Functions — Paper 2 · IGCSE Mathematics 0580

2 x + 1 For 22 f(x) = 4x + 1 g(x) = x3 + 1 h(x) = Examiner's 3 Use (a) Find the value of gf(0). Answer(a) [2] (b) Find fg(x). Simplify your answer. Answer(b) [2] (c) Find h -1(x). Answer(c) [2]

6 marks

Mark scheme: 22 (a) 2 2 M1 f(0) = 1 (b) 4x3 + 5 2 M1 4(x3 + 1) + 1 (3 x − )1 (c) 2 M1 rearranging y = (2x + 1)/3 to make x the subject 2 and interchanging x and y. Allow any one error in the working 70

This question in 0580/21 Oct/Nov 2009

2 x + 1 For 22 f(x) = 4x + 1 g(x) = x3 + 1 h(x) = Examiner's 3 Use (a) Find the value of gf(0). Answer(a) [2] (b) Find fg(x). Simplify your answer. Answer(b) [2] (c) Find h -1(x). Answer(c) [2]

6 marks

Mark scheme: 22 (a) 2 2 M1 f(0) = 1 (b) 4x3 + 5 2 M1 4(x3 + 1) + 1 (3 x − )1 (c) 2 M1 rearranging y = (2x + 1)/3 to make x the 2 subject and interchanging x and y. Allow any one error in the working 70

This question in 0580/22 Oct/Nov 2009

20 f(x) = (x – 1)3 g(x) = (x – 1)2 h(x) = 3x + 1 Examiner's Use (a) Work out fg(-1). Answer(a) [2] (b) Find gh(x) in its simplest form. Answer(b) [2] (c) Find f -1(x). Answer(c) [2] Question 21 is printed on the next page.

6 marks

Mark scheme: 20 (a) 27 2 M1 g(–1) = 4 seen or ((x – 1)2 – 1)3 (b) 9x2 cao 2 M1 (3x + 1 – 1)2 or better (c) 3√x + 1 2 M1 interchange x, y & rearrange formula

This question in 0580/21 May/June 2010

18 (a) f(x) = 1 – 2x. Examiner's Use (i) Find f(-5). Answer(a)(i) [1] (ii) g(x) = 3x – 2. Find gf(x). Simplify your answer. Answer(a)(ii) [2] (b) h(x) = x2 – 5x – 11. Solve h(x) = 0. Show all your working and give your answer correct to 2 decimal places. Answer(b) x = or x = [4] Question 19 is printed on the next page.

7 marks

Mark scheme: 18 (a) (i) 11 1 (ii) 1 – 6x 2 M1 3(1 – 2x) – 2 5 ± k (b) –1.65, 6.65 4 M1 M1 √[(–5)2 – 4 × 1 × (–11)] 2 or better A1 A1

This question in 0580/22 May/June 2010

18 f(x) = x2 + 2 g(x) = (x + 2)2 h(x) = 3x – 5 Examiner's Use Find (a) gf(–2), Answer(a) [2] (b) h –1(22) . Answer(b) [2]

4 marks

Mark scheme: 18 (a) 64 2 B1 for evidence of f(–2) = 6 x + 5 (b) 9 2 M1 for 3x – 5 = 22 or seen 3 3

This question in 0580/23 May/June 2010

1 25 f : x → 2 x − 7 g : x → x Find  1  (a) fg   ,  2  Answer(a) [2] (b) gf (x), Answer(b) gf (x) = [1] (c) f –1 (x). Answer(c) f–1 (x) = [2]

5 marks

Mark scheme: 1 2 25 (a) –3 2 B1 g( ) = 2 or fg(x) = – 7 oe 2 x 1 (b) 1 2 x − 7 x + 7 (c) 2 M1 for y + 7 = 2x or x = 2y – 7 2

This question in 0580/23 Oct/Nov 2010

1 For 17 f(x) = (x ≠ O4) Examiner's x + 4 Use g(x) = x2 – 3x h(x) = x3 + 1 (a) Work out fg(1). Answer(a) [2] (b) Find hO1(x). Answer(b) h O1(x) = [2] (c) Solve the equation g(x) = O2. Answer(c) x = or x = [3] Question 18 is printed on the next page.

7 marks

Mark scheme: 1 17 (a) 2 B1 f(–2) seen 2 (b) 3√(x – 1) or 3 x − 1 2 M1 x –1 = y3 or 3√(y – 1) (c) 1 2 3 M2 (x – 1)(x – 2) = 0 or M1 (x + a)(x + b) = 0 where ab = 2 or a + b = –3 If 0 scored give M1 for x2 – 3x + 2 = 0 1

This question in 0580/21 Oct/Nov 2011

x 3 For 20 f(x) = 4(x + 1) g(x) = O 1 Examiner's 2 Use (a) Write down the value of x when f O1(x) = 2. Answer(a) x = [1] (b) Find fg(x). Give your answer in its simplest form. Answer(b) fg(x)= [2] (c) Find gO1(x). Answer(c) g O1(x) = [3] Question 21 is printed on the next page.

6 marks

Mark scheme: 20 (a) 12 1 (b) 2x3 cao 2 M1 clear evidence of adding 1 then multiplying by 4 to g(x) (c) 3 2 ( x+ )1 oe 3 M1 each correct move

This question in 0580/21 Oct/Nov 2012

x + 2 19 f(x) = x2 + 1 g(x) = 3 (a) Work out ff(O1). Answer(a) [2] (b) Find gf(3x), simplifying your answer as far as possible. Answer(b) gf(3x) = [3] (c) Find gO1(x). Answer(c) g O1(x) = [2] Question 20 is printed on the next page.

7 marks

Mark scheme: 19 (a) 5 2 M1 f(2) = seen (b) 3x2 + 1 3 M1 9x2 + 1 M1 ( “9x2 + 1” + 2)/3 seen (c) 3x – 2 2 y + 2 M1 for 3y = x + 2 or x = 3

This question in 0580/22 Oct/Nov 2012

1 1 x Examiner′s , x ¸ 0 h(x) =21 f(x) = 5x + 4 g(x) = 2x c 2 m Use Find (a) fg(5) , Answer(a) … [2] (b) gg(x) in its simplest form, Answer(b) gg(x) = … [2] (c) f –1(x) , Answer(c) f –1(x) = … [2] (d) the value of x when h(x) = 8. Answer(d) x = … [2]

8 marks

Mark scheme: 21 (a) 4.5 oe 2 B1 for [g(5)=] 0.1 oe 1 seen oe M1 for (b) x 2 2 1 ( 2 x ) x − 4 (c) oe 2 M1 for a correct first step 5 4 y e.g. y − 4 = 5x or = x + or 5 5 x = 5y + 4 (d) or − 3 2 M1 for 8 x 1 −x 3 or 2 = oe or 2 = 2 8

This question in 0580/22 May/June 2013

x 2 Examiner′s – 3, x ¸ 0 g(x) = – 516 f(x) = x + x 2 Use Find (a) fg(18), Answer(a) … [2] (b) g–1(x). Answer(b) g–1(x) = … [2] _____________________________________________________________________________________

4 marks

Mark scheme: 16 (a) 1.5 2 B1 for [g(18) =] 4 y (b) 2(x + 5) or 2x + 10 2 M1 for correct first step e.g. x = − 5 or 5 x = + 5 or 2y = x – 10 y 2 IGCSE – May/June 2013 0580 23

This question in 0580/23 May/June 2013

19 f(x) = 2x + 3 g(x) = x2 For Examiner′s Use (a) Find fg(6). Answer(a) … [2] (b) Solve the equation gf(x) = 100. Answer(b) x = … or x = … [3] (c) Find f –1(x). Answer(c) f –1(x) = … [2] (d) Find ff –1(5). Answer(d) … [1]

8 marks

Mark scheme: 19 (a) 75 2 B1 for [g(6) =] 36 (b) 3.5 –6.5 3 M1 for (2x + 3)2 = 100 M1 for 2x + 3 = [±]10 If 0 scored, SC1 for one correct value as answer x − 3 3 y (c) oe final answer 2 M1 for x = 2y + 3 or y – 3 = 2x or = x + 2 2 2 or better (d) 5 1

This question in 0580/23 Oct/Nov 2013

220 f(x) = 3x – 2 g(x) = , x ≠ –1 x + 1 (a) Find gf(2). Answer(a) … [2] (b) Solve g(x) = 10. Answer(b) x = … [2] (c) Simplify. f(2x) – f(x + 2) Answer(c) … [3]

7 marks

Mark scheme: 20 (a) 0.4 or 52 2 B1 for [f(2) =] 4 2 or M1 for or better (3 x − 2 ) + 1 (b) –0.8 or − 54 2 M1 for 2 = 10( x + )1 or better (c) 3 x − 6 or 3( x − 2 ) nfww 3 M2 for 3(2 x ) − 2 − (3( x + 2 ) − 2 ) or M1 for [f (2 x ) = ]3(2 x ) − 2 or [f ( x + 2 )] = 3( x + 2 ) − 2

This question in 0580/21 Oct/Nov 2014

x - 116 f(x) = (x – 3)2 g(x) = h(x) = x3 4 Find (a) hf(1), Answer(a) … [2] (b) g–1(x), Answer(b) g–1(x) = … [2] (c) gh(x), Answer(c) gh(x) = … [1] (d) the solution to the equation f(x) = 0. Answer(d) x = … [1] __________________________________________________________________________________________

6 marks

Mark scheme: 16 (a) 64 2 B1 for [f(1) =] 4 or M1 for ((x – 3)2)3 or better y − 1 (b) 4x + 1 oe 2 M1 for x = or 4y = x – 1 4 x 3 − 1 (c) oe final answer 1 4 (d) 3 nfww 1

This question in 0580/23 Oct/Nov 2014

22 f(x) = 5x – 3 g(x) = x2 (a) Find fg(–2). Answer(a) … [2] (b) Find gf(x), in terms of x, in its simplest form. Answer(b) … [2] (c) Find f –1(x). Answer(c) f –1(x) = … [2]

6 marks

Mark scheme: 22 (a) 17 2 M1 for [g(−2) =] 4 seen or for 5x2 – 3 (b) 2 M1 for g(5 x − 3) 25 x 2 − 30 x + 9 or (5x – 3)2 as final answer x + 3 (c) 2 M1 for 5 x = y + 3 or x = 5 y − 3 or 5 y 3 = x − 5 5

This question in 0580/22 Feb/March 2015

24 f(x) = 3x + 5 g(x) = x2 (a) Find g(3x). Answer(a) … [1] (b) Find f −1(x), the inverse function. Answer(b) f −1(x) = … [2] (c) Find ff(x). Give your answer in its simplest form. Answer(c) … [2] __________________________________________________________________________________________

5 marks

Mark scheme: 24 (a) 9x2 1 (b) x − 5 2 M1 for correct first algebraic step e.g. 3 y 5 y – 5 = 3x or = x + or better 3 3 or for interchanging x and y, e.g. x = 3y + 5, this does not need to be the first step (c) 9x + 20 cao final answer 2 M1 for 3(3x + 5) + 5

This question in 0580/22 May/June 2015

21 f(x) = x3 g(x) = 3x – 5 h(x) = 2x + 1 Work out (a) ff(2), Answer(a) … [2] (b) gh(x) and simplify your answer, Answer(b) … [2] (c) h–1(x), the inverse of h(x). Answer(c) h–1(x) = … [2] __________________________________________________________________________________________

6 marks

Mark scheme: 21 (a) 512 2 B1 for [f ( 2 ) = ]8 or M1 for (x 3) 3 or better (b) 6 x − 2 or 2(3 x − )1 final answer 2 B1 for (3 2 x + )1 − 5 or better (c) 1 M1 for correct first step ( x − )1 oe 2 2 y 1 eg y − 1 = 2 x or = x + 2 2 or x = 2 y + 1 or better

This question in 0580/21 Oct/Nov 2015

x - 318 f(x) = x2 g ( x) = 2 Find (a) f(– 5), … [1] (b) gf(x), … [1] (c) g –1(x). g –1(x) = … [2]

4 marks

Mark scheme: 18 (a) 25 1 x 2 − 3 (b) oe final answer 1 2 y − 3 (c) 2x + 3 final answer 2 M1 for correct first step, e.g. x = 2 or 2y = x – 3

This question in 0580/23 Oct/Nov 2016

x21 f (x) = - 3 g ( x) = 6x - 7 h (x) = 2x 4 (a) Work out the value of x when f(x) = -0.5 . x = … [2] (b) Find g−1(x). g−1(x) = … [2] (c) Work out the value of x when h(x) = f(13). x = … [2]

6 marks

Mark scheme: x 21 (a) 10 2 M1 for − 3 = − 0.5 4 x + 7 y 7 (b) final answer 2 M1 for y + 7 = 6 x or = x − or 6 6 6 x = 6 y − 7 1 (c) –2 2 M1 for [f(13) =] 4

This question in 0580/22 Feb/March 2017

12 f(x) = 3 + 4x g(x) = 6x + 7 Find, in its simplest form, (a) f(3x), … [1] (b) fg(x). … [2]

3 marks

Mark scheme: 12(a) 3 + 12x final answer 1 12(b) 24 x + 31 final answer 2 M1 for 3 + 4(6 x + 7)

This question in 0580/23 May/June 2017

22 f(x) = 5 - 2x g(x) = x2 + 8 (a) Calculate ff(-3). … [2] (b) Find (i) g(2x), … [1] (ii) f -1(x). f -1(x) = … [2]

5 marks

Mark scheme: 22(a) −17 2 M1 for f (11) seen or 5 − 2(5 − 2x) or better 22(b)(i) 4 x 2 + 8 oe 1 22(b)(ii) 5 − x 2 M1 for x = 5 − 2 y or 2 x = 5 − y or oe final answer 2 y 5 y – 5 = –2x or = − x 2 2

This question in 0580/21 May/June 2018

23 f (x) = 7 + 3x g (x) = x4 h (x) = 3x (a) h (3x) = k x Find the value of k. k = … [2] (b) Find the value of x when f (x) = g (2) . x = … [2] (c) Find f -1 (x) . f -1 (x) = … [2]

6 marks

Mark scheme: 23(a) 27 2 M1 for 33x seen 23(b) 3 2 M1 for 7 + 3x = 24 23(c) x − 7 2 M1 for x = 7 + 3y oe final answer y 7 3 or y – 7 = 3x or –3x = 7 – y or = + x 3 3

This question in 0580/21 Oct/Nov 2018

25 (a) f (x) = x 3 g ()x = 5x + 2 (i) Find gf ()x . … [1] (ii) Find g -1 ()x . g -1 ()x = … [2] (b) h (x) = ax 2 + 1 Find the value of a when h (- 2) = 21. a = … [2]

5 marks

Mark scheme: 25(a)(i) 5x3 + 2 final answer 1 25(a)(ii) x − 2 2 M1 for correct first step final answer y 2 5 e.g. y – 2 = 5x, x = 5y + 2, = x + 5 5 25(b) 5 2 M1 for a × (–2)2 + 1 = 21

This question in 0580/22 Oct/Nov 2018

10 f(x) = 2x + 3 Find f(1 - x) in its simplest form. … [2]

2 marks

Mark scheme: 10 5 – 2x final answer 2 M1 for 2(1 – x) + 3 oe

This question in 0580/21 May/June 2019

24 f (x) = 3x - 5 g (x) = 2x (a) Find fg(3). … [2] (b) Find f -1 (x) . f -1 (x) = … [2]

4 marks

Mark scheme: x24(a) 19 2 M1 for 3(2 ) − 5 soi or for f(8) 24(b) x + 5 2 M1 for correct first step oe final answer y 5 3 y + 5 = 3x or = x − or x = 3y – 5 3 3

This question in 0580/21 Oct/Nov 2019

14 (a) f ( x) = 4 x + 3 g ( x) = 5 x - 4 fg ( x) = 20 x + p Find the value of p. p = … [2] 5x - 1 (b) h ( x) = 3 Find h -1 ( x) . h -1 ( x) = … [3]

5 marks

Mark scheme: 14(a) [p = ] –13 2 M1 for 4(5x – 4) + 3 or better 14(b) 3 x + 1 3 3 y + 1 1 3x M2 for x = , 5y = 3x + 1 or y − = 5 5 5 5 5 y –1 1 5 x M1 for x = , 3y = 5x − 1 or y + = 3 3 3

This question in 0580/21 May/June 2020

17 (a) f ( x) = 3x 2 + a where a is an integer. f ( - 2) = 19 Find the value of a. a = … [2] (b) g ( x) = 2 x + 7 h ( x) = 3 x - 8 (i) Find gh(x) in its simplest form. … [2] (ii) Find g -1 ( x) . g -1 ( x) = … [2]

6 marks

Mark scheme: 17(a) [a =] 7 2 M1 for 3(–2)2 + a = 19 or better 17(b)(i) 6x – 9 or 3(2x – 3) final answer 2 M1 for 2(3x – 8) + 7 or better 17(b)(ii) x − 7 2 M1 for a correct first step final answer x = 2 y + 7 or y − 7 = 2 x 2 y 7 or = x + 2 2

This question in 0580/21 Oct/Nov 2020

x - 3 520 f ( )x = 2 g ( )x = 2 x - 1 h ( )x = x - 4 (a) Find ff(6). … [2] (b) Find g -1 g ( x + 21) . … [1] (c) Find x when f ( x) = h ( 84) . x = … [2]

5 marks

Mark scheme: 20(a) 32 2 M1 for f(6) = 8 ( 2 x – 3 ) – 3 or ff( x ) = 2 oe 20(b) x + 21 1 20(c) –1 2 1 M1 for oe or 2–4 oe 16

This question in 0580/22 Oct/Nov 2021

4 x19 f ( )x = 7 x - 8 g ( )x = + 5 h ( )x = 2 + 1 x (a) Find f -1 ( )x . f -1 ( )x = … [2] 1 (b) Find the value of x when h ( )x = g e 3 o. x = … [2]

4 marks

Mark scheme: 19(a) x 8 2 y 8 final answer M1 for x  7 y  8 or y  8  7 x or  x  7 7 7 19(b) 4 2 1 M1 for 4 ÷ + 5 oe or better 3

This question in 0580/22 May/June 2022

1 7x - 2 3 - 10x19 f( )x = kx2 g ( x) = h ( x) = j( )x = x 5 14 (a) f(- 5)k = 675 Find the value of k. k = … [2] (b) Find gh ( x) . … [1] (c) Find h -1 ( x) + j( x) . Give your answer in its simplest form. … [4]

7 marks

Mark scheme: 19(a) 3 2 M1 for k(–5k)2 = 675 or better 19(b) 5 1 final answer 7 x  2 19(c) 1 4 7 or 0.5 B3 for answer 2 14 OR 5 x  2 B2 for 7 or M1 for correct first step for h –1(x) 7 y  2 e.g. x = 5 y  7 x  2 5 2 7 x y +  5 5 2  5 x  2  3  10 x M1FT for  oe with 14 14 common denominator

This question in 0580/23 May/June 2022

x + 517 f ( x) = x2 g ( x) = h ( x) = 7 x - 3 2 (a) Find f ( - 3) . … [1] (b) Find g -1 ( x) . g -1 ( x) = … [2] (c) Solve gf ( x) = hh -1 ( 63) where x 2 0 . x = … [3]

6 marks

Mark scheme: 17(a) 9 1 17(b) 2x – 5 final answer 2 M1 for correct first step e.g. y + 5 5 x x = or 2y = x + 5 or y – = or 2 2 2 better 17(c) 11 3 x 2 + 5 M1 for 2 M1 for hh–1(63) = 63 soi

This question in 0580/22 Oct/Nov 2022

19 f ( )x = 5 x - 3 , x 2 1 10 g ( )x = , x ! 2 x - 2 (a) Find gf(x). Give your answer in its simplest form. … [2] (b) Find g -1 ( )x . g -1 ( )x = … [3] (c) Find ff -1 ( x - 1) . … [1]

6 marks

Mark scheme: 19(a) 2 2 10 final answer M1 for or better x − 1 5 x −−3 2 19(b) 10 10 + 2 x 3 10 10 + 2 y + 2 or final answer M2 for y − 2 = or x = oe x x x y or yx = 10 + 2x oe 10 or M1 for x = or y(x – 2) = 10 oe y − 2 or better 19(c) x – 1 1

This question in 0580/23 Oct/Nov 2022

25 f ( x) = x 3 + 1 Find f -1 ( x) . f -1 ( x) = … [2]

2 marks

Mark scheme: 25 1 2 M1 for x = y 3 + 1 or for y −=1 x 3 3 3 x − 1 or ( x − 1) or better

This question in 0580/22 Feb/March 2023

14 f ( x) = 5x + 2 Find f -1 ( x) . f -1 ( x) = … [2]

2 marks

Mark scheme: 14 x  2 2 M1 for a correct first step oe final answer 5 y 2 x  5 y  2 or y  2  5 x or  x  5 5

This question in 0580/22 May/June 2023

20 f ( x) = 6 x - 7 g ( x) = x -3 (a) Find f ( x + 2) . Give your answer in its simplest form. … [2] (b) Find f -1 ( x) . f - 1 ( x) = … [2] (c) Find x when g(x) = f(22) . x = … [2]

6 marks

Mark scheme: 20(a) 6 x  5 cao final answer 2 M1 for 6  x  2   7 oe 20(b) x  7 x 7 2 y 7 or  final answer M1 for x  6 y  7 or y  7  6 x or  x  6 6 6 6 6 20(c) 1 2 M1 for x −3 = 6  22  7 or better or 0.2 5

This question in 0580/23 May/June 2023

20 f( )x = 3 x + 2 (a) Find x when f ( )x = 245 . x = … [2] (b) Find x when f - 1 ( )x = 7 . x = … [2]

4 marks

Mark scheme: 20(a) 5 2 M1 for 3x + 2 = 245 20(b) 2189 2 M1 for x = f(7) or 37 + 2

This question in 0580/23 May/June 2024

18 g ( x) = 4x + 3 (a) Find x when g ( x) = 1. … [1] -1 1 (b) Find g e o . 16 … [2]

3 marks

Mark scheme: 18(a) –3 1 18(b) –5 2 1 M1 for or 4–2 4 2

This question in 0580/21 Oct/Nov 2024

14 f ( x) = 3x - 4 g ( x) = 4x + 1 (a) Find f ( - 2 ) . … [1] (b) Find f -1 ( x) . f -1 ( x) = … [2] (c) fg ( x) = ax + b Find the value of a, and the value of b. a = … b = … [2] (d) Simplify. 2 5 - f ( x) g ( x) Give your answer as a single fraction in terms of x. … [3]

8 marks

Mark scheme: 14(a) –10 1 14(b) x + 4 2 M1 for correct first step oe y + 4 = 3x or x = 3y – 4 3 y 4 or = x − 3 3 14(c) a = 12, b = –1 2 B1 for either a or b correct or M1 for 3(4x + 1) – 4 14(d) 22 − 7 x 22 − 7 x 3 B1 for 2(4x + 1) – 5(3x – 4) oe or better isw or ( 3 x − 4 )( 4 x + 1) 12 x 2 − 13 x − 4 B1 for common denominator (3x – 4)(4x + 1) final answer oe isw

This question in 0580/21 May/June 2025

22 f ( )x = 2x + 5 g ( )x = x - 4 h ( )x = 5 x (a) Find f ( 3 ) . … [1] (b) Find f -1 ( )x . f -1 ( )x = … [2] (c) Solve fg ( )x = 25 . x = … [3] (d) Find x when h -1 ( )x = 2 . x = … [2]

8 marks

Mark scheme: 22(a) 11 1 22(b) x − 5 2 M1 for correct first step y − 5 2 x = 2y + 5 or y – 5 = 2x or = x 2 y 5 or = x + 2 2 22(c) 14 3 M2 for x – 4 = (25 – 5) ÷ 2 oe or better or 2x – 8 = 25 – 5 oe or better or M1 for 2(x – 4) + 5 = 25 22(d) 25 2 M1 for h(2) or 52

This question in 0580/23 May/June 2025

21 1 20 f ( )x = , x ! g ( )x = 3x + 4 2x - 1 2 (a) Find (i) g(2) … [1] (ii) gf(-1) … [2] (iii) f -1 ( )x . f -1 ( )x = … [3] (b) Solve f ( x) = g ( x) . x = … or x = … [5]

11 marks

Mark scheme: 20(a)(i) 10 1 20(a)(ii) –17 2 M1 for g(–7) oe  21 or 3  + 4 oe soi  2 x − 1  20(a)(iii) 21 + x 21 1 3 M2 for one step from answer or + oe final answer 21 + y 21 1 2 x 2 x 2 e.g. x = or x = + 2 y 2 y 2 21 or 2 y = + 1 x 21 + x or 2xy = 21 + x or 2 y = x or M1 for correct first step e.g. 21 x = or better 2 y − 1 or y (2x – 1) = 21 or better 20(b) 5 5 5 B3 for 6x2 + 5x – 25 [= 0] and − oe 3 2 or M1 for 21 = (2x – 1)(3x + 4) oe B1 for 6x2 – 3x + 8x – 4 or better and M1 for correct method to solve their three-term quadratic e.g. (3x – 5)(2x + 5) [= 0] −5 5 2 −−4 6 25 or oe 12

This question in 0580/22 Oct/Nov 2025