Functions — Paper 2 · A Level Mathematics 9709

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

10 marks

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

This question in 9709/21 May/June 2005

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

9 marks

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

This question in 9709/21 May/June 2007

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

3 The sequence x1, x2, x3, . . . defined by x2n 6 x1 1, 12 3p = xn+1 = + converges to the value α. (i) Find the value of α correct to 3 decimal places. Show your working, giving each calculated value of the sequence to 5 decimal places. [3] (ii) Find, in the form ax3 bx2 c 0, an equation of which α is a root. [2] + + = 2

5 marks

Mark scheme: 3 (i) Use the iteration process correctly at least once M1 Obtain at least two correct iterates to 5 decimal places A1 Conclude α = 0.952 A1 [3] [1 → 0.95647 → 0.95257 → 0.95223 → 0.95220] 1 3 2 (ii) State or imply equation is x = x + 6 B1 2 Obtain 8x3 – x2 – 6 = 0 B1 [2] 1

This question in 9709/22 May/June 2011

3 The sequence x1, x2, x3, . . . defined by x2n 6 x1 1, 12 3p = xn+1 = + converges to the value α. (i) Find the value of α correct to 3 decimal places. Show your working, giving each calculated value of the sequence to 5 decimal places. [3] (ii) Find, in the form ax3 bx2 c 0, an equation of which α is a root. [2] + + = 2

5 marks

Mark scheme: 3 (i) Use the iteration process correctly at least once M1 Obtain at least two correct iterates to 5 decimal places A1 Conclude α = 0.952 A1 [3] [1 → 0.95647 → 0.95257 → 0.95223 → 0.95220] 1 3 2 (ii) State or imply equation is x = x + 6 B1 2 Obtain 8x3 – x2 – 6 = 0 B1 [2] 1

This question in 9709/23 May/June 2011

2 It is given that x satisfies the equation x 1 4. Find the possible values of + = x 4 x . [4] + − −4 … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 Solve 3-term quadratic equation or a pair of M1 For M1, must square both sides linear equations when attempting a quadratic equation Obtain x = − 5 and x = 3 A1 Substitute (at least) one value of x (less than 4) M1 into x + 4 − x − 4 , showing correct evaluation of modulus and producing only one answer in each case Obtain –8 and 6 and no others A1 4

This question in 9709/22 Oct/Nov 2017

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

9 marks

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

This question in 9709/21 Oct/Nov 2018

2 Given that x satisfies the equation 2x 3 2x , find the value of + = −1 4x 6x . [4] −3 − … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 Solve non-modular equation 2 2 (2 3) (2 1) x x + = − or linear equation with signs of 2x different M1 Obtain 1 2 x = − A1 Substitute negative value into expression and show correct evaluation of modulus at least once M1 Obtain 5 3 2 − = with no errors seen A1 4

This question in 9709/22 Feb/March 2019

3 2 ln x3 A curve has equation y + . Find the exact gradient of the curve at the point for which y 4. = 1 ln x = + [5] … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 Use quotient rule (or product rule) to find first derivative *M1 Must have correct u and v Obtain 2 1 (1 ln ) x x − + or (unsimplified) equivalent A1 Use 4 y = to obtain 1 2 ln x = − or exact equivalent for x B1 Substitute for x in their first derivative DM1 Obtain 1 2 4e − or exact equivalent A1 Must be simplified to contain a single exponential term 5

This question in 9709/21 Oct/Nov 2019

3 2 ln x3 A curve has equation y + . Find the exact gradient of the curve at the point for which y 4. = 1 ln x = + [5] … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 Use quotient rule (or product rule) to find first derivative *M1 Must have correct u and v Obtain 2 1 (1 ln ) x x − + or (unsimplified) equivalent A1 Use 4 y = to obtain 1 2 ln x = − or exact equivalent for x B1 Substitute for x in their first derivative DM1 Obtain 1 2 4e − or exact equivalent A1 Must be simplified to contain a single exponential term 5

This question in 9709/23 Oct/Nov 2019

5 (a) Sketch, on the same diagram, the graphs of y x 2k and y 2x , where k is a positive constant. = + = −3k Give, in terms of k, the coordinates of the points where each graph meets the axes. [3] (b) Find, in terms of k, the coordinates of each of the two points where the graphs intersect. [4] … … … … … … … … … … … … … … … … … … … … (c) Find, in terms of k, the largest value of t satisfying the inequality 2t . [2] 2k + ≥ 2t+1 −3k … … … … … … … … … … … … …

9 marks

Mark scheme: 5(a) Draw two V-shaped graphs with one vertex on negative x-axis and one vertex on positive x-axis M1 Draw correct graphs related correctly to each other A1 State correct coordinates 3 2 2 , 2 , , 3 k k k k − A1 Either given on axes or stated separately 3 5(b) State or imply non-modulus equation 2 2 ( 2 ) (2 3 ) x k x k + = − or pair of linear equations B1 Attempt solution of 3-term quadratic equation or pair of linear equations M1 Obtain 1 3 , 5 x k x k = = A1 Obtain 7 3 , 7 y k y k = = A1 If A0A0, award A1 for one pair of correct coordinates 4 5(c) Relate 2t to larger value of x from part (b) M1 Apply logarithms to obtain ln(5 ) ln2 k t = A1 OE such as 10 10 log (5 ) log 2 k or 2 log (5 ) k 2

This question in 9709/22 Feb/March 2020

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

10 marks

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

This question in 9709/22 Feb/March 2020

5 (a) Sketch, on the same diagram, the graphs of y 2x and y 3x 5. [2] = −3 = + (b) Solve the inequality 3x 5 2x . [3] + < −3 … … … … … … … … … … … …

5 marks

Mark scheme: 5(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw (more or less) correct graph of 3 5 = + y x B1 2 5(b) State equation 3 5 (2 3) + = − − x x or corresponding inequality B1 Attempt solution of linear equation / inequality where signs of 3x and 2x are different M1 State answer 2 5 < − x A1 Alternative method for question 5(b) Square both sides of equation / inequality and attempt solution of 3-term quadratic equation / inequality M1 Obtain (eventually) only 2 5 − A1 State answer 2 5 < − x A1 3

This question in 9709/22 May/June 2020

5 (a) Sketch, on the same diagram, the graphs of y 2x and y 3x 5. [2] = −3 = + (b) Solve the inequality 3x 5 2x . [3] + < −3 … … … … … … … … … … … …

5 marks

Mark scheme: 5(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw (more or less) correct graph of 3 5 = + y x B1 2 5(b) State equation 3 5 (2 3) + = − − x x or corresponding inequality B1 Attempt solution of linear equation / inequality where signs of 3x and 2x are different M1 State answer 2 5 < − x A1 Alternative method for question 5(b) Square both sides of equation / inequality and attempt solution of 3-term quadratic equation / inequality M1 Obtain (eventually) only 2 5 − A1 State answer 2 5 < − x A1 3

This question in 9709/23 May/June 2020

. . 3 (a) Sketch, on a single diagram, the graphs of y 12x and y 32x 2a, where a is a positive constant. = −a = −1 [2] (b) Find the coordinates of the point of intersection of the two graphs. [3] … … … … … … … … … . . (c) Deduce the solution of the inequality 12x 32x 2a. [1] −a > −1 … … … … …

6 marks

Mark scheme: 3(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw straight line graph correctly positioned with greater gradient B1 2 3(b) Solve linear equation with signs of 1 2 x and 3 2 x different or solve non-modulus equation 2 2 1 3 1 2 2 2 x a x a     − = −         to obtain x = M1 Obtain 3 4 x a = A1 Obtain 5 8 y a = A1 And no other point 3 Question Answer Marks Guidance 3(c) State 3 4 x a < B1 FT Following their (single) x-coordinate from part (b) 1

This question in 9709/22 Oct/Nov 2020

1 (a) Sketch, on the same diagram, the graphs of y 3x and y x 2. [2] = −5 = + (b) Solve the equation 3x x 2. [3] −5 = + … … … … … … … … … … … … … …

5 marks

Mark scheme: 1(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw correct graph of 2 y x = + with smaller positive gradient B1 Crossing y-axis between 0 and y-intercept of first graph. 2 1(b) Solve 3 5 2 x x − = + to obtain 7 2 x = B1 Attempt solution of linear equation where signs of 3x and x are different. M1 Obtain 3 4 x = A1 Alternative method for question 1(b) State or imply non-modulus equation 2 2 (3 5) ( 2) x x − = + B1 Attempt solution of 3-term quadratic equation M1 Obtain 3 4 and 7 2 A1 3

This question in 9709/22 Feb/March 2021

2 The solutions of the equation 5 x 5 are x a and x b, where a b. = −2x = = < Find the value of 3a 7b . [5] −1 + −1 … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 Solve 5 5 2 = − x x to obtain 5 7 = x Attempt solution of linear equation where signs of 5x and 2x are the same M1 Obtain 5 3 = − x A1 Allow AWRT –1.67 Substitute their values correctly M1 Substitution must be seen unless implied by a correct answer. Their values must come from consideration of 5 5 2 = − x x Obtain 6 4 − + and hence 10 A1 Alternative method for Question 2 State or imply non-modulus equation 2 2 25 (5 2 ) = − x x B1 Attempt solution of 3-term quadratic equation M1 Obtain 5 3 − and 5 7 A1 Allow AWRT 0.714 and AWRT -1.67 Substitute their values correctly M1 Substitution must be seen unless implied by a correct answer. Their values must come from consideration of 5 5 2 = − x x Obtain 6 4 − + and hence 10 A1 5

This question in 9709/22 May/June 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

7 y x O P Q The diagram shows the curve with parametric equations x = 3 cos 21, y = 4 sin 1, for π ≤1 ≤32π. Points P and Q lie on the curve. The gradient of the curve at P is 2. The straight line 3x + y = 0 meets the curve at Q. (a) Find the value of 1 at P, giving your answer correct to 3 significant figures. [5] … … … … … … … … … … … … … … (b) Find the gradient of the curve at Q, giving your answer correct to 3 significant figures. [6] … … … … … … … … … … … … … … … … … … … … … … … … …

11 marks

Mark scheme: 7(a) dx dy B1 State −6sin2 and 4cos d= d= dy dy dx M1 Use = / and equate to 2 dx d d Use sin2= 2sincos and attempt value of sin M1 Obtain sin= − 16 A1 7(a) Obtain = 3.31 only A1 AWRT; and no second answer. Alternative method for Question 7(a) 2 2 2 2 M1 Using x = 3 2cos − 1 , x = 3 1 − 2sin  or x = 3 cos − sin  to ( ) ( ) ( ) dx obtain a sincos d=  dy  4cos A1 Obtain  =  = 2  dx  −12sincos Attempt value of sin M1 Obtain sin= − 16 A1 Obtain = 3.31 only in the given range A1 5 7(b) State or imply 9cos2+ 4sin= 0 and use identity to obtain quadratic in M1 sin Obtain 18sin 2 − 4sin− 9 = 0 A1 OE Attempt solution to find negative value of sin DM1 Obtain sin= −0.604... A1 2 − 166 Or , = 3.79... 18 Substitute value of sin (or their  between  and 32 ) in expression for M1 first derivative Obtain 0.551 A1 AWRT 7(b) Alternative method for question 7(b) y 2 x M1 Must be a complete method, allow unsimplified. Cartesian equation of curve 1 − = oe 8 3 Intersection of line and curve 27 x 2 + 8 x − 24 = 0 oe M1 x = 0.8062... A1 = 3.791... A1 Substitute value of  (or their  between  and 32 ) in expression for first M1 derivative Obtain 0.551 A1 AWRT 6

This question in 9709/22 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

4 (a) Sketch, on the same diagram, the graphs of y 2x and y 3x [2] = −11 = −3. (b) Solve the inequality 2x 3x [3] −11 < −3. … … … … … … … … … … … … … … … … … (c) Find the smallest integer N satisfying the inequality 2 ln N 3 ln N [2] −11 < −3. … … … … … … … … … … … … … …

7 marks

Mark scheme: 4(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw approximately correct graph of y = 3 x − 3 with greater B1 Crossing x-axis between origin and vertex of first graph. gradient 2 4(b) Attempt solution of linear equation where signs of 2x and 3x are M1 different Solve −2 x + 11 = 3x − 3 to obtain x = 145 A1 OE Conclude x  145 A1 OE Alternative method for Question 4(b) Attempt solution of 3-term equation (2 x − 11) 2 = (3 x − 3) 2 to M1 Or equivalent inequality. obtain at least one value of x Obtain at least x = 145 A1 OE Conclude x  145 A1 OE 3 4(c) Attempt value of N (maybe non-integer at this stage) using M1 logarithms and their answer to part (b). Conclude with single integer 17 A1 2

This question in 9709/22 Feb/March 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) Sketch, on the same diagram, the graphs of y 3x and y 2x 7. [2] = −5 = + (b) Solve the equation 3x 2x 7. [3] −5 = + … … … … … … (c) Hence solve the equation 2 3y 7, giving your answer correct to 3 significant figures. 3y+1 −5 = × + [2] … … … … … …

7 marks

Mark scheme: 4(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw (more or less) correct graph of y = 2 x + 7 with smaller gradient B1 And crossing y-axis above y-intercept of modulus graph. 2 4(b) Solve 3x −=5 2 x + 7 to obtain x = 12 B1 Attempt solution of linear equation where signs of 3x and 2x are M1 3x −=5 −2 x − 7 OE. different 2 A1 Obtain x = − 5 Alternative solution for question 4(b) State or imply non-modulus equation (3 x − 5) 2 = (2 x + 7) 2 B1 Must be working with (3 x − 5) 2 = (2 x + 7) 2 Attempt solution of 3-term quadratic equation M1 2 A1 Obtain − and 12 5 3 4(c) Apply logarithms and use power law for 3y = k where k  0 or M1 Using their positive answer from part (b) correct equivalent or greater accuracy; and no other values. Obtain 2.26 A1 2

This question in 9709/21 Oct/Nov 2023

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

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