form a + b cos 2x, where ab ≠ 0, correctly M1 1 1 State correct integral x + sin 2 x , or equivalent A1 2 4 1 Obtain correct integral (for sin 2x term) of − cos 2 x B1 2 Attempt to substitute limits, using exact values M1 Obtain given answer correctly A1 [5]

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

7 (i) Show that tan2x cos2x 1 cos 2x and hence find the exact value of + ≡sec2x + 2 −12 14π dx. ã 0 (tan2x + cos2x) [7] (ii) y 1 x O 4p 0 is shown in The region enclosed by the curve y tan x cos x and the lines x 0, x = + = = 14π and y = the diagram. Find the exact volume of the solid produced when this region is rotated completely about the x-axis. [4]

11 marks

Mark scheme: 7 (i) Replace tan2 x by sec2 x – 1 B1 Express cos2 x in the form ± 12 ± 12 cos 2 x M1 Obtain given answer sec 2 x + 12 cos 2 x − 12 correctly A1 Attempt integration of expression M1 Obtain tan x + 14 sin 2 x − 12 x A1 Use limits correctly for integral involving at least tan x and sin 2x M1 Obtain 54 − 18 π or exact equivalent A1 [7] (ii) State or imply volume is ∫ π (tan x + cos x 2) dx B1 Attempt expansion and simplification M1 Integrate to obtain one term of form k cos x M1 Obtain π ( 54 − 18 π ) + π ( 2 − 2 ) or equivalent A1 [4]

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

7 (i) Show that sin x cos can be written in the form 5 2 sin 2x cos 2x. [5] 2 2 (2 + x)2 + −3 14π (ii) Hence find the exact value of sin x cos dx. [4] ã 0 (2 + x)2

9 marks

Mark scheme: 7 (i) Expand to obtain 4 sin2 x + 4 sin x cos x + cos2 x B1 Use 2 sin x cos x = sin 2x B1 Attempt to express sin2 x or cos2 x (or both) in terms of cos 2x M1 Obtain correct 12 k 1( − cos 2 x ) for their k sin2 x or equivalent A1√ Confirm given answer 52 + 2 sin 2 x − 32 cos 2 x A1 [5] (ii) Integrate to obtain form px + q cos 2x + r sin 2x M1 Obtain 52 x − cos 2 x − 34 sin 2 x A1 Substitute limits in integral of form px + q cos 2x + r sin 2x and attempt simplification DM1 Obtain 85 π + 14 or exact equivalent A1 [4]

This question in 9709/23 May/June 2012

5 y B (q, cos q ) C R x O A 12p The diagram shows the curve y cos x, for 0 2π. A rectangle OABC is drawn, where B is the = ≤x ≤1 point on the curve with x-coordinate θ, and A and C are on the axes, as shown. The shaded region R is bounded by the curve and by the lines x θ and y 0. = = (i) Find the area of R in terms of θ. [2] (ii) The area of the rectangle OABC is equal to the area of R. Show that 1 θ θ . −sinθ cos = [1] 1 θn (iii) Use the iterative formula −sin , with initial value θ1 0.5, to determine the value θn+1 = cos θn = of θ correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

6 marks

Mark scheme: 5 (i) Attempt to integrate and use limits θ and π M1 Obtain 1– sin θ A1 [2] (ii) State that area of rectangle = θcos θ, equate area of rectangle to area of R and rearrange to given equation B1 [1] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 0.56 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (0.555, 0.565) B1 [3]

This question in 9709/21 Oct/Nov 2012

6 (a) Use the trapezium rule with two intervals to estimate the value of 1 1 dx, 6 2ex ä 0 + giving your answer correct to 2 decimal places. [3] (b) Find dx. [4] (ex −2)2 ä e2x

7 marks

Mark scheme: 6 (a) State or imply correct ordinates 0.125, 0.08743…, 0.21511… B1 Use correct formula, or equivalent, correctly with h = 0.5 and three ordinates M1 Obtain answer 0.11 with no errors seen A1 [3] (b) Attempt to expand brackets and divide by e2x M1 Integrate a term of form ke−x or ke−2x correctly A1 Obtain 2 correct terms A1 Fully correct integral x + 4e−x - 2e−2x + c A1 [4]

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 y x O 1p 2 The diagram shows the part of the curve y for 0 2π. = √(2 −sin x) ≤x ≤1 (i) Use the trapezium rule with 2 intervals to estimate the value of 12π dx, ã 0 √(2 −sin x) giving your answer correct to 2 decimal places. [3] (ii) The line y x intersects the curve y at the point P. Use the iterative formula = = √(2 −sin x) xn+1 = √(2 −sin xn) to determine the x-coordinate of P correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

6 marks

Mark scheme: 4 (i) State or imply correct ordinates 1.4142…, 1.1370…, 1 B1 π Use correct formula, or equivalent, correctly with h = and three ordinates M1 4 Obtain answer 1.84 with no errors seen A1 [3] (ii) Use the iterative formula correctly at least once M1 Obtain final answer 1.06 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (1.055, 1.065) B1 [3]

This question in 9709/22 Oct/Nov 2012

5 y B (q, cos q ) C R x O A 12p The diagram shows the curve y cos x, for 0 2π. A rectangle OABC is drawn, where B is the = ≤x ≤1 point on the curve with x-coordinate θ, and A and C are on the axes, as shown. The shaded region R is bounded by the curve and by the lines x θ and y 0. = = (i) Find the area of R in terms of θ. [2] (ii) The area of the rectangle OABC is equal to the area of R. Show that 1 θ θ . −sinθ cos = [1] 1 θn (iii) Use the iterative formula −sin , with initial value θ1 0.5, to determine the value θn+1 = cos θn = of θ correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

6 marks

Mark scheme: 5 (i) Attempt to integrate and use limits θ and π M1 Obtain 1– sin θ A1 [2] (ii) State that area of rectangle = θcos θ, equate area of rectangle to area of R and rearrange to given equation B1 [1] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 0.56 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (0.555, 0.565) B1 [3]

This question in 9709/23 Oct/Nov 2012

6 (a) Use the trapezium rule with two intervals to estimate the value of 1 1 dx, 6 2ex ä 0 + giving your answer correct to 2 decimal places. [3] (b) Find dx. [4] (ex −2)2 ä e2x

7 marks

Mark scheme: 6 (a) State or imply correct ordinates 0.125, 0.08743…, 0.21511… B1 Use correct formula, or equivalent, correctly with h = 0.5 and three ordinates M1 Obtain answer 0.11 with no errors seen A1 [3] (b) Attempt to expand brackets and divide by e2x M1 Integrate a term of form ke−x or ke−2x correctly A1 Obtain 2 correct terms A1 Fully correct integral x + 4e−x - 2e−2x + c A1 [4]

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

3 (i) Show that 12 sin2x cos2x [3] ≡32(1 −cos 4x). (ii) Hence show that 13π π 3√3 12 sin2x cos2x dx . ã 1 = 8 + 16 [3] 4π

6 marks

Mark scheme: 3 (i) Either Use sin 2x = 2sin x cos x to convert integrand to k sin2 2x M1 Use cos 4x = 1 – 2 sin2 2x M1 1 1 State correct expression − cos 4 x or equivalent A1 2 2 Or 1 − cos 2 x 1 − cos 2 x Use cos2 x = and/or x = to obtain an equation in cos 2 x only M1 2 2 1 + cos 4 x Use cos2 2 x = M1 2 1 1 State correct expression − cos 4 x or equivalent A1 [3] 2 2 3 3 (ii) State correct integral x − sin 4 x , or equivalent B1 2 8 Attempt to substitute limits, using exact values M1 Obtain given answer correctly A1 [3] 3

This question in 9709/21 May/June 2013

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

9 marks

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

This question in 9709/22 May/June 2013

3 (i) Show that 12 sin2x cos2x [3] ≡32(1 −cos 4x). (ii) Hence show that 13π π 3√3 12 sin2x cos2x dx . ã 1 = 8 + 16 [3] 4π

6 marks

Mark scheme: 3 (i) Either Use sin 2x = 2sin x cos x to convert integrand to k sin2 2x M1 Use cos 4x = 1 – 2 sin2 2x M1 1 1 State correct expression − cos 4 x or equivalent A1 2 2 Or 1 − cos 2 x 1 − cos 2 x Use cos2 x = and/or x = to obtain an equation in cos 2 x only M1 2 2 1 + cos 4 x Use cos2 2 x = M1 2 1 1 State correct expression − cos 4 x or equivalent A1 [3] 2 2 3 3 (ii) State correct integral x − sin 4 x , or equivalent B1 2 8 Attempt to substitute limits, using exact values M1 Obtain given answer correctly A1 [3] 3

This question in 9709/23 May/June 2013

2 1 (i) Find dx. [2] Ô 4x −1 7 2 (ii) Hence find dx, expressing your answer in the form ln a, where a is an integer. [3] 4x Ô1 −1

5 marks

Mark scheme: 1 (i) State indefinite integral of the form k ln (4x – 1), where k = 2, 4, or ½ M1 State correct integral ½ ln (4x – 1) A1 [2] (ii) Substitute limits correctly M1 Use law for the logarithm of a power or a quotient M1 Obtain ln 3 correctly A1 [3]

This question in 9709/22 Oct/Nov 2013

6 (a) Find sin x x 2 dx. [4] Ó −cos (b) (i) Use the trapezium rule with 2 intervals to estimate the value of 1 20 cosec x dx, 1 Ó 40 giving your answer correct to 3 decimal places. [3] (ii) Using a sketch of the graph of y cosec x for 0 x explain whether the trapezium rule gives an under-estimate or an=over-estimate of< the≤1true20, value of the integral in part (i). [2]

9 marks

Mark scheme: 6 (a) Expand brackets and use sin2 x + cos2 x = 1 M1 Obtain 1 – sin 2x A1 Integrate and obtain term of form ±k cos 2x, where k = ½, 1 or 2 M1 cos 2 x State correct integral x + ( + c ) A1 [4] 2 (b) (i) State or imply correct ordinates 1.4142…, 1.0823…, 1 B1 π Use correct formula, or equivalent, correctly with h = and three ordinates M1 8 Obtain answer 0.899 with no errors seen A1 [3] (ii) Make a recognisable sketch of y = cosec x for 0 < x Y 12 π B1 Justify statement that the trapezium rule gives an over-estimate B1 [2] GCE AS LEVEL – October/November 2013 9709 22

This question in 9709/22 Oct/Nov 2013

7 y R x O a The diagram shows part of the curve y 8x 12ex. The shaded region R is bounded by the curve and = + 1 by the lines x 0, y 0 and x a, where a is positive. The area of R is equal to 2. = = = (i) Find an equation satisfied by a, and show that the equation can be written in the form O@2 A a . −ea 8 = O@2 A (ii) Verify by calculation that the equation a has a root between 0.2 and 0.3. [2] −ea 8 = _P2 Q (iii) Use the iterative formula to determine this root correct to 2 decimal places. −ean 8 an+1 = Give the result of each iteration to 4 decimal places. [3]

5 marks

Mark scheme: 7 (i) Integrate to obtain terms 4x2 and 12 xe B1 + B1 Substitute limits correctly M1 2 1 a 1 1 Obtain correct equation in any form 4 a + e − = A1 2 2 2 Rearrange to given answer correctly A1 [5] 2 − ea (ii) Consider sign of − a , or equivalent M1 8 Complete the argument correctly with appropriate calculations A1 [2] (f (0.2 ) = .0112, f (0.3) = −.0015) (iii) Use the iterative formula correctly at least once M1 Obtain final answer 0.29 A1 Show sufficient iterations to justify its accuracy to 2 d.p. B1 0x = 0.2 0x = 0.25 0x = 0.3 0.3120 0.2992 0.2851 0.2815 0.2853 0.2894 0.2905 0.2894 0.2879 or show there is a sign change in the interval (0.285, 0.295) [3]

This question in 9709/22 Oct/Nov 2013

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 (a) Find 4 cos2 1 [3] Ó 21 d1. 6 1 (b) Find the exact value of dx. [4] Ô 2x 3 −1 +

7 marks

Mark scheme: 3 (a) Express integrand in the form p cos θ + 2 M1 State correct 2 cos θ + 2 A1 Integrate to obtain 2 sin θ + 2θ (+ c) A1 [3] (b) Integrate to obtain form k ln (2 x + 3) M1 1 Obtain correct ln (2 x + 3) A1 2 Apply limits correctly DM1 1 Obtain ln 15 A1 [4] 2

This question in 9709/23 Oct/Nov 2014

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

6 y P x O The diagram shows part of the curve with equation y = 4 sin2x + 8 sin x + 3 and its point of intersection P with the x-axis. (i) Find the exact x-coordinate of P. [3] (ii) Show that the equation of the curve can be written y = 5 + 8 sin x −2 cos 2x, and use integration to find the exact area of the shaded region enclosed by the curve and the axes. [6]

9 marks

Mark scheme: 6 (i) Solve three-term quadratic equation for sin x M1 Obtain at least sin x = − 12 and no errors seen A1 Obtain x = 76 π A1 [3] (ii) State sin 2 x = 12 − 12 cos2 x B1 Obtain given 5 + 8sin x − 2cos2 x with necessary detail seen B1 Integrate to obtain expression of form ax + b cos x + c sin 2 x M1 Obtain correct 5 x − 8cos x − sin 2 x A1 Apply limits 0 and their x-value correctly M1 depM Obtain 356 π + 72 3 + 8 or exact equivalent A1 [6] 4 dy

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

6 y P x O The diagram shows part of the curve with equation y = 4 sin2x + 8 sin x + 3 and its point of intersection P with the x-axis. (i) Find the exact x-coordinate of P. [3] (ii) Show that the equation of the curve can be written y = 5 + 8 sin x −2 cos 2x, and use integration to find the exact area of the shaded region enclosed by the curve and the axes. [6]

9 marks

Mark scheme: 6 (i) Solve three-term quadratic equation for sin x M1 Obtain at least sin x = − 12 and no errors seen A1 Obtain x = 76 π A1 [3] (ii) State sin 2 x = 12 − 12 cos2 x B1 Obtain given 5 + 8sin x − 2cos2 x with necessary detail seen B1 Integrate to obtain expression of form ax + b cos x + c sin 2 x M1 Obtain correct 5 x − 8cos x − sin 2 x A1 Apply limits 0 and their x-value correctly M1 depM Obtain 356 π + 72 3 + 8 or exact equivalent A1 [6] 4 dy

This question in 9709/23 May/June 2015

cos 2x 9 cosx 5 5 (i) Show that 2 cos x 1. [3] + + cosx 4  + + 0 cos 4x 9 cos 2x 5 (ii) Hence find the exact value of dx. [4] + + Ô cos 2x 4 −0 +

7 marks

Mark scheme: 5 (i) Use cos2 x = 2cos 2 x − 1 and attempt factorisation of numerator M1 Obtain (2cos x + 1)(cos x + 4) A1 Confirm given result 2cos x + 1 A1 [3] (ii) Express integrand as 2cos2 x + 1 B1 Integrate to obtain sin 2 x + x B1 Apply limits correctly to integral of form k1 sin 2 x + k 2 x M1 Obtain 2π A1 [4] dy

This question in 9709/21 Oct/Nov 2016

5 y R x O 6 1 The diagram shows the curve y 1 e for 0 The region bounded by the curve and the = ?0 + 3x1 ≤x ≤6. lines x 0, x 6 and y 0 is denoted by R. = = = (i) Use the trapezium rule with 2 strips to find an estimate of the area of R, giving your answer correct to 2 decimal places. [3] (ii) With reference to the diagram, explain why this estimate is greater than the exact area of R. [1] (iii) The region R is rotated completely about the x-axis. Find the exact volume of the solid produced. [4]

8 marks

Mark scheme: 5 (i) State or imply correct ordinates 2, 1 + e, 1 + e 2 or decimal equivalents B1 Use correct formula, or equivalent, correctly with h = 3 and three ordinates M1 Obtain answer 12.25 with no errors seen A1 [3] (ii) Refer to top of each trapezium being above curve or equivalent B1 [1] (1 + e 1 (iii) State or imply volume is 3 x ) dx B1 ∫ π 3 x with or without π M1 Integrate to obtain form k1 x + k 2e 1 3 x ) or x + 3e 1 Obtain correct π( x + 3e 1 3 x A1 Obtain π(3 + 3e 2 ) or exact equivalent A1 [4]

This question in 9709/23 Oct/Nov 2016

6 y x O 160 The diagram shows the curve y tan 2x for 0 The shaded region is bounded by the curve = ≤x ≤160. and the lines x 1 and y 0. = 60 = (i) Use the trapezium rule with two intervals to find an approximation to the area of the shaded region, giving your answer correct to 3 significant figures. [3] … … … … … … … … … … … … … … … … (ii) Find the exact volume of the solid formed when the shaded region is rotated completely about the x-axis. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 6(i) State or imply correct y-values 0, 1 2 6 6 tan , tan π π B1 degree mode when working out π tan 6 etc. this gives 0.00915 and 0.0183. Allow B1. Use correct formula, or equivalent, with 1 12 h π = and y–values M1 Must be convinced they have considered 3 values for y for M1 Obtain 0.378 A1 Total: 3 Question Answer Marks Guidance 6(ii) State or imply ( ) 2 sec 2 1 d x x π − ∫ B1 Integrate to obtain 1 2 tan2 k x k x + , any non-zero constants including π or not M1 Obtain 1 2 tan2x x − or ( ) 1 2 tan2x x π − A1 Obtain ( ) 1 1 2 6 3 π π − or equivalent A1 Total: 4

This question in 9709/21 May/June 2017

7 (a) Find 2 cos cos 1 [4] Ó 1 −3 1 + d1. … … … … … … … … … … … … … … … … … … … … … … … … … @ A 4 1 (b) (i) Find dx. [2] 2x Ô 2x 1 + + … … … … … … … … … … … … … … 4 @ A 4 1 (ii) Hence find dx, giving your answer in the form ln k. [3] 2x 1 2x + Ô 1 + … … … … … … … …

9 marks

Mark scheme: 7(a) 2 2cos cos 3 d θ θ θ − − ∫ B1 Attempt use of identity to obtain integrand involving cos2θ and cosθ M1 Integrate to obtain form 1 2 3 sin2 sin k k k θ θ θ + + for non-zero constants M1 Obtain 1 2 sin2 sin 2 c θ θ θ − − + A1 Total: 4 Question Answer Marks Guidance 7(b)(i) Integrate to obtain form ( ) ( ) 1 2 ln 2 1 ln k x k x + + or ( ) ( ) 1 2 ln 2 1 ln 2 k x k x + + M1 Obtain ( ) 1 2 2ln 2 1 ln x x + + or ( ) ( ) 1 2 2ln 2 1 ln 2 x x + + A1 Total: 2 7(b)(ii) Use relevant logarithm power law for expression obtained from application of limits M1 Use relevant logarithm addition / subtraction laws M1 Obtain ln18 A1 Total: 3

This question in 9709/22 May/June 2017

7 (a) Find 2 cos cos 1 [4] Ó 1 −3 1 + d1. … … … … … … … … … … … … … … … … … … … … … … … … … @ A 4 1 (b) (i) Find dx. [2] 2x Ô 2x 1 + + … … … … … … … … … … … … … … 4 @ A 4 1 (ii) Hence find dx, giving your answer in the form ln k. [3] 2x 1 2x + Ô 1 + … … … … … … … …

9 marks

Mark scheme: 7(a) 2 2cos cos 3 d θ θ θ − − ∫ B1 Attempt use of identity to obtain integrand involving cos2θ and cosθ M1 Integrate to obtain form 1 2 3 sin2 sin k k k θ θ θ + + for non-zero constants M1 Obtain 1 2 sin2 sin 2 c θ θ θ − − + A1 Total: 4 Question Answer Marks Guidance 7(b)(i) Integrate to obtain form ( ) ( ) 1 2 ln 2 1 ln k x k x + + or ( ) ( ) 1 2 ln 2 1 ln 2 k x k x + + M1 Obtain ( ) 1 2 2ln 2 1 ln x x + + or ( ) ( ) 1 2 2ln 2 1 ln 2 x x + + A1 Total: 2 7(b)(ii) Use relevant logarithm power law for expression obtained from application of limits M1 Use relevant logarithm addition / subtraction laws M1 Obtain ln18 A1 Total: 3

This question in 9709/23 May/June 2017

4 4 (a) Find [4] + sin21 Ô 1 d1. −sin21 … … … … … … … … … … … a 2 (b) Given that dx ln 16, find the value of the positive constant a. [4] 3x 1 = Ô0 + … … … … … … … … … … …

8 marks

Mark scheme: 4(a) Obtain integrand of form a sec 2 θ+ b M1 Obtain correct 5sec 2 θ− 1 A1 Integrate to obtain form a tanθ+ bθ M1 Obtain 5tanθ− θ+ c A1 4 4(b) Obtain integral of form k ln(3 x + 1) *M1 Apply limits and obtain 23 ln(3a + 1) = ln16 A1 Obtain equation with no presence of ln DM1 Obtain 21 A1 4

This question in 9709/21 Oct/Nov 2017

4 4 (a) Find [4] + sin21 Ô 1 d1. −sin21 … … … … … … … … … … … a 2 (b) Given that dx ln 16, find the value of the positive constant a. [4] 3x 1 = Ô0 + … … … … … … … … … … …

8 marks

Mark scheme: 4(a) Obtain integrand of form a sec 2 θ+ b M1 Obtain correct 5sec 2 θ− 1 A1 Integrate to obtain form a tanθ+ bθ M1 Obtain 5tanθ− θ+ c A1 4 4(b) Obtain integral of form k ln(3 x + 1) *M1 Apply limits and obtain 23 ln(3a + 1) = ln16 A1 Obtain equation with no presence of ln DM1 Obtain 21 A1 4

This question in 9709/23 Oct/Nov 2017

a 1 2x 26 It is given that 1 e dx 10, where a is a positive constant. Ó 0 + = P Q 15 (i) Show that a 2 ln −a1 . [6] = 2a 4 e + … … … … … … … … … … … … … … … … … … … … … … … (ii) Use the equation in part (i) to show by calculation that 1.5 a 1.6. [2] < < … … … … … … … … … … … … (iii) Use an iterative formula based on the equation in part (i) to find the value of a correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … …

11 marks

Mark scheme: 6(i) Rewrite integrand as 1 2 1 2e e + + x B1 Integrate to obtain form 1 2 1 2 e e + + x x x k k M1 Obtain 1 2 4e e + + x x x A1 Use limits to obtain 1 2 4e e 5 10 + + − = a a a A1 Rearrange as far as 1 2e ... = a including use of 1 1 1 2 2 2 4e e e (4 e ) + = + a a a a M1 Confirm 1 2 15 2ln 4 e   − =   +   a a a A1 AG; necessary detail needed 6 6(ii) Consider sign of 1 2 15 2ln 4 e   − −   +   a a a for 1.5 and 1.6 or equivalent M1 Obtain 0.08... − and 0.06… or equivalents and justify conclusion A1 2 6(iii) Use iterative process correctly at least once M1 Obtain final answer 1.56 A1 Show sufficient iterations to 5 sf to justify answer or show sign change in interval (1.555,1.565) A1 3

This question in 9709/23 May/June 2018

6 y R x 0 O The diagram shows the curve with equation 1 for 0 y The region R is 3 cos2 12x ! = / + ≤x ≤0. bounded by the curve, the axes and the line x = 0. (i) Use the trapezium rule with two intervals to find an approximation to the area of R, giving your answer correct to 3 significant figures. [3] … … … … … … … … … … … … … … … … … … (ii) The region R is rotated completely about the x-axis. Without using a calculator, find the exact volume of the solid produced. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(i) Use y values 2, 2.5 , 1 or equivalents B1 Use correct formula, or equivalent, with 1 2 h π = and three y values M1 Obtain 1 1 2 2 (2 2 2.5 1) π × + + or equivalent and hence 4.84 A1 3 Question Answer Marks Guidance 6(ii) State or imply volume is 2 1 2 (1 3cos ) d x x π + ∫ B1 Allow if π appears later; condone omission of dx Use appropriate identity to express integrand in form 1 2 cos k k x + M1 Obtain 5 3 2 2 ( cos ) d x x π + ∫ or 5 3 2 2 ( cos ) d x x + ∫ A1 Condone omission of dx Integrate to obtain 5 3 2 2 ( sin ) x x π + or 5 3 2 2 sin x x + A1 Obtain 2 5 2 π with no errors seen A1 5

This question in 9709/21 Oct/Nov 2018

6 y R x 0 O The diagram shows the curve with equation 1 for 0 y The region R is 3 cos2 12x ! = / + ≤x ≤0. bounded by the curve, the axes and the line x = 0. (i) Use the trapezium rule with two intervals to find an approximation to the area of R, giving your answer correct to 3 significant figures. [3] … … … … … … … … … … … … … … … … … … (ii) The region R is rotated completely about the x-axis. Without using a calculator, find the exact volume of the solid produced. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(i) Use y values 2, 2.5 , 1 or equivalents B1 Use correct formula, or equivalent, with 1 2 h π = and three y values M1 Obtain 1 1 2 2 (2 2 2.5 1) π × + + or equivalent and hence 4.84 A1 3 Question Answer Marks Guidance 6(ii) State or imply volume is 2 1 2 (1 3cos ) d x x π + ∫ B1 Allow if π appears later; condone omission of dx Use appropriate identity to express integrand in form 1 2 cos k k x + M1 Obtain 5 3 2 2 ( cos ) d x x π + ∫ or 5 3 2 2 ( cos ) d x x + ∫ A1 Condone omission of dx Integrate to obtain 5 3 2 2 ( sin ) x x π + or 5 3 2 2 sin x x + A1 Obtain 2 5 2 π with no errors seen A1 5

This question in 9709/23 Oct/Nov 2018

18 3 6 (a) Show dx ln 27. [4] 2x = thatÔ2 … … … … … … … … … 1 60 3 (b) Find the exact value of 4 sin2 2x dx. Show all necessary working. [5] Ó 0 … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) Obtain 3 2 ln x or 3 2 ln(2 ) x or 3 2 ln kx Use subtraction law of logarithms correctly, showing sufficient detail M1 216 ln216 ln8 ln 8   − =     Use power law of logarithms correctly M1 ( ) ( ) ln ln n n kx kx = Confirm ln27 with sufficient working and no incorrect working A1 AG 4 6(b) Use appropriate identity to express integrand in form 1 2 cos3 k k x + *M1 1 0 k ≠ . Allow 3 2 2 x × for 3x Obtain correct 2 2cos3x − A1 Integrate to obtain form 3 4 sin3 k x k x + DM1 Obtain correct 2 3 2 sin3 x x − A1 Use limits to obtain 1 2 3 3 π − or exact equivalent A1 5

This question in 9709/22 Oct/Nov 2019

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

a 4 7 It is given that 8x dx 10, where a is a positive constant. 2x 1 + = Ô0 + (a) Show that a 2.5 ln 2a 1 . [4] = −0.5 + … … … … … … … … … … … … … … … … … … … … … … … (b) Using the equation in part (a), show by calculation that 1 a 2. [2] < < … … … … … … … … … (c) Use an iterative formula, based on the equation in part (a), to find the value of a correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3] … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(a) Integrate to obtain the form 2 1 2 ln(2 1) + + k x k x *M1 Obtain correct 2 2ln(2 1) 4 + + x x A1 Use limits correctly and attempt rearrangement DM1 Confirm 2.5 0.5ln(2 1) = − + a a AG A1 4 Question Answer Marks 7(b) Consider sign of 2.5 0.5ln(2 1) − − + a a or equivalent for 1 and 2 M1 Obtain 0.3... − and 0.6... or equivalents and justify conclusion A1 2 7(c) Use iteration process correctly at least once M1 Obtain final answer 1.358 A1 Show sufficient iterations to 6 sf to justify answer or show a sign change in the interval [1.3575, 1.3585] A1 3

This question in 9709/22 May/June 2020

a 4 7 It is given that 8x dx 10, where a is a positive constant. 2x 1 + = Ô0 + (a) Show that a 2.5 ln 2a 1 . [4] = −0.5 + … … … … … … … … … … … … … … … … … … … … … … … (b) Using the equation in part (a), show by calculation that 1 a 2. [2] < < … … … … … … … … … (c) Use an iterative formula, based on the equation in part (a), to find the value of a correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3] … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(a) Integrate to obtain the form 2 1 2 ln(2 1) + + k x k x *M1 Obtain correct 2 2ln(2 1) 4 + + x x A1 Use limits correctly and attempt rearrangement DM1 Confirm 2.5 0.5ln(2 1) = − + a a AG A1 4 Question Answer Marks 7(b) Consider sign of 2.5 0.5ln(2 1) − − + a a or equivalent for 1 and 2 M1 Obtain 0.3... − and 0.6... or equivalents and justify conclusion A1 2 7(c) Use iteration process correctly at least once M1 Obtain final answer 1.358 A1 Show sufficient iterations to 6 sf to justify answer or show a sign change in the interval [1.3575, 1.3585] A1 3

This question in 9709/23 May/June 2020

3 y A B x O 1 2 The diagram shows the curve y The curve crosses the y-axis at the point A, and the point B = + e−2x. on the curve has x-coordinate 1. The shaded region is bounded by the curve and the line segment AB. Find the exact area of the shaded region. [5] … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 Integrate to obtain form 2 e x ax b − + Obtain correct 2 1 2 e 2 x x − − A1 Apply limits to obtain 2 5 1 e 2 2 − − A1 Attempt to find area of relevant trapezium M1 Obtain 2 5 1 e 2 2 − + and subtract to obtain 2 e− or exact equivalent A1 5

This question in 9709/21 Oct/Nov 2020

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

7 marks

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

This question in 9709/22 Oct/Nov 2020

3 y A B x O 1 2 The diagram shows the curve y The curve crosses the y-axis at the point A, and the point B = + e−2x. on the curve has x-coordinate 1. The shaded region is bounded by the curve and the line segment AB. Find the exact area of the shaded region. [5] … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 Integrate to obtain form 2 e x ax b − + M1 Obtain correct 2 1 2 e 2 x x − − A1 Apply limits to obtain 2 5 1 e 2 2 − − A1 Attempt to find area of relevant trapezium M1 Obtain 2 5 1 e 2 2 − + and subtract to obtain 2 e− or exact equivalent A1 5

This question in 9709/23 Oct/Nov 2020

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

4 6 6 (a) Use the trapezium rule with three intervals to find an approximation to dx. Give your 1 x Ô1 + answer correct to 5 significant figures. [3] … … … … … … … … … … … 4 1 (b) Find the exact value of Ó 1 2e 2x−2 dx. [3] … … … … … … … … … … … (c) y y = 2e 2x−21 6 y = 1 + x x O 1 4 6 1 The diagram shows the curves y and y 2e 2x−2 which meet at a point with x-coordinate 4. 1 x = = The shaded region is bounded by the+ two curves and the line x 1. = Use your answers to parts (a) and (b) to find an approximation to the area of the shaded region. Give your answer correct to 3 significant figures. [2] … … … … … … … … (d) State, with a reason, whether your answer to part (c) is an over-estimate or under-estimate of the exact area of the shaded region. [1] … … … … … …

9 marks

Mark scheme: 6(a) Use y-values 6 6 3, , , 2 1 2 1 3 + + or 3, 6 2 6, 3 3 3, 2 − − or 3, 2.4853..., 2.19615..., 2 B1 Use correct formula, OE, with 1 = h M1 Allow 3 separate trapezia of width 1. Obtain 7.1814 A1 AWRT 3 6(b) Integrate to obtain form 1 2 2 e − x k M1 1 ≠ k If 2 = k , integration must be implied by use of square bracket notation or by substitution of limits. Obtain correct 1 2 2 4e − x A1 or exact equivalents Obtain 3 2 4 4e − − A1 or exact equivalents 3 6(c) Evaluate answer to part (a) minus answer to part (b) M1 Obtain 4.07 A1 2 6(d) State over-estimate with reference to top of each trapezium being above the [first] curve, or clear equivalent, e.g. concave up so over- estimate or convex down so over-estimate. B1 1

This question in 9709/21 May/June 2021

a+14 1 4 Given that dx ln 2, find the value of the positive constant a. [5] 3x = Ôa … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 4 k x Apply limits and obtain 1 1 3 3 ln( 14) ln ln2 + − = a a or 1 1 3 3 ln(3 42) ln(3 ) ln 2 + − = a a A1 OE Use one relevant logarithm property correctly M1 Apply correct process to obtain equation without logarithms M1 OE. M0 if incorrect logarithm property used. Obtain 14 8 + = a a or equivalent and hence 2 = a A1 5

This question in 9709/22 Oct/Nov 2021

3 The diagram shows the curve with equation y 3 sin x sin 2x for 0 The curve meets the x-axis at the origin and at the points with x-coordinates= −3a and ≤x ≤π. π. (a) Find the exact value of a. [3] … … … … … … … (b) Find the area of the shaded region. [4] … … … … … … … … …

7 marks

Mark scheme: 3(a) Attempt to find x-value from 3sin 3sin 2 0 x x using identity for sin 2x M1 Obtain at least 1 2 cos  x A1 Obtain 1 3  A1 SC B3 can be spotted from sin sin 2  x x 3 3(b) Integrate to obtain form 1 2 cos cos2  k x k x *M1 non-zero constants 1 2 , k k M0 for 3cos 6cos2  x x Obtain correct 3 2 3cos cos2   x x A1 Attempt value of integral using their lower limit (in radians) and  correctly DM1 Allow one sign error Obtain 27 4 A1 OE 4

This question in 9709/21 May/June 2022

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

9 marks

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

This question in 9709/21 Oct/Nov 2022

6 y x O 4 6 The diagram shows the curves y = and y = 3e−x −3 for values of x between 0 and 4. The 3x + 2 shaded region is bounded by the two curves and the lines x = 0 and x = 4. Find the exact area of the shaded region, giving your answer in the form ln a + b + ced. [9] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6 6 *M1 for any constant 1k . Integrate to obtain form k1ln(3 x + 2) 3 x + 2 Obtain correct 2ln(3 x + 2) A1 Apply limits correctly DM1 Obtain 2ln14 − 2ln2 and hence ln49 A1 at this stage or later. Integrate 3e −−x 3 to obtain form k 2 e −+x k 3 x M1 for any non-zero constants 2k , 3k . Obtain correct −3e − x − 3 x A1 ( y1 − y2 )dx approach used. Apply limits to obtain −3e −4 − 12 + 3 A1 OE; implied if  Use correct procedure to find exact total area M1 Obtain ln49 + 9 + 3e−4 A1 9

This question in 9709/22 Oct/Nov 2022

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

1 2π 11 Find the exact value of 2 tan2 2x dx. [4] Ó 0 … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 1 Express integrand as 2sec 2 12 x − 2 B1 Integrate to obtain form a tan 12 x + bx M1 where ab  0 . Obtain correct 4tan 12 x − 2 x A1 Obtain 4 − π A1 4

This question in 9709/22 Feb/March 2023

a 4 3 5 It is given that dx ln 10, where a is a constant greater than 1. 1 2x x + = Ô1 + 3/ (a) Show that a 90 1 2a [5] = + −2. … … … … … … … … … … … … … … … … … … … … … … … (b) Use an iterative formula, based on the equation in (a), to find the value of a correct to 3 significant figures. Use an initial value of 1.7 and give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(a) Integrate to obtain form k1 ln(1 + 2 x ) + k 2 ln x *M1 k1  0, k 2  0 . Obtain correct 2ln(1 + 2 x ) + 3ln x A1 Use limits correctly and equate to ln10 DM1 Apply relevant logarithm properties correctly and arrange as far as DM1 a 3 = ... 3 −2 A1 AG Confirm given result a = 90(1 + 2a ) with sufficient detail 5 5(b) Use iteration process correctly at least once M1 Need to see 1.6848 . Obtain final answer 1.68 A1 Answer required to exactly 3 sf. Show sufficient iterations to 5 sf to justify answer or show a sign A1 change in the interval [1.675, 1.685] 3

This question in 9709/22 Feb/March 2023

a 3 It is given that 3e2x dx 12, where a is a positive constant. Ó 0 −1 = (a) Show that a 1 ln 9 23a . [4] 2 = + … … … … … … … … … … … … … (b) Use an iterative formula, based on the equation in (a), to find the value of a correct to 4 significant figures. Use an initial value of 1 and give the result of each iteration to 6 significant figures. [3] … … … … … … … … …

7 marks

Mark scheme: 3(a) Integrate to obtain the form 2 1 2 e  x k k x 1 2 0 k k . Obtain correct 2 3 2 e  x x A1 Use limits correctly and attempt rearrangement at least as far as 2 e ...  a DM1 For DM1, must be equated to 12 and simplified using a correct method. Do not condone verification. Confirm given result 1 2 2 3 ln(9 )   a a with sufficient detail A1 AG 4 3(b) Use iteration process correctly at least once M1 Need to see 1.13434 and 1.13895 . Obtain final answer 1.139 A1 Final answer needed to exactly 4 sf. Show sufficient iterations to 6 sf to justify answer or show a sign change in interval [1.1385, 1.1395] A1 3

This question in 9709/21 May/June 2023

1 @ A 3π 1 3 16 Show that 4 cos2 2x dx 3 [7] 4 Ô 1 + cos2x = + 6π −1. 4π … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 6 Express 2 4cos 2x in the form 1 2 cos4  k x k M1 where 1 2 0 k k . Obtain correct 2cos4 2  x A1 Allow unsimplified. State or imply 2 2 1 sec cos  x x B1 Maybe implied by integration. Integrate to obtain 3 4 sin 4 tan   k x k x x *M1 where 3 4 0  k k . Obtain correctly 1 2 sin 4 2 tan   x x x A1 Use limits correctly with correct values of 4 3 sin π and 1 3 tan π indicated DM1 Confirm given result 3 1 4 6 3 π 1   with sufficient detail A1 AG 7

This question in 9709/21 May/June 2023

3 y 2 x O 6 6 The diagram shows part of the curve y The shaded region is bounded by the curve and the = 2x 3. + lines x 6 and y 2. = = Find the exact area of the shaded region, giving your answer in the form a b, where a and b are −ln integers. [5] … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 Integrate to obtain the form ln(2 3) k x Obtain correct 3ln(2 3)  x A1 Allow unsimplified. Apply limits 0 and 6 correctly to obtain ln15 ln 3  k k *DM1 Allow unsimplified. Apply relevant logarithm properties correctly to obtain form ln b DM1 Obtain 12 ln125  A1 5

This question in 9709/22 May/June 2023

3 y 2 x O 6 6 The diagram shows part of the curve y The shaded region is bounded by the curve and the = 2x 3. + lines x 6 and y 2. = = Find the exact area of the shaded region, giving your answer in the form a b, where a and b are −ln integers. [5] … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 Integrate to obtain the form ln(2 3) k x Obtain correct 3ln(2 3)  x A1 Allow unsimplified. Apply limits 0 and 6 correctly to obtain ln15 ln3  k k *DM1 Allow unsimplified. Apply relevant logarithm properties correctly to obtain form lnb DM1 Obtain 12 ln125  A1 5

This question in 9709/23 May/June 2023

10 4 3 (a) Find dx, giving your answer in the form ln a, where a is an integer. [4] 2x Ô 4 −5 … … … … … … … … … … … … … … 10 (b) Find the exact value of [2] e2x−5 dx. Ó 4 … … … … … … … …

6 marks

Mark scheme: 3(a) Obtain 2ln(2 x − 5) B1 Apply limits correctly M1 For integral of form k ln(2 x − 5) . Use one relevant logarithm property correctly M1 For integral of form k ln(2 x − 5) . Apply second logarithm property correctly and obtain ln25 A1 4 3(b) 1 2 x− 5 B1 Integrate to obtain e 2 1 15 1 3 B1FT or exact equivalent, FT on their ke 2 x −.5 Obtain final answer e − e 2 2 2

This question in 9709/21 Oct/Nov 2023

6 (a) Show that cosec 3 sin 2 4 sin3 4 6 cos 4 cos2 . [3] … … … … … … … … … … … … … … … … (b) Solve the equation cosec 3 sin 2 4 sin3 3 0 + + = for 0. [3] −π < < … … … … … … … … … … … … … … … (c) Find cosec 3 sin 2 4 sin3 d . [3] + … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) 1 B1 Use cosec= sin Express in terms of sin and cos only M1 Dependent on B1. Obtain given result 4 + 6cos− 4cos 2  with sufficient detail A1 AG 3 6(b) Attempt use of formula to solve 3-term quadratic equation as far as M1 where −1 k1 .1 cos= k1 Solve 4cos 2 − 6cos− 7 = 0 to obtain at least cos= − 0.770... A1 6 − 148 or exact equivalent cos= . 8 Obtain −2.45 A1 or greater accuracy; and no others between − π and 0. 3 6(c) Express cos 2  term in the form k 2 + k3 cos2 M1 where k 2 k3  0 . Obtain integrand 6cos+ 2 − 2cos2 A1 Following the 3-term expression in cos from part (a). Integrate to obtain correct 6sin+ 2− sin2 A1 Condone absence of + c , but all in terms of . 3

This question in 9709/23 Oct/Nov 2023

4 y O 6 x The diagram shows the curve with equation y = 1 + e 0 .5 x . The shaded region is bounded by the curve and the straight lines x = 0, x = 6 and y = 0 . (a) Use the trapezium rule with three intervals to find an approximation to the area of the shaded region. Give your answer correct to 3 significant figures. [3] … … … … … … … … … … … … … … … … … … … (b) The shaded region is rotated completely about the x-axis. Find the exact volume of the solid produced. [4] … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(a) 2 3 B1 Or decimal equivalents Use y-values 2, 1 + e, 1 + e , 1 + e Use correct formula, or equivalent, with h = 2 M1 Obtain 15.7 A1 Or greater accuracy. 3 4(b) 0.5 x B1 Implied if π appears only at the end. State or imply that volume is π  (1 + e ) dx Integrate to obtain form k1 x + k 2e 0.5 x M1 Where k1k 2  0 with or without π . Obtain π( x + 2e 0.5 x ) or x + 2e 0.5 x A1 Obtain 4π + 2πe 3 A1 Or exact equivalent. 4

This question in 9709/22 Feb/March 2024

6 y x O 1 r 6 The diagram shows the curve with equation y = sin 2x + sin 2 2x for 0 G x G 16 r. The shaded region is r and y = 0 . bounded by the curve and the straight lines x = 16 (a) 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] … … … … … … … … … … … … … … … … … … … (b) The shaded region is rotated completely about the x-axis. Find the exact volume of the solid produced. [6] … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) Use y-values 2 2 1 1 1 1 6 6 3 3 (0), sin π sin π, sin π sin π   or decimal equivalents 0 , 0.75 or 0.866, 1.616 or 1.271. Use correct formula, or equivalent, with 1 12 π h  M1 Must be using ‘y’ values. May do as 2 separate trapezia (0.113359 + 0.27976). Obtain 0.39 A1 Allow 0.393 but not greater accuracy. 3 Question Answer Marks Guidance 6(b) Use  2 π (sin2 sin 2 ) d x x x   M1 OE Express integrand in the form 1 2 3 sin 2 cos4 k x k k x   *M1 Where 1 2 3 0 k k k  . Obtain correct 1 1 2 2 sin 2 cos 4 x x   A1 Or π times this. Integrate to obtain 1 1 1 2 2 8 cos 2 sin 4 x x x    A1 FT Following their integrand only if of correct form. Apply limits correctly to obtain exact value DM1 Obtain volume = 2 1 1 1 4 12 16 π π π 3   A1 Or exact equivalent. 6

This question in 9709/21 May/June 2024

7 The polynomial p ( )x is defined by p ( )x = 9 x 3 + 6x 2 + 12x + k , where k is a constant. (a) Find the quotient when p ( )x is divided by ( 3x + 2) and show that the remainder is ( k - 8 ) . [3] … … … … … … … … … … … 6 p ( x) (b) It is given that dx = a + ln 64 , where a is an integer. + 2 y1 3 x Find the values of a and k. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … Additional page If you use the following page to complete the answer to any question, the question number must be clearly shown. … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(a) Carry out division at least as far as 2 1 3x n  M1 Or equivalent (inspection, …). Obtain quotient 2 3 4 x  A1 Confirm remainder is 8 k  A1 Answer given – necessary detail needed. SC B1 for correct use of factor theorem to show remainder is 8 k  . Alternative Method for Question 7(a) Synthetic division –2/3 9 6 12 k –6 0 –8 9 0 12 8 k  (M1) Allow one sign error. Obtain quotient 2 3 4 x  (A1) Confirm remainder is 8 k  (A1) 3 Question Answer Marks Guidance 7(b) Integrate to obtain at least a term in 3 x and term of form 2 ln(3 2) n x  *M1 Need to be using their answer to part (a). Obtain 3 1 3 4 ( 8)ln(3 2) x x k x     A1 FT on a quotient of 2 9 12 x  . Apply limits correctly to expression with three terms DM1 Obtain 235 a  A1 FT on a quotient of 2 9 12 x  . Equate logarithm term to ln64 and apply appropriate logarithm properties DM1 Obtain 17 k  A1 6 SC 2 marks for use of quotient 2 3 4 x  or 2 9 12 x  to obtain either 235 or 705 if no other marks are available.

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

5 The polynomial p ( )x is defined by p ( )x = 9 x 3 + 18x 2 + 5x + 4 . (a) Find the quotient when p ( )x is divided by ( 3x + 2) , and show that the remainder is 6. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … 2 p ( x) (b) Find the value of dx , giving your answer in the form a + ln b where a and b are integers. + 2 y0 3x [5] … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(a) Carry out division at least as far as 2 1 3  x k x M1 Or equivalent (inspection, …). Obtain quotient 2 3 4 1   x x A1 Confirm remainder is 6 A1 Answer given – necessary detail needed. SC B1 for use of the factor theorem to show remainder is 6 if no other marks are awarded. Alternative Method for Question 5(a) Synthetic division –2/3 9 18 5 4 –6 8 –2 9 12 –3 6 (M1) Obtain quotient 2 3 4 1   x x (A1) Confirm remainder is 6 (A1) 3 Question Answer Marks Guidance 5(b) Identify integrand as 2 6 3 4 1 3 2    x x x B1FT Following their quotient. Integrate to obtain at least 3 x and 2 ln(3 2)  k x terms *M1 Obtain 3 2 2 2ln(3 2)     x x x x A1 Apply limits and appropriate logarithm properties DM1 Obtain 14 ln16  A1 5

This question in 9709/22 May/June 2024

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

8 marks

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

This question in 9709/23 May/June 2024

5 The polynomial p ( )x is defined by p ( )x = 9 x 3 + 18x 2 + 5x + 4 . (a) Find the quotient when p ( )x is divided by ( 3x + 2) , and show that the remainder is 6. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … 2 p ( x) (b) Find the value of dx , giving your answer in the form a + ln b where a and b are integers. y 0 3x + 2 [5] … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(a) Carry out division at least as far as 2 1 3  x k x M1 Or equivalent (inspection, …). Obtain quotient 2 3 4 1   x x A1 Confirm remainder is 6 A1 Answer given – necessary detail needed. SC B1 for use of the factor theorem to show remainder is 6 if no other marks are awarded. Alternative Method for Question 5(a) Synthetic division –2/3 9 18 5 4 –6 8 –2 9 12 –3 6 (M1) Obtain quotient 2 3 4 1   x x (A1) Confirm remainder is 6 (A1) 3 Question Answer Marks Guidance 5(b) Identify integrand as 2 6 3 4 1 3 2    x x x B1 FT Following their quotient. Integrate to obtain at least 3 x and 2 ln(3 2)  k x terms *M1 Obtain 3 2 2 2ln(3 2)     x x x x A1 Apply limits and appropriate logarithm properties DM1 Obtain 14 ln16  A1 5

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

a 10 5 It is given that d x = 7 , where a is a constant greater than 1. y 2x + 1 a (a) Show that a = 3 0.5e 1 .4 ( 2 a + 1 ) - 0. 5 . [5] … … … … … … … … … … … … … … … … … … … … … … … … … (b) Use an iterative formula, based on the equation in part (a), to find the value of a correct to 3 significant figures. Use an initial value of 2 and give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 5(a) Obtain integral of form k ln(2 x + 1) *M1 Obtain correct 5ln(2 x + 1) A1 Apply limits correctly and equate to 7 DM1 Apply appropriate logarithm property to reach at least a 3 = ... DM1 3 1.4 A1 AG – necessary detail needed. Confirm a = 0.5e (2a + 1) − 0.5 5 5(b) Use iterative process correctly at least once M1 Obtain final answer 2.18 A1 Answer required to exactly 3 sf. Show sufficient iterations to 5 sf to justify answer or show a sign change in the A1 interval [2.175, 2.185] 3

This question in 9709/21 Oct/Nov 2024

6 y 3 2 y = 5x + 7 B 27 A y = 2 x + 5 x O 2 3 2 27 The diagram shows the curves with equations y = 5x + 7 and y = for x H 0 . 2x + 5 The curves meet at the point ( 2, 3) . 3 2 Region A is bounded by the curve y = 5x + 7 and the straight lines x = 0 , x = 2 and y = 0 . Region B is bounded by the two curves and the straight line x = 0 . (a) Use the trapezium rule with two intervals to find an approximation to the area of region A. Give your answer correct to 3 significant figures. [3] … … … … … … … … … … … … … … … … … (b) Find the exact total area of regions A and B. Give your answer in the form k ln m, where k and m are constants. [3] … … … … … … … … … … (c) Deduce an approximation to the area of region B. Give your answer correct to 3 significant figures. [1] … … … … … (d) State, with a reason, whether your answer to part (c) is an over-estimate or an under-estimate of the area of region B. [2] … … … … … … … … …

9 marks

Mark scheme: 6(a) Use y-values 3 7, 3 12 and 3 27, or decimal equivalents B1 1.913, 2.289, 3 Use correct formula, or equivalent, with h = 1 M1 May see 2 separate trapezia (2.1….+ 2.65…). Obtain 4.75 A1 3 6(b) Integrate to obtain form k ln(2 x + 5) M1 Obtain correct 272 ln(2 x + 5) A1 Condone inclusion of working for area in part (a). Apply limits 0 and 2 to obtain 272 ln 95 or exact equivalent of required form and A1 no extra terms 3 6(c) Obtain 3.19 B1 FT Following answers to (b) and (a). If incorrect answers for (a) and/or (b), then need to see their (b) (with no extra terms) – their (a). 1 6(d) State under-estimate … *B1 … because (a) is over-estimate due to tops of trapezia being above curve DB1 Or similarly detailed comment. 2

This question in 9709/22 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

3 y x O The diagram shows the curves y = e 2 x and y = 8e -x . The shaded region is bounded by the two curves and the y-axis. (a) Show that the x-coordinate of the point of intersection of the two curves is ln2. [2] … … … … … … … (b) Find the area of the shaded region. [3] … … … … … … … … … … …

5 marks

Mark scheme: 3(a) State 2e x = 8e− x and obtain 3x = ln8 or e x = 2 or equivalent B1 Confirm x = ln2 B1 AG Necessary detail needed. Alternative Method for Question 3(a) Substitute x = ln2 in either 2e x or 8e−x and obtain value 4 B1 Substitute in other expression, obtain value 4 and conclude B1 AG appropriately Necessary detail needed. 2 3(b) x 2 x − x 1 2 x B1 OE, involving separate integrations. Integrate 8e −− e to obtain −8e − 2 e Apply limits correctly and simplify to eliminate e and ln M1 Obtain −8 12 − 12  4 + 8 + 12 or equivalent, and hence 52 or A1 equivalent 3

This question in 9709/22 Feb/March 2025

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

7 (a) Prove that sin 2 2x + 4 cos 2 x cos 2x / 4 cos 4 x . [3] … … … … … … … … … … … … … … … (b) Find the set of possible values of the constant k for which the equation sin 2 2x + 4 cos 2 x cos 2x + 5 = k has no real solutions. [2] … … … … … … … … … 1 r 3 (c) Find the exact value of y 1 sin 2 t + 4 cos 2 b 12 tl cos t td . [4] - 3 r … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(a) Use correct identities to express in terms of sin x and cos x M1 Allow for 2sin 2 x cos 2 x + 4cos 2 x cos 2 x − sin 2 x , OE. ( ) Obtain 4sin 2 x cos 2 x + 4cos 2 x (cos 2 x − sin 2 x ) A1 OE 4 A1 Confirm 4cos x 3 7(b) Use identity from part (a) to obtain k  5 or k  9 B1 Obtain k ,5 k  9 B1 2 7(c) 2 1 4 t M1* State or imply integrand is 2cos 2 t or 4cos 2 Use double-angle identity to express in terms of cost M1 Obtain 1 + cost and integrate to obtain t + sin t A1 Use limits to obtain 23 π + 3 A1 4

This question in 9709/21 May/June 2025

11 8 1 Show that dx = ln a , where a is an integer to be found. [3] y 2 4x + 1 … … … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: Question Answer Marks Guidance 1 Integrate to obtain 2ln(4 x + 1) B1 Apply limits correctly to k ln(4 x + 1) and use at least one relevant logarithm property M1 Obtain 2ln45 − 2ln9 = 2ln5 or equivalent, and conclude ln25 A1 3

This question in 9709/22 May/June 2025

4 y P O x The diagram shows parts of the curves with equations y = 4e -2 x and y = 1 + 0.5 sin 3x . Point P is a point of intersection of the curves, and the shaded region is bounded by the two curves and the y-axis. (a) Show that the x-coordinate of P satisfies the equation x =-0. 5 ln ( 0. 25 + 0. 125 sin 3 x) . [1] … … … … … … (b) Use an iterative formula, based on the equation in part (a), to find the x-coordinate of P correct to 4 significant figures. Use an initial value of 0.5 and give the result of each iteration to 6 significant figures. [3] … … … … … … … … … … … (c) Hence find the area of the shaded region. Give your answer correct to 2 significant figures. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4(a) State 4e−2 x = 1 + 0.5sin3x and confirm x = −0.5ln(0.25 + 0.125sin3 x ) B1 AG Necessary detail needed. 1 4(b) Use iterative process correctly at least once M1 Obtain final answer 0.4912 A1 Answer required to exactly 4 significant figures. Show sufficient iterations to 6 significant figures to justify answer A1 or show a sign change in the interval [0.49115, 0.49125] 3 4(c) Integrate to obtain the form k1e −2 x + k 2 x + k3 cos3 x *M1 k1 , k2 , k3  0 OE, with separate integrals or ‘reversed’. Obtain correct −2e −2 x − x + 1 cos3 x A1 OE 6 −2 x 1 Allow for 2e + x − cos3 x 6 Apply limits 0 and their part (b) answer correctly and attempt evaluation DM1 Obtain 0.61 A1 4

This question in 9709/22 May/June 2025

4 y P O x The diagram shows parts of the curves with equations y = 4e -2 x and y = 1 + 0.5 sin 3x . Point P is a point of intersection of the curves, and the shaded region is bounded by the two curves and the y-axis. (a) Show that the x-coordinate of P satisfies the equation x =-0. 5 ln ( 0. 25 + 0. 125 sin 3 x) . [1] … … … … … … (b) Use an iterative formula, based on the equation in part (a), to find the x-coordinate of P correct to 4 significant figures. Use an initial value of 0.5 and give the result of each iteration to 6 significant figures. [3] … … … … … … … … … … … (c) Hence find the area of the shaded region. Give your answer correct to 2 significant figures. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4(a) State 4e−2 x = 1 + 0.5sin3x and confirm x = −0.5ln(0.25 + 0.125sin3 x ) B1 AG Necessary detail needed. 1 4(b) Use iterative process correctly at least once M1 Obtain final answer 0.4912 A1 Answer required to exactly 4 significant figures. Show sufficient iterations to 6 significant figures to justify answer A1 or show a sign change in the interval [0.49115, 0.49125] 3 4(c) Integrate to obtain the form k1e −2 x + k 2 x + k3 cos3 x *M1 k1 , k2 , k3  0 OE, with separate integrals or ‘reversed’. Obtain correct −2e −2 x − x + 1 cos3 x A1 OE 6 −2 x 1 Allow for 2e + x − cos3 x 6 Apply limits 0 and their part (b) answer correctly and attempt evaluation DM1 Obtain 0.61 A1 4

This question in 9709/23 May/June 2025

4 y P O x The diagram shows parts of the curves with equations y = 4e -2 x and y = 1 + 0.5 sin 3x . Point P is a point of intersection of the curves, and the shaded region is bounded by the two curves and the y-axis. (a) Show that the x-coordinate of P satisfies the equation x =-0. 5 ln ( 0. 25 + 0. 125 sin 3 x) . [1] … … … … … … (b) Use an iterative formula, based on the equation in part (a), to find the x-coordinate of P correct to 4 significant figures. Use an initial value of 0.5 and give the result of each iteration to 6 significant figures. [3] … … … … … … … … … … … (c) Hence find the area of the shaded region. Give your answer correct to 2 significant figures. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4(a) State 4e−2 x = 1 + 0.5sin3x and confirm x = −0.5ln(0.25 + 0.125sin3 x ) B1 AG Necessary detail needed. 1 4(b) Use iterative process correctly at least once M1 Obtain final answer 0.4912 A1 Answer required to exactly 4 significant figures. Show sufficient iterations to 6 significant figures to justify answer A1 or show a sign change in the interval [0.49115, 0.49125] 3 4(c) Integrate to obtain the form k1e −2 x + k 2 x + k3 cos3 x *M1 k1 , k2 , k3  0 OE, with separate integrals or ‘reversed’. Obtain correct −2e −2 x − x + 1 cos3 x A1 OE 6 −2 x 1 Allow for 2e + x − cos3 x 6 Apply limits 0 and their part (b) answer correctly and attempt evaluation DM1 Obtain 0.61 A1 4

This question in 9709/25 May/June 2025

1 Find 6 sin 2 x dx . [3] y … … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: Question Answer Marks Guidance 1 Use identity to express integrand in the form k1 + k 2 cos2 x M1 Obtain correct 3 − 3cos2 x A1 Integrate to obtain 3 x − 32 sin 2 x A1 Condone omission of + .c 3

This question in 9709/21 Oct/Nov 2025

5 y A B x O - 21 x The diagram shows the curve with equation y = 8e - 1. The curve meets the axes at the points A and B. The shaded region is bounded by the curve and the line segment AB. (a) Show that the x-coordinate of B is 6 ln 2. [2] … … … … … (b) Find the area of the shaded region. Give your answer in the form p ln 2- q , where p and q are positive integers. [5] … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(a) − 12 x M1 1 Attempt solution of 8e −=1 0 with use of a relevant logarithm property E.g. ln x = ln8 2 Confirm x = 6ln2 A1 AG – necessary detail needed. Need to see ln8 = 3ln2 or 64 = 2 6 OE used. Alternative Method for Question 5(a) − 12 x M1 Substitute x = 6ln2 in 8e − 1 and show use of a relevant logarithm property −3ln2 ln 18 A1 AG – necessary detail needed. Obtain 8e − 1 and hence 8e − 1 or equivalent and verify answer 0 2 5(b) − 12 x 12 x M1 Integrate 8e − 1 to obtain the form k e− − x 2 x − x A1 Obtain correct −16e − 1 2 6ln2 − 6ln2 + 16 A1 OE Apply limits 0 and 6ln2 to obtain −16e −1 14 − 6ln2 Attempt area of triangle minus area under curve M1 Allow decimals for this mark (14.56). Obtain 12  6ln 2  7 − (14 − 6ln 2) or equivalent, and hence 27ln2 − 14 A1 5

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

a 6 (a) Given that y b 12 e 2 x + 14 e – xl dx = 5 , where a is a positive constant, show that – 2 a a = 1 ln b10 + 1 e – a + 1 e – 4 al. [4] 2 2 2 … … … … … … … … … … … … … … … … … … … … … … … … … (b) Hence show by calculation that the value of a lies between 1.0 and 1.2. [2] … … … … … … … … … (c) Use the iterative formula a = 1 ln b10 + 1 e – a n + 1 e – 4 a nl n + 1 2 2 2 to find the value of a correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3] … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) Integrate to obtain expression of form k1e 2 x + k 2 e− x *M1 Obtain correct 1 4 e2 x − 14 e− x A1 Apply limits and arrange at least as far as 1 2 e 2 a = ... or 2e a = ... DM1 Confirm a = 12 ln(10 + 12 e− a + 12 e−4 a ) A1 AG – necessary detail needed. 4 16(b) Consider sign of a − 2 ln(10 + 12 e − a + 12 e −4 a ) or equivalent for values 1.0 and 1.2 M1 May use an intermediate equation from part (a). Example 1: consider 1 2 ln(10 + 12 e− a + 12 e −4 a ) for values 1.0 and 1.2. 1 2 a 1 − a 1 −4 a Example 2: consider e − e − e − 5 2 4 4 for values 1.0 and 1.2. Obtain –0.16… and 0.041… or equivalents and justify conclusion A1 Example 1: 1.0  1.1608... and 1.2  1.1589... (allow to 2dp and truncation), so 1.0   1.2 OE. Example 2: f (1.0 ) = −1.40... f (1.2 ) = 0.43... and change of sign so 1.0   1.2 OE. 2 6(c) Use iteration process correctly at least once M1 Obtain final answer 1.159 A1 Required to precisely 4 significant figures. Show sufficient iterations to justify answer or show a sign change in the interval A1 [1.1585, 1.1595] 3

This question in 9709/22 Oct/Nov 2025

1 Find 6 sin 2 x dx . [3] y … … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: Question Answer Marks Guidance 1 Use identity to express integrand in the form k1 + k 2 cos2 x M1 Obtain correct 3 − 3cos2 x A1 Integrate to obtain 3 x − 32 sin 2 x A1 Condone omission of + .c 3

This question in 9709/23 Oct/Nov 2025

5 y A B x O - 21 x The diagram shows the curve with equation y = 8e - 1. The curve meets the axes at the points A and B. The shaded region is bounded by the curve and the line segment AB. (a) Show that the x-coordinate of B is 6 ln 2. [2] … … … … … (b) Find the area of the shaded region. Give your answer in the form p ln 2- q , where p and q are positive integers. [5] … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(a) − 12 x M1 1 Attempt solution of 8e −=1 0 with use of a relevant logarithm property E.g. ln x = ln8 2 Confirm x = 6ln2 A1 AG – necessary detail needed. Need to see ln8 = 3ln2 or 64 = 2 6 OE used. Alternative Method for Question 5(a) − 12 x M1 Substitute x = 6ln2 in 8e − 1 and show use of a relevant logarithm property −3ln2 ln 18 A1 AG – necessary detail needed. Obtain 8e − 1 and hence 8e − 1 or equivalent and verify answer 0 2 5(b) − 12 x 12 x M1 Integrate 8e − 1 to obtain the form k e− − x 2 x − x A1 Obtain correct −16e − 1 2 6ln2 − 6ln2 + 16 A1 OE Apply limits 0 and 6ln2 to obtain −16e −1 14 − 6ln2 Attempt area of triangle minus area under curve M1 Allow decimals for this mark (14.56). Obtain 12  6ln 2  7 − (14 − 6ln 2) or equivalent, and hence 27ln2 − 14 A1 5

This question in 9709/23 Oct/Nov 2025

1 Find 6 sin 2 x dx . [3] y … … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: Question Answer Marks Guidance 1 Use identity to express integrand in the form k1 + k 2 cos2 x M1 Obtain correct 3 − 3cos2 x A1 Integrate to obtain 3 x − 32 sin 2 x A1 Condone omission of + .c 3

This question in 9709/25 Oct/Nov 2025

5 y A B x O - 21 x The diagram shows the curve with equation y = 8 e - 1. The curve meets the axes at the points A and B. The shaded region is bounded by the curve and the line segment AB. (a) Show that the x-coordinate of B is 6 ln 2. [2] … … … … … (b) Find the area of the shaded region. Give your answer in the form p ln 2- q , where p and q are positive integers. [5] … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(a) − 12 x M1 1 Attempt solution of 8e −=1 0 with use of a relevant logarithm property E.g. ln x = ln8 2 Confirm x = 6ln2 A1 AG – necessary detail needed. Need to see ln8 = 3ln2 or 64 = 2 6 OE used. Alternative Method for Question 5(a) − 12 x M1 Substitute x = 6ln2 in 8e − 1 and show use of a relevant logarithm property −3ln2 ln 18 A1 AG – necessary detail needed. Obtain 8e − 1 and hence 8e − 1 or equivalent and verify answer 0 2 5(b) − 12 x 12 x M1 Integrate 8e − 1 to obtain the form k e− − x 2 x − x A1 Obtain correct −16e − 1 2 6ln2 − 6ln2 + 16 A1 OE Apply limits 0 and 6ln2 to obtain −16e −1 14 − 6ln2 Attempt area of triangle minus area under curve M1 Allow decimals for this mark (14.56). Obtain 12  6ln 2  7 − (14 − 6ln 2) or equivalent and hence 27ln2 − 14 A1 5

This question in 9709/25 Oct/Nov 2025