Coordinate geometry — 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

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

7 marks

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

This question in 9709/21 Oct/Nov 2010

5 The parametric equations of a curve are x y e2t 2t. = ln(t + 1), = + dy (i) Find an expression for in terms of t. [4] dx (ii) Find the equation of the normal to the curve at the point for which t 0. Give your answer in = the form ax by c 0, where a, b and c are integers. [4] + + =

8 marks

Mark scheme: d x 1 5 (i) State = B1 d t t + 1 dy 2 t State = 2e + 2 B1 dt dy Attempt expression for M1 dx dy 2 t Obtain = ( 2e + 2)(t + )1 or equivalent A1 [4] dx (ii) Substitute t = 0 and attempt gradient of normal M1 dy 1 Obtain − 4 following their expression for A1√ dx Attempt to find equation of normal through point (0, 1) M1 Obtain x + 4y – 4 = 0 A1 [4] GCE AS/A LEVEL – May/June 2012 9709 22

This question in 9709/22 May/June 2012

5 The parametric equations of a curve are x e2t, y 4tet. = = dy (i) Show that 2(t + 1) . [4] dx = et (ii) Find the equation of the normal to the curve at the point where t 0. [4] =

8 marks

Mark scheme: 5 (i) Use product rule to differentiate y M1 Obtain correct derivative in any form A1 dy dy dx Use = ÷ M1 dx dt dt Obtain given answer correctly A1 [4] d y (ii) Substitute t = 0 in and both parametric equations B1 d x d y Obtain = 2 and coordinates (1, 0) B1 d x dy Form equation of the normal at their point, using negative reciprocal of their M1 dx 1 1 State correct equation of normal y = − x + or equivalent A1 [4] 2 2

This question in 9709/23 May/June 2013

7 The parametric equations of a curve are x t3 6t 1, y t4 4t2 5. = + + = −2t3 + −12t + dy dy (i) Find and use division to show that can be written in the form at b, where a and b are dx dx + constants to be found. [5] … … … … … … … … … … … … … … … … … … … … … … (ii) The straight line x 9 0 is the normal to the curve at the point P. Find the coordinates −2y + = of P. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 7(i) Differentiate x and y and form d d y x M1 Obtain 3 2 2 4 6 8 12 3 6 t t t t − + − + A1 First 2 marks may be implied by an attempt at division Carry out division at least as far as kt or equivalent M1 For M1, it must be division by a quadratic factor. Allow attempt at factorisation with same conditions as for division Obtain 4 3 t A1 Obtain 4 3 2 t − with complete division shown and no errors seen A1 Total: 5 Question Answer Marks Guidance 7(ii) State or imply gradient of straight line is 1 2 B1 Allow B1 if 1 9 2 2 y x = + is seen Attempt value of t from their d d y x = their negative reciprocal of gradient of line M1 Obtain 0 t = and hence ( ) 1,5 A1 Total: 3

This question in 9709/21 May/June 2017

6 The parametric equations of a curve are x 2e2t 4et, y 5te2t. = + = dy (i) Find in terms of t and hence find the coordinates of the stationary point, giving each coordinate dx correct to 2 decimal places. [6] … … … … … … … … … … … … … … … … … … … … … … (ii) Find the gradient of the normal to the curve at the point where the curve crosses the x-axis. [3] … … … … … … … … … … … … … … … … … … … … … … … … …

9 marks

Mark scheme: 6(i) Obtain ddxt = 4e 2 t + 4e t B1 Use product rule to find ddyt M1 dy 5e 2 t + 10te 2 t A1 Obtain = or equivalent dx 4e 2 t + 4e t ae 2 t + bte 2 t M1 Equate first derivative of the form ce 2 t + de t to zero and solve to find t Obtain t = − 12 from completely correct work A1 Obtain (3.16, − 0.92) A1 6 6(ii) Identify t = 0 B1 Substitute t = 0 in expression for first derivative M1 and find negative reciprocal Obtain − 85 or equivalent A1 3

This question in 9709/23 Oct/Nov 2017

4 (a) Sketch, on the same diagram, the graphs of y 3x 2a and y 3x , where a is a positive constant. = + = −4a Give the coordinates of the points where each graph meets the axes. [3] (b) Find the coordinates of the point of intersection of the two graphs. [3] … … … … … … … … (c) Deduce the solution of the inequality 3x 2a 3x . [1] + < −4a … … … …

7 marks

Mark scheme: 4(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 2 4 , 2 , , 4 3 3 − a a a a A1 3 4(b) Solve linear equation with signs of 3x different or solve non-modulus equation 2 2 (3 2 ) (3 4 ) + = − x a x a M1 Obtain 1 3 = x a A1 Obtain 3 = y a A1 3 Question Answer Marks 4(c) State 1 3 < x a (FT from part (b)) B1FT 1

This question in 9709/21 May/June 2020

2 (a) Sketch, on the same diagram, the graphs of y 3x and y x . [2] = = −3 (b) Find the coordinates of the point where the two graphs intersect. [3] … … … … … … … … … … … … (c) Deduce the solution of the inequality 3x x . [1] < −3 … … … …

6 marks

Mark scheme: 2(a) Draw V-shaped graph with vertex on positive x-axis B1 Must be straight lines. Draw straight line through origin with positive gradient greater than gradient of first graph, together with a V shaped graph B1 Must have the first B1. 2 2(b) Solve linear equation with signs of 3x and x different or solve non-modulus equation 2 2 (3 ) ( 3) = − x x M1 Obtain 3 4 = x A1 Obtain 9 4 = y A1 And no other point. 3 2(c) State 3 4 < x B1 FT Following their (single) x-coordinate from part (b). 1

This question in 9709/21 Oct/Nov 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

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) 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

2 (a) Sketch the graph of y = 3 x - 7 , stating the coordinates of the points where the graph meets the axes. [2] (b) Hence find the set of values of the constant k for which the equation 3 x - 7 = k ( x - 4) has exactly two real roots. [2] … … … … … … … … … … … … … … …

4 marks

Mark scheme: 2(a) Draw V-shaped graph with vertex on positive x-axis B1 State 3(7 , 0) and (0, 7) B1 Allow if only 73 and 7 shown on relevant axes. 2 2(b) State or imply that gradient of left-hand part of graph is –3 B1 State −3 k  0 B1 Using < and not ⩽. 2

This question in 9709/22 Feb/March 2024

3 (a) Sketch on the same diagram the graphs of y = 3x - 8 and y = 5 - x . [2] (b) Solve the inequality 3x - 8 1 5 - x . [4] … … … … … … … … … … … … … … … (c) Hence determine the largest integer N satisfying the inequality 3 e 0 .1 N - 8 1 5 - e 0 .1 N . [2] … … … … … … … … … … … … … … … … … … … … … …

8 marks

Mark scheme: 3(a) Draw V-shaped graph with vertex on positive x-axis in the first quadrant. B1 Draw correct graph of 5 y x   correctly positioned with respect to modulus graph. B1 Two points of intersection. 2 3(b) Solve 3 8 5 x x    to obtain 13 4 B1 Or inequality. Solve linear equation or inequality with signs of 3x and x the same M1 Obtain 3 2 A1 Conclude 3 13 2 4 x   or 3 2 x  and 13 4 x  A1 Allow alternative notation e.g.   3 13 2 4 , . Alternative Method for Question 3(b) State or imply non-modulus equation (or inequality) 2 2 (3 8) (5 ) x x    (B1) Attempt solution of three-term equation (or inequality) (M1) Obtain 3 2 and 13 4 (A1) Conclude 3 13 2 4 x   or 3 2 x  and 13 4 x  (A1) Allow alternative notation e.g.   3 13 2 4 , . 4 3(c) Attempt value of N (maybe non-integer at this stage) for 0.1 13 4 e N their  M1 Allow 0.1 13 4 e N their  (or inequality). Conclude with single integer 11 A1 2

This question in 9709/21 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

2 (a) Sketch on the same diagram the graphs of y = 2x - 9 and y = 4x - 5 . [2] (b) Solve the inequality 2x - 9 1 4 x - 5 . [3] … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw correct graph of y = 4 x − 5 B1 Correctly placed with reference to modulus graph and with steeper gradient. 2 2(b) Solve linear equation or inequality with signs of 2x and 4x different M1 Obtain −2 x + 9 = 4 x − 5 and hence x = 73 A1 OE Conclude x  7 A1  7   7  3 OE, e.g. ,  , or ,  .      3   3  Alternative Method for Question 2(b) State or imply non-modulus equation (or inequality) (2 x − 9) 2 = (4 x − 5) 2 B1* Attempt solution of three-term quadratic equation (or inequality) DM1 12 x 2 − 4 x − 56 = 0 leading to 73 and −2. Obtain at least 7 and conclude x  7 A1 Must be from correct work. 3 3  7   7  OE, e.g. ,  or ,       3   3  3

This question in 9709/22 May/June 2025

2 (a) Sketch on the same diagram the graphs of y = 2x - 9 and y = 4x - 5 . [2] (b) Solve the inequality 2x - 9 1 4 x - 5 . [3] … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw correct graph of y = 4 x − 5 B1 Correctly placed with reference to modulus graph and with steeper gradient. 2 2(b) Solve linear equation or inequality with signs of 2x and 4x different M1 Obtain −2 x + 9 = 4 x − 5 and hence x = 73 A1 OE Conclude x  7 A1  7   7  3 OE, e.g. ,  , or ,  .      3   3  Alternative Method for Question 2(b) State or imply non-modulus equation (or inequality) (2 x − 9) 2 = (4 x − 5) 2 B1* Attempt solution of three-term quadratic equation (or inequality) DM1 12 x 2 − 4 x − 56 = 0 leading to 73 and −2. Obtain at least 7 and conclude x  7 A1 Must be from correct work. 3 3  7   7  OE, e.g. ,  or ,       3   3  3

This question in 9709/23 May/June 2025

2 (a) Sketch on the same diagram the graphs of y = 2x - 9 and y = 4x - 5 . [2] (b) Solve the inequality 2x - 9 1 4 x - 5 . [3] … … … … … … … … … … … … … … …

5 marks

Mark scheme: 2(a) Draw V-shaped graph with vertex on positive x-axis B1 Draw correct graph of y = 4 x − 5 B1 Correctly placed with reference to modulus graph and with steeper gradient. 2 2(b) Solve linear equation or inequality with signs of 2x and 4x different M1 Obtain −2 x + 9 = 4 x − 5 and hence x = 73 A1 OE Conclude x  7 A1  7   7  3 OE, e.g. ,  , or ,  .      3   3  Alternative Method for Question 2(b) State or imply non-modulus equation (or inequality) (2 x − 9) 2 = (4 x − 5) 2 B1* Attempt solution of three-term quadratic equation (or inequality) DM1 12 x 2 − 4 x − 56 = 0 leading to 73 and −2. Obtain at least 7 and conclude x  7 A1 Must be from correct work. 3 3  7   7  OE, e.g. ,  or ,       3   3  3

This question in 9709/25 May/June 2025