Numerical solution of equations — Paper 2 · A Level Mathematics 9709

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

5 marks

Mark scheme: 3 (i) Use the given iterative formula correctly at least once M1 Obtain final answer α = 1 . 68 A1 Show sufficient iterations to justify the answer to 2 dp B1 3 3 2 (ii) State equation, e.g. x = x + , in any correct form B1 4 x 3 Derive the exact answer α (or x) = 4 8 , or equivalent B1 2

This question in 9709/21 May/June 2005

5 (i) By sketching a suitable pair of graphs, show that the equation secx = 3 −x, 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 1.0 and 1.2. [2] (iii) Show that this root also satisfies the equation 1 x = cos−1 . [1] 3 −x (iv) Use the iterative formula 1 xn+1 = cos−1 , 3 −xn with initial value x1 = 1.1, to calculate 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, e.g. y = secx B1 Sketch an appropriate second graph, e.g. y = 3 –x, correctly and justify the given statement B1 [2] (ii) Consider sign of secx – (3 – x) at x = 1 and x = 1.2, or equivalent M1 Complete the argument correctly with appropriate calculations A1 [2] (iii) Show that the given equation is equivalent to secx = 3 –x, or vice versa B1 [1] (iv) Use the iterative formula correctly at least once M1 Obtain final answer 1.04 A1 Show sufficient iterations to justify its accuracy to 2 d.p., or show there is a sign change in the interval (1.035, 1.045) B1 [3] 1 1

This question in 9709/21 May/June 2007

2 The sequence of values given by the iterative formula 2xn 4 xn+1 = + , 3 x2 n with initial value x1 = 2, converges to α. (i) Use this iterative formula to determine α correct to 2 decimal places, giving the result of each iteration to 4 decimal places. [3] (ii) State an equation that is satisfied by α and hence find the exact value of α. [2]

5 marks

Mark scheme: 2 (i) Use the iterative formula correctly at least once M1 Obtain final answer 2.29 A1 Show sufficient iterations to justify its accuracy to 2 d.p. (must be working to 4 d.p.) – 3 iterations are sufficient B1 [3] 2 4 (ii) State equation x = x + 2 , or equivalent B1 3 x Derive the exact answer α (or x) = 3 12 , or equivalent B1 [2]

This question in 9709/21 Oct/Nov 2007

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

10 marks

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

This question in 9709/21 May/June 2008

7 (i) By sketching a suitable pair of graphs, show that the equation cos x = 2 −2x, where x is in radians, has only one root for 0 ≤x ≤12π. [2] (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 xn+1 = 1 −12 cos xn converges, then it converges to the root of the equation in part (i). [1] (iv) Use this iterative formula, with initial value x1 = 0.6, 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 a recognizable sketch of a relevant graph, e.g. y = cos x or y = 2 – 2x B1 Sketch a second relevant graph and justify the given statement B1 [2] (ii) Consider sign of cos x – (2 – 2x) at x = 0.5 and x = 1, or equivalent M1 Complete the argument correctly with appropriate calculations A1 [2] (iii) Show that the given equation is equivalent to x = 1 – 1 cos x, or vice versa B1 [1] 2 (iv) Use the iterative formula correctly at least once Ml Obtain final answer 0.58 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (0.575, 0.585) B1 [3]

This question in 9709/21 Oct/Nov 2008

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

9 marks

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

This question in 9709/21 May/June 2009

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

8 marks

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

This question in 9709/21 Oct/Nov 2009

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

9 marks

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

This question in 9709/22 Oct/Nov 2009

7 (i) By sketching a suitable pair of graphs, show that the equation e2x 2 = −x has only one root. [2] (ii) Verify by calculation that this root lies between x 0 and x 0.5. [2] = = (iii) Show that, if a sequence of values given by the iterative formula 1 2 xn+1 = ln(2 −xn) converges, then it converges to the root of the equation in part (i). [1] (iv) Use this iterative formula, with initial value x1 0.25, to determine the root correct to 2 decimal = places. Give the result of each iteration to 4 decimal places. [3]

8 marks

Mark scheme: 7 (i) Make a recognisable sketch of a relevant graph, e.g. y = 2 – x B1 Sketch an appropriate second graph, e.g. y = e2x, and justify the given statement B1 [2] (ii) Consider sign of e2x – (2 – x) at x = 0 and x = 0.5, or equivalent M1 Complete the argument correctly with correct calculations A1 [2] (iii) Show that e2x = 2 – x is equivalent to x = 1 ln(2 – x), or vice versa B1 [1] 2 (iv) Use the iterative formula correctly at least once M1 Obtain final answer 0.27 A1 Show sufficient iterations to justify its accuracy to 2 d.p., or show there is a sign change in the interval (0.265, 0.275) A1 [3]

This question in 9709/21 May/June 2010

6 (i) By sketching a suitable pair of graphs, show that the equation ln x 2 = −x2 has only one root. [2] (ii) Verify by calculation that this root lies between x 1.3 and x 1.4. [2] = = (iii) Show that, if a sequence of values given by the iterative formula xn+1 = √(2 −ln xn) converges, then it converges to the root of the equation in part (i). [1] (iv) Use the iterative formula xn+1 = √(2 −ln xn) 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) Make a recognisable sketch of a relevant graph, e.g. y = ln x or y = 2 – x2 B1 Sketch a second relevant graph and justify the given statement B1 [2] (ii) Consider sign of In x – (2 – x2) at x = 1.3 and x = 1.4, or equivalent M1 Complete the argument correctly with appropriate calculations A1 [2] GCE AS/A LEVEL – May/June 2010 9709 22 (iii) Show that given equation is equivalent to x = ( 2 − ln x ) or vice versa B1 [1] (iv) Use the iterative formula correctly at least once M1 Obtain final answer 1.31 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (1.305, 1.315) B1 [3]

This question in 9709/22 May/June 2010

6 (i) By sketching a suitable pair of graphs, show that the equation ln x 2 = −x2 has only one root. [2] (ii) Verify by calculation that this root lies between x 1.3 and x 1.4. [2] = = (iii) Show that, if a sequence of values given by the iterative formula xn+1 = √(2 −ln xn) converges, then it converges to the root of the equation in part (i). [1] (iv) Use the iterative formula xn+1 = √(2 −ln xn) 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) Make a recognisable sketch of a relevant graph, e.g. y = ln x or y = 2 – x2 B1 Sketch a second relevant graph and justify the given statement B1 [2] (ii) Consider sign of In x – (2 – x2) at x = 1.3 and x = 1.4, or equivalent M1 Complete the argument correctly with appropriate calculations A1 [2] GCE AS/A LEVEL – May/June 2010 9709 23 (iii) Show that given equation is equivalent to x = ( 2 − ln x ) or vice versa B1 [1] (iv) Use the iterative formula correctly at least once M1 Obtain final answer 1.31 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (1.305, 1.315) B1 [3]

This question in 9709/23 May/June 2010

6 6 The curve with equation y intersects the line y x 1 at the point P. = x2 = + (i) Verify by calculation that the x-coordinate of P lies between 1.4 and 1.6. [2] (ii) Show that the x-coordinate of P satisfies the equation q 6 x . = x 1 [2] + (iii) Use the iterative formula r 6 xn+1 = , xn 1 + with initial value x1 1.5, to determine the x-coordinate of P correct to 2 decimal places. Give = the result of each iteration to 4 decimal places. [3]

7 marks

Mark scheme: 6 6 (i) Consider sign of 2 −x − 1 at x = 1.4 and x = 1.6, or equivalent M1 x Complete the argument correctly with appropriate calculations A1 [2] 6 (ii) State 2 = x + 1 B1 x Rearrange equation to given equation or vice versa B1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.54 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (1.535, 1.545) B1 [3]

This question in 9709/21 Oct/Nov 2010

6 6 The curve with equation y intersects the line y x 1 at the point P. = x2 = + (i) Verify by calculation that the x-coordinate of P lies between 1.4 and 1.6. [2] (ii) Show that the x-coordinate of P satisfies the equation q 6 x . = x 1 [2] + (iii) Use the iterative formula r 6 xn+1 = , xn 1 + with initial value x1 1.5, to determine the x-coordinate of P correct to 2 decimal places. Give = the result of each iteration to 4 decimal places. [3]

7 marks

Mark scheme: 6 6 (i) Consider sign of 2 −x − 1 at x = 1.4 and x = 1.6, or equivalent M1 x Complete the argument correctly with appropriate calculations A1 [2] 6 (ii) State 2 = x + 1 B1 x Rearrange equation to given equation or vice versa B1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.54 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (1.535, 1.545) B1 [3]

This question in 9709/22 Oct/Nov 2010

2 The sequence of values given by the iterative formula 7xn 5 8 xn+1 = , + 2x4n with initial value x1 1.7, converges to α. = (i) Use this iterative formula to determine α correct to 2 decimal places, giving the result of each iteration to 4 decimal places. [3] (ii) State an equation that is satisfied by α and hence show that α [2] 5√20. =

5 marks

Mark scheme: 2 (i) Use the iterative formula correctly at least once M1 Obtain final answer 1.82 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (1.815, 1.825) B1 [3] 7 x 5 (ii) State equation x = + 4 , or equivalent B1 8 2 x Derive the exact answer α (or x) = 5 20 B1 [2]

This question in 9709/23 Oct/Nov 2010

7 (i) By sketching a suitable pair of graphs, show that the equation e2x 14 = −x2 has exactly two real roots. [3] (ii) Show by calculation that the positive root lies between 1.2 and 1.3. [2] (iii) Show that this root also satisfies the equation x 1 = 2 ln(14 −x2). [1] (iv) Use an iteration process based on the equation in part (iii), with a suitable starting value, to find the root correct to 2 decimal places. Give the result of each step of the process to 4 decimal places. [3]

9 marks

Mark scheme: 7 (i) Draw correct sketch of y = e2x B1 Draw correct sketch of y = 14 – x2 B1 Indicate two real roots only from correct sketches B1 [3] (ii) Consider sign of e2x + x2 – 14 for 1.2 and 1.3 or equivalent M1 Justify conclusion with correct calculations ( f(1.2) = –1.54, f(1.3) = 1.15 ) A1 [2] 1 2 (iii) Confirm given answer x = ln (14 − x ) B1 [1] 2 (iv) Use the iteration process correctly at least once M1 Obtain final answer 1.26 A1 Show sufficient iterations to 4 decimal places to justify answer or show a sign change in the interval (1.255, 1.256) A1 [3] [1.2 → 1.2653 → 1.2588 → 1.2595 ; 1.25 → 1.2604 → 1.2593 → 1.2594 ; 1.3 → 1.2522 → 1.2598 → 1.2594 ]

This question in 9709/21 May/June 2011

6 (i) Verify by calculation that the cubic equation x3 5x 0 −2x2 + −3 = has a root that lies between x 0.7 and x 0.8. [2] = = (ii) Show that this root also satisfies an equation of the form ax2 3 x , + = x2 b + where the values of a and b are to be found. [2] (iii) With these values of a and b, use the iterative formula ax2n 3 + xn+1 = x2n b + to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

7 marks

Mark scheme: 6 (i) Consider sign of x3 – 2x2 + 5x – 3 at x = 0.7 and x = 0.8 M1 Complete the argument correctly with appropriate calculations A1 [2] (ii) Rearrange equation to given equation or vice versa B1 State a = 2 and b = 5 B1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 0.74 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (0.735, 0.745) B1 [3]

This question in 9709/21 Oct/Nov 2011

5 (i) By sketching a suitable pair of graphs, show that the equation 1 sin x, x = [2] where x is in radians, has only one root for 0 x < ≤12π. (ii) Verify by calculation that this root lies between x 1.1 and x 1.2. [2] = = 1 (iii) Use the iterative formula xn+1 = sin xn to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

7 marks

Mark scheme: 1 5 (i) Make a recognisable sketch of a relevant graph, e.g. y = sin x or y = B1 x Sketch a second relevant graph and justify the given statement B1 [2] 1 (ii) Consider sign of − sin x at x = 1.1 and x = 1.2, or equivalent M1 x Complete the argument correctly with appropriate calculations A1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.11 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (1.105, 1.115) B1 [3] GCE AS/A LEVEL – October/November 2011 9709 22 dx dy 2

This question in 9709/22 Oct/Nov 2011

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

9 marks

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

This question in 9709/23 Oct/Nov 2011

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

9 marks

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

This question in 9709/21 May/June 2012

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

9 marks

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

This question in 9709/22 May/June 2012

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

9 marks

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

This question in 9709/23 May/June 2012

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

6 marks

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

This question in 9709/21 Oct/Nov 2012

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

6 marks

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

This question in 9709/22 Oct/Nov 2012

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

6 marks

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

This question in 9709/23 Oct/Nov 2012

6 (i) By sketching a suitable pair of graphs, show that the equation cot x 4x = −2, where x is in radians, has only one root for 0 2π. [2] ≤x ≤1 (ii) Verify by calculation that this root lies between x 0.7 and x 0.9. [2] = = (iii) Show that this root also satisfies the equation 1 2 tan x x + . = 4 tan x [1] 1 2 tan xn (iv) Use the iterative formula + to determine this root correct to 2 decimal places. xn+1 = 4 tan xn Give the result of each iteration to 4 decimal places. [3]

8 marks

Mark scheme: 6 (i) Make a recognisable sketch of a relevant graph, e.g. y = cot x or y = 4x – 2 B1 Sketch a second relevant graph and justify the given statement B1 [2] (ii) Consider sign of 4x – 2 – cot x at x = 0.7 and x = 0.9, or equivalent M1 Complete the argument correctly with appropriate calculations A1 [2] 1+ 2 tan x (iii) Show that given equation is equivalent to x = , or vice versa B1 [1] 4 tan x (iv) Use the iterative formula correctly at least once M1 Obtain final answer 0.76 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (0.755, 0.765) B1 [3] GCE AS LEVEL – May/June 2013 9709 21

This question in 9709/21 May/June 2013

6 (i) By sketching a suitable pair of graphs, show that the equation 3ex = 8 −2x has only one root. [2] (ii) Verify by calculation that this root lies between x = 0.7 and x = 0.8. [2] (iii) Show that this root also satisfies the equation @8 −2x A x = ln . 3 @8 −2xn A to determine this root correct to 3 decimal places. (iv) Use the iterative formula xn+1 = ln 3 Give the result of each iteration to 5 decimal places. [3]

7 marks

Mark scheme: 6 (i) Make a recognisable sketch of a relevant graph, e.g. y = 3ex or y = 8 – 2x B1 Sketch a second relevant graph and justify the given statement B1 [2] GCE AS LEVEL – May/June 2013 9709 22 (ii) Consider sign of 3ex – 8 + 2x at x = 0.7 and x = 0.8, or equivalent M1 Complete the argument correctly with appropriate calculations A1 [2] (f (0.7) = –0.559, f (0.8) = 0.277 or equivalent)  8 − 2 x  (iii) Show that given equation is equivalent to x = ln   , or vice versa B1 [1]  3  (iv) Use the iterative formula correctly at least once M1 Obtain final answer 0.768 A1 Show sufficient iterations to justify its accuracy to 3 d.p. xo = 0.7 xo = 0.75 xo = 0.8 0.78846 0.77319 0.75769 0.76129 0.76603 0.77082 0.76971 0.76825 0.76676 0.76711 0.76756 0.76802 0.76791 0.76763 0.76766 or show there is a sign change in the interval (0.7675, 0.7685) B1 [3] 2 1

This question in 9709/22 May/June 2013

6 (i) By sketching a suitable pair of graphs, show that the equation cot x 4x = −2, where x is in radians, has only one root for 0 2π. [2] ≤x ≤1 (ii) Verify by calculation that this root lies between x 0.7 and x 0.9. [2] = = (iii) Show that this root also satisfies the equation 1 2 tan x x + . = 4 tan x [1] 1 2 tan xn (iv) Use the iterative formula + to determine this root correct to 2 decimal places. xn+1 = 4 tan xn Give the result of each iteration to 4 decimal places. [3]

8 marks

Mark scheme: 6 (i) Make a recognisable sketch of a relevant graph, e.g. y = cot x or y = 4x – 2 B1 Sketch a second relevant graph and justify the given statement B1 [2] (ii) Consider sign of 4x – 2 – cot x at x = 0.7 and x = 0.9, or equivalent M1 Complete the argument correctly with appropriate calculations A1 [2] 1+ 2 tan x (iii) Show that given equation is equivalent to x = , or vice versa B1 [1] 4 tan x (iv) Use the iterative formula correctly at least once M1 Obtain final answer 0.76 A1 Show sufficient iterations to justify its accuracy to 2 d.p. or show there is a sign change in the interval (0.755, 0.765) B1 [3] GCE AS LEVEL – May/June 2013 9709 23

This question in 9709/23 May/June 2013

2 y x O P The diagram shows the curve y x4 2x The curve cuts the positive x-axis at the point P. = + −9. (i) Verify by calculation that the x-coordinate of P lies between 1.5 and 1.6. [2] (ii) Show that the x-coordinate of P satisfies the equation 3O@9 A x . x = −2 (iii) Use the iterative formula _P Q 3 9 xn+1 = xn −2 to determine the x-coordinate of P correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

5 marks

Mark scheme: 2 (i) Consider sign of x4 + 2x – 9 at x = 1.5 and x = 1.6 M1 Complete the argument correctly with appropriate calculations A1 [2] (f (1.5 ) = −.0 9375f, (1.6 ) = .0 7536 ) (ii) Rearrange x4 + 2x – 9 = 0 to given equation or vice versa B1 [1] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.56 A1 Show sufficient iterations to justify its accuracy to 2 d.p. B1 [3] xo = 1.5 xo = 1.55 xo = 1.6 1.5874 1.5614 1.5362 1.5424 1.5556 1.5685 1.5653 1.5520 1.5536 1.5604 1 5595 1.5561 1.5565 or show there is a sign change in the interval (1.555, 1.565) 2

This question in 9709/21 Oct/Nov 2013

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

5 marks

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

This question in 9709/22 Oct/Nov 2013

2 y x O P The diagram shows the curve y x4 2x The curve cuts the positive x-axis at the point P. = + −9. (i) Verify by calculation that the x-coordinate of P lies between 1.5 and 1.6. [2] (ii) Show that the x-coordinate of P satisfies the equation 3O@9 A x . x = −2 (iii) Use the iterative formula _P Q 3 9 xn+1 = xn −2 to determine the x-coordinate of P correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

5 marks

Mark scheme: 2 (i) Consider sign of x4 + 2x – 9 at x = 1.5 and x = 1.6 M1 Complete the argument correctly with appropriate calculations A1 [2] (f (1.5 ) = −.0 9375f, (1.6 ) = .0 7536 ) (ii) Rearrange x4 + 2x – 9 = 0 to given equation or vice versa B1 [1] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 1.56 A1 Show sufficient iterations to justify its accuracy to 2 d.p. B1 [3] xo = 1.5 xo = 1.55 xo = 1.6 1.5874 1.5614 1.5362 1.5424 1.5556 1.5685 1.5653 1.5520 1.5536 1.5604 1 5595 1.5561 1.5565 or show there is a sign change in the interval (1.555, 1.565) 2

This question in 9709/23 Oct/Nov 2013

4 (i) By sketching a suitable pair of graphs, show that the equation 4 ln x = −12x has exactly one real root, [2] !. (ii) Verify by calculation that 4.5 5.0. [2] < ! < (iii) Use the iterative formula xn+1 = 8 −2 ln xn to find ! correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

7 marks

Mark scheme: 4 (i) Make a recognisable sketch of y = ln x B1 Draw straight line with negative gradient crossing positive y-axis and justify one real root B1 [2] 1 (ii) Consider sign of ln x + x − 4 at 4.5 and 5.0 or equivalent M1 2 Complete the argument correctly with appropriate calculations A1 [2] (iii) Use the iterative formula correctly at least once M1 Obtain final answer 4.84 A1 Show sufficient iterations to justify accuracy to 2 d.p. or show sign change in interval (4.835, 4.845) A1 [3]

This question in 9709/21 Oct/Nov 2015

2 The sequence of values given by the iterative formula 4 2 xn+1 = + x2n 2xn 4, + + with initial value x1 2, converges to = !. (i) Determine the value of correct to 3 decimal places, giving the result of each iteration to 5 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 2.289 A1 Show sufficient iterations to justify accuracy to 3 d.p. or show sign change in interval (2.2885, 2.2895) A1 [3] 4 (ii) State x = 2 + 2 or equivalent B1 x + 2 x + 4 Obtain 3 12 B1 [2]

This question in 9709/22 Oct/Nov 2015

4 The sequence of values given by the iterative formula , / n ! 12x2n + 4x−3 xn+1 = with initial value x1 1.5, converges to = !. (i) Use this iterative formula to find correct to 3 decimal places. Give the result of each iteration ! to 5 decimal places. [3] (ii) State an equation that is satisfied by and hence find the exact value of [2] ! !.

5 marks

Mark scheme: 4 (i) Use the iterative formula correctly at least once M1 Obtain final answer 1.516 A1 Show sufficient iterations to justify accuracy to 3 dp or show sign change in interval (1.5155,1.5165) B1 [3] (ii) State equation x = 12 x 2 + 4 x −3 or equivalent B1 Obtain exact value 5 8 or 80.2 B1 [2]

This question in 9709/22 Feb/March 2016

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

10 marks

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

This question in 9709/21 May/June 2016

1 5 The equation of a curve is y 6xe 3x. At the point on the curve with x-coordinate p, the gradient of the curve is 40. = @ A 20 (i) Show that p 3 ln . [4] p 3 = + (ii) Show by calculation that 3.3 p 3.5. [2] < < (iii) Use an iterative formula based on the equation in part (i) to find the value of p correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3]

9 marks

Mark scheme: 1 3 x 13 x5 (i) Use product rule to obtain form k1e + k 2 xe *M1 1 3 x 13 x Obtain correct 6e + 2 xe A1 Equate first derivative to 40 and obtain equation without e present, dep *M DM1 Confirm p = 3ln p20+ 3 or x = 3ln x20+ 3 A1 [4] (ii) Consider sign of p − 3ln p20+ 3 at 3.3 and 3.5 or equivalent M1 Complete argument correctly with appropriate calculations A1 [2] (iii) Carry out iterative process correctly at least once M1 Obtain final answer 3.412 A1 Show sufficient iterations to justify accuracy to 3 dp or show sign change in interval (3.4115, 3.4125) B1 [3] 2

This question in 9709/22 May/June 2016

1 The sequence of values given by the iterative formula 4 2xn xn+1 = x2 + 3 , n with initial value x1 2, converges to = !. (i) Use this iterative formula to find correct to 3 decimal places. Give the result of each iteration ! to 5 decimal places. [3] (ii) State an equation that is satisfied by and hence find the exact value of [2] !, !.

5 marks

Mark scheme: 1 (i) Use the iterative formula correctly at least once M1 Obtain final answer 2.289 A1 Show sufficient iterations to justify accuracy to 3 dp or show sign change in interval (2.2885, 2.2895) B1 [3] 4 2 (ii) State equation x = 2 + 3 x or equivalent B1 x Obtain exact value 12 1 3 or 3 12 B1 [2]

This question in 9709/23 Oct/Nov 2016

4 The sequence of values given by the iterative formula 2x2 n xn 9 + + , xn+1 = 1 xn 2 + with x1 2, converges to = !. (i) Find the value of correct to 2 decimal places, giving the result of each iteration to 4 decimal ! places. [3] … … … … … … … … (ii) Determine the exact value of [3] !. … … … … … … … … … … … …

6 marks

Mark scheme: 4(i) Use iteration correctly at least once M1 Obtain final answer 2.08 A1 Show sufficient iterations to 4 dp to justify answer or show sign change in interval ( ) 2.075, 2.085 A1 Total: 3 4(ii) State or clearly imply equation ( ) 2 2 2 9 1 + + = + x x x x or same equation using α B1 Carry out relevant simplification M1 Obtain 3 9 A1 3

This question in 9709/21 May/June 2017

3 (i) By sketching a suitable pair of graphs, show that the equation x3 11 = −2x has exactly one real root. [2] … … … (ii) Use the iterative formula xn+1 = 3 11 −2xn to find the root correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3] … … … … … … … … … … …

5 marks

Mark scheme: 3(i) Draw sketch of 3 y x = *B1 Draw straight line with negative gradient crossing positive y-axis and indicate one intersection DB1 dep *B Total: 2 3(ii) Use iterative formula correctly at least once M1 Obtain final answer 1.926 A1 Show sufficient iterations to justify 4 sf or show sign change in interval ( ) 1.9255,1.9265 A1 Total: 3

This question in 9709/22 May/June 2017

3 (i) By sketching a suitable pair of graphs, show that the equation x3 11 = −2x has exactly one real root. [2] … … … (ii) Use the iterative formula xn+1 = 3 11 −2xn to find the root correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3] … … … … … … … … … … …

5 marks

Mark scheme: 3(i) Draw sketch of 3 y x = *B1 Draw straight line with negative gradient crossing positive y-axis and indicate one intersection DB1 dep *B Total: 2 3(ii) Use iterative formula correctly at least once M1 Obtain final answer 1.926 A1 Show sufficient iterations to justify 4 sf or show sign change in interval ( ) 1.9255,1.9265 A1 Total: 3

This question in 9709/23 May/June 2017

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

9 marks

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

This question in 9709/21 Oct/Nov 2017

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

9 marks

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

This question in 9709/22 Oct/Nov 2017

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

9 marks

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

This question in 9709/23 Oct/Nov 2017

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

8 marks

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

This question in 9709/21 May/June 2018

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

11 marks

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

This question in 9709/22 May/June 2018

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

11 marks

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

This question in 9709/23 May/June 2018

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

10 marks

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

This question in 9709/22 May/June 2019

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

10 marks

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

This question in 9709/23 May/June 2019

4 The sequence x1, x2, x3, defined by à xn x1 1, ln 2xn = xn+1 = converges to the value !. (i) Use the iterative formula to find the value of correct to 4 significant figures. Give the result of each iteration to 6 significant figures. ! [3] … … … … … … … … … … (ii) State an equation satisfied by and hence determine the exact value of [2] ! !. … … … … … … … … … …

5 marks

Mark scheme: 4(i) Use iteration correctly at least once M1 Must see correct attempt at 3x Obtain final answer 1.359 A1 Show sufficient iterations to 6 sf to justify answer or show sign change in interval [1.3585, 1.3595] A1 Answer required to exactly 4 sf Must see to at least 5x 3 4(ii) Form correct equation in x (or α ) B1 ln 2 x x x = OE Obtain 1 2 e B1 2

This question in 9709/22 Oct/Nov 2019

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

9 marks

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

This question in 9709/22 Feb/March 2020

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

8 marks

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

This question in 9709/21 May/June 2020

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

9 marks

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

This question in 9709/22 May/June 2020

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

9 marks

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

This question in 9709/23 May/June 2020

6 8xn5 The sequence of values given by the iterative formula + with initial value 2 converges xn+1 = 8 x2 x1 = n + to !. (a) Use the iterative formula to find the value of correct to 4 significant figures. Give the result of ! each iteration to 6 significant figures. [3] … … … … … … … … … … … (b) State an equation satisfied by and hence determine the exact value of [2] ! !. … … … … … … … … … …

5 marks

Mark scheme: 5(a) Use iteration correctly at least once M1 Need to see 3 values including the starting values Obtain final answer 1.817 A1 Answer required to exactly 4 significant figures Show sufficient iterations to 6 significant figures to justify answer or show sign change in interval [1.8165, 1.8175] A1 3 5(b) State equation 2 6 8 8 x x x + = + or equivalent using α B1 Obtain 3 6 or exact equivalent B1 2

This question in 9709/21 Oct/Nov 2020

9 5 (a) Given that 2 ln x 1 ln x ln x 9 , show that x x 2. [3] + + = + = + … … … … … … … … … … … … … … … … … … … … … … … … … _ 9 (b) It is given that the equation x has a single root. x 2 = + Show by calculation that this root lies between 1.5 and 2.0. [2] … … … … … … … … … … … (c) Use an iterative formula, based on the equation in part (b), to find the root correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … …

8 marks

Mark scheme: 5(a) Use the power law correctly *M1 Use correct process to obtain equation with no logarithms DM1 Confirm 9 2 x x = + A1 AG; condone absence of justification for choice of positive root. 3 5(b) Consider sign of 9 2 x x − + or equivalent for 1.5 and 2 M1 Obtain 0.1... − and 0.5 or equivalents and justify conclusion A1 AG 2 Question Answer Marks Guidance 5(c) Use iteration process correctly at least once M1 Obtain final answer 1.58 A1 Final answer required to exactly 3 sf. Show sufficient iterations to 5 sf to justify answer or show a sign change in interval [1.575, 1.585] A1 3

This question in 9709/22 Feb/March 2021

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

8 marks

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

This question in 9709/21 May/June 2021

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

11 marks

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

This question in 9709/22 May/June 2021

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

11 marks

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

This question in 9709/23 May/June 2021

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

8 marks

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

This question in 9709/21 Oct/Nov 2021

6 (a) By sketching a suitable pair of graphs on the same diagram, show that the equation ln x = 2e−x has exactly one root. [2] (b) Verify by calculation that the root lies between 1.5 and 1.6. [2] … … … … … … … … … … … … … (c) Show that if a sequence of values given by the iterative formula xn+1 = e2e−xn converges, then it converges to the root of the equation in part (a). [1] … … … … … … … (d) Use the iterative formula in part (c) to determine the root correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … … … … … …

8 marks

Mark scheme: 6(a) Draw correct sketch of ln = y x or 2e− = x y B1 = 2e− = x y must extend into 1st and 2nd quadrants. Draw correct sketch of second curve and indicate one root B1 Point of intersection must be circled/identified or a statement such that ‘there is only one point of intersection so one root only’ or similar. 2 Question Answer Marks Guidance 6(b) Consider sign of ln 2e− − x x , or equivalent, for 1.5 and 1.6 M1 Obtain 0.04... − and 0.06... or equivalents and justify conclusion A1 2 6(c) Replace 1 + nx and nx by x and apply logarithms to confirm result B1 AG Allow if done ‘in reverse’ but nx and 1 + nx need to be seen in the final statement. 1 6(d) Use iteration process correctly at least once M1 Need to see 3 correct values. Obtain final answer 1.54 A1 Answer required to exactly 3sf Show sufficient iterations to 5sf to justify answer or show sign change in interval [1.535, 1.545] A1 3

This question in 9709/22 Oct/Nov 2021

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

8 marks

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

This question in 9709/23 Oct/Nov 2021

5 (a) By sketching the graphs of y 5 and y 3 ln x = −2x = on the same diagram, show that the equation 5 3 ln x has exactly two roots. [3] −2x = (b) Show that the value of the larger root satisfies the equation x 2.5 1.5 ln x. [1] = + … … … … … … … (c) Show by calculation that the value of the larger root lies between 4.5 and 5.0. [2] … … … … … … … … … … … … (d) Use an iterative formula, based on the equation in part (b), to find the value of the larger root correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … …

9 marks

Mark scheme: 5(a) Draw correct sketch of 5 2   y x *B1 with vertex on positive x-axis Draw correct sketch of 3ln  y x *B1 Indicate the two roots either on the diagram or by a statement DB1 3 Question Answer Marks Guidance 5(b) State 2 5 3ln   x x and rearrange to confirm 2.5 1.5ln   x x B1 AG – necessary detail needed 1 5(c) Consider sign of 2.5 1.5ln   x x , or equivalent, for 4.5 and 5.0 M1 Obtain 0.25...  and 0.08... or equivalents and justify conclusion A1 AG – necessary detail needed Alternative method for question 5(c) Consider sign of 5 2 3ln   x x , or equivalent, for 4.5 and 5.0 M1 Obtain 0.51...  and 0.17... or equivalents and justify conclusion A1 AG – necessary detail needed 2 5(d) Use iteration process correctly at least once M1 Obtain final answer 4.88 A1 Answer required to exactly 3 s.f. Show sufficient iterations to 5 s.f. to justify answer or show sign change in interval [4.875, 4.885] A1 3

This question in 9709/21 May/June 2022

4 (a) By sketching a suitable pair of graphs on the same diagram, show that the equation 2x x5 e−1 = has exactly one real root. [2] 5? 2xn to determine the root correct to 4 significant figures. Give (b) Use the iterative formula the result of each iterationxn+1to 6=significante−1 figures. [3] … … … … … … … … … …

5 marks

Mark scheme: 4(a) − 12 x B1 with some curve in second quadrant as well as first. Draw approximately correct sketch of y = e Draw approximately correct sketch of y = x 5 and confirm one root B1 with some curve in third quadrant as well as first. Alternative method for question 4(a) Draw approximately correct sketch of y = 5ln x or y = ln x 5 B1 x B1 Must have intersection in the 4th quadrant. Draw approximately correct sketch of y = − and confirm one root 2 2 4(b) Use iteration process correctly at least once M1 Obtain final answer 0.9128 A1 answer required to exactly 4 s.f. Show sufficient iterations to 6 s.f. to justify answer or show sign change in A1 the interval [0.91275, 0.91285] 3

This question in 9709/21 Oct/Nov 2022

4 (a) By sketching a suitable pair of graphs on the same diagram, show that the equation 2x x5 e−1 = has exactly one real root. [2] 5? 2xn to determine the root correct to 4 significant figures. Give (b) Use the iterative formula the result of each iterationxn+1to 6=significante−1 figures. [3] … … … … … … … … … …

5 marks

Mark scheme: 4(a) − 12 x B1 with some curve in second quadrant as well as first. Draw approximately correct sketch of y = e Draw approximately correct sketch of y = x 5 and confirm one root B1 with some curve in third quadrant as well as first. Alternative method for question 4(a) Draw approximately correct sketch of y = 5ln x or y = ln x 5 B1 x B1 Must have intersection in the 4th quadrant. Draw approximately correct sketch of y = − and confirm one root 2 2 4(b) Use iteration process correctly at least once M1 Obtain final answer 0.9128 A1 answer required to exactly 4 s.f. Show sufficient iterations to 6 s.f. to justify answer or show sign change in A1 the interval [0.91275, 0.91285] 3

This question in 9709/23 Oct/Nov 2022

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

8 marks

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

This question in 9709/22 Feb/March 2023

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

7 marks

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

This question in 9709/21 May/June 2023

4 (a) y x O 3 2x. The diagram shows the graph of y = −e−1 2x On the diagram, sketch the graph of y 5x , and show that the equation 3 5x = −4 −e−1 = −4 has exactly two real roots. [2] 2x It is given that the two roots of 3 5x are denoted by and where −e−1 = −4 ! ", ! < ". (b) Show by calculation that lies between 0.36 and 0.37. [2] ! … … … … … 1 7 to find correct to 4 significant figures. Give the 5 −e−12xn! " (c) Use the iterative formula xn+1 = result of each iteration to 6 significant figures. [3] … … … … …

7 marks

Mark scheme: 4(a) Draw (more or less) correct sketch with vertex on positive x-axis *B1 crossing y-axis above given graph, may be implied by extrapolation. Indicate in some way the two roots DB1 2 4(b) Consider sign of 1 2 3 e 5 4     x x or of 1 2 3 e 5 4     x x for 0.36 and 0.37 M1 but not for sign of 1 2 3 e 5 4     x x . May be implied by 0.035...  and 0.018..., or equivalents. Obtain 0.035...  and 0.018..., or equivalents, and justify conclusion A1 AG necessary detail needed. 2 Question Answer Marks Guidance 4(c) Use iteration process correctly at least once M1 Obtain final answer 1.295 A1 answer required to exactly 4 sf. Show sufficient iterations to 6 sf to justify answer or show sign change in interval [1.2945, 1.2955] A1 3

This question in 9709/22 May/June 2023

4 (a) y x O 3 2x. The diagram shows the graph of y = −e−1 2x On the diagram, sketch the graph of y 5x , and show that the equation 3 5x = −4 −e−1 = −4 has exactly two real roots. [2] 2x It is given that the two roots of 3 5x are denoted by and where −e−1 = −4 ! ", ! < ". (b) Show by calculation that lies between 0.36 and 0.37. [2] ! … … … … … 1 7 to find correct to 4 significant figures. Give the 5 −e−12xn! " (c) Use the iterative formula xn+1 = result of each iteration to 6 significant figures. [3] … … … … …

7 marks

Mark scheme: 4(a) Draw (more or less) correct sketch with vertex on positive x-axis *B1 crossing y-axis above given graph, may be implied by extrapolation. Indicate in some way the two roots DB1 2 4(b) Consider sign of 1 2 3 e 5 4     x x or of 1 2 3 e 5 4     x x for 0.36 and 0.37 M1 but not for sign of 1 2 3 e 5 4     x x . May be implied by 0.035...  and 0.018..., or equivalents. Obtain 0.035...  and 0.018..., or equivalents, and justify conclusion A1 AG necessary detail needed. 2 Question Answer Marks Guidance 4(c) Use iteration process correctly at least once M1 Obtain final answer 1.295 A1 answer required to exactly 4 sf. Show sufficient iterations to 6 sf to justify answer or show sign change in interval [1.2945, 1.2955] A1 3

This question in 9709/23 May/June 2023

7 The curve with equation e2x y3 y 11 has a stationary point at p, q . −18x + + = (a) Find the exact value of p. [4] … … … … … … … … … … … … … … … … … … … … … … … … (b) Show that q 3 2 18 ln 3 [2] = + −q. … … … … … … … … (c) Show by calculation that the value of q lies between 2.5 and 3.0. [2] … … … … … (d) Use an iterative formula, based on the equation in (b), to find the value of q correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3] … … … … … … … … …

11 marks

Mark scheme: 7(a) 3 2 d y B1 Differentiate y to obtain 3 y d x Differentiate complete equation to produce at least one term involving M1 d y using implicit differentiation. d x 2 x 2 dy dy A1 Obtain 2e − 18 + 3 y + = 0 dx dx dy 1 A1 Substitute = 0 to obtain either p = 2 ln9 or p = ln3 dx 4 7(b) Substitute value of p in original equation and rearrange as far as y 3 = ... M1 Allow in terms of ln9 . or q3 = … Obtain given result q = 3 2 + 18ln3 − q or y = 3 2 + 18ln3 − y with A1 AG sufficient detail 2 7(c) Consider sign of q − 3 2 + 18ln3 − q or equivalent for 2.5 and 3.0 M1 Obtain −0.18... and 0.34... with sufficient detail and justify A1 OE conclusion 2 7(d) Use iteration process correctly at least once M1 Obtain final answer q = 2.673 A1 Answer required to exactly 4 s.f. Show sufficient iterations to 6 sf to justify answer or show sign change A1 in the interval [2.6725, 2.6735] 3

This question in 9709/21 Oct/Nov 2023

6 y B A x O The diagram shows the curve with parametric equations x 3 ln 2t , y 4t ln t. = −3 = The curve crosses the y-axis at the point A. At the point B, the gradient of the curve is 12. (a) Find the exact gradient of the curve at A. [5] … … … … … … … … … … … … … (b) Show that the value of the parameter t at B satisfies the equation 9 3 t [2] 1 ln t 2. = + + … … … … … … … … … … … … … (c) Use an iterative formula, based on the equation in (b), to find the value of t at B, giving your answer correct to 3 significant figures. Use an initial value of 5 and give the result of each iteration to 5 significant figures. [3] … … … … … … … …

10 marks

Mark scheme: 6(a) dx 6 B1 Obtain = dt 2t − 3 d y M1 Allow unsimplified. Use product rule to find d t dy (4ln t + 4)(2t − 3) A1 OE Obtain = dx 6 Attempt to find t corresponding to point A using a complete and correct M1 method Obtain t = 2 and hence gradient is 23 ln 2 + 23 A1 Or exact equivalent. 5 6(b) d y k M1 Equate to 12 and attempt rearrangement to 2t −=3 d x 4ln t + 4 9 3 A1 AG Confirm given result t = + with sufficient detail 1 + ln t 2 2 6(c) Use iteration process correctly at least once M1 Need to see 4.9626 . Obtain final answer 4.96 A1 Answer required to exactly 3 s.f. Show sufficient iterations to 5 sf to justify answer or show sign change in A1 interval [4.955, 4.965] 3

This question in 9709/22 Oct/Nov 2023

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

9 marks

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

This question in 9709/22 Feb/March 2024

5 A curve has equation y = . The curve has exactly one stationary point P. 1 + 3x dy 1 1 -2 x (a) Find and hence show that the x-coordinate of P satisfies the equation x = 6 + 2 e . [4] dx … … … … … … … … … … … … … … … … … … … … … … … … … (b) Show by calculation that the x-coordinate of P lies between 0.35 and 0.45 . [2] … … … … … … … … … … … … … (c) Use an iterative formula based on the equation in part (a) to find the x-coordinate of P correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … … … …

9 marks

Mark scheme: 5(a) Differentiate using quotient rule *M1 OE Obtain 2 2 2 (1 3 ) 2e (1 e ) 3 (1 3 ) x x x x     A1 OE Equate first derivative to zero and arrange as far as ... x  DM1 Confirm 2 1 1 6 2 e x x    A1 Answer given – necessary detail needed. Must have exact terms. 4 5(b) Consider sign of 2 1 1 6 2 e x x    M1 OE Obtain 0.06...  (– 0.064959...) and 0.08... (0.0800…) or equivalents and justify conclusion A1 Answer given – necessary detail needed. Alternative Method 1 for Question 5(b) Consider  2 1 1 6 2 f e x x    and obtain   f 0.35 0.42 (0.4149…) and   f 0.45 0.37 (0.36995….) (M1) Conclude   f 0.35 0.45  and   f 0.45 0.35  so root lies in given interval. (A1) Alternative Method 2 for Question 5(b) Consider the sign of their d d y x from part (a) (M1) Obtain 0.19...  (– 0.187...) and 0.21... (0.2139…) or equivalents and justify conclusion (A1) 2 Question Answer Marks Guidance 5(c) Use iterative process correctly at least once M1 Obtain final answer 0.394 A1 Answer required to exactly 3sf. Show sufficient iterations to 5 sf to justify answer or show sign change in interval [0.3935, 0.3945] A1 3

This question in 9709/21 May/June 2024

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

9 marks

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

This question in 9709/22 May/June 2024

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

8 marks

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

This question in 9709/21 Oct/Nov 2024

4 (a) Sketch the graphs of y = 1 + e 2 x and y = x - 4 on the same diagram. [2] (b) The two graphs meet at the point P. Show that the x-coordinate of P satisfies the equation x = 1 ln ( 3 - x) . [2] 2 … … … … … … … … … … … … … … (c) Use an iterative formula, based on the equation in part (b), to find the x-coordinate of P correct to 3 significant figures. Use an initial value of 0.45 and give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(a) Show increasing curve above x-axis for y = 1 + e 2 x *B1 And appearing in first and second quadrants. Show V-shaped graph with vertex on positive x-axis and only one point of DB1 With modulus graph crossing y-axis above first graph. intersection with first curve 2 4(b) State or clearly imply 1 + e 2 x = 4 − x B1 Arrange to confirm x = 12 ln(3 − x ) B1 AG – necessary detail needed. Do not condone incorrect use of logs. 2 4(c) Use iterative process correctly at least once M1 Obtain final answer 0.465 A1 Answer required to exactly 3 sf. Show sufficient iterations to justify answer or show a sign change in the A1 interval [0.4645, 0.4655] 3

This question in 9709/22 Oct/Nov 2024

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

8 marks

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

This question in 9709/23 Oct/Nov 2024

5 (a) Sketch on the same diagram the graphs of y = 2 x - 3 and y = ln ( x + 1 ) . [2] The x-coordinates of the points where the graphs intersect are denoted by a and b, where a 1 b. (b) Show that a = 1.5 - 0 .5 ln ( a + 1 ) . [1] … … … (c) Use an iterative formula, based on the equation in part (b), to find the value of a correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] … … … … … … (d) Show by calculation that 2.055 1 b 1 2.065 . [2] … … … … … …

8 marks

Mark scheme: 5(a) Show an increasing curve through the origin for y = ln( x + 1) B1 Appearing in first and third quadrants. Show V-shaped graph with vertex on positive x-axis and showing B1 two intersections 2 5(b) Equate − (2 x − 3) and ln( x + 1) or equivalent and confirm result B1 AG (using x or ) Necessary detail needed. 1 5(c) Use iterative process correctly at least once M1 Obtain final answer 1.12 A1 Answer required to 3 significant figures only. Show sufficient iterations to justify answer or show a sign change A1 in the interval [1.115, 1.125] 3 5(d) Consider sign of 2 x −−3 ln( x + 1) or equivalent for 2.055 and M1 But not for − (2 x − 3) − ln( x + 1), nor for calculations based on 2.065 equation in part (b). Obtain − 0.006... and 0.009... or equivalents and justify A1 conclusion 2

This question in 9709/22 Feb/March 2025

3 (a) Sketch, on a single diagram, the graphs of y = 3e -2 x and y = sec x for values of x such that 0 G x 1 1 r . [2] 2 (b) Show that the x-coordinate of the point of intersection of the two graphs satisfies the equation x = 1 ln ( 3 cos x) . [2] 2 … … … … … … (c) Use an iterative formula, based on the equation in part (b), to find the x-coordinate of the point of intersection correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3] … … … … … … … …

7 marks

Mark scheme: 3(a) Sketch decreasing positive curve for y = 3e −2 x B1 Sketch curve for y = sec x B1 Correctly placed with reference to first sketch or correct with ‘1’ marked on y-axis. 2 3(b) −2 x x M1 Equate and arrange at least as far as e = ... or 2e = ... with cos x present Confirm x = 12 ln(3cos x ) A1 AG – necessary detail needed. 2 3(c) Use iterative process correctly at least once M1 Calculator must be in radian mode. Obtain final answer 0.487 A1 Required to precisely 3 decimal places. Show sufficient iterations to justify answer, or show a sign change in the interval A1 Allow iterations to greater accuracy. [0.4865, 0.4875] 3

This question in 9709/21 May/June 2025

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

8 marks

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

This question in 9709/22 May/June 2025

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

8 marks

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

This question in 9709/23 May/June 2025

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

8 marks

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

This question in 9709/25 May/June 2025

7 The polynomial p ( )x is defined by p ( )x = 2 x 4 + kx 3 + kx 2 + 17x + 18 , where k is a constant. It is given that ( x + 2 ) is a factor of p ( )x . (a) Find the value of k. [2] … … … … … … It is given that the equation p ( )x = 0 has exactly two real roots, denoted by a and b, where a is an integer and b is not an integer. (b) State the value of a and show that b satisfies the equation x = 3 - 2x - 4 .5 . [4] … … … … … … … … … … … … … … … … (c) Show by calculation that -1.4 1 b 1 -1 .0 . [2] … … … … … … … … … … (d) Use an iterative formula, based on the equation in part (b), to find the value of b correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(a) Substitute x = −2 , equate to zero and attempt solution M1 Obtain 32 − 8k + 4k − 34 + 18 = 0 or equivalent and hence k = 4 A1 2 7(b) State = −2 B1 Divide by x + 2 at least as far as 2 x 3 + mx M1 Or equivalent method, e.g. inspection. Obtain 2 x 3 + 4 x + 9 A1 Rearrange 2 x 3 + 4 x + 9 = 0 to confirm x = 3 −2 x − 4.5 A1 AG – necessary detail needed. 4 7(c) Consider sign of x − 3 −2 x − 4.5 or equivalent for −1.4 and −1.0 M1 Or Example 1: f ( x ) = 2 x3 + 4 x + 9, finding f ( − 1.4 ) and f ( − 1) . Example 2: f ( x ) = 3 −2 x − 4.5, finding f ( − 1.4 ) and f ( − 1) . Obtain −0.2... and 0.3… or equivalents and justify conclusion A1 Example 1: f ( −1) = 3, f ( −1.4 ) = −1.2... with a correct conclusion. Example 2: f ( −1) = − 1.357 … so −−1 1.357... f ( −1.4 ) = −1.193 … so −1.4 −1.19... with a correct conclusion. AG – necessary detail needed. 2 7(d) Use iterative process correctly at least once M1 Obtain final answer −1.26 A1 Answer required to exactly 3 significant figures. Show sufficient iterations to 5 sf to justify answer or show sign change in interval A1 [ −1.265, − 1.255] 3

This question in 9709/21 Oct/Nov 2025

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

9 marks

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

This question in 9709/22 Oct/Nov 2025

7 The polynomial p ( )x is defined by p ( )x = 2 x 4 + kx 3 + kx 2 + 17x + 18 , where k is a constant. It is given that ( x + 2 ) is a factor of p ( )x . (a) Find the value of k. [2] … … … … … … It is given that the equation p ( )x = 0 has exactly two real roots, denoted by a and b, where a is an integer and b is not an integer. (b) State the value of a and show that b satisfies the equation x = 3 - 2x - 4 .5 . [4] … … … … … … … … … … … … … … … … (c) Show by calculation that -1.4 1 b 1 -1 .0 . [2] … … … … … … … … … … (d) Use an iterative formula, based on the equation in part (b), to find the value of b correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(a) Substitute x = −2 , equate to zero and attempt solution M1 Obtain 32 − 8k + 4k − 34 + 18 = 0 or equivalent and hence k = 4 A1 2 7(b) State = −2 B1 Divide by x + 2 at least as far as 2 x 3 + mx M1 Or equivalent method, e.g. inspection. Obtain 2 x 3 + 4 x + 9 A1 Rearrange 2 x 3 + 4 x + 9 = 0 to confirm x = 3 −2 x − 4.5 A1 AG – necessary detail needed. 4 7(c) Consider sign of x − 3 −2 x − 4.5 or equivalent for −1.4 and −1.0 M1 Or Example 1: f ( x ) = 2 x3 + 4 x + 9, finding f ( − 1.4 ) and f ( − 1) . Example 2: f ( x ) = 3 −2 x − 4.5, finding f ( − 1.4 ) and f ( − 1) . Obtain −0.2... and 0.3… or equivalents and justify conclusion A1 Example 1: f ( −1) = 3, f ( −1.4 ) = −1.2... with a correct conclusion. Example 2: f ( −1) = − 1.357 … so −−1 1.357... f ( −1.4 ) = −1.193 … so −1.4 −1.19... with a correct conclusion. AG – necessary detail needed. 2 7(d) Use iterative process correctly at least once M1 Obtain final answer −1.26 A1 Answer required to exactly 3 significant figures. Show sufficient iterations to 5 sf to justify answer or show sign change in interval A1 [ −1.265, − 1.255] 3

This question in 9709/23 Oct/Nov 2025

7 The polynomial p ( )x is defined by p ( )x = 2 x 4 + kx 3 + kx 2 + 17x + 18 , where k is a constant. It is given that ( x + 2 ) is a factor of p ( )x . (a) Find the value of k. [2] … … … … … … It is given that the equation p ( )x = 0 has exactly two real roots, denoted by a and b, where a is an integer and b is not an integer. (b) State the value of a and show that b satisfies the equation x = 3 - 2x - 4.5 . [4] … … … … … … … … … … … … … … … … (c) Show by calculation that -1. 4 1 b 1 -1.0 . [2] … … … … … … … … … … (d) Use an iterative formula, based on the equation in part (b), to find the value of b correct to 3 significant figures. Give the result of each iteration to 5 significant figures. [3] … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(a) Substitute x = −2 , equate to zero and attempt solution M1 Obtain 32 − 8k + 4k − 34 + 18 = 0 or equivalent and hence k = 4 A1 2 7(b) State = −2 B1 Divide by x + 2 at least as far as 2 x 3 + mx M1 Or equivalent method, e.g. inspection. Obtain 2 x 3 + 4 x + 9 A1 Rearrange 2 x 3 + 4 x + 9 = 0 to confirm x = 3 −2 x − 4.5 A1 AG – necessary detail needed. 4 7(c) Consider sign of x − 3 −2 x − 4.5 or equivalent for −1.4 and −1.0 M1 Or Example 1: f ( x ) = 2 x3 + 4 x + 9, finding f ( − 1.4 ) and f ( − 1) . Example 2: f ( x ) = 3 −2 x − 4.5, finding f ( − 1.4 ) and f ( − 1) . Obtain −0.2... and 0.3… or equivalents and justify conclusion A1 Example 1: f ( −1) = 3, f ( −1.4 ) = −1.2... with a correct conclusion. Example 2: f ( −1) = − 1.357 … so −−1 1.357... f ( −1.4 ) = −1.193 … so −1.4 −1.19... with a correct conclusion. AG – necessary detail needed. 2 7(d) Use iterative process correctly at least once M1 Obtain final answer −1.26 A1 Answer required to exactly 3 significant figures. Show sufficient iterations to 5 sf to justify answer or show sign change in interval A1 [ −1.265, − 1.255] 3

This question in 9709/25 Oct/Nov 2025