Algebraic fractions — Paper 4 · IGCSE Mathematics 0580

10 (a) Simplify. For Examiner′s x2 - 3 x Use x2 - 9 Answer(a) … [3] (b) Solve. 15 20 – = 2 x x + 1 Answer(b) x = … or x = … [7]

10 marks

Mark scheme: x 10 (a) cao 3 B1 for (x + 3)(x – 3) x + 3 B1 for x(x – 3) 3 (b) and –5 7 M2 for 15(x + 1) – 20x = 2x(x + 1) 2 or M1 for multiplication by one denominator only 15( x + )1 − 20 x or x ( x + )1 and B2 for 2x2 + 7x – 15 [= 0] or B1 for 15x + 15 – 20x or 2x2 + 2x and M2 for (2x – 3)(x + 5) or their correct factors or formula or M1 for (2x + a)(x + b) where ab = –15 or a + 2b = 7 3 A1 for x = and –5 2

This question in 0580/43 Oct/Nov 2013

11 (a) Make x the subject of the formula. xr A - x = t Answer(a) x = … [4] (b) Find the value of a and the value of b when x2 – 16x + a = (x + b)2. Answer(b) a = … b = … [3] (c) Write as a single fraction in its simplest form. 6 5 - x - 4 3x - 2 Answer(c) … [3]

10 marks

Mark scheme: At 11 (a) final answer oe nfww 4 B1 for t (A – x) = xr t + r or tA – tx = xr xr or A = + x t M1 for correctly completing multiplication by t (eliminating any bracket) and x terms isolated M1 for correct factorisation M1 dep for correct division (b) [a = ] 64 3 B1 for 2b = –16 or (x – 8)2 [b = ] −8 B1 for a = (their b)2 If 0 scored, SC1 for x2 +2bx + b2 soi 13 x + 8 (c) final answer nfww 3 B1 for 6(3x – 2) – 5(x – 4) or better seen ( x − 4 )(3 x − 2 ) B1 for (x – 4)(3x – 2) oe seen as denom 13 x − 32 or SC2 for final answer ( x − 4 )(3 x − 2 )

This question in 0580/41 May/June 2015

8 (a) Factorise x2 – 3x – 10. Answer(a) … [2] x + 2 3 (b) (i) Show that + = 3 simplifies to 2x2 – 2x – 3 = 0. x + 1 x Answer(b)(i) [3] (ii) Solve 2x2 – 2x – 3 = 0. Give your answers correct to 3 decimal places. Show all your working. Answer(b)(ii) x = … or x = … [4] 2x + 3 x (c) Simplify – . x + 2 x + 1 Answer(c) … [4] __________________________________________________________________________________________

13 marks

Mark scheme: 8 (a) (x – 5)(x + 2) final answer 2 B1 for (x – 5)(x + 2) seen and then spoiled or M1 for (x + a)(x + b) where a + b = – 3 or ab = –10 [a, b integers] (b) (i) x(x + 2) + 3(x + 1) = 3x(x + 1) or M2 M1 for x(x + 2) + 3(x + 1) or better seen x2 + 2x + 3x + 3 = 3x2 + 3x Allow recovery of omitted brackets for M marks but not A mark 0 = 2x2 – 2x – 3 A1 Brackets expanded correctly and/or no errors or omission of brackets seen (ii) [ −− ]2 ± ([ − ] 2) 2 − 4( 2)( −3) B2 B1 for ([ − ]2 ) 2 − 4 ( 2 )( −3) or 28 2( 2) or .175 oe in completion of square p + q p − q or 0.5 ± .175 and B1 for in form or , r r p = – –2 and r = 2(2) or better or (x – 0.5)2 oe in completion of square – 0.823 and 1.823 final answer B1 B1 If B0B0 for answers, SC1 for – 0.82 or – 0.822… and 1.82 or 1.822.. as final answers or – 0.823 and 1.823 seen or –1.823 and 0.823 as final answers

This question in 0580/41 Oct/Nov 2015

5 (a) The area of shape ABCDEF is 24 cm2. All lengths are in centimetres. 3x – 9 F E 2x NOT TO C SCALE D 3x + 13 A 4x B (i) Show that 5x 2 + 17x - 12 = 0 . Answer(a)(i) [3] (ii) Solve, by factorising, the equation 5x 2 + 17x - 12 = 0 . You must show all your working. Answer(a)(ii) x = … or x = … [3] (b) Solve the simultaneous equations. You must show all your working. 3x – 2y = 23 –4x – y = –5 Answer(b) x = … y = … [3] (c) Solve the equation. 2 ^t + 3h t - = 1 t t + 3 Answer(c) t = … [5]

14 marks

Mark scheme: 5 (a) (i) 4 x (3 x + 13 ) − 2 x (4 x − {3 x − 9}) = 24 M1 oe 12 x 2 + 52 x − 2 x 2 − 18 x M1 Correct removal of all their brackets Dep on two areas added or subtracted 5 x 2 + 17 x − 12 = 0 A1 with no errors or omissions seen and at least one more line of working showing collection of like terms or division by 2 (ii) (5 x − 3 )( x + 4 ) [= 0] M2 M1 for (5 x + a )( x + b ) where ab = −12 or 5b + a = 17 [a, b integers] 3 A1 If zero scored SC1 for correct answers with no oe , − 4 5 working or from other methods. (b) For correctly eliminating one M1 variable x = 3 A1 SC1 if no working shown, but 2 correct answers y = − 7 A1 given If zero scored SC1 for 2 values satisfying one of the original equations (c) t = − 2 nfww 5 M1 for 2(t + 3)(t + 3) − t 2 or better seen M1 for denominator[s] t (t + 3 ) isw or for t (t + 3 ) isw on RHS M1dep for 2t 2 + 12 t + 18 − t 2 = t 2 + 3t oe dependent on both numerators and denominator expanding to give quadratics A1 for 9t + 18 = 0 oe

This question in 0580/42 Oct/Nov 2015

8 (a) y is directly proportional to the positive square root of ^ x + 2h. When x = 7, y = 9. Find y when x = 23. y = … [3] (b) Simplify. x 2 + 12x + 36 x 2 + 4x - 12 … [5] X - a(c) W = a Make a the subject of the formula. a = … [5] (d) Write as a single fraction in its simplest form. x - 2 x + 3 - x + 1 x - 1 … [5]

18 marks

Mark scheme: 8 (a) 15 nfww 3 M1 for y = k ( x + 2 ) oe A1 for k = 3 x + 6 2 (b) nfww final answer 5 B2 for ( x + 6 ) oe x − 2 or SC1 for ( x + a )( x + b ) where ab = 36 or a + b = 12 or x(x + 6) + 6( x + 6) B2 for ( x − 2 )( x + 6 ) or SC1 for ( x + a )( x + b ) where ab = − 12 or a + b = 4 or x(x + 6) – 2( x + 6) or x(x – 2) + 6(x – 2) X 2 X − a (c) 2 nfww final answer 5 M1 for W = or W a = X − a W + 1 a M1 for next productive step M1 for 2nd productive step M1 for 3rd productive step M1 for final step leading to a = −7 x − 1 −7 x − 1 (d) 2 or 5 M1 for common denominator ( x − 1)( x + 1) isw x − 1 ( x − 1)( x + 1) final answer M1 for ( x − 2 )( x − 1) − ( x + 3 )( x + 1) B2 for x 2 − 2 x − x + 2 − ( x 2 + 3 x + x + 3) oe or B1 for either expansion Qu. Answers Mark Part Marks

This question in 0580/42 Feb/March 2016

5 (a) (i) Factorise 3x 2 + 11x - 4 . … [2] (ii) Solve the equation 3x 2 + 11x - 4 = 0 . x = … or x = … [1] 2 1 1 2 (b) (i) Show that - = simplifies to 2x + 3x - 6 = 0 . 2x + 11 x - 4 2 [4] (ii) Solve the equation 2x 2 + 3x - 6 = 0 . You must show all your working and give your answers correct to 2 decimal places. x = … or x = … [4]

11 marks

Mark scheme: 5 (a) (i) ( 3 x − 1)( x + 4 ) 2 M1 for ( 3 x + b )( x + c ) with bc = − 4 or 3c + b = 11 or for 3x(x + 4) – 1(x + 4) or for x(3x – 1) + 4(3x – 1) 1 (ii) oe and −4 1 3 (b) (i) 2 × 2 ( x − 4 ) − 2 ( 2 x + 11) = ( 2 x + 11)( x − 4 ) M2 M1 for common denom or better 2( 2 x + 11)( x − 4 ) seen or attempt to multiply through by denoms 2( x − 4) − (2 x + 11)  1  or for =   (2 x + 11)( x − 4)  2  2 x 2 + 11x − 8 x − 44 or better B1 or for other correct relevant 2 bracket expansion if alt method used 4 x − 16 − 4 x − 22 = 2 x 2 − 8 x + 11x − 44 2 A1 correct solution reached with all brackets 2 x + 3 x − 6 = 0 expanded and no errors or omissions seen −±3 ( 3 ) 2 − 4 ( 2 )( − 6 ) 2 (ii) 2 B1 for ( 3 ) − 4 ( 2) ( −6 ) ) or better 2 × 2  3  2 or  x +  oe  4  −+3 q −−3 q and B1 for or or better 2(2) 2(2) 3 57 3 57 or − + oe or − − oe 4 16 4 16 −2.64 and 1.14 final ans cao B1B1 SC1 for −2.6 or –2.637... and 1.1 or 1.137... or −2.64 and 1.14 seen in working or 2.64 and −1.14 as final answers

This question in 0580/42 Feb/March 2017

7 (a) Solve the simultaneous equations. You must show all your working. 2x + 3y = 11 3x - 5y = -50 x = … y = … [4] (b) x 2 - 12x + a = x + b 2 ^ h Find the value of a and the value of b. a = … b = … [3] (c) Write as a single fraction in its simplest form. x 3x + 2 + 2x - 5 x - 1 … [4]

11 marks

Mark scheme: 7(a) [x =] −5 4 M1 for correctly equating one set of coefficients [y =] 7 M1 for correct method to eliminate one with correct working variable OR M1 for correctly rearranging one equation M1 for correct method to eliminate one variable A1 x = −5 A1 y = 7 both dep on M2 If zero scored, SC1 for 2 values satisfying one of the original equations SC1 if no correct working shown, but 2 correct answers given 7(b) [a =] 36 3 B2 for either correct [b =] −6 or M1 for a = b 2 or for x 2 + bx + bx + b 2 or better or for (x – 6)2 seen and M1 for 2b = − 12 soi 7(c) 7 x 2 − 12 x − 10 4 B1 for common denom ( 2 x − 5 )( x − 1) oe final answer nfww ( 2 x − 5 )( x − 1) seen oe isw M1 for x ( x − 1) + ( 3 x + 2 )( 2 x − 5 ) soi isw B1 for 6 x 2 − 15 x + 4 x − 10 soi

This question in 0580/43 May/June 2017

3 (a) Solve. 11x + 15 = 3x – 7 x = … [2] (b) (i) Factorise. x2 + 9x – 22 … [2] (ii) Solve. x2 + 9x – 22 = 0 x = … or x = … [1] 2 x - a (c) Rearrange y = ^ h to make x the subject. x x = … [4] (d) Simplify. 2 x - 6x x 2 - 36 … [3]

12 marks

Mark scheme: 3(a) 3 2 M1 for 11x – 3x = –7 – 15 or better –2.75 or – 2 4 3(b)(i) (x + 11)(x – 2) final answer 2 M1 for (x + a)(x + b) where ab = –22 or a + b = 9 3(b)(ii) –11 and 2 final answer 1 3(c) 2 a − 2 a 4 M1 for clearing the x term in the denominator [x] = or nfww M1 for correctly removing the bracket (expand 2 − y y − 2 or divide by 2) final answer M1 for factorising to obtain single x term M1 for their factor and division Incorrect answer scores 3 out of 4 maximum 3(d) x 3 M1 for x(x – 6) nfww final answer M1 for (x + 6)(x – 6) x + 6

This question in 0580/41 Oct/Nov 2017

2 (a) Solve. x = 49 7 x = … [1] (b) Simplify. (i) x0 … [1] (ii) x 7 # x 3 … [1] 6 2 3x (iii) ^ -4h x … [2] (c) (i) Factorise completely. 2x 2 - 18 … [2] (ii) Simplify. 2x 2 - 18 x 2 + 7x - 30 … [3]

10 marks

Mark scheme: 2(a) 343 1 2(b)(i) 1 1 2(b)(ii) x10 final answer 1 2(b)(iii) 9x16 final answer 2 B1 for x12 or x16 or (3x8)2 seen 2(c)(i) 2(x – 3)(x + 3) final answer 2 M1 for (2x + 6)(x – 3) or (2x – 6)(x + 3) or (x – 3)(x + 3) 2(c)(ii) 2( x + 3) 2 x + 6 3 M2 for (x + 10)(x – 3) or or x + 10 x + 10 M1 for (x + a)(x + b) where ab = –30 final answer nfww or a + b = 7

This question in 0580/43 Oct/Nov 2017

5 (a) Factorise. (i) 2 mn + m 2 - 6 n - 3m … [2] (ii) 4y 2 - 81 … [1] (iii) t 2 - t6 + 8 … [2] (b) Rearrange the formula to make x the subject. 2m - x k = x x = … [4] (c) Solve the simultaneous equations. You must show all your working. 1 2 x - 3y = 9 5x + y = 28 x = … y = … [3] 3 4(d) - = 6 m + 4 m (i) Show that this equation can be written as 6m 2 + 25m + 16 = 0 . [3] (ii) Solve the equation 6m 2 + 25m + 16 = 0 . Show all your working and give your answers correct to 2 decimal places. m = … or m = … [4]

19 marks

Mark scheme: 5(a)(i) ( 2 n + m )( m − 3 ) final answer 2 M1 for m ( 2 n + m ) − 3 ( 2 n + m ) or 2 n ( m − 3 ) + m ( m − 3 ) 5(a)(ii) ( 2 y − 9 )( 2 y + 9 ) final answer 1 5(a)(iii) ( t − 4 )( t − 2 ) final answer 2 B1 for ( t − 4 )( t − 2 ) seen and spoiled or M1 for t(t – 2) – 4(t – 2) or t(t – 4) – 2(t – 4) or (t + a)(t + b) where a + b = – 6 or ab = +8 5(b) 2 m 4 2 m [ x = ] M1 for xk = 2 m − x or k = − 1 k + 1 x 2 m M1 for xk + x = 2 m or k + 1 = x M1 for x ( k + )1 = 2 m 5(c) correctly eliminating one variable M1 [x = ] 6 A1 [y = ] −2 A1 If 0 scored SC1 for 2 values satisfying one of the original equations or SC1 if no working shown, but 2 correct answers given 5(d)(i) 3m − 4 ( m + 4 ) = 6 m ( m + 4 ) M1 3m − 4( m + or 4)[ = 6] oe m ( m + 4) 3m − 4 m − 16 = 6 m 2 + 24 m M1 removes brackets correctly 6 m 2 + 25 m + 16 = 0 A1 with no errors or omissions 5(d)(ii) 2 2 2 −25 ± ( 25 ) − 4 ( 6 )(16 ) B1 for ( 25 ) − 4 ( 6) (16 ) ) or better 2 × 6 2  25  or or B1 for  m +  2  12  −25  25  16 ±   − p + q p − q 12  12  6 and if in form or r r B1 for p = −25 and r = 2(6) −0.79 and −3.38 2 B1 for each final ans cao SC1 for −0.8 and −3.4 or for − 0.78 and − 3.37 or −0.789... and −3.377... or 0.79 and 3.38 or −0.79 and −3.38 seen in working

This question in 0580/41 May/June 2018

10 x8 f ()x = 8 - 3x g (x) = , x ! - 1 h ()x = 2 x + 1 (a) Find 8 (i) hf c 3 m, … [2] (ii) gh(-2), … [2] (iii) g -1 ()x , g -1 ()x = … [3] (iv) f -1 f (5 ) . … [1] (b) Write f(x) + g(x) as a single fraction in its simplest form. … [3]

11 marks

Mark scheme: 8(a)(i) 1 2 M1 for h(0) or for 28–3x 8(a)(ii) 8 2 10 M1 for g(¼) or for 2 x + 1 8(a)(iii) 10 −x 10 3 10 −y or − 1 final answer M2 for x = or better or x x y xy = 10 – x or better 10 or y + 1 = x or M1 for x(y + 1) = 10 or y(x + 1) = 10 10 10 or x = or x + 1 = y + 1 y 8(a)(iv) 5 1 8(b) −3 x 2 + 5 x + 18 3 (8 − 3 x )( x + 1) + 10 final answer M1 for x + 1 x + 1 B1 for – 3x2 – 3x + 8x + 8 [+10]

This question in 0580/42 May/June 2018

3 28 f (x) = , x ! - 2 g (x) = 8x - 5 h (x) = x + 6 x + 2 1 (a) Work out g e 4 o. … [1] (b) Work out ff(2). … [2] (c) Find gg(x), giving your answer in its simplest form. … [2] (d) Find g -1 (x) . g -1 ( x) = … [2] (e) Write g (x) - f (x) as a single fraction in its simplest form. … [3] (f) (i) Show that hg(x) = 19 simplifies to 16x 2 - 20x + 3 = 0 . [3] (ii) Use the quadratic formula to solve 16x 2 - 20x + 3 = 0 . Show all your working and give your answers correct to 2 decimal places. x = … or x = … [4]

17 marks

Mark scheme: 8(a) −3 1 8(b) 12 2 3 oe M1 for soi 11 3 + 2 x + 2 8(c) 64 x − 45 final answer 2 M1 for 8 ( 8 x − 5 ) − 5 isw 8(d) x + 5 2 M1 for a correct first step y + 5 = 8 x , oe final answer 8 y 5 = x − or x = 8 y − 5 8 8 8(e) 8 x 2 + 11x − 13 3 M1 for ( 8 x − 5 )( x + 2 ) − 3 oe isw final answer x + 2 B1 for common denominator ( x + 2 ) 8(f)(i) ( 8 x − 5 ) 2 + 6 = 19 M1 64 x 2 − 40 x − 40 x + 25 B1 64 x 2 − 40 x − 40 x + 25 + 6 = 19 oe A1 with no errors and must show 2 leading to 16 x 2 − 20 x + 3 = 0 ( 8 x − 5 ) + 6 = 19 with no omissions after this 8(f)(ii) 2 2 2 [ −− ]20 ± ( [ − ]20 ) − 4 (16 )( 3 ) B1 for ( [ − ]20 ) − 4 (16 )( 3 ) or better oe 2 × 16 [ −− ]20 + q or B1 for oe or 2(16) [ −− ]20 − q 2(16) 0.17 and 1.08 final ans 2 B1 for each If 0 scored, SC1 for answer 0.2 and 1.1 or answer − 0.17 and −1.08 or 0.174... and 1.075 to 1.076 seen or 0.17 and 1.08 seen in working

This question in 0580/42 Feb/March 2019

10 Solve. 1 2 - = 3 x x + 1 Show all your working and give your answers correct to 2 decimal places. x = … or x = … [7]

7 marks

Mark scheme: 10 x + 1 – 2x = 3x(x + 1) M2 M1 for a common denominator of x(x + 1) seen or attempt to multiply through by denominators x + 1 − 2 x or for = 3 x ( x + 1) 3x2 + 4x – 1[= 0] oe nfww A1 2 B2 2 −±4 4 −×4 3 × ( −1) B1FT for 4 − 4 × 3 × ( − 1) or better [ x = ] 2 2 × 3  2  or for x +    3  −+4 q −−4 q B1FT for or 2 × 3 2 × 3 2 1  2  2 or for − ± +   3 3  3  –1.55 and 0.22 final answers B2 B1 for each or B1 for –1.548 to –1.549 and 0.215… or for –1.55 and 0.22 seen in working or for –0.22 and 1.55 as final answer or for –1.5 or –1.54 and 0.2 or 0.21 as final answer

This question in 0580/43 Oct/Nov 2019

5 (a) Write as a single fraction in its simplest form. x + 3 x - 2 - x - 3 x + 2 … [4] 12 2k (b) 2 ' 2 = 32 Find the value of k. k = … [2] (c) Expand and simplify. ( y + 3 )( y - 4 )( 2y - 1) … [3] (d) Make x the subject of the formula. 3 + x x = y x = … [3]

12 marks

Mark scheme: 5(a) 10 x 10 x 4 M1 for common denominator( x − 3 )( x + 2 ) or ( x − 3 )( x + 2 ) x 2 − x − 6 isw final answer M1 for ( x + 3 )( x + 2 ) − ( x − 2 )( x − 3 ) isw B1 for correct numerator in terms of x only 5(b) 14 2 k 12 k 2 2 M1 for 12 = 5 or 2 = oe 5 −2 2 4096 142 or or 12 – 5 or 212 ÷ 2 [= 32] seen 32 5(c) 2 y 3 − 3 y 2 − 23 y + 12 final answer 3 B2 for correct unsimplified expanded expression or for simplified four-term expression of correct form with 3 terms correct or B1 for correct expansion of 2 of the brackets with at least 3 terms correct 5(d) 3 3 M1 for xy = 3 + x [ x = ] final answer y − 1 x 3 M1 for xy – x = 3 or x – = y y M1 for factorising and dividing

This question in 0580/42 Feb/March 2020

3 (a) s = ut + 12 at 2 Find the value of s when u = 5.2 , t = 7 and a = 1.6 . s = … [2] (b) Simplify. (i) 3a - 5b - a + 2b … [2] 5 9x (ii) # 3x 20 … [2] (c) Solve. 15 (i) =- 3 x x = … [1] (ii) 4 ( 5 - 3)x = 23 x = … [3] (d) Simplify. 2 ( 27x 9) 3 … [2] (e) Expand and simplify. (3x - 5y)(2x + y) … [2]

14 marks

Mark scheme: 3(a) 75.6 2 1 M1 for 5.2 × 7 + × 1.6 × 72 2 3(b)(i) 2a – 3b final answer 2 B1 for answer 2a + kb or ka – 3b or for 2a – 3b seen in working 3(b)(ii) 3 2 45 x B1 for oe single fraction 4 60 x 3(c)(i) −5 1 3(c)(ii) 1 3 23 −0.25 or – M1 for 20 – 12x = 23 or for 5 – 3x = 4 4 M1 for correct completion to ax = b FT their first step 3(d) 9x6 2 B1 for 9xk or kx6 3(e) 6x2 – 7xy – 5y2 2 M1 for 3 terms out of 4 from 6x2 – 10xy + 3xy – 5y2

This question in 0580/41 May/June 2020

6 f ( x) = 3x + 2 g ( x) = x 2 + 1 h ( x) = 4x (a) Find h(4). … [1] (b) Find fg(1). … [2] (c) Find gf(x) in the form ax 2 + bx + c . … [3] (d) Find x when f ( x) = g ( 7) . x = … [2] (e) Find f -1 ( x) . f -1 ( x) = … [2] g (x) (f) Find + x . f (x) Give your answer as a single fraction, in terms of x, in its simplest form. … [3] (g) Find x when h -1 ( x) = 2 . x = … [1]

14 marks

Mark scheme: 6(a) 256 1 6(b) 8 2 M1 for 3(x2 + 1) + 2 or for 3(2) + 2 6(c) 9 x 2 + 12 x + 5 3 M1 for (3x + 2)2 + 1 B1 for [(3x + 2)2 =] 9 x 2 + 6 x + 6 x + 4 oe 6(d) 16 2 M1 for 3x + 2 = 72 + 1 or better 6(e) x− 2 2 M1 for x = 3y + 2 or for y – 2 = 3x or for oe final answer y 2 3 = x + 3 3 6(f) 4 x 2 + 2 x + 1 3 B1 for x2 + 1 + x (3x + 2) or better seen final answer M1 for common denominator 3x + 2 3 x + 2 6(g) 16 1

This question in 0580/42 May/June 2020

4 (a) Solve the inequality. 3m + 12 G 8m - 5 … [2] (b) Solve the equation. 2x + 5 14 = 3 - x 15 x = … [3] (c) Solve the simultaneous equations. You must show all your working. y = 4 - x x 2 + 2y 2 = 67 x = … , y = … x = … , y = … [6]

11 marks

Mark scheme: 4(a) m ≥ 3.4 oe final answer 2 M1 for 12 + 5 ≤ 8m – 3m or better or 3m – 8m ≤ –5 – 12 or better 4(b) x = − 0.75 oe 3 M1 for 15 ( 2 x + 5 ) = 14 ( 3 − x ) B1 for 30 x + 75 = 42 − 14 x or better 4(c) 3 x 2 − 16 x − 35[ = 0] or M3 M1 for x 2 + 2 ( 4 − x ) 2 = 67 3 y 2 − 8 y − 51[ = 0] 2 2 or ( 4 − y ) + 2 y = 67 seen B1 for 16 − 8x + x 2 or 16 − 8y + y 2 (3x + 5)(x – 7) [= 0] M1 or for correct factors for their equation or (3y – 17)(y + 3)[= 0] or for correct use of quadratic formula or completing the square for their equation x = 7, y = −3 B2 5 B1 for x = 7, x = − 3 5 2 2 x = − , y = 5 or for y = −3, y = 5 3 3 3 or for a correct pair of x and y values

This question in 0580/43 May/June 2020

2x11 f ( x) = 7 x - 4 g ( x) = , x ! 3 h ( x) = x2 x - 3 (a) Find g(6). … [1] (b) Find fg(4). … [2] (c) Find fh(x). … [1] f ( x) (d) Find + g ( x) . 2 Give your answer as a single fraction, in terms of x, in its simplest form. … [3] (e) Find the value of x when f ( x + 2) =- 11. x = … [2] (f) Find the values of p that satisfy h(p) = p. … [2]

11 marks

Mark scheme: 11(a) 4 1 11(b) 52 2 2 x M1 for f( 8 ) seen or 7 × − 4 x − 3 11(c) 7x2 – 4 1 11(d) 7 x 2 − 21x + 12 7 x 2 − 21x + 12 3 M1 for ( 7 x − 4 )( x − 3 ) + 2 × 2 x or 2( x − 3) 2 x − 6 B1 for denominator 2 ( x − 3 ) or 2x – 6 final answer 11(e) −3 2 M1 for 7 x + 14 − 4 = −11 11(f) [p =] 0 and [p =] 1 2 B1 for each

This question in 0580/43 May/June 2020

8 (a) Factorise completely. 3a 2 b - ab 2 … [2] (b) Solve the inequality. 3x + 12 1 5x - 3 … [2] (c) Simplify. 3 3x 2 y 4 ` j … [2] (d) Solve. 2 6 = x 2 - x x = … [3] (e) Expand and simplify. ( x - 2)( x + 5)( 2x - 1) … [3] (f) Alan invests $200 at a rate of r% per year compound interest. After 2 years the value of his investment is $206.46 . (i) Show that r 2 + 200r - 323 = 0 . [3] (ii) Solve the equation r 2 + 200r - 323 = 0 to find the rate of interest. Show all your working and give your answer correct to 2 decimal places. r = … [3]

18 marks

Mark scheme: 8(a) ab(3a – b) final answer 2 B1 for a(3ab – b2) or b(3a2 – ab) or ab(3a – b) seen 8(b) x > 7.5 final answer 2 B1 for 12+3 < 5x – 3x oe 8(c) 27x6y12 2 B1 for two of 27, x6 and y12 correct 8(d) 1 3 M2 for 4 = 6x + 2x or better 0.5 or 2 or M1 for 2(2 – x) = 6x oe 8(e) 2x3 + 5x2 – 23x + 10 final answer 3 B2 for correct expansion of three brackets unsimplified B1 for correct expansion of two brackets with at least 3 terms correct 8(f)(i) 2 M1  r  200  1 +  = 206.46 oe  100  2r r 2 M1 1 + + oe 100 100 2 r2 + 200r – 323 = 0 A1 Correct solution reached with no errors or omissions seen If 0 scored, SC1 for 200( n ) 2 = 206.46 8(f)(ii) 2 B2 2 −200 + 200 − 4(1)( −323) B1 for 200 − 4(1)( − 323) or (r + 100)2 2 × 1 −200 + q 2 B1 for or r = 323 + 100 – 100 2 × 1 OR  206.46  B2 for 100  − 1     200  206.46 or B1 for 200 1.60 cao final answer B1

This question in 0580/41 Oct/Nov 2020

7 (a) Simplify. x 2 - 25 x 2 - x - 20 … [3] (b) Write as a single fraction in its simplest form. x + 5 x + 8 + x x - 1 … [3] (c) A curve has equation y = 2 x 3 - 4x 2 + 6 . dy (i) Find , the derived function of y. dx … [2] (ii) Calculate the gradient of the curve y = 2x 3 - 4x 2 + 6 at x = 4. … [2] (iii) Find the coordinates of the two stationary points on the curve. ( … , … ) and ( … , … ) [4]

14 marks

Mark scheme: 7(a) x + 5 3 B1 for ( x − 5 )( x + 5 ) final answer x + 4 B1 for ( x − 5 )( x + 4 ) 7(b) 2 x 2 + 12 x − 5 2 x 2 + 12 x − 5 3 B1 for common denominator x ( x − 1) oe or x ( x − 1) x 2 − x B1 for ( x − 1)( x + 5 ) + x ( x + 8 ) or better final answer 7(c)(i) 6 x 2 − 8 x final answer 2 B1 for each term in final answer or M1 for correct answer seen and spoilt 7(c)(ii) 64 2 FT their (c)(i) correctly evaluated provided at least 2 terms but not the original equation M1 for substituting x = 4 into their (c)(i) 7(c)(iii) (0, 6) 4 dy M1 for their derivative = 0 or = 0 soi  4 98  dx  ,  oe  3 27  4 B1 for x = 0 and x = 3 M1dep for substituting one of their x values into y = 2 x 3 − 4 x 2 + 6 soi

This question in 0580/41 May/June 2021

2 (a) y = px 2 + t (i) Find the value of y when p = 3, x = 2 and t = -13. y = … [2] (ii) Rearrange the formula to write x in terms of p, t and y. x = … [3] (b) (i) Factorise. 15x 2 - 2x - 8 … [2] (ii) Solve the equation. 15x 2 - 2x - 8 = 0 x = … or x = … [1] (c) Factorise completely. x 3 - 16xy 2 … [3] (d) Simplify. 2x - 1 - 4ax + 2a 2x 2 - x … [4]

15 marks

Mark scheme: 2(a)(i) –1 2 M1 for 3 × 22 – 13 oe 2(a)(ii) y − t 3 M1 for correct rearrangement to isolate x2 [±] oe final answer term p M1 for correct division by p M1 for correct square root Incorrect answer scores a maximum of M2 If 0 scored, SC1 for a correctly rearranged formula with p = 3 and t = – 13 substituted 2(b)(i) (5 x − 4)(3 x + 2) oe final answer 2 B1 for ( ax + b )( cx + d ) where either ac = 15 and bd = –8 or ad + bc = –2 or 5x(3x + 2) – 4(3x + 2) or 3x(5x – 4) + 2(5x – 4) or correct factors seen and spoiled 2(b)(ii) 4 2 1 FT a factorised quadratic oe and − oe 5 3 2(c) x ( x + 4 y )( x − 4 y ) final answer 3 B2 for ( x 2 + 4 xy )( x − 4 y ) or ( x + 4 y )( x 2 − 4 xy ) or answer in the form x(a + b)(a – b) or correct answer seen and spoiled or B1 for x ( x 2 − 16 y 2 ) oe or ( x + 4 y )( x − 4 y ) 2(d) 1 −a2 4 B2 for (2x – 1)(1 – 2a) oe oe final answer or B1 for 2x – 1 – 2a(2x – 1) x or 2x(1 – 2a) – (1 – 2a) B1 for x(2x – 1)

This question in 0580/43 May/June 2021

8 Darpan runs a distance of 12 km and then cycles a distance of 26 km. His running speed is x km / h and his cycling speed is 10 km / h faster than his running speed. He takes a total time of 2 hours 48 minutes. 12 (a) An expression for the time, in hours, Darpan takes to run the 12 km is . x Write an equation, in terms of x, for the total time he takes in hours. … [3] (b) Show that this equation simplifies to 7x 2 - 25x - 300 = 0 . [4] (c) Use the quadratic formula to solve 7x 2 - 25x - 300 = 0 . You must show all your working. x = … or x = … [4] (d) Calculate the number of minutes Darpan takes to run the 12 km. … min [2]

13 marks

Mark scheme: 8(a) 12 26 3 12 26 + = 2.8 oe isw B2 for + oe isw x x + 10 x x + 10 OR 26 B1 for seen x + 10 168 48 B1 for time = 2.8 or or 2 oe 60 60 8(b) 12 ( x + 10 ) + 26 x = 2.8 x ( x + 10 ) or M2 FT their time, provided 2 algebraic fractions one in x and other in ± x ± 10 better M1 for 12 ( x + 10 ) + 26 x seen or better 12 x + 120 + 26 x = 2.8 x 2 + 28 x M1 FT their equation dep on M2 2.8 x 2 − 10 x − 120 = 0 oe A1 or 30x + 300 + 65x = 7x2 + 70x or better 2 with no errors or omissions leading to 7 x − 25 x − 300 = 0 8(c) 2 B2 2 [ −− ]25 ± ( [ − ]25 ) − 4 × 7 × −300 B1 for ( [ − ]25 ) − 4(7)( −300) or better 2 × 7 [ −− ]25 + q [ −− ]25 − q oe or for or 2 × 7 2 × 7 − 5 and 8.57 or 8.571… B2 B1 for each or SC1 for final answers 5 and –8.57 8(d) 84 to 84.01… 2 720 FT to 3 sf or better their positive answer 12 M1 for [× 60 ] oe their positive answer

This question in 0580/42 Feb/March 2022

10 (a) Expand and simplify. ( x + 1)( x - 2)( x + 3) … [3] (b) Make g the subject of the formula. 2fg M = g - c g = … [4] (c) Simplify. 4x 2 - 16 x x 2 - 16 … [3]

10 marks

Mark scheme: 10(a) x 3 + 2 x 2 − 5 x − 6 final answer 3 B2 for correct expansion of three brackets unsimplified or for simplified expression of correct form with 3 out of 4 terms correct or B1 for correct expansion of 2 of the 3 given brackets with at least 3 terms out of four correct 10(b) Mc − Mc 4 M1 for clearing g – c from denominator or final answer e.g. M(g – c) = 2fg M − 2 f 2 f − M M1 for correctly isolating terms in g in numerator on one side M1 for correctly factorising or simplifying, to single term in g in an equation M1 for correctly dividing by bracket to final answer 10(c) 4 x 3 B1 for 4x(x – 4) final answer B1 for (x + 4) (x – 4) x + 4

This question in 0580/42 Feb/March 2022

14 f ( x) = 2 x - 1 g ( x) = 3 x - 2 h ( x) = , x ! 0 j ( x) = 5x x (a) Find (i) f ( 2) , … [1] (ii) gf ( 2) . … [1] (b) Find g -1 ( x) . g -1 ( x) = … [2] (c) Find x when h ( x) = j (- 2) . x = … [2] (d) Write f ( x) - h ( x) as a single fraction. … [2] (e) Find the value of jj ( 2 ) . … [1] (f) Find x when j -1 ( x) = 4 . x = … [2]

11 marks

Mark scheme: 4(a)(i) 3 1 4(a)(ii) 7 1 FT their (i) 3×their (i) 2  4(b) 2 3  x oe final answer 2 M1 for y + 2 = 3x or 2 3 3   y x or x = 3y – 2 4(c) 25 2 M1 for 2 1 5  x oe 4(d) 2 2 1   x x x final answer 2 M1 for 2x – 1 – 1 x 4(e) 2.98 × 1017 or 2.980... × 1017 1 4(f) 625 2 M1 for x = j(4)

This question in 0580/41 May/June 2022

8 (a) Solve. 10 - 3p = 3 + 11p p = … [2] (b) Make m the subject of the formula. mc 2 - 2k = mg m = … [3] (c) Solve. 1 4 + = 1 x - 3 2x + 3 x = … or x = … [5] (d) Solve the simultaneous equations. You must show all your working. x + 2y = 12 5 x + y 2 = 39 x = …………….. y = ……………… x = …………….. y = ……………… [5] (e) Expand and simplify. ( 2x - 3)( x + 6)( x - 4) … [3]

18 marks

Mark scheme: 8(a) 1 2 M1 for 10  3  11 p  3 p oe or better or 0.5 oe 2 8(b) 2 k 3 M1 for correctly isolating m terms [ m  ] oe final answer 2 M1 for correctly factorising c  g M1 for dividing by a bracket with two terms to the final answer Maximum mark M2 if final answer incorrect 8(c) 0 4.5 oe 5 B4 for 2 x 2  9 x [  0] or 9x – 2x2 [= 0] or better OR M2 for  2 x  3   4  x  3    x  3  2 x  3  or better or M1 for  2 x  3   4  x  3  seen oe or common denominator  x  3  2 x  3  oe B1 for 2 x 2  6 x  3 x  9 or better seen 8(d) y 2  10 y  21[  0] or M2 M1 for y 2  5 12  2 y   39 oe x 2  4 x  12[  0] 12  x  2 or 5 x   39 seen oe 2 2 (y – 3)(y – 7) [= 0] M1 or for correct factors for their 3– term quadratic or (x + 2)(x – 6) [= 0] equation or for correct substitution into quadratic formula or correctly completing the square for their 3– term quadratic equation x = − 2 y = 7 B2 B1 for x = − 2, x = 6 or for y = 7, y = 3 x = 6 y = 3 or for one correct pair of x and y values 8(e) 2 x 3  x 2  54 x  72 final answer 3 B2 correct expansion of three brackets unsimplified or for final answer of correct form with 3 out of 4 terms correct or B1 correct expansion of two brackets with at least three terms out of four correct

This question in 0580/42 May/June 2022

6 (a) Simplify. a - 2b - 3a + 7b … [2] (b) Expand and simplify. 4 ( x - 5) - ( 3 - 2x) … [2] (c) Write as a single fraction in its simplest form. 3 7 - x - 5 2x … [3] (d) Solve. 13 - 4x = 6 - x 3 x = … [3] (e) Make x the subject of the formula. 5 ( p - 2x) y = x x = … [4]

14 marks

Mark scheme: 6(a) 5b – 2a final answer 2 B1 for 5b or – 2a in final answer or for 5b – 2a seen 6(b) 6x – 23 final answer nfww 2 M1 for 4x – 20 or –3 + 2x 6(c) 35  x 35  x 3 B1 for 3(2x) – 7(x – 5) or better isw or oe final answer 2 B1 for 2x(x – 5) as common denominator 2 x ( x  5) 2 x  10 x isw, allow expanded nfww 6(d) –5 3 M1 for 13 – 4x = 18 – 3x oe 4 x 13 or  x  6  oe 3 3 M1FT for 4x +3x = 18 – 13 oe  x 5 or for  3 3 6(e) 5 p 4 M1 for correctly clearing the x from the [x =] oe final answer denominator y  10 M1 for correctly expanding the brackets or (dealing with the 5 correctly throughout) M1 for correctly isolating terms in x M1 for correctly factorising and dividing by the bracket Max 3 marks if answer is incorrect

This question in 0580/43 May/June 2022

3 (a) x – 2 4 Write down the inequality shown by the number line. … [1] (b) - 3 G 2x + 3 1 9 (i) Solve the inequality. … [3] (ii) Write down all the integer values of x that satisfy the inequality. … [2] (c) Solve the equations. 2 ( x + 2) (i) 3 ( 3 - x) - = 1 5 x = … [4] 5 3 (ii) = x + 3 x + 5 x = … [3]

13 marks

Mark scheme: 3(a) –2 < x ⩽ 4 oe 1 3(b)(i) –3 ⩽ x < 3 final answer 3 M2 for –3 < x < k or for k ⩽ x < 3 or for –6 ⩽ 2x < 6 3 3 9 3 or for − − ⩽ x < − 2 2 2 2 or M1 for – 3 – 3 ⩽ 2x < 9 – 3 3 3 9 or for − ⩽ x +  2 2 2 After 0 scored SC1 for -3 ⩽ x or for x < 3 3(b)(ii) –3, –2, –1, 0, 1, 2 final answer 2 FT their (i) as long as negative and positive values B1FT for one error or omission 3(c)(i) 36 4 B3 for –15x–2x = 5 + 4 – 45 or better oe OR 17 B2 for 45 - 15x – 2x – 4 = 5 oe OR M1 for correct removal of fraction or M1 for correct removal of brackets 3(c)(ii) –8 3 B2 for 5x – 3x = 9 – 25 or better or M1 for 5(x + 5) = 3(x + 3) oe or better

This question in 0580/41 Oct/Nov 2022

27 f ( x) = 10 - x g ( x) = , x ! 0 h ( x) = 2x j ( x) = 5 - 2 x x 1 (a) (i) Find g b 2 l. … [1] 1 (ii) Find hg b 2 l. … [1] (b) Find x when f ( x) = 7 . x = … [1] (c) Find x when g ( x) = h ( 3) . x = … [2] (d) Find j -1 ( x) . j -1 ( x) = … [2] (e) Write f ( x) + g ( x) + 1 as a single fraction in its simplest form. … [3] 2 2(f) f ( x) - ff ( x) = ax + bx + c ` j Find the values of a, b and c. a = … b = … c = … [4] (g) Find x when h -1 ( x) = 10 . x = … [2]

16 marks

Mark scheme: 7(a)(i) 4 1 7(a)(ii) 16 1 FT 2their 4 7(b) 3 1 7(c) 1 2 2 3 oe M1 for = 2 or better 4 x 7(d) 5 −x 2 M1 for oe final answer x = 5 – 2y or y + 2x = 5 oe 2 y 5 or = − x oe 2 2 7(e) 11x − x 2 + 2 3 x (10 − x ) + 2 + x final answer B2 for oe single fraction x x or B1 for x(10 – x) + 2 + x oe 2 or M1 for 10 − x + + 1 x 7(f) [a =] 1 4 B3 for x 2 − 21x + 100 [b =] –21 OR [c =] 100 2 M1 for (10 − x ) − (10 − (10 − x ) ) oe or better 2 2 B2 for [(10 − x ) ] = 100 − 10 x − 10 x + x or B1 for three out of four terms of [(10 − x ) 2 ] = 100 − 10 x − 10 x + x 2 correct 7(g) 1024 2 M1 for [x =] h(10) oe or better

This question in 0580/41 Oct/Nov 2022

6 (a) Solve. 4x + 15 = 9 x = … [2] (b) Factorise. a 2 - 9 … [1] (c) Write as a single fraction in its simplest form. 4a 3ad ' 5 10c … [3] (d) 5 n + 5 n + 5 n + 5 n + 5 n = 5 m Find an expression for m in terms of n. m = … [2] (e) Solve by factorisation. 4x 2 + 8x - 5 = 0 x = … or x = … [3] (f) (i) y is directly proportional to ( x + 3) 3 . When x = 2 , y = 13.5 . Find x when y = 108 . x = … [3] (ii) g is inversely proportional to the square of d. When d is halved, the value of g is multiplied by a factor n. Find n. n = … [2] (g) Expand and simplify. ( 2x + 3)( x - 1)( x + 3) … [3] dy 2(h) Find the derivative, , of y = 3x + 4x - 1. dx … [2]

21 marks

Mark scheme: 6(a) 1 3 2 15 9 –1.5 or –1 or – M1 for 4x = 9 – 15 or x + = 2 2 4 4 6(b) (a – 3)(a + 3) final answer 1 6(c) 8c 3 8 ac 40 c final answer B2 for or 3d 3ad 15 d 4 2 or c seen 1 3d or for correct answer seen then spoiled 4 a 10c 8 ac 3ad or M1 for  or  oe 5 3ad 10c 10c 6(d) n + 1 final answer 2 M1 for 5  5 n or 5n+1 seen 6(e) (2x – 1)(2x + 5) [= 0] oe B2 M1 for 2x(2x + 5) – [1](2x + 5) [ = 0] or 2x(2x – 1) + 5(2x – 1) [ = 0] or for (2x + m)(2x + n) [ = 0] with and mn = –5 or n + m = 4 1 1 5 B1 or 0.5 and –2.5 or –2 or – 2 2 2 6(f)(i) 7 3 M1 for y = k(x + 3)3 or better M1 for 108 = their k(x + 3)3 6(f)(ii) 4 2 2  1  M1 for   oe  2  k or oe seen or better 1 2 d 4 6(g) 2x3 + 7x2 – 9 final answer 3 B2 for correct expansion unsimplified or for simplified 4 term expression of correct form with 3 terms correct or B1 for one pair of brackets expanded with at least 3 terms out of 4 correct 6(h) 6x + 4 2 B1 for 6x or 4 or 6x + 4 with one extra term seen

This question in 0580/42 Oct/Nov 2022

1 211 f ( x) = 2 x - 1 g ( x) = 3x + 2 h ( x) = , x ! 0 j ( x) = x x (a) Find j ( - 1) . … [1] (b) Find x when f ( x) + g ( x) = 0 . x = … [2] (c) Find gg(x), giving your answer in its simplest form. … [2] (d) Find hf ( x) + gh ( x) , giving your answer as a single fraction in its simplest form. … [4] (e) When pp ( x) = x, p ( x) is a function such that p -1 ( x) = p ( x) . Draw a ring around the function that has this property. 1 2 f ( x) = 2 x - 1 g ( x) = 3x + 2 h ( x) = , x ! 0 j ( x) = x x [1]

10 marks

Mark scheme: 11(a) 1 1 11(b) 1 2 M1 for 2x – 1 + 3x + 2 = 0 oe isw − or –0.2 5 11(c) 9x + 8 final answer 2 M1 for 3(3x + 2) + 2 11(d) 4 x 2 + 5 x − 3 4 final answer x (2 x − 1) 1  1 M1 for and 3 + 2 oe   2 x − 1  x  B1 for x + 3(2 x − 1) + 2 x (2 x − 1) oe or better isw B1 for common denominator = x(2x – 1) isw 4 x 2 + 9 x + 3 If 0 scored, SC1 for answer x (2 x + 1) 11(e) h(x) indicated 1

This question in 0580/42 Feb/March 2023

9 (a) Simplify. (i) ( 3x 2 y 4 ) 3 … [2] 3 16 - 2 (ii) 16 8 e x y o … [3] (b) (i) Factorise. x 2 - 9 … [1] (ii) Simplify. x 2 - 9 2 xy - 6 y + 5 x - 15 … [3] (c) Solve the simultaneous equations. You must show all your working and give your answers correct to 2 decimal places. 2x + y = 7 y = 5x 2 + 2x - 13 x = … , y = … x = … , y = … [6]

15 marks

Mark scheme: 9(a)(i) 27x6y12 final answer 2 B1 for two terms correct in answer e.g. 27x6yk or 27xky12 or kx6y12 or for correct answer seen then spoilt 9(a)(ii) x 24 y12 3 B2 for final answer with two correct final answer elements 64 64 641 or final answer or or x 24 y12 x 24 y  12 better or for correct answer seen or B1 for 64 or x24 or y12 seen in final answer k or final answer x 24 y 12 or M1 for first correct step seen 3  3  x16 y 8   2  4  eg   or  8 4  or  16   x y   1  2   4096     48 24   x y  9(b)(i) (x + 3)(x – 3) final answer 1 9(b)(ii) x  3 3 M2 for (x – 3)(2y + 5) final answer or M1 for 2y(x – 3) + 5(x – 3) 2 y  5 or x (2y + 5) – 3( 2y + 5) 9(c) 5x2 + 4x – 20 [= 0] oe M2 M1 for 7 – 2x = 5x2 + 2x – 13 oe seen or 2  7  y   7  y  5y2 – 78y + 221 [= 0] oe or y  5    2    13 oe seen  2   2  2 M2 FT their 3-term quadratic 4 4  4(5)( 20) oe 2(5) 2 or M1 for (4)  4(5)( 20) or better or 2 4 q 4 q 4  4  or for or   4   oe 2  5 2  5 10  10  2  4  or for x  oe    10  x = 1.64 y = 3.72 B2 B1 for one correct pair or both x-values and correct or both y – values correct x = – 2.44 y = 11.88

This question in 0580/42 May/June 2023

2 311 f ( )x = 1 - 3 x g ( x) = ( x - 1) h ( x) = , x ! 0 x (a) Find g(3). … [1] (b) Find f ( x - 2 ) , giving your answer in its simplest form. … [2] (c) Find f -1 ( )x . f -1 ( )x = … [2] (d) gf ( x) - g ( x) f ( x) = 3x 3 + ax 2 + bx + c Find the value of each of a, b and c. a = … b = … c = … [5] (e) Find h ( x) - f ( x) , giving your answer as a single fraction in its simplest form. … [3] (f) h ( x n ) = 3x 7 Find the value of n. n = … [1]

14 marks

Mark scheme: 11(a) 4 1 11(b) 7 – 3x final answer 2 M1 for 1 – 3(x – 2) 11(c) 1 −x 2 oe final answer 3 M1 for x = 1 – 3y or y – 1 = –3x or 1 – y y 1 = 3x or = − x 3 3 11(d) a = 2, b = 5, c = –1 5 B4 for two correct values only after correct substitution seen i.e. (1 – 3x – 1)2 – (x – 1)2(1 – 3x) or for correct unsimplified expansion or a correct simplified expansion. OR M1 for (1 – 3x – 1)2 – (x – 1)2(1 – 3x) B2 for correct expansion of [–](x – 1)2(1 – 3x) [–]( x2 – x – x + 1 – 3x3 + 3x2 + 3x2 – 3x) or better or B1 for expansion of one pair of brackets  ( x − 1) 2 =  x 2 − x − x + 1 or   better or [ (x – 1)(1 – 3x) =] – 3x2 + x + 3x – 1 11(e) 3 − x + 3 x 2 3 B1 for 3 − x(1 − 3 x) or better final answer x B1 for common denominator x isw 11(f) –7 1

This question in 0580/42 Oct/Nov 2023

x 26 f ( )x = 5 x - 3 g ( )x = 64 h ( )x = , x !- 1 x + 1 (a) Find the value of (i) f ( 2) … [1] (ii) gf ( 0.5) . … [2] (b) Find h -1 ( )x . h -1 ( )x = … [3] 1 (c) Find x when g ( )x = 5 . 2 x = … [2] 1 (d) Write as a single fraction in its simplest form - h ( x) . f ( x) … [4]

12 marks

Mark scheme: 6(a)(i) 7 1 6(a)(ii) 1 2 M1 for g(–0.5) oe 5(x) – 3 or for 64 or better 8 6(b) 2 −x 2 3 2 or − 1 final answer M1 for y(x + 1) = 2 or x = or better x x y + 1 2 −y M1 for or xy = 2 – x oe y 6c 5 2 M1 for [64x =] 26x or (26)x or 6x = –5 − –0.833 or better 6 6(d) 7 − 9 x 7 − 9 x 4 1 2 or or B1 for − 2 (5 x − 3)( x + 1) 5 x + 2 x − 3 5 x − 3 x + 1 9 x − 7 − final answer M1 for x + 1 – 2 (5x – 3) seen isw 2 5 x + 2 x − 3 M1 for (5x – 3)(x + 1) seen isw

This question in 0580/43 Oct/Nov 2023

5 (a) (i) Factorise. x 2 - x - 12 … [2] (ii) Simplify. x 2 - 16 x 2 - x - 12 … [2] (b) Simplify. 2 2 2x - 3 - x + 1 ` j ` j … [3] (c) Write as a single fraction in its simplest form. 2x + 4 x - x + 1 x - 3 … [4] (d) Expand and simplify. ( x - 3)( x - 5)( 2x + 1) … [3] (e) Solve the simultaneous equations. You must show all your working. x - 3y = 13 2x 2 - 9y = 116 x = … y = … x = … y = … [6]

20 marks

Mark scheme: 5(a)(i) ( x − 4 )( x + 3 ) final answer 2 M1 for ( x + a )( x + b ) where ab = −12 or a + b = −1 or for x ( x + 3 ) − 4 ( x + 3 ) or x ( x − 4 ) + 3 ( x − 4 ) 5(a)(ii) x + 4 2 M1 for( x − 4 )( x + 4 ) seen final answer x + 3 5(b) 3 x 2 − 14 x + 8 or ( x − 4 )( 3 x − 2 ) final 3 M2 for ( ( 2 x − 3) − ( x + 1) ) ( ( 2 x − 3) + ( x + 1) ) answer 2 2 or 4 x − 6 x − 6 x + 9 − x + x + x + 1 or ( ) ( ) better or correct answer seen or M1 for ( x − 4 ) ( ax + b ) or ( 3 x − 2 ) ( x + c ) 4 x 2 − 6 x − 6 x + 9 or x 2 + x + x + 1 oe or( ) ±( ) 5(c) x 2 − 3 x − 12 x 2 − 3 x − 12 4 or final B1 for common denominator ( x + 1)( x − 3 ) x 2 − 2 x − 3 ( x + 1)( x − 3 ) oe isw answer B1 for ( 2 x + 4 )( x − 3 ) − x ( x + 1) or better seen B1 for 2 x 2 − 6 x + 4 x − 12 or − x 2 − x seen 5(d) 2 x 3 − 15 x 2 + 22 x + 15 final answer 3 B2 for correct expansion of three brackets unsimplified or for simplified four-term expression of correct form with 3 terms correct in final answer or B1 for correct expansion of two brackets with at least 3 terms out of 4 correct 5(e) 2 x 2 − 3 x − 77[ = 0] oe M2 6 x 2 − 9 x − 231[ = 0] ( ) M1 for correct method to eliminate one or variable e.g. 2 (13 + 3 y ) 2 − 9 y = 116 18 y 2 + 147 y + 222[ = 0] oe 2 or 2 x − 3 ( x − 13) = 116 oe 6 y 2 + 49 y + 74[ = 0] ( ) ( 2 x + 11)( x − 7 ) [ = 0] M2 FT their 3-term quadratic in x or y , correct oe factors, correct substitution into formula or [ −−]3  ([ −]3) 2 −−4 2 77 for correctly completing square or oe 2  2 or ( 6 y + 37 )( 3 y + 6 ) [ = 0] −147  147 2 − 4  18  222 M1 for a pair of factors giving 2 correct or oe 2  18 terms when expanded their quadratic or for e.g. ([ −]3) 2 −−4 2 77 oe [ −− ]3  p or oe 2  2 x =7 and y = − 2 B2 B1 for both x-values or both y-values or for 1 correct pair 1 1 x = − 5 oe and y = − 6 oe 2 6

This question in 0580/42 Feb/March 2024

3 5 (a) Simplify 25x 6 2 . ` j … [2] (b) These are the first five terms of a sequence. 1 1 6 36 216 6 Find the nth term of the sequence. … [2] (c) Expand and simplify. ( x + 4)( x - 3)( 3x - 1) … [3] 7 2(d) (i) Show that ( 3x + 5) + = x simplifies to 2x + x - 3 = 0 . x - 2 [4] (ii) Solve by factorisation 2x 2 + x - 3 = 0 . x = … or x = … [3] (e) A solid cylinder has base radius x and height 3x. The total surface area of the cylinder is the same as the total surface area of a solid hemisphere of radius 5y. 2 75y 2 Show that x = . 8 [The surface area, A, of a sphere with radius r is A = 4rr 2 .] [4]

18 marks

Mark scheme: 5(a) 125x 9 final answer 2 B1 for answer 125 kx or m x 9 or for correct answer seen then spoilt 5(b) 6 n  2 oe final answer 2 B1 for answer of form 6k oe  k  1  or answer of the form   oe  6  or for correct answer seen 5(c) 3x3 + 2x2 –37x + 12 final answer 3 B2 for correct expansion of three brackets unsimplified or for simplified four-term expression of correct form with 3 terms correct or B1 for correct expansion of two brackets with at least 3 terms out of 4 correct 5(d)(i) eliminates the fraction correctly M1 eg (3x + 5) (x – 2) + 7 = x (x – 2) 3x2 + 5x –6x – 10 + 7 = x2 – 2x oe B2 B1 for 3x2 + 5x –6x –10 [ + 7] oe seen with at least 3 terms correct leading to 2x2 + x – 3 = 0 A1 dep on M1 B2 with no errors or omissions 5(d)(ii)  2 x  3  x  1 M2 or M1 for (2x + a)(x + b) where ab = −3 or 2b + a = [+]1 or for partial factors 2x(x – 1) + 3(x – 1) or x(2x + 3) –[1](2x + 3) −1.5 oe and +1 B1 5(e) 2 M1 [TSA cylinder =] 2x  2x  3 x 2 4(5 y ) 2 M1 [TSA hemisphere=] (5 y )  2 Leading to M1 dep M1M1 2x 2  6x 2  50y 2  25y 2 oe 2 75 y 2 A1 dep on M1M1M1 x  8

This question in 0580/42 May/June 2024

1 x 11 f ( )x = 2x + 5 g ( )x = 1 - 2x h ( x) = , x ! -1 j ( )x = 2 x + 1 (a) Find g(-3). … [1] (b) Find f ( x) g ( x) + fg ( x) + 1. Give your answer in its simplest form. … [4] (c) Find g -1 ( )x . g -1 ( )x = … [2] (d) Find hh(1). … [2] 1 (e) Simplify - h ( x) . f ( x) Give your answer as a single fraction in its simplest form. … [3] 1 (f) Find x when j ( )x = . 32 x = … [1] (g) Find x when j -1 ( )x = 0 . x = … [1]

14 marks

Mark scheme: 11(a) 7 1 11(b) −4 x 2 − 12 x + 13 final answer 4 B1 for (2x + 5)(1 – 2x) B1 for 2x – 4x2 + 5 – 10x oe B1 for 2(1 – 2x) + 5 11(c) 1 −x 2 M1 for oe final answer y 1 2 x = 1 − 2 y or 2 x = 1 − y or = − x 2 2 11(d) 2 2 oe 3  1  1 M1 for h   or oe  2  1 + 1 x + 1 11(e) −−x 4 −−x 4 3 M1 for x + 1 − (2 x + 5) oe or or (2 x + 5)( x + 1) 2 x 2 + 7 x + 5 x + 4 − 2 M1 for common denominator 2 x + 7 x + 5 (2 x + 5)( x + 1) seen final answer 11(f) –5 1 11(g) 1 1

This question in 0580/43 Oct/Nov 2024

23 Simplify. h 2 + 4 h h 2 - 16 … [3]

3 marks

Mark scheme: 23 h 3 B1 for h(h + 4) isw final answer h − 4 B1 for (h + 4)(h – 4) isw

This question in 0580/42 May/June 2025

24 Martha walks a distance of 10 km at a speed of x km/h. She then runs a distance of 5 km at a speed of ( x + 4 ) km/h. The total time taken for the whole journey is 3.5 hours. (a) Write down an expression in terms of x for the time Martha is walking. … h [1] (b) Show that 7x 2 - 2 x - 80 = 0 . [4] (c) Solve 7x 2 - 2 x - 80 = 0 , giving your answers correct to 2 decimal places. You must show all your working. x = … or x = … [3] (d) Calculate the difference between the time Martha is walking and the time she is running. Give your answer in hours and minutes correct to the nearest minute. … h … min [3]

11 marks

Mark scheme: 24(a) 10 1 x 24(b) their10 + 5 = 7 oe M1 x x + 4 2 20 x + 80 + 10 x = 7 x 2 + 28 x oe M2 Strict FT for correctly clearing fractions from their three-term equation with two algebraic denominators in x and x + 4 and expanding all brackets Strict M1FT for correctly expressing their two algebraic fractions with two denominators in x and x + 4 as a single fraction or with a common denominator within a correct equation or for correctly clearing fractions from their three-term equation with two algebraic denominators in x and x + 4 but not all brackets expanded Leading to 7 x 2 − 2 x − 80 = 0 A1 No errors or omissions 24(c) 2 B2 2 −−( 2 )  ([ − ]2) − 4 ( 7 )( −80 ) or B1 for ([ −]2) − 4 ( 7 )( −80 ) oe or for oe 2 ( 7 ) −−( 2) − p −−( 2) + p oe or for oe 2(7) 2(7)  2  2 or  x −   14  –3.24 and 3.53 B1 24(d) 2h 10min 3 B2 for 2.168 to 2.18 [h] or for 130.08 to 130.8 [min] or for 2hours 10.08 min to 2 hours 10.8 min OR 10 5 M2 for − their positive x their positive x + 4 or 10 5 M1 for or their positive x their positive x + 4

This question in 0580/41 Oct/Nov 2025

13 Simplify. 7 3 + 2m 8 m … [2]

2 marks

Mark scheme: 13 31 2 28 3 final answer M1 for + oe 8m 8m 8m

This question in 0580/42 Oct/Nov 2025