Algebraic fractions — Paper 4 · IGCSE Mathematics - International 0607

12 f(x) = 2x + 1 g(x) = 4 – 3x h(x) = 2x – 1 (a) Find h(-2). … [1] (b) Find g-1(x). g-1(x) = … [2] (c) Find g(f(3)). … [2] (d) Find and simplify g(g(x)). … [2] (e) Find h-1(7). … [2] (f) Write as a single fraction in its simplest form. 1 1 + f x g x ^ h ^ h … [3]

12 marks

Mark scheme: 12(a) 3 1 – or – 0.75 4 12(b) 4 − x 2 M1 for x = 4 – 3y or y + 3x = 4 or oe final answer 3 y 4 y – 4 = –3x or = – x 3 3 12(c) –17 2 B1 for [f(3)] = 7 or M1 for 4 – 3(2x + 1) soi 12(d) 9x – 8 final answer 2 M1 for 4 – 3(4 – 3x) 12(e) 3 2 M1 for 2x – 1 = 7 or log2(x + 1) oe 12(f) 5 − x 3 M1 for 4 – 3x + 2x + 1 oe final answer M1 for common denominator (2 x + 1)(4 − 3x ) (2x + 1)(4 – 3x)

This question in 0607/43 May/June 2018

12 f x = 10 - x g x = x 2 + 1 h x = j x = log 3 x x (a) Find g(3). … [1] (b) Find f(h(2)). … [2] (c) Find g(f(x)) in the form ax 2 + bx + c . … [3] (d) For some functions, p-1(x) = p(x). Write down which two functions, f(x), g(x), h(x) or j(x), have this property. … and … [2] 1 (e) Write h x - as a single fraction in its simplest form. ` j f x ` j … [3] (f) (i) Find j(243). … [1] (ii) Find x when j(x) = 1.5 . x = … [1] (iii) Find j-1(x). j – 1(x) = … [2]

15 marks

Mark scheme: 12(a) 10 1 12(b) 9.5 oe 2 1 1 M1 for 10 − soi e.g. 10 – x 2 12(c) x 2 − 20 x + 101 3 M1 for (10 −x ) 2 + 1 B1 for 100 – 10x – 10x + x2 oe 12(d) f(x) and h(x) 2 B1 for each 12(e) 10 − 2 x 3 M1 for common denominator x(10 – x) oe oe B1 for (10 – x) – x oe seen x (10 − x ) 12(f)(i) 5 1 12(f)(ii) 3 1 3 3 oe or 3 2 or 5.2[0] or 5.196... 12(f)(iii) 3x 2 M1 for x = log 3 y or x = 3 y

This question in 0607/42 May/June 2019

9 (a) Lionel runs 10.6 km in 94 minutes. Calculate his average speed in km/h. … km/h [2] (b) Monika walks 2 km at a speed of 4 km/h and then 3 km at a speed of 3 km/h. Calculate Monika’s overall average speed. … km/h [3] (c) A train is travelling at v metres per second. Find an expression, in terms of v, for the speed of the train in kilometres per hour. Give your answer in its simplest form. … km/h [2] (d) (i) A car travels 50 km at x km/h and then 80 km at (x + 10) km/h. Find an expression, in terms of x, for the total time taken, T hours. Give your answer as a single fraction, in its simplest form. T = … h [3] (ii) When T = 2 , show that x 2 - 55x - 250 = 0 . [2] (iii) When T = 2 , find the value of x. x = … [3]

15 marks

Mark scheme: 9(a) 6.77 or 6.765 to 6.766 2 M1 for 10.6 ÷ 94 or 94 ÷ 60 9(b) 1 3 2 + 3 3.33 or 3.333... or 3 M2 for 3 2 3 + 4 3 2 3 or M1 for oe or oe 4 3 9(c) 18 v 3 2 M1 for × (60 × 60) oe or for ÷ 1000 or 3.6v or v 5 35 9(d)(i) 130 x + 500 130 x + 500 3 50 80 or M1 for + x ( x + 10) x 2 + 10 x x x + 10 B1 for common denominator x(x + 10) oe 9(d)(ii) 130 x + 500 = 2 x ( x + 10) oe M1 i.e. fraction with linear numerator and quadratic denominator removed correctly 130 x + 500 = 2 x 2 + 20 x A1 i.e. equation with four terms 2 no errors or omissions leading to x − 55 x − 250 = 0 9(d)(iii) 59.2 or 59.22... only 3 M2 for correct graph of quadratic showing positive root −−( 55) ± ( − 55) 2 − 4(1)( − 250) or for oe 2(1) or M1 for appropriate quadratic graph or for ( −55) 2 − 4(1)( −250) oe −−( 55) or for oe in correct formula 2(1)

This question in 0607/41 Oct/Nov 2019

13 f ()x = 2x + 5 g ()x = 1 - 2x (a) Find g (- 4) . … [1] (b) Find f - 1 (- 7) . … [2] (c) Find g (f (3)) . … [2] (d) Find and simplify f (g (x)) . … [2] (e) Find and simplify g -1 ()x . g -1 ()x = … [2] (f) Write as a single fraction, simplifying your answer. 3 2 + f ()x … [2]

11 marks

Mark scheme: 13(a) 9 1 13(b) –6 2 x− 5 M1for f(x) = –7 or for f–1(x) = 2 13(c) –21 2 B1 for 11 seen or M1 for 1 – 2(2x + 5) 13(d) 7 – 4x 2 M1 for 2(1 – 2x) + 5 13(e) 1 − x 2 M1 for 2x = 1 – y or x = 1 – 2y oe 2 y 1 or = – x 2 2 13(f) 4 x + 13 2 2(2 x + 5) + 3 final answer M1 for 2 x + 5 2 x + 5

This question in 0607/43 Oct/Nov 2019

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

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

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

12 (a) Solve. 2 (i) 9 = 5 - x x = … [3] 6 (ii) 2 3 x - 4 … [3] (b) (i) Solve the equation, giving your answers correct to 3 significant figures. 2x 2 - 5x + 1 = 0 x = … or x = … [3] (ii) Use your answers to part (b)(i) to solve 2 ( tan y) 2 - 5 ( tan y) + 1 = 0 for 0° G y G 180° . y = … or y = … [2]

11 marks

Mark scheme: 12(a)(i) –0.5 oe 3 x 1 M2 for = − or 4 x = − 2 2 4 2 or M1 for = 5 − 9 oe or 9 x = 5 x − 2 oe x 12(a)(ii) 4 < x < 6 3 B2 for x < 6 seen and not spoiled or B1 for [x =] 6 seen OR 6 − 3 x + M2 for 12[ > 0] x − 4 3( x − 4) or M1 for soi x − 4 OR M2 for correct graph showing answers or M1 for appropriate graph 12(b)(i) 0.219 3 B2 for 0.2192... or 0.22 and 2.280 to 2.281 2.28 or M1 for correct curve or correct use of formula 12(b)(ii) 12.4 or 12.35 to 12.36... 2 B1 for each 66.3 or 66.31 to 66.33 FT their (b)(i) 13 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 0607/43 Oct/Nov 2021

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

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

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

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

10 (a) Simplify. 3x - 5y + 4x - 6y … [2] (b) Expand. x ( x + 2) … [1] (c) Factorise. 10ab + 8ac - 15b 2 - 12bc … [2] 2 5 (d) - = 3 2x + 1 x - 3 (i) Show that 6x 2 - 7x + 2 = 0 . [4] (ii) Solve 6x 2 - 7x + 2 = 0 . You must show all your working. x = … or x = … [3]

12 marks

Mark scheme: 10(a) 7x – 11y Final answer 2 B1 for 7x – ky or kx – 11y k not zero 10(b) x2 + 2x Final answer 1 10(c) (5b + 4c)(2a – 3b) Final answer 2 M1 for 2a(5b + 4c) – 3b(5b + 4c) or 5b(2a – 3b) + 4c(2a – 3b) or correct answer seen but spoiled 10(d)(i) 2(x – 3) – 5(2x +1) = 3(2x + 1)(x – 3) M1 oe or better 2x – 6 – 10 x – 5 or better B1 [3](2x2 – 6x + x – 3) oe or better B1 completion to 6x2 – 7x + 2 [ = 0] A1 at least one step with no errors or omissions 10(d)(ii) (2x – 1)(3x – 2) [= 0] M2 M1 for pair of brackets giving two terms or sketch of parabola showing two correct positive solutions or sketch of any parabola for +ve x2 2 ( 7) ( 7)  (  7)  4(6)(2) or correct formula with or or 2  6 2  6 ( 7) 2  4(6)(2) seen 1 2 B1 , oe 2 3

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

10 (a) Simplify. k t (i) # 2p 3 … [1] u 2u (ii) + 7 21 … [2] (b) Simplify. x 2 - x - 42 2x 2 - 98 … [4] (c) Write as a single fraction in its simplest form. g - 1 2g - + 4 g + 1 5 … [3]

10 marks

Mark scheme: 10(a)(i) kt 1 final answer 6 p 10(a)(ii) 5u 2 M1 for correct use of common final answer 21 3uk 2uk denominator + 21k 21k 10(b) x + 6 x + 6 4 B2 for (x + 6)(x – 7) or final answer or 2( x + 7) 2 x + 14 B1 for (x + a)(x + b) with ab = –42 or a + b = –1 or for x(x + 6) – 7(x + 6) or x(x – 7) + 6(x – 7) B1 for 2(x + 7)(x – 7) or (2x + 14)(x – 7) or (2x – 14) (x + 7) 10(c) −2 g 2 + 23 g + 15 3 B1 for 5(g – 1) – 2g(g + 1) + 5  4(g + 1) final answer oe or better 5( g + 1) B1 for common denominator seen 5(g + 1) or 5g + 5

This question in 0607/41 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

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

11 (a) Solve. 3x + 2 2 7x - 8 … [2] (b) Factorise fully. 75x 2 - 3 … [2] (c) Simplify. 2 1 1 (i) + - 3x 6x 5x … [2] 2 x 2 + 3 x - 2bx - 3b (ii) 2 2x - 7x - 15 … [4]

10 marks

Mark scheme: 11(a) x < 2.5 oe final answer 2 M1 for 2 + 8 > 7x – 3x oe or B1 for x * 2.5 where * is =, >, ≤ or ≥ 11(b) 3(5x + 1)(5x – 1) final answer 2 B1 for 3(25x2 – 1) or (15x + 3)(5x – 1) or (15x – 3)(5x + 1) 11(c)(i) 19 2 B1 for any equivalent cao final answer or M1 for correct use of common 30x denominator 20 + 5 − 6 20 x + 5 x − 6 x e.g. , etc. oe 30 x 30 x 2 11(c)(ii) x − b 4 B3 for (x – b)(2x + 3) and (x – 5)(2x + 3) final answer or B2 for (x – b)(2x + 3) x − 5 or for (x – 5)(2x + 3) or B1 for x(2x + 3) – b(2x + 3) or 2x(x – b) + 3(x – b) or x(2x + 3) – 5 (2x + 3) or 2x(x – 5) + 3 (x – 5) or (2x + c)(x + d) where c + 2d = –7 or cd = –15

This question in 0607/42 Feb/March 2024

11 (a) Simplify. 9x 2 - 4y 2 9x 2 - 6xy … [3] 5 7 (b) - = 2 2x - 3 4 - x (i) Show that 4x 2 - 41x + 65 = 0 . [3] 5 7 (ii) Solve - = 2 , giving your answers correct to 2 decimal places. 2x - 3 4 - x You must show all your working. x = … or x = … [4]

10 marks

Mark scheme: 11(a) 3 x  2 y 2 y 3 B1 for (3 x  2 y )(3 x  2 y ) isw or 1  final answer 3 x 3 x B1 for 3 x (3 x  2 y ) isw 11(b)(i) 5(4  x )  7(2 x  3)  2(2 x  3)(4  x ) M1 Correctly clearing fractions or better 2  3 x B1  (2 x  3)(4  x )   8 x  12  2 x or 2  6 x 2  (2 x  3)(4  x )   16 x  24  4 x Leading to 4 x 2  41x  65  0 with no A1 errors or omissions 11(b)(ii) 2 M2 2 ( 41)  ( 41) 4 4 65 M1 for (  41)  4 4 65 or 2  4 ( 41)  or  p or sketch with both answers indicated 2  4 or suitable sketch which would lead to answers 1.96 , 8.29 cao B2 B1 for each or for 1.960... and 8.289 to 8.290

This question in 0607/42 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

6 f ( )x = 5x - 1 g ( )x = x 2 + x h ( x) = ( x - 1) 3 The domain for all three functions is x 2 2 . (a) Find f ( 3 ) . … [1] (b) Find the range of f ( )x . … [1] (c) Find g ( f ( 4)) . … [2] (d) Find h -1 ( )x . h -1 ( )x = … [2] (e) Simplify fully. 10h ( x) f ( x) - 4 … [3]

9 marks

Mark scheme: 6(a) 14 1 6(b) f(x) > 9 1 6(c) 380 2 M1 for g(19) or better or (5 x − 1) 2 + (5 x − 1) 3 y = x − 1 or x = ( y − 1)36(d) 1+ 3 x oe 2 M1 for 6(e) 2( x − 1) 2 nfww 3 10( x − 1) 3 M2 for or better seen 5( x − 1) Or M1 for 5 x −−1 4 or better seen

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

20 Simplify. 3x - 5 3 xy - 3x - 5y + 5 … [3]

3 marks

Mark scheme: 20 1 −1 3 B2 for ( y − 1)(3x − 5) or (1 − y )( 5 − 3 x ) or final answer y − 1 1 −y or B1 for 3 x ( y − 1) − 5( y − 1) or y (3 x − 5) − [1](3 x − 5)

This question in 0607/43 May/June 2025

12 Solve. 2x - 5 5 1 - 4x - = 3 6 2 x = … [4]

4 marks

Mark scheme: 12 9 1 4 M1 for correctly eliminating fractions or 1 or 1.125 M1 for correctly expanding their brackets 8 8 M1 for correctly collecting their terms into ax = b

This question in 0607/42 Oct/Nov 2025