Equations — Paper 4 · IGCSE Mathematics - International 0607

7 y 10 x –4 0 4 –10 f x = 9 - x 2 ^ h (a) On the diagram, sketch the graph of y = f x for values of x between -4 and 4. ^ h, [4] (b) Solve f x = 7 . ^ h … [2] (c) The equation 9 - x 2 = k has two solutions. Find the range of values of k. … [2]

8 marks

Mark scheme: 7(a) Correct Graph 4 B1 for maximum point on or close to y-axis B1 for correct shape between their –3 and 3 10 y f(x)=abs(9-x^2) B1 for mod graph x -4 4 -10 7(b) [x =] ±4, ± 2 2 B1 for any 2 correct answers or ± 1.41 or ± 1.414... 7(c) k > 9 2 B1 for each k = 0

This question in 0607/41 May/June 2017

11 Vito lives in Sicily. Table A shows the distances, in km, between different towns. Table B shows the average speed, in km/h, that Vito drives his car between towns. Table A (distances, in km) Agrigento Catania Messina Palermo Trapani Agrigento 175 275 155 170 Catania 175 100 215 325 Messina 275 100 225 330 Palermo 155 215 225 110 Trapani 170 325 330 110 Table B (average speeds, in km/h) Agrigento Catania Messina Palermo Trapani Agrigento 90 110 75 100 Catania 90 120 95 90+ x Messina 110 120 105 80 Palermo 75 95 105 30 + 2x Trapani 100 90+ x 80 30 + 2x (a) (i) Write down the distance from Agrigento to Messina. … km [1] (ii) Find the time taken for Vito to drive from Agrigento to Messina. … hours [2] (b) On another day, Vito drives from Agrigento to Trapani. He arrives at Trapani at 10 42. At what time did he leave Agrigento? … [3] (c) One day Vito drives from Catania to Palermo. Vito’s car uses fuel at the rate of 12.5 km/litre. The cost of fuel is 1.432 euros per litre. Find the cost of this journey. … euros [3] (d) The time for Vito to drive from Catania to Trapani is 1 12 hours longer than the time for Vito to drive from Palermo to Trapani. (i) Show that x 2 - 75x + 1400 = 0 . [5] (ii) Find the two possible average speeds that Vito drives from Catania to Trapani. … km/h, … km/h [3]

17 marks

Mark scheme: 11(a)(i) 275 1 11(a)(ii) 2.5 oe 2 M1 for 275 ÷ 110 11(b) 09 00 oe 3 B2 for 1 h 42 mins or 102 mins soi or M1 for 170 ÷ 100 oe If 0 scored, SC1 for correct conversion of their decimal time into hours and mins 11(c) 24.6 or 24.63... 3 215 M2 for × 1.432 12.5 215 1.432 or M1 for or 215 × 1.432 or soi 12.5 12.5 11(d)(i) 325 110 3 M2 325 110 – = oe or M1 for or 90 + x 30 + 2x 2 90 + x 30 + 2x 650(30 + 2x) – 220(90 + x) M1 Dependent on first equation containing the three = 3(90 + x)(30 + 2x) oe terms. Correctly eliminating fractions Correct completion to A2 B1 for 2700 + 180x + 30x + 2x2 soi x2 – 75x +1400 with no errors or omissions 11(d)(ii) 125 and 130 3 B2 for one or for 35 and 40 or B1 for 35 or 40 −−( 75) ± ( − 75) 2 − ( 4 ) (1)(1400) or M1 for 2 × 1 or sketch of parabola with two positive zeros or (x – 35)(x – 40)

This question in 0607/41 May/June 2017

6 (a) (i) x is proportional to v. Write down an expression for x in terms of v and a constant c. x = … [1] (ii) y is proportional to v2. Write down an expression for y in terms of v and a constant k. y = … [1] (iii) d = x + y Write down an expression for d in terms of v, c and k. d = … [1] (b) The table shows two values of v and the corresponding values of d. v d 12 750 20 2050 Using your answer to part (a)(iii), (i) show that 125 = 2c + 24k, [1] (ii) write down a second equation connecting c and k. … [1] (c) Solve the simultaneous equations in part (b) to find the value of c and the value of k. c = … k = … [3] (d) Find the value of d when v = 40. d = … [2]

10 marks

Mark scheme: 6(a)(i) [x =] cv oe 1 6(a)(ii) [y =] kv2 oe 1 6(a)(iii) [d =] cv + kv2 or v(c + kv) oe 1 FT 6(b)(i) 750 = 12c + 122 k oe M1 isw any cancelling 6(b)(ii) 2050 = 20c + 202 k oe 1 isw any cancelling 6(c) [c =] 2.5 oe cao 3 M1 for correctly eliminating one variable from [k =] 5 cao their equations in this part. or sketches of lines A1 for either solution If zero scored SC1 for their values satisfying one equation. 6(d) 8100 2 M1 for correct substitution of 40 into their (a)(iii) containing their values of c and k.

This question in 0607/42 May/June 2017

9 C x cm NOT TO SCALE 60° A B (x + 2) cm In the diagram AC = x cm, AB = (x + 2) cm and angle A = 60°. (a) (i) Find an expression, in terms of x, for the area of triangle ABC. Give your answer in surd form. … cm2 [2] (ii) The area of triangle ABC = 18 3 cm2. Show that x2 + 2x – 72 = 0. [2] (b) (i) Solve the equation x2 + 2x – 72 = 0. x = … or x = … [2] (ii) Find the shortest distance between the line AB and the point C. … cm [2]

8 marks

Mark scheme: 9(a)(i) 1 3 2 1 x + 2 ) × sin60 × x × ( x + 2 ) × oe or better M1 for × x × ( 2 2 2 final answer 9(a)(ii) equating to 18 3 and correct M1 Dependent on correct answer used from (a)(i) or answer to (a)(i) contains sin60 but is elimination of 3 otherwise correct. Completion with at least one step A1 No errors or omissions 9(b)(i) 7.54 or 7.544... , –9.54 or –9.544... 2 B1 for each If 0 scored, M1 for substitution in formula or sketch or (x + 1)2 – 73 or better 9(b)(ii) 6.53 or 6.54 or 6.529 to 6.536... 2 [ ] M1 for sin 60 = oe their 7.54

This question in 0607/42 May/June 2017

6 y 6 x 0 10 f(x) = x - 5 log x (a) On the diagram, sketch the graph of y = f(x) for 0 1 x G 10 . [2] (b) Find the co-ordinates of the local minimum point. ( … , … ) [2] (c) Find the range of f(x) for the domain 1 G x G 5 . … [2] (d) Solve the equation f(x) = 2. x = … or x = … [2] (e) Solve the inequality f(x) 1 2. … [1] (f) (i) Find f(0.001), f(0.000 01) and f(0.000 000 1). f(0.001) = … , f(0.000 01) = … , f(0.000 000 1) = … [1] (ii) Complete the statement. The y-axis is … to the graph of y = f(x). [1]

11 marks

Mark scheme: 6(a) Correct sketch 2 B1 for correct shape 6666 5555 4444 3333 2222 1111 0000 0000 2222 4444 6666 8888 10101010 6(b) (2.17, 0.488) or (2.171…, 0.4877…) 2 B1 for each 6(c) 0.488 - f ( x ) - 1.51 2 FT their 0.488 or 0.4877... - f ( x ) - 1.505... B1 for 0.488 - f ( x ) oe or f ( x ) - 1.51 oe 6(d) 0.502 or 0.5015… 2 B1 for each 5.83 or 5.827… 6(e) 0.502 < x < 5.83 1 FT their (d) or 0.5015... < x < 5.827... 6(f)(i) 15.[0] or 15.00… 1 25.[0] or 25.00… 35. [0] or 35.00… 6(f)(ii) [an] asymptote oe 1

This question in 0607/43 May/June 2017

10 (a) The time, t hours, taken by each of 200 cars to complete a journey of 200 km is recorded. The results are shown in the table. Time (t hours) 2.5 1 t G 3 3 1 t G 3.25 3.25 1 t G 3.75 Frequency 60 100 40 (i) Calculate an estimate of the mean. … h [2] (ii) On the grid, draw the histogram to show the information in the table. 450 400 350 300 Frequency 250 density 200 150 100 50 0 t 2.5 2.6 2.7 2.8 2.9 3.0 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 Time (hours) [3] (b) One car completes the 200 km journey at an average speed of x km/h. Another car completes the 200 km journey at an average speed of (x + 10) km/h. The difference between the times taken by the two cars is 20 minutes. (i) Show that x 2 + 10x - 6000 = 0 . [4] (ii) Find the time taken for the slower journey. Give your answer in hours and minutes correct to the nearest minute. … h … min [4] Question 11 is printed on the next page.

13 marks

Mark scheme: 10(a)(i) 3.0875 2 M1 for 2.75, 3.125, 3.5 soi 10(a)(ii) Correct histogram 3 B1 correct widths B1 for two correct heights 10(b)(i) 200 200 20 B2 200 200 − = oe B1 for or x x + 10 60 x x + 10 60 × 200( x + 10) − 60 × 200 x = 20 x ( x + 10) M1 i.e. correctly clearing fractions or all over oe common denominator x 2 + 10 x − 6000 = 0 A1 completion with at least one interim line and without any errors or omissions 10(b)(ii) 2 h 45 min 4 B2 for 72.6 or 72.62… or M1 for correct use of formula or correct sketch M1 for 200 ÷ their positive x, implied by 2.75 …

This question in 0607/43 May/June 2017

7 (a) Ali walks for 1 hour at x km/h and then for 2 hours at x + km/h. 4 He walks a total distance of 8 km. Write an equation and solve it to find the value of x. x = … [3] (b) NOT TO SCALE x (x – 2) (x – 2) x 2x x The volume of the cube is equal to the volume of the cuboid. (i) Show that x 3 - 8x 2 + 8x = 0 . [3] (ii) y 40 0 x 7.5 –40 On the diagram, sketch the graph of y = x 3 - 8x 2 + 8x for 0 G x G 7.5 . [2] (iii) Find the volume of the cuboid. … [2]

10 marks

Mark scheme: 7(a)  1  M2  1  x + 2  x +  = 8 oe M1 for 2  x +  oe seen  4   4  2.5 B1 7(b)(i) x 3 = 2 x ( x − 2)( x − 2) oe M1 [( x − 2) 2 = ] x 2 − 2 x − 2 x + 4 B1 Allow – 4x for – 2x – 2x 3 2 2 Allow – 8x2 for – 4x2 – 4x2 or 2 x − 4 x − 4 x + 8 x leading to x 3 − 8 x 2 + 8 x = 0 A1 Final equation reached without any errors or omissions 7(b)(ii) Correct sketch 2 B1 for correct shaped cubic with max 40404040 before min 20202020 0000 0000 2222 4444 6666 -20-20-20-20 -40-40-40-40 7(b)(iii) 318 or 319 or 318.3 to 318.7 2 B1 for 6.83 or 6.828… seen isw use of other values (1.1715…)

This question in 0607/42 Oct/Nov 2017

10 (a) Solve the equation 4x 2 = 12 - 3x . Give your answers correct to 2 decimal places. You must show all your working. x = … or x = … [4] (b) Solve the inequality 4x 2 2 12 - 3x . … [2] (c) Solve the inequality 4x 2 + 5 G 12 - 3x . … [4]

10 marks

Mark scheme: 10(a) appropriate sketch giving one M2 M1 for sketch of parabola or parabola and positive and one negative answer or 2 −±3 … fully correct use of formula straight line or 3 − 4(4)( −12) or 2(4) oe 1.4[0] and –2.15 final answers B2 B1 for each If 0 scored B1 for 1.397… and –2.147... or SC1 for 2.15 and –1.4[0] 10(b) x > 1.40 and x < −2.15 2 FT [ x ] > their max(a), [ x ] < their min(a) B1 for each 10(c) −1.75 - x - 1 nfww 4 B3 for 1, − 1.75 oe B2 for 1 inequality correct B1 for 1 correct value seen or M2 for appropriate sketch or correct factorising or correct use of formula or M1 for 4 x 2 + 3 x − 7 - 0

This question in 0607/43 Oct/Nov 2017

11 (a) Solve the simultaneous equations. You must show all your working. 3x - 2y = 11 4x - 5y = 10 x = … y = … [4] (b) Use your answers to part (a) to solve the simultaneous equations. 3a - 2b = 22 4a - 5b = 20 a = … b = … [2] (c) (i) Use your answers to part (a) to find the exact answers to these simultaneous equations. 3 # 10 p - 2 # 10 q = 11 4 # 10 p - 5 # 10 q = 10 p = … q = … [3] (ii) Find the value of p + q . … [1]

10 marks

Mark scheme: 11(a) [x = ] 5 4 M1 for correctly equating one set of coefficients [y = ] 2 with correct working M1 for correct method to eliminate one variable OR M1 for equation x = or y = from one equation M1 for correct substitution into other equation B1 for x = 5 B1 for y = 2 If zero scored SC1 for correct subst into one of original equs and evaluation to find other variable 10(b) [ a = ]10 2 B1 for each FT their (a) × 2 [b = ] 4 10(c)(i) [ p = ] log 5 and [ q = ] log 2 3 B2 FT their (a) for either seen Final answers or B1 FT for each correct decimal answer 0.699 or 0.6989 to 0.6990 0.301 or 0.3010... or M1 for 10 p = their 5 or 10 q = their 2 10(c)(ii) 1 cao 1

This question in 0607/43 Oct/Nov 2017

11 f ()x = 10 x g ()x = 2x - 1 (a) Find the value of g(3). … [1] (b) Find the range of f (x) for the domain {-1, 0, 1, 2}. { … } [2] (c) Find x when g ()x = 12 . x = … [2] 2 (d) The graph of y = g (x) is translated by the vector onto the graph of h(x). e3o Find h(x). Give your answer in its simplest form. h(x) = … [3] (e) Find f -1 ()x . f -1 ()x = … [2] (f) tan (g (x)) = 1 and 0° G x G 180° . Find the two values of x. x = … or x = … [3]

13 marks

Mark scheme: 11(a) 5 1 11(b) 0.1oe , 1, 10, 100 2 B1 for 3 correct or all correct seen and spoilt. 11(c) 6.5 oe 2 M1 for 2x – 1 = 12 11(d) 2x – 2 or 2(x – 1) 3 B2 for correct unsimplified answer OR M1 for substituting x − 2 for x M1 for adding 3 to a function in x oe OR M1 for y = 2x + c (c ≠ – 1) leading to answer with gradient 2 M1 for substituting coords of valid point into y = 2x + c 11(e) log x 2 M1 for log y = x or x = 10y 11(f) 23, 113 3 B2 for 23 or B1 for [g(x) =] 45 soi

This question in 0607/41 May/June 2018

2 y 5 x −5 0 5 −5 x f (x) = 1 - 2 (x - 9 ) (a) On the diagram, sketch the graph of y = f ( x) , for values of x between -5 and 5. [3] (b) Write down the equations of the three asymptotes. … , … , … [3] x (c) The line y = x intersects the curve y = 1 - 2 three times. (x - 9 ) Find the values of the x co-ordinates of the points of intersection. x = … or x = … or x = … [3]

9 marks

Mark scheme: 2(a) Correct sketch 3 B1 for each branch 2(b) y = 1, x = 3, x = − 3 3 B1 for each 2(c) –2.87 or –2.874 to –2.873 3 B1 for each 1.15 or 1.149 to 1.150 If 0 scored SC1 for –2.9, 1.1 and 2.7 2.72 or 2.723 to 2.724

This question in 0607/42 May/June 2018

5 y 10 x –6 0 6 –10 2x 2 - x + 5 f(x) = ^ 2 h x + x - 6 ^ h (a) On the diagram, sketch the graph of y = f(x) for values of x between -6 and 6. [3] (b) Find the co-ordinates of the local maximum. ( … , … ) [2] (c) Find the equations of the three asymptotes to the graph of y = f(x) . … , … , … [3] (d) The equation f(x) = k has no solutions. Find the range of values of k. … [2] (e) g(x) = x + 1 (i) Solve f(x) = g(x). x = … or x = … [2] (ii) Solve the inequality f(x) 2 g(x). … [2]

14 marks

Mark scheme: 5(a) Correct sketch 3 10 y f(x)=(2x^2 - x + 5)/((x-2)(x+3)) 5 B1 for each branch x -6 -4 -2 2 4 6 -5 -10 5(b) (0.0295, – 0.833) 2 or (0.02948 to 0.02949, –0.8329...) B1 for each 5(c) x = –3, x = 2, y = 2 3 B1 for each 5(d) –0.833 < k ⩽ 2 2 FT their (b) B1 for each inequality 5(e)(i) –5.13, 2.81 2 –5.131..., 2.812 to 2.813 B1 for each 5(e)(ii) –5.13 < x < –3, 2 –5.131..., 2.812 to 2.813 2 < x < 2.81 B1 for each FT their (c) and (e)(i)

This question in 0607/43 May/June 2018

10 Isabel drives from Geneva to Rome, a distance of 930 km. Her average speed is x km/h. (a) Write down an expression, in terms of x, for the time, in hours, the journey takes. … h [1] (b) She returns from Rome to Geneva along the same route at an average speed of (x + 5) km/h. 1 The journey takes hour less than the journey from Geneva to Rome. 2 (i) Write down an equation, in terms of x, and show that it simplifies to x2 + 5x – 9300 = 0. [3] (ii) Solve this equation. Give your answers correct to 1 decimal place. x = … or x = … [3] (iii) Find the time taken for the journey from Rome to Geneva in hours and minutes. … h … min [2]

9 marks

Mark scheme: 10(a) 930 1 x 10(b)(i) 930 930 M1 – soi x x+ 5 1860(x + 5) – 1860x = x(x + 5) oe M1 FT dep on 1st M1with 12 as separate term in equation Completion to x2 + 5x – 9300 = 0 A1 with no errors 10(b)(ii) 94[.0] or 93.96 to 93.97 3 2 5 ± 5 − 4(1)( − 9300) –99[.0] or –98.97 to –98.96 M1 for 2 or sketch of parabola B1 for 1 correct 10(b)(iii) 9 [h] 24 or 23 to 24 [min] 2 M1 for 930 ÷ (their 94 + 5) oe If 0 scored SC1 for 9 h 53min to 9 h 54 min

This question in 0607/43 May/June 2018

1 (a) Solve the following equations. (i) 12 - x = 4 x = … [1] (ii) 9x - 4 = 6x + 8 x = … [2] 12 (iii) + 5 = 9 x x = … [2] (b) (i) Solve 6x 2 - 5x + 1 = 0 . x = … or x = … [3] (ii) Use your answer to part (b)(i) to solve 6 sin 2 x - 5 sin x + 1 = 0 for 0° G x G 90 ° . x = … or x = … [3]

11 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 8 1 1(a)(ii) 4 2 M1 for correct 1st step 1(a)(iii) 3 2 M1 for correct 1st step 1(b)(i) 1 1 3 M2 for (3 x − 1)(2 x − 1) [ = 0] [ x = ] , 2 3 or M1 for ( ax ± 1)(bx ± 1) where ab = 6 or a + b = –5 or 3 x (2 x − 1) − 1(2 x − 1) or 2 x (3 x − 1) − 1(3 x − 1) OR M2 for correct sketch or M1 for any U-shaped parabola crossing x-axis twice OR 5 ± ( −5) 2 − 4 × 6[× 1] M2 for 2 × 6 b 2 or M1 for or b − 4 ac correct 2 a 1(b)(ii) 30, 19.5 or 19.47… 3 B2 FT for one correct answer 1  or M1 for sin x = their   2  1  sin x = their    3 

This question in 0607/41 Oct/Nov 2018

5 y 30 x –4 0 4 –20 f ()x = x 3 - 12x + 6 (a) On the diagram, sketch the graph of y = f ( x) for -4 G x G 4 . [2] (b) Find the positive zeros of f(x). … [2] (c) Find the co-ordinates of (i) the local maximum, ( … , … ) [1] (ii) the local minimum. ( … , … ) [1] (d) Describe fully the symmetry of the graph of y = f(x). … … [3]

9 marks

Mark scheme: 5(a) Correct sketch 2 y f(x)=x^3-12x+6 30 20 B1 for any cubic with max on left of min 10 x -4 4 -10 -20 5(b) 0.511 or 0.5111... 2 B1 for each 3.18 or 3.180... 5(c)(i) (–2, 22) 1 5(c)(ii) (2, –10) 1 5(d) Rotation[al] 3 B1 for each [Order] 2 [About] (0, 6)

This question in 0607/42 Oct/Nov 2018

9 In this question all lengths are in centimetres. x NOT TO SCALE 15 20 x x x x x The diagram shows a picture frame with three pictures. The frame and the pictures are rectangles. Each picture measures 20 cm by 15 cm. The width of the borders between each picture and between each picture and the frame are all x cm. The total area of the frame is 2208 cm2. (a) Show that 4x 2 + 85x - 654 = 0 . [3] (b) Solve the equation 4x 2 + 85x - 654 = 0 . You must show all your working. x = … or x = … [3] (c) Find the dimensions of the picture frame. Length … cm Height … cm [2]

8 marks

Mark scheme: 9(a) (45 + 4x)(20 + 2x) = 2208 M1 900 + 90x + 80x + 8x2 B1 For expansion Completion to 4x2 + 85x – 654 = 0 A1 with no errors or omissions 9(b) 2 M1 or (x – 6)(4x + 109) − 85 ± 85 − 4(4)( − 654) or sketch of parabola (+x2) with one positive 2 × 4 zero and one negative 6, –27.25 oe B2 B1 for each 9(c) Length = 69 B2 B1FT for each Height = 32

This question in 0607/41 May/June 2019

11 y 6 –2 0 7 x –6 (x + 2) f (x) = (x - 1)(x - 4) (a) On the diagram, sketch the graph of y = f ( x) for values of x between -2 and 7. [3] (b) Write down the co-ordinates of the local maximum. ( … , … ) [2] (c) Write down the equation of each of the three asymptotes. … , … , … [3] (d) g ()x = x - 5 (i) Solve the equation f (x) = g (x) . x = … or x = … or x = … [3] (ii) Solve the inequality f (x) 2 g (x) . … [3]

14 marks

Mark scheme: 11(a) Correct sketch 3 B1 for each branch y f(x)=(x+2)/((x-1)(x-4)) 6 4 2 x -2 2 4 6 -2 -4 -6 11(b) (2.24, –1.94) 2 or (2.242 to 2.243, –1.943 to –1.942) B1 for each co-ordinate 11(c) x = 1, x = 4, y = 0 3 B1 for each 11(d)(i) 1.34 or 1.344 to 1.345 3 B1 for each 2.79 or 2.789... 5.87 or 5.866... If 0 scored, SC1 for 1.3, 2.8 and 5.9 11(d)(ii) x < 1 3 B1 for each 1.34 < x < 2.79 FT dep on two solutions to (i) between 1 and 4 < x < 5.87 4. FT dep on solution to (i) > 4

This question in 0607/41 May/June 2019

10 (a) Amy buys 3 pencils and 1 ruler and pays 67 cents. Ben buys 2 pencils and 3 rulers and pays 96 cents. Find the cost of 1 pencil and the cost of 1 ruler. You must show all your working. Pencil … cents Ruler … cents [5] (b) In this part, all measurements are in centimetres. NOT TO SCALE x x – 4 x + 1 5x + 3 4 The area of the triangle is the same as the area of the rectangle. (i) Show that 3x 2 - 10x - 48 = 0 . [4] (ii) Factorise 3x 2 - 10x - 48 . … [2] (iii) Find the area of the triangle. … cm2 [2]

13 marks

Mark scheme: 10(a) 3p + r = 67 oe B2 B1 for each 2p + 3r = 96 oe Accept words in equations, using + and =. correctly eliminating one variable M1 [pencil = ] 15 B1 [ruler =] 22 B1 If M0 scored in addition to B0 (for answers) scored then award SC1 for answers satisfying one of their two original equations in 2 variables 10(b)(i) 1  5 x  M2 1  5 x  ( x + )1 x =  + 3  ( x − 4) oe M1 for ( x + )1 x or  + 3  ( x − 4) 2  4  2  4  5 x 2 B1 i.e. correct expansion for rectangle − 5 x + 3 x − 12 oe 4 3 x 2 − 10 x − 48 = 0 reached with no A1 Dependent on B1 errors or omissions including at least one line of working 10(b)(ii) (3 x + 8)( x − 6 ) 2 B1 for 3 x ( x − 6 ) + 8( x − 6 ) or x (3 x + 8) − 6(3 x + 8) or (3 x + a )( x + b ) with ab = – 48 or a + 3b = – 10 10(b)(iii) 21 2 FT 1 2 × (their positive x ) × ( their positive x + 1) if x > 4 M1 12 × (their positive x) × (their positive x + 1) if x > 4

This question in 0607/42 May/June 2019

9 (a) Solve the following equations. 135 (i) = 5 x x = … [1] (ii) 3x + 5 = 7x + 25 x = … [2] (iii) 8x 2 = 11 - 2x x = … or x = … [4] (b) Solve the following inequalities. (i) 6 - 2x H 10 … [2] 1 (ii) 2 3 x - 2 … [3] (c) Solve the simultaneous equations. You must show all your working. 3x + 5y =-3 5x - 2y = 26 x = … y = … [4] (d) Solve the equation. log x + 4 log 2 = log 13 x = … [3]

19 marks

Mark scheme: 9(a)(i) 27 1 9(a)(ii) –5 2 M1 for 5 − 25 = 7 x − 3x or better 9(a)(iii) 1.05 or 1.054… 4 2 −±2 2 −×4 8 ×−11 –1.3[0] or –1.304… M3 for 2 × 8 or correct sketch which would lead to solution. b 2 or M2 for correct or b − 4 ac correct 2a or M1 for 8 x 2 + 2 x − 11 or − 8 x 2 − 2 x + 11 or sketch of 8 x 2 or 11 − 2x 9(b)(i) x - − 2 oe 2 M1 for 6 − 10 . 2x or − 2 x . 10 − 6 or 3 − x . 5 or better If 0 scored SC1 for x . − 2 or x = –2 9(b)(ii) 1 3 7 2 < x < 2 oe M2 for x = 2 and x = 3 3 or correct sketch which would lead to solution. or M1 for 1 > 3( x − 2) or better or sketch of 1 y = x − 2 1 or B1 for x < 23 or for x > 2 9(c) Correctly equating one set of M1 coefficients oe Correct method to eliminate one M1 variable [x=] 4 B1 [y=] –3 B1 If 0 scored SC1 for correct substitution into one of original equations and evaluation to find other variable. 9(d) 13 3 4 or 0.8125 M1 for log2 or better 16 p M1 for correct use of log p − log q = log q or use of log p + log q = log pq

This question in 0607/43 May/June 2019

2 f ()x = , x ! 2 g ()x = x + 2 h ()x = x 2 x - 2 (a) Find f (6 ). … [1] (b) Solve f (x) =- 2. x = … [2] (c) Find h (g (x)). … [1] (d) Solve h (g (x)) = h (x) + 2. x = … [4] (e) Find f - 1 (x). f - 1 ()x = … [3] (f) y 9 x –3 0 3 –3 (i) On the diagram, sketch the graph of y = f (x) and the graph of y = h (x) for values of x between - 3 and 3. [3] (ii) Write down the equation of the line of symmetry of y = h (x). … [1] (iii) Solve f (x) 2 h (x). … [2]

17 marks

Mark scheme: 2(a) 0.25 oe 1 2(b) 1.5 2 1 M1 for 1 = –2(x –2) or −= x – 22 2(c) (x + 2)2 1 2(d) 1 4 B1 for x 2 + 4 x + 4 –0.5 or − 2 M2 for 4 x + 4 = 2 or M1 for their ( c ) = x 2 + 2 or M2 for correct sketch or M1 for any U-shaped parabola 2(e) 1 3 1 1 + 2 oe final answer M2 for x = + 2 or xy = 1 + 2 y or y – 2 = x y x 1 or M1 for x − 2 = or y ( x − 2) = 1 y 1 or x = y − 2 2(f)(i) Correct sketches 3 B1 for correct quadratic shape through origin B2 for correct rectangular hyperbola shape or B1 for one branch 2(f)(ii) x = 0 1 2(f)(iii) 2 < x < 2.21 or 2.205 to 2.206 2 B1 for each part or 2 and 2.21 or 2.205 to 2.206 seen

This question in 0607/42 Oct/Nov 2019

4 (a) Solve the following equations. (i) 2x - 3 =- 11 x = … [2] 36 (ii) =- 4 x x = … [2] (iii) 6x + 13 = 17 - 2x x = … [2] (b) Solve the simultaneous equations. You must show all your working. 5x + 3y =- 19 3x + 5y =- 21 x = … y = … [4]

10 marks

Mark scheme: 4(a)(i) –4 2 3 11 M1 for 2x = –11 + 3 or x – = oe 2 2 4(a)(ii) –9 2 36 M1 for 36 = –4x or = x oe −4 4(a)(iii) 0.5 oe 2 M1 for 6 x + 2 x = 17 − 13 oe 4(b) Correctly equating one set of M1 Allow correct sketches, i.e. two lines with negative coefficients gradients OR x = … or y = … from one equation Correct method for eliminating one M1 variable OR correct substitution into other equation [x = ] –2 B2 B1 for each [y = ] –3 If 0 scored, SC1 for correct substitution in one of the original equations to find other variable

This question in 0607/42 Oct/Nov 2019

12 y 4 x –3 0 3 –4 (a) On the diagram, sketch the graph of y = f ( x), where 1 f (x) = for values of x between - 3 and 3. x (x - 1)(x + 1) [4] (b) Write down the equations of the asymptotes. … , … , … , … [3] (c) Write down the co-ordinates of the local maximum. ( … , … ) [2] (d) The line y = 2x + 1 intersects the curve y = f (x) twice. Find the value of the x co-ordinate of each point of intersection. x = … or x = … [2]

11 marks

Mark scheme: 12(a) Correct sketch 4 B1 for each branch 12(b) x = 0 3 B2 for three correct x = 1 or B1 for one correct x = –1 y = 0 12(c) (0.577, –2.6[0]) 2 B1 for each or (0.5773 to 0.5774, –2.598…) 12(d) [x = ] –1.24 or –1.242 to –1.241 2 B1 for each [x =] 1.13 or 1.127 to 1.128

This question in 0607/42 Oct/Nov 2019

3 y 15 x –1.5 0 3 –15 f ()x = 2x 3 - 5x 2 + 3 for - 1.5 G x G 3 (a) On the diagram, sketch the graph of y = f (x). [2] (b) Find the zeros of f(x). … [3] (c) Find the co-ordinates of the local maximum. ( … , … ) [1] (d) Find the co-ordinates of the local minimum. ( … , … ) [2] (e) The equation 2x 3 - 5x 2 + 3 = k has three solutions. Find the range of values of k. … [2]

10 marks

Mark scheme: 3(a) Correct sketch 2 B1 for cubic curve (+ x3) with 2 15 y f(x)=2x^3 -5x^2+3 turning points x -2 -1 1 2 3 -15 3(b) –0.686 or –0.6861..., 3 B1 for each 1, 2.19 or 2.186... If 0 scored, SC1 for three correct but in coordinate form (…, 0) 3(c) (0, 3) 1 3(d) (1.67, –1.63) or 2 B1 for each co-ordinate (1.666 to 1.667, –1.630 to –1.629) 3(e) –1.63 < k < 3 2 FT their y co-ords from (c) and (d) B1 for each

This question in 0607/43 Oct/Nov 2019

10 All lengths in this question are in metres and all areas are in square metres. 2x + 3 NOT TO SCALE The length of this rectangle is (2x + 3) and the area is 840. (a) Write down an expression, in terms of x, for the width of the rectangle. … [1] (b) The perimeter of the rectangle is 118. Show that 2x 2 - 53x + 336 = 0. [3] (c) Solve the equation 2x 2 - 53x + 336 = 0. Show all your working. x = … or … [3] (d) Find the length and the width of the rectangle. Length = … m Width = … m [2]

9 marks

Mark scheme: 10(a) 840 1 2 x + 3 10(b) 840 M1 2(2x + 3) + 2 their × = 118 oe 2 x + 3 2(2x + 3)2 + 1680 = 118(2x + 3) oe M1 Clearing fractions Correct completion to A1 No errors or omissions 2x2 – 53x + 336 = 0 10(c) (2x – 21)(x – 16) = 0 M1 2 −−( 53) ± ( − 53) − 4(2)(336) or x = 2 × 2 or sketch of parabola (+ve x2, +ve zeros) 10.5, 16 B2 B1 for each 10(d) 35 2 B1 for each 24 If 0 scored, SC1 for a pair of values with a product of 840 or a sum of 59

This question in 0607/43 Oct/Nov 2019

7 y 9 x – 6 0 2 – 3 1 (a) f ( )x = 2 + x + 2 (i) On the diagram, sketch the graph of y = f ( x) for values of x between - 6 and 2. [2] (ii) Write down the coordinates of the points where the graph crosses the axes. ( … , … ) and ( … , … ) [2] (iii) Write down the equations of the asymptotes of the graph. … , … [2] (b) g ( x) = ( x + 4) 2 On the diagram, sketch the graph of y = g ( x) for - 6 G x G - 1 . [2] (c) Solve the equation. f ( x) = g ( x) … [3] (d) Solve the inequality. f ( x) H g ( x) … [2]

13 marks

Mark scheme: 7(a)(i) 8 y f(x)=2+1/(x+2) 2 B1 for correct ‘hyperbolic shape’ 7 6 B1 for intersects with axes correct 5 4 (approx.) 3 2 1 x -5 -4 -3 -2 -1 1 -1 -2 7(a)(ii) (–2.5, 0) 2 B1 for each (0, 2.5) 7(a)(iii) x = –2 2 B1 for each y = 2 y f(x)=2+1/(x+2)f(x)=(x+4)^2 5 7(b) 4 2 B1 for correct ‘quadratic shape’ 3 B1 for min point at (–4, 0) (approx.) 2 1 -5 -4 -3 -2 -1 1 x -1 -2 -3 -4 -5 7(c) [x =] – 5.30 3 B1 for each correct answer [x =] –3 [x =] –1.70 7(d) −5.30 ≤ x ≤−3 2 B1 for each and −<2 x ≤−1.70

This question in 0607/42 May/June 2020

11 (a) Solve the equations. (i) 5 + 2x = 1 x = … [2] 10 (ii) 6 - = 1 x x = … [2] (iii) 3 ( 1 - 2x) = 2 - 4 ( x - 7) x = … [3] (b) (i) Solve 6x 2 = 7 - 3 x . Give your answers correct to 3 decimal places. You must show all your working. x = … or x = … [4] (ii) Solve 6y 4 = 7 - 3 y 2 . Give your answers correct to 3 decimal places. y = … or y = … [2] (c) Solve 2 log x + log 5 = 1. x = … [4]

17 marks

Mark scheme: 11(a)(i) –2 2 5 1 M1 for 2 x = 1 − 5 or + x = 2 2 11(a)(ii) 2 2 10 M1 for − = 1 − 6 oe or 6 x − 10 = x x 11(a)(iii) –13.5 3 M1 for correct expansion 3 − 6 x = 2 − 4 x + 28 M1 for correct collection of their terms 3 − 30 = 6 x − 4 x their (3 − 30) M1 for their (6 − 4) 11(b)(i) 6 x 2 + 3 x − 7 = 0 B1 Correct sketch M2 M1 for any U-shaped parabola OR OR b −±3 32 − 4 × 6 × −7 for or b 2 − 4 ac correct 2 a 2 × 6 0.859, –1.359 B1 11(b)(ii) 0.927, 2 FT their (b)(i) –0.927 B1 for each 11(c) 1.41 or 1.414… cao 4 M3 for 5 x 2 = 10 or M2 for log5x 2 [=1] or M1 for logx2 + log5 [=1]

This question in 0607/42 May/June 2020

4 y 30 x – 3 0 5 – 40 f ( )x = x 3 - 4x 2 - 3x + 18 (a) On the diagram, sketch the graph of y = f ( x) for - 3 G x G 5 . [2] (b) Solve the equation f ( )x = 10 . x = … , or x = … , or x = … [3] (c) Write down the coordinates of (i) the local maximum, ( … , … ) [2] (ii) the local minimum. ( … , … ) [1] (d) f ( )x = k has only 1 solution. Find the ranges of values of k . … [2]

10 marks

Mark scheme: 4(a) Correct Sketch 2 With maximum in second quadrant 30 y f(x)=x^3-4x^2-3x+18 and minimum on positive x-axis 20 B1 for cubic graph for +ve x3 10 x -3 -2 -1 1 2 3 4 5 -10 -20 -30 -40 4(b) –1.51 or –1.508 to –1.507 3 B1 for each 1.24 or 1.244... 4.26 or 4.263 to 4.264 4(c)(i) (–0.333, 18.5) or 2 B1 for each coordinate (–0.3333..., 18.51 to 18.52) 4(c)(ii) (3, 0) 1 4(d) k < 0, 2 B1FT for each k > 18.5

This question in 0607/43 May/June 2020

10 In this question, all lengths are in centimetres. NOT TO SCALE 2x + 4 2x + 1 30° 4x 4x + 5 The areas of the two triangles are equal. (a) Show that 8x 2 + 18 x - 5 = 0 . [5] (b) Solve 8x 2 + 18 x - 5 = 0 . You must show all your working. x = … or x = … [3] (c) Find the area of each of the triangles. … cm2 [2]

10 marks

Mark scheme: 10(a) 1 M2 M1 for either area × 4 x ( 2 x + 4 ) = 2 1 ( 2 x + 1)( 4 x + 5 ) sin30 2 1 M1 sin30 = and eliminating fractions 2 Expanding brackets M1 FT Completion to 8x2 + 18x – 5 = 0 A1 with no errors 10(b) (4x – 1)(2x + 5) = 0 M1 2 − 18 ± 18 − 4 × 8 × ( − 5) or x = 2 × 8 or sketch of parabola (U shaped) with one +ve and one –ve zero. 1 1 A2 A1 for each. , – 2 oe 1 1 4 2 If 0 scored, SC1 for , – 2 4 2 10(c) 2.25 2 M1 for substituting their positive solution in either area formula.

This question in 0607/43 May/June 2020

12 f ( x) = 2x + 3 g ( )x = 5 - 3 x (a) Find f ( 4) . … [1] (b) Solve f ( x) - g ( x) = 5 . x = … [2] (c) Find g -1 ( )x . g -1 ( )x = … [2] (d) Find and simplify f ( g ( x)) . … [2] 2 3 (e) Simplify + . f ( x) g ( x) … [3]

10 marks

Mark scheme: 12(a) 11 1 12(b) 1.4 oe 2 M1 for 2x + 3x = 5 – 3 + 5 12(c) 5 − x 2 M1 for x = 5 – 3y or y – 5 = – 3x oe or oe 3 y 5 = – x oe 3 3 12(d) 13 – 6x 2 M1 for 2(5 – 3x) + 3 12(e) 19 3 M1 for 2(5 – 3x) + 3(2x + 3) final answer M1 for common denominator (2 x + 3)(5 − 3 x ) (2x + 3)(5 – 3x)

This question in 0607/43 May/June 2020

3 (a) Solve the simultaneous equations. You must show all your working. 2x + 5y =- 12 7x - 3y =- 1 x = … y = … [4] (b) Solve ( 4x - 1)( 2x + 3) =- 5 . You must show all your working. x = … or x = … [5]

9 marks

Mark scheme: 3(a) Correctly equating one set of M1 coefficients or correctly expressing one variable in terms of the other Correct method to eliminate one M1 variable [x =] –1 A1 [y =] –2 A1 If 0 scored SC1 for correct substitution into one of original equations and evaluation to find other variable, or for 2 correct answers with no working 3(b) Algebraic method. M1 Correct Expansion 8 x 2 + 12 x − 2 x − 3 [ = −5] 8 x 2 + 10 x + 2 = 0 or better M1 Correctly equating their quadratic to zero (4 x + 1)(2 x + 2) [ = 0] oe M1 or correct use of formula with their 3 (8x + 2)(x + 1) [= 0] term quadratic or correct sketch of parabola [x =] –1 and [x =] –0.25 oe B2 A1 dep on 3rd M1 for either OR GDC method M3 M2 for parabola intersecting y = –5 twice or M1 for parabola OR M3 for correct graph of their rearranged equation [x=] –1and [x=] –0.25 oe B2 A1 for either

This question in 0607/41 Oct/Nov 2020

7 y 5 – 1.5 0 1.5 x – 5 3 1 f ( )x = x - x (a) On the diagram, sketch the graph of y = f ( x) , for values of x between - .15 and 1.5 . [3] (b) Write down the equation of the asymptote of the graph. … [1] (c) Solve the equation f ( )x = 2 for values of x between - .15 and 0. x = … or x = … [2] (d) Solve the inequality f ( )x + x 2 G 2 for values of x between - .15 and 1.5 . … [3]

9 marks

Mark scheme: 7(a) Correct sketch 3 B1 for modulus graph B1 for correct for x > 1, or –1 < x < 0 4444 B1 for x = –1 and 1 when y = 0 plotted 2222 correctly. .5.5.5.5 -1-1-1-1 -0.5-0.5-0.5-0.5 0000 0000 0.50.50.50.5 1111 1.51.51.51.5 Maximum 2 marks if sketch not fully -2-2-2-2 correct -4-4-4-4 7(b) x = 0 1 7(c) –1.4[0] or –1.395… 2 B1 for each –0.475 or –0.4746… 7(d) –1.15 ⩽ x ⩽ –0.536 3 B2 for one fully correct inequality or ––1.154 to –1.153...⩽ x ⩽ –0.5357 or B1 for –1.15 ⩽ x ⩽ – k to –0.5356 – k ⩽ x ⩽ –0.536 or 0.536 ⩽ x ⩽ k AND k ⩽ x ⩽ 1.15 0.536 ⩽ x ⩽ 1.15 or M1 for suitable sketch, or 0.5356 to 0.5357 ⩽ x ⩽ 1.153 to e.g. f(x) + x2 ⩽ 2 1.154 or B1 for 4 correct solutions seen

This question in 0607/41 Oct/Nov 2020

12 Solve the equations. 2 (a) 6 - =- 2 x x = … [3] (b) 3 + 2 ( 4x + 5) = 1 - 2 ( x + 8) x = … [3] (c) 3 log x + 2 log 3 = 2 log 6 + log 2 x = … [3] (d) 2 x = 10 x = … [3]

12 marks

Mark scheme: 12(a) 0.25 oe 3 M2 for 8 x = 2 or − 2 = −8 x or better −2 or M1 for 6 x − 2 = −2 x or = −8 oe x OR M2 for correct sketch that could lead to correct answer or M1 for appropriate but incomplete 2 sketch e.g. 6 −x 12(b) –2.8 oe 3 M2 for 8 x + 2 x = 1 − 16 − 3 − 10 oe or M1 for 3 + 8 x + 10 or 1 − 2 x − 16 12(c) 2 3 M1 for log x 3 or log32 or log6 2 or better M1 for correct use of p log p − log q = log q or log p + log q = log pq 12(d) 3.32 or 3.321 to 3.322 3 log10 1 B2 for or log 2 10 or log2 log2 or M1 for x log2 = log10 OR M2 for correct sketch that could lead to correct answer or M1 for appropriate but incomplete sketch e.g. y = 2 x

This question in 0607/41 Oct/Nov 2020

11 f ( x) = x3 g ( x) = 3x (a) Find g ( 2) - f ( 2) . … [2] 1 (b) Find x when g ( x) = . 9 … [1] 1 (c) Write x - in terms of x. f ( x) Give your answer as a single fraction. … [2] (d) Find f - 1 ( x) . f -1 ( x) = … [1]

6 marks

Mark scheme: 11(a) 1 2 B1 for 32 or 23 11(b) –2 1 11(c) x 4 − 1 2 1 final answer M1 for x − 3 3 x x 11(d) 3 x oe final answer 1

This question in 0607/43 Oct/Nov 2020

5 (a) B 49 mm NOT TO 91 mm SCALE A C Calculate the length of AC. AC = … mm [2] (b) 305° NOT TO O SCALE B 16 cm A The diagram shows a circle with centre O and radius 16 cm. Calculate the length of the major arc AB. … cm [2] (c) NOT TO SCALE 12 cm The diagram shows a prism with length 12 cm. The cross-section of the prism is a quarter of a circle. The radius of the circle is 6 cm. Calculate the volume of the prism. … cm3 [2] (d) C NOT TO (2x + 4) cm SCALE B D (x + 1) cm A E (x – 3) cm Shape ABCDE is made by joining rectangle ABDE and triangle BCD. The perpendicular height of triangle BCD is (2x + 4) cm. The total area of ABCDE is 11 cm2. (i) Show that 2x 2 - 3x - 20 = 0 . [3] (ii) Factorise 2x 2 - 3x - 20 . … [2] (iii) Use your answer to part (ii) to solve the equation 2x 2 - 3x - 20 = 0 . x = … or x = … [1] (iv) Find the perpendicular height of triangle BCD. … cm [1]

13 marks

Mark scheme: 5(a) 103 or 103.3 to 103.4 2 M1 for 492 + 912 oe 5(b) 85.2 or 85.17 to 85.18 2 305 M1 for × π × 2 × 16 360 5(c) 339 or 339.2 to 339.3… 2 1 2 M1 for × π × 6 × 12 4 5(d)(i) 1 M1 (x – 3)(x + 1) + (x – 3)(2x + 4) 2 [=11] x2 – 3x + x – 3 B1 one correct expansion seen 1 or (2x2 – 6x + 4x – 12)  2 or x2 – 3x + 2x – 6 At least one more line of A1 no errors or omissions working leading to 2x2 – 3x – 20 = 0 5(d)(ii) (2x + 5)(x – 4) 2 M1 for (2x + a)(x + b) where ab = –20 or a + 2b = –3 or 2x(x – 4) + 5(x – 4) or x(2x + 5) – 4(2x + 5) 5(d)(iii) 4 , –2.5 1 Strict FT their factors Dep on factors in part (ii) 5(d)(iv) 12 1 FT 2 × (their positive root (d)(iii)) + 4

This question in 0607/42 Feb/March 2021

11 f ( )x = 3 x + 1 g ( )x = x 2 - 5 h ( )x = 3 x (a) Find g(3). … [1] (b) Find f(h(2)). … [2] (c) Find the value of r when f(r) = r. r = … [2] (d) Solve g(f(x)) = 20. x = … or x = … [3] (e) Find h -1 ( )x . h -1 ( )x = … [2]

10 marks

Mark scheme: 11(a) 4 1 11(b) 28 2 B1 for f(32) seen or M1 for 3 × 3x + 1 oe 11(c) 1 oe 2 M1 for 3r + 1 = r − 2 11(d) 4 3 M1 for (3x + 1)2 – 5 oe , −2 M1 for (3x + 1) = ±5 or [3](x + 2)(3x – 4) = 0 oe 3 or correct substitution in formula for 3x2 + 2x – 8 or 9x2 + 6x – 24 or correct and suitable sketch 11(e) log x 2 M1 for log y = log 3x oe or correct answer seen log 3 x or final answer log3 or x = 3y or log 3 y = x

This question in 0607/42 Feb/March 2021

5 f ( )x = 2 x - 1 g ( )x = 3 - x h ( )x = x 2 (a) Find (i) f ( - 2) , … [1] (ii) h ( g ( - 2)) . … [2] (b) Solve f ( )x = 7 . x = … [2] (c) Find f ( g ( x)) . … [1] (d) Solve f (x) # g (x) + 2h ( x) = 0 . x = … [3] (e) Find g -1 ( )x . g -1 ( )x = … [2] (f) y 10 – 3 0 3 x – 2 (i) On the diagram, sketch the graph of y = h ( x) for values of x between - 3 and 3. [2] (ii) Write down the equation of the line of symmetry of the graph of y = h ( x) . … [1] (iii) On the diagram, sketch the graph of y = g ( x) for values of x between - 3 and 3. [1] (iv) Solve g ( x) 2 h ( x) . … [2]

17 marks

Mark scheme: 5(a)(i) –5 1 5(a)(ii) 25 2 2 M1 for g(–2) = 5 or ( 3 − x ) soi 5(b) 4 2 M1 for 2x – 1 = 7 5(c) 5 – 2x oe Final answer 1 5(d) 3 3 2 2 oe B2 for 6 x −+3 x − 2 x + 2 x [= 0] or 7 better or B1 for 6 x −+3 x − 2 x 2 or M1 for ( 2 x − 1)( 3 − x ) + 2 x 2 [ = 0 ] 5(e) 3 − x Final answer 2 M1 for x = 3 – y or y + x = 3 5(f)(i) Correct sketch 2 B1 for any quadratic graph 5(f)(ii) x = 0 cao 1 5(f)(iii) Correct sketch 1 5(f)(iv) –2.3[0] < x < 1.3[0] 2 For x, do not allow f(x) or y for full or –2.303 to –2.302 < x < 1.302 to 1.303 marks B1 for either inequality or for 1.3[0] and –2.3[0] y range included scores 0

This question in 0607/41 May/June 2021

7 (a) Solve the simultaneous equations. You must show all your working. 7x + 2y = 8 2x - 3y = 13 x = … y = … [4] (b) Solve. (i) 3x - 4 =- 19 x = … [2] (ii) 15 - 5x = 7 - 3x x = … [2] 28 (iii) =- 4 ( x + 1) x = … [2] (c) 3 log p - log q - log 8 = 2 log x Find x in terms of p and q. x = … [3]

13 marks

Mark scheme: 7(a) Correctly equating one set of coefficients M1 or making x or y the subject of one equation Correct method to eliminate one variable M1 May be intersection of two straight line graphs [x =] 2 A2 A1 for each [y =]–3 If 0 scored for whole question, SC1 for answers that satisfy one equation 7(b)(i) –5 2 M1 for 3x = 4 – 19 oe 7(b)(ii) 4 2 M1 for 5x – 3x = 15 – 7 oe 7(b)(iii) –8 2 M1 for 28 = –4(x + 1) oe 28 or = − ( x + 1) oe 4 7(c) 3 3 M1 for log p3 or log x2 or better p oe final answer M1 for correct use of 8q log a + log b = log ab a or log a − log b = log b

This question in 0607/41 May/June 2021

11 (a) Using a suitable sketch, solve 5 x = 10 . y 12 0 x – 1 3 – 1 (b) Solve. 5 + x 6x - 1 = 2x + 3 You must show all your working. x = … or x = … [5]

8 marks

Mark scheme: 11(a) Correct sketch M2 or M1 for exponential graph 1.43 or 1.430 to 1.431 B1 11(b) Algebraic method (6 x − 1)(2 x + 3) = [5 + x ] or better M1 12 x 2 + 15 x −=8 0 A1 2 M1 Correct use of formula −15 ± 15 − 4 × 12 × −8 x = or correct sketch of parabola 2 × 12 0.403 or 0.4032… B1 –1.65 or –1.653… B1 11(b) Graphical method (1) Correct sketch of y = 6x – 1 M1 5 + x M2 or M1 for hyperbolic graph Correct sketch of y = 2 x + 3 0.403 or 0.4032… B1 –1.65 or –1.653… B1 11(b) Graphical method (2) Correct sketch M3 M2 for parabola y = (6x –1)(2x + 3) oe or M1 for parabola and M1 for y = 5 + x 0.403 or 0.4032... B1 –1.65 or –1.653... B1

This question in 0607/41 May/June 2021

4 y 15 – 5 0 5 x – 15 f ( )x = 10 - x 2 (a) On the diagram, sketch the graph of y = f(x) for - 5 G x G 5 . [2] (b) Solve the equation f(x) = 6. … [2] (c) Solve f ( )x 2 6 . … [3] (d) Find the values of k for which f(x) = k has exactly two solutions. … [2]

9 marks

Mark scheme: 4(a) Correct sketch 2 B1 for correct middle section 4(b) ± 4 2 B1 for 2 correct solutions ± 2 4(c) x < –4 3 B1 for each –2 < x < 2 x > 4 4(d) 0 2 B1 for each [k ] > 10

This question in 0607/42 May/June 2021

7 Roisin drives 250 km. She drives the first 200 km at an average speed of x km/h. (a) Write down an expression for the time, in hours, it takes to drive the 200 km. … h [1] (b) For the remainder of the journey, Roisin is in heavy traffic and her average speed is 40 km/h less than for the first 200 km. 1 The total time for the journey is 3 hours. 2 Show that 7x 2 - 780 x + 16000 = 0 . [4] (c) Solve the equation 7x 2 - 780x + 16000 = 0 to find the time taken to travel the first 200 km. Give your answer in hours and minutes correct to the nearest minute. … h … min [5]

10 marks

Mark scheme: 7(a) 200 1 x 7(b) 200 50 7 M1 + = oe x x − 40 2 400(x – 40) +100x = 7x(x – 40) oe M2 FT only an equation of the correct form with equivalent difficulty 200 ( x − 40 ) + 50 x M1FT for x ( x − 40 ) or better Completion to 7x2 – 780x + 16 000 = 0 A1 With at least one intermediate step with no errors or omissions 7(c) 2 [h] 22 [min] 5 B4 for 2.37 or 2.371 to 2.372 or 2h 22 to 22.3... min or 142 to 142.3... or B3 for 27.1 or 27.10 to 27.11 and 84.3 or 84.3... , or M2 for 2 −−( 780 ) ± ( −780 ) − 4 × 7 × 16000 2 × 7 or sketch of parabola (positive x2 ) with two positive zeros or M1 for ( − 780 ) 2 − 4 × 7 × 16000 −−( 780 ) ± p or 2 × 7 and M1 for 200 ÷ their solution from their quadratic if > 40

This question in 0607/42 May/June 2021

12 f ( )x = 2 - 3 x g ( )x = 2 - 3x (a) Find f(4). … [1] (b) Solve g(x) = 4. … [3] (c) Find f -1 ( )x . f -1 ( )x = … [2] (d) Find g ( f ( x)) . Write your answer as a single fraction in its simplest form. … [2] (e) Find f(x) - g(x). Write your answer as a single fraction in its simplest form. … [3]

11 marks

Mark scheme: 12(a) –10 1 12(b) 1 3 M2 for 5 = 8 – 12x oe oe 5 4 or M1 for = 4 2 − 3x 12(c) 2 − x 2 M1 for 3x + y = 2 or x = 2 – 3y oe y 2 3 or = − x or better 3 3 12(d) 5 2 5 oe final answer M1 for −+4 9x 2 − 3(2 − 3 )x 12(e) 9 x 2 − 12 x − 1 3 ( 2 − 3 x )( 2 − 3 x ) − 5 oe final answer M1 for 2 − 3 x 2 − 3 x B1 for 4 – 6x – 6x + 9x2

This question in 0607/42 May/June 2021

13 y 5 – 5 0 5 x – 5 x 2 + 3 f ( x) = ( 1 - x)( x + 3) (a) On the diagram, sketch the graph of y = f(x) for values of x between -5 and 5. [3] (b) Find the equations of the asymptotes parallel to the y-axis. … [2] (c) Solve f(x) = 2x + 3. … [3]

8 marks

Mark scheme: 13(a) Correct sketch 3 B1 for each branch 13(b) x = 1, 2 B1 for each x = –3 13(c) –3.79 or –3.791... 3 B1 for each –1 0.791 or 0.7912 to 0.7913 If 0 scored SC1 for y = 2x + 3 sketched and cutting both axes

This question in 0607/42 May/June 2021

1 y 5 – 5 0 5 x – 5 4 f ( )x = x - x (a) On the diagram, sketch the graph of y = f(x) for values of x between -5 and 5. [2] (b) Find the zeros of f(x). x = … or x = … [2] (c) Solve the equation f(x) = 2. x = … or x = … [2] (d) g(x) = f(x + 2) (i) On the same diagram, sketch the graph of y = g(x) for values of x between -5 and 5. [2] (ii) Describe fully the single transformation that maps the graph of y = f(x) onto the graph of y = g(x). … … [2]

10 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) Correct sketch 2 B1 for each branch 4444 or B1 for correct but branches joined 2222 -4-4-4-4 -2-2-2-2 0000 0000 2222 4444 -2-2-2-2 -4-4-4-4 1(b) –2, 2 2 B1 for each 1(c) –1.24 or –1.236... 2 B1 for each 3.24 or 3.236... 1(d)(i) Correct sketch 2 B1 for each branch 4444 or B1 FT their f(x) translated in x direction 2222 -4-4-4-4 -2-2-2-2 0000 0000 2222 4444 -2-2-2-2 -4-4-4-4 1(d)(ii)  − 2  2 B1 for each Translation    0 

This question in 0607/43 May/June 2021

7 In this question all lengths are in centimetres. (a) C B 8x° ( x + 5 )° NOT TO SCALE A In triangle ABC, AC = BC, angle ABC = ( x + 5)° and angle ACB = 8x° . Find the value of x. x = … [3] (b) NOT TO ( p - 2) SCALE ( p + 1) The diagram shows a rectangle with sides of length ( p + 1) and ( p - 2) . The area of the rectangle is 90 cm2 . Find the value of p. p = … [4] (c) ( y - 1) ( y - 4) NOT TO SCALE 30° The diagram shows a right-angled triangle. Find the value of y. y = … [3] (d) 13 ( w + 1 ) NOT TO SCALE ( 2w + 3) The diagram shows a right-angled triangle with sides of length ( w + 1) , ( 2w + 3) and 13. Work out the area of the triangle. … cm2 [6]

16 marks

Mark scheme: 7(a) 17 3 M2 for x + 5 + 8 x + x + 5 = 180 oe or M1 for angle A = x + 5 7(b) 10.1 or 10.10... 4 B3 for correct sketch indicating roots −−( 1) ± ( − 1) 2 − 4(1)( − 92) or for oe 2(1) or B2 for p 2 − 2 p + p − 2 [ = 90] or better or M1 for ( p + 1)( p − 2) [ = 90] 7(c) 7 3 M2 for 2(y – 4) = y – 1 or better y − 4 or M1 for = sin30 y − 1 If 0 scored SC1 for sin 30 = 0.5 7(d) 2.04 oe 6 B4 for (5 w − 1)( w + 3) or correct sketch indicating roots − 14 ± 14 2 − 4(5)( −3) or 2(5) or B3 for 5 w 2 + 14 w − 3 = 0 and M1 for correct calculation of area of triangle with their positive w OR 2 2 M1 for ( w + 1) + ( 2 w + 3 ) = 13 B1 for w 2 + w + w + 1 oe or 4 w 2 + 6 w + 6 w + 9 oe and M1 for correct calculation of area of triangle with their positive w

This question in 0607/43 May/June 2021

9 y 10 0 x 2.5 f ( x) = x x, x 2 0 (a) On the diagram, sketch the graph of y = f(x) for 0 1 x G 2.5 . [2] (b) Find the coordinates of the local minimum point. ( … , … ) [2] (c) (i) Find x when f(x) = 3x. … [3] (ii) Solve f ( )x H 3x . … [2]

9 marks

Mark scheme: 9(a) Correct sketch 2 B1 for correct shape but cutting either axis or without minimum 9(b) (0.368, 0.692) 2 B1 for each or (0.3678 to 0.3679, 0.6922...) 9(c)(i) 0.237 or 0.2369 to 0.2370 3 M1 for correct line sketched 2.31 or 2.311 … B1 for one correct 9(c)(ii) [0 <] x ⩽ 0.237 or 0.2369 to 2 B1 for each 0.2370 x ⩾ 2.31 or 2.311…

This question in 0607/43 May/June 2021

8 y 5 – 2 0 2 x – 5 f ( x) = 3 x - x 3 for - 2 G x G 2 (a) On the diagram, sketch the graph of y = f ( x) . [2] (b) Find the coordinates of the local maximum. ( … , … ) [1] (c) Write down the x-coordinates of the points where the curve meets the x-axis. x = … , x = … , x = … [2] (d) (i) Describe fully the single transformation that maps y = f ( x) onto y = f ( x + 1) . … … [2] (ii) Solve f ( x) = f ( x + 1) for - 2 G x G 2 . … [2] (iii) Solve f ( x) H f ( x + 1) for - 2 G x G 2 . … [2]

11 marks

Mark scheme: 8(a) Correct sketch 2 B1 for negative cubic graph with 2 turning y f(x)=3x-x^3 points x 8(b) (1, 2) 1 8(c) –1.73 or –1.732… oe 2 B1 for two correct 0 1.73 or 1.732… oe 8(d)(i) Translation 2 B1 for each  − 1     0  8(d)(ii) –1.46 or –1.457… 2 B1 for each but without y-coords. 0.457 or 0.4574… or M1 for graph of y = 3( x + 1) − ( x + 1) 3 oe 8(d)(iii) [–2 ⩽] x ⩽ –1.46 2 B1 for each 0.457 ⩽ x [⩽ 2] or for x ⩽ –1.46 and 0.457 ⩽ x

This question in 0607/43 Oct/Nov 2021

6 In this question all lengths are in centimetres. NOT TO SCALE 2x – 1 7 5x + 1 13 – x The area of the larger rectangle is 84 cm 2 greater than the area of the smaller rectangle. (a) Show that 5x 2 + 2x - 88 = 0 . [4] (b) Factorise 5x 2 + 2x - 88 . … [2] (c) Find the area of the smaller rectangle. … cm2 [2]

8 marks

Mark scheme: 6(a) (5x + 1)(2x – 1) – 7(13 – x) = 84 oe M1 Correct first statement without brackets expanded 10x2 – 5x + 2x – 1 B1 – 91 + 7x B1 on LHS or 91 – 7x on RHS 10 x 2 + 4 x − 176 = 0 oe A1 leading to 5 x 2 + 2 x − 88 = 0 with no errors or omissions 6(b) ( 5 x + 22 )( x − 4 ) 2 B1 for (5x + a)(x + b) with ab = –88 or a + 5b = 2 or for 5x(x – 4) + 22(x – 4) or for x(5x + 22) – 4(5x + 52) 6(c) 63 2 M1 for solving their factorised quadratic, allowing omission of negative root. FT (13 – their positive root) × 7 1 if < x < 13 2

This question in 0607/42 Feb/March 2022

8 f ( x) = 2 x + 1 g ( x) = 3 - 2 x h ( x) = log ( x + 1) (a) Find the value of (i) f(12), … [1] (ii) g(f(12)). … [1] (b) Find the value of x when f ( x) = g ( x) . x = … [2] (c) Find f(g(x)), giving your answer in its simplest form. … [2] (d) Find g -1 ( x) . g -1 ( x) = … [2] (e) Find x when h (x) = f ( 0 .5 ) . x = … [2] (f) Find h -1 ( x) . h -1 ( x) = … [2]

12 marks

Mark scheme: 8(a)(i) 25 1 8(a)(ii) –47 1 FT 3 – 2 × their 25 8(b) 1 2 M1 for 2x + 1 = 3 – 2x or better oe 2 8(c) 7 – 4x final answer 2 M1 for 2(3 – 2x) + 1 8(d) 3 −x 2 M1 for y + 2x = 3 or better oe final answer y 3 2 or = − x 2 2 or x = 3 – 2y 8(e) 99 2 M1 for log( x + 1) = 2(0.5) + 1 or better 8(f) 10 x − 1 2 M1 for 10 y = x + 1 or x = log( y + 1)

This question in 0607/42 Feb/March 2022

5 A sequence of patterns is made using grey tiles and white tiles. Pattern 1 Pattern 2 Pattern 3 (a) Complete the table. Pattern number 1 2 3 4 n Number of grey tiles 6 10 Number of white tiles 0 2 [6] (b) Find and simplify an expression for the total number of tiles in Pattern n. … [1] (c) Pattern k has a total of 600 tiles. Find the number of grey tiles in Pattern k. … [4] (d) The tiles in a pattern are put in a bag. 5 The probability of taking a grey tile from the bag at random is . 12 A tile is taken from the bag at random and replaced. This is repeated 3 times. Find the probability that all 3 tiles are white. … [2] (e) All the grey tiles from Pattern 4 are put in a bag. Two tiles are taken from the bag at random without replacement. Find the probability that one tile came from a corner of the pattern and the other did not. … [3]

16 marks

Mark scheme: 5(a) 6 B2 for all four numbers correct 14 18 4n + 2 oe or B1 for at least two correct B2 for 4 n  2 oe 6 12 n2 – n oe or M1 for 4 n  k B2 for n 2  n oe or M1 for any quadratic or for second differences of 2 seen 5(b) n2 + 3n + 2 or (n + 1)(n + 2) 1 FT their grey + white if both in terms of n 5(c) 94 4 M1 for their k 2  3k  2 = 600 oe M1 for correct method for solving their quadratic M1 for substituting their positive integer (from a quadratic) k into their 4n+2 5(d) 343 2  5  3 oe M1 for 1  oe 1728    12  5(e) 56 3 FT their 18 from (a) for M marks only oe 153 4 14 14 4 M2 for    oe 18 17 18 17 M1 for one product

This question in 0607/41 May/June 2022

10 (a) Simplify fully. 4 x 2 y x ' 3 12 y … [2] (b) Write as a single fraction in its simplest form. 1 x - 3 - x - 3 2 … [3] (c) The nth term of a sequence is an 2 + bn - 5 . The second term of this sequence is - 3 and the third term is 4. Find the value of a and the value of b. You must show all your working. a = … b = … [6]

11 marks

Mark scheme: 10(a) 16xy 2 Final answer 2 4 x 2 y 12 y M1 for  or better 3 x 10(b)  x 2  6 x  7 3 B1 for 2[ 1]  ( x  3)( x  3) Final answer B1 for 2( x  3) as denominator 2( x  3) 10(c) 2 2  a  2  b  5 3 oe M2 M1 for either 32  a 3 b  5  4 oe correctly equating one set of coefficients M1 FT or making a or b subject of one equation correct method for eliminating one M1 FT variable or correctly substituting in other equation [a=] 2 B2 B1 for each [b=] −3

This question in 0607/41 May/June 2022

4 y 4 – 4 0 4 x – 4 (a) On the diagram, sketch the graph of y = f( x) , where f( )x = 4 - 2 x for values of x between - 4 and 4. [3] (b) Write down the x-coordinates of the points where the graph meets the x-axis. x = … and x = … [1] (c) On the diagram, sketch the graph of y = g( x) , where g ( x) = 0 .25 x 2 for values of x between - 4 and 4. [2] (d) Write down the equation of the line of symmetry of the graph of y = g( x) . … [1] (e) Find the value of the x-coordinate of each point of intersection of the two graphs. x = … and x = … [2] (f) On your diagram shade the region defined by f( x) H g( x). [1]

10 marks

Mark scheme: 4(a) Correct sketch 3 B1 for inverted ‘v’ B1 for symmetrical about y-axis 4 4(b) –2, 2 1 4(c) Correct sketch 2 Must touch x-axis at origin and without serious curl backs B1 for a u-shaped parabola 4(d) x = 0 1 4(e) –1.66 or –1.657 to –1.656 2 B1 for each 1.66 or 1.656 to 1.657 4(f) Correct region shaded 1 Dependent on at least B2 in (a) and at least B1 in (c)

This question in 0607/42 May/June 2022

10 (a) P = 5 Work out the value of P when x =- 18 and y = 28 . P = … [3] (b) Simplify fully. 5 y 4 x # 2 x 3 … [2] (c) Factorise fully. (i) 15ab - 25bc … [2] (ii) 6 x 2 y 5 - 16 x 3 y 3 … [2] (iii) 6cd - 3 - 9d + 2c … [2] (d) Make x the subject of the formula. 2 x 3ax = 1 - a + 2 x = … [4] (e) Solve the inequality. 3 - x 2 1 2 + x … [3]

18 marks

Mark scheme: 10(a) – 84 3 M1 for correct substitution B1 for answer 84 10(b) 10 y 1 2 20 xy 10 xy 20 y 5 y  2 or 3 3 y or 3.3 (or 3.33 or 3.333…)y B1 for or or or 3 6 x 3 x 6 3 final answer or correct answer seen 10(c)(i) 5b(3a – 5c) final answer 2 M1 for b(15a – 25c) or 5(3ab – 5bc) or correct answer seen 10(c)(ii) 2x2y3(3y2 – 8x) final answer 2 M1 for x2y3(6y2 – 16x) or 2y3(3x2y2 – 8x3) or 2x2(3y5 – 8xy3) or better i.e. answers which are correct and have only one common factor left inside brackets e.g. 2x2y(3y4 – 8xy2) or correct answer seen 10(c)(iii) (2c – 3)(3d + 1) final answer 2 M1 for 2c(3d + 1) – 3(3d + 1) or 3d(2c – 3) + 2c – 3 or correct answer seen 10(d) a  2 4 M1 for correctly eliminating fractions [ x  ] oe 2 M1 for correctly expanding brackets 3a  6 a  2 final answer M1 for correctly collecting all terms in x on one side and other terms on other side of equation M1 for correctly isolating x by factorising and dividing Max 3 marks only if final answer is incorrect 10(e) –2 < x < 0.5 final answer 3 M2 for –2 and 0.5 SOI or M1 for correct graph(s) sketched 1  2 x or M1 for  0 oe 2  x or B1 for 0.5 soi

This question in 0607/42 May/June 2022

4 (a) Solve 4x - 3 = 7 . x = … [2] 3x + 1 (b) y = z Find the value of y when x = 4.3 and z =- 2 . y = … [2] (c) Solve the simultaneous equations. You must show all your working. 4x - 3y = 14 3x + 5y = 25 x = … y = … [4] 2 x 2 + 4 x x 2 - 4 (d) Simplify 2 ' . 5 y 10y … [4]

12 marks

Mark scheme: 4(a) 2.5 oe 2 M1 for 4x = 7 + 3 oe 4(b) –6.95 2 M1 for correct substitution or B1 for 13.9 seen 4(c) correctly equating one set of M1 Allow one incorrect number coefficients or making x or y the subject of 1 equation Correct method to eliminate one M1 e.g. Adding, subtracting substitution variable x = 5, y = 2 A2 A1 for either nfww. If 0 scored, SC1 for a pair of numbers that satisfy either equation or for correct solutions with no working. 4(d) 4 x 4 B1 for 2x(x + 2) oe final answer B1 for (x – 2)(x + 2) y ( x  2) M1 for inverting and changing sign to multiplication at any stage

This question in 0607/43 May/June 2022

11 A tank has a capacity of 400 litres. Water from Tap A flows at x litres per minute. Water from Tap B flows at 2 litres per minute less than the water from tap A. (a) Write down an expression in terms of x for the time, in minutes, for tap A to fill the tank. … [1] (b) Tap B takes 10 minutes longer to fill the tank than tap A. Write down an equation in terms of x and show that it simplifies to x 2 - 2x - 80 = 0 . [4] (c) Solve x 2 - 2x - 80 = 0 and find the time it takes to fill the tank when both taps are turned on. Give your answer in minutes and seconds, correct to the nearest second. … minutes … seconds [4]

9 marks

Mark scheme: 11(a) 400 1 x 11(b) 400 400 M1 – = 10 oe x  2 x 400x – 400(x – 2) = 10x(x – 2) M1 FT clearing fractions. Dep on equation in the right form –400x + 800 or 400x – 800 M1 FT Expansion of both brackets of correct form and 10x2– 20x Completion to x² – 2x – 80 = 0 A1 with no errors or omissions 11(c) 22 min 13 sec 4 B2 for x = 10 isw other value of x or M1 for (x – 10)(x + 8) ( 2)  ( 2) 2  4(1)( 80) or 2  1 or sketch of parabola with one positive zero and one negative zero 400 M1 for their 10  their10  2

This question in 0607/43 May/June 2022

5 y 4 x 0 –1 5 – 4 (a) On the diagram, sketch the graph of y = f ( x) , where 1 f ( x) = for values of x between - 1 and 5. [3] ( x - 1)( x - 2)( x - 3) (b) Write down the y‑coordinate of the point where the curve meets the y‑axis. y = … [1] (c) Write down the equations of all the asymptotes to the graph of y = f ( x) . … [3] (d) On the diagram, sketch the graph of y = g ( x) , where g ( x) = x - 1 , for values of x between - 1 and 5 . [1] (e) Find the x‑coordinate of each point of intersection of the two graphs. x = … or x = … [2] (f) Solve the inequality f ( x) 2 g ( x) . … [3]

13 marks

Mark scheme: 5(a) Correct sketch f(x)=1/((x-1)(x-2)(x-3)) 3 B1 for graph in 4 sections B1 for rectangular hyperbola type on outside 2 sections not crossing x-axis B1 for 2 quadratic type sections (one inverted) Max 2 marks if not fully correct 5(b) 1 1 –0.167 or –0.1667 to –0.1666 or − 6 5(c) x = 1, x = 2, x = 3, y = 0 3 B2 for 3 correct or B1 for 1 correct If 0 scored, SC1 for all four with  5(d) 1 Can be good freehand, cutting negative y-axis and positive x-axis 5(e) x = 0.487 or 0.4871… 2 B1 for each x = 3.18 or 3.178 to 3.179 5(f) [–1 < ] x < 0.487 3 B1 FT their(e) for each 1 < x < 2 3 < x < 3.18

This question in 0607/41 Oct/Nov 2022

7 (a) Solve the simultaneous equations. You must show all your working. 4x + 3y = - 21 6x - 2y = 1 x = … y = … [4] 1 3(b) f ( x) = 5 x - 2 g ( x) = , x ! 0.5 h ( x) = ( x - 1 ) 2x - 1 (i) Find f ( 3) . … [1] (ii) Find h ( f ( 2)) . … [2] (iii) Solve f ( h ( x)) = - 7 . x = … [3] (iv) Find g ( g ( x)) in terms of x. Give your answer in its simplest form. … [3]

13 marks

Mark scheme: 7(a) Correctly equating coefficients M1 or correctly isolating x or y Correct method to eliminate one variable M1 [x = ] –1.5 A1 [y = ] –5 A1 If second M is M0, SC1 for answers that satisfy one equation or if 2 correct answers and no working shown 7(b)(i) 13 1 7(b)(ii) 343 2 M1 for ((5 x − 2) − 1) 3 or ((5(2) − 2) − 1) 3 or for h(8) used correctly 7(b)(iii) 0 3 B2 for ( x − 1) 3 = − 1 or M1 for 5( x − 1) 3 − 2 [ = − 7] oe 7(b)(iv) 2 x − 1 3 1 oe final answer B2 for or better 3 − 2 x  2 − 2 x + 1     2 x − 1  1 or M1 for  1 2  − 1  2 x − 1  8 In parts (a), (b) and (c), marks can only be earned with an increasing curve or plots

This question in 0607/41 Oct/Nov 2022

9 NOT TO SCALE The diagram shows a square of side 2a cm inside a square of side ( 2a + 2x)cm . (a) (i) Find an expression, in terms of a and x, for the area of the shaded region. Give your answer in the form px 2 + qax , where p and q are integers. … [2] (ii) Calculate the area of the shaded region when a = 6 and x = 1. … cm2 [1] (b) Find an expression, in terms of a and x, for the total perimeter of the shaded region. Give your answer in its simplest form. … [2] (c) The numerical value of the shaded area is equal to the numerical value of the perimeter of the shaded region. Find x when a = 10 . You must show all your working. x = … [4]

9 marks

Mark scheme: 9(a)(i) 4 x 2 + 8ax final answer 2 M1 for (2 a + 2 x ) 2 − (2 a ) 2 oe 9(a)(ii) 52 1 FT their (i) if answer positive and if a and x both used 9(b) 8x + 16a or 8(x + 2a) oe final answer 2 M1 for 4(2a + 2 x) + 4(2a ) oe 9(c) 2 cao 4 M1 for their (a)(i) = their (b) M1 for rearranging to 3-term quadratic[=0] with a = 10 substituted, or for appropriate sketch M1 for correct method to solve their 3 term quadratic e.g. ( x + 20)( x − 2) [ = 0] or sketch If 0 scored, SC1 for their (a)(ii) = their (b) 10 For all parts accept decimals or percentages with the usual rules for 3sf Do not penalise incorrect cancelling or converting Do not accept ratios or words

This question in 0607/41 Oct/Nov 2022

12 (a) a b = 1 where a 2 0 (i) When b = 13 , write down the value of a. a = … [1] (ii) When a = 17 , write down the value of b. b = … [1] (b) Write down the solution to each equation. (i) 3 x - 5 = 1 x = … [1] (ii) ( x - 5) 3 = 1 x = … [1] (c) Use part (a) to find all the solutions to the following equations. (i) ( 4x - 1) ( x - 3 ) = 1 … [3] (ii) ( x 2 - 4x + 4) ( x 2 - 9 x + 20 ) = 1 … [4]

11 marks

Mark scheme: 12(a)(i) 1 1 12(a)(ii) 0 1 12(b)(i) 5 1 12(b)(ii) 6 1 12(c)(i) 1 3 B2 for one correct [ x =] , 3 or M1 for 4 x −=1 1 or x −=3 0 2 12(c)(ii) x 2 − 4 x + 4 = 1 oe and x 2 − 9 x + 20[ = 0] M2 M1 for x 2 − 4 x + 4 = 1 oe or x 2 − 9 x + 20[ = 0] 1, 3, 4, 5 A2 A1 for 1 and 3 or 4 and 5

This question in 0607/41 Oct/Nov 2022

6 (a) Simplify. (i) 5 ( 2a + 3) - 3 ( a - 7) … [2] 2x x - 1 (ii) - 3 2 … [2] ab + 3 (b) x = b - 2 Rearrange the formula to make (i) a the subject, a = … [3] (ii) b the subject. b = … [2] (c) Solve. (i) x 12 = 1200 x = … [1] (ii) .12 x = 12 x = … [2] (iii) x + 3 = 7 … [2] (d) Solve by factorising. 6x 2 - 11 x - 10 = 0 x = … or x = … [3]

17 marks

Mark scheme: 6(a)(i) 7a + 36 Final answer 2 B1 for ka + 36 or 7a + k or 10a + 15 – 3a + 21 6(a)(ii) x + 3 2 2  2 x − 3( x − 1) Final answer M1 for orbetter 6 6 6(b)(i) bx − 2 x − 3 3 M1 for x(b – 2) = ab + 3 oe Final answer M1FT for bx – 2x – 3 = ab oe b 6(b)(ii) 2 x + 3 2 M1FT for bx – ab = 2x + 3 oe Final answer OR x − a M1FT for factorising and dividing Max 1 mark if answer incorrect 6(c)(i) 1.81 or 1.805 to 1.806 1 6(c)(ii) 13.6 or 13.62 to 13.63 2 M1 for xlog1.2 = log12 or log1.2 12 or a suitable sketch leading to answer 6(c)(iii) 4, –10 final answer 2 B1 for either seen 6(d) (2x – 5)(3x + 2) [= 0] B2 B1 for (ax + b)(cx + d) where ac = 6 and bd = –10 or ad + bc = –11 or for 3x(2x – 5) + 2(2x – 5) or for 2x(3x + 2) –5(3x + 2) 5 2 B1 oe − oe 2 3

This question in 0607/42 Oct/Nov 2022

11 y 8 – 5 0 5 x – 9 5 f ( x) = x + ( x - 2)( x + 3) (a) Sketch the graph of y = f ( x) for values of x between - 5 and 5. [4] (b) Write down the equations of the asymptotes parallel to the y-axis. … [2] (c) (i) Find the coordinates of the local maximum. ( … , … ) [2] (ii) Find the coordinates of the local minimum. ( … , … ) [2] (iii) Write down the range of values of k for which f ( x) = k has exactly one solution. … [2] (d) g ( x) =- 4 - x (i) Solve the equation f ( x) = g ( x) . … [3] (ii) Find the solutions to the inequality f ( x) 2 g ( x) . … [3] Question 12 is printed on the next page.

18 marks

Mark scheme: 11(a) Correct sketch 4 B1 for each outside branch B2 for middle branch with offset maximum or B1 if not offset or if offset crosses x- axis 11(b) x = 2, x = –3 2 B1 for each 11(c)(i) (1.03, –0.249) 2 1.029... –0.2491 to –0.2490 B1 for each coordinate 11(c)(ii) (2.99, 3.83) 2 2.986... 3.833... B1 for each coordinate 11(c)(iii) –0.249 < k < 3.83 2 B1FT for each 11(d)(i) –3.35 or –3.347..., –1.52 or –1.520... 3 B1 for each 1.87 or 1.867... If 0 scored, SC1 for y = –4 – x sketched on diagram or for –3.3, –1.5, and 1.9 or if y-coordinates also given 11(d)(ii) –3.35 < x < –3, 3 B1FT for each –1.52 < x < 1.87, x > 2

This question in 0607/42 Oct/Nov 2022

2 y 1 – 5 0 5 x – 3 1 1 f ( x) = - 2 x x (a) On the diagram, sketch the graph of y = f ( x) for values of x between - 5 and 5. [2] (b) Find f ( - 2) . … [1] (c) Solve the equation f ( x) = 0 . x = … [1] (d) Find the maximum value of f(x). … [1] (e) Write down the equation of each asymptote. … [2] (f) (i) Solve the equation. 1 1 2 - = x - 2 2 x x … [3] 1 1 2 4 2 (ii) The equation - 2 = x - 2 can be rearranged to the form x + ax + bx + c = 0 . x x Find the values of a, b and c. a = … b = … c = … [2]

12 marks

Mark scheme: 2(a) Correct sketch 2 No intersections with y-axis B1 for each branch with no large curl back or feathering. Right hand branch with a maximum or level. 2(b) –[0].75 oe 1 2(c) 1 1 2(d) 0.25 oe 1 Not coordinates 2(e) x = 0 2 B1 for each y = 0 2(f)(i) 0.525 or 0.5248 to 0.5249 3 B2 for one correct or M1 for sketch of y = x2 – 2 added 1.49 or 1.490... to diagram 2(f)(ii) [a =] –2 2 B1 for x −=1 x 4 − 2 x 2 oe [b =] –1 [c =] 1

This question in 0607/43 Oct/Nov 2022

3 (a) Amira buys a magazine that costs $n and a book that costs $(2n + 5). She pays with a $20 note and receives $1.62 change. Find the cost of a magazine. $ … [3] (b) The cost of a bar of chocolate is $x and the cost of a bag of sweets is $y. Bruce buys 2 bars of chocolate and 1 bag of sweets for a total of $3.60 . Charlie buys 3 bars of chocolate and 2 bags of sweets for a total of $6.05 . Find the total cost of 1 bar of chocolate and 3 bags of sweets. You must show all your working. $ … [5]

8 marks

Mark scheme: 3(a) 4.46 3 M1 for n + 2n + 5 = 20 – 1.62 oe M1 for an = b from their equation including n and 2n + 5 3(b) 2x + y = 3.6[0] oe B1 3x + 2y = 6.05 oe B1 Correctly eliminating one variable M1 With working seen 5.05 A2 A1 [x =] 1.15 or [y =]1.3[0] If M0 scored, SC1 for their values satisfying one of their original equations

This question in 0607/43 Oct/Nov 2022

5 (a) X = 3A + 5B Work out the value of B when X = 48 and A = 4. B = … [2] (b) Solve 6 ( 1 - 2x) = 2 + 4 ( x - 1) . x = … [3] 3x - 2 3 + 2x (c) Solve = - 2 . 5 4 x = … [3] (d) Solve 4 log 2 - 2 log x + log 4 = 2 . You must show your working. x = … [4] (e) Solve x = 16 - 6x 2 . Give your answers correct to 2 decimal places. … [3]

15 marks

Mark scheme: 5(a) 7.2 oe 2 M1 for 48 = 3 +4 5B 5(b) 0.5 oe 3 M1 for 6 − 12 x or 2 + 4 x − 4 M1 for correctly collecting their terms e.g. −12 x − 4 x = 2 − 4 − 6 oe 5(c) –8.5 oe 3 M1 for eliminating fractions M1 for expanding brackets and collecting their terms M1 for correctly solving their equation of the form ax = b Max 2 marks for incorrect answer 5(d) 0.8 oe 4 B1 for 2 = 2log10 or log100 M1 for a correct use of log a + log b = log ab a or log a − log b = log b M1 for a correct use of log a b = b log a 5(e) x = –1.72 3 M2 for sketch indicating correct roots x = 1.55 2 −1 1 −−4 6 ( 16) or x = 2  6 or M1 for 6 x 2 + x − 16 [ = 0] or reverse signs If 0 scored, SC1 for one correct answer

This question in 0607/42 Feb/March 2023

6 y 5 – 5 0 5 x – 5 x 2 f ( )x = 2 - 2 x - x - 2 (a) On the diagram, sketch the graph of y = f ( x) for values of x between -5 and 5. [4] (b) Write down the equations of the two vertical asymptotes. … , … [2] (c) Write down the coordinates of the local minimum point. ( … , … ) [1] (d) On the diagram, sketch the graph of y = g ( x) , where g ( )x = 3 - x for - 2 G x G 5 . [1] (e) (i) Solve the equation f ( x) = g ( x) . … [2] (ii) Solve the inequality f ( x) 2 g ( x) . … [3]

13 marks

Mark scheme: 6(a) 4 B4 for fully correct curve or B3 for ‘correct’ curve with overlaps. or B2 for 2 sections correct or B1 for 1 section correct 6(b) x = –1, x = 2 2 B1 for each 6(c) (0, 2) 1 6(d) 1 Must intersect curve 3 times 6(e)(i) x = –0.861 or –0.8608… 2 B1 for one correct x = 0.746 or 0.7458… x = 3.11 or 3.114 to 3.115 If 0 scored SC1 for –0.86, 0.75, 3.1 6(e)(ii) –1 < x < –0.861 3 FT their (i) 0.746 < x < 2 B1 for each x > 3.11

This question in 0607/42 Feb/March 2023

8 (a) The cost of a television is $t and the cost of a computer is $c. The total cost of 2 televisions and 1 computer is $1470. The total cost of 3 televisions and 2 computers is $2480. Use simultaneous equations to find the cost of a television. You must show all your working. $ … [4] (b) Jono spends $9.69 on bags of potatoes. When the cost of a bag is x cents he can buy 2 more bags than when the cost of a bag is ( x + 6) cents. (i) Show that x 2 + 6x - 2907 = 0 . [3] (ii) Solve the equation x 2 + 6x - 2907 = 0 . x = … or x = … [2] (iii) Find the number of bags Jono can buy for $9.69 when the cost of one bag is x cents. … [1]

10 marks

Mark scheme: 8(a) 2t  c = 1470 B1 3t  2c = 2480 B1 Correctly eliminating one variable M1 460 B1 8(b)(i) 969 969 2 oe M1 x x  6 969(x 969x x(x  6) oe M1 dep fractions cleared from equation with two algebraic fractions with linear denominators Leading to x2  6x – 2907 [ = 0] A1 With at least one step and no errors or omissions 8(b)(ii) –57 and 51 B2 M1 for (x (x – 51)  6 ± (  6) 2  4(1)( 2907) or 2×1 or sketch of parabola (ve x2) with one positive x intercept and one negative 8(b)(iii) 19 B1 969 FT their 51 FT only if their 51 and answer are positive integers

This question in 0607/41 May/June 2023

11 (a) (i) Write 0.000 021 in standard form. … [1] (ii) Calculate 7.3 # 10 -11 # 4.7 # 10 -7 giving your answer in standard form. ` j ` j, … [1] (iii) Calculate .32 # 10 -200 ' 4 # 10 - 100 giving your answer in standard form. ` j ` j, … [2] 2 (iv) Simplify 5 # 10 p , giving your answer in standard form. ` j … [2] (b) y = 10x Write x in terms of y. x = … [1] (c) Solve 7 x = 14 . x = … [1] 1(d) log y = 1 + 3 log x - log w 2 Find y in terms of x and w. y = … [4]

12 marks

Mark scheme: 11(a)(i) 2.1  105 1 11(a)(ii) 3.431  1017 1 11(a)(iii) 8  10101 2 B1 for 0.8  10100 seen 11(a)(iv) 2.5  10 2 p1 2 B1 for 25  10 2 p or 2.5  10  10 2 p seen 11(b) logy or log10 y final answer 1 11(c) log14 1 1.36 or 1.356… or or log7 14 log7 final answer 11(d) 10x 3 10x 3 3  12 10x 3 w 4 M1 log10 soi or or 10 x w or 1 w 1 w 2 2 M1 for log w or log w or log x3 w final answer M1 for correct use of p log p – log q = q or correct use of log p  log q  log pq

This question in 0607/41 May/June 2023

6 y 9 x – 6 0 6 – 9 x 2 + 3 x f ( x) = ( x - 2)( x + 1) (a) On the diagram sketch the graph of y = f ( x) for values of x between - 6 and 6. [3] (b) Write down the equations of the asymptotes parallel to the y‑axis. … [2] (c) Find the zeros of the graph of y = f ( x) . … [2] (d) g ( x) = x - 3 (i) On the diagram sketch the graph of y = g ( x) for - 6 G x G 6 . [1] (ii) Use your graphs to solve f ( x) = g ( x) . … [3] (iii) Solve g ( x) 2 f ( x) . … [3]

14 marks

Mark scheme: 6(a) Correct Sketch 3 B1 for each branch correct 6(b) x = 2, x = –1 2 B1 for each 6(c) –3, 0 2 B1 for each 6(d)(i) Correct Sketch 1 6(d)(ii) –1.16 or –1.162... 3 B1 for each 1 5.16 or 5.162... 6(d)(iii) –1.16 < x < –1 3 B1FT from their (d)(ii) and their (b) for each. 1< x < 2 FT dep on answers to (d)(ii) that lead to three x > 5.16 equivalent inequalities Same accuracy as (d)(ii)

This question in 0607/42 May/June 2023

11 f ( x) = 2 x + 5 g ( x) = 1 - 3 x (a) Find f ( - 2) . … [1] (b) Solve f ( g ( x)) = 19 . … [3] (c) Find g -1 ( x) . g -1 ( x) = … [2] g ( x) (d) y = f ( x) Find x in terms of y. x = … [3]

9 marks

Mark scheme: 11(a) 1 1 11(b) –2 3 B2 for –6x = 12 oe or better or M1 for 2(1 – 3x) + 5 = 19 11(c) 1  x 2 y 1 oe Final answer M1 for x = 1 – 3y or y + 3x = 1 or   x 3 3 3 or y – 1 = –3x 11(d) 1  5 y 3 M1 for y(2x + 5) = 1 – 3x oe oe Final answer M1FT dep for 2xy + 3x = 1 – 5y dependent on 2 y  3 4 term equation with 2 terms in x. M1FT for factorising and dividing to form a  by c  dy Max 2 marks if final answer is incorrect.

This question in 0607/42 May/June 2023

6 (a) Solve. 7x - 5 = 3x + 13 x = … [2] (b) Solve. 4 ( 2x - 3) = 3 ( 1 - 2 x) x = … [3] (c) Solve. 3x + 2 2 = 8 3x + 2 x = … or x = … [3] (d) Solve. 1 - 2 x 2 = 5x - 1 Give your answer correct to two decimal places. x = … or x = … [3] (e) log x = 1 + 4 log y Find x in terms of y. x = … [3] (f) There are 12 balls in a bag, n of them are blue. A ball is taken from the bag at random and replaced. The probability that the ball is blue is p. 6 more blue balls are added to the bag. A ball is taken from the bag at random. The probability that this ball is blue is 2p. Find the value of p. p = … [4]

18 marks

Mark scheme: 6(a) 4.5 oe 2 B1 for 7 x  3x = 13  5 oe 6(b) 15 3 B1 for 8 x  12 3 6 x oe oe 14 M1 for correctly collecting terms in an equation 6(c) 2 3 B2 for 3 x  2 4 oe ,  2 oe 3 or for 3 3 x  2  x  2   0  oe 4 4(3)( 4) or for oe 2(3) or M1 for  3 x  2  2  8  2 oe 6(d) 0.35 –2.85 3 B2 for –2.851 to –2.850 and 0.350 to 0.351 OR M2 for correct sketch indicating both roots 5 5 2  4(2)(  2) or for 2(2) or M1 for 2 x 2  5 x  2   0  or  2 x 2  5 x  2   0  6(e)  x  10 y 4 3 M1 for logy4 B1 for 1 = log10 6(f) 1 4 B3 for n = 3 oe 4 n n  6 or M2 for 212  18 or for 12p = 36p – 6 oe n n  6 or M1 for p  or 2 p  12 18 n n  6 or for and seen 12 18

This question in 0607/43 May/June 2023

7 y 2 – 90 0 90 x – 2 f ( x) = 2 cos ( x - 45)° - 1 for values of x between - 90 and 90. (a) On the diagram, sketch the graph of y = f ( x) . [3] (b) Write down the x‑coordinates of the points where the curve meets the x‑axis. x = … or x = … [2] (c) Write down the coordinates of the local maximum point. ( … , … ) [1] (d) The line y = 0.005 x intersects the curve y = 2 cos ( x - 45 )° - 1 three times. (i) Find the x‑coordinates of the points of intersection. x = … or x = … or x = … [3] (ii) Solve the inequality. 2 cos ( x - 45)° - 1 2 0.005x … [2]

11 marks

Mark scheme: y f(x)=abs(2cos(x-0.785))-1 1.8 7(a) 1.6 3 Correct graph 1.4 1.2 1 0.8 B1 for 1 max in first quadrant 0.6 0.4 0.2 x B1 for V shape at approx x = –45 -1.7 -1.6 -1.5 -1.4 -1.3 -1.2 -1.1 -1 -0.9 -0.8 -0.7 -0.6 -0.5 -0.4 -0.3 -0.2 -0.1 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 -0.2 -0.4 -0.6 -0.8 -1 -1.2 -1.4 -1.6 -1.8 7(b) –75, 2 B1 for each –15 7(c) (45, 1) 1 7(d)(i) –64.8 or –64.76... 3 B1 for each –17.9 or – 17.92... or for –65, –18, 89 88.8 or 88.78 … If 0 scored SC1 for sketch of y = 0.005x 7(d)(ii) x < –64.8 2 FT –17.9 < x < 88.8 B1 for one correct

This question in 0607/43 May/June 2023

8 (a) Solve the simultaneous equations. 5x - 4y = 13 3x + 2y =- 1 You must show all your working. x = … y = … [3] 1 (b) f ( x) = 3 x + 1 g ( x) = , x ! 1.5 2x - 3 (i) Find f ( - 2) . … [1] (ii) Find f ( f ( x)) , giving your answer in its simplest form. … [2] 1 (iii) Solve g ( f ( x)) = . 5 x = … [3]

9 marks

Mark scheme: 8(a) Correct method to eliminate one variable M1 [ x  ] 1 A2 If 0 scored, SC1 for answers that [ y ] 2 satisfy one equation 8(b)(i) –5 1 8(b)(ii) 9x + 4 final answer M1 for 3  3 x  1  1 oe 2 8(b)(iii) 1 3 M2 for 2  3 x  1  3  5 oe 1 1 or M1 for  oe 2  3 x  1  3 5 OR M2 for f(x) = 4 oe 1 1 or M1 for  2f x  3 5

This question in 0607/43 May/June 2023

5 f ( x) = 2 x - 5 g ( x) = x 2 + x + 3 h ( x) = x3 j ( x) = 3x (a) The domain of f ( x) is 0 G x G 10 . Find the range of f ( x) . … [2] (b) Solve. (i) f ( x) =- 2 x = … [2] (ii) g ( x) = 3 - x x = … or x = … [3] (c) Find g ( f ( 4)) . … [2] (d) Find h ( 2) - j ( 2) . … [2] (e) Find h -1 ( x) . h -1 ( x) = … [1] (f) Find j -1 ( x) . j -1 ( x) = … [2]

14 marks

Mark scheme: 5(a) –5 ⩽ f(x) ⩽ 15 2 B1 for each If 0 scored, SC1 for -5 and 15 seen 5(b)(i) 1.5 oe 2 M1 for 2x = –2 + 5 5(b)(ii) –2 and 0 3 B2 for x2 + 2x = 0 oe or M1 for x2 + x +3 = 3 – x 5(c) 15 2 B1 for f(4) = 3 stated or used twice or M1 for (2x – 5)2 + (2x–5) + 3 oe 5(d) –1 2 M1 for 23 – 32 oe 5(e) 3 x oe 1 5(f) log x 2 log y log 3 x or M1 for x = log 3 y or x = log3 log3 or for x = 3y

This question in 0607/41 Oct/Nov 2023

7 (a) Marcus runs for 1 hour at x km/h and then walks for 2 hours at ( x - 5)km/h . He travels a total distance of 14 km. Find his running speed. … km/h [3] (b) Nina runs 5 km at y km/h and then walks 7 km at ( y - 7)km/h . She takes a total of 2 hours. (i) Show that 2y 2 - 26 y + 35 = 0 . [3] (ii) Solve 2y 2 - 26 y + 35 = 0 . y = … or y = … [3] (iii) Find Nina’s walking speed. … km/h [1]

10 marks

Mark scheme: 7(a) 8 3 B2 for x + 2x = 10 + 14 or better or M1 for x + 2(x – 5) = 14 oe 7(b)(i) 5 7 M1 + = 2 oe y y − 7 5(y – 7) + 7y = 2y(y – 7) M1 dep fractions correctly cleared dependent on equation with two different linear denominators leading to 2y2 – 26y + 35 = 0 A1 No errors or omissions 7(b)(ii) 1.53 or 1.525... AND 3 2 26  ( −26 ) − 4(2)(35) 11.5 or 11.47... M2 for 2(2) or correct shaped graph with 2 intersections on the positive x-axis or M1 for ( −26) 2 − 4(2)(35) 26 + (or −) p or for 2(2) If 0 scored, SC2 for 2 correct solutions given in surd form or SC1 for 1 correct solution 7(b)(iii) 4.5 or 4.47... 1 FT their positive solution – 7, provided final answer positive

This question in 0607/41 Oct/Nov 2023

9 (a) A bag contains 3 black discs and 5 white discs. Jani takes a disc from the bag at random. When the disc is black, he does not replace it. When the disc is white, he replaces it in the bag. Jani then takes a second disc at random. (i) Complete the tree diagram for the first and second discs. First disc Second disc Third disc Black … Black … … White Black … … White … White [3] (ii) Jani takes a third disc from the bag at random. Find the probability that he takes 2 black discs and 1 white disc. … [4] (b) Another bag contains 10 discs. x are red and the rest are green. (i) Write down an expression for the number of green discs. … [1] (ii) y blue discs are added to the bag. A disc is taken from the bag at random. (a) The probability of taking a red disc from the bag is 1. 3 Show that 3x = 10 + y . [1] (b) The probability of taking a green disc is 2. 9 Write another equation in x and y and find the number of red discs and the number of blue discs. number of red discs, x = … number of blue discs, y = … [5]

14 marks

Mark scheme: 9(a)(i) 3 5 2 5 3 5 3 B1 for each correct pair in correct position , , , , , 8 8 7 7 8 8 9(a)(ii) 365 4 FT their tree diagram probabilities for method or 0.233 marks 1568 3 2 5 3 5 2 5 3 2 M3 for   [ + ]   [ + ]   soi 8 7 6 8 7 7 8 8 7 without extras or M2 for two correct products soi or M1 for one correct product soi or for clear indication on tree diagram of all three combinations. or list of the options 9(b)(i) 10−x 1 9(b)(ii)(a) x 1 1 = oe seen 10 + y 3 leading to 3x = 10 + y 9(b)(ii)(b) [x = or red =] 6 5 B2 for 9 x + 2 y = 70 oe [y = or blue =] 8 10 − x 2 or B1 for = 10 + y 9 M1 for correct method to eliminate one variable A1 for [x = or red =] 6 or [y = or blue =] 8

This question in 0607/42 Oct/Nov 2023

11 (a) Simplify fully ( 64x 6 y 3 ) 3 . … [3] (b) 3 x # 2 x = 279 936 Find the value of x. x = … [2] (c) B NOT TO SCALE 15 x + 2 A C 2 x In triangle ABC, AB = BC . The perimeter of triangle ABC is 16 cm. (i) Show that 4x 2 - 1 = 0 . [5] (ii) Find the length of AB. AB = … cm [2]

12 marks

Mark scheme: 11(a) 16x 4 y 2 final answer 3 B2 for final answer kx 4 y 2 or 16 kx y 2 or 16 x 4 y k 2 or 4x 2 y ( ) B1 for 16 or x4 or y2 correct in 3 term final answer or M1 for 4 x 2  y or 4096 x12  y 6 seen 11(b) 7 nfww 2 M1 for 128 or 6x or 2187 seen OR log279936 M1 for x = log6 11(c)(i) 15 15 2 M1 + + = 16 x + 2 x + 2 x 30 2 or + = 16 x + 2 x 30 x + 2( x + 2)[ = 16] or better M2 M1 for 30 x + 2( x + 2) x ( x + 2) M1 for common denominator x ( x + 2) oe 30 x + 2 x + 4 = 16 x( x + 2) M1 FT their numerator with correct denominator to fraction removed rearranging to get to 4 x 2 −=1 0 A1 no errors or omissions 11(c)(ii) 6 2 1 M1 for x = or for 6 and 10 as answers 2

This question in 0607/42 Oct/Nov 2023

4 y 40 0 x -3 4 -40 f ( x) = 2x 3 - 3x 2 - 12x + 7 for -3 G x G 4 (a) Sketch the graph of y = f ( x) . [2] (b) Solve f ( x) = 0 . … [3] (c) Find the values of k for which f ( x) = k has exactly two solutions. k = … or k = … [2] (d) Find the range of values of x for which the gradient of f ( x) is negative. … [2]

9 marks

Mark scheme: 4(a) Correct sketch 2 B1 for any cubic with positive x3 4(b) –2.12 or –2.116... 3 B1 for each 0.537 or 0.5370... 3.08 or 3.079... 4(c) 14 and –13 cao 2 B1 for each or B1 for both14 and -13 seen 4(d) –1 < x < 2 2 B1 for each

This question in 0607/43 Oct/Nov 2023

8 (a) v = u + at Find v when u = 60, a =-32 and t = 3 . v = … [2] (b) Solve. (i) 6x + 2 = 9 - 4x x = … [2] (ii) 2x - 3 = 7 … [3] (c) Solve by factorisation. 3x 2 - 11x + 6 = 0 x = … or x = … [2] ax + 3b(d) Rearrange y = to make x the subject. 5x x = … [3] (e) Simplify. ax - 2bx + 3ay - 6by x 2 - 9y 2 … [4]

16 marks

Mark scheme: 8(a) –36 2 M1 for 60 + (– 32) × 3 oe or B1 for 96 seen 8(b)(i) 0.7 oe 2 M1 for 6x + 4x = 9 – 2 or better 8(b)(ii) 5 and -2 nfww 3 B2 for –2 nfww B1 for 5 or M1 for 2x – 3 = –7 or 2x – 3 = ±7 or M1 for a correct diagram 8(c) (3x – 2)(x – 3) M1 2 B1 [ x = ]3 oe , 3 8(d) 3b 3 M1 for 5xy = ax + 3b oe final answer M1FT for 5xy – ax = 3b 5 y − a M1FT for factorising and division Incorrect answers score M2 maximum. 8(e) a − 2b 4 B2 for (x + 3y)(a – 2b) oe final answer or B1 for x(a – 2b) + 3y(a – 2b) oe x − 3 y B1 for (x + 3y)(x – 3y)

This question in 0607/43 Oct/Nov 2023

2 Asif, Basheera and Chelsea make baskets. (a) The selling price of a basket increases by 8%. The new selling price is $4.86 . Find the original selling price of a basket. $ … [2] (b) Asif earns $4.70 per hour plus $1.21 for each basket he makes. Each week he works 8 hours a day for 5 days. Each day Asif makes 18 baskets. Calculate the total amount Asif earns in one week. $ … [3] (c) One day Basheera and Chelsea make a total of 36 baskets. They each work for 8 hours. Basheera takes x minutes to make a basket. Basheera takes 6 minutes longer than Chelsea to make a basket. (i) Write down an expression in terms of x for the number of baskets Chelsea makes. … [1] (ii) Write down an equation in terms of x and show that it simplifies to 3x 2 - 98x + 240 = 0 . [3] (iii) Solve the equation 3x 2 - 98 x + 240 = 0 . x = … or … [2] (iv) Find the number of baskets Chelsea makes. … [2]

13 marks

Mark scheme: 2(a) 4.5[0] 2  8  M1 for A   1 +  = 4.86 oe  100  2(b) 296.9[0] 3 M1 for 8 [× 5] × 4.7[0] oe M1 for [5 ×] 18 × 1.21 oe 2(c)(i) 480 1 oe x − 6 2(c)(ii) 480 480 M1 480 + = 36 oe Allow + their (i) = 36 x x − 6 x 480(x – 6) + 480x = 36x(x – 6) oe M1 Clearing fractions correctly (Dep on two fractions with different algebraic denominators) Completion to A1 3x2 – 98x + 240 [= 0] with no errors 2(c)(iii) 8 2 M1 for suitable sketch of parabola with 2 30, oe positive solutions 3 or (3x – 8)(x – 30) = 0 −−( 98)  ( −98) 2 − 4(3)(240) or 2  3 2(c)(iv) 20 2 480 M1 for their 30 − 6

This question in 0607/42 Feb/March 2024

1 (a) Solve the equations. (i) 3x - 2 = - 14 x = … [2] (ii) 7x + 11 = 26 - 3 x x = … [2] (b) Solve the simultaneous equations. You must show all your working. 5x + 3y = - 15 3x + 5y = - 17 x = … y = … [4] (c) Solve the inequality. 2x + 1 2 9 … [4]

12 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) –4 2 2 14 M1 for 3x = –14 + 2 or x   3 3 1(a)(ii) 15 2 M1 for 7x + 3x = 26 – 11 oe or better 1.5 or oe 10 1(b) Correctly equating coefficients M1 or sketch of one equation with negative slope and y intercept Correct method to eliminate one variable M1 or sketch of other equation with negative slope and y intercept If 0 scored, SC1 for answers that satisfy one x = –1.5 oe A1 equation. SC1 for two correct answers with no working y = –2.5 oe A1 1(c) x > 4 and x < –5 4 M2 for correct sketch showing both answers Mark final answer or M1 for appropriate sketch of y = 2 x  1 OR M2 for 2x + 1 > 9 oe and 2x + 1 < –9 oe or M1 for either correct inequality or for 2x + 1 = 9 and 2x + 1 = –9 oe B1 for x > 4 or x < –5 Mark final answer

This question in 0607/41 May/June 2024

2 (a) The heights, x cm, of 100 plants are shown in the table. Height (x cm) 0 1 x G 20 20 1 x G 35 35 1 x G 40 40 1 x G 60 60 1 x G 80 Frequency 7 13 20 32 28 (i) Calculate an estimate of the mean height of the plants. … cm [2] (ii) (a) Complete the cumulative frequency table for the plants. Height (x cm) x G 20 x G 35 x G 40 x G 60 x G 80 Cumulative 7 100 frequency [1] (b) On the grid, draw the cumulative frequency curve. 100 90 80 70 60 Cumulative 50 frequency 40 30 20 10 0 x 0 10 20 30 40 50 60 70 80 Height (cm) [3] (c) Use your cumulative frequency curve to find an estimate for the interquartile range. … cm [2] (b) The heights, h cm, of 50 different plants are shown in the table, where k is an integer. Height (h cm) Frequency 0 1 h G 20 25 20 1 h G k 15 k 1 h G 80 10 An estimate of the mean height of these plants is 27 cm. Find the value of k. k = … [3]

11 marks

Mark scheme: 2(a)(i) 47.4 or 47.375 2 M1 for at least 4 correct mid-points soi 2(a)(ii)(a) [7], 20, 40, 72, [100] 1 In parts (a)(ii)(b) and (a)(ii)(c), marks can only be earned with an increasing curve (a)(ii)(b) Correct curve 3 B1 for horizontal plot correct (20, 7), (35, 20), (40, 40), (60, 72), B1FT for at least 4 vertical plots correct (80, 100) (a)(ii)(c) 24 to 28 2 M1FT for [UQ =] 61 to 63 or [LQ =] 35 to 37 2(b) 44 3 M2 for (20  k ) ( k  80) 10  25   15   10 [  27  50] 2 2 oe (20  k ) ( k  80) or M1 for or 2 2

This question in 0607/41 May/June 2024

6 y 3 – 3 0 3 x – 3 f( x) = 2 - 1 - 0 .5 x 2 (a) On the diagram, sketch the graph of y = f( x) , for values of x between - 3 and 3. [3] (b) The graph cuts the x-axis at points A and B. Work out the length AB. AB = … [2] (c) Solve f( x) = 0 .5 . … [2] (d) Write down the coordinates of the minimum point of the graph. ( … , … ) [1] (e) The equation f( )x = k has two solutions. Find the range of values of k. … [2]

10 marks

Mark scheme: 6(a) Correct sketch 3 B2 if peaks below y = 1 (by eye) or rounded at peaks not cusps or outside branches convex from below. or B1 for graph symmetrical about y-axis and in all 4 quadrants 6(b) 4.9[0] or 4.898 to 4.899 2 B1 for 2.45 or –2.45 or 2.449... or –2.449… seen 6(c) –2.24 or –2.236… 2 B1 for each or for both values seen 2.24 or 2.236… 6(d) (0, 1) 1 6(e) k =2 2 B1 for each k < 1

This question in 0607/41 May/June 2024

11 f( )x = 3 x - 1 g( )x = 5 - 2 x h( x) = , x ! 1 .5 2x - 3 (a) Find f( 4) . … [1] (b) Solve f( )x = - 7 . … [2] (c) Find g -1 ( )x . g -1 ( )x = … [2] (d) Solve g( x) = 7 h(f( x)) . You must show all your working. x = … [6]

11 marks

Mark scheme: 11(a) 11 1 11(b) –2 2 M1 for 3x = – 7 + 1 11(c) 5 x 2 y 5 oe final answer M1 for x = 5 – 2y or   x or 2x = 5 – y 2 2 2 11(d) 1 M1  h(f ( x ))   2(3 x  1)  3 (5  2 x )(6 x  5)  7 A1 All further FTs dep on second stage in correct form (5 – 2x)(ax + b) = k where a, and b are integers or sketch of rectangular hyperbola Correct expansion of brackets M1 30x – 25 – 12x2 + 10x [= 7] or sketch of straight line with negative gradient Correct rearrangement to 3 term quadratic M1 12 x 2  40 x  32  0 oe on one side or graphs intersecting twice in 1st quadrant Correct factorisation M1 (6 x  8)(2 x  4)  0 oe or correct use of formula or correct sketch of the quadratic or solutions indicated at points of intersection 4 B1 Both answers correct 2, oe 3

This question in 0607/41 May/June 2024

7 (a) Solve 63 = 8 ( 3 - 2 a) . a = … [3] (b) Solve the simultaneous equations. You must show all your working. p 5 - q = 3 12 q 7 2 p + = 2 8 p = … q = … [3] (c) (i) Factorise c 2 - c - 56 . … [2] (ii) Solve c 2 - c - 56 = 0 . c = … or c = … [1]

9 marks

Mark scheme: 7(a) 39 3 B1 for [63 = ] 24 – 16a  oe or –2.44 16 M1 for their 16a = their 24 – 63 oe or better 7(b) Correctly equating coefficients M1 Allow 1 arithmetic slip Correct method to eliminate M1 Allow 1 further arithmetic slip one variable 1 B1 If 0 scored, SC1 for answers that satisfy one [p = ] oe equation 2 1 [q = ]  oe 4 7(c)(i) ( c  7)( c  8) oe 2 B1 for ( c  a )( c  b ) where ab = –56 or a + b = –1 or c ( c  7)  8( c  7) or c ( c  8)  7( c  8) 7(c)(ii) –7, 8 1 FT their (i)

This question in 0607/42 May/June 2024

4 Line L has equation 3y + 2x = 8 . (a) Find the gradient and the y-intercept of line L. gradient … y-intercept … [3] (b) Line P passes through the point (2, 10) and is perpendicular to line L. Show that the equation of line P is 2y - 3x = 14 . [3] (c) Find the coordinates of the point where line L and line P intersect. You must show all your working. ( … , … ) [4]

10 marks

Mark scheme: 4(a) 2 3 B2 for one correct  oe or M1 for correctly isolating y oe 3 8 2 or 2 oe 3 3 4(b) 3 M1 2 gradient = FT 1 ÷ their  2 3 substituting (2, 10) into M1 2 FT their m ≠  y = their m + c 3 completing to 2y – 3x = 14 with at A1 least one line of working and no errors 4(c) Correctly equating coefficients M1 or sketch of one equation with positive slope and positive y-intercept Correct method to eliminate one M1 variable or sketch of other equation with negative slope and positive y-intercept x = –2 in correct answer space A1 y = 4 in correct answer space A1 If 0 scored, SC1 for correct answer with no working

This question in 0607/43 May/June 2024

5 y 20 – 5 0 2 x – 20 f ( x) = 5 + 2x - 4x 2 - x 3 for - 5 G x G 2 (a) On the diagram, sketch the graph of y = f ( x) . [2] (b) Find the zeros of f ( x) . … [3] (c) Write down the coordinates of the local minimum. ( … , … ) [2] (d) The point ( a, b) lies on the graph of y = f ( x) where the gradient is positive. Find the range of values for a. … [2] (e) The equation 5 + 2 x - 4 x 2 - x 3 = k has exactly one solution. Write down a possible value of the integer k. … [1]

10 marks

Mark scheme: 5(a) Correct sketch 2 With minimum in 3rd quadrant and maximum in 1st quadrant B1 for any cubic with negative x3 5(b) –4.19 or –4.193 to –4.192 3 B1 for each –1 1.19 or 1.192 to 1.193 or B1 for –1 and B1 for –4.2 and 1.2 5(c) (–2.9[0], –10.1) 2 B1 for each coordinate or (–2.897 to –2.896, –10.05...) 5(d) their –2.9[0] < a < 0.23[0] 2 –2.896 to 2.897 , 0.2301... B1 for 0.23[0] seen or their –2.9[0] < a < k 5(e) Integer ⩽ –11 or ⩾ 6 1

This question in 0607/43 May/June 2024

9 (a) Solve. (i) 2x + 3 = 1 - 5x x = … [2] (ii) x + 3 = 2 … [2] (b) Factorise completely. 6x 3 y 2 - 3x 2 y 3 … [2] 5 2 (c) Write - as a single fraction in its simplest form. 2x + 3 x - 5 … [3] (d) Solve 2x 2 + 3x = 7 . You must show all your working and give your answers correct to 2 decimal places. x = … or x = … [3]

12 marks

Mark scheme: 9(a)(i) 2 2 M1 for 2x + 5x = 1 – 3 or better  7 9a)(ii) –5 2 B1 for each –1 or M1 for x + 3 = ±2 9(b) 3x2y2(2x – y) Final answer 2 B1 for correctly extracting 2 or more factors 9(c) x  31 x  31 3 B1 for 5(x – 5) – 2(2x + 3) oe ISW or 2 B1 for denominator (2x + 3)(x – 5) oe (2 x  3)( x  5) 2 x  7 x  15 Final answer 9(d) M2 2 M1 for 3  4 2 7 3 32 4 2 x  7 3  p 3 p or M1 for or 2  2 2  2 2  2 or suitable sketch(es) with both Denominator must be shown as 2  2 to earn the answers indicated second M1 but a denominator of 4 is condoned for M2 1.27 and –2.77 cao B1

This question in 0607/43 May/June 2024

10 f ( x) = 5 - x g ( x) = 3 ( x + 1) h ( x) = sin xc for 0 G x G 180 2 (a) Find f ( 3) . … [1] (b) Solve f ( x) = 2 . x = … [2] (c) Find and simplify f ( g ( x) ) . … [2] (d) Find g -1( )x . g -1( )x = … [2] (e) Find h ( g ( 29)) . … [2] (f) Using a graphical method, solve h ( g ( x)) = 1 - 0. 01x . y 2 0 x 180 – 2 … [5]

14 marks

Mark scheme: 10(a) 1 1 3 oe 2 10(b) 6 2 1 M1 for 5 – x = 2 2 10(c) 1 1 7 x3 2 1 3  1 x or oe Final answer M1 for 5  (3( x  1)) oe 2 2 2 2 10(d) x  3 2 y oe Final answer M1 for x = 3(y + 1) or x + 1 = or y – 3 = 3x 3 3 10(e) 1 2 B1 for h(90) or M1 for sin(3(x + 1)) oe 10(f) sin(3(x + 1)) soi 1 Correct sketches e.g. 2 or a single graph of h(g(x)) – 1 + 0.01x B1 for each graph 17.5 or 17.52... 2 B1 for 1 correct. 48.7 or 48.71... 115.9 or 115.94...

This question in 0607/43 May/June 2024

1 (a) Solve the equations. (i) 6x + 5 =-19 x = … [2] (ii) 8x - 13 = 11 - 4x x = … [2] 8 (iii) =- 5 2x - 3 x = … [3] (b) Solve the equation 6x 2 - 2 x - 1 = 0 . Give your answers correct to 2 decimal places. You must show all your working. x = … or x = … [3]

10 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) –4 2 5 19 M1 for 6 x = −19 − 5 or x + = − 6 6 1(a)(ii) 2 2 M1 for 8x + 4 x = or px + qx = 11 + 13 1(a)(iii) 7 3 M1 for correctly eliminating fractions 0.7 or M1 for correct rearrangement into the form 10 ax = b 1(b) Correct sketch M1 or Correct substitution into formula 0.61 –0.27 A2 A1 for each or both correct to a greater degree of accuracy SC1 for two correct answers with no or incorrect working

This question in 0607/41 Oct/Nov 2024

9 y 4 x 0 -2 4 - 4 1 f ( x) = ( 2x - 3)( 2x + 1) (a) On the diagram, sketch the graph of y = f ( x) for values of x between - 2 and 4. [3] (b) Write down the equations of the asymptotes parallel to the y-axis. … [2] (c) Write down the coordinates of the local maximum. ( … , … ) [2] (d) The line y = x - 2 intersects the curve y = f ( x) three times. Find the x-coordinate of each point of intersection. x = … or x = … or x = … [3] (e) Solve the inequality f ( )x H x - 2 . … [3]

13 marks

Mark scheme: 9(a) Correct sketch 3 B1 for correct shape with 3 branches B1 for the local maximum in correct position, not above x-axis B1 for graph with no excessive overlaps, gaps or curl backs, the upper branches not crossing the x-axis 9(b) x = –0.5 x = 1.5 2 B1 for each 9(c) (0.5, –0.25) 2 B1 for each 9(d) –0.448 1.3[0] 2.15 3 B1 for each If 0 scored, SC1 for –0.45, 1.3 and 2.1 9(e) [ −2 ] x −0.5 3 B1 for each, strict inequality on the asymptote −0.448 x 1.30 values – only penalised once 1.5 x 2.15

This question in 0607/41 Oct/Nov 2024

2 (a) Work out 24% of $15.50 . $ … [2] (b) The price of a bookcase is $123. This price is increased by 7%. Calculate the new price. $ … [2] (c) An amount of money is shared between Ali, Kat and Lena in the ratio 5 : 3 : 4. Lena’s share is $76. Work out the total amount of money. $ … [3] (d) A library has 32 800 books. Each year the number of books in the library increases by 300. Calculate the number of years it takes until there are 40 000 books in the library. … [2] (e) A different library has 32 695 books at the end of 2024. Each year the number of books increases by 0.6% of the number of books in the library at the end of the previous year. (i) Calculate the number of books the library had at the end of 2023. … [2] (ii) Calculate the number of complete years from 2024 that it takes for the number of books to first be greater than 40 000. … [4]

15 marks

Mark scheme: 2(a) 3.72 2 24 M1 for  15.5 oe 100 2(b) 131.61 final answer 2 100 + 7 M1 for  123 oe 100 or B1 for 8.61 2(c) 228 3 76 M2 for  ( 5 + 3 + 4 ) oe 4 76 or M1 for 4 2(d) 24 2 M1 for 300 +x 32 800 = 40 000 oe 2(e)(i) 32500 2  100 + 0.6 M1 for [...]  = 32 695 oe  100  2(e)(ii) 34 nfww 4 B3 for 33.71… or 33.7 OR 0.6 40 000 M3 for n log(1 + ) = log oe 100 32 695 or a good sketch indicating value between 33 and 34 or correct trials reaching 33 and 34  0.6  n 40 000 or M2 for 1 + = oe    100  32 695 or suitable graph or at least three correct trials  0.6  n or M1 for 32 695 1 + = 40 000 oe soi    100  or at least 2 trials with n > 1

This question in 0607/43 Oct/Nov 2024

4 y 17 0 x -3 3 -13 3 3 2 f ( )x = x - 4x + 2 g ( )x = + x x (a) On the diagram, sketch the graph of y = f ( x) for values of x between -3 and 3. [2] (b) Find the solutions of f ( )x = 0 . x = … , x = … , x = … [3] (c) On the diagram, sketch the graph of y = g ( x) for values of x between -3 and 3. [3] (d) Write down the equation of the asymptote of the graph of y = g ( x) . … [1] (e) Solve f ( x) G g ( x) . … [4]

13 marks

Mark scheme: 4(a) correct sketch 2 M1 for positive cubic shape 4(b) –2.21 0.539 1.68 3 B1 for each correct or –2.214… 0.5391… 1.675… penalise 1 mark if y co-ordinates included if 0 scored SC1 for –2.2, 0.54 and 1.7 4(c) correct sketch 3 For full marks there must be exactly one intersection in the first quadrant B2 for both branches but joined or touching the y axis or B1 for one correct branch on either side of y axis 4(d) x = 0 1 4(e) [ −3] ⩽ x ⩽ –1.96 or –1.959…. 4 B2 for x ⩽ –1.96 0 < x ⩽ 2.48 or 2.482…to 2.483 or B1 for –1.96 seen B2 for 0 < x ⩽ 2.48 or B1 for 2.48 seen

This question in 0607/43 Oct/Nov 2024

7 (a) Find the next term and the nth term for each of these sequences. (i) 19 16 11 4 next term = … nth term = … [3] (ii) 20 10 5 2.5 next term = … nth term = … [3] (b) The nth term of a sequence is 2n 2 - 3n + 1 . The kth term is 465. Work out the value of k. k = … [3]

9 marks

Mark scheme: 7(a)(i) -5 1 20 −n 2 oe 2 M1 for expression involving −n 2 or for 2nd differences of –2 or 2 seen 7(a)(ii) 1.25 oe 1 n n −1 2 n  1   1   1  40   or 20   oe M1 for an expression involving   or 2−n oe  2   2   2  7(b) 16 3 2 3  ( −3) −−4 2 ( 464) M2 oe 2  2 or ( k − 16)(2k + 29) or for a correct sketch indicating solutions or M1 for correct use of formula but with one error or for 2 k 2 − 3k − 464 = 0 or for a correct sketch

This question in 0607/43 Oct/Nov 2024

8 (a) The amount charged for electricity in one month is $E. $E is the sum of a fixed charge $f and a cost of $d for each unit of electricity used. Find a formula for the amount charged in one month when u units of electricity are used. … [2] (b) Write as a single fraction in its simplest form. x 2 x 5x - + 2 3 18 … [2] (c) Solve 7n - 9 2 21 + 2 n . … [2] (d) Solve the simultaneous equations. You must show all your working. 2x + 15y = –57 20x + 3y = 18 x = … y = … [3] (e) y is proportional to the square of ( x - 3) . y = 5 when x = 7 . Find the value of y when x = 27 . y = … [3]

12 marks

Mark scheme: 8(a) E = du + f final answer 2 M1 for du + f 8(b) x 2 M1 for correct use of common denominator eg final answer 9 9 x 12 x 5 x − + 18 18 18 8(c) n  6 final answer 2 M1 for 7n − 2n *21 + 9 or better * can be = or any inequality 8(d) correctly equating one set of M1 coefficients Or correctly making x or y the subject of an equation and correct substitution x = 1.5 A2 A1 for each y = −4 If M0 scored SC1 for correct substitution and evaluation to find the other variable. or SC1 if no working shown, but 2 correct answers given. 8(e) 180 3 5 2 M2 for y = their ( x − 3) oe 16 OR M1 for y = k ( x − 3) 2 5 A1 for k = 16

This question in 0607/43 Oct/Nov 2024

18 - = 1 2x - 5 x + 1 (a) Show that 2x 2 - x - 45 = 0 . [4] (b) Solve by factorising. 2x 2 - x - 45 = 0 x = … or x = … [3]

7 marks

Mark scheme: 18(a) 10(x + 1) – 6(2x – 5) or better M1 2 x 2 − 5 x + 2 x − 5 M1 Correct method for clearing M1 fractions Leading to 2 x 2 − x − 45 = 0 A1 No errors or omissions 18(b) (2x + 9)(x – 5) [= 0] M2 M1 for (2x + a)(x + b) where ab = –45 or a + 2b = −1 or 2x(x – 5) + 9(x – 5) [=0] or x(2x + 9) – 5(2x + 9) [=0] [x =] 5 B1 [x =] –4.5 oe

This question in 0607/42 Feb/March 2025

13 Pierre drives from town P to town Q and then to town R. The distance from P to Q is 112 km. The distance from Q to R is 87 km. The average speed from P to Q is x km/h. The average speed from Q to R is ( x + 10 ) km/h. The total time for the journey from P to Q to R is 3 hours. (a) Show that 3x 2 - 169 x - 1120 = 0 . [4] (b) (i) Solve 3x 2 - 169 x - 1120 = 0 . You must show your working. x = … or x = … [3] (ii) Find the average speed from Q to R. … km/h [1]

8 marks

Mark scheme: 13(a) 112 87 M2 112 87 + = 3 oe B1 for or x x + 10 x x + 10 112(x + 10) + 87x = 3x(x + 10) oe M1 Clearing two algebraic fractions Completion to 3x2 – 169x – 1120 = 0 A1 with no errors or omissions 13(b)(i) Correct sketch of 3x2 – 169x – 1120 M1 With negative and positive x-intercept or ( −−)( 169)  ( −169) 2 − 4(3)( −1120) 2  3 62.3 or 62.32… B2 B1 for each –5.99 or –5.990… 13(b)(ii) 72.3 or 72.32… 1 FT 10 + their positive answer to (i)

This question in 0607/41 May/June 2025

8 y 10 x -3 0 3 -10 3 2 f ( )x = - x x (a) On the diagram, sketch the graph of y = f ( x) for values of x between -3 and 3. [3] (b) Find the zero of f ( )x . … [1] (c) Find the equation of the asymptote to the graph. … [1] (d) Find the coordinates of the local maximum point. ( … , … ) [2] (e) The equation f ( )x = k has one solution. Find the range of values of k. … [1]

8 marks

Mark scheme: 8(a) Correct sketch 10101010 3 Both branches correct, without excessive gaps, curl backs, overlaps or feathering. 5555 Max point in 3rd quadrant and intersection with x-axis between 1 and 2. 3333 -2-2-2-2 -1-1-1-1 0000 0000 1111 2222 3333 B1 each branch correct shape (if joined B1 B1) -5-5-5-5 8(b) 1.44 or 1.442... 1 8(c) x = 0 1 8(d) (–1.14, –3.93) 2 B1 for each or for (–1.1, –3.9) 8(e) k > –3.93 1 FT their y-coordinate in part (d)

This question in 0607/43 May/June 2025

14 Priya buys pieces of fruit. The pieces of fruit are apples and oranges. The price of 1 apple is x cents. The price of 1 orange is ( x + 10 ) cents. Priya spends 420 cents on apples. (a) Write down an expression, in terms of x, for the number of apples she buys. … [1] (b) Priya also spends 420 cents on oranges. She buys 13 pieces of fruit. (i) Show that 13x 2 - 710 x - 4200 = 0 . [3] (ii) Factorise 13x 2 - 710 x - 4200 . … [2] (iii) Find the cost of one orange. … cents [2]

8 marks

Mark scheme: 14(a) 420 1 x 14(b)(i) 420 420 M1 + = 13 x x + 10 420( x + 10) + 420 x = 13x ( x + 10) M1 Correctly clearing 2 fractions with linear denominators Correctly expanding brackets and A1 No errors or omissions leading to 13 x 2 − 710 x − 4200 = 0 14(b)(ii) (13 x + 70 )( x − 60 ) final answer 2 B1 for (13 x + a )( x + b ) where ab = –4200 or a + 13b = –710 or for 13 x( x − 60) + 70( x − 60) or for x (13 x + 70) − 60(13 x + 70) 14(b)(iii) 70 2 FT their +ve x + 10 B1 for correct positive x from their factors

This question in 0607/43 May/June 2025

8 y 5 x 0 – 5 5 – 5 2x f ( x) = x + 2 (a) On the diagram, sketch the graph of y = f ( x) for values of x between -5 and 5. [3] (b) Find the coordinates of the local minimum. ( … , … ) [2] (c) Solve f ( x) = 3 . … [2] (d) Write down a value of k for which f ( x) = k has only one solution. … [1] (e) Find the range of values of k for which f ( x) = k has no solutions. … [2]

10 marks

Mark scheme: 8(a) Correct sketch 3 B1 for either branch with correct shape B1 for two branches of the correct shape not crossing the x-axis B1 for no excessive overlaps, gaps, curlbacks or feathering If sketch not fully correct, a maximum of 2 marks 8(b) (– 0.557, 0.471) or 2 B1 for each coordinate (– 0.5573…, 0.4710…) or B1 for (– 0.56, 0.47) 8(c) – 1.91 or –1.911… 2 B1 for each 4.22 or 4.222… or B1 for both correct with y-coordinates included or both correct in an inequality or B1 for – 1.9 and 4.2 8(d) their 0.471 or for k < 0 1 8(e) 0 ⩽ k < their 0.471 oe 2 B1 for either inequality

This question in 0607/41 Oct/Nov 2025