Logarithmic and exponential functions — Paper 2 · A Level Mathematics 9709

2 The variables x and y satisfy the relation 3y = 4x+2. (i) By taking logarithms, show that the graph of y against x is a straight line. Find the exact value of the gradient of this line. [3] (ii) Calculate the x-coordinate of the point of intersection of this line with the line y = 2x, giving your answer correct to 2 decimal places. [3]

6 marks

Mark scheme: 2 (i) State or imply y ln 3 = ( x + 2 ) ln 4 B1 State that this is of the form ay = bx + c and thus a straight line, or equivalent B1 ln 4 State gradient is , or equivalent (allow 1.26) ln 3 B1 [3] (ii) Substitute y = 2x and obtain a linear equation in x M1* Solve for x M1(dep*) Obtain answer 3.42 A1 [3]

This question in 9709/21 May/June 2007

3 (i) Solve the inequality |y −5| < 1. [2] (ii) Hence solve the inequality |3x −5| < 1, giving 3 significant figures in your answer. [3]

5 marks

Mark scheme: 3 (i) Obtain critical values 4 and 6 B1 State answer 4 < y < 6 B1 [2] (ii) Use correct method for solving an equation of the form 3x = a, where a > 0 M1 Obtain one critical value, i.e. either 1.26 or 1.63 A1 State answer 1.26 < x <1.63 A1 [3] 2

This question in 9709/21 Oct/Nov 2007

2 Use logarithms to solve the equation 4x = 2(3x), giving your answer correct to 3 significant figures. [4] 16π

4 marks

Mark scheme: 2 Use law for the logarithm of a product, a quotient or a power M1* Obtain xln 4 = ln 2 + xln 3, or equivalent A1 Solve for x M1 (dep*) Obtain answer x = 2.41 A1 [4] 1

This question in 9709/21 May/June 2008

3 ln y (0, 1.3) (1.6, 0.9) x O The variables x and y satisfy the equation y = A(b−x), where A and b are constants. The graph of ln y against x is a straight line passing through the points (0, 1.3) and (1.6, 0.9), as shown in the diagram. Find the values of A and b, correct to 2 decimal places. [5]

5 marks

Mark scheme: 3 State or imply ln y = ln A – xln b B1 State ln A = 1.3 B1 Obtain A = 3.67 B1 Form a numerical expression for the gradient of the line M1 Obtain b = 1.28 A1 [5]

This question in 9709/21 Oct/Nov 2008

x 1 Given that use logarithms to find the value of correct to 3 significant figures. [3] y (1.25)x = (2.5)y,

3 marks

Mark scheme: 1 Use logarithms to linearise an equation M1 x ln 5.2 Obtain = , or equivalent A1 y ln .125 Obtain answer 4.11 A1√ [3] 2 2

This question in 9709/21 May/June 2009

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

2 Solve the equation 2 ln x, giving your answer correct to 3 significant figures. [4] ln(3 −x2) =

4 marks

Mark scheme: 2 Use lnx2 = 2lnx B1 Obtain 3 – x2 = x2, or equivalent B1 Solve for x M1 Obtain answer x – 1.22, having rejected x = –1.22 A1 [4] 1

This question in 9709/21 Oct/Nov 2009

2 It is given that y 2 ln x. Express y in terms of x, in a form not involving logarithms. ln(y + 5) −ln = [4]

4 marks

Mark scheme: 2 State or imply 2 ln x = ln(x2) B1 Use law for the logarithm of a quotient or a product M1 Remove logarithms and obtain yx2 = y + 5, or equivalent A1 5 Obtain answer y = A1 [4] x 2 − 1

This question in 9709/22 Oct/Nov 2009

5 (i) Given that y 2x, show that the equation = 2x 4 + 3(2−x) = can be written in the form y2 3 0. −4y + = [3] (ii) Hence solve the equation 2x 4, + 3(2−x) = giving the values of x correct to 3 significant figures where appropriate. [3]

6 marks

Mark scheme: 1 5 (i) State or imply 2–x = , or 2–x = y–1 B1 y Substitute and obtain a 3-term quadratic in y M1 Obtain the given answer correctly A1 [3] (ii) Solve the given quadratic and carry out correct method for solving an equation of the form 2x = a, where a > 0 M1 Obtain answer x = 1.58 or 1.585 A1 Obtain answer x = 0 B1 [3] GCE AS/A LEVEL – May/June 2010 9709 21 2 dy 2

This question in 9709/21 May/June 2010

1 Given that 13x use logarithms to show that y kx and find the value of k correct to 3 significant = (2.8)y, = figures. [3]

3 marks

Mark scheme: 1 State or imply y log 2.8 = x log 13 B1 log 13 Rearrange into form y = x or equivalent B1 log 8.2 Obtain answer k = 2.49 B1 [3]

This question in 9709/22 May/June 2010

1 Given that 13x use logarithms to show that y kx and find the value of k correct to 3 significant = (2.8)y, = figures. [3]

3 marks

Mark scheme: 1 State or imply y log 2.8 = x log 13 B1 log 13 Rearrange into form y = x or equivalent B1 log 8.2 Obtain answer k = 2.49 B1 [3]

This question in 9709/23 May/June 2010

giving your answer correct to 3 significant figures.2 Use logarithms to solve the equation 5x = 22x+1, [4]

4 marks

Mark scheme: 2 Use law for the logarithm of a product, a quotient or a power M1* Obtain x log 5 = (2 x + 1) log 2 , or equivalent A1 Solve for x, via correct manipulative technique(s) M1(dep*) Obtain answer x = .311 . Allow x ∈ [.3 10, .3 11] A1 [4] 1 2

This question in 9709/21 Oct/Nov 2010

giving your answer correct to 3 significant figures.2 Use logarithms to solve the equation 5x = 22x+1, [4]

4 marks

Mark scheme: 2 Use law for the logarithm of a product, a quotient or a power M1* Obtain x log 5 = (2 x + 1) log 2 , or equivalent A1 Solve for x, via correct manipulative technique(s) M1(dep*) Obtain answer x = .311 . Allow x ∈ [.3 10, .3 11] A1 [4] 1 2

This question in 9709/22 Oct/Nov 2010

5 ln y (2.2, 1.2) (1.4, 0.8) x O The variables x and y satisfy the equation y where A and b are constants. The graph of ln y against x is a straight line passing through= theA(bx),points and as shown in the diagram. Find the values of A and b, correct to 2 decimal places.(1.4, 0.8) (2.2, 1.2), [6]

6 marks

Mark scheme: 5 State or imply ln y = ln A + x ln b B1 Form a numerical expression for the gradient of the line M1 Obtain b = 1.65 A1 Use gradient and one point correctly to find In A M1 Obtain ln A = 0.1 A1 Obtain A = 1.11 A1 [6] GCE A/AS LEVEL – October/November 2010 9709 23

This question in 9709/23 Oct/Nov 2010

3 ln y (6, 10.2) (0, 2.0) ln x O The variables x and y satisfy the equation y Kxm, where K and m are constants. The graph of = ln y against ln x is a straight line passing through the points and as shown in the (0, 2.0) (6, 10.2), diagram. Find the values of K and m, correct to 2 decimal places. [5]

5 marks

Mark scheme: 3 State or imply that ln y = ln K + m ln x B1 Equate intercept on axis for ln y to ln K M1 Obtain 7.39 for K A1 Attempt calculation of gradient of line M1 Obtain 1.37 for m A1 [5]

This question in 9709/21 May/June 2011

1 Use logarithms to solve the equation 3x giving your answer correct to 3 significant figures. = 2x+2, [4]

4 marks

Mark scheme: 1 Attempt use of power law for logarithms M1* Obtain xlog3 = xlog2 + 2log2 or equivalent A1 Attempt solution for x of linear equation M1 dep* Obtain 3.42 A1 [4]

This question in 9709/22 May/June 2011

1 Use logarithms to solve the equation 3x giving your answer correct to 3 significant figures. = 2x+2, [4]

4 marks

Mark scheme: 1 Attempt use of power law for logarithms M1* Obtain xlog3 = xlog2 + 2log2 or equivalent A1 Attempt solution for x of linear equation M1 dep* Obtain 3.42 A1 [4]

This question in 9709/23 May/June 2011

4 Solve the equation 32x 10 0, giving your answers correct to 3 significant figures. [5] −7(3x) + =

5 marks

Mark scheme: 4 Carry out recognizable solution method for quadratic in 3x M1 Obtain 3x = 5 and 3x = 2 A1 Use logarithmic method to solve an equation of the form 3x = k, where k > 0 M1 State answer 1.46 A1 State answer 0.631 A1 [5] 1

This question in 9709/21 Oct/Nov 2011

giving your answer correct to 3 significant figures.2 Use logarithms to solve the equation 4x+1 = 52x−3, [4]

4 marks

Mark scheme: 2 Use law for the logarithm of a product, a quotient or a power M1* Obtain (x +1)log4 = (2x – 3)log5, or equivalent A1 Solve for x M1(dep*) Obtain answer x = 3.39 A1 [4]

This question in 9709/22 Oct/Nov 2011

2 ln y (5, 4.49) (0, 2.14) x O The variables x and y satisfy the equation y where A and b are constants. The graph of ln y = A(bx), against x is a straight line passing through the points and as shown in the diagram. (0, 2.14) (5, 4.49), Find the values of A and b, correct to 1 decimal place. [5]

5 marks

Mark scheme: 2 State or imply that ln y = ln A + x ln b B1 Equate intercept on y-axis to ln A M1 Obtain ln A = 2.14 and hence A = 8.5 A1 Attempt gradient of line or equivalent (or use of correct substitution) M1 Obtain 0.47 = ln b or equivalent and hence b = 1.6 A1 [5]

This question in 9709/21 May/June 2012

2 (i) Given that 52x 5x 12, find the value of 5x. [3] + = (ii) Hence, using logarithms, solve the equation 52x 5x 12, giving the value of x correct to + = 3 significant figures. [2]

5 marks

Mark scheme: 2 (i) State or imply equation in the form (5x)2 + 5x – 12 = 0 B1 Attempt solution of quadratic equation for 5x M1 Obtain 5x = 3 only A1 [3] (ii) Use logarithms to solve equation of the form 5x = k where k > 0 M1 Obtain 0.683 A1 [2]

This question in 9709/22 May/June 2012

2 ln y (5, 4.49) (0, 2.14) x O The variables x and y satisfy the equation y where A and b are constants. The graph of ln y = A(bx), against x is a straight line passing through the points and as shown in the diagram. (0, 2.14) (5, 4.49), Find the values of A and b, correct to 1 decimal place. [5]

5 marks

Mark scheme: 2 State or imply that ln y = ln A + x ln b B1 Equate intercept on y-axis to ln A M1 Obtain ln A = 2.14 and hence A = 8.5 A1 Attempt gradient of line or equivalent (or use of correct substitution) M1 Obtain 0.47 = ln b or equivalent and hence b = 1.6 A1 [5]

This question in 9709/23 May/June 2012

5 ln y (1, 2.9) (3.5, 1.4) x O The variables x and y satisfy the equation y where A and b are constants. The graph of ln y against x is a straight line passing through the= A(b−x),points and as shown in the diagram. Find the values of A and b, correct to 2 decimal places.(1, 2.9) (3.5, 1.4), [6]

6 marks

Mark scheme: 5 State or imply ln y = ln A − x ln b B1 Form a numerical expression for the gradient of the line M1 Obtain b = 1.82 A1 Use gradient and one point correctly to find ln A M1 Obtain ln A = 3.5 A1 Obtain A = 33.12 A1 [6] GCE AS LEVEL – October/November 2012 9709 22 1 − x

This question in 9709/22 Oct/Nov 2012

1 Solve the equation 1, giving answers correct to 2 decimal places where appropriate. [5] |2x −7| =

5 marks

Mark scheme: x 21 Either State or imply non-modular equation (2 − 7 ) = 12, or corresponding pair of equations M1 Obtain 2x = 8 and 2x = 6 A1 State answer 3 B1 Use logarithmic method to solve an equation of the form 2x = k, where k > 0 M1 State answer 2.58 A1 Or State or imply one value for 2x, e.g. 8, by solving an equation or by inspection B1 State answer 3 B1 State second value for 2x B1 Use logarithmic method to solve an equation of the form 2x = k, where k > 0 M1 State answer 2.58 A1 [5] 2

This question in 9709/21 May/June 2013

2 Solve the equation ln x ln 5. [5] ln(3 −2x) −2 =

5 marks

Mark scheme: 2 Use 2 ln x = ln(x2) M1 Use law for addition or subtraction of logarithms M1 Obtain correct quadratic equation in x A1 Make reasonable solution attempt at a 3-term quadratic DM1 (dependent on previous M marks) 3 State x = and no other solutions A1 [5] 5

This question in 9709/21 May/June 2013

4 The variables x and y satisfy the equation 5y+1 = 23x. (i) By taking logarithms, show that the graph of y against x is a straight line. [2] (ii) Find the exact value of the gradient of this line and state the coordinates of the point at which the line cuts the y-axis. [2]

4 marks

Mark scheme: 4 (i) State or imply (y + 1) log 5 = 3x log 2 M1 State that this is of the form ay = bx + c and thus a straight line, or equivalent A1 [2] 3 ln 2 (ii) State gradient is , or equivalent, e.g. 3log52 B1 ln 5 State (0, –1) B1 [2] d y

This question in 9709/22 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

1 Solve the equation 1, giving answers correct to 2 decimal places where appropriate. [5] |2x −7| =

5 marks

Mark scheme: x 21 Either State or imply non-modular equation (2 − 7 ) = 12, or corresponding pair of equations M1 Obtain 2x = 8 and 2x = 6 A1 State answer 3 B1 Use logarithmic method to solve an equation of the form 2x = k, where k > 0 M1 State answer 2.58 A1 Or State or imply one value for 2x, e.g. 8, by solving an equation or by inspection B1 State answer 3 B1 State second value for 2x B1 Use logarithmic method to solve an equation of the form 2x = k, where k > 0 M1 State answer 2.58 A1 [5] 2

This question in 9709/23 May/June 2013

2 Solve the equation ln x ln 5. [5] ln(3 −2x) −2 =

5 marks

Mark scheme: 2 Use 2 ln x = ln(x2) M1 Use law for addition or subtraction of logarithms M1 Obtain correct quadratic equation in x A1 Make reasonable solution attempt at a 3-term quadratic DM1 (dependent on previous M marks) 3 State x = and no other solutions A1 [5] 5

This question in 9709/23 May/June 2013

1 (i) Solve the equation x 2 x . [2] + = −13 (ii) Hence solve the equation 3y 2 3y , giving your answer correct to 3 significant figures. + = −13 [2]

4 marks

Mark scheme: 1 (i) Either Square both sides to obtain linear equation M1 Obtain x = 16530 or 336 or 112 A1 [2] Or Solve linear equation in which, initially, signs of x are different M1 Obtain x + 2 = − x + 13 or equivalent and hence 112 or equivalent A1 [2] (ii) Apply logarithms and use power law M1 Obtain y log 3 = log 112 and hence y = .155 A1 [2]

This question in 9709/22 May/June 2014

1 (i) Solve the equation x 2 x . [2] + = −13 (ii) Hence solve the equation 3y 2 3y , giving your answer correct to 3 significant figures. + = −13 [2]

4 marks

Mark scheme: 1 (i) Either Square both sides to obtain linear equation M1 Obtain x = 16530 or 336 or 112 A1 [2] Or Solve linear equation in which, initially, signs of x are different M1 Obtain x + 2 = − x + 13 or equivalent and hence 112 or equivalent A1 [2] (ii) Apply logarithms and use power law M1 Obtain y log 3 = log 112 and hence y = .155 A1 [2]

This question in 9709/23 May/June 2014

4 (a) Find the value of x satisfying the equation 2 ln x x ln 2. [5] −4 −ln = (b) Use logarithms to find the smallest integer satisfying the inequality 1.4y 1010. >

5 marks

Mark scheme: 4 (a) Use power law to produce ln ( x − 4 )2 B1 Apply logarithm laws to produce equation without logarithms M1 Obtain ( x − 4 )2 = 2 x or equivalent A1 Solve 3-term quadratic equation DM1 Obtain (finally) x = 8 only A1 [5] (b) Apply logarithms and use power law (once) M1 ln 1010 Obtain or equivalent as part of inequality or equation A1 ln 4.1 Conclude with single integer 69 A1 [3]

This question in 9709/22 Oct/Nov 2014

1 (i) Solve the equation 3x 4 3x . [3] + = −11 (ii) Hence, using logarithms, solve the equation 3 2y 4 3 2y , giving the answer correct × + = × −11 to 3 significant figures. [2]

5 marks

Mark scheme: 1 (i) State or imply equation (3 x + 4) 2 = (3 x − 11) 2 or 3 x + 4 = − (3 x − 11) B1 Attempt solution of ‘quadratic’ equation or linear equation M1 7 Obtain x = or equivalent (and no other solutions) A1 [3] 6 (ii) Use logarithms to solve equation of form 2y = their answer to (i) ( must be + ve) M1 Obtain 0.222 (and no other solutions) A1 [2]

This question in 9709/21 May/June 2015

1 (i) Use logarithms to solve the equation 2x = 205, giving the answer correct to 3 significant figures. [2] (ii) Hence determine the number of integers n satisfying 20−5 < 2n < 205.

2 marks

Mark scheme: 1 (i) Introduce logarithms and use power law M1 Obtain x = 21.6 A1 [2] (ii) Obtain or imply –21.6 or –21 as lower value B1 State 43 B1 [2]

This question in 9709/22 May/June 2015

1 (i) Use logarithms to solve the equation 2x = 205, giving the answer correct to 3 significant figures. [2] (ii) Hence determine the number of integers n satisfying 20−5 < 2n < 205.

2 marks

Mark scheme: 1 (i) Introduce logarithms and use power law M1 Obtain x = 21.6 A1 [2] (ii) Obtain or imply –21.6 or –21 as lower value B1 State 43 B1 [2]

This question in 9709/23 May/June 2015

1 Use logarithms to solve the equation 5x+3 = 7x−1, giving the answer correct to 3 significant figures. [4]

4 marks

Mark scheme: 1 Introduce logarithms and use power law twice M1* Obtain ( x + 3) log 5 = ( x − )1 log 7 or equivalent A1 Solve linear equation for x M1 dep Obtain 20.1 A1 [4]

This question in 9709/21 Oct/Nov 2015

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

1 (i) Solve the equation 3x 5. [3] −2 = (ii) Hence, using logarithms, solve the equation 3 5y 5, giving the answer correct to 3 significant figures. × −2 = [2]

5 marks

Mark scheme: 1 (i) Either Square both sides to obtain three-term quadratic equation M1 Solve three-term quadratic equation to obtain two values M1 Obtain –1 and 73 A1 Or Obtain 73 from graphical method, inspection or linear equation B1 Obtain –1 similarly B2 [3] (ii) Use logarithmic method to solve an equation of the form 5 y = k where k > 0 M1 Obtain 0.526 and no others A1 [2]

This question in 9709/22 Oct/Nov 2015

2 (i) Solve the equation 2x 3 x 8 . [3] + = + (ii) Hence, using 3 2y 8 . Give the answer correct to logarithms, solve the equation 2y+1 + = + 3 significant figures. [2]

5 marks

Mark scheme: 2 (i) Either State or imply non-modulus equation ( 2 x + 3) 2 = ( x + 8) 2 or corresponding pair of linear equations B1 Solve 3-term quadratic equation or 2 linear equations M1 Obtain x = − 113 and x = 5 A1 Or Obtain x = 5 from graphical method, inspection, equation, … B1 Obtain x = − 113 similarly B2 [3] (ii) Use logarithms to solve equation of form 2 y = k where k > 0 M1 Obtain 2.32 A1 [2] dx t t

This question in 9709/23 Oct/Nov 2015

3 Given that 3ex 14, find the possible values of ex and hence solve the equation 3ex 14 correct to 3 significant+ 8e−x =figures. + 8e−x = [6]

6 marks

Mark scheme: 3 Rearrange to 3e 2 x − 14e x + 8 = 0 or equivalent involving substitution B1 Solve quadratic equation in ex to find two values of ex *M1 Obtain 23 and 4 A1 Use natural logarithms to solve equation of form ex = k where k > 0 dep on DM1 Allow M mark if left in exact form M1 Obtain − 0.405 A1 Obtain 1.39 A1 [6] 2

This question in 9709/21 May/June 2016

x 1 Given that 53x 74y, use logarithms to find the value of correct to 4 significant figures. [3] y =

3 marks

Mark scheme: 1 Use power law for logarithms correctly at least once M1 Obtain 3 x log5 = 4 y log7 or 3 x ln5 = 4 y ln7 or equivalent A1 Obtain 1.612 A1 [3]

This question in 9709/22 May/June 2016

x 1 Given that 53x 74y, use logarithms to find the value of correct to 4 significant figures. [3] y =

3 marks

Mark scheme: 1 Use power law for logarithms correctly at least once M1 Obtain 3 x log5 = 4 y log7 or 3 x ln5 = 4 y ln7 or equivalent A1 Obtain 1.612 A1 [3]

This question in 9709/23 May/June 2016

3 (i) Solve the inequality 2x x 3 . [4] −5 < + … … … … … … … … … … … … … … … … … (ii) Hence find the largest integer y satisfying the inequality 2 ln y ln y 3 . [2] −5 < + … … … … … … …

6 marks

Mark scheme: 3(i) State or imply non-modulus inequality (2 x − 5) 2 < ( x + 3) 2 or B1 corresponding equation or pair of linear equations Attempt solution of 3-term quadratic inequality or equation M1 or of 2 linear equations Obtain critical values 23 and 8 A1 State answer 23 < x < 8 A1 Total: 4 3(ii) Attempt to find y from ln y = upper limit of answer to part (i) M1 Obtain 2980 A1 Total: 2

This question in 9709/22 Feb/March 2017

1 Given that 5x 34y, use logarithms to show that y mx and find the value of the constant m correct = = to 3 significant figures. [3] … … … … … … … … … … … …

3 marks

Mark scheme: 1 Take logarithms of both sides and apply power law to both sides M1 Allow log5 4log3 = y for M1 A1 Rearrange to the form ln5 4ln3 = y x or equivalent A1 Obtain 0.366 = m A1 Total: 3 State or imply non-modulus inequality ( ) ⩽ ( )

This question in 9709/21 May/June 2017

2 Use logarithms to solve the equation 52x, giving your answer correct to 3 significant figures. 3x+4 = [4] … … … … … … … … … … …

4 marks

Mark scheme: 2 Apply logarithms to both sides and apply power law *M1 Obtain ( ) 4 log3 2 log5 x x + = or equivalent A1 Solve linear equation for x DM1 dep *M Obtain 2.07 A1 Allow greater accuracy Total: 4 May be implied by part graph in first quadrant

This question in 9709/22 May/June 2017

2 Use logarithms to solve the equation 52x, giving your answer correct to 3 significant figures. 3x+4 = [4] … … … … … … … … … … …

4 marks

Mark scheme: 2 Apply logarithms to both sides and apply power law *M1 Obtain ( ) 4 log3 2 log5 x x + = or equivalent A1 Solve linear equation for x DM1 dep *M Obtain 2.07 A1 Allow greater accuracy Total: 4 May be implied by part graph in first quadrant

This question in 9709/23 May/June 2017

3 It is given that the variable x is such that 1.32x 80 and 3x 3x . < −1 > −10 Find the set of possible values of x, giving your answer in the form a x b where the constants a and b are correct to 3 significant figures. < < [7] … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 3 Take logarithms of both sides and apply power M1 Condone incorrect inequality law signs until final answer. The first 6 marks are for obtaining the correct critical values. ln80 A1 Obtain 2 x < or equivalent using log10 ln1.3 Obtain x = 8.35... A1 State or imply non-modulus inequality B1 (3 x − 1) 2 > (3 x − 10) 2 or corresponding equation or linear equation 3 x −=1 − (3 x − 10) Attempt solution of inequality or equation M1 (obtaining 3 terms when squaring each bracket or solving linear equation with signs of 3x different) Obtain x = 116 or x = 1.83... A1 Conclude 1.83 < x < 8.35 A1 7

This question in 9709/21 Oct/Nov 2017

1 Use logarithms to solve the equation 24x, giving your answer correct to 3 significant figures. 53x−1 = [4] … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 1 Introduce logarithms to both sides and use *M1 power law Obtain (3 x − 1)log5 = 4 x log2 or equivalent A1 Allow A1 for poor use of brackets if recovered later Solve linear equation for x DM1 dep *M Obtain 0.783 A1 Allow 3 sf or better 4

This question in 9709/22 Oct/Nov 2017

3 It is given that the variable x is such that 1.32x 80 and 3x 3x . < −1 > −10 Find the set of possible values of x, giving your answer in the form a x b where the constants a and b are correct to 3 significant figures. < < [7] … … … … … … … … … … … … … … … … … … … … … … …

7 marks

Mark scheme: 3 Take logarithms of both sides and apply power M1 Condone incorrect inequality law signs until final answer. The first 6 marks are for obtaining the correct critical values. ln80 A1 Obtain 2 x < or equivalent using log10 ln1.3 Obtain x = 8.35... A1 State or imply non-modulus inequality B1 (3 x − 1) 2 > (3 x − 10) 2 or corresponding equation or linear equation 3 x −=1 − (3 x − 10) Attempt solution of inequality or equation M1 (obtaining 3 terms when squaring each bracket or solving linear equation with signs of 3x different) Obtain x = 116 or x = 1.83... A1 Conclude 1.83 < x < 8.35 A1 7

This question in 9709/23 Oct/Nov 2017

1 (i) Solve the equation 9x 3x 2 . [3] −2 = + … … … … … … … … … … … … … … … (ii) Hence, using logarithms, solve the equation 2 , giving your answer correct to 3 significant figures. 3y+2 −2 = 3y+1 + [2] … … … … … … … …

5 marks

Mark scheme: 1(i) State or imply non-modular equation 2 2 (9 2) (3 2) x x − = + or pair of linear equations B1 Attempt solution of quadratic equation or of 2 linear equations M1 Obtain 0 and 2 3 A1 SC: B1 for one correct solution 3 1(ii) Apply logarithms and use power law for 3y k = where 0 k > M1 Must be using their answers to part (i) Obtain 0.369 − A1 2

This question in 9709/21 Oct/Nov 2018

2 Given that 9x 3x 240, find the value of 3x and hence, using logarithms, find the value of x correct to 4 significant+ figures.= [5] … … … … … … … … … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2 2 (3 ) x or 2 B1 May be implied by 3 3 1 240 + = x x Attempt solution of quadratic equation in 3x *M1 Perhaps using substitution 3 = x u Obtain, finally, 3 15 = x only A1 Apply logarithms and use power law for 3 = x k where 0 > k M1 Dependent *M, need to see ln3 ln = x k , 3 log = x k oe Obtain 2.465 A1 May be done using 2 9 , x same processes 5

This question in 9709/22 Oct/Nov 2018

1 (i) Solve the equation 9x 3x 2 . [3] −2 = + … … … … … … … … … … … … … … … (ii) Hence, using logarithms, solve the equation 2 , giving your answer correct to 3 significant figures. 3y+2 −2 = 3y+1 + [2] … … … … … … … …

5 marks

Mark scheme: 1(i) State or imply non-modular equation 2 2 (9 2) (3 2) x x − = + or pair of linear equations B1 Attempt solution of quadratic equation or of 2 linear equations M1 Obtain 0 and 2 3 A1 SC: B1 for one correct solution 3 1(ii) Apply logarithms and use power law for 3y k = where 0 k > M1 Must be using their answers to part (i) Obtain 0.369 − A1 2

This question in 9709/23 Oct/Nov 2018

2 (i) Solve the equation 4 2x 3 . [3] + = −5x … … … … … … … … … … … … … (ii) Hence solve the equation 4 2e3y 3 , giving the answer correct to 3 significant figures. + = −5e3y [2] … … … … … … … … … …

5 marks

Mark scheme: 2(i) State or imply non-modular equation 2 2 (4 2 ) (3 5 ) x x + = − or pair of linear equations Attempt solution of 3-term quadratic eqn or pair of linear equations M1 Obtain 7 1 7 3 , − A1 SC B1 for 1 7 x = − from one linear equation 3 2(ii) Attempt correct process to solve 3e y k = where 0 k > from (i) M1 Obtain 0.282 and no others A1 2

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 9709/22 May/June 2019

2 (i) Solve the equation 4 2x 3 . [3] + = −5x … … … … … … … … … … … … … (ii) Hence solve the equation 4 2e3y 3 , giving the answer correct to 3 significant figures. + = −5e3y [2] … … … … … … … … … …

5 marks

Mark scheme: 2(i) State or imply non-modular equation 2 2 (4 2 ) (3 5 ) x x + = − or pair of linear equations Attempt solution of 3-term quadratic eqn or pair of linear equations M1 Obtain 7 1 7 3 , − A1 SC B1 for 1 7 x = − from one linear equation 3 2(ii) Attempt correct process to solve 3e y k = where 0 k > from (i) M1 Obtain 0.282 and no others A1 2

This question in 9709/23 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

1 (i) Solve the inequality 2x 2x . [3] −7 < −9 … … … … … … … … … … … … … (ii) Hence find the largest integer n satisfying the inequality 2 ln n 2 ln n . [2] −7 < −9 … … … … … … … … … … …

5 marks

Mark scheme: 1(i) State or imply non-modular inequality 2 2 (2 7) (2 9) x x − < − or corresponding equation or linear equation (with signs of 2x different) M1 Obtain critical value 4 A1 State 4 x < only A1 3 1(ii) Attempt to find n from lnn = their critical value from part (i) M1 Obtain or imply 4 e n < and hence 54 A1 2

This question in 9709/21 Oct/Nov 2019

2 (i) Solve the equation 4x 5 x . [3] + = −7 … … … … … … … … … … … … … (ii) Hence, using logarithms, solve the equation 5 2y , giving the answer correct to 3 significant figures. 2y+2 + = −7 [2] … … … … … … … … … …

5 marks

Mark scheme: 2(i) State or imply non-modular equation 2 2 (4 5) ( 7) x x + = − or pair of different linear equations Attempt solution of 3-term quadratic equation or pair of linear equations M1 Obtain 2 5 and 4 − A1 SC For 4 x = −only, from correct work, allow B1 3 2(ii) Apply logarithms and use power law for 2y k = where 0 k > from (i) M1 Obtain –1.32 only A1 AWRT 2

This question in 9709/22 Oct/Nov 2019

1 (i) Solve the inequality 2x 2x . [3] −7 < −9 … … … … … … … … … … … … … (ii) Hence find the largest integer n satisfying the inequality 2 ln n 2 ln n . [2] −7 < −9 … … … … … … … … … … …

5 marks

Mark scheme: 1(i) State or imply non-modular inequality 2 2 (2 7) (2 9) x x − < − or corresponding equation or linear equation (with signs of 2x different) M1 Obtain critical value 4 A1 State 4 x < only A1 3 1(ii) Attempt to find n from lnn = their critical value from part (i) M1 Obtain or imply 4 e n < and hence 54 A1 2

This question in 9709/23 Oct/Nov 2019

1 Given that 2y 93x, use logarithms to show that y kx and find the value of k correct to 3 significant figures. = = [3] … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 1 Apply logarithms to both sides and apply power law at least once M1 Rearrange to the form 3ln9 ln 2 = y x OE A1 Obtain 9.51 = k A1 3

This question in 9709/22 May/June 2020

1 Given that 2y 93x, use logarithms to show that y kx and find the value of k correct to 3 significant figures. = = [3] … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 1 Apply logarithms to both sides and apply power law at least once M1 Rearrange to the form 3ln9 ln 2 = y x OE A1 Obtain 9.51 = k A1 3

This question in 9709/23 May/June 2020

4 (a) Solve the equation 2x x 6 . [3] −5 = + … … … … … … … … … … … … … 6 . Give your answer correct to 3 significant (b) Hence find the value of y such that 21−y −5 = 2−y + figures. [2] … … … … … … … … … …

5 marks

Mark scheme: 4(a) State or imply non-modulus equation 2 2 (2 5) ( 6) x x − = + or pair of linear equations B1 Attempt solution of 3-term quadratic equation or of pair of linear equations M1 Obtain 1 3 − and 11 A1 3 Question Answer Marks Guidance 4(b) Apply logarithms and use power law for 2 y k − = where 0 k > from (a) M1 Obtain 3.46 − A1 AWRT 2

This question in 9709/21 Oct/Nov 2020

4 (a) Solve the equation 2x x 6 . [3] −5 = + … … … … … … … … … … … … … 6 . Give your answer correct to 3 significant (b) Hence find the value of y such that 21−y −5 = 2−y + figures. [2] … … … … … … … … … …

5 marks

Mark scheme: 4(a) State or imply non-modulus equation 2 2 (2 5) ( 6) x x − = + or pair of linear equations B1 Attempt solution of 3-term quadratic equation or of pair of linear equations M1 Obtain 1 3 − and 11 A1 3 Question Answer Marks Guidance 4(b) Apply logarithms and use power law for 2 y k − = where 0 k > from (a) M1 Obtain 3.46 − A1 AWRT 2

This question in 9709/23 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

6 The polynomial p x is defined by p x x3 ax b, = + + where a and b are constants. It is given that x 2 is a factor of p x and that the remainder is 5 when p x is divided by x . + −3 (a) Find the values of a and b. [5] … … … … … … … … … … … … … … … … … … … … … … (b) Hence find the exact root of the equation p e2y 0. [5] = … … … … … … … … … … … … … … … … … … … … … … … … …

10 marks

Mark scheme: 6(a) Substitute 2 x = − and equate to zero *M1 Substitute 3 x = and equate to 5 *M1 Obtain 8 2 0 a b −− + = and 27 3 5 a b + + = or equivalents A1 Solve a pair of relevant linear simultaneous equations for a or b DM1 Dependent at least one M mark. Obtain 6 a = − and 4 b = − A1 5 Question Answer Marks Guidance 6(b) Attempt division by 2 x + at least as far as 2 x kx + M1 Obtain 2 2 2 x x − − A1 Obtain (at least) the positive root 2 12 2 + or exact equivalent A1 Equate 2 e y to positive root, apply logarithms and use power law M1 Obtain 1 2 12 ln 2 2   +       or 1 ln(1 3) 2 + or exact equivalent A1 5

This question in 9709/22 Feb/March 2021

4 A curve has parametric equations x ln 2t 6 t, y t ln t. = + −ln = (a) Find the value of t at the point P on the curve for which x ln 4. [3] = … … … … … … … … … … … … … … … … … … … … … … … (b) Find the exact gradient of the curve at P. [5] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 4(a) Equate x to ln4 and use relevant logarithm property M1 Obtain equation with no logarithm present, 2 6 4 + = t t A1 OE Obtain 3 = t A1 3 4(b) Obtain d 2 1 d 2 6 = − + x t t t B1 Use product rule to find d d y t M1 Obtain 1 ln + × t t t A1 Divide to obtain d d y x using their d d y t and d d x t correctly DM1 Must have at least B1 or M1. Do not condone incorrect inverting of terms unless a correct statement is seen initially. Obtain 6(ln3 1) − + A1 or exact equivalents 5

This question in 9709/21 May/June 2021

5 (a) Find the quotient when x4 55 is divided by x 2 and show that the remainder is 7. −32x + −2 [3] … … … … … … … … … … … … … … … … … … … … … … … … … (b) Factorise x4 48. [2] −32x + … … … … … … … … … … … … 48 0, giving your answer in an exact form. [2](c) Hence solve the equation e−12y −32e−3y + = … … … … … … … … … … … …

7 marks

Mark scheme: 5(a) Carry out division at least as far as x kx or equivalent … M1 OE, e.g. comparing coefficients with coefficient of 2 x equal to 1 and attempt at a second coefficient. Obtain quotient 2 4 12 + + x x A1 Confirm remainder is 7 A1 AG 3 5(b) Include 2 ( 2) − x as a factor M1 Must be a product of factors only SC B1 for ( )( ) 2 2 4 4 4 12 − + + + x x x x Conclude 2 2 ( 2) ( 4 12) − + + x x x A1 isw any attempt to factorise the quotient. 2 5(c) Apply logarithms and use power law for 3 e− = y k where 0 > k M1 Obtain 1 1 1 ln2, ln 3 3 2 = − y A1 Or exact equivalent Must be simplified e.g. not lne or 6 3 ISW extra solutions but A0 if undefined solutions are included. 2

This question in 9709/22 May/June 2021

5 (a) Find the quotient when x4 55 is divided by x 2 and show that the remainder is 7. −32x + −2 [3] … … … … … … … … … … … … … … … … … … … … … … … … … (b) Factorise x4 48. [2] −32x + … … … … … … … … … … … … 48 0, giving your answer in an exact form. [2](c) Hence solve the equation e−12y −32e−3y + = … … … … … … … … … … … …

7 marks

Mark scheme: 5(a) Carry out division at least as far as x kx or equivalent … M1 OE, e.g. comparing coefficients with coefficient of 2 x equal to 1 and attempt at a second coefficient. Obtain quotient 2 4 12 + + x x A1 Confirm remainder is 7 A1 AG 3 5(b) Include 2 ( 2) − x as a factor M1 Must be a product of factors only SC B1 for ( )( ) 2 2 4 4 4 12 − + + + x x x x Conclude 2 2 ( 2) ( 4 12) − + + x x x A1 isw any attempt to factorise the quotient. 2 5(c) Apply logarithms and use power law for 3 e− = y k where 0 > k M1 Obtain 1 1 1 ln 2, ln 3 3 2 = − y A1 Or exact equivalent Must be simplified e.g. not lne or 6 3 ISW extra solutions but A0 if undefined solutions are included. 2

This question in 9709/23 May/June 2021

2 (a) Sketch, on the same diagram, the graphs of y x 3 and y 2x . [2] = + = −1 (b) Solve the equation x 3 2x . [3] + = −1 … … … … … … … … … … 12y 12y 3 2 5 . Give your answer correct to 3 significant (c) Find the value of y such that 5 + = . × −1. figures. [2] … … … … … …

7 marks

Mark scheme: 2(a) Draw V-shaped graph with vertex on positive x-axis *B1 Must be straight lines. Draw (more or less) correct graph of 3 = + y x with smaller gradient together with a V shaped graph DB1 And crossing y-axis above y-intercept of first graph. Intersection in the first quadrant may be implied. 2 Question Answer Marks Guidance 2(b) Solve 3 2 1 + = − x x to obtain 4 = x B1 Attempt solution of linear equation where signs of 2x and xare different M1 Obtain 2 3 = − x A1 Alternative method for question 2(b) State or imply non-modulus equation 2 2 ( 3) (2 1) + = − x x B1 Attempt solution of 3-term quadratic equation obtained from squaring both terms. M1 Must have B1. Obtain 2 3 − and 4 A1 3 2(c) Apply logarithms and use power law for 1 2 5 = y k where 0 > k M1 Using their positive root from part (b). Allow M1 for 5 2log 4 = y . Obtain 1.72 = y A1 AWRT; and no other values. 2

This question in 9709/22 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

3 y 2.58, 9.00 1.03, 6.36 ln x O The variables x and y satisfy the equation ay kx, where a and k are constants. The graph of y against = ln x is a straight line passing through the points 1.03, 6.36 and 2.58, 9.00 , as shown in the diagram. Find the values of a and k, giving each value correct to 2 significant figures. [5] … … … … … … … … … … … … … … … …

5 marks

Mark scheme: 3 State or imply equation is ln ln ln = + y a k x Equate gradient of line to 1 ln a M1 Or eliminate lnk from simultaneous equations. Obtain 1 2.64 ln 1.55 = a or equivalent and hence 1.8 = a A1 AWRT Substitute appropriate values to find lnk M1 Obtain ln 2.7... = k and hence 15 = k A1 AWRT Alternative method for question 3 6.36 1.03 e = a k and 9 2.58 e = a k B1 For both. Elimination of k to obtain an equation in a only ( ) 2.64 1.55 e = a M1 Must have previous B1. Use of a correct method to obtain a M1 Allow for 0.59 e = a . 1.8 = a A1 15 = k A1 5

This question in 9709/23 Oct/Nov 2021

3 The variables x and y satisfy the equation y 32aax, where a is a constant. The graph of ln y against x is a straight line with gradient 0.239. = (a) Find the value of a correct to 3 significant figures. [3] … … … … … … … … … … … … (b) Hence find the value of x when y 36. Give your answer correct to 3 significant figures. [2] = … … … … … … … … … …

5 marks

Mark scheme: 3(a) State or imply equation is 2 ln ln3 ln = + a y x a B1 Equate gradient of line involving a to 0.239 M1 Obtain ln 0.239 = a and hence 1.27 = a A1 3 3(b) Substitute 36 = y in ln ... = y equation and solve for x M1 Or substitute in original equation with necessary manipulation Obtain 3.32 A1 2

This question in 9709/22 Feb/March 2022

1 The variables x and y satisfy the equation y where a is an integer. As shown in the diagram, the graph of ln y against x is a straight line =passing42x−a, through the point 0, , where the second coordinate is given correct to 3 significant figures. −20.8 (a) Show that the gradient of the straight line is ln 16. [2] … … … … … … … (b) Determine the value of a. [2] … … … … … … …

4 marks

Mark scheme: 1(a) State or imply equation is ln (2 )ln 4 y x a State gradient is 2ln4 and confirm ln16 B1 AG – necessary detail needed 2 1(b) Substitute for ln y and attempt value of a M1 Allow if ln 2 ln4   y x a Obtain 15  a A1 Integer answer required, but condone 15.0 2

This question in 9709/21 May/June 2022

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

2 Use logarithms to solve the equation giving your answer correct to 3 significant figures. 14e−2x = 5x+1, [4] … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 Apply logarithms correctly to both sides and apply power law at least once *M1 Obtain ln14 − 2 x = ( x + 1)ln5 A1 OE with x no longer part of a power. Attempt solution of linear equation DM1 Must have ln14 − ln5 = x ( 2 + ln5 ) . Obtain 0.285 A1 4

This question in 9709/21 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

1 Use logarithms to solve the equation 12x Give your answer correct to 3 significant figures. = 32x+1. [4] … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 1 Apply logarithms to both sides and apply power law correctly at least once *M1 OE with x not in a power. Obtain ln12 (2 1)ln3   x x A1 Attempt solution of linear equation DM1 Obtain 3.82 A1 Do not condone incorrect use of logarithms or greater accuracy. 4

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 9709/23 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

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

4 (a) Sketch, on the same diagram, the graphs of y 3 and y 9 [2] = −x = −2x. (b) Solve the inequality 3 9 [3] −x > −2x. … … … … … (c) Use logarithms to solve the inequality 500. Give your answer in the form x a, where 23x−10 <figures. < [3] the value of a is given correct to 3 significant … … … … … (d) List the integers that satisfy both of the inequalities 3 9 and 500. [1] −x > −2x 23x−10 < … …

9 marks

Mark scheme: 4(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw (more or less) correct graph of y = 9 − 2 x with steeper negative B1 Dependent on first B mark, appropriately positioned gradient with respect to first graph. 2 4(b) Solve linear equation or inequality with signs of x and 2x different M1 Obtain critical value 4 A1 Conclude x  4 only A1 6 must be discounted. Alternative Method for Question 4(b) State or imply non-modulus equation (or inequality) (3 − x ) 2 = (9 − 2 x ) 2 B1 Attempt solution of three-term quadratic equation (or inequality) M1 Dependent on previous B1. Conclude x  4 only A1 6 must be discounted. 3 4(c) State or imply (3 x − 10)ln2  ln500 B1 Or equivalent perhaps involving different logarithm base. Obtain critical value 6.32 B1 Obtain x  6.32 B1 Or greater accuracy. 3 4(d) State 5 and 6 only B1 1

This question in 9709/22 Oct/Nov 2023

4 (a) Sketch, on the same diagram, the graphs of y 3x 5 and y 2x 7. [2] (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 9709/23 Oct/Nov 2023

1 Use logarithms to solve the equation 3 4 x + 3 = 5 2 x + 7 . Give your answer correct to 3 significant figures. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: Question Answer Marks Guidance 1 Apply logarithms to both sides and apply power law at least once *M1 Obtain (4 x + 3)ln3 = (2 x + 7)ln5 A1 Or equivalent with x not in a power. Attempt solution of linear equation DM1 Obtain 6.78 A1 Or greater accuracy. 4

This question in 9709/22 Feb/March 2024

2 Use logarithms to solve the equation 6 2 x - 1 = 5e 3 x + 2 . Give your answer correct to 4 significant figures. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 2 1 ln6  x ln5 3 2   x *M1 Attempt solution of linear equation DM1 Obtain 9.256 A1 Or greater accuracy. 4

This question in 9709/22 May/June 2024

2 Use logarithms to solve the equation 6 2 x - 1 = 5e 3 x + 2 . Give your answer correct to 4 significant figures. [4] … … … … … … … … … … … … … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2 2 1 ln6  x ln5 3 2   x *M1 Attempt solution of linear equation DM1 Obtain 9.256 A1 Or greater accuracy. 4

This question in 9709/23 May/June 2024

1 The variables x and y satisfy the equation a 2y = e 3x + k , where a and k are constants. The graph of y against x is a straight line. 3 (a) Use logarithms to show that the gradient of the straight line is . [1] 2 lna … … … … … … … (b) Given that the straight line passes through the points ( 0 .4 , 0 .95) and ( 3 .3 , 3 .80) , find the values of a and k. [4] … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: Question Answer Marks Guidance 1(a) 3 B1 AG – necessary detail needed. State or imply 2 y ln a = 3 x + k and conclude that gradient is 2ln a 1 1(b) 3 M1 Equate to gradient of line 2ln a 3 2.85 1929 A1 Allow greater accuracy. Obtain = or equivalent and hence obtain a = 4.6 or a = e 2ln a 2.9 Substitute appropriate values to find value of k M1 Obtain k = 1.7 A1 Alternative Method for Question 1(b) Obtain 0.95 ( 2ln a ) = 3 ( 0.4 ) + k M1 OE or a1.9 = e1.2 + k Obtain 3.80 ( 2ln a ) = 3 ( 3.3 ) + k M1 OE or a 7.6 = e 9.9 + k 29 A1 Allow greater accuracy. 19 Obtain a = 4.6 or a = e Obtain k = 1.7 A1 4

This question in 9709/21 Oct/Nov 2024

1 Use logarithms to show that the equation 5 8 y = 6 7 x can be expressed in the form y = kx . Give the value of the constant k correct to 3 significant figures. [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: Question Answer Marks Guidance 1 Apply logarithms correctly and apply power law at least once *M1 7ln6 DM1 May be implied by 0.97, 0.973 seen. Obtain y = x or equivalent perhaps involving decimals 8ln5 May use a different base. Obtain k = 0.974 or state y = 0.974 x A1 Or greater accuracy. 3

This question in 9709/22 Oct/Nov 2024

1 The variables x and y satisfy the equation a 2y = e 3x + k , where a and k are constants. The graph of y against x is a straight line. 3 (a) Use logarithms to show that the gradient of the straight line is . [1] 2 lna … … … … … … … (b) Given that the straight line passes through the points ( 0 .4 , 0 .95) and ( 3 .3 , 3 .80) , find the values of a and k. [4] … … … … … … … … … … … … … … … … …

5 marks

Mark scheme: Question Answer Marks Guidance 1(a) 3 B1 AG – necessary detail needed. State or imply 2 y ln a = 3 x + k and conclude that gradient is 2ln a 1 1(b) 3 M1 Equate to gradient of line 2ln a 3 2.85 1929 A1 Allow greater accuracy. Obtain = or equivalent and hence obtain a = 4.6 or a = e 2ln a 2.9 Substitute appropriate values to find value of k M1 Obtain k = 1.7 A1 Alternative Method for Question 1(b) Obtain 0.95 ( 2ln a ) = 3 ( 0.4 ) + k M1 OE or a1.9 = e1.2 + k Obtain 3.80 ( 2ln a ) = 3 ( 3.3 ) + k M1 OE or a 7.6 = e 9.9 + k 29 A1 Allow greater accuracy. 19 Obtain a = 4.6 or a = e Obtain k = 1.7 A1 4

This question in 9709/23 Oct/Nov 2024

1 Solve the equation ln ( 3x + 1) - ln ( x - 5) = ln 7 . [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: Question Answer Marks Guidance 1 Apply appropriate logarithm property M1 3 x + 1 A1 OE Obtain = 7 x − 5 Solve to obtain x = 9 A1 3

This question in 9709/22 Feb/March 2025

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

5 marks

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

This question in 9709/22 Feb/March 2025

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

6 A curve has equation ( x 2 - 3 )ln y + 6 x = 14 . (a) Show that there is no point on the curve at which the y-coordinate is e -1 . [3] … … … … … … … … … … … … … … … (b) Find the equation of the tangent to the curve at the point ( 2, e2 ) . Give your answer in the form y = mx + c , where m and c are exact constants. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) Substitute y = e − 1 in equation and simplify to quadratic equation in x *M1 Evaluate discriminant DM1 OE, such as completion of square or use of formula. Obtain − x 2 + 6 x − 11 = 0, giving discriminant −8 and confirm result A1 OE, such as ( x − 3) 2 + 2 = 0, etc. AG – necessary detail needed. 3 6(b) Attempt use of product rule for differentiation of ( x 2 − 3)ln y *M1 May see in part (a) x 2 − 3 dy A1 OE Obtain 2 x ln y + y dx x 2 − 3 dy A1 FT OE Obtain complete 2 x ln y + + 6 = 0 2 Following their derivative of ( x − 3)ln y. y dx Substitute x = 2 and y = e 2 to find value of gradient of tangent DM1 Need to see attempt at substitution unless correct. −103 implies M1. Obtain gradient −14e 2 A1 Obtain equation y = −14e 2 x + 29e 2 or equivalent of required form A1 6

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

6 A curve has equation ( x 2 - 3 )ln y + 6 x = 14 . (a) Show that there is no point on the curve at which the y-coordinate is e -1 . [3] … … … … … … … … … … … … … … … (b) Find the equation of the tangent to the curve at the point ( 2, e2 ) . Give your answer in the form y = mx + c , where m and c are exact constants. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) Substitute y = e − 1 in equation and simplify to quadratic equation in x *M1 Evaluate discriminant DM1 OE, such as completion of square or use of formula. Obtain − x 2 + 6 x − 11 = 0, giving discriminant −8 and confirm result A1 OE, such as ( x − 3) 2 + 2 = 0, etc. AG – necessary detail needed. 3 6(b) Attempt use of product rule for differentiation of ( x 2 − 3)ln y *M1 May see in part (a) x 2 − 3 dy A1 OE Obtain 2 x ln y + y dx x 2 − 3 dy A1 FT OE Obtain complete 2 x ln y + + 6 = 0 2 Following their derivative of ( x − 3)ln y. y dx Substitute x = 2 and y = e 2 to find value of gradient of tangent DM1 Need to see attempt at substitution unless correct. −103 implies M1. Obtain gradient −14e 2 A1 Obtain equation y = −14e 2 x + 29e 2 or equivalent of required form A1 6

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

6 A curve has equation ( x 2 - 3 )ln y + 6 x = 14 . (a) Show that there is no point on the curve at which the y-coordinate is e -1 . [3] … … … … … … … … … … … … … … … (b) Find the equation of the tangent to the curve at the point ( 2, e2 ) . Give your answer in the form y = mx + c , where m and c are exact constants. [6] … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(a) Substitute y = e − 1 in equation and simplify to quadratic equation in x *M1 Evaluate discriminant DM1 OE, such as completion of square or use of formula. Obtain − x 2 + 6 x − 11 = 0, giving discriminant −8 and confirm result A1 OE, such as ( x − 3) 2 + 2 = 0, etc. AG – necessary detail needed. 3 6(b) Attempt use of product rule for differentiation of ( x 2 − 3)ln y *M1 May see in part (a) x 2 − 3 dy A1 OE Obtain 2 x ln y + y dx x 2 − 3 dy A1 FT OE Obtain complete 2 x ln y + + 6 = 0 2 Following their derivative of ( x − 3)ln y. y dx Substitute x = 2 and y = e 2 to find value of gradient of tangent DM1 Need to see attempt at substitution unless correct. −103 implies M1. Obtain gradient −14e 2 A1 Obtain equation y = −14e 2 x + 29e 2 or equivalent of required form A1 6

This question in 9709/25 May/June 2025

2 Solve the equation e 2 x `e 2 x - 8j = 48 . [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 2 Attempt solution of quadratic equation in e 2 x to obtain at least one value of e 2 x M1 Obtain (e 2 x − 12)(e 2 x + 4) = 0 OE, and hence at least e 2 x = 12 A1 Obtain x = 12 ln12 or 1.24 and no other solution A1 3

This question in 9709/21 Oct/Nov 2025

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

7 marks

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

This question in 9709/21 Oct/Nov 2025

2 Solve the equation e 2 x `e 2 x - 8j = 48 . [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 2 Attempt solution of quadratic equation in e 2 x to obtain at least one value of e 2 x M1 Obtain (e 2 x − 12)(e 2 x + 4) = 0 OE, and hence at least e 2 x = 12 A1 Obtain x = 12 ln12 or 1.24 and no other solution A1 3

This question in 9709/23 Oct/Nov 2025

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

7 marks

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

This question in 9709/23 Oct/Nov 2025

2 Solve the equation e 2 x `e 2 x - 8j = 48 . [3] … … … … … … … … … … … … … … … … … … … … … … … … … … …

3 marks

Mark scheme: 2 Attempt solution of quadratic equation in e 2 x to obtain at least one value of e 2 x M1 Obtain (e 2 x − 12)(e 2 x + 4) = 0 OE, and hence at least e 2 x = 12 A1 Obtain x = 12 ln12 or 1.24 and no other solution A1 3

This question in 9709/25 Oct/Nov 2025

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

7 marks

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

This question in 9709/25 Oct/Nov 2025