Functions — Paper 3 · A Level Mathematics 9709

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

8 marks

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

This question in 9709/31 Oct/Nov 2004

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

8 marks

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

This question in 9709/31 Oct/Nov 2007

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

12 marks

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

This question in 9709/31 Oct/Nov 2008

9 y M A x O 4 ln x The diagram shows the curve y and its maximum point M. The curve cuts the x-axis at the = √x point A. (i) State the coordinates of A. [1] (ii) Find the exact value of the x-coordinate of M. [4] (iii) Using integration by parts, show that the area of the shaded region bounded by the curve, the x-axis and the line x 4 is equal to 8 ln 2 [5] = −4.

10 marks

Mark scheme: 9 (i) State coordinates (1, 0) B1 [1] (ii) Use correct quotient or product rule M1 Obtain derivative in any correct form A1 Equate derivative to zero and solve for x M1 Obtain x = e2 correctly A1 [4] GCE A/AS LEVEL – October/November 2009 9709 31 1 (iii) Attempt integration by parts reaching a x ln x ± a ∫ x x dx M1* 1 Obtain 2 x ln x − 2 ∫ dx A1 x Integrate and obtain 2 x ln x − 4 x A1 Use limits x = 1 and x = 4 correctly, having integrated twice M1(dep*) Justify the given answer A1 [5] dA

This question in 9709/31 Oct/Nov 2009

2 8 (i) Express (x + 1)(x + 3) in partial fractions. [2] (ii) Using your answer to part (i), show that 2 2 1 1 1 1 x 1 x 3 ≡ − + + [2] (x + 1)(x + 3) (x + 1)2 + + (x + 3)2. 1 4 7 3 (iii) Hence show that dx 12 2. [5] ä 0 = −ln (x + 1)2(x + 3)2

9 marks

Mark scheme: A B 8 (i) State or imply the form + and use a relevant method to find A or B M1 x + 1 x + 3 Obtain A = 1, B = −1 A1 [2] (ii) Square the result of part (i) and substitute the fractions of part (i) M1 Obtain the given answer correctly A1 [2] 1 1 (iii) Integrate and obtain − − ln ( x + 1) + ln( x + 3) − B3 x + 1 x + 3 Substitute limits correctly in an integral containing at least two terms of the correct form M1 Obtain given answer following full and exact working A1 [5] GCE AS/A LEVEL – May/June 2010 9709 31

This question in 9709/31 May/June 2010

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

7 marks

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

This question in 9709/31 May/June 2016

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

10 marks

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

This question in 9709/33 Oct/Nov 2016

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

7 marks

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

This question in 9709/32 Feb/March 2017

4 The variables x and y satisfy the differential equation dy 1 y sin x. −cosx dx = It is given that y 4 when x = = π. (a) Solve the differential equation, obtaining an expression for y in terms of x. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … (b) Sketch the graph of y against x for 0 x [1] < < 2π.

7 marks

Mark scheme: 4(a) Separate variables correctly and attempt integration of at least one side M1 Obtain term ln y A1 Obtain term of the form ln(1 cos ) ± − x M1 Obtain term ( ) ln 1 cos − x A1 Use π = x , y = 4 to evaluate a constant, or as limits, in a solution containing terms of the form ln a y and ln(1 cos ) − b x M1 Obtain final answer 2(1 cos ) = − y x A1 OE 6 Question Answer Marks Guidance 4(b) Show a correct graph for 0 2π x < < with the maximum at x = π B1 FT The FT is for graphs of the form (1 cos ) = − y a x , where a is positive. 1

This question in 9709/32 Feb/March 2021

1 (a) Sketch the graph of y 4x . [1] = −2 (b) Solve the inequality 1 3x 4x . [4] + < −2 … … … … … … … … … … … … …

5 marks

Mark scheme: Question Answer Marks Guidance 1(a) B1 Show a recognizable sketch graph of y = 4 x − 2 . y Roughly symmetrical. Should extend into the second quadrant. Ignore y = 4 x − 2 below the axis if intention is clear e.g. dashed or the required lines are clearly bolder. Some indication of scale on both axes – accept dashes. 2 Must go beyond (0, 2) and (1, 2). Ignore any attempt to sketch y = 1 + 3 x . x 1 2 1 1(b) Obtain critical value x = 3 B1 Allow incorrect inequality. Allow if later rejected. Allow 217 . Solve the linear equation 1 + 3x = 2 − 4 x M1 Or corresponding linear inequality. Obtain critical value 17 A1 Allow 0.143 or better. Allow incorrect inequality. Allow if later rejected. Obtain final answer x  17 [or] x  3 A1 Or equivalent. Allow with a comma, or nothing between. Strict inequalities only. Exact values. A0 for 17  x  3 A0 for x  17 and x .3 Alternative method for question 1(b) Solve the quadratic inequality ( 4 x − 2 ) 2  (1 + 3 x ) 2 , or corresponding M1 e.g. 7 x 2 − 22 x + 3 = 0 . Available if they start with the correct equation / quadratic equation inequality, have a correct method for squaring 2 2 2 (i.e. not ( a + b ) = a + b ) and a correct method for solving. Need to obtain at least one critical value. Obtain critical value x = 3 A1 Allow incorrect inequality. Allow if later rejected. Allow 217 . Obtain critical value 17 A1 Allow 0.143 or better. Allow incorrect inequality. Allow if later rejected. Obtain final answer x  17 [or] x  3 A1 Or equivalent. Strict inequalities only. Allow with a comma, or nothing between. Exact values. A0 for 17  x  3 A0 for x  17 and x .3 4

This question in 9709/32 Oct/Nov 2023

1 (a) Sketch the graph of y = x - 2a , where a is a positive constant. [1] (b) Solve the inequality 2x - 3a 1 x - 2a . [2] … … … … … … … … … … … … … …

3 marks

Mark scheme: 1(a) B1 Correct shape, roughly symmetrical. Both sections should be solid straight lines. If not drawn with a ruler the intention must be clear. Allow construction lines if dashed or clearly fainter. 2a marked on each axis (must be 2a, not just 2). Needs to extend into negative x. If a is given a value, then B0. Ignore y = 2x – 3a if seen. 1 1(b) Solve linear equation or inequality to obtain critical value 5 3 x a  or exact equivalent. B1 Ignore x a  if seen. Obtain 5 3 x a  or exact equivalent B1 Accept 10 6 x a  or   5 3, . a  Must be strict inequality. Need a clear final solution: x a  or x a  must be rejected if seen as part of the working. Rejection can be implied, e.g. if only the correct inequality is underlined. B0 B0 if a is given a value. Alternative Method for Question 1(b) Solve quadratic equation     2 2 2 3 2 x a x a    to obtain critical value 5 3 x a  or exact equivalent (B1)   2 2 3 8 5 0 x ax a    Ignore x a  if seen. Obtain 5 3 x a  or exact equivalent (B1) Accept 10 6 x a  or   5 3, . a  Must be strict inequality. Need a clear final solution: x a  or x a  must be rejected if seen as part of the working. Rejection can be implied, e.g. if only the correct inequality is underlined. B0 B0 if a is given a value. 2 2a 2a y x O

This question in 9709/32 May/June 2024

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

7 marks

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

This question in 9709/31 Oct/Nov 2024

6 (a) By sketching a suitable pair of graphs, show that the equation x - 2 = 2 sin 1 x 2 has only one root in the interval 0 1 x 1 r. [2] (b) Show by calculation that this root lies between 1 and 1.5. [2] … … … … … … … … … … … … … (c) Use the iterative formula x = 2 - 2 sin 1 x with an initial value of 1.03 to calculate the root n + 1 2 n correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3] … … … … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 6(a) Sketch a relevant graph for 0  x  π B1 2 For y = |x – 2| graph should be symmetrical and have correct intercepts on the axes π For y = 2 sin 12 x , graph should pass through the origin, have correct curvature Ignore anything outside 0  x  π. Ignore what happens in y < 0. and max y = 2 when x = π Sketch second relevant graph and confirm root. B1 Needs to mark intersection with a dot, a cross, or say roots at points of intersection, OE. The vertex of the modulus graph in roughly correct position relative to π and/or If the intersection is highlighted in some way, then 1 2 π they do not need to make a comment. If no mark on the graph, check to see if they have written something below the graph. SC A sketch y = x − 2 and y = 2sin 12 x (above and below the x-axis) scores B1. A clear indication of the root scores second B1. 2 6(b) Calculate the values of a relevant expression or pair of expressions at x = 1 and M1 Or comparing x – 2 and 2sin 12 x. x = 1.5 1 Using 1: x – 2 = 1, 2sin 2 x = 0.958 1  1 > 0.958… e.g. f(x) = |x – 2| – 2sin 2 x Using 1.5: x – 2 = 0.5 2sin 12 x = 1.36  f(1) = 0.0411… > 0 f(1.5) = –0.863… < 0  0.5 < 1.36 2 2 1 Need all values but condone one error. e.g. f ( x ) = ( x − 2 ) − 4sin ( 2 x )  f (1) = 0.0806..., f (1.5 ) = −1.60... If the solution involves 4 values, the pairing must be clear. Embedded values are not sufficient, e.g. f1(1) = … and f2(1) =… etc. M0 if working in degrees (gives 0.98… and 0.47… if using f(x) = 0). Allow if working on a smaller interval. Complete the argument correctly with correct calculated values A1 Values correct. They must have a conclusion in words or symbols, but they do not need to say that the function is continuous. A correct statement with correct inequalities is sufficient. 2 6(c) Use the iterative process correctly at least once starting at 1.03 (get as far as M1 M0 if working in degrees. 1.0281) Obtain final answer 1.02 A1 No working seen at all scores 0/3. Show sufficient iterations to 4 d.p. to justify 1.02 to 2 d.p. A1 1.03, 1.0149, 1.0281, 1.0166, 1.0266, 1.0179, 1.0255, 1.0189, 1.0246 or show there is a sign change in the interval (1.015, 1.025) (or a smaller interval Incorrect starting point is M0. containing the root) Once the convergence is established ISW. 3

This question in 9709/32 May/June 2025

1 (a) Sketch the graph of y = 3 x - 2 a , where a is a positive constant. [1] (b) Hence or otherwise solve the inequality 3x - 2a 1 x + 5a . [3] … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 1(a) y B1 Symmetrical. In correct position. Lines intended to be straight. Must be in both first and second quadrants. 2a Key coordinates must be correct. Ignore y = x + 5 a if seen. x O 2a 3 1 1(b) 7 a B1 Allow if seen in an inequality. Obtain critical value from x + 5a = 3x − 2a 2 3 a B1 Allow if seen in an inequality. Obtain critical value − from x + 5a = 2a − 3x 4 3a 7 a B1 3a 7 a State final answer −  x  SC B1 only for −  x  with their a from 4 2 4 2 part (a). Allow any equivalent notation. 3a 7 a Allow −  x and x  . 4 2 Alternative Method for Question 1(b) Solve quadratic equation ( 3 x − 2a ) 2 = ( x + 5a ) 2 M1 8 x 2 − 22ax − 21a 2 = 0 3a 7 a A1 Obtain critical values − and 4 2 3a 7 a A1 3a 7 a State final answer −  x  SC B1 only for −  x  with their a from 4 2 4 2 part (a). Allow any equivalent notation. 3a 7 a Allow −  x and x  . 4 2 3

This question in 9709/33 May/June 2025

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

10 marks

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

This question in 9709/33 Oct/Nov 2025