Introduction to algebra — Paper 2 · IGCSE Mathematics 0580

21 For Examiner's Use  x 6   2 3  A =   B =    4 3   2 1  (a) Find AB.     Answer(a)       [2] (b) When AB = BA, find the value of x. Answer(b) x = [3] Question 22 is on the next page

5 marks

Mark scheme: 21 (a)  2 x + 12 3 x + 6  2 M1 for any correct row or column    14 15  Allow 2(x + 6), 3(x + 2) (b) 5 3  2 x + 12 21  M1   one row (or column) correct  2 x + 4 15  M1 2x + 4 = 14 or 3x + 6 = 21

This question in 0580/21 May/June 2009

19 The position vector r is given by r = 2p + t(p + q). Examiner's Use (a) Complete the table below for the given values of t. Write each vector in its simplest form. One result has been done for you. t 0 1 2 3 r 4p + 2q [3] (b) O is the origin and p and q are shown on the diagram. (i) Plot the 4 points given by the position vectors in the table. q O p [2] (ii) What can you say about these four points? Answer(b)(ii) [1]

6 marks

Mark scheme: 19 (a) 2p 3p + q ……….. 5p + 3q cao 1, 1, 1 (b) (i) all 4 plotted correctly ft 2 B1 2 or 3 correct (ii) a (straight) line 1 Allow linear, collinear 2 3

This question in 0580/21 May/June 2010

20 f(x) = x3 g(x) = 2x − 3 ForFor Examiner'sExaminer's UseUse (a) Find (i) g(6), Answer(a)(i) [1] (ii) f(2x). Answer(a)(ii) [1] (b) Solve fg(x) = 125. Answer(b) x = [3] (c) Find the inverse function g−1(x) . Answer(c) g –1(x) = [2]

7 marks

Mark scheme: 20 (a) (i) 9 1 (ii) 8x3 cao 1 (b) 4 www 3 M1 for (2x – 3)3 = 125 M1 2x – 3 = 5 (c) x + 3 2 M1 for x ± 3 = 2y or x = y ± 3 2 2

This question in 0580/21 May/June 2011

23 f(x) = 3x + 5 g(x) = 4x O 1 For Examiner's Use (a) Find the value of gg(3). Answer(a) [2] (b) Find fg(x), giving your answer in its simplest form. Answer(b)fg(x) = [2] (c) Solve the equation. f –1(x) = 11 Answer(c) x = [1] Question 24 is printed on the next page.

5 marks

Mark scheme: 23 (a) 43 2 M1 for g(11) or 4[4(3) – 1] –1 (b) 12x + 2 2 M1 for 3(4x – 1) + 5 (c) 38 1 2 2 2

This question in 0580/23 Oct/Nov 2012

2 3 2 1 5 17 M = N = e 3 6 o e 1 7 2 o (a) Work out MN. Answer(a) [2] (b) Find M–1, the inverse of M. Answer(b) [2] _____________________________________________________________________________________

4 marks

Mark scheme: 17 (a)  7 23 16  2 B1 for any one row or column correct, must   be in a 2 by 3 matrix 12 45 27   1  6 − 3  (b)  6 − 3  1  a b    or 3   B1 for k    − 3 2  2 c d 3    − 3 2 

This question in 0580/23 May/June 2013

23 f(x) = 5 – 3x (a) Find f(6). Answer(a) … [1] (b) Find f(x + 2). Answer(b) … [1] (c) Find ff(x), in its simplest form. Answer(c) … [2] (d) Find f –1(x), the inverse of f(x). Answer(d) f –1(x) = … [2]

6 marks

Mark scheme: 23 (a) −13 1 (b) −3x − 1 or 5 − 3( x + 2 ) 1 (c) 9x − 10 cao 2 M1 for 5 − 3( 5 − 3x) 5 − x (d) final answer oe 2 M1 for correct first step e.g. 3 y 5 y + 3 x = 5 or = − x or y − 5 = − 3 x or 3 3 better or for interchanging x and y, e.g. x = 5 − 3 y , this does not need to be the first step

This question in 0580/21 May/June 2015

3 - 2 8 Find the inverse of the matrix c- 8 7m. [2] f p

2 marks

Mark scheme: 1  7 2  1  a b   7 2  8   oe isw 2 M1 for   soi or k   k ≠ 0 5  8 3  5  c d   8 3  or det = 5 soi 35( or 95) 39 35k ( or 95k ) 39 k

This question in 0580/22 Feb/March 2016

14 H C D NOT TO G F SCALE J A E B The diagram shows a rectangular garden divided into different areas. FG is the perpendicular bisector of BC. The arc HJ has centre D and radius 20 m. CE is the bisector of angle DCB. Write down two more statements using loci to describe the shaded region inside the garden. The shaded region is • nearer to C than to B • … • … [2]

2 marks

Mark scheme: 14 More than 20m from D oe 2 B1 for each Nearer to CD than to CB oe

This question in 0580/21 May/June 2016

2 - 3 19 (a) Find the inverse of . c5 - 4m [2] f p w - 9 (b) The matrix does not have an inverse. f 4 w - 12p Calculate the value of w. w = … [4]

6 marks

Mark scheme: 1  4 3   4 3  19 (a)   oe isw 2 B1 for k   or det = 7 soi 7  − 5 2   −5 2  (b) 6 nfww 4 M3 for ( w − 6 ) 2 = 0 or M2 for w 2 − 12 w + 36[ = 0] or M1 for w ( w − 12 ) − 4 × ( −9 ) [ = 0] oe or clear attempt at determinant = 0 oe

This question in 0580/22 Oct/Nov 2016

7 Find the value of 5a - 3b when a = 7 and b =- 2 . … [2]

2 marks

Mark scheme: 7 41 2 M1 for 5(7) – 3(–2)

This question in 0580/22 May/June 2017

4 Find the value of 7x + 3y when x = 12 and y = -6. … [2]

2 marks

Mark scheme: 4 66 2 B1 for 84 or −18 seen

This question in 0580/21 May/June 2018

4 Complete these statements. (a) When w = … , 10w = 70. [1] (b) When 5x = 15, 12x = … [1]

2 marks

Mark scheme: 4(a) [w =] 7 1 4(b) [12x =] 36 1

This question in 0580/22 May/June 2018

1 1 0 1 1 1 1 0 20 A = B = C = I = c 9 9m c 9 8m c 3 3m c 0 1m (a) Here are four matrix calculations. AI IA C2 B + I 1 1 Work out which matrix calculation does not give the answer c 9 9m. … [2] (b) Find B . … [1] (c) Explain why matrix A has no inverse. … [1]

4 marks

Mark scheme: 20(a) C2 2 B1 for any correct matrix calculation evaluated 20(b) –9 1 20(c) The determinant is 0 oe 1 e.g. it is singular.

This question in 0580/22 May/June 2018

5 y = mx + c Find the value of y when m =- 3 , x =- 2 and c =- 8 . y = … [2]

2 marks

Mark scheme: 5 –2 2 M1 for (–3)(–2) + (–8)

This question in 0580/22 May/June 2020

11 x = 2 y (-1, 4) NOT TO SCALE (-1, 1) 0 x The diagram shows a rectangle with a line of symmetry at x = 2 . Two vertices of the rectangle are at (-1, 1) and (-1, 4). The shaded region is defined by the inequalities a G x G b and c G y G d . Find the values of a, b, c and d. a = … b = … c = … d = … [2]

2 marks

Mark scheme: 11 [a =] –1 2 B1 for two or three correct [b =] 5 [c =] 1 or SC1 for [d =] 4 [a =] x ⩾ –1 [b =] x ⩽ 5 [c =] y ⩾ 1 [d =] y ⩽ 4

This question in 0580/21 Oct/Nov 2020

20 The table shows some values for y = 3x 2 - 2x - 1. x -1 -0.5 0 0.5 1 1.5 y 4 -1 0 2.75 (a) Complete the table. [1] (b) On the grid, draw the graph of y = 3x 2 - 2x - 1 for - 1 G x G 1.5 . y 4 3 2 1 – 1 – 0.5 0 0.5 1 1.5 x – 1 – 2 [3] (c) By drawing a suitable straight line, solve the equation 3x 2 - 4x - 2 = 0 for - 1 G x G 1.5 . x = … [3] Question 21 is printed on the next page.

7 marks

Mark scheme: 20(a) 0.75 and –1.25 1 20(b) Correct curve 3 B2 FT for 6 or 5 correct plots or B1 FT for 4 or 3 correct plots 20(c) ruled line y  2 x  1 B2 B1 for correct equation  y  2 x  1 soi or y  2 x  k or y  kx  1 drawn −0.35 to −0.45 B1

This question in 0580/21 May/June 2023