Numerical solution of equations — Paper 3 · A Level Mathematics 9709

5 The diagram shows a sector OAB of a circle with centre O and radius r. The angle AOB is α radians, where 0 < α < 12π. The point N on OA is such that BN is perpendicular to OA. The area of the triangle ONB is half the area of the sector OAB. (i) Show that α satisfies the equation sin 2x = x. [3] (ii) By sketching a suitable pair of graphs, show that this equation has exactly one root in the interval 0 < x < 12π. [2] (iii) Use the iterative formula xn+1 = sin(2xn), with initial value x1 = 1, to find α correct to 2 decimal places, showing the result of each iteration. [3]

8 marks

Mark scheme: 1 25 (i) Obtain area of ONB in terms of r and α e.g. r cos α sin α B1 2 1  1 2  Equate area of triangle in terms of r and α to  r α  or equivalent M1 2  2  Obtain given form, sin 2α = α, correctly A1 3 [Allow use of OA and/or OB for r.] (ii) Make recognisable sketch in one diagram over the given range of two suitable graphs, e.g. y = sin 2x and y = x B1 State or imply link between intersections and roots and justify the given answer B1 2 [Allow a single graph and its intersection with y = 0 to earn full marks.] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 0.95 A1 Show sufficient iterations to justify its accuracy to 2d.p., or show there is a sign change in (0.945, 0.955) A1 3 [SR: Allow the M mark if calculations are attempted in degree mode.]

This question in 9709/31 Oct/Nov 2004

7 (i) By sketching a suitable pair of graphs, show that the equation cosec x = 12x + 1, where x is in radians, has a root in the interval 0 < x < 12π. [2] (ii) Verify, by calculation, that this root lies between 0.5 and 1. [2] (iii) Show that this root also satisfies the equation 2 x = sin−1 . [1] x + 2 (iv) Use the iterative formula 2 xn+1 = sin−1 , xn + 2 with initial value x1 = 0.75, to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

8 marks

Mark scheme: 7 (i) Make recognisable sketch of a relevant graph over the given range, e.g. y = cosec x B1 Sketch the other relevant graph, e.g. y = ½ x + 1, and justify the given statement B1 2 (ii) Consider sign of cosec x − ½ x − 1 at x = 0.5 and x = 1, or equivalent M1 Complete the argument correctly with appropriate calculations A1 2 (iii) Rearrange cosec x = 1 x + 1 in the given form, or vice versa B1 1 2 (iv) Use the iterative formula correctly at least once M1 Obtain final answer x = 0.80 A1 Show sufficient iterations to at least 3 d.p. to justify its accuracy to 2 d.p., or show there is a sign change in the interval (0.795, 0.805) A1 3

This question in 9709/31 May/June 2005

4 The equation x3 −x −3 = 0 has one real root, α. (i) Show that α lies between 1 and 2. [2] Two iterative formulae derived from this equation are as follows: xn+1 = x3n −3, (A) 1 xn+1 = (xn + 3) 3. (B) Each formula is used with initial value x1 = 1.5. (ii) Show that one of these formulae produces a sequence which fails to converge, and use the other formula to calculate α correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [5]

7 marks

Mark scheme: 4 (i) Consider sign of x 3 −x − 3 , or equivalent M1 Justify the given statement A1 [2] (ii) Apply an iterative formula correctly at least once, with initial value x1 = 5.1 M1 Show that (A) fails to converge A1 Show that (B) converges A1 Obtain final answer 1.67 A1 Show sufficient iterations to justify its accuracy to 2 d.p., or show there is a sign change in the interval (1.665, 1.675) A1 [5]

This question in 9709/31 Oct/Nov 2005

6 (i) By sketching a suitable pair of graphs, show that the equation 2 −x = ln x has only one root. [2] (ii) Verify by calculation that this root lies between 1.4 and 1.7. [2] (iii) Show that this root also satisfies the equation x = 13(4 + x −2 ln x). [1] (iv) Use the iterative formula xn+1 = 13(4 + xn −2 ln xn), with initial value x1 = 1.5, to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

8 marks

Mark scheme: 6 (i) Make a recognisable sketch of an appropriate graph, e.g. y = ln x B1 Sketch an appropriate second graph, e.g. y = 2 –x, correctly and justify the given statement B1 [2] (ii) Consider sign of 2 –x –ln x when x = 1.4 and x = 1.7, or equivalent M1 Complete the argument with correct calculations A1 [2] (iii) Rearrange the equation x = 13 ( 4 + x − 2ln x ) as 2 –x = ln x, or vice versa B1 [1] (iv) Use the iterative formula correctly at least once M1 Obtain final answer 1.56 A1 Show sufficient iterations to 4 d.p. to justify its accuracy to 2 d.p., or show there is a sign change in the interval (1.555, 1.565) A1 [3]

This question in 9709/31 Oct/Nov 2007

3 N D C a r x M A B 3a In the diagram, ABCD is a rectangle with AB = 3a and AD = a. A circular arc, with centre A and radius r, joins points M and N on AB and CD respectively. The angle MAN is x radians. The perimeter of the sector AMN is equal to half the perimeter of the rectangle. (i) Show that x satisfies the equation sin x = 14(2 + x). [3] (ii) This equation has only one root in the interval 0 < x < 12π. Use the iterative formula 2 + xn xn+1 = sin−1 , 4 with initial value x1 = 0.8, to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

6 marks

Mark scheme: 3 (i) State or imply r = a cosec x, or equivalent B1 Using perimeters, obtain a correct equation in x, e.g. 2a cosec x + ax cosec x = 4a, or 2r + rx = 4a B1 Deduce the given form of equation correctly B1 [3] (ii) Use the iterative formula correctly at least once M1 Obtain final answer 0.76 A1 Show sufficient iterations to 4 d.p. to justify its accuracy to 2 d.p., or show that there is a sign change in the value of sin x − 1 (2 + x ) in the interval (0.755, 0.765) A1 [3] 4

This question in 9709/31 May/June 2008

a 1 2x9 The constant a is such that xe dx = 6. 0 (i) Show that a satisfies the equation −1 x = 2 + e 2x. [5] (ii) By sketching a suitable pair of graphs, show that this equation has only one root. [2] (iii) Verify by calculation that this root lies between 2 and 2.5. [2] (iv) Use an iterative formula based on the equation in part (i) to calculate the value of a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

12 marks

Mark scheme: 2 x9 (i) Integrate by parts and reach kxe 2 x d x M1 − k ∫ e 2 x − 2 e 1 Obtain 2 xe 1 ∫ 2 x d x A1 1 2 x 12 x Complete the integration, obtaining 2 xe − 4e , or equivalent A1 Substitute limits correctly and equate result to 6, having integrated twice M1 −a12 Rearrange and obtain a = e + 2 A1 [5] −x12 (ii) Make recognizable sketch of a relevant exponential graph, e.g. y = e + 2 B1 Sketch a second relevant straight line graph, e.g. y = x, or curve, and indicate the root B1 [2] −x12 (iii) Consider sign of x − e − 2 at x = 2 and x = 2.5, or equivalent M1 Justify the given statement with correct calculations and argument A1 [2] − 12 x n (iv) Use the iterative formula x n +1 = 2 + e correctly at least once, with 2 ≤ x n ≤ 5.2 M1 Obtain final answer 2.31 A1 Show sufficient iterations to at least 4 d.p. to justify its accuracy to 2 d.p., or show there is a sign change in the interval (2.305, 2.315) A1 [3]

This question in 9709/31 Oct/Nov 2008

4 The equation x3 0 has one real root. −2x −2 = (i) Show by calculation that this root lies between x 1 and x 2. [2] = = (ii) Prove that, if a sequence of values given by the iterative formula 2x3 2 n + xn+1 = 3x2 n −2 converges, then it converges to this root. [2] (iii) Use this iterative formula to calculate the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

7 marks

Mark scheme: 4 (i) Compare signs of x3 – 2x – 2 when x = 1 and x = 2, or equivalent M1 Complete the argument with correct calculations A1 2 (ii) State or imply the equation x = (2x3 + 2) / (3x2 – 2) B1 Rearrange this in the form x 3 – 2x – 2 = 0, or work vice versa B1 2 (iii) Use the iterative formula correctly at least once with xn > 0 M1 Obtain final answer 1.77 A1 Show sufficient iterations to 4 d.p. to justify its accuracy to 2 d.p., or show there is a sign change In the interval (1.765, 1.775) A1 3 GCE A/AS LEVEL – May/June 2009 9709 03 2 ( ) 2

This question in 9709/31 May/June 2009

3 The sequence of values given by the iterative formula 3xn 15 xn+1 = 4 + x3 , n with initial value x1 3, converges to α. = (i) Use this iterative formula to find α correct to 2 decimal places, giving the result of each iteration to 4 decimal places. [3] (ii) State an equation satisfied by α and hence find the exact value of α. [2]

5 marks

Mark scheme: 3 (i) Use the iterative formula correctly at least once M1 State final answer 2.78 A1 Show sufficient iterations to at least 4 d.p. to justify its accuracy to 2 d.p., or show there is a sign change in an appropriate function in (2.775, 2.785) A1 [3] 3 15 (ii) State a suitable equation, e.g. x = x + B1 4 x 3 State that the exact value of α is 4 60 , or equivalent B1 [2] GCE A/AS LEVEL – October/November 2009 9709 31

This question in 9709/31 Oct/Nov 2009

2 The equation x3 0 has one real root. −8x −13 = (i) Find the two consecutive integers between which this root lies. [2] (ii) Use the iterative formula 1 xn+1 = (8xn + 13) 3 to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

5 marks

Mark scheme: 2 (i) Evaluate, or consider the sign of, x3 – 8x – 13 for two integer values of x, or equivalent M1 Conclude x = 3 and x = 4 with no errors seen A1 [2] (ii) Use the iterative formula correctly at least once M1 Obtain final answer 3.43 A1 Show sufficient iterations to at least 4 d.p. to justify its accuracy to 2 d.p., or show there is a sign change in the interval (3.425, 3.435) A1 [3] 2 dy 2

This question in 9709/32 Oct/Nov 2009

6 C r x rad A O B The diagram shows a semicircle ACB with centre O and radius r. The angle BOC is x radians. The area of the shaded segment is a quarter of the area of the semicircle. (i) Show that x satisfies the equation x 34π x. = −sin [3] (ii) This equation has one root. Verify by calculation that the root lies between 1.3 and 1.5. [2] (iii) Use the iterative formula xn 34π xn+1 = −sin to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

8 marks

Mark scheme: 6 (i) Using the formulae 12 r 2θ and 12 r 2 sin θ , or equivalent, form an equation M1 Obtain a correct equation in r and x and/or x/2 in any form A1 Obtain the given equation correctly A1 [3] (ii) Consider the sign of x − ( 34 π − sin x ) at x = 1.3 and x = 1.5, or equivalent M1 Complete the argument with correct calculations A1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.38 A1 Show sufficient iterations to at least 4 d.p. to justify its accuracy to 2 d.p., or show there is a sign change in the interval (1.375, 1.385) A1 [3]

This question in 9709/31 May/June 2010

4 y 1 x O p 2p M sin x The diagram shows the curve y for 0 x and its minimum point M. x = < ≤2π, (i) Show that the x-coordinate of M satisfies the equation x tan x. = [4] (ii) The iterative formula π xn+1 = tan−1(xn) + can be used to determine the x-coordinate of M. Use this formula to determine the x-coordinate of M correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

7 marks

Mark scheme: 4 (i) Use correct quotient or product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and solve for x M1 Obtain the given answer correctly A1 [4] (ii) Use the iterative formula correctly at least once M1 Obtain final answer 4.49 A1 Show sufficient iterations to at least 4 d.p. to justify its accuracy to 2 d.p., or show that there is a sign change in the interval (4.485, 4.495) A1 [3]

This question in 9709/32 May/June 2010

ln x 6 The curve y has one stationary point. = x 1 + (i) Show that the x-coordinate of this point satisfies the equation x 1 x + , = ln x and that this x-coordinate lies between 3 and 4. [5] (ii) Use the iterative formula xn 1 + xn+1 = ln xn to determine the x-coordinate correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

8 marks

Mark scheme: 6 (i) Use correct quotient or product rule M1 1 ln x Obtain correct derivative in any form, e.g. − A1 x ( x + )1 ( x + 2)1 Equate derivative to zero and obtain the given equation correctly A1 ( x + )1 Consider the sign of x − at x = 3 and x = 4, or equivalent M1 ln x Complete the argument with correct calculated values A1 [5] (ii) Use the iterative formula correctly at least once, using or reaching a value in the interval (3, 4) M1 Obtain final answer 3.59 A1 Show sufficient iterations to at least 4 d.p. to justify its accuracy to 2 d.p., or show there is a sign change in the interval (3.585, 3.595) A1 [3]

This question in 9709/33 May/June 2010

4 (i) By sketching suitable graphs, show that the equation 4x2 cotx −1 = has only one root in the interval 0 x 12π. [2] < < (ii) Verify by calculation that this root lies between 0.6 and 1. [2] (iii) Use the iterative formula 1 2 xn+1 = √(1 + cotxn) to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

7 marks

Mark scheme: 4 (i) Make recognisable sketch of a relevant graph over the given range B1 Sketch the other relevant graph on the same diagram and justify the given statement B1 [2] (ii) Consider sign of 4x2 – 1 – cot x at x = 0.6 and x = 1, or equivalent M1 Complete the argument correctly with correct calculated values A1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 0.73 A1 Show sufficient iterations to at least 4 d.p. to justify its accuracy to 2 d.p., or show there is a sign change in the interval (0.725, 0.735) A1 [3] GCE A/AS LEVEL – October/November 2010 9709 31 dx

This question in 9709/31 Oct/Nov 2010

4 (i) By sketching suitable graphs, show that the equation 4x2 cotx −1 = has only one root in the interval 0 x 12π. [2] < < (ii) Verify by calculation that this root lies between 0.6 and 1. [2] (iii) Use the iterative formula 1 2 xn+1 = √(1 + cotxn) to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

7 marks

Mark scheme: 4 (i) Make recognisable sketch of a relevant graph over the given range B1 Sketch the other relevant graph on the same diagram and justify the given statement B1 [2] (ii) Consider sign of 4x2 – 1 – cot x at x = 0.6 and x = 1, or equivalent M1 Complete the argument correctly with correct calculated values A1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 0.73 A1 Show sufficient iterations to at least 4 d.p. to justify its accuracy to 2 d.p., or show there is a sign change in the interval (0.725, 0.735) A1 [3] GCE A/AS LEVEL – October/November 2010 9709 32 dx

This question in 9709/32 Oct/Nov 2010

6 A q B O 10 cm The diagram shows a circle with centre O and radius 10 cm. The chord AB divides the circle into two regions whose areas are in the ratio 1 : 4 and it is required to find the length of AB. The angle AOB is θ radians. (i) Show that θ 25π sin θ. [3] = + (ii) Showing all your working, use an iterative formula, based on the equation in part (i), with an initial value of 2.1, to find θ correct to 2 decimal places. Hence find the length of AB in centimetres correct to 1 decimal place. [5]

8 marks

Mark scheme: 6 (i) State or imply area of segment is 12 r2θ – 12 r2 sinθ or 50θ – 50 sinθ B1 Attempt to form equation from area of segment = 15 of area of circle, or equivalent M1 Confirm given result θ = 52 π + sinθ A1 [3] (ii) Use iterative formula correctly at least once M1 Obtain value for θ of 2.11 A1 Show sufficient iterations to justify value of θ or show sign change in interval (2.105, 2.115) A1 Use correct trigonometry to find an expression for the length of AB M1 e.g. 20 sin 1.055 or 200 − 200 cos .211 Hence 17.4 A1 [5] [2.1 → 2.1198 → 2.1097 → 2.1149 → 2.1122] GCE AS/A LEVEL – May/June 2011 9709 31

This question in 9709/31 May/June 2011

4 C r x A O B T The diagram shows a semicircle ACB with centre O and radius r. The tangent at C meets AB produced at T. The angle BOC is x radians. The area of the shaded region is equal to the area of the semicircle. (i) Show that x satisfies the equation tan x = x + π. [3] (ii) Use the iterative formula xn+1 = tan−1(xn + π) to determine x 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 CT = r tan x or OT = r sec x , or equivalent B1 Using correct area formulae, form an equation in r and x M1 Obtain the given answer correctly A1 [3] (ii) Use the iterative formula correctly at least once M1 Obtain the final answer 1.35 A1 Show sufficient iterations to 4 d.p. to justify its accuracy to 2 d.p. , or show there is a sign change in the interval (1.345, 1.355) A1 [3] GCE AS/A LEVEL – May/June 2011 9709 32 dx 2

This question in 9709/32 May/June 2011

6 (i) By sketching a suitable pair of graphs, show that the equation cotx 1 x2, = + where x is in radians, has only one root in the interval 0 x 12π. [2] < < (ii) Verify by calculation that this root lies between 0.5 and 0.8. [2] (iii) Use the iterative formula xn+1 = tan−1 1 1 x2n + to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

7 marks

Mark scheme: 6 (i) Make recognisable sketch of a relevant graph over the given range B1 Sketch the other relevant graph and justify the given statement B1 [2] 2 x = 0.5 and x = 0.8, or equivalent M1 (ii) Consider the sign of cot x − (1 + x ) at Complete the argument with correct calculated values A1 [2] (iii) Use the iterative formula correctly at least once with 5.0 ≤ x n ≤ 8.0 M1 Obtain final answer 0.62 A1 Show sufficient iterations to 4 d.p. to justify its accuracy to 2 d.p., or show there is a sign change in the interval (0.615, 0.625) A1 [3]

This question in 9709/33 May/June 2011

5 (i) By sketching a suitable pair of graphs, show that the equation secx 3 2x2, = −1 where x is in radians, has a root in the interval 0 x 12π. [2] < < (ii) Verify by calculation that this root lies between 1 and 1.4. [2] (iii) Show that this root also satisfies the equation . x [1] = cos−1 6 2 −x2 (iv) Use an iterative formula based on the equation in part (iii) to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

8 marks

Mark scheme: 5 (i) Make recognisable sketch of a relevant graph over the given interval B1 Sketch the other relevant graph and justify the given statement B1 [2] (ii) Consider the sign of sec x – (3 – 1 x2) at x = 1 and x = 1.4, or equivalent M1 2 Complete the argument with correct calculated values A1 [2] (iii) Convert the given equation to sec x = 3 – 1 x2 or work vice versa B1 [1] 2 (iv) Use a correct iterative formula correctly at least once M1 Obtain final answer 1.13 A1 Show sufficient iterations to 4 d.p. to justify 1.13 to 2 d.p., or show there is a sign change in the interval (1.125, 1.135) A1 [3] [SR: Successive evaluation of the iterative function with x = 1, 2, … scores M0.]

This question in 9709/31 Oct/Nov 2011

5 (i) By sketching a suitable pair of graphs, show that the equation secx 3 2x2, = −1 where x is in radians, has a root in the interval 0 x 12π. [2] < < (ii) Verify by calculation that this root lies between 1 and 1.4. [2] (iii) Show that this root also satisfies the equation . x [1] = cos−1 6 2 −x2 (iv) Use an iterative formula based on the equation in part (iii) to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

8 marks

Mark scheme: 5 (i) Make recognisable sketch of a relevant graph over the given interval B1 Sketch the other relevant graph and justify the given statement B1 [2] (ii) Consider the sign of sec x – (3 – 1 x2) at x = 1 and x = 1.4, or equivalent M1 2 Complete the argument with correct calculated values A1 [2] (iii) Convert the given equation to sec x = 3 – 1 x2 or work vice versa B1 [1] 2 (iv) Use a correct iterative formula correctly at least once M1 Obtain final answer 1.13 A1 Show sufficient iterations to 4 d.p. to justify 1.13 to 2 d.p., or show there is a sign change in the interval (1.125, 1.135) A1 [3] [SR: Successive evaluation of the iterative function with x = 1, 2, … scores M0.]

This question in 9709/32 Oct/Nov 2011

a 5 It is given that x ln x dx 22, where a is a constant greater than 1. ã 1 = r 87 (i) Show that a . [5] = 2 ln a −1 (ii) Use an iterative formula based on the equation in part (i) to find the value of a correct to 2 decimal places. Use an initial value of 6 and give the result of each iteration to 4 decimal places. [3]

8 marks

Mark scheme: 5 (i) Either Use integration by parts and reach an expression kx2 lnx ± n ∫ x2. 1x dx M1 Obtain 12 x2 ln x – ∫ 12 x dx or equivalent A1 Obtain 12 x2 ln x – 14 x2 A1 Or Use Integration by parts and reach an expression kx(xlnx – x) ± m ∫ xlnx – xdx M1 Obtain I = (x2 lnx – x2) – I + ∫ xdx A1 Obtain 12 x2 ln x – 14 x2 A1 Substitute limits correctly and equate to 22, having integrated twice DM1* 87 Rearrange and confirm given equation a = A1 [5] 2ln a − 1 (ii) Use iterative process correctly at least once M1 Obtain final answer 5.86 A1 Show sufficient iterations to 4 d.p. to justify 5.86 or show a sign change in the A1 interval (5.855, 5.865) (6 → 5.8030 → 5.8795 → 5.8491 → 5.8611 → 5.8564) [3] GCE AS/A LEVEL – October/November 2011 9709 33 2 3

This question in 9709/33 Oct/Nov 2011

10 (i) It is given that 2 tan 2x 5 tan2x 0. Denoting tan x by t, form an equation in t and hence show that either t 0 or t + = [4] = = 3√(t + 0.8). (ii) It is given that there is exactly one real value of t satisfying the equation t Verify by calculation that this value lies between 1.2 and 1.3. = 3√(t + 0.8). [2] to find the value of t correct to 3 decimal places. Give (iii) Use the iterative formula tn+1 = 3√(tn + 0.8) the result of each iteration to 5 decimal places. [3] (iv) Using the values of t found in previous parts of the question, solve the equation 2 tan 2x 5 tan2x 0 + = for [3] −π ≤x ≤π.

12 marks

Mark scheme: 10 (i) Use correct identity for tan 2 x and obtains at4 + bt3 + ct2 + dt = 0, where b may be zero M1 Obtain correct horizontal equation, e.g. 4t + 5t2 – 5t4 = 0 A1 Obtain kt(t3 + et + f) = 0 or equivalent M1 Confirm given results t = 0 and t = 3 t + 8.0 A1 [4] (ii) Consider sign of t − 3 t + 8.0 at 1.2 and 1.3 or equivalent M1 Justify the given statement with correct calculations (–0.06 and 0.02) A1 [2] (iii) Use the iterative formula correctly at least once with 1.2 < tn < 1.3 M1 Obtain final answer 1.276 A1 Show sufficient iterations to justify answer or show there is a change of sign in interval (1.2755, 1.2765) A1 [3] (iv) Evaluate tan–1 (answer from part (iii)) to obtain at least one value M1 Obtain –2.24 and 0.906 A1 State –π, 0 and π B1 [3] [SR If A0, B0, allow B1 for any 3 roots]

This question in 9709/31 May/June 2012

2 C q M a A B In the diagram, ABC is a triangle in which angle ABC is a right angle and BC a. A circular arc, with centre C and radius a, joins B and the point M on AC. The angle ACB is θ radians.= The area of the sector CMB is equal to one third of the area of the triangle ABC. (i) Show that θ satisfies the equation tan θ 3θ. = [2] (ii) This equation has one root in the interval 0 θ 12π. Use the iterative formula < < θn+1 = tan−1(3θn) to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

5 marks

Mark scheme: 1 2 1 2 (i) Using the formulae r θ and bh , form an equation an a and θ M1 2 2 Obtain given answer A1 [2] (ii) Use the iterative formula correctly at least once M1 Obtain answer θ = 1.32 A1 Show sufficient iterations to 4 d.p. to justify 1.32 to 2 d.p., or show there is a sign change in the interval (1.315, 1.325) A1 [3] GCE AS/A LEVEL – May/June 2012 9709 32 1 1 2

This question in 9709/32 May/June 2012

7 y R x O p2 The diagram shows part of the curve y for x where x is in radians. The shaded region = cos(√x) ≥0, between the curve, the axes and the line x p2, where p 0, is denoted by R. The area of R is equal = > to 1. p2 3 cos p (i) Use the substitution x u2 to find dx. Hence show that sin p −2 . [6] = ã 0 cos(√x) = 2p 3 cos pn (ii) Use the iterative formula sin−1 −2 , with initial value p1 1, to find the value of pn+1 = 2pn = p correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

9 marks

Mark scheme: 7 (i) Substitute for x and dx throughout the integral M1 Obtain ∫ 2u cos u d u A1 Integrate by parts and obtain answer of the form au sin u + b cos u , where ab ≠ 0 M1 Obtain 2u sin u + 2 cos u A1 Use limits u = 0, u = p correctly and equate result to 1 M1 Obtain the given answer A1 [6] (ii) Use the iterative formula correctly at least once M1 Obtain final answer p = 1.25 A1 Show sufficient iterations to 4 d.p. to justify its accuracy to 2 d.p., or show there is a sign change in the interval (1.245, 1.255) A1 [3] GCE AS/A LEVEL – May/June 2012 9709 33 B C

This question in 9709/33 May/June 2012

6 y x a O b The diagram shows the curve y x4 2x3 2x2 which crosses the x-axis at the points = + + −4x −16, (α, 0) and where α β. It is given that α is an integer. (β, 0) < (i) Find the value of α. [2] (ii) Show that β satisfies the equation x [3] = 3√(8 −2x). (iii) Use an iteration process based on the equation in part (ii) to find the value of β correct to 2 decimal places. Show the result of each iteration to 4 decimal places. [3]

8 marks

Mark scheme: 6 (i) Find y for x = –2 M1 Obtain 0 and conclude that ~== –2 A1 [2] (ii) Either Find cubic factor by division or inspection or equivalent M1 Obtain x 3 + 2 x − 8 A1 Rearrange to confirm given equation x = 3 8 − 2 x A1 Or Derive cubic factor from given equation and form product with (x – ~) M1 ( x + 2 )(x 3 + 2 x − 8 ) A1 Obtain quartic x 4 + 2 x 3 + 2 x 2 − 4 x − 16 ( = 0) A1 Or Derive cubic factor from given equation and divide the quartic by the cubic M1 (x 4 + 2 x 3 + 2 x 2 − 4 x − 16 ) ÷ (x 3 + 2 x − 8 ) A1 Obtain correct quotient and zero remainder A1 [3] (iii) Use the given iterative formula correctly at least once M1 Obtain final answer 1.67 A1 Show sufficient iterations to at least 4 d.p. to justify answer 1.67 to 2 d.p. or show there is a change of sign in interval (1.665, 1.675) A1 [3] GCE A LEVEL – October/November 2012 9709 33

This question in 9709/33 Oct/Nov 2012

2 The sequence of values given by the iterative formula xn x3n 100 , + xn+1 = 2 x3n 25 + with initial value x1 3.5, converges to = !. (i) Use this formula to calculate correct to 4 decimal places, showing the result of each iteration to 6 decimal places. ! [3] (ii) State an equation satisfied by and hence find the exact value of [2] ! !.

5 marks

Mark scheme: 2 (i) Use the iterative formula correctly at least once M1 Obtain final answer 3.6840 A1 Show sufficient iterations to at least 6 d.p. to justify 3.6840, or show there is a sign change in the interval (3.68395, 3.68405) A1 [3] x ( x 3 + 100) (ii) State a suitable equation, e.g. x = B1 2( x 3 + 25) State that the value of α is 3 50 , or exact equivalent B1 [2] 2

This question in 9709/32 May/June 2013

6 O B C r q A In the diagram, A is a point on the circumference of a circle with centre O and radius r. A circular arc with centre A meets the circumference at B and C. The angle OAB is radians. The shaded region is bounded by the circumference of the circle and the arc with centre A 1joining B and C. The area of the shaded region is equal to half the area of the circle. 2 sin (i) Show that cos . [5] 21 −0 21 = 41 (ii) Use the iterative formula P2 sin Q 1 , 2 cos−1 21n −0 1n+1 = 41n with initial value 1, to determine correct to 2 decimal places, showing the result of each iteration to 4 decimal11 =places. 1 [3]

8 marks

Mark scheme: 6 (i) State or imply AB = 2r cos θ or AB 2 = 2 r 2 − 2 r 2 cos(π − 2θ ) B1 Use correct formula to express the area of sector ABC in terms of r and θ M1 Use correct area formulae to express the area of a segment in terms of r and θ M1 State a correct equation in r and θ in any form A1 Obtain the given answer A1 [5] [SR: If the complete equation is approached by adding two sectors to the shaded area above BO and OC give the first M1 as on the scheme, and the second M1 for using correct area formulae for a triangle AOB or AOC, and a sector AOB or AOC.] (ii) Use the iterative formula correctly at least once M1 Obtain final answer 0.95 A1 Show sufficient iterations to 4 d.p. to justify 0.95 to 2 d.p., or show there is a sign change in the interval (0.945, 0.955) A1 [3] GCE A LEVEL – October/November 2013 9709 31 A Bx + C

This question in 9709/31 Oct/Nov 2013

6 O B C r A In the diagram, A is a point on the circumference of a circle with centre O and radius r. A circular arc with centre A meets the circumference at B and C. The angle OAB is radians. The shaded region is bounded by the circumference of the circle and the arc with centre A joining B and C. The area of the shaded region is equal to half the area of the circle. 2 sin 2 − (i) Show that cos 2 = . [5] 4 (ii) Use the iterative formula 2 sin 2 n − 1 n+1 = 2 cos−1 , 4 n with initial value 1 = 1, to determine correct to 2 decimal places, showing the result of each iteration to 4 decimal places. [3]

8 marks

Mark scheme: 6 (i) State or imply AB = 2r cosθ or AB 2 = 2 r 2 − 2 r 2 cos (π − 2θ ) B1 Use correct formula to express the area of sector ABC in terms of r and θ M1 Use correct area formulae to express the area of a segment in terms of r and θ M1 State a correct equation in r and θ in any form A1 Obtain the given answer A1 [5] [SR: If the complete equation is approached by adding two sectors to the shaded area above BO and OC give the first M1 as on the scheme, and the second M1 for using correct area formulae for a triangle AOB or AOC, and a sector AOB or AOC.] (ii) Use the iterative formula correctly at least once M1 Obtain final answer 0.95 A1 Show sufficient iterations to 4 d.p. to justify 0.95 to 2 d.p., or show there is a sign change in the interval (0.945, 0.955) A1 [3] GCE A LEVEL – October/November 2013 9709 32 A Bx + C

This question in 9709/32 Oct/Nov 2013

p 2x5 It is given that Ó 0 4xe−1 dx = 9, where p is a positive constant. @8p 16 A (i) Show that p 2 ln . [5] + 7 = (ii) Use an iterative process based on the equation in part (i) to find the value of p correct to 3 significant figures. Use a starting value of 3.5 and give the result of each iteration to 5 significant figures. [3]

8 marks

Mark scheme: 5 (i) Use integration by parts to obtain axe 2 + ∫ b e 2 dx M1* 1 1 − x − x Obtain − 8 xe 2 + ∫ 8e 2 d x or unsimplified equivalent A1 1 1 − x − x Obtain − 8 xe 2 − 16 e 2 A1 Use limits correctly and equate to 9 M1 (d*M)  8 p + 16  Obtain given answer p = 2 ln   correctly A1 [5]  7  GCE A LEVEL – October/November 2013 9709 33 (ii) Use correct iteration formula correctly at least once M1 Obtain final answer 3.77 A1 Show sufficient iterations to 5sf or better to justify accuracy 3.77 or show sign change in interval (3.765, 3.775) A1 [3] [ 5.3 → .3 6766 → .3 7398 → .3 7619 → .3 7696 → .3 7723 ]

This question in 9709/33 Oct/Nov 2013

6 A B C x r O In the diagram, A is a point on the circumference of a circle with centre O and radius r. A circular arc with centre A meets the circumference at B and C. The angle OAB is equal to x radians. The shaded region is bounded by AB, AC and the circular arc with centre A joining B and C. The perimeter of the shaded region is equal to half the circumference of the circle. 0 1 0 (i) Show that x = cos−1 . [3] 4 + 4x (ii) Verify by calculation that x lies between 1 and 1.5. [2] (iii) Use the iterative formula @ A 0 xn+1 = cos−1 4 + 4xn to determine the value of x correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

8 marks

Mark scheme: 6 (i) Use correct arc formula and form an equation in r and x M1 Obtain a correct equation in any form A1 Rearrange in the given form A1 3 (ii) Consider sign of a relevant expression at x = 1 and x = 1.5, or compare values of relevant expressions at x = 1 and x = 5.1 M1 Complete the argument correctly with correct calculated values A1 2 (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.21 A1 Show sufficient iterations to 4 d.p. to justify 1.21 to 2 d.p., or show there is a sign change in the interval (1.205,1.215) A1 3 3

This question in 9709/32 May/June 2014

10 4 The equation x has one positive real root, denoted by = e2x !. −1 (i) Show that lies between x 1 and x 2. [2] ! = = (ii) Show that if a sequence of positive values given by the iterative formula P Q 1 10 2 + xn+1 = ln 1 xn converges, then it converges to [2] !. (iii) Use this iterative formula to determine correct to 2 decimal places. Give the result of each ! iteration to 4 decimal places. [3]

7 marks

Mark scheme: 4 (i) Consider sign of x − 10 /( e 2 x − )1 at x = 1 and x = 2 M1 Complete the argument correctly with correct calculated values A1 2 (ii) State or imply α = 12 ln(1 + 10 / α ) B1 Rearrange this as α = 10 /( e 2α − )1 or work vice versa B1 2 (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.14 A1 Show sufficient iterations to 4 d.p. to justify 1.14 to 2 d.p., or show there is a sign change in the interval (1.135, 1.145) A1 3

This question in 9709/33 May/June 2014

10 y O x P The diagram shows part of the curve with parametric equations x 2 ln t 2 , y t3 2t 3. = + = + + (i) Find the gradient of the curve at the origin. [5] (ii) At the point P on the curve, the value of the parameter is p. It is given that the gradient of the curve at P is 12. 1 (a) Show that p [1] = 3p2 2 −2. + (b) By first using an iterative formula based on the equation in part (a), determine the coordinates of the point P. Give the result of each iteration to 5 decimal places and each coordinate of P correct to 2 decimal places. [4]

10 marks

Mark scheme: dx 2 dy 210 (i) Obtain = and = 3t + 2 B1 dt t + 2 dt dy dy dx Use = ÷ M1 dx dt dt dy 1 2 Obtain = (3t + 2)(t + 2) A1 dx 2 Identify value of t at the origin as –1 B1 5 Substitute to obtain as gradient at the origin A1 [5] 2 1 1 (ii) (a) Equate derivative to and confirm p = − 2 B1 [1] 2 2 3 p + 2 (b) Use the iterative formula correctly at least once M1 Obtain value p = − .1924 or better (–1.92367…) A1 Show sufficient iterations to justify accuracy or show a sign change in appropriate interval A1 Obtain coordinates (–5.15, –7.97) A1 [4]

This question in 9709/31 May/June 2015

a 6 It is given that x cosx dx 0.5, where 0 a 1 Ó 0 = < < 20. 1.5 (i) Show that a satisfies the equation sin a −cosa . [4] = a (ii) Verify by calculation that a is greater than 1. [2] (iii) Use the iterative formula P1.5 −cosan Q sin−1 an+1 = an to determine the value of a correct to 4 decimal places, giving the result of each iteration to 6 decimal places. [3]

9 marks

Mark scheme: 6 (i) Integrate and reach ± x sin x m ∫ sin x dx M1* Obtain integral x sin x + cos x A1 Substitute limits correctly, must be seen since AG, and equate result to 0.5 M1(dep*) Obtain the given form of the equation A1 4 (ii) EITHER: Consider the sign of a relevant expression at a = 1 and at another relevant value, π e.g. a = 1.5 Y M1 2 OR: Using limits correctly, consider the sign of [x sin x + cos x ]0a − 5.0 , or compare the value of [x sin x + cos x ]a0 with 0.5, for a =1 AND for another relevant value, π e.g a = 1.5 Y . M1 2 Complete the argument, so change of sign, or above and below stated, both with correct calculated values A1 2 (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.2461 A1 Show sufficient iterations to 6 d.p. to justify 1.2461 to 4 d.p., or show there is a sign change in the interval (1.24605, 1.24615) A1 3

This question in 9709/33 May/June 2015

4 The equation x3 −x2 −6 = 0 has one real root, denoted by !. (i) Find by calculation the pair of consecutive integers between which ! lies. [2] (ii) Show that, if a sequence of values given by the iterative formula _P Q 6 xn+1 = xn + xn converges, then it converges to !. [2] (iii) Use this iterative formula to determine ! correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3]

7 marks

Mark scheme: 4 (i) Evaluate, or consider the sign of, x 3 −x 2 − 6 for two integer values of x, or equivalent M1 Obtain the pair x = 2 and x = 3, with no errors seen A1 [2] (ii) State a suitable equation, e.g. x = ( x + ( 6 / x )) B1 Rearrange this as x 3 −x 2 − 6 = 0 , or work vice versa B1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 2.219 A1 Show sufficient iterates to 5 d.p. to justify 2.219 to 3 d.p., or show there is a sign change in the interval (2.2185, 2.2195) A1 [3]

This question in 9709/31 Oct/Nov 2015

4 The equation x3 −x2 −6 = 0 has one real root, denoted by !. (i) Find by calculation the pair of consecutive integers between which ! lies. [2] (ii) Show that, if a sequence of values given by the iterative formula _P Q 6 xn+1 = xn + xn converges, then it converges to !. [2] (iii) Use this iterative formula to determine ! correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3]

7 marks

Mark scheme: 4 (i) Evaluate, or consider the sign of, x 3 −x 2 − 6 for two integer values of x, or equivalent M1 Obtain the pair x = 2 and x = 3, with no errors seen A1 [2] (ii) State a suitable equation, e.g. x = ( x + ( 6 / x )) B1 Rearrange this as x 3 −x 2 − 6 = 0 , or work vice versa B1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 2.219 A1 Show sufficient iterates to 5 d.p. to justify 2.219 to 3 d.p., or show there is a sign change in the interval (2.2185, 2.2195) A1 [3]

This question in 9709/32 Oct/Nov 2015

3 The equation x5 x2 0 has one positive root. −3x3 + −4 = (i) Verify by calculation that this root lies between 1 and 2. [2] (ii) Show that the equation can be rearranged in the form O@ A 3 4 x 3x . [1] = + x2 −1 (iii) Use an iterative formula based on this rearrangement to determine the positive root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

6 marks

Mark scheme: 3 (i) Consider sign of x 5 − 3 x 3 + x 2 − 4 at x = 1 and x = 2, or equivalent M1 Complete the argument correctly with correct calculated values A1 [2] (ii) Rearrange the given quintic equation in the given form, or work vice versa B1 [1] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.78 A1 Show sufficient iterations to 4 d.p. to justify 1.78 to 2 d.p., or show there is a sign change in the interval (1.775, 1.785) A1 [3]

This question in 9709/32 Feb/March 2016

6 (i) By sketching a suitable pair of graphs, show that the equation x 5e−x = has one root. [2] (ii) Show that, if a sequence of values given by the iterative formula 1 @25 A xn+1 = 2ln xn converges, then it converges to the root of the equation in part (i). [2] (iii) Use this iterative formula, with initial value x1 1, to calculate the root correct to 2 decimal = places. Give the result of each iteration to 4 decimal places. [3]

7 marks

Mark scheme: 6 (i) Make recognizable sketch of a relevant graph B1 Sketch the other relevant graph and justify the given statement B1 [2] 1 (ii) State x = ln(25 / x ) B1 2 Rearrange this in the form5e −=x x B1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.43 A1 Show sufficient iterations to 4 d.p. to justify 1.43 to 2 d.p., or show there is a sign change in the interval (1.425, 1.435) A1 [3] 2 dy 2

This question in 9709/31 May/June 2016

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

10 marks

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

This question in 9709/32 May/June 2016

6 The curve with equation y x2 cos 2x1 has a stationary point at x p in the interval 0 x = = < < 0. 1 4 (i) Show that p satisfies the equation tan 2p p. [3] = (ii) Verify by calculation that p lies between 2 and 2.5. [2] @ A 4 (iii) Use the iterative formula 2 to determine the value of p correct to 2 decimal tan−1 pn+1 = pn places. Give the result of each iteration to 4 decimal places. [3]

8 marks

Mark scheme: 6 (i) Use the product rule M1 Obtain correct derivative in any form A1 Equate 2-term derivative to zero and obtain the given answer correctly A1 [3] (ii) Use calculations to consider the sign of a relevant expression at p = 2 and p = 2.5, or compare values of relevant expressions at p= 2 and p = 2.5 M1 Complete the argument correctly with correct calculated values A1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 2.15 A1 Show sufficient iterations to 4 d.p. to justify 2.15 to 2 d.p., or show there is a sign change in the interval (2.145,2.155) A1 [3]

This question in 9709/33 May/June 2016

6 (i) By sketching a suitable pair of graphs, show that the equation 1 cosec 2x1 = 13x + has one root in the interval 0 x [2] < ≤0. (ii) Show by calculation that this root lies between 1.4 and 1.6. [2] (iii) Show that, if a sequence of values in the interval 0 x given by the iterative formula < ≤0 @ A 3 2 sin−1 xn+1 = xn 3 + converges, then it converges to the root of the equation in part (i). [2] (iv) Use this iterative formula to calculate the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3]

9 marks

Mark scheme: 6 (i) Make recognizable sketch of a relevant graph B1 Sketch the other relevant graph and justify the given statement B1 [2] (ii) Use calculations to consider the value of a relevant expression at x = 1.4 and x = 1.6, or the values of relevant expressions at x = 1.4 and x = 1.6 M1 Complete the argument correctly with correct calculated values A1 [2] −1 3  (iii) State x = 2sin   B1  x + 3  Rearrange this in the form cosec 12 x = 13 x + 1 B1 [2] x  3  If working in reverse, need sin = for first B1   2  x + 3  (iv) Use the iterative formula correctly at least once M1 Obtain final answer 1.471 A1 Show sufficient iterations to 5 d.p. to justify 1.471 to 3 d.p., or show there is a sign change in the interval (1.4705, 1.4715) A1 [3]

This question in 9709/31 Oct/Nov 2016

9 y 1 x O a 20 k The diagram shows the curves y x cosx and y x, where k is a constant, for 0 x The curves = = < ≤120. touch at the point where x a. = 2 (i) Show that a satisfies the equation tan a a. [5] = @ A 2 (ii) Use the iterative formula to determine a correct to 3 decimal places. Give the tan−1 an+1 = an result of each iteration to 5 decimal places. [3] (iii) Hence find the value of k correct to 2 decimal places. [2] [Question 10 is printed on the next page.]

10 marks

Mark scheme: 9 (i) Differentiate both equations and equate derivatives M1* k Obtain equation cos a − a sin a = − 2 A1 + A1 a k State a cos a = and eliminate k DM1 a Obtain the given answer showing sufficient working A1 [5] (ii) Show clearly correct use of the iterative formula at least once M1 Obtain answer 1.077 A1 Show sufficient iterations to 5 d.p. to justify 1.077 to 3 d.p., or show there is a sign change in the interval (1.0765, 1.0775) A1 [3] (iii) Use a correct method to determine k M1 Obtain answer k = 0.55 A1 [2]

This question in 9709/33 Oct/Nov 2016

2x3 (i) By sketching 4 has one positive root and one suitable graphs, show that the equation e−1 = −x2 negative root. [2] (ii) Verify by calculation that the negative root lies between and [2] −1 −1.5. … … … … … … … … … … … … 2xn(iii) Use the iterative formula xn+1 = − 4 −e−1 to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 3(i) − 12 x B1 Sketch a relevant graph, e.g. y = e Sketch a second relevant graph, e.g. y = 4 − x 2 , and justify the given statement B1 Total: 2 3(ii) Calculate the value of a relevant expression or values of a pair of expressions at M1 x = – 1 and x = – 1.5 complete the argument correctly with correct calculated values A1 Total: 2 3(iii) Use the iterative formula correctly at least once M1 Obtain final answer – 1.41 A1 Show sufficient iterations to 4 d.p. to justify – 1.41 to 2 d.p., or show there is a sign A1 change in the interval ( – 1.415, – 1.405) Total: 3

This question in 9709/32 Feb/March 2017

10 y M x O p 140 The diagram shows the curve y x2 cos 2x for 0 The curve has a maximum point at M where x p. = ≤x ≤140. = 1 @1 A (i) Show that p satisfies the equation p . [3] 2 tan−1 p = … … … … … … … … @ A 1 1 (ii) Use the iterative formula to determine the value of p correct to 2 decimal 2 tan−1 pn+1 = pn places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … (iii) Find, showing all necessary working, the exact area of the region bounded by the curve and the x-axis. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 10(i) Use correct product rule M1 2 A1 Obtain correct derivative in any form y ′ = 2 x cos2 x − 2 x sin 2 x ( ) Equate to zero and derive the given equation A1 Total: 3 10(ii) Use the iterative formula correctly at least once e.g. M1 0.5 → 0.55357 → 0.53261 → 0.54070 → 0.53755 Obtain final answer 0.54 A1 Show sufficient iterations to 4 d.p. to justify 0.54 to 2 d.p., or show there is a sign change in A1 the interval (0.535, 0.545) Total: 3 10(iii) 2 *M1 Integrate by parts and reach ax sin 2 x + b ∫ x sin 2 x dx 1 2 1 A1 Obtain x sin 2 x −∫ 2 x. sin 2 x dx 2 2 1 2 1 1 A1 Complete integration and obtain x sin 2 x + x cos2 x − sin 2 x , or equivalent 2 2 4 1 DM1 Substitute limits x = 0, x = 4π , having integrated twice 1 2 A1 Obtain answer (π − 8) , or exact equivalent 32 Total: 5

This question in 9709/32 May/June 2017

6 The equation cot x 1 has one root in the interval 0 x denoted by = −x < < 0, !. (i) Show by calculation that is greater than 2.5. [2] ! … … … … … … … … … … … … … … (ii) Show that, if a sequence of values in the interval 0 x given by the iterative formula @ A 1 < < 0 converges, then it converges to [2] tan−1 1 xn+1 = 0 + !. −xn … … … … … … … … … … … … … … … … (iii) Use this iterative formula to determine correct to 3 decimal places. Give the result of each iteration to 5 decimal places. ! [3] … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 6(i) Calculate the value of a relevant expression or expressions at x = 2.5 and at another M1 relevant value, e.g. x = 3 Complete the argument correctly with correct calculated values A1 Total: 2 6(ii) State a suitable equation, e.g. x = π + tan − 1 (1/ (1 − x ) ) without suffices B1 Rearrange this as cot x = 1 − x , or commence working vice versa B1 Total: 2 6(iii) Use the iterative formula correctly at least once M1 Obtain final answer 2.576 only A1 Show sufficient iterations to 5 d.p. to justify 2.576 to 3 d.p., or show there is a sign A1 change in the interval (2.5755, 2.5765) Total: 3

This question in 9709/33 May/June 2017

a 1 9 It is given that x 2 ln x dx 2, where a 1. Ó 1 = > 3 3 7 2a 2 (i) Show that a 2 . [5] + 3 ln a = … … … … … … … … … … … … … … … … … … … … … … … (ii) Show by calculation that a lies between 2 and 4. [2] … … … … … … … … … … … (iii) Use the iterative formula ` 3 a23 7 2a2n + 3 ln an an+1 = to determine a correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] … … … … … … … … … …

10 marks

Mark scheme: 3 9(i) 3 2 2 1 *M1 x . dx Integrate by parts and reach ax ln x + b ∫ x 2 32 2 12 A1 Obtain 3 x ln x 3 x d x −∫ 2 32 4 32 A1 Obtain integral 3 x ln x − 9 x , or equivalent Substitute limits correctly and equate to 2 DM1 Obtain the given answer correctly AG A1 5 9(ii) Evaluate a relevant expression or pair of expressions at x = 2 and x = 4 M1 Complete the argument correctly with correct calculated values A1 2 9(iii) Use the iterative formula correctly at least once M1 Obtain final answer 3.031 A1 Show sufficient iterations to 5 d.p. to justify 3.031 to 3 d.p., or show there is a sign A1 change in the interval (3.0305, 3.0315) 3

This question in 9709/32 Oct/Nov 2017

7 (i) By sketching suitable graphs, show that the equation e2x 6 has exactly one real root. [2] = + e−x (ii) Verify by calculation that this root lies between 0.5 and 1. [2] … … … … … … … … … … … … (iii) Show that if a sequence of values given by the iterative formula 3 + xn+1 = 1 ln 1 6exn converges, then it converges to the root of the equation in part (i). [2] … … … … … … … … … … (iv) Use this iterative formula to calculate the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] … … … … … … … … … … … …

9 marks

Mark scheme: 7(i) Sketch a relevant graph, e.g. y = 2e x B1 Sketch a second relevant graph, e.g. y = 6 + e− x , and justify the given statement B1 2 7(ii) Calculate the value of a relevant expression or values of a pair of relevant M1 expressions at x = 0.5 and x = 1 Complete the argument correctly with correct calculated values A1 2 7(iii) 1 x B1 State a suitable equation, e.g. x = ln 1 + 6e ( ) 3 Rearrange this as 2e x = 6 + e− x , or commence working vice versa B1 2 7(iv) Use the iterative formula correctly at least once M1 Obtain final answer 0.928 A1 Show sufficient iterations to 5 d.p. to justify 0.928 to 3 d.p., or show there is a sign A1 change in the interval (0.9275, 0.9285) 3

This question in 9709/32 Feb/March 2018

6 A a 1 rad a B C The diagram shows a triangle ABC in which AB AC a and angle BAC radians. Semicircles are drawn outside the triangle with AB and AC as= diameters.= A circular arc= 1with centre A joins B and C. The area of the shaded segment is equal to the sum of the areas of the semicircles. (i) Show that 1 sin [3] 1 = 20 + 1. … … … … … … … … … … … … … … … … … … (ii) Verify by calculation that lies between 2.2 and 2.4. [2] 1 … … … … … … … … … … … … (iii) Use an iterative formula based on the equation in part (i) to determine correct to 2 decimal places. Give the result of each iteration to 4 decimal places. 1 [3] … … … … … … … … … … …

8 marks

Mark scheme: 6(i) Use correct method for finding the area of a segment and area of semicircle and form an equation in θ M1 e.g. 2 2 2 1 1 sin 4 2 2 π θ θ = − a a a State a correct equation in any form A1 Given answer so check working carefully Obtain the given answer correctly A1 3 6(ii) Calculate values of a relevant expression or pair of expressions at 2.2 θ = and θ = 2.4 M1 e.g. ( ) ( ) ( ) f 2.2 2.37... 2.2 f sin f 2.4 2.24... 2.4 2 π θ θ  = >  = +  = <  or ( ) ( ) ( ) f 2.2 0.17... 0 f sin f 2.4 0.15... 0 2 π θ θ θ  = − <  = − −  = + >  Complete the argument correctly with correct calculated values A1 2 Question Answer Marks Guidance 6(iii) Use 1 1 sin 2 θ π θ + = + n n correctly at least once M1 e.g. 2.2 2.3 2.4 2.3793 2.3165 2.2463 2.2614 2.3054 2.3512 2.3417 2.3129 2.2814 2.2881 2.3079 2.3288 2.3244 2.2970 2.3000 2.3185 2.3165 2.3041 2.3054 2.3138 2.3129 2.3072 Obtain final answer 2.31 A1 Show sufficient iterations to 4 d.p. to justify 2.31 to 2 d.p. or show there is a sign change in the interval (2.305, 2.315) A1 3

This question in 9709/32 May/June 2018

ln x 4 The curve with equation y has a stationary point at x p. 3 x = = + 3 (i) Show that p satisfies the equation ln x 1 . [3] x = + … … … … … … … … … … … … … … … … … … … … … … … (ii) By sketching suitable graphs, show that the equation in part (i) has only one root. [2] 3 x (iii) It is given that the equation in part (i) can be written in the form x . Use an iterative + ln x = formula based on this rearrangement to determine the value of p correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … … …

8 marks

Mark scheme: 4(i) Use the quotient or product rule M1 Obtain correct derivative in any form A1 Equate derivative to zero and obtain the given equation A1 Total: 3 4(ii) Sketch a relevant graph, e.g. y = ln x B1 3 B1 Sketch a second relevant graph, e.g. y = 1 + , and justify the given statement x Total: 2 4(iii) 3 + x M1 Use iterative formula nx +1 = correctly at least once ln x n Obtain final answer 4.97 A1 Show sufficient iterations to 4 d.p.to justify 4.97 to 2 d.p. or show there is a sign A1 change in the interval (4.965, 4.975) Total: 3

This question in 9709/33 May/June 2018

3 (i) By sketching a suitable pair of graphs, show that the equation x3 3 has exactly one real = −x root. [2] (ii) Show that if a sequence of real values given by the iterative formula 2x3 3 n + xn+1 = 3x2 1 n + converges, then it converges to the root of the equation in part (i). [2] … … … … … … … … … … … … … … … … … … … … … … … … (iii) Use this iterative formula to determine the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] … … … … … … … … … … … …

7 marks

Mark scheme: 3 (i) Sketch a relevant graph, e.g. 3 y x = B1 Sketch a second relevant graph, e.g. y = 3 – x, and justify the given statement B1 Consideration of behaviour for x < 0 is needed for the second B1 2 3(ii) State or imply the equation ( ) ( ) 3 2 2 3 / 3 1 x x x = + + B1 Rearrange this in the form 3 3 x x = − , or commence work vice versa B1 2 Question Answer Marks Guidance 3(iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.213 A1 Show sufficient iterations to 5 d.p. or more to justify 1.213 to 3 d.p., or show there is a sign change in the interval (1.2125, 1.2135) A1 3

This question in 9709/31 Oct/Nov 2018

3 (i) By sketching a suitable pair of graphs, show that the equation x3 3 has exactly one real = −x root. [2] (ii) Show that if a sequence of real values given by the iterative formula 2x3 3 n + xn+1 = 3x2 1 n + converges, then it converges to the root of the equation in part (i). [2] … … … … … … … … … … … … … … … … … … … … … … … … (iii) Use this iterative formula to determine the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] … … … … … … … … … … … …

7 marks

Mark scheme: 3 (i) Sketch a relevant graph, e.g. 3 y x = B1 Sketch a second relevant graph, e.g. y = 3 – x, and justify the given statement B1 Consideration of behaviour for x < 0 is needed for the second B1 2 3(ii) State or imply the equation ( ) ( ) 3 2 2 3 / 3 1 x x x = + + B1 Rearrange this in the form 3 3 x x = − , or commence work vice versa B1 2 Question Answer Marks Guidance 3(iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.213 A1 Show sufficient iterations to 5 d.p. or more to justify 1.213 to 3 d.p., or show there is a sign change in the interval (1.2125, 1.2135) A1 3

This question in 9709/33 Oct/Nov 2018

2 The sequence of values given by the iterative formula 2x6 n 12xn + , xn+1 = 3x5 8 n + with initial value x1 2, converges to = !. (i) Use the formula to calculate correct to 4 decimal places. Give the result of each iteration to ! 6 decimal places. [3] … … … … … … … … … … (ii) State an equation satisfied by and hence find the exact value of [2] ! !. … … … … … … … … … …

5 marks

Mark scheme: 2(i) Use the iterative formula correctly at least once M1 Obtain answer 1.3195 A1 Show sufficient iterations to 6 d.p. to justify 1.3195 to 4 d.p., or show there is a sign change in (1.31945, 1.31955) A1 3 2(ii) State x = 6 5 2 12 3 8 + + x x x , or equivalent B1 State answer 5 4 , or exact equivalent B1 2

This question in 9709/32 Feb/March 2019

7 y x O a 4 1 1 for 0 The diagram shows the curves y 4. When x = ≤x < = 4 cos 2x and y = 4 −x, a, the tangents to the curves are perpendicular. 4 (i) Show that a / 2 sin 12a . [4] = − … … … … … … … … … … … … … … … (ii) Verify by calculation that a lies between 2 and 3. [2] … … … … … … … … … … … (iii) Use an iterative formula based on the equation in part (i) to determine a correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] … … … … … … … … … … … …

9 marks

Mark scheme: 7(i) State at least one correct derivative B1 ( ) 2 1 1 2sin , 2 4 − − x x Equate product of derivatives to – 1 M1 or equivalent Obtain a correct equation, e.g. ( ) 2 1 2sin 4 2 = − x x A1 Rearrange correctly to obtain 4 2sin 2 = − a a AG A1 4 7(ii) Calculate values of a relevant expression or pair of expressions at a = 2 and a = 3 M1 e.g. 2 2 2.7027.. 3 3 2.587.. = < = > a a 0.703 2.317 0.412 0.995         − −     Values correct to at least 2 dp Complete the argument correctly with correct calculated values A1 2 7(iii) Use the iterative formula 1 ) 1 4 (2sin 2 n n a a + = − correctly at least once M1 Obtain final answer 2.611 A1 Show sufficient iterations to 5 d.p. to justify 2.611 to 3 d.p., or show there is a sign change in the interval (2.6105, 2.6115) A1 2, 2.70272, 2.60285, 2.61152, 2.61070, 2.61077 2.5, 2.62233, 2.60969, 2.61087, 2.61076 3, 2.58756, 2.61301, 2.61056, 2.61079 Condone truncation. Accept more than 5 dp 3

This question in 9709/31 May/June 2019

6 A x rad B C r O In the diagram, A is the mid-point of the semicircle with centre O and radius r. A circular arc with centre A meets the semicircle at B and C. The angle OAB is equal to x radians. The area of the shaded region bounded by AB, AC and the arc with centre A is equal to half the area of the semicircle. (i) Use triangle OAB to show that AB 2r cosx. [1] = … … … … … O@ 0 A. [2] (ii) Hence show that x cos−1 = 16x … … … … … … … … … … (iii) Verify by calculation that x lies between 1 and 1.5. [2] … … … … … … … … … (iv) Use an iterative formula based on the equation in part (ii) to determine x correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(i) Correct use of trigonometry to obtain B1 AG 1 Question Answer Marks Guidance 6(ii) Use correct method for finding the area of the sector and the semicircle and form an equation in x M1 ( ) 2 2 1 1 1 2 cos 2 2 2 2 π × = r r x x Obtain 1 cos 16 π − = x x correctly AG A1 Via correct simplification e.g. from 2 cos 16 π = x x 2 6(iii) Calculate values of a relevant expression or pair of expressions at x = 1 and x = 1.5 Must be working in radians M1 e.g. 1 1 1.11 1.5 1.5 1.20 = → = → x x Accept ( ) ( ) f 1 1.11 f 1.5 1.20 = = ( ) 1 f cos 16 π − = − x x x : ( ) ( ) f 1 0.111., f 1.5 0.3.. = − = f( ) cos 16 π = − x x x :f (1) 0.097.,f (1.5) 0.291. = = − For 2 16 cos π − x x f (1) 1.529..,f (1.5) 3.02.. = = − Must find values. M1 if at least one value correct Correct values and complete the argument correctly A1 2 Question Answer Marks Guidance 6(iv) Use 1 1 π cos 16 n n x x − +   =     correctly at least twice Must be working in radians M1 1,1.11173,1.13707,1.14225,1.14329,1.14349, 1.14354,1.14354 1.25,1.16328,1.14742,1.14432,1.14370 1.5,1.20060,1.15447,1.14570,1.14397,1.14363 Obtain final answer 1.144 A1 Show sufficient iterations to at least 5 d.p. to justify 1.144 to 3 d.p. or show there is a sign change in the interval (1.1435, 1.1445) A1 3

This question in 9709/32 May/June 2019

6 y x a O b The diagram shows the curve y x4 The curve intersects the x-axis at the points a, 0 = −2x3 −7x −6. and b, 0 , where a b. It is given that b is an integer. < (i) Find the value of b. [1] … … … … … … … (ii) Hence show that a satisfies the equation a 2 a2 a3 . [4] = −13 + + … … … … … … … … … … … … … … … … … … (iii) Use an iterative formula based on the equation in part (ii) to determine a correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(i) State b = 3 B1 1 6(ii) Commence division by x – b and reach partial quotient 3 2 + x kx M1 Obtain quotient 3 2 3 2 + + + x x x A1 There being no remainder Equate quotient to zero and rearrange to make the subject a M1 Obtain the given equation A1 4 6(iii) Use the iterative formula ( ) 2 3 1 1 2 3 + = − + + n n n a a a correctly at least once M1 Obtain final answer –0.715 A1 Show sufficient iterations to 5 d.p. to justify –0.715 to 3 d.p., or show there is a sign change in the interval (–0.7145, –0.7155) A1 3

This question in 9709/33 May/June 2019

a 39 It is given that 3, where the constant a is such that 0 a x cos 3x1 dx = < < 20. Ó 0 (i) Show that a satisfies the equation 4 cos 13a −3 . [5] a = sin 13a … … … … … … … … … … … … … … … … … … … … … … (ii) Verify by calculation that a lies between 2.5 and 3. [2] … … … … … … … … … … … (iii) Use an iterative formula based on the equation in part (i) to calculate a correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] … … … … … … … … … … … …

10 marks

Mark scheme: 9(i) Commence integration by parts, reaching 1 1 sin sin d 3 3 ax x b x x −∫ *M1 Obtain 1 1 3 sin 3 sin d 3 3 x x x x −∫ A1 Complete integration and obtain 1 1 3 sin 9cos 3 3 x x x + A1 Substitute limits correctly and equate result to 3 in an integral of the form 1 1 sin cos 3 3 px x q x + DM1 ( ) 3 3 sin 9cos 0 9 3 3 a a a = + − − Obtain 4 3cos 3 sin 3 a a a − = correctly A1 With sufficient evidence to show how they reach the given equation 5 9(ii) Calculate values at a = 2.5 and a = 3 of a relevant expression or pair of expressions. M1 2.5 2.679 and 3 2.827 < > If using 2.679 and 2.827 must be linked explicitly to 2.5 and 3. Solving f(a) = 0, f(2.5) = 0.179. and f(3) = –0.173 or if 1 1 f ( ) sin 3cos 4 f (2.5) 0.13..,f (3) 0.145... 3 3 a a a a = + − ⇒ = − = Complete the argument correctly with correct calculated values A1 Accept values to 1 sf. or better 2 Question Answer Marks Guidance 9(iii) Use the iterative process 1 na + = 1 3 1 1 3 4 3cos sin n n n a a a + − correctly at least once M1 Show sufficient iterations to at least 5 d.p. to justify 2.736 to 3d.p., or show a sign change in the interval (2.7355, 2.7365) A1 Obtain final answer 2.736 A1 3

This question in 9709/32 Oct/Nov 2019

2 has exactly one root3 (a) By sketching a suitable pair of graphs, show that the equation sec x = −12x in the interval 0 1 [2] ≤x < 2π. … … (b) Verify by calculation that this root lies between 0.8 and 1. [2] … … … … … @ A 2 (c) Use the iterative formula xn+1 = cos−1 4 −xn to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … …

7 marks

Mark scheme: 3(a) Sketch the graph y = sec x M1 Sketch the graph 1 2 2 = − y x , and justify the given statement A1 2 3(b) Calculate the values of a relevant expression or pair of expressions at x = 0.8 and x = 1 M1 Complete the argument correctly with correct calculated values A1 2 3(c) Use the iterative formula correctly at least once M1 Obtain final answer 0.88 A1 Show sufficient iterations to 4 d.p. to justify 0.88 to 2 d.p., or show there is a sign change in the interval (0.875, 0.885) A1 3

This question in 9709/32 Feb/March 2020

9 y 1 x O p 12π k The diagram shows the curves y cos x and y where k is a constant, for 0 The = = 1 x, ≤x ≤12π. + curves touch at the point where x p. = 1 (a) Show that p satisfies the equation tan p [5] = 1 p. + … … … … … … … … … … … … … … … … @ A 1 (b) Use the iterative formula pn+1 = tan−1 1 pn + to determine the value of p correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] … … … … … … … … … … … … … … (c) Hence find the value of k correct to 2 decimal places. [2] … … … … … … … … …

10 marks

Mark scheme: 9(a) State cos 1 = + k p p B1 Differentiate both equations and equate derivatives at x = p M1 Obtain a correct equation in any form, e.g. ( ) 2 sin 1 − = − + k p p A1 Eliminate k M1 Obtain the given answer showing sufficient working A1 5 9(b) Use the iterative formula correctly at least once M1 Obtain final answer p = 0.568 A1 Show sufficient iterations to justify 0.568 to 3 d.p., or show there is a sign change in the interval (0.5675, 0.5685) A1 3 9(c) Use a correct method to find k M1 Obtain answer k = 1.32 A1 2

This question in 9709/32 May/June 2020

6 (a) By sketching a suitable pair of graphs, show that the equation x5 2 x has exactly one real = + root. [2] … … (b) Show that if a sequence of values given by the iterative formula 4x5 2 n + xn+1 = 5x4 n −1 converges, then it converges to the root of the equation in part (a). [2] … … … … … … … … … … … … … … … … (c) Use the iterative formula with initial value x1 1.5 to calculate the root correct to 3 decimal = places. Give the result of each iteration to 5 decimal places. [3] … … … … … … … … … … … … … …

7 marks

Mark scheme: 6(a) Sketch a second relevant graph, e.g. y = x + 2 and justify the given statement B1 2 6(b) State a suitable equation, e.g. 5 4 4 2 5 1 x x x + = − B1 Rearrange this as x5 = 2 + x or commence working vice versa B1 2 6(c) Use the iterative formula correctly at least once M1 Obtain final answer 1.267 A1 Show sufficient iterations to 5 d.p. to justify 1.267 to 3 d.p., or show there is a sign change in the interval (1.2665, 1.2675) A1 3

This question in 9709/33 May/June 2020

2x5 (a) By sketching a suitable pair of graphs, show that the equation cosec x = 1 + e−1 has exactly two roots in the interval 0 < x < π. [2] … … (b) The sequence of values given by the iterative formula P Q 1 xn+1 = π −sin−1 , e−12xn + 1 with initial value x1 = 2, converges to one of these roots. Use the formula to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … …

5 marks

Mark scheme: 5(a) Sketch a relevant graph, e.g. y = cosec x B1 cosec x, U shaped, roughly symmetrical about π π , 1 2 2 x y  = =     and domain at least π 5π , 6 6       . Sketch a second relevant graph, e.g. 1 2 1 e − = + x y , and justify the given statement B1 Exponential graph needs y(0) = 2, negative gradient, always increasing, and y(π) > 1 Needs to mark intersections with dots, crosses, or say roots at points of intersection, or equivalent 2 5(b) Use the iterative formula correctly at least twice M1 2, 2.3217, 2.2760, 2.2824… Need to see 2 iterations and following value inserted correctly Obtain final answer 2.28 A1 Must be supported by iterations Show sufficient iterations to at least 4 d.p. to justify 2.28 to 2 d.p., or show there is a sign change in the interval (2.275, 2.285) A1 3

This question in 9709/31 Oct/Nov 2020

10 y R 3 x O a 2π M The diagram shows the curve y x cos x, for 0 and its minimum point M, where x a. = ≤x ≤32π, = The shaded region between the curve and the x-axis is denoted by R. 1 (a) Show that a satisfies the equation tan a [3] = 2a. … … … … … … @ A 1 , with initial value (b) The sequence of values given by the iterative formula an+1 = π + tan−1 2an x1 3, converges to a. = Use this formula to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … (c) Find the volume of the solid obtained when the region R is rotated completely about the x-axis. Give your answer in terms of [6] π. … … … … … … … … … … … … … … … … … … … … … … … … …

12 marks

Mark scheme: 10(a) Use correct product rule M1 Obtain correct derivative in any form A1 e.g. d 1 cos sin d 2 y x x x x x = − . Accept in a or in x Equate derivative to zero and obtain 1 tan 2 a a = A1 Obtain given answer from correct working. The question says ‘show that ..’ so there should be an intermediate step e.g. cos 2 sin x x x = . Allow 1 tan 2 x x = 3 10(b) Use the iterative process correctly at least once (get one value and go on to use it in a second use of the formula) M1 Must be working in radians Degrees gives 1, 12.6039, 5.4133, ... M0 Obtain final answer 3.29 A1 Clear conclusion Show sufficient iterations to at least 4 d.p.to justify 3.29, or show there is a sign change in the interval (3.285, 3.295) A1 3, 3.3067, 3.2917, 3.2923 Allow more than 4d.p. Condone truncation. 3 Question Answer Marks Guidance 10(c) State or imply the indefinite integral for the volume is ( ) 2 π cos d x x x  B1 [If π omitted, or 2π or 1 π 2 used, give B0 and follow through. 4/6 available] Use correct cos 2A formula, commence integration by parts and reach ( sin2 ) sin2 d x ax b x ax b x x + ±  + *M1 Alternative: 2 1 sin 2 sin 2 d 4 4 4 x x x x x + − Obtain 1 1 1 1 ( sin2 ) sin2 d 2 4 2 4 x x x x x x + − + , or equivalent A1 Complete integration and obtain 2 1 1 1 sin2 cos2 4 4 8 x x x x + + A1 OE Substitute limits x = 0 and x = 1 π 2 , having integrated twice DM1 2 π π 1 1 0 0 0 2 8 4 4   + − − − −     Obtain answer ( ) 2 1 π π 4 16 − , or exact equivalent A1 CAO 6

This question in 9709/32 Oct/Nov 2020

2x5 (a) By sketching a suitable pair of graphs, show that the equation cosec x = 1 + e−1 has exactly two roots in the interval 0 < x < π. [2] … … (b) The sequence of values given by the iterative formula P Q 1 xn+1 = π −sin−1 , e−12xn + 1 with initial value x1 = 2, converges to one of these roots. Use the formula to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … …

5 marks

Mark scheme: 5(a) Sketch a relevant graph, e.g. y = cosec x B1 cosec x, U shaped, roughly symmetrical about π π , 1 2 2 x y  = =     and domain at least π 5π , 6 6       . Sketch a second relevant graph, e.g. 1 2 1 e − = + x y , and justify the given statement B1 Exponential graph needs y(0) = 2, negative gradient, always increasing, and y(π) > 1 Needs to mark intersections with dots, crosses, or say roots at points of intersection, or equivalent 2 5(b) Use the iterative formula correctly at least twice M1 2, 2.3217, 2.2760, 2.2824… Need to see 2 iterations and following value inserted correctly Obtain final answer 2.28 A1 Must be supported by iterations Show sufficient iterations to at least 4 d.p. to justify 2.28 to 2 d.p., or show there is a sign change in the interval (2.275, 2.285) A1 3

This question in 9709/33 Oct/Nov 2020

7 y M x O a tan−1x The diagram shows the curve y and its maximum point M where x a. x = = (a) Show that a satisfies the equation @ A 2a a tan . [4] = 1 a2 + … … … … … … … … … … … … … … … … … (b) Verify by calculation that a lies between l.3 and 1.5. [2] … … … … … … … … … (c) Use an iterative formula based on the equation in part (a) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … … … … …

9 marks

Mark scheme: 7(a) Use correct quotient rule or correct product rule M1 e.g. 1 2 1 1 . tan . d 1 2 d x x y x x x x − − + = Obtain correct derivative in any form A1 Equate derivative to zero and remove inverse tangent M1 Obtain 2 2 tan 1 a a a   =   +   from correct working A1 AG. Accept with x in place of a. 4 Question Answer Marks Guidance 7(b) Calculate the value of a relevant expression or pair of expressions at a = 1.3 and a = 1.5 M1 Must be using radians Complete the argument correctly with correct calculated values A1 e.g.1.3 1.448, 1.5 1.322 < > ( ) 0.148, 0.178 − 2 7(c) Use the iterative process 1 + = na tan 2 2 1     +   n n a a correctly at least twice M1 Obtain final answer 1.39 A1 Show sufficient iterations to at least 4 d.p. to justify 1.39 to 2 d.p. or show there is a sign change in the interval (1.385, 1.395) A1 Allow recovery 3

This question in 9709/31 May/June 2021

6 (a) By sketching a suitable pair of graphs, show that the equation cot 1 1 has exactly one 2x = + e−x [2] root in the interval 0 x < ≤π. (b) Verify by calculation that this root lies between 1 and 1.5. [2] … … … … … … … … … … … @ A 1 (c) Use the iterative formula 2 to determine the root correct to 2 decimal tan−1 1 xn+1 = e−xn places. Give the result of each iteration to 4 +decimal places. [3] … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 6(a) Sketch a relevant graph, e.g. 1 cot 2 = y x B1 Sketch a second relevant graph, e.g. 1 e− = + x y , and justify the given statement B1 2 6(b) Calculate values of a relevant expression or pair of expressions at x = 1 and x = 1.5 M1 Complete the argument correctly with correct calculated values A1 2 6(c) Use the iterative formula correctly at least once M1 Obtain final answer 1.34 A1 Show sufficient iterations to 4 d.p. to justify 1.34 to 2 d.p. or show there is a sign change in the interval (1.335, 1.345) A1 3

This question in 9709/33 May/June 2021

10 A large plantation of area 20 km2 is becoming infected with a plant disease. At time t years the area infected is x km2 and the rate of increase of x is proportional to the ratio of the area infected to the area not yet infected. dx When t 0, x 1 and 1. = = dt = (a) Show that x and t satisfy the differential equation dx 19x [2] dt = 20 −x. … … … … … … … (b) Solve the differential equation and show that when t 1 the value of x satisfies the equation = [5] x = e0.9+0.05x. … … … … … … … … … … … … … … … … … … … … (c) Use an iterative formula based on the equation in part (b), with an initial value of 2, to determine x correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … (d) Calculate the value of t at which the entire plantation becomes infected. [1] … … … …

11 marks

Mark scheme: 10(a) State or imply equation of the form d d x t = k 20 x x − M1 Obtain k = 19 A1 AG 2 10(b) Separate variables and integrate at least one side M1 Obtain terms 20 ln x – x and 19t, or equivalent A1 A1 Evaluate a constant or use t = 0 and x = 1 as limits in a solution containing terms a ln x and bt M1 Substitute t = 1 and rearrange the equation in the given form A1 AG 5 10(c) Use 1 nx + = 0.9 0.05 e n x + correctly at least once M1 Obtain final answer x = 2.83 A1 Show sufficient iterations to 4 decimal places to justify 2.83 to 2 d.p. or show there is a sign change in the interval (2.825, 2.835) A1 3 10(d) Set x = 20 and obtain answer t = 2.15 B1 1

This question in 9709/33 Oct/Nov 2021

7 (a) By sketching a suitable pair of graphs, show that the equation 4 sec 2x1 has exactly one root in the interval 0 −x2 = [2] ≤x < π. (b) Verify by calculation that this root lies between 1 and 2. [2] … … … … … … 2xn (c) Use the iterative formula xn+1 = ?4 −sec 1 to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … …

7 marks

Mark scheme: 7(a) Sketch a relevant graph, e.g. 2 4 = − y x Sketch a second relevant graph, e.g. 1 sec 2 = y x , and justify the given statement B1 Needs (0, 1) or mark on axis and (π, 0) Asymptote NOT required, but must NOT reach x = π. Sec graph must exist over at least interval 3π 0, 4 é ù ê ú ê ú ë û and quadratic graph over [0, 2.5]. 2 7(b) Calculate the value of a relevant expression or values of a pair of relevant expressions at x = 1 and x = 2. M1 Need all 4 values or the 2 values correct for M1. Angles in degrees score M0. Complete the argument with correct calculated values A1 2 Question Answer Marks Guidance 7(c) Use the iterative process correctly at least twice M1 Obtain final answer 1.60 A1 Must be 2 d.p. Show sufficient iterations to 4 d.p.to justify 1.60 to 2 d.p. or show there is a sign change in the interval (1.595, 1.605) A1 3

This question in 9709/32 Feb/March 2022

10 y x O a π The curve y x sin x has one stationary point in the interval 0 x where x a (see diagram). = < < π, = (a) Show that tan a 2a. [4] = −1 … … … … … … … … … … … … … … … … … … (b) Verify by calculation that a lies between 2 and 2.5. [2] … … … … … … … (c) Show that if a sequence of values in the interval 0 x given by the iterative formula 1 converges, then it converges to a,<the root< π of the equation in part (a). [2] xn+1 = π −tan−1 2xn … … … … … … … (d) Use the iterative formula given in part (c) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … …

11 marks

Mark scheme: 10(a) Use correct product rule M1 Condone incorrect / missing chain rule Obtain correct derivative in any form A1 e.g. d cos sin d 2 sin y x x x x x   or 2 d 2 2 sin cos d y y x x x x x   Equate derivative to zero and obtain an equation in tan x or tan a M1 Obtain 1 2 tana a  correctly A1 AG 4 10(b) Calculate the value of a relevant expression or pair of expressions at a = 2 and a = 2.5 M1 Must be working in radians At least one correct Complete the argument correctly with correct calculated values A1 e.g. 1 2.18 and 1.25 0.747    2 10(c) State a suitable equation, e.g. 1 1 π tan 2 x x         B1 A correct equation without subscripts or quote   tan tan      Using   tan A B  formula, or otherwise, rearrange this as 1 tan 2 x x  B1 Complete argument correctly 2 Question Answer Marks Guidance 10(d) Use the iterative process correctly at least once M1 Must be working in radians Obtain answer a = 2.29 A1 Show sufficient iterations to 4 dp to justify 2.29 to 2 dp or show there is a sign change in the interval (2.285, 2.295) A1 e.g. 2.25, 2.2974, 2.2871, 2.2893, 2.2888, … 3

This question in 9709/31 May/June 2022

5 (a) By sketching a suitable pair of graphs, show that the equation ln x 3x has one real root. = −x2 [2] (b) Verify by calculation that the root lies between 2 and 2.8. [2] … … … … … … … (c) Use the iterative formula /3xn xn xn+1 = −ln to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … …

7 marks

Mark scheme: 5(a) Sketch a relevant graph, e.g. y = ln x B1 ln( ) x : sketch should imply y-axis is an asymptote. Through (1, 0) if marked. Correct shape. 2 3x x  : Symmetrical. Through (0, 0) and (3, 0) if marked. If ln(x) correct accept parabola for +ve y only. If ln(x) incorrect then need parabola in 3 quadrants. Sketch a second relevant graph, e.g. 2 3 y x x   , and justify the given statement by marking the root on the sketch or by use of a suitable comment B1 2 5(b) Calculate the values of a relevant expression or pair of expressions at x = 2 and x = 2.8 M1 Allow for a smaller interval. At least one value correct if comparing with 0. If using pairs then the pairing must be clear. Complete the argument correctly with correct calculated values A1 e.g. 0.693 2 and1.03 0.56   or 1.307 0, 0.47 0    using 3 ln x x  0.304 0, 0.085 0    . Need to have calculated values to at least 2 sf. 2 2 2 Question Answer Marks Guidance 5(c) Use the iterative process correctly at least once M1 Obtain final answer 2.63 A1 Show sufficient iterations to at least 4 dp to justify 2.63 to 2 dp or show there is a sign change in the interval (2.625, 2.635) A1 SC Allow M1 A1 A0 to a candidate who starts at a point in the interval and reaches a premature conclusion 3

This question in 9709/32 May/June 2022

a 10 The constant a is such that x2 ln x dx 4. Ó 1 = @ A1 35 3 (a) Show that a . [5] = 3 ln a −1 … … … … … … … … … … … … … … … … … … … … … … … (b) Verify by calculation that a lies between 2.4 and 2.8. [2] … … … … … … … … … … … … (c) Use an iterative formula based on the equation in part (a) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … …

10 marks

Mark scheme: 10(a) Commence integration and reach 3 3 ln   ax x b x . 1 x dx Obtain 3 3 1 1 ln . 3 3   x x x 1 x dx A1 OE Allow omission of dx. Complete integration and obtain 3 3 1 1 ln 3 9  x x x A1 Allow  1 3 3 1 3       x . Use limits correctly and equate to 4, having integrated twice DM1 3 3 1 1 ln 3 9  a a a  (0  1 9 ) = 4 allow one sign error OR one numerical error, but 0 may be absent or expressed as 3 ln1 3 a . Allow  1 3 3 1 3       ax and  1 3 1 3       . Obtain given result correctly A1 1 3 35 3ln 1        a a AG After substitution, any errors even if corrected A0. Need to see at least one line of working between substitution and the given answer. 5 10(b) Calculate the values of a relevant expression or pair of expressions at a = 2.4 and a = 2.8 All values must be correct for M1 (numerical question) M1 Justify the given statement with correct calculated values A1 2.4 < 2.7(8) and 2.8 > 2.5(6) sign change here insufficient OR –0.3(8) and 0.2(4) < 0, > 0 or change of sign. 2 Question Answer Marks Guidance 10(c) Use the iterative process 1  n a 1 3 35 3ln 1        n a correctly at least twice M1 Obtain final answer a = 2.64 A1 Must be 2 dp. Show sufficient iterations to 4 dp to justify 2.64 to 2 dp, or show there is a sign change in (2.635, 2.645) A1 2.635 (35/(3lna  1))1/3 a = 0.0029(4) > 0 2.645 (35/(3lna  1))1/3 a =  0.012 < 0 3

This question in 9709/33 May/June 2022

9 C r 1 rad A O B The diagram shows a semicircle with diameter AB, centre O and radius r. The shaded region is the minor segment on the chord AC and its area is one third of the area of the semicircle. The angle CAB is radians. 1 (a) Show that 1 sin . [4] 1 = 3 π −1.5 21 … … … … … … … … … … … … … … … … … … (b) Verify by calculation that 0.5 0.7. [2] < 1 < … … … … … … … … … … … … (c) Use an iterative formula based on the equation in part (a) to determine correct to 3 decimal 1 places. Give the result of each iteration to 5 decimal places. [3] … … … … … … … … … … …

9 marks

Mark scheme: 9(a) State or imply angle AOC = π − 2 B1 Might be seen on the printed diagram. Use correct formulae for the area of a sector and triangle, or of a segment, and M1 1 2 1 2 r sin ( π − 2) or 2 r ( π − 2) − 2 find the area of the shaded region 1 1 r 2 ( 2) + 2 πr 2 − 12 2 r 2 sin ( π − 2)  M0 if subtraction the wrong way round. 1 A1 1 e.g. 1 π − 2) − r 2 sin ( π − 2) . 6 πr 2 = 12 r 2 ( 2 Equate to π r 2 and obtain a correct equation in any form 6 Obtain  = 13 ( − 1.5sin 2) correctly A1 AG Condone if state / imply sin ( π − 2) = sin 2 . 4 9(b) Evaluate a relevant expression or pair of expressions at = 0.5 and = 0.7 M1 Allow work on a smaller interval. Need to evaluate for both limits, with at least one correct. When using x = f ( x ) embedded values are not sufficient e.g. f(0.5) …. is accepted but 1 3 (− 1.5sin 2  0.5 ) = ... is not. Complete the argument correctly with correct calculated values A1 e.g. 0.5  0.626, 0.7  0.554 or 0.126  0, − 0.146  0 If using pairs then the pairing must be clear. Need to see the inequalities or an appropriate comment. Need to see values calculated to at least 2 sf. 2 9(c) M1 i.e obtain one value and use that value to obtain a second Use the iterative process n +1 = 1(π − 1.5sin2n ) correctly at least once value. Must be working in radians. 3 Obtain final answer 0.586 A1 Show sufficient iterations to 5 d.p. to justify 0.586 to 3 d.p., or show there is a A1 0.5,0.62646,0.57225,0.59195,0.58416,0.58715, sign change in the interval (0.5855, 0.5865). e.g. 0.58599,0.58644 0.6,0.58118,0.58833,0.58553,0.58661,0.58619,0.58636 0.7,0.55447,0.59958,0.58133,0.58827,0.58556, 0.58661,0.58620,0.58636 Allow working to more than 5 dp, but not less. 3

This question in 9709/32 Oct/Nov 2022

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

8 marks

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

This question in 9709/33 Oct/Nov 2022

7 B r O x rad A The diagram shows a circle with centre O and radius r. The angle of the minor sector AOB of the circle is x radians. The area of the major sector of the circle is 3 times the area of the shaded region. (a) Show that x 3 sin x 1 [4] = 4 + 2π. … … … … … … … … … … … … … … … … (b) Show by calculation that the root of the equation in (a) lies between 2 and 2.5. [2] … … … … … … … … … … … … (c) Use an iterative formula based on the equation in (a) to calculate this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … …

9 marks

Mark scheme: 7(a) 1 2 B1 OE State or imply area of major sector = r (2− x ) 2 1 2 1 2 B1 OE State or imply area of shaded segment = r x − r sin x r2 sin(x/2) cos(x/2) B0 until changed to (1/2)r2 sin x. 2 2 1 2  1 2 1 2  M1 OE State r (2− x ) = 3  r x − r sin x  Area of major sector = 3 times (area of minor sector – area of 2  2 2  triangle). Allow r2 sin(x/2) cos(x/2). 3 1 A1 AG Allow rectified slip if before penultimate line. Obtain the given answer x = sin x + after full and correct working 4 2 4 7(b) Calculate the values of a relevant expression or pair of expressions at x = M1 x= 2 x = 2.5 2 and x = 2.5 (3/4) sin x + (1/2)π 2.2(5277) 2.0(197) 2 < 2.2 or 2.3 2.5 > 2.0 x – (3/4) sin x – (1/2)π − 0.2(5277) < 0 + 0.4(803) > 0 or change of sign Attempt both values and one correct for M1. Complete the argument correctly with correct calculated values A1 Degrees award 0/2 2 7(c) Use the iterative formula correctly at least twice M1 Obtain final answer 2.18 A1 Show sufficient iterations to 4 d.p. to justify 2.18 to 2 d.p. or show there is A1 a sign change in the interval (2.175, 2.185) 2 2.25 2.5 2.2528 2.1543(5) 2.0196(5) 2.1530 2.1967 2.2465 2.1972 2.1786 2.1560 2.1784 2.1865 2.1960 2.1866 2.1831 2.1789 2.1830 2.1845 2.1863 2.1846 2.1831 2.1845 Degrees award 0/3 3

This question in 9709/32 Feb/March 2023

a 19 The constant a is such that xe−2x dx 8. Ó 0 = (a) Show that a 1 ln 4a 2 . [5] = 2 + … … … … … … … … … … … … … … … … … … … … … … … … (b) Verify by calculation that a lies between 0.5 and 1. [2] … … … … … … … … … … … (c) Use an iterative formula based on the equation in (a) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … … …

10 marks

Mark scheme: 9(a) Commence integration and reach 2 2 e e d x x px q x    Obtain 2 2 1 1 2 2 e e d x x x x     A1 OE Complete integration and obtain 2 2 1 1 e e 2 4 x x x     A1 Use limits correctly and equate to 1 , 8 having integrated twice DM1 2 2 1 1 1 1 e e 2 4 4 8 a a a       . Obtain   1 ln 4 2 2 a a   correctly A1 AG 5 9(b) Calculate the values of a relevant expression or pair of expressions at a = 0.5 and a = 1 M1 Justify the given statement with correct calculated values A1 e.g. 0.5 < 0.69…, 1 > 0.89… 0.193 > 0, –1.105 < 0 0.066 < 0.125, 0.148 > 0.125 if put limits in the integral. Condone if they use calculator for the definite integral. 2 9(c) Use the iterative process   1 1 2 ln 4 2 n n a a   correctly at least once. M1 Obtain final answer 0.84 A1 Show sufficient iterations to at least 4 d.p. to justify 0.84 to 2 d.p. or show that there is a sign change in (0.835, 0.845) A1 e.g. 0.75, 0.8047, 0.8261, 0.8343, 0.8373, 0.8385 1, 0.8959, 0.8599, 0.8469, 0.8420, 0.8402 . 3

This question in 9709/31 May/June 2023

denoted by 3x has one root in the interval 0 x6 The equation cot 2x1 = < < π, !. (a) Show by calculation that lies between 0.5 and 1. [2] ! … … … … … … … … … … … (b) Show that, if a sequence of positive values given by the iterative formula P @ AQ 1 1 4 tan−1 xn + xn+1 = 3 3xn converges, then it converges to [2] !. … … … … … … … … … … (c) Use this iterative formula to calculate correct to 2 decimal places. Give the result of each ! iteration to 4 decimal places. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 6(a) Calculate the values of a relevant expression or pair of expressions at x = 0.5 and x = 1 M1 Need to evaluate at both points, but M1 still available if one value incorrect. Use of degrees is M0. Correct use of a smaller interval is M1. If using g(x) – f(x), there needs to be a clear indication of the comparison being made e.g. by listing values in a table. Embedded values 0.5 and 1 are not sufficient. 3.92 and 1.83 alone are not sufficient. Complete the argument correctly with conclusion about change of sign or change of inequalities and with correct calculated values. Can all be in symbols – an explanation in words is not required. A1 e.g. 3.92 > 1.5, 1.83 < 3 or 2.42 > 0, –1.17 < 0. 2 6(b) State 1 1 1 4tan 3 3 x x x          M1 Or rearrange cot 3 2 x x       as far as 1 1 2 4tan 3 x x        Rearrange to the given equation cot 3 2 x x       Need intermediate step between 1 1 tan 2 3 x x   and cot 3 2 x x       A1 Or continue rearrangement to 1 1 1 4tan 3 3 x x x          and state iterative formula of 1 1 1 1 4tan 3 3 n n n x x x           AG 2 Question Answer Marks Guidance 6(c) Use the iterative process correctly at least once M1 Obtain one value and substitute that back in to obtain a second value. Working in degrees is M0. Obtain final answer 0.79 A1 Must be to 2 d.p. Show sufficient iterations to at least 4 d.p. to justify 0.79 to 2 d.p. or show there is a sign change in the interval (0.785, 0.795) A1 e.g. 1, 0.7623, 0.8037, 0.7921, 0.7951, 0.7943, 0.7945 or 0.5, 0.9506, 0.7665, 0.8024, 0.7924, 0.7950, 0.7944, 0.7945 or 0.75, 0.8076, 0.7911, 0.7954, 0.7943, 0.7946, 0.7945 . Condone truncation. Allow recovery. Condone minor differences in the final d.p. 3 If they do the iteration in (b) but restate the conclusion here, no marks in (b) but could score 3/3 for (c).

This question in 9709/32 May/June 2023

5 y M x O a 6π1 The diagram shows the part of the curve y x2 cos 3x for 0 and its maximum point M, where = ≤x ≤16π, x a. = @ A 1 2 (a) Show that a satisfies the equation a = 3 tan−1 3a . [3] … … … … … … … … … … … … … … … … … … (b) Use an iterative formula based on the equation in (a) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

6 marks

Mark scheme: 5(a) Use correct product rule M1 d dx (x2)cos(3x) + x2 d dx (cos 3x). Obtain correct derivative in any form A1 e.g. 2 2 cos3 3 sin3  x x x x . Equate derivative to zero and obtain 1 1 2 tan . 3 3        a a A1 AG Condone 1 1 2 tan 3 3   a a . Must at least reach expression 2x = 3x2 tan(3x) or better before final answer to gain A1. Final answer must be in terms of a. Can work with x and switch to a at very end. Look for 2 3 a or 2 3 x in working not immediately corrected or as penultimate line A0. 3 5(b) Use the iterative process 1 1 1 2 tan 3 3          n n a a correctly at least twice during successive iterations in the numerous iterations M1 Degrees 0/3. Obtain final answer 0.36 A1 Must be 2d.p. Show sufficient iterations to 4 or more d.p. to justify 0.36 to 2 d.p. or show there is a sign change in the interval   0.355, 0.365 A1 Allow small errors in 4th d.p. Allow errors at start if self corrects later. 0.5 0.4 0.3 0.2 0.1 /6 /12 0.3091 0.3435 0.3826 0.4264 0.4740 0.3017 0.3989 0.3789 0.3650 0.3499 0.3339 0.3176 0.3820 0.3439 0.3513 0.3566 0.3625 0.3688 0.3754 0.3502 0.3649 0.3619 0.3599 0.3576 0.3552 0.3526 0.3624 0.3567 0.3578 0.3604 0.3614 0.3576 0.3580 3

This question in 9709/33 May/June 2023

6 (a) By sketching a suitable pair of graphs, show that the equation cotx 2 = −cosx has one root in the interval 0 x [2] < ≤12π. (b) Show by calculation that this root lies between 0.6 and 0.8. [2] … … … … … … … … … … … P Q 1 (c) Use the iterative formula to determine the root correct to 2 decimal tan−1 2 xn xn+1 = −cos places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 6(a) Sketch a relevant graph. B1 y e.g. y = cot x: x intercept should be correct. Not touching the y-axis. No incorrect curvature. y = cotx Ignore anything outside 0 < x ⩽12 π . Sketch a second relevant graph and justify the given statement B1 e.g. y = 2 − cos x : Condone if looks almost straight, but not if drawn with a 2 ruler and not incorrect curvature. Correct y intercept. y = 2 - cosx Needs to be drawn for 0 < x ⩽12 π . Ignore outside this. 2 1 1 x 2π 2nd B1 requires a mark at the point of intersection or a suitable comment for the justification. 6(b) Calculate the value of a relevant expression or values of a pair of expressions M1 e.g. 1.17  1.46, 1.30  0.971 , −0.29  0, 0.33  0 . at x = 0.6 and x = 0.8 Must be working in radians. Values correct to at least 2 −0.20  0, 0.342  0 from tan x ( 2 − cos x ) −=1 0 . significant figures. 0.80  1, 1.34 from1 tan x ( 2 − cos x ) = 1 . Need all relevant values but only one (pair) needs to be correct to award M1. 1 . Complete set of values for their expression. If not comparing with 0 or 1 then 0.146  0, − 0.105  0 from x − tan −1 ( 2 − cos x ) the pairing must be clear, not just embedded values. Complete the argument correctly with correct calculated values (awrt 2 s.f.). A1 Accept truncated values. If comparing with 0 can either Clear comparison for their expression. Allow work on a smaller interval. indicate different signs or a negative product. 2 6(c) Use the iterative process correctly at least once. Must be working in radians M1 Obtain final answer 0.68 A1 Must be a clear conclusion. Show sufficient iterations to at least 4 d.p. to justify 0.68 to 2 d.p. or show A1 e.g. 0.7, 0.6806, 0.6855, 0.6843, 0.6846 there is a sign change in the interval (0.675, 0.685). 0.6, 0.7053, 0.6792, 0.6858, 0.6842, 0.6846 Allow recovery. 0.8, 0.6545, 0.6920, 0.6826, 0.6850, 0.6844, 0.6845. Allow truncation. Allow small differences in the 4th s.f. 3 Question Answer Marks Guidance

This question in 9709/32 Oct/Nov 2023

7 y a x O M The diagram shows the curve y = xe 2 x - 5x and its minimum point M, where x = a . 1 5 (a) Show that a satisfies the equation a = ln [3] 2 b 1 + 2a l. … … … … … … … … … … … … … … … … (b) Verify by calculation that a lies between 0.4 and 0.5 . [2] … … … … … … … … … (c) Use an iterative formula based on the equation in part (a) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 7(a) Use correct product rule M1 y Obtain correct derivative in any form A1 e.g. d = e 2 x + 2 xe 2 x − 5 d x 1  5  A1 Given answer – need to see e2x= 5/(1 + 2x) Equate derivative to zero and obtain α= ln   or ln e2x = ln (5/(1 + 2x)) in working. 2  1 + 2α Must be in terms of α not x. Allow α to be used before equating to 0. 3 7(b) Calculate the value of a relevant expression or values of a pair of expressions at M1 Need to attempt BOTH values and have one x = 0.4 and x = 0.5 correct. Complete the argument correctly with correct calculated values A1 e.g. 0.4 < 0.51[ 08 ] and 0.5 > 0.458 or 0.46 or 0.45 or – 0.11[08] < 0 and 0.042 > 0 If use original derivative −0.994 (0.4) and 0.437 (0.5). 2 7(c) 1  5  M1 Obtain one value and then substitute it into the Use the iterative process αn +1 = ln   correctly at least twice anywhere in formula to obtain a second value. 2  1 + 2αn  iteration process Obtain final answer 0.47 A1 Show sufficient iterations to 4 d.p. to justify 0.47 to 2 d.p. or show there is a sign A1 0.4,0.5108,0.4528,0.4823,0.4670,0.4749 change in the interval ( 0.465, 0.475 ) 0.45,0.4838,0.4663,0.4753,0.4707,0.4730 0.5,0.4581,0.4795,0.4685,0.4742 Allow self correction. 3 SC B1 No working 0.47

This question in 9709/32 Feb/March 2024

= ex - 3 has exactly one6 (a) By sketching a suitable pair of graphs, show that the equation cosec 12 x root, denoted by a, in the interval 0 1 x 1 r . [2] (b) Verify by calculation that a lies between 1 and 2. [2] … … … … … … … … … … … … … (c) Show that if a sequence of values in the interval 0 1 x 1 r given by the iterative formula x n + 1 1 = ln ( cosec 2 x n + 3) converges, then it converges to a. [1] … … … … … … … (d) Use this iterative formula with an initial value of 1.4 to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … … … … (e) State the minimum number of calculated iterations needed with this initial value to determine a correct to 2 decimal places. [1] … …

9 marks

Mark scheme: 6(a) Correct shape, correct vertical intercept B1 Sketch a second relevant graph, e.g. cosec 2  x y (correct shape, minimum above the axis) and justify the given statement. Need to mark intersection with a dot, a cross, or say root at points of intersection, or equivalent B1 2 6(b) Calculate the values of a relevant expression or pair of expressions at x = 1 and x = 2 M1 Use of degrees is M0. Complete the argument correctly with correct calculated values A1 E.g. 0.282 2.086, 4.389 1.188    1 < 1.626, 2 > 1.432 2.36 > 0, –3.2 < 0 At least 2sf. Condone truncation. 2 6(c) State 1 ln cosec 3 2         x x and rearrange to the given equation cosec e 3 2   x x B1 AG. Or vice versa and obtain the iterative formula. 1 6(d) Use the iterative formula correctly at least twice M1 Use of degrees in M0 (might see 1.38….). Obtain final answer 1.50 A1 Show sufficient iterations to 4 dp to justify 1.50 to 4 dp or show there is a sign change in the interval (1.495, 1.505) A1 1.5156, 1.4940, 1.4978, 1.4971. 3 1 -2 y x O π Question Answer Marks Guidance 6(e) 4 B1 1

This question in 9709/31 May/June 2024

5 (a) It is given that the equation e 2 x = 5 + cos 3x has only one root. Show by calculation that this root lies in the interval 0.7 1 x 1 0.8 . [2] … … … … … … (b) Show that if a sequence of values in the interval 0.7 1 x 1 0.8 given by the iterative formula 1 x n + 1 = ln 5 + cos 3x n 2 ` j converges then it converges to the root of the equation in part (a). [1] … … … … … … (c) Use this iterative formula to determine the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] … … … … … … … … … …

6 marks

Mark scheme: 5(a) Calculate the value of a relevant expression or values of a pair of expressions at 0.7 x  and 0.8 x  Need all relevant values but condone one error. Pairings must be clear for solutions involving four values (do not accept embedded values). M0 if working in degrees e.g -1.94…, 1.04… Complete the argument correctly with correct calculated values. (can be using the equation in the rubric or the equation in (b) or equivalent) A1 E.g. 4.95... 4.26...  and 4.06... 4.49...  -0.439…< 0, 0.690…> 0 1.4 < 1.5029…, 1.6 > 1.449… 1.1 > 1, 0.86 < 1 0.0515 > 0, -0.075 < 0. Allow values rounded or truncated to 2sf. 2 5(b) State   2 ln 5 cos3 x x   and take exponential of both sides to obtain 2e 5 cos3 x x   B1 Given answer requires fully correct working or work vice versa. If working in reverse, must get to the iterative formula, including subscripts. 1 5(c) Use the iterative process correctly at least once M1 M0 if working in degrees (e.g. values heading for 0.89….). Obtain final answer 0.740 A1 Show sufficient iterations to at least 5dp to justify 0.740 to 3dp, or show that there is a sign change in the interval   0.7395, 0.7405 A1 E.g. 0.7,0.75150,0.73719,0.74105,0.74000,0.74028 0.75,0.73759,0.74094,0.74003,0.74028 0.8,0.72494,0.74443,0.73909,0.74053,0.74014,0.74025 Allow recovery. Allow truncation or rounding and condone small differences in the final decimal place. 3

This question in 9709/32 May/June 2024

5 (a) By sketching a suitable pair of graphs, show that the equation 2 + e -0 .2 x = ln ( 1 + x) has only one root. [2] (b) Show by calculation that this root lies between 7 and 9. [2] … … … … … … … … … … … … (c) Use the iterative formula x = exp `2 + e -0 .2 x nj - 1 n + 1 to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [exp(x) is an alternative notation for ex.] [3] … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 5(a) Sketch a relevant graph, e.g. y = 2 + e− 0.2 x B1 y 1 For the sketches: y=2+exp(- 5x) y=ln(1+x) Correct curvature Intersections with the y-axis approximately correct 3 Horizontal asymptote approximately correct – need not draw in 2 Allow scale not marked and implied by their sketch O x Sketch a second relevant graph, e.g. y = ln (1 + x ) and justify the given B1 statement 2 5(b) Calculate the value of a relevant expression or values of a relevant pair of M1 expressions at x = 7 and x = 9 Complete the argument correctly with correct calculated values A1 E.g. 2.079  2.246 and 2.302  2.165, or 0.167  0 and −0.137  0. 2 5(c) Use the iterative process correctly at least once M1 I.e., obtain one value and substitute that value back into the formula. Obtain final answer 8.03 A1 Show sufficient iterations to at least 4 decimal places to justify 8.03 to 2 A1 E.g. decimal places, or show that there is a sign change in the interval 7, 8.4555, 7.8846, 8.0849, 8.0115, 8.0380, 8.0283, 8.0318 ( 8.025,8.035 ) 8, 8.0421, 8.0268, 8.0324, 9, 7.7172, 8.1490, 7.9887, 8.0463, 8.0253, 8.0329 3

This question in 9709/31 Oct/Nov 2024

2 (a) By sketching a suitable pair of graphs, show that the equation cot 2x = sec x has exactly one root in the interval 0 1 x 1 1 r . [2] 2 (b) Show that if a sequence of real values given by the iterative formula 1 -1 x = tan ( cos x ) n + 1 2 n converges, then it converges to the root in part (a). [1] … … … … … … … … … … … … …

3 marks

Mark scheme: 2(a) Sketch a relevant graph, e.g. y = cot 2x B1 Alt: use tan2xand cosx . And only one root in π range. (also cross at 2 ) Sketch a second relevant graph on the same axes, e.g. y = sec x and justify the given B1 Need to mark intersection with a dot, a cross, or statement say roots at points of intersection, OE. 2 2(b) 1 −1 B1 Should see tan2 x = cos x before the given State x = tan ( cos x ) 2 conclusion. and rearrange to the given equation cot 2x = sec x 1 −1 Or rearrange cot 2x = sec x to x = tan ( cos x ) 2 and state iterative formula 1 −1 xn +1 = tan ( cos xn ) . Note: If using the alternative approach in (a), can stop at tan2 x = cos x 2 1

This question in 9709/32 Oct/Nov 2024

2 Let f ( )x = 2x 3 - 5 x 2 + 4 . (a) Show that if a sequence of values given by the iterative formula 4 x = n + 1 5 - 2x n converges, then it converges to a root of the equation f ( )x = 0 . [2] … … … … … … … … … … … (b) The equation has a root close to 1.2 . Use the iterative formula from part (a) and an initial value of 1.2 to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … … …

5 marks

Mark scheme: 2(a) 4 *B1 Could work with nx +1 throughout or with nx State or imply the equation x = and square the equation 5 − 2 x throughout instead of x. Rearrange this with at least one intermediate step in the form 2 x 3 − 5 x 2 + 4 = 0 DB1 Alternative Method 1 for Question 2(a) Rearrange 2 x 3 − 5 x 2 + 4 = 0 to x2(5 – 2x) = 4 (or a different intermediate step) B2 2 4 4 and to either x = or x = 5 − 2 x 5 − 2 x 4 and then obtain the iterative formula xn +1 = 5 − 2 xn Alternative Method 2 for Question 2(a) Rearrange 2 x 3 − 5 x 2 + 4 = 0 to x2(5 – 2x) = 4 (or a different intermediate step) and *B1 Must have introduced nx +1 and nx . 2 4 2 4 to x = and to xn +1 = 5 − 2 x 5 − 2 xn 4 DB1 Obtain the iterative formula xn +1 = 5 − 2 xn 2 2(b) Use the iterative process correctly at least once M1 The question specifies initial value 1.2, so must use the formula to obtain a value and then use this value in the formula. Obtain final answer 1.28 A1 Can gain this mark even if less than 4 dp shown in iteration. Show sufficient iterations to at least 4 dp to justify 1.28 to 2 dp or show that there is A1 1.2, 1.2403, 1.2601, 1.2700,1.2752,1.2778,... a sign change in the interval (1.275,1.285 ) Allow small errors, truncation and recovery. 3

This question in 9709/33 Oct/Nov 2024

9 The constant a is such that ; 6x ln x dx = 4 . 1 1 5 (a) Show that a = exp f e 2 + 3op, where exp(x) denotes ex. [5] 6 a … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Verify by calculation that a lies between 2 and 2.1. [2] … … … … … … … … … … … (c) Use an iterative formula based on the equation in part (a) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … … … … … …

10 marks

Mark scheme: 9(a) M1* x 2 x dx  d x. Allow M1 with q Commence integration and reach px 2 ln x + q   x A1 x 2 3 x dx must be simplified. Obtain 3 x 2 ln x − x Complete integration and obtain 3 x 2 ln x − 32 x 2 A1 Use limits correctly in an expression of the form ax 2 ln x − bx 2 and equate to 4, having DM1 3a 2 ln a − 32 a 2 − 0 + 32 = 4 integrated twice  1  5   + 3 Obtain a = exp   correctly A1 AG   2  6  a   5 9(b) Calculate the values of a relevant expression or pair of expressions at M1 Not from using calculator to evaluate the original a = 2 and a = 2.1 integral. Justify the given statement with correct calculated values A1 E.g., using f( x ) = 6 x 2 ln x − 3 x 2 f(2) = 4.636  5 and f(2.1) = 6.402  5, or using the exponential, 2  2.0306 and 2.1  1.9917 or 0.0306 > 0 and −0.1083  0. . 2 9(c)  1  5   M1 Use the iterative process a n +1 = exp   2 + 3   correctly at least once.  6  a n   Obtain final answer 2.02 A1 Show sufficient iterations to at least 4 d.p. to justify 2.02 to 2 d.p. A1 E.g. 2 → 2.0306 → 2.0180 → 2.0231 … or show that there is a sign change in ( 2.015, 2.025 ) 2.05 → 2.0103 → 2.0263 → 2.0197 → 2.0224.. 2.1 → 1.9917 → 2.0342 → 2.0166 → 2.0237.. 3

This question in 9709/31 May/June 2025

11 y M O a 1 x r 2 The diagram shows the curve y = x sin 2x for 0 G x G 1 r. The curve has a maximum point at M, 2 where x = a . (a) Show that tan 2a =-4a [4] … … … … … … … … … … … … … (b) Show by calculation that 0.9 1 a 1 0. 95 . [2] … … … … … … (c) Show that if a sequence of values given by the iterative formula = x 1 -1 n + 1 2 br - tan `4x njl converges, then it converges to a. [2] … … … … … … … … … … … (d) Use the iterative formula in part (c) to calculate a correct to 4 decimal places. Give the result of each iteration to 6 decimal places. [3] … … … … … … … … … … … … …

11 marks

Mark scheme: 11(a) Use the correct product rule to differentiate *M1 Could be working in terms of a. p Obtain the form sin2 x + q x cos2 x. x dy 1 A1 Obtain = sin2 x + 2 x cos2 x dx 2 x Equate the derivative to zero and form an equation without surds DM1 E.g. sin2a + 4a cos2a = 0. Could be working in terms of x. Obtain tan2a = −4a from correct work A1 AG 4 11(b) Calculate the values of a relevant expression or pair of expressions at x = 0.9 and M1 Allow smaller interval, provided it contains the x = 0.95 root. Must be working in radians. Complete the argument correctly with correct calculated values A1 E.g.  tan1.8 + 3.6 = −0.686...  0   tan1.9 + 3.8 = 0.873...  0 2 11(c) 1 −1 −1 B1 Or work from right to left. π − tan 4 a and rearrange to π − 2a = tan 4a State a = 2 ( ) Allow working in x or a. State tan ( π − 2 a ) = 4 a and rearrange to tan2a = −4a B1 Allow working in x or a. 2 11(d) Use the iterative process correctly at least once M1 M0 if working in degrees. Obtain final answer 0.9183 A1 Show sufficient iterations to at least 6 d.p. to justify 0.9183 to 4 d.p. A1 E.g. or show there is a sign change in the interval ( 0.91825, 0.91835 ) 0.9,0.920872,0.917944,0.918347,0.918292... 3

This question in 9709/33 May/June 2025

8 (a) By sketching a suitable pair of graphs, show that the equation sec 2x =- 2x - 1 has exactly one 2 root in the interval 0 G x G 1 r . [2] 2 (b) Show by calculation that this root lies between 0.8 and 1.2. [2] … … … … … … … … … … … … … … … (c) Show that, if a sequence of real values given by the iterative formula 1 -1 - 2 x = cos e o n + 1 2 4x + 1 n converges, then it converges to the root of the equation in part (a). [2] … … … … … … … … … … … (d) Use this iterative formula to calculate this root correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] … … … … … … … … … … …

9 marks

Mark scheme: 8(a) Sketch y = sec 2x for 0 ⩽ x ⩽ 12 π M1 Need 1 or –1 and 14 π or 12 π. Ignore regions outside 0 ⩽ x ⩽ 12 π. Sketch y = –2x – 12 for 0 ⩽ x ⩽ 12 π and justify the given statement A1 Need a dot at the intersection of graphs, or dotted line parallel to the y-axis from where graphs cross to the x-axis, or state only one point of intersection OE. Do not allow, e.g. ‘only one root’. Ignore regions outside 0 ⩽ x ⩽ 12 π. Diagram for reference 2 8(b) Calculate the values of a relevant expression or pair of expressions at x = 0.8 and M1 M1 two values attempted and at least one correct in x = 1.2 f(x) = sec 2x +2 x + 12 : f(0.8) = –32.1 < 0, f(1.2) = 1.54 > 0. Can use smaller interval provided it contains root M1 four values attempted and at least three correct M0 if working in degrees 1 when comparing sec 2 x and –2 x – 2 : 1 Using 0.8: sec2 x = –34.2 –2 x – = –2.1 , 2 so –34.2 < –2.1. 1 Using 1.2: sec 2 x = –1.36 –2 x – = –2.9, 2 so –1.36 > –2.9. M1 two values attempted and at least one correct in 1 −1 −2 f(x) = cos − x : 2 4 x + 1 f(0.8) = 0.234 > 0, f(1.2) = – 0.239 < 0. M1 four values attempted (must see 0.8 and 1.2 explicitly, not just embedded) and at least three correct when 1 −1 −2 comparing x and 2 cos 4 x + 1: 1 −1 −2 Using 0.8: cos = 1.03, so 0.8 < 1.03. 2 4 x + 1 1 −1 −2 Using x = 1.2 cos = 0.961 , so 1.2 > 0.961 2 4 x + 1 Complete the argument correctly with correct calculated values A1 If accurate to only 1sf, M1A0. < 0 and > 0 or change of sign is sufficient For A1, answers must be correct to at least 2 sf 2 8(c) 1 −1 −2 1 −1 −2 M1 Need consistent variable, could be xn or xn+1. Express xn +1 = cos as x = cos 2 4 xn + 1 2 4 x + 1 1 −1 −2 1 A1 AG Rearrange x = cos to sec2 x = −2 x − with full and correct working, 2 4 x + 1 2 −2 Full working should include cos2x = or no slips allowed 4 x + 1 1 1 1 −2 x − = or cos2 x = . 2 cos2x 1 −2 x − 2 Alternative Method for Question 8(c) 1 1 −1 −2 M1 −2 1 Rearrange sec2 x = −2 x − to x = cos after full and correct working, Full working should include = or 2 2 4 x + 1 4 x + 1 sec2x no slips allowed −2 1 = cos 2x or cos2 x = . 4 x + 1 1 −2 x − 2 1 −1 −2 1 −1 −2 A1 AG Express x = cos as xn +1 = cos 2 4 x + 1 2 4 xn + 1 2 8(d) Use the iterative formula correctly at least twice (consecutive) even if only 3 M1 M0 if first value is not between 0.8 and 1.2 inclusive. decimal places M0 if working in degrees. Obtain final answer [x = or α = ] 0.992 A1 A0 if state nx = 0.992. Show sufficient iterations to at least 5 d.p. to justify 0.992 to 3 d.p. or show there A1 Iterations for each starting value: is a sign change in the interval (0.9915, 0.9925) 0.8, 1.03356, 0.98547, 0.99373, 0.99226, 0.99252, 0.99247, 0.99248 0.9, 1.1030, 0.98938, 0.99303, 0.99238, 0.99250, 0.99248, 0.99248 1, 0.99116, 0.99271, 0.99244, 0.99249 1.1, 0.97510, 0.99560, 0.99193, 0.999258, 0.99246, 0.99248 1.2, 0.96143, 0.99813, 0.99148, 0.99265, 0.99245, 0.99248 3

This question in 9709/35 May/June 2025

9 (a) By sketching a suitable pair of graphs, show that the equation sec 2x =- ex has only one root in the interval 0 1 x 1 1 r . [2] 2 (b) Show by calculation that this root lies between 0.9 and 1. [2] … … … … … … … … … … … … (c) Show that if a sequence of values given by the iterative formula nj x = 1 cos -1 `-e -x n + 1 2 converges, then it converges to the root of the equation in part (a). [1] … … … … … … … … … … (d) Use the iterative formula given in part (c) to calculate x correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] … … … … … … … … … … … … … …

8 marks

Mark scheme: 9(a) Sketch a relevant graph, e.g. y = sec 2x B1 2 May be sketching cos2x and –e–x 2 π 2 π 2 2 For y = sec 2x correct shape, asymptote in correct position. Cuts at (0, 1) For −xe correct shape, cuts at (0, 1) 4 For y = cos2 x correct shape, correct max/min, cuts axis at π4 Complete scale not needed, but should have enough x to imply the key points for both graphs. For − e− correct shape, cuts at (0, -1) Sketch a second relevant graph, e.g. y = –ex and justify the given statement B1 Needs to mark intersection with a dot, a cross, or say roots at points of intersection, OE. 2 9(b) Calculate the values of a relevant expression or pair of expressions at M1 −4.401 −2.460 − 1.94  0 E.g. or x = 0.9 and x = 1 −2.403 −2.718 0.315  0 0.179  0 (or for the reciprocal graphs). −0.048  0 Complete the argument correctly with correct calculated values A1 2 9(c) 1 −1 − x B1 and rearrange to obtain sec 2x = –ex −e State the equation x = 2 cos ( ) −e − x −e − nx Or rearrange sec 2x = –ex as x = 12 cos −1 ( ) and state nx +1 = 12 cos −1 ( ) 1 9(d) Use the iterative process correctly at least once M1 Obtain final answer 0.978 A1 Show sufficient iterations to at least 5 d.p. to justify 0.978 to 3 d.p. or show there A1 E.g. is a sign change in the interval (0.9775, 0.9785) 1, 0.97376, 0.97903,0.97796, 0.97813 0.95, 0.98395, 0.97697, 0.97838, 0.97809 0.9, 0.99475, 0.97480, 0.97882, 0.97800, 0.97817 3

This question in 9709/31 Oct/Nov 2025

8 The curve with equation y = e -5 x ln 5x has a stationary point at x = p. 1 (a) Show that p satisfies the equation ln 5p = . [3] 5p … … … … … … … … … … … (b) By sketching a suitable pair of graphs, show that the equation in part (a) has only one root. [2] (c) Show by calculation that 0.2 1 p 1 0. 6 . [2] … … … … … … … … … … … … 1 1 (d) It is given that the equation in part (a) can be written in the form p = exp e o, where exp (x) 5 5p denotes ex. Use an iterative formula based on this rearrangement to calculate p correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … … …

10 marks

Mark scheme: 8(a) d d M1 M0 if y = e−5p ln 5p seen prior to differentiation. Use the correct product or quotient rule, e.g. e−5x (ln 5x) + ln 5x (e−5x) Accept if only seen in actual derivative = 0. dx dx 1 −5 x −5 x A1 Obtain the correct derivative in any form e.g. e − 5e ln5 x x 1 A1 AG Obtain the given answer ln5 p = after full and correct working 1 −5 x −5 x 5 p May go from e − 5e ln5 x = 0 , or x 1 −5 x −5 x e = 5e ln5 x to the given answer without x intermediate working. 3 8(b) 1 M1 For both marks: Sketch an acceptable graph, e.g. y = ln 5x or y = Note: Allow without scale on either axis, but if 5x y = ln5x = 0 identified to be not x = 0.2, then 0 marks for y = ln 5x. Allow graphs not labelled, or labelled with p instead of x. Allow ln 5x starting at the x-axis. If either graph shown in other quadrants, must be correct. 1 For y = , asymptotic behaviour needed for at 5x least one axis. Must not touch axes. 1 A1 Sketch a second acceptable graph, e.g. y = or ln 5x, and justify the given 5x statement by dot, cross or statement only one intersection. 2 8(c) Calculate the values of a relevant expression or pair of expressions at p = 0.2 M1 1 f(p) = ln5 p − and p = 0.6 5 p f(0.2) = –1 < 0, f(0.6) = 0.765 > 0 Note can use, e.g., p = 0.3 and p = 0.5, or any smaller interval which works At least one correct value to at least 2sf. 1 Or comparing ln5 p and . 5 p At least 3 correct values to at least 2sf. Complete the argument correctly with correct calculated values A1 2 8(d) Use the iterative formula correctly at least twice M1 M0 for 0.3526, 0.3526, 0.3526… Obtain final answer p = 0.35, Answer = 0.35, or just 0.35 stated A1 Allow, e.g., a1, a2, a3 … or x1, x2, x3 … or answer1 , answer2, answer3 … for M1 and second A1. For first A1, must be p = 0.35 or answer = 0.35 unless just 0.35 is stated, e.g. not x = 0.35, p7 =…, p∞ = … etc. Show sufficient iterations to 4 dp to justify 0.35 to 2 dp or show there is a sign A1 E.g. 0.4, 0.3297, 0.3668, 0.3450, 0.3571, 0.3502, change in the interval (0.345, 0.355) 0.3541. Allow M1(A1 or A0) A1 if more values are to at 0.2, 0.5437, 0.2889, 0.3996, 0.3299, 0.3667, 0.3451, 0.3571, 0.3502, 0.3541 least 4dp than to 3dp. 0.25, 0.4451, 0.3135, 0.3786, 0.3392, 0.3607, 0.3482, 0.3552, 0.3512, 0.3535 SC B1 for starting from either 0.3526 or 0.3527 and 0.3, 0.3895, 0.3342,0.3639, 0.3465, 0.3562, 0.3507, 0.3538 0.3520, 0.3530 using iterative formula correctly at least twice if the 0.45, 0.3119, 0.3797, 0.3387, 0.3610, 0.3480, 0.3553, 0.3512, 0.3535 sequence shows a correct change in the 4th decimal 0.5, 0.2984,0.3910, 0.3336, 0.3643, 0.3463, 0.3563, 0.3506, 0.3538 place (and SC DB1 for getting p = 0.35), but 0 0.55, 0.2877, 0.4008, 0.3294, 0.3670, 0.3449, 0.3572, 0.3501, 0.3541 marks otherwise. 0.6, 0.2791, 0.4095, 0.3260, 0.3694, 0.3437, 0.3579, 0.3497, 0.3543 3

This question in 9709/33 Oct/Nov 2025

a 1 10 The constant a is such that xe 2 x dx = 6. y 0 - 21 a (a) Show that a = 2 + e . [5] … … … … … … … … … … … … … … … … … … … … … … … … … (b) Verify by calculation that a lies between 2.2 and 2.4. [2] … … … … … … … … … (c) Use an iterative formula based on the equation in part (a) to determine a correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 10(a) 12 x 12 x *M1 e dx Use integration by parts to obtain pxe + q  1 2 x 12 x A1 Obtain 2 xe 2 e dx − 1 2 x 12 x A1 Complete the integration to obtain 2 xe − 4 e 1 Use limits 0 and a correctly and equate the answer to 6 DM1 2ae 2 a − 4e 12 a ( −0 ) + 4 = 6 − 12 a A1 AG Obtain a = 2 + e from full and correct working 5 10(b) Calculate the value of a relevant expression or values of a relevant pair of M1 Example 1 2.2 < 2.33 and 2.4 > 2.30 expressions at 2.2 and 2.4. Values to at least 2 sf. − 1 a 2 OE Need all relevant values but only one (pair) needs to be correct to score M1. Example 2 f ( a ) = a −−2 e Complete set of values for their expression. If not comparing with 0, then the f ( 2.2 ) = −0.132... comparison must be clear. f ( 2.4 ) = 0.0988... 1 1 a a 2 − 4e 2 + 4 Example 3 f ( a ) = 2ae f ( 2.2 ) = 5.20... f ( 2.4 ) = 6.65... 1 1 a a 2 − 4e 2 − 2 Example 4 f ( a ) = 2ae f ( 2.2 ) = −0.80... f ( 2.4 ) = 0.65... Complete the argument correctly with correct calculated values. Accept truncated A1 Example 1 A clear explanation is needed. values. Allow work on a smaller interval. Example 2 f ( 2.2 ) = −0.132...  0 f ( 2.4 ) = 0.0988...  0 OE Example 3 f ( 2.2 ) = 5.20...  6 f ( 2.4 ) = 6.65...  6 Example 4 f ( 2.2 ) = −0.80...  0 f ( 2.4 ) = 0.65...  0 2 10(c) Use iterative process correctly at least once M1 Obtain final answer 2.31 A1 Show sufficient iterations to at least 4 dp to justify 2.31 to 2 dp A1 Allow recovery. Allow truncation. E.g. 2.3, 2.3166, 2.3140, 2.3144. 3

This question in 9709/35 Oct/Nov 2025