Graphs of functions — Paper 4 · IGCSE Mathematics 0580

3 The table shows some values for y = 1.5 x - 1. x –2 –1 0 1 2 3 4 5 y – 0.56 – 0.33 2.38 4.06 6.59 (a) Complete the table. [3] (b) Draw the graph of y = 1.5 x - 1 for - 2 G x G 5 . y 7 6 5 4 3 2 1 x –2 –1 0 1 2 3 4 5 –1 [4] (c) Use your graph to solve the equation 1.5 x - 1 = 3. 5 . x = … [2] (d) By drawing a suitable straight line, solve the equation 1.5 x - x - 2 = 0 . x = … or x = … [3] (e) (i) On the grid, plot the point A at (5, 5). [1] (ii) Draw the tangent to the graph of y = 1.5 x - 1 that passes through the point A. [1] (iii) Work out the gradient of this tangent. … [2]

16 marks

This question in 0580/42 Feb/March 2017

(a) Find f(1). … [1] (b) Solve f (x) = 3 . x = … [1] (c) The equation f ( x) = k has only one solution for - 2.5 G x G 2 . Write down the range of values of k for which this is possible. … [2] (d) By drawing a suitable straight line, solve the equation f(x) = x – 5. x = … or x = … or x = … [3] (e) Draw a tangent to the graph of y = f (x ) at the point where x = 1. Use your tangent to estimate the gradient of y = f (x) when x = 1. … [3]

10 marks

Mark scheme: 4(a) –1.6 to − 1.4 1 4(b) –0.5 1 4(c) k > –4 2 B1 for identifying the –4 or for horizontal line drawn y = –4 4(d) y = x – 5 ruled 3 B2 for correct line and 2 correct values or and no line and 3 correct values –2.3 to –2.1 or B1 for no line and 2 correct values –1.2 to –1.1 or B1 for correct line 1.3 to 1.4 4(e) Tangent ruled at x = 1 B1 No daylight at point of contact. Consider point of contact as midpoint between two vertices of daylight, the midpoint must be between x = 0.8 and 1.2 –6 to –4 2 Dep on B1 or close attempt at tangent at x = 1 M1 for rise/run for their tangent at x = 1

This question in 0580/41 May/June 2017

4 f (x) = 2 x 2 - 1 The graph of y = f (x) , for - 2 G x G 2 , is drawn on the grid. y 8 7 6 5 4 3 2 1 x –2 –1 0 1 2 –1 –2 (a) Use the graph to solve the equation f (x) = 5 . x = … or x = … [2] (b) (i) Draw the tangent to the graph of y = f (x) at the point (-1.5, 3.5) . [1] (ii) Use your tangent to estimate the gradient of y = f (x ) when x = -1.5 . … [2] (c) g (x) = 2x (i) Complete the table for y = g (x) . x - 2 - 1 0 1 2 y 0.25 0.5 2 4 [1] (ii) On the grid opposite, draw the graph of y = g (x) for - 2 G x G 2 . [3] (d) Use your graphs to solve (i) the equation f (x) = g ( x) , x = … or x = … [2] (ii) the inequality f (x) 1 g (x) . … [1] (e) (i) Write down the three values. g (-3) = … g ( -5) = … g ( -10) = … [1] (ii) Complete the statement. As x decreases, g(x) approaches the value … [1]

14 marks

Mark scheme: 4(a) –1.75 to –1.7 1 1.7 to 1.75 1 4(b)(i) Correct ruled solid tangent at 1 (–1.5, 3.5) 4(b)(ii) –7 to –5 2 dep dep on close attempt at ruled solid tangent at x = –1.5 in part (b)(i) M1 for rise/run dep on close attempt at ruled solid tangent at x = –1.5 4(c)(i) 1 1 4(c)(ii) Correct curve 3 B2 for 4 or 5 correct points or B1 for 2 or 3 correct points 4(d)(i) –0.95 to –0.8 1 1.1 to 1.45 1 4(d)(ii) their (–0.95 to –0.8 )< x < 1FT correct or FT their (d)(i) their( 1.1 to 1.45) oe 4(e)(i) 0.125 oe and 0.03125 oe and 1 0.000976 to 0.000977 oe 4(e)(ii) 0 1 accept zero, nought, etc

This question in 0580/42 May/June 2017

3 The table shows some values for y = 2x 3 + 4x 2 . x –2.2 –2 –1.5 –1 –0.5 0 0.5 0.8 y –1.94 0.75 0 3.58 (a) Complete the table. [4] (b) Draw the graph of y = 2x 3 + 4x 2 for - 2.2 G x G 0.8 . y 4 3 2 1 x –2.5 –2 –1.5 –1 –0.5 0 0.5 1 –1 –2 [4] (c) Find the number of solutions to the equation 2x 3 + 4x 2 = 3 . … [1] (d) (i) The equation 2x 3 + 4x 2 - x = 1 can be solved by drawing a straight line on the grid. Write down the equation of this straight line. y = … [1] (ii) Use your graph to solve the equation 2x 3 + 4x 2 - x = 1. x = … or x = … or x = … [3] (e) The tangent to the graph of y = 2x 3 + 4x 2 has a negative gradient when x = k . Complete the inequality for k. … 1 k 1 … [2]

15 marks

Mark scheme: 3(a) 0 2.25 2 1.25 4 B1 for each 3(b) Fully correct smooth curve 4 B3 FT for 7 or 8 points or B2 FT for 5 or 6 points or B1 FT for 3 or 4 points 3(c) 1 1 3(d)(i) [y =] x + 1 1 3(d)(ii) −2.2 to −2.1 1 −0.45 to −0.4 1 0.51 to 0.6 1 If zero scored, SC1 for their line in (d)(i) drawn. It must be of the form y = mx + c (m ≠ 0) and drawn ‘fit for purpose’ 3(e) −1.33 < k < 0 to 0.1 2FT FT Strict ft of their max point and min point dep on cubic graph or accept correct answer from calculus B1 for each If zero scored, SC1 for two correct values reversed

This question in 0580/43 May/June 2017

4 f(x) = x3 – 4x2 + 15 (a) Complete the table of values for y = f(x). x –2 –1 –0.5 0 1 2 2.5 3 3.5 4 4.5 y –9 13.9 15 12 5.6 6 8.9 15 25.1 [2] (b) On the grid, draw the graph of y = f(x) for –2 G x G 4.5 . y 30 25 20 15 10 5 x –2 –1 0 1 2 3 4 –5 –10 [4] (c) Use your graph to solve the equation f(x) = 0. x = … [1] (d) By drawing a suitable tangent, estimate the gradient of the graph of y = f(x) when x = 3.5 . … [3] (e) By drawing a suitable straight line on the grid, solve the equation x3 – 4x2 – 2x + 5 = 0. x = … or x = … or x = … [4]

14 marks

Mark scheme: 4(a) 10, 7 2 B1 for each value 4(b) Correct curve 4 B3 FT for 10 or 11 correct points B2 FT for 8 or 9 correct points B1 FT for 6 or 7 correct points FT their table 4(c) –1.7 to –1.55 1 FT their graph if one answer 4(d) Tangent ruled at x = 3.5 B1 No daylight between tangent and curve at point of contact 6.5 to 11 B2 dep on tangent drawn or close attempt at tangent at x = 3.5 M1 for rise/run also dep on tangent or close attempt at x = 3.5 4(e) line y = 2x + 10 ruled 4 B3 for correct line (could be short) and 1 correct AND value −1.3 to −1.1 or B2 for correct line (could be short) 1 or B1 for [y = ] 2x + 10 seen 4.1 to 4.25 If zero scored, SC1 for no/wrong line and 3 correct values

This question in 0580/41 Oct/Nov 2017

5 y = - 2 , x =Y 0 8 x (a) Complete the table of values. x 0.5 1 1.5 2 2.5 3 3.5 y – 8.0 – 1.9 – 0.5 0.5 1.6 [2] (b) y 6 5 4 3 2 1 x –3 –2 –1 0 1 2 3 –1 –2 –3 –4 –5 –6 –7 –8 –9 x 3 2 The graph of y = - 2 for - 3.5 G x G - 0. 5 has already been drawn. 8 x x 3 2 On the grid, draw the graph of y = - 2 for 0.5 G x G 3.5 . [4] 8 x x 3 2(c) Use your graph to solve the equation - 2 = 0 . 8 x x = … [1] x 3 2(d) - 2 = k and k is an integer. 8 x x 3 2 Write down a value of k when the equation - 2 = k has 8 x (i) one answer, k = … [1] (ii) three answers. k = … [1] (e) By drawing a suitable tangent, estimate the gradient of the curve where x =- 3 . … [3] x 3 2(f) (i) By drawing a suitable line on the grid, find x when - 2 = 6 - x . 8 x x = … [3] x 3 2 5 3 2 (ii) The equation - 2 = 6 - x can be written as x + ax + bx + c = 0 . 8 x Find the values of a, b and c. a = … b = … c = … [4]

19 marks

Mark scheme: 5(a) 3.2 or 3.15 or 3.152 to 3.153 2 B1 for each 5.2 or 5.19 or 5.20 or 5.196… 5(b) Correct graph for 0.5 ⩽ x ⩽ 3.5 4 B3FT for 6 or 7 correct points or B2FT for 4 or 5 correct points or B1FT for 2 or 3 correct points 5(c) 1.7 to 1.8 1FT FT their graph if one answer 5(d)(i) Any integer k ⩾ −1 1 5(d)(ii) Any integer k < –1 1 5(e) Tangent ruled at x = –3 B1 2.5 to 4 B2 dep on tangent drawn at x = –3 or close attempt at tangent at x = –3 M1 for rise/run also dep on tangent at x = –3 or close attempt at tangent at x = –3 5(f)(i) y = 6 – x ruled accurately M2 M1 for correct line but freehand or ruled line gradient –1.1 to –0.9, or through (0, 6) but not y = 6 2.85 ⩽ x ⩽ 3 A1 5(f)(ii) [a = ] 8 4 B3 for 2 correct [b = ] –48 or x 5 + 8 x 3 − 48 x 2 − 16 = 0 seen [c = ] –16 5 3 2 or − x − 8 x + 48 x + 16 = 0 seen or M2 for correct multiplication by 8x2 or B1 for answers ± 8, ± 48, ± 16 x 2 × x 3 −×8 2 or M1 for = 6 − x x 2 × 8 or M1 for correct multiplication by 8 or M1 for correct multiplication by x2

This question in 0580/42 Oct/Nov 2017

7 The table shows some values of y = 2x 2 + 5x - 3 for -4 G x G 1.5 . x -4 -3 -2 -1 0 1 1.5 y 0 -5 -3 4 (a) Complete the table. [3] (b) On the grid, draw the graph of y = 2x 2 + 5x - 3 for -4 G x G 1.5 . y 10 9 8 7 6 5 4 3 2 1 x –4 –3 –2 –1 0 1 –1 –2 –3 –4 –5 –6 –7 [4] (c) Use your graph to solve the equation 2x2 + 5x – 3 = 3. x = … or x = … [2] (d) y = 2x 2 + 5x - 3 can be written in the form y = 2 x + a 2 + b . ^ h Find the value of a and the value of b. a = … b = … [3]

12 marks

Mark scheme: 7(a) 9, – 6, 9 3 B1 for each 7(b) Correct graph 4 B3FT for 6 or 7 correct points or B2FT for 4 or 5 correct points or B1FT for 2 or 3 correct points 7(c) –3.5 to –3.35 and 0.8 to 0.9.. 2FT FT their graph B1FT for either 7(d) 5 1 3 B2 for either correct a = or 1 or 1.25 2 4 4  5  or M1 for [2]  x +  seen isw 49 1  4  b = − 8 or − 68 or –6.125 or for 2x2 + 4ax + 2a2 + b

This question in 0580/43 Oct/Nov 2017

(a) Write down the equation of the line of symmetry of the graph. … [1] (b) On the grid opposite, draw the tangent to the curve at the point where x = 0.5 . Find the gradient of this tangent. … [3] (c) The table shows some values for y = x 3 + 3x + 4 . x - .15 - 1 - .05 0 0.5 1 1.5 y - .39 5.6 8 11.9 (i) Complete the table. [3] (ii) On the grid opposite, draw the graph of y = x 3 + 3x + 4 for - 1.5 G x G 1.5 . [4] (d) Show that the values of x where the two curves intersect are the solutions to the equation x 3 + 8x 2 + 3x - 6 = 0 . [1] (e) By drawing a suitable straight line, solve the equation x 3 + 5x + 2 = 0 for - 1.5 G x G 1.5 . x = … [3]

15 marks

Mark scheme: 7(a) x = 0 1 7(b) Tangent ruled at x = 0.5 B1 No daylight between tangent and curve at point of contact −9 to −6.5 2 dep on ruled tangent or close attempt at tangent at x = 0.5 M1 for rise/run also dep on tangent or close attempt at tangent at x = 0.5 7(c)(i) 0 2.4 or better 4 3 B1 for each 7(c)(ii) Correct smooth curve 4 B3FT for 6 or 7 correct plots or B2 FT for 4 or 5 correct plots or B1 FT for 2 or 3 correct plots FT their table 7(d) x 3 + 3 x + 4 = 10 − 8 x 2 and correctly 1 completed 7(e) line y = −2 x + 2 drawn and 3 B2 for ruled y = −2 x + 2 −0.45 to −0.35 nfww or B1 for − 2 x + 2 seen or for line y = –2x + c drawn or for y = cx + 2 (c ≠ 0) drawn and B1 for −0.45 to − 0.35 nfww

This question in 0580/41 May/June 2018

(c) (i) By drawing a suitable tangent, find an estimate of the gradient of the curve at x = - 2. … [3] (ii) Write down the equation of the tangent to the curve at x = - 2. Give your answer in the form y = mx + c. y = … [2] (d) Use your graph to solve the equations. x 3 1 (i) - 2 = 0 3 2x x = … [1] x 3 1 (ii) - 2 + 4 = 0 3 2x x = … or x = … or x = … [3] x 3 1 -3 + bxn - 3 = 0.(e) The equation - 2 + 4 = 0 can be written in the form axn 3 2x Find the value of a, the value of b and the value of n. a = … b = … n = … [3]

20 marks

Mark scheme: 6(a) – 2[.0], – 0.2, 2.5 3 B1 for each 6(b) Fully correct curve 5 B4 for correct curve, but branches joined or B3FT for 9 or 10 correct plots or B2FT for 7 or 8 correct plots or B1FT for 5 or 6 correct plots and B1 indep two separate branches not touching or cutting y-axis 6(c)(i) Correct tangent and 3 B2 for close attempt at tangent to curve 3 ⩽ grad ⩽ 5 at x = – 2 and answer in range OR B1 for ruled tangent at x = – 2, no daylight at x = –2 and M1dep (dep on B1 or close attempt rise at tangent) [at x = –2] for run 6(c)(ii) [y =] their(c)(i) x + their y-intercept final 2 Strict FT their y-intercept for their line answer M1 for y = their(c)(i) x + any value or ‘c’ oe seen or for y = any value(non-zero) x or ‘mx’ + their y-intercept seen oe 6(d)(i) 1.05 to 1.25 1 6(d)(ii) – 2.3 to – 2.2 3 B1 for each – 0.4 to – 0.3 After 0 scored B1 for y = –4 ruled 0.3 to 0.4 6(e) [a =] 2 3 B2 for 2 correct or for [b =] 24 2x5 + 24x2 [–3 = 0] [n =] 5 or B1 for 1 correct or for 2 x 5 − 3 + 4(6 x 2 ) [ = 0] oe 6 x 2 If 0 scored SC1 for 2x5 seen in final line of algebra

This question in 0580/42 May/June 2018

2 (a) (i) y = 2x Complete the table. x 0 1 2 3 4 y 2 4 8 [2] (ii) y = 14 - x 2 Complete the table. x 0 1 2 3 4 y 13 10 5 [2] (b) On the grid, draw the graphs of y = 2x and y = 14 - x 2 for 0 G x G 4 . y 16 14 12 10 8 6 4 2 0 x 1 2 3 4 −2 [6] (c) Use your graphs to solve the equations. (i) 2 x = 12 x = … [1] (ii) 2 x = 14 - x 2 x = … [1] (d) (i) On the grid, draw the line from the point (4, 2) that has a gradient of - 4 . [1] (ii) Complete the statement. This straight line is a … to the graph of y = 14 - x 2 at the point ( … , … ). [2]

15 marks

Mark scheme: 2(a)(i) 1, ….., ….., …., 16 2 B1 for each 2(a)(ii) 14, ….., ….., …., – 2 2 B1 for each 2(b) Fully correct smooth curves 6 B3 for correct curve of y = 2 x or B2FT for 4 or 5 correct points or B1FT for 2 or 3 correct points B3 for correct curve of y = 14 − x 2 or B2FT for 4 or 5 correct points or B1FT for 2 or 3 correct points 2(c)(i) 3.5 to 3.7 1 2(c)(ii) 2.65 to 2.8 1 2(d)(i) Correct line 1 Ruled, through (4, 2) and gradient −4 2(d)(ii) Tangent 2 B1 for each (2, 10)

This question in 0580/43 May/June 2018

3 The table shows some values of y = x 3 - 3x 2 + x . x -0.75 -0.5 -0.25 0 0.5 1 1.5 2 2.5 2.75 y -2.9 -1.4 -0.5 -0.1 -1 -1.9 -0.6 (a) Complete the table. [3] (b) On the grid, draw the graph of y = x 3 - 3x 2 + x for - 0.75 G x G 2.75 . [4] y 1 0 x –1 1 2 3 –1 –2 –3 (c) Use your graph to complete the inequalities in x for which y 2- 1. … 1 x 1 … and x 2 … [3] (d) The equation x 3 - 3x 2 + 2x - 1 = 0 can be solved by drawing a straight line on the grid. (i) Write down the equation of this line. … [2] (ii) On the grid, draw this line and use it to solve the equation x 3 - 3x 2 + 2x - 1 = 0 . x = … [3] (e) By drawing a suitable tangent, find an estimate for the gradient of the graph of y = x 3 - 3x 2 + x at x =- 0.25 . … [3]

18 marks

Mark scheme: 3(a) 0 −2 0.9 3 B1 for each 3(b) Correct curve 4 B3 FT for 9 or 10 points or B2 FT for 7 or 8 points or B1 FT for 5 or 6 points 3(c) −0.45 to –0.35 3 FT their graph 1 B1 for each in the correct position 2.35 to 2.45 If zero scored, SC1FT for 3 correct values 3(d)(i) y =1 − x oe 2 B1 for y =1 − kx oe, k ≠ 0 or y = k − x oe or 1−x 3(d)(ii) Correct ruled line and 3 B2FTdep for correct ruled line 2.25 to 2.4 or B1 dep for line through (0, 1) when extended but not y = 1 or with gradient –1.1 to –0.9 or correct line but freehand or SC2 for y = x – 1 ruled after answer [y =] x – 1 in (d)(i) and B1 for 2.25 to 2.4 3(e) Correct tangent and 3 No daylight between tangent and curve at 1.7 to 3.7 x = –0.25. Point of contact is the midpoint between two vertices of daylight and this point of contact must be between –0.35 and –0.15 B2 for close attempt at tangent at x = −0.25 and answer in range OR B1 for ruled tangent at x = −0.25, no daylight Consider point of contact as midpoint between two vertices of daylight, the midpoint must be between x = −0.35 and −0.15 and M1 dep on B1 or close attempt at rise tangent at x = –0.25 for run

This question in 0580/41 Oct/Nov 2018

4 f (x) = - , x =Y 0 4 x (a) Complete the table for f ()x . x 0.5 1 2 3 4 5 6 f ()x –7.9 –3.8 0.9 5.5 8.3 [2] (b) The graph of y = f (x) for - 6 G x G - 0.5 is drawn on the grid. y 10 8 6 4 2 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 x –2 –4 –6 –8 –10 On the same grid, draw the graph of y = f (x) for 0.5 G x G 6 . [3] (c) By drawing a suitable tangent, estimate the gradient of the graph of y = f (x) at the point (– 4, 5). … [3] 9(d) g (x) = , x =Y 0 x Complete the table for g ()x . x –4 –3 –2 –1 1 2 3 4 g ()x –2.3 –4.5 –9 9 4.5 2.3 [1] (e) On the same grid, draw the graph of y = g (x) for - 4 G x G - 1 and 1 G x G 4 . [4] (f) (i) Use your graphs to find the value of x when f (x) = g ( x) . x = … [1] (ii) Write down an inequality to show the positive values of x for which f (x) 2 g (x) . … [1] (g) The exact answer to part (f)(i) is 3 k . Use algebra to find the value of k. k = … [2]

17 marks

Mark scheme: 4(a) –1, 3 2 B1 for each 4(b) Correct graph 3 B2FT for 6 or 7 correct points or B1FT for 4 or 5 correct points 4(c) Correct ruled tangent 3 B2 for close attempt at tangent at x = –4 and and –2 ⩽ gradient ⩽ –1.5 answer in range OR B1 for ruled tangent at x = –4 with no daylight and M1 for rise/run also dep on close attempt at tangent. Must see correct or implied calculation from a drawn tangent. 4(d) –3, 3 1 4(e) Correct graph 4 B3FT for 7 or 8 correct points or B2FT for 5 or 6 correct points or B1FT for 3 or 4 correct points 4(f)(i) 3.6 to 3.85 1 4(f)(ii) x > their (f)(i) 1 FT 4(g) x 2 9 4 x 3 M1 13 9 4 = + or − 4 = 9 Allow for + 4 x x 4 x x x 52 A1

This question in 0580/43 Oct/Nov 2018

5 The table shows some values for y = x 3 - 2x for - 3 G x G 3 . 10 x −3 −2 −1.5 −1 0 1 1.5 2 3 y 2.0 1.7 0 −2.0 −1.6 (a) Complete the table. [3] 3 3 (b) On the grid, draw the graph of y = x - 2x for - 3 G x G 3 . 10 y 2 1 –3 –2 –1 0 1 2 3 x –1 –2 [4] 3 3 1(c) On the grid opposite, draw a suitable straight line to solve the equation x - 2x = (1 - x) for 10 2 - 3 G x G 3 . x = … or x = … or x = … [4] 3 3(d) For - 3 G x G 3 , the equation x - 2x = 1 has n solutions. 10 Write down the value of n. n = … [1]

12 marks

Mark scheme: 5(a) −2.1, 1.6, −1.7, 2.1 3 B2 for 3 correct or B1 for 2 correct 5(b) Fully correct curve 4 B3FT for 8 or 9 correct plots or B2FT for 6 or 7 correct plots or B1FT for 4 or 5 correct plots 5(c) 1 M2 line y = (1 − x ) ruled M1 for line with gradient −1 2 2 1 1 M1 for line through (0, ) but not y = 2 2 −2.15 to −2.01 B2 B1 for two correct −0.45 to −0.2 2.25 to 2.45 5(d) number of intersections of their curve and 1 strict FT for their curve the line y = 1

This question in 0580/42 Feb/March 2019

2 The table shows some values for y = x 3 + 3x 2 + 2 . x -3.5 -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 1.5 y -4.1 5.1 6 5.4 4 2.6 2.9 12.1 (a) Complete the table. [3] (b) On the grid, draw the graph of y = x 3 + 3x 2 + 2 for - 3.5 G x G 1.5 . y 15 10 5 – 3.5 – 3 – 2.5 – 2 – 1.5 – 1 – 0.5 0 0.5 1 1.5 x – 5 [4] (c) Use your graph to solve the equation x 3 + 3x 2 + 2 = 0 for - 3.5 G x G 1.5 . x = … [1] (d) By drawing a suitable straight line, solve the equation x 3 + 3x 2 + 2x + 2 = 0 for - 3.5 G x G 1.5 . x = … [2] (e) For - 3.5 G x G 1.5 , the equation x 3 + 3 x 2 + 2 = k has three solutions and k is an integer. Write down a possible value of k. k = … [1]

11 marks

Mark scheme: 2(a) 2, 2, 6 3 B1 for each 2(b) Correct graph 4 B3FT for 10 or 11 correct plots or B2FT for 8 or 9 correct plots or B1FT for 6 or 7 correct plots 2(c) –3.3 to –3.1 1 FT their graph 2(d) y = –2x ruled M1 or B1 for y = −2 x stated –2.6 to –2.45 A1 2(e) 3 or 4 or 5 1 FT their graph Allow more than one correct value

This question in 0580/41 May/June 2019

(a) Use the graph to find (i) f (1 ) , … [1] (ii) ff (- 2) . … [2] (b) On the grid opposite, draw a suitable straight line to solve the equation 2 2 x - - 7 =- 3x for - 3 G x G 3 . x x = … or x = … [4] (c) By drawing a suitable tangent, find an estimate of the gradient of the curve at x = - 2. … [3] (d) (i) Complete the table for y = g (x) where g ()x = 2 -x for - 3 G x G 3 . x -3 -2 -1 0 1 2 3 y 2 1 0.5 0.125 [3] (ii) On the grid opposite, draw the graph of y = g (x) . [3] (iii) Use your graph to find the positive solution to the equation f (x) = g ( x) . x = … [1]

17 marks

Mark scheme: 5(a)(i) –3 1 5(a)(ii) 6.2 to 6.4 oe 2 M1 for 3 seen or used 5(b) y = 5 – 3x ruled 2 B1 for y = 5 – 3x soi or ruled line with gradient – 3 or with y – intercept at 5 (but not y = 5) or B1FT for incorrect line equation/expression shown in working and their line correctly drawn – 0.3 to – 0.2 2 B1 for each, dep on y = 5 – 3x drawn 1.65 to 1.8 or FT their line provided equation/expression shown in working, dep on B1FT for line 5(c) Tangent ruled at x = −2 1 B1 for correct tangent –4.5 to –2.5 2 Dep on B1 for tangent or close attempt at tangent at x = –2 M1 for rise/run also dep on tangent drawn or close attempt at correct tangent Must see correct or implied calculation from a drawn tangent 5(d)(i) 8, 4, 0.25 oe 3 B1 for each 5(d)(ii) Correct graph 3 B2FT for 6 or 7 correct plots or B1FT for 4 or 5 correct plots 5(d)(iii) 1.8 to 1.9 1

This question in 0580/43 May/June 2019

x 2 1 2(c) Use your graph to solve + 2 - G 0 . 2 x x … G x G … [2] x 2 1 2(d) Find the smallest positive integer value of k for which + 2 - = k has two solutions 2 x x for - 3 G x G - 0.3 and 0.2 G x G 3 . … [1] x 2 1 2(e) (i) By drawing a suitable straight line, solve + 2 - = 3x + 1 for - 3 G x G - 0.3 and 2 x x 0.2 G x G 3 . x = … [3] x 2 1 2 4 3 2 (ii) The equation + 2 - = 3x + 1 can be written as x + ax + bx + cx + 2 = 0 . 2 x x Find the values of a, b and c. a = … b = … c = … [3]

17 marks

Mark scheme: 5(a) 3.5, 15, 3.9 3 B1 for each 5(b) Correct graph 5 B4 for correct curves but branches joined or touching y-axis or B3FT 10 or 11 points or B2FT for 8 or 9 points or B1FT for 6 or 7 points B1indep two separate branches not touching or crossing y-axis 5(c) 0.5 to 0.6 and 1.3 to 1.6 2 B1 for each or both correct but in reverse order 5(d) 1 1 5(e)(i) y = 3x + 1 ruled 3 B2 for correct ruled line that crosses their and 0.3 to 0.49 curve or B1 for y = 3x + 1 soi or freehand line or ruled line with gradient 3 or with y – intercept at 1 (but not y = 1) 5(e)(ii) [a = ] –6 3 M2 for x4 + 2 – 4x = 6x3 + 2x2 or better seen [b = ] –2 [c = ] –4 or B1 for each correct value to a maximum of 2 marks If 0 scored, SC1 for answer [a = ] 6,[b = ] 2 and [c = ] 4 or for x5 + 2x – 4x2 = 6x4 + 2x3 or better

This question in 0580/42 Oct/Nov 2019

3 The table shows some values for y = x 3 + x 2 - 5x . x -3 -2 -1.5 -1 0 1 1.5 2 2.5 3 y -3 6 6.4 0 -1.9 2 9.4 (a) Complete the table. [3] (b) On the grid, draw the graph of y = x 3 + x 2 - 5x for - 3 G x G 3 . y 25 20 15 10 5 x - 3 - 2 - 1 0 1 2 3 - 5 [4] (c) Use your graph to solve the equation x 3 + x 2 - 5x = 0 . x = … or x = … or x = … [2] (d) By drawing a suitable tangent, find an estimate of the gradient of the curve at x = 2 . … [3] (e) Write down the largest value of the integer, k, so that the equation x 3 + x 2 - 5x = k has three solutions for - 3 G x G 3 . k = … [1]

13 marks

Mark scheme: 3(a) 5, –3, 21 3 B1 for each 3(b) Fully correct curve 4 B3 FT for 9 or 10 points or B2 FT for 7 or 8 points or B1 FT for 5 or 6 points 3(c) –2.9 to –2.7 2 B1 for 2 correct values 0 1.7 to 1.9 3(d) Tangent ruled at x = 2 B1 10 to 14 B2 Dep on correct tangent or close attempt at tangent at x = 2 M1 for rise/run also dep on correct tangent drawn or close attempt at tangent Must see correct or implied calculation from a drawn tangent 3(e) 6 1

This question in 0580/43 Oct/Nov 2019

2 (a) The table shows some values for y = 2x 3 - 4x 2 + 3 . x -1 - 0.5 0 0.5 1 1.5 2 y -3 1.75 0.75 3 (i) Complete the table. [3] (ii) On the grid, draw the graph of y = 2x 3 - 4x 2 + 3 for - 1 G x G 2 . y 3 2 1 – 1 0 1 2 x – 1 – 2 – 3 [4] (iii) Use your graph to solve the equation 2x 3 - 4x 2 + 3 = 1 .5 . x = … or x = … or x = … [3] (iv) The equation 2x 3 - 4x 2 + 3 = k has only one solution for - 1 G x G 2 . Write down a possible integer value of k. … [1] (b) y 6 5 4 3 2 1 – 1 0 1 x – 1 – 2 – 3 – 4 (i) On the grid, draw the tangent to the curve at x = 1. [1] (ii) Use your tangent to estimate the gradient of the curve at x = 1. … [2] (iii) Write down the equation of your tangent in the form y = mx + c . y = … [2]

16 marks

Mark scheme: 2(a)(i) 3 2.25 1 3 B1 for each 2(a)(ii) Fully correct smooth curve 4 B3FT for 7 or 6 correct plots B2FT for 5 or 4 correct plots B1FT for 3 correct plots 2(a)(iii) −0.6 to −0.51, 0.75 to 0.85, 3 B1 for each 1.7 to 1.85 If 0 scored, SC1 for y = 1.5 drawn 2(a)(iv) −3 or −2 or −1 or 0 1 2(b)(i) Tangent ruled at x = 1 1 2(b)(ii) 4.4 to 5.6 2 Dep on tangent at x = 1 or close attempt M1 for rise/run for their line 2(b)(iii) y = (4.4 to 5.6)x – (1.8 to 2.2) 2 FT for any line but not horizontal or vertical or line for 2 marks or B1 [y =] their (b)(ii)x + their(y-intercept) B1FT for [m =] their 5 or for their y-intercept

This question in 0580/42 Feb/March 2020

12 (a) A curve has equation y = 4 x 3 - 3x + 3 . (i) Find the coordinates of the two stationary points. ( … , … ) and ( … , … ) [5] (ii) Determine whether each of the stationary points is a maximum or a minimum. Give reasons for your answers. [3] (b) The graph of y = x 2 - x + 1 is shown on the grid. y 7 6 5 4 3 2 1 – 3 – 2 – 1 0 1 2 3 4 x – 1 By drawing a suitable line on the grid, solve the equation x 2 - 2x - 2 = 0 . x = … or x = … [3]

11 marks

Mark scheme: 12(a)(i)  1   1  5 B2 for 12 x 2 − 3[ = 0]  − , 4  and  , 2   2   2  or B1 for 12x2 or – 3 M1 for their derivative = 0 or dy/dx = 0 B1 for [x =] – ½ and ½ or one coordinate pair correct 12(a)(ii)  1  3 B2 for one correct with reason  − , 4  Max with reason or M1 for correct attempt to find  2  e.g. 2nd derivative/gradients/sketch  1   , 2  Min with reason  2  12(b) line y = x + 3 ruled M2 B1 for [ y = ] x + 3 identified or rules y = x + k or y = px + 3 −0.7 to −0.8 A1 2.7 to 2.8

This question in 0580/43 May/June 2020

5 (a) The diagram shows the graph of y = f ( x) for - 3 G x G 3 . y 20 16 12 8 4 – 3 – 2 – 1 0 1 2 3 x – 4 – 8 – 12 (i) Solve f ( x) = 14 . x = … [1] (ii) By drawing a suitable tangent, find an estimate of the gradient of the graph at the point (-2, 4). … [3] (iii) By drawing a suitable straight line on the grid, solve f ( x) = 2 x - 2 for - 3 G x G 3 . . x = … [3] (b) y A NOT TO B SCALE O x The diagram shows a curve with equation y = 2x 2 - 2x - 7 . The straight line with equation y = 3x + 5 intersects the curve at the points A and B. Find the coordinates of the points A and B. A ( … , … ) B ( … , … ) [5]

12 marks

Mark scheme: 5(a)(i) 2.7 to 2.8 1 5(a)(ii) tangent ruled at x = –2 B1 6 to 10 2 dep on B1 or a close attempt at tangent at x = –2 or M1 for rise/run for their tangent, or close attempt, at any point Must see correct or implied calculation from a drawn tangent After M0, SC1 for gradient of tangent (or close attempt) in range embedded in y = mx + c 5(a)(iii) y = 2x – 2 ruled 3 B2 for correct ruled line and x = –2.9 to –2.8 cao or B1 for short line or for freehand line or broken line or ruled line with gradient 2 or with y-intercept at –2 (but not y = –2) 5(b) A (4, 17) B (–1.5, 0.5) 5 B4 for (–1.5, 0.5) and (4, 17), or for x = 4 and x = –1.5 OR B3 for A(4, 17) or B(–1.5, 0.5) OR M1 for 2x2 –2x – 7 = 3x + 5 oe AND either M2 for (2x + 3)(x – 4) or M1 for 2x(x – 4) + 3(x – 4) or x(2x + 3) – 4(2x + 3) or (2x +c)(x + d) where cd = –12 or c + 2d = –5 [c and d are integers] OR M2 for − their b ± (theirb ) 2 − 4( their a )( their c ) 2( their a ) or M1 for ( their b ) 2 − 4( their a )( their c ) or for p = –their b, r = 2(their a) if in the ௣ ା √௤ ௣ ି √௤ form or ௥ ௥

This question in 0580/41 Oct/Nov 2020

10 (a) y NOT TO SCALE B A O x C The diagram shows a sketch of the curve y = x 2 + 3x - 4 . (i) Find the coordinates of the points A, B and C. A ( … , … ) B ( … , … ) C ( … , … ) [4] (ii) Differentiate x 2 + 3x - 4 . … [2] (iii) Find the equation of the tangent to the curve at the point (2, 6). … [3] (b) y 0 x 90° 180° 270° 360° (i) On the diagram, sketch the graph of y = tan x for 0° G x G 360° . [2] (ii) Solve the equation 5 tanx =- 7 for 0° G x G 360° . x = … or x = … [3]

14 marks

Mark scheme: 10(a)(i) A(–4, 0) 4 B3 for A and B correct B(1, 0) Or B2 for B (–4, 0) and A (1, 0) C(0, –4) Or B1 for (x + 4)(x – 1) or for −±3 32 −×4 1 ×−4 oe 2 and B1 for A or B correct B1 for C(0, –4) OR SC2 for –4, 1 and –4 in correct positions on the graph 10(a)(ii) 2x + 3 [ ± 0] final answer 2 B1 for answer 2x +c or for ax + 3, a ≠ 0 or for correct answer seen 10(a)(iii) y = 7x – 8 oe 3 B2 for answer 7x – 8 OR M1 for [gradient =] 2(2) + 3 FT their part (a)(ii) of the form ax + b M1dep for substitution of (2, 6) into y = their mx + c oe 10(b)(i) Correct sketch 2 B1 for one correct section out of 4 OR B1 for two properties correct from • Crosses x-axis at (0, 0) (180, 0) and (360, 0) only • Correct curvature in each section of 90o • Asymptotes at x = 90 and x = 270 0 90 180 270 360 10(b)(ii) 125.5 or 125.53 to 125.54 3 B2 for one correct angle and or B1 for –54.5 or –54.46… or for 2 angles 305.5 or 305.53 to 305.54 with a difference of 180.

This question in 0580/41 Oct/Nov 2020

7 y = x 2 + , x ! 0 x (a) Complete the table. x 0.2 0.3 0.5 1 1.5 2 2.5 y 5.0 3.4 2.3 2.9 6.7 [2] 2 1 (b) On the grid, draw the graph of y = x + for 0.2 G x G 2.5 . x 2 1 The graph of y = x + for - 2.5 G x G - 0.2 has been drawn for you. x y 7 6 5 4 3 2 1 – 2.5 – 2 – 1.5 – 1 – 0.5 0 0.5 1 1.5 2 2.5 x – 1 – 2 – 3 – 4 – 5 [4] (c) By drawing suitable straight lines on the grid, solve the following equations. 2 1 (i) x + =- 2 x x = … [1] 2 1 (ii) x + + x - 1 = 0 x x = … [2] 2 1(d) k is an integer and the equation x + = k has three solutions. x Write down a possible value of k. k = … [1]

10 marks

Mark scheme: 7(a) 2, 4.5 2 B1 for each 7(b) Correct graph 4 B3 FT for 6 or 7 correct points FT their table or B2 FT for 4 or 5 correct points FT their 6666 table 4444 or B1 FT for 2 or 3 correct points FT their 2222 table -2-2-2-2 -1-1-1-1 0000 0000 1111 2222 -2-2-2-2 -4-4-4-4 7(c)(i) –0.5 to –0.4 1 7(c)(ii) y = 1 – x ruled 2  2 1  M1 for [y =] 1 – x or x + = 1 − x soi and    x  –1.9 to –1.75 or B1 for –1.9 to –1.75 7(d) Any integer > 2 1

This question in 0580/42 Oct/Nov 2020

6 y 5 4 3 2 1 x – 2 – 1 0 1 2 – 1 – 2 (a) The grid shows the graph of y = a + bx2 . The graph passes through the points with coordinates (0, 4) and (1, 1). (i) Find the value of a and the value of b. a = … b = … [2] (ii) Write down the equation of the tangent to the graph at (0, 4). … [1] (iii) The equation of the tangent to the graph at x =- 1 is y = 6x + 7 . Find the equation of the tangent to the graph at x = 1. … [2] 5(b) The table shows some values for y = 1 + for - 2 G x G 1.5 . 3 - x x – 2 – 1.5 – 1 – 0.5 0 0.5 1 1.5 y 2 2.11 2.43 3 4.33 (i) Complete the table. [3] 5 (ii) On the grid, draw the graph of y = 1 + for - 2 G x G 1.5 . [4] 3 - x (c) (i) Write down the values of x where the two graphs intersect. x = … or x = … [2] (ii) The answers to part(c)(i) are two solutions of a cubic equation in terms of x. Find this equation in the form ax 3 + bx 2 + cx + d = 0 , where a, b, c and d are integers. … [4]

18 marks

Mark scheme: 6(a)(i) [a = ] 4 2 B1 for [a = ] 4 B1 for [b = ] – 3 nfww [b = ] – 3 nfww 6(a)(ii) y = 4 oe 1 6(a)(iii) y = − 6x + 7 oe final answer 2 B1 for answer − 6x + 7 or answers y = − 6x + c or y = kx + 7 (k < 0) 6(b)(i) 2.25 2.67 3.5 3 B1 for each 6(b)(ii) correct curve 4 B3 FT for 7 or 8 points or B2 FT for 5 or 6 points or B1 FT for 3 or 4 points 6(c)(i) –0.78 to –0.72 and 0.55 to 0.59 2 B1 for each 6(c)(ii) 3 x 3 − 9 x 2 − 3 x + 4 [ = 0] final answer 4 B3FT for 3 out of 4 correct terms or for bx3 – 3bx2 + (a – 1)x + 8 – 3a [ = 0] oe or B2FT for 2 out of 4 correct terms or for 3 out of 4 terms from bx3 – 3bx2 + (a – 1)x + 8 – 3a [ = 0] 5 2 or M1 for 1 + = their 4 + (their ( −3)) x oe 3 − x

This question in 0580/42 Feb/March 2021

10 The table shows some values of y = 3 + 4x - x 2 for - 1 G x G 5 . x -1 -0.5 0 1 2 3 4 4.5 5 y -2 6 6 -2 (a) Complete the table. [3] (b) On the grid, draw the graph of y = 3 + 4x - x 2 for - 1 G x G 5 . y 8 6 4 2 – 1 0 1 2 3 4 5 x – 2 [4] (c) Write down an integer value of k for which the equation 3 + 4x - x 2 = k has no solutions. … [1] 9 2 (d) By drawing a suitable straight line on the grid, solve the equation - 1 + x - x = 0 . 2 x = … or x = … [4]

12 marks

Mark scheme: 10(a) 0.75 3 7 3 0.75 3 B2 for 4 or 3 correct or B1 for 2 correct 10(b) correct curve 4 B3FT for 8 or 9 correct plots B2FT for 6 or 7 correct plots B1FT for 4or 5 correct plots 10(c) Accept any integer ≥ 8 1 10(d) 1 B3 1 line y = 4 − x ruled B2 for [ y = ] 4 − x identified 2 2 1 or B1 for ruled line with gradient − 2 or B1 for ruled line through (0, 4) but not y = 4 0.2 to 0.3 4.2 to 4.3 B1

This question in 0580/41 May/June 2021

2 The table shows some values for y = 2 # 0.5 x - 1. x -1 -0.5 0 0.5 1 1.5 2 y 3 1.83 0.41 0 -0.29 (a) (i) Complete the table. [2] (ii) On the grid, draw the graph of y = 2 # 0.5 x - 1 for - 1 G x G 2 . y 3 2 1 x – 1 – 0.5 0 0.5 1 1.5 2 – 1 – 2 [4] (b) By drawing a suitable straight line, solve the equation 2 # 0.5 x + 2x - 3.5 = 0 for - 1 G x G 2 . x = … [3] (c) There are no solutions to the equation 2 # 0.5 x - 1 = k where k is an integer. Complete the following statements. The highest possible value of k is … The equation of the asymptote to the graph of y = 2 # 0.5 x - 1 is … [2]

11 marks

Mark scheme: 2(a)(i) 1, –0.5 oe 2 B1 for each 2(a)(ii) Correct curve 4 B3FT for 6 or 7 correct plots or B2FT for 4 or 5 correct plots or B1FT for 2 or 3 correct plots 2(b) y = 2.5 − 2x ruled B2 B1 for y = k − 2 x or y = px + 2.5 ruled (p ≠ 0) or for [y =] 2.5 − 2x oe identified 1.3 to 1.4 B1 2(c) –1 B1 y = –1 B1 FT their k (must be negative)

This question in 0580/42 May/June 2021

2 3 24(b) By drawing a suitable straight line on the grid, solve the equation x - = - 2x 2x 5 for - 3 G x G - 0.2 and 0.2 G x G 3 . x = … or x = … [4] 2 3 24(c) The solutions to the equation x - = - 2x are also the solutions to an equation of the 2x 5 form ax 3 + bx 2 + cx - 15 = 0 where a, b and c are integers. Find the values of a, b and c. a = … b = … c = … [4]

16 marks

Mark scheme: 6(a)(i) 9.5, 4.8 and 8.5 3 B1 for each 6(a)(ii) correct curve 5 B4 for correct curve, but branches joined or touching y axis or B3FT for 9 or 10 correct plots or B2FT for 7 or 8 correct plots or B1FT for 5 or 6 correct plots AND B1 indep two separate branches not touching or cutting y-axis 6(b) 24 4 B2 for correct ruled line crossing curve y = − 2 x ruled twice 5 and or B1 for correct freehand or for short – 0.4 to – 0.2 and 1.45 to 1.7 ruled line or for line with negative gradient through (0, 4.8) or for line with gradient – 2 B1 for each value 6(c) [a =] 10 4 B3 for 10x3 – 15 = 48x – 20x2 oe or better [b =] 20 or B2 for 2 correct values [c =] – 48 or B1 for 1 correct value 2 15 or for 5 x − = 24 − 10 x or better 2 x 3 48 2 or for 2 x − 3 = x − 4 x or better 5 3 3 24 2 or for x − = x − 2 x 2 5 After 0 scored SC1 for correct elimination of a denominator of 5, x or 2x from a four term expression.

This question in 0580/41 Oct/Nov 2021

5 The table shows some values for y = x 3 - 3x 2 + 3 . x - 1 - .05 0 0.5 1 1.5 2 2.5 3 y 2.125 3 2.375 1 - 1 - .0125 (a) Complete the table. [3] (b) On the grid, draw the graph of y = x 3 - 3x 2 + 3 for - 1 G x G 3 . y 3.5 3 2.5 2 1.5 1 0.5 – 1 – 0.5 0 0.5 1 1.5 2 2.5 3 x – 0.5 – 1 – 1.5 [4] (c) By drawing a suitable straight line on the grid, solve the equation x 3 - 3x 2 + x + 1 = 0 . x = … or x = … or x = … [4]

11 marks

Mark scheme: 5(a) –1, –0.375, 3 3 B1 for each 5(b) Correct graph 4 B3FT for 8 or 9 correct points or B2FT for 6 or 7 correct points or B1FT for 4 or 5 correct points 5(c) y = 2 – x ruled correctly 4 B2 for y = 2 – x ruled AND or B1 for [y = ] 2 – x soi –0.45 to –0.35 or y = k – x ruled 1 or y = kx + 2 ruled, but not y = 2 2.35 to 2.45 B2 for all three values or B1 for any two values

This question in 0580/42 Oct/Nov 2021

9 f ( x) = x ( x - 1)( x - 2) (a) Find the coordinates of the points where the graph of y = f ( x) crosses the x-axis. ( … , … ) ( … , … ) ( … , … ) [2] (b) Show that (f x) = x 3 - 3x 2 + 2x . [2] (c) Find the coordinates of the turning points of the graph of y = f ( x) . Show all your working and give your answers correct to 1 decimal place. ( … , … ) ( … , … ) [8] (d) Sketch the graph of y = f ( x) . y O x [2]

14 marks

Mark scheme: 9(a) (0, 0), (1, 0), (2, 0) 2 B1 for any two correct If 0 scored, SC1 for all three x values clearly identified 9(b) 2 2 2 2 x x − x − 2 x + 2 or x − x x − x − 2 x + 2 ( ) ( )( x − 2 ) B1 for x ( ) or x − 2 ) or ( x − 1)( x 2 − 2 x ) ( x 2 − x )( leading to x 3 − 3 x 2 + 2 x with no errors or or ( x − 1)( x 2 − 2 x ) omissions 9(c) 3 x 2 − 6 x + 2 B2 B1 for 2 correct terms dy M1 their = 0 dx 2 M2 2 −−( 6) ± ( −6 ) − 4(3)(2) M1 for ( −6) − 4(3)(2) or for their 2(3) p ± q p = –(–6) and r = 2(3) if in form r (0.4, 0.4) B3 B2 for 0.4 or 0.42... and 1.6 or 1.57 to (1.6, –0.4) 1.58 or for one correct pair of coordinates or B1 for 0.4 or 0.42... or 1.6 or 1.57 to 1.58 1 1 If 0 scored SC1 for 1 + and 1 – 3 3 or better or for one correct pair of coordinates in any form 9(d) Correct2222 sketch 2 FT their (c) but must be cubic i.e. correct shape cubic through origin and 1111 max and min in correct quadrants .5.5.5.5 0000 0000 0.50.50.50.5 1111 1.51.51.51.5 2222 2.52.52.52.5 -1-1-1-1 B1 for cubic shape sketch -2-2-2-2

This question in 0580/43 Oct/Nov 2021

2 The table shows some values for y = x 2 - , x ! 0 . 3x The y-values are rounded to 1 decimal place. x -2 -1.5 -1 -0.75 -0.5 -0.25 -0.1 y 4.2 2.5 1.3 1.4 3.3 (a) Complete the table. [2] 2 1 (b) On the grid, draw the graph of y = x - for - 2 G x G - 0 .1 . 3x 2 1 The graph of y = x - for x 2 0 has been drawn for you. 3x y 4 3 2 1 – 2 – 1 0 1 2 x – 1 – 2 [4] 2 1 (c) By drawing a suitable line on the grid, solve the equation x - + 1 = 0 . 3x x = … [2]

8 marks

Mark scheme: 2(a) 1[.0] 0.9 2 B1 for each 2(b) correct curve 4 B3 FT for 6 or 7 points B2 FT for 4 or 5 points B1 FT for 2 or 3 points 2(c) ruled line at y = − 1 B1 0.3 to 0.32 B1

This question in 0580/42 Feb/March 2022

The diagram shows the graph of y = f ( x) for - 1.5 G x G 5 . (i) Find f ( 2) . … [1] (ii) Solve the equation f ( x) = 0 for - 1.5 G x G 5 . x = … or x = … or x = … [3] (iii) f ( x) = k has three solutions for - 1.5 G x G 5 where k is an integer. Find the smallest possible value of k. k = … [1] (iv) On the grid, draw a line y = mx so that f ( x) = mx has exactly one solution for - 1.5 G x G 5 . [2] (b) y = 3 x 2 - 12x + 7 dy (i) Find the value of when x = 5 . dx … [3] (ii) Find the coordinates of the point on the graph of y = 3x 2 - 12x + 7 where the gradient is 0. ( … , … ) [2] p 2 dy 6(c) When y = 2x + qx , = 14x + 6x . dx Find the value of p and the value of q. p = … q = … [2]

14 marks

Mark scheme: 6(a)(i) –3 1 6(a)(ii) –1 1.55 to 1.6 4.4 to 4.45 3 B1 for each 6(a)(iii) –8 1 6(a)(iv) Ruled line through origin intersecting curve once 2 B1 for ruled line through origin 6(b)(i) 18 3 B2 for 6x – 12 or B1 for 6x or –12 6(b)(ii) (2, –5) 2 B1 for each. If 0 scored, M1 for their 6x – 12 = 0 or states 0  dy dx 6(c) [p = ] 7 [q = ] 3 2 B1 for each

This question in 0580/41 May/June 2022

10 (a) y C (1.2, 10) A (– 1.5, 7.9) B (– 1, 7) E (6, 5.5) NOT TO SCALE D (5, 3) 0 x The diagram shows a sketch of the graph of y = f ( x) for - 1.5 G x G 6 . The coordinates of five points on the graph of y = f ( x) are shown on the diagram. (i) f ( x) = k has two solutions in the interval - 1.5 G x G 6 . Write down a possible integer value of k. k = … [1] (ii) f ( x) = j has no solutions in the interval - 1.5 G x G 6 when j 1 a or j 2 b . Find the maximum value of a and the minimum value of b. a = … b = … [2] (b) Find the coordinates of the two stationary points on the graph of y = x 6 - 6x 5 . You must show all your working. ( … , … ) ( … , … ) [5]

8 marks

Mark scheme: 10(a)(i) 4 or 5 or 7 or 8 or 9 1 10(a)(ii) [a =] 3, [b =] 10 2 B1 for each or for a and b transposed 10(b) 6 x 5 − 30 x 4 B2 B1 for 6x 5 or −30 x 4 their derivative = 0. M1 (0, 0) and (5, –3125) B2 B1 for (5, –3125) or for x = 0 and x = 5

This question in 0580/41 Oct/Nov 2022

2 All the lengths in this question are measured in centimetres. NOT TO 9 – x SCALE x x The diagram shows a solid cuboid with a square base. (a) The volume, V cm 3, of the cuboid is V = x 2 ( 9 - x) . The table shows some values of V for 0 G x G 9 . x 0 1 2 3 4 5 6 7 8 9 V 0 8 54 80 100 108 98 64 0 (i) Complete the table. [1] (ii) On the grid on the opposite page, draw the graph of V = x 2 ( 9 - x) for 0 G x G 9 . [4] (iii) Find the values of x when the volume of the cuboid is 44 cm 3. x = … or x = … [2] V 110 100 90 80 70 60 50 40 30 20 10 0 x 0 1 2 3 4 5 6 7 8 9 (b) (i) Show that the total surface area of the cuboid is ( 36x - 2x 2 )cm 2 . [2] (ii) Find the surface area when the volume of the cuboid is a maximum. … cm2 [3]

12 marks

Mark scheme: 2(a)(i) 28 1 2(a)(ii) Correct curve 4 B3FT for 9 or 10 correct points or B2FT for 7 or 8 correct points or B1FT for 5 or 6 correct points 2(a)(iii) 2.5 to 2.8 8.2 to 8.5 2 B1 for each value 2(b)(i) 2x2 + 4x(9 – x) oe M1 Accept the sum of individual areas if done in smaller parts 2x2 + 36x – 4x2 oe A1 With intermediate step shown and brackets removed with no Leading to 36x – 2x2 errors or omissions 2(b)(ii) 144 3 B1 for x = 6 identified from graph or using calculus M1 for 36  their6 – 2  (their 6)2

This question in 0580/42 Oct/Nov 2022

9 (a) Sketch the following graphs. On each sketch, indicate any intercepts with the axes. (i) 3x - 4y = 12 y O x [2] (ii) y = x 2 - 3x - 4 y O x [4] (iii) y = 6x y O x [2] dy 4 3(b) (i) Find the derivative, , of y = 5 + 8x - x . dx 3 … [2] 4 3 (ii) Find the gradient of y = 5 + 8 x - x at x =- 1. 3 … [2] 4 3 (iii) A tangent is drawn to the graph of y = 5 + 8x - x . 3 The gradient of the tangent is - 28 . Find the coordinates of the two possible points where this tangent meets the graph. ( … , … ) ( … , … ) [5]

17 marks

Mark scheme: 9(a)(i) Correct sketch of 3x – 4y = 12 with 2 B1 for line with positive gradient y = –3 and x = 4 indicated on axes 4 –3 9(a)(ii) Correct sketch of y = x2 – 3x – 4 4 B3 for correct sketch with one value omitted or incorrect or for a with (0, – 4) indicated as y – intercept and x = – 1 and x = 4 poor sketch with all 3 intercepts correct. indicated as roots or B2 for roots x = – 1 and x = 4 soi with no extra roots -1 4 or for correct shape with y = – 4 indicated or B1 for correct shape -4 or for (x – 4) (x + 1) shown or for incorrect sketch with (0, – 4 ) indicated as y – intercept Minimum in fourth quadrant, not at x = 0 9(a)(iii) Correct sketch of y = 6x 2 B1 for increasing exponential graph seen on both sides of the with y-intercept indicated at (0, 1) y-axis. 1 9(b)(i) 8 – 4x2 [+ 0] 2 B1 for two terms correct and one extra incorrect term or for one of two terms correct or for correct answer seen and spoilt 9(b)(ii) 4 2 M1 for substitution of x = –1 into their (b)(i) 9(b)(iii) (3, –7) and (–3, 17) 5 B4 for (3, –7) or (–3, 17) or B3 for x = ± 3 or M2 for x2 = 9 or k(x – 3)(x + 3) = 0 oe or for correct method for solving their (b)(i) = – 28 or M1 for their (b)(i) = – 28

This question in 0580/43 Oct/Nov 2022

8 A tailor makes x dresses and y shirts in one week. In one week • he makes at least 4 dresses • he makes no more than 7 shirts • he makes less than 14 dresses and shirts altogether 2 • the number of shirts he makes is more than of the number of dresses. 3 One of the inequalities that shows this information is x H 4 . (a) Write down the other three inequalities in x and/or y. … … … [3] (b) y 14 13 12 11 10 9 8 7 6 5 4 3 2 1 x 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 On the grid, draw 4 straight lines and shade the unwanted regions to show these inequalities. Label the region R that satisfies the 4 inequalities. [6] (c) Use your diagram to find the smallest number of dresses and the smallest number of shirts the tailor makes in one week. … dresses and … shirts [1] (d) The profit the tailor makes on one dress is $10 and the profit on one shirt is $6. Use your diagram to find the largest profit the tailor can make in one week. $ … [2]

12 marks

Mark scheme: 8(a) y  7 oe 3 B1 for each x  y  14 oe 2 y  x oe 3 8(b) x  4 solid M4 B1 for each y  7 solid x  y  14 dashed 2 y  x dashed 3 correct shading everywhere but A2 M1dep (dependent on M4 or B1B1B1B0 where region R the only error is wrong use of solid/dashed lines) for shading the correct side of 3 of the 4 lines. R 8(c) 4 dresses and 3 shirts 1 8(d) 106 2 M1 for 10 x  6 y evaluated for (x, y) in their region R or B1 for (7, 6) After 0 scored, SC1 for answer 112 or 116

This question in 0580/41 May/June 2023

7 (a) The diagram shows the graph of a function. y x O Put a ring around the word which correctly identifies the type of function. reciprocal quadratic cubic exponential linear [1] (b) (i) y x O 1 On the diagram, sketch the graph of y = , x ! 0 . [2] 2 x 1 (ii) Solve the equation = 2x . 2 x x = … and x = … [2] (c) (i) y 1 0 180° 360° x – 1 On the diagram, sketch the graph of y = sin x for 0° G x G 360 ° . [2] (ii) Solve the equation 3 sinx + 1 = 0 for 0° G x G 360° . x = … and x = … [3]

10 marks

Mark scheme: 7(a) Cubic 1 7(b)(i) Correct sketch 2 B1 for one branch correct or an attempt at the correct shape y Maximum 1 mark if sketch crosses x- axis or y-axis O x 7(b)(ii) 1 2 M1 for 4 x 2  1 oe  nfww 2 1 1 or B1 for or  nfww 2 2 7(c)(i) Correct sketch through (0, 0) (180, 0) and 2 B1 for correct sine curve shape, starting (360, 0) with max and min at 1 and –1 resp. at the origin, with minimum of 1 cycle. 180 360 7(c)(ii) 199.5 or 199.47... 3 B2 for one correct 1 and or M1 for sin x =  oe 3 340.5... If 0 scored, SC1 for two reflex angles with a sum of 540 or 2 non-reflex angles with a sum of 180

This question in 0580/43 May/June 2023

10 The table shows some values for y = 2 x - 3 . x -2 -1 0 0.5 1 1.5 2 2.5 y -2.75 -1.58 -0.17 1 2.66 (a) Complete the table. [3] (b) On the grid, draw the graph of y = 2 x - 3 for - 2 G x G 2.5 . y 3 2 1 – 2 – 1 0 1 2 x – 1 – 2 – 3 [4] (c) Use your graph to solve the equation 2 x - 3 = 2 . x = … [1] (d) By drawing a suitable straight line, solve the equation 2 x - x - 1.5 = 0 . x = … or x = … [4]

12 marks

Mark scheme: 10(a) 2.5 2 1 3 B1 for each 10(b) Correct curve 4 B3 FT for 8 or 7 correct plots B2 FT for 6 or 5 correct plots B1 FT for 4 or 3 correct plots y O x 10(c) 2.3 to 2.4 1 10(d) ruled line y  x  1.5 M2 M1 for y = x – 1.5 soi or for 2x – 3 = x – 1.5 seen. or y  x  k or y  kx  1.5 drawn Do not accept y = –1.5 1 and 1.55 to 1.7 A2 A1 for each

This question in 0580/43 May/June 2023

7 (a) Complete the table of values for y = 3 cos 2x° . Values are given correct to 1 decimal place. x 0 10 20 30 40 45 50 60 70 80 90 y 3.0 2.8 2.3 1.5 0.5 - .05 - .23 - .30 [3] (b) Draw the graph of y = 3 cos 2x° for 0 G x G 90 . y 3 2 1 0 x 10 20 30 40 50 60 70 80 90 - 1 - 2 - 3 [4] (c) Use your graph to solve the equation 3 cos 2x° =- 2 for 0 G x G 90 . x = … [1] (d) By drawing a suitable straight line, solve the equation 120 cos 2x° = 80 - x for 0 G x G 90 . x = … [3]

11 marks

Mark scheme: 7(a) 0, –1.5 oe, –2.8 3 B1 for each 7(b) Correct graph 4 B3 FT for 10 or 11 correct points FT their table or B2 FT for 8 or 9 correct points FT their table or B1 FT for 6 or 7 correct points FT their table 7(c) 65 to 67 1 FT intersection of their graph with y = – 2 7(d) M2 y = 2 −x oe ruled M1 for [ y =] 2 −x oe soi 40 40 or for 3 cos 2x = 2 −x oe soi 40 32 to 36 B1

This question in 0580/43 Oct/Nov 2023

10 The table shows some values for y = 2x 3 + 6x 2 - 2.5 . x -3 -2.5 -2 -1.5 -1 -0.5 0 0.5 1 y 3.75 5.5 4.25 1.5 -2.5 -0.75 (a) Complete the table. [3] (b) On the grid, draw the graph of y = 2x 3 + 6x 2 - 2.5 for - 3 G x G 1 . y 6 5 4 3 2 1 – 3 – 2 – 1 0 1 x – 1 – 2 – 3 – 4 [4] (c) By drawing a suitable line on the graph, solve the equation 2x 3 + 6x 2 = 4.5 . x = … or x = … or x = … [3] (d) The equation 2x 3 + 6x 2 - 2.5 = k has exactly two solutions. Write down the two possible values of k. k = … or k = … [2]

12 marks

Mark scheme: 10(a) −2.5 −1.25 5.5 3 B1 for each 10(b) Correct graph 4 B3FT for 8 or 9 correct points or B2FT for 6 or 7 correct points or B1FT for 4 or 5 correct points 10(c) y = 2 drawn M1 −2.75 to –2.65 A2 A1 for 1 solution –1.1 to −1.05 0.75 to 0.85 10(d) –2.5 5.5 2 B1 for each

This question in 0580/42 Feb/March 2024

(iii) The equation f ( x) = k has exactly two solutions. Write down the value of k. k = … [1] (iv) tangent asymptote root perpendicular Choose the correct word from the box to complete the statement. The line x = 0 is the … to the graph of y = f ( x) . [1] (b) (i) On the grid, draw the graph of y = x - 2 for values of x from -3 to 3. [2] (ii) Find x when f ( x) = x - 2 . x = … [1] 2 c (c) f ( x) = x - , x ! 0 x Use the graph to show that c = 2 . [2] (d) The equation f ( x) = x - 2 can be written as x 3 + px 2 + qx = 2 . Find the value of p and the value of q. p = … q = … [2]

13 marks

Mark scheme: 5(a)(i) 3 cao 1 5(a)(ii) –2, –0.45 to –0.4, 2.40 to 2.45 3 B1 each 5(a)(iii) 3 cao 1 5(a)(iv) Asymptote 1 5(b)(i) Correct ruled line 2 B1 for ruled line through (0, –2) but not y = –2 or for ruled line with gradient 1 5(b)(ii) 1 cao 1 5(c) Substituting values of x and y into M1 2 c y = x  for an exact point on graph x of y = f(x) or substituting their value of x from 5b(ii) 2 c into x  = x – 2 x leading to c = 2 with no errors A1 5(d) [p = ] –1 and [q = ] 2 nfww 2 M1 for x3 – x2 + 2x = 2 seen or B1 for each nfww

This question in 0580/43 May/June 2024

2 The table shows some values for y = x 3 + 4x 2 - 4 . x -4.5 -4 -3 -2 -1 0 1 1.5 y -14.1 5 4 -4 1 8.4 (a) Complete the table. [2] (b) On the grid, draw the graph of y = x 3 + 4x 2 - 4 for - 4.5 G x G 1. 5 . y 10 5 x –5 – 4 –3 –2 –1 0 1 2 –5 –10 –15 [4] (c) (i) Draw the tangent to the graph at the point (1, 1). [1] (ii) Use your tangent to estimate the gradient of the curve at the point (1, 1). … [2] (d) By drawing a suitable straight line on the grid, solve the equation x 3 + 4x 2 - x - 6 = 0 . x = … or x = … or x = … [4]

13 marks

Mark scheme: 2(a) –4, –1 2 B1 for each correct value 2(b) Correct graph 4 B3FT for 7 or 8 correct points or B2FT for 5 or 6 correct points or B1FT for 3 or 4 correct points 2(c)(i) Ruled tangent at x = 1 1 2(c)(ii) 6 to 14 nfww 2 dep on correct tangent or a close attempt at the tangent at x = 1 M1 for rise/run for their tangent, or close attempt at tangent at any point. Must see correct or implied calculation from a drawn tangent. 2(d) y = x + 2 ruled M2 M1 for [y =] x + 2 soi or y = x + k ruled or y = kx + 2 ruled, but not y = 2 x = –3.95 to –3.75 A2 A1 for any two values x = –1.4 to –1.25 x = 1.1 to 1.25 If A0, SC1 for three correct values

This question in 0580/42 Oct/Nov 2024

11 The table shows some values for y = x 3 - 2x + 3 . Where appropriate, values of y are given correct to 2 decimal places. x -2 -1.5 -1 -0.5 0 0.5 1 1.5 2 y -1 4 3.88 3 2.13 2 7 (a) Complete the table. [2] (b) Draw the graph of y = x 3 -2x + 3 for - 2 G x G 2 . y 7 6 5 4 3 2 1 x 0 – 2 – 1 1 2 – 1 [4] (c) By drawing a suitable straight line on the grid, solve the equation x 3 - 2.5 x + 1 = 0 . x = … or x = … or x = … [4]

10 marks

Mark scheme: 11(a) 2.63 and 3.38 2 B1 for each 11(b) Correct graph 4 B3FT for 8 correct plots or B2FT for 6 correct plots or B1FT for 4 correct plots 11(c) y = 2 + 0.5x ruled M2 M1 for [y =] 2 + 0.5x oe soi e.g. x3 – 2x + 3 = 2 + 0.5x or y = k + 0.5x ruled or y = kx + 2 ruled, but not y = 2 –1.85 to –1.7 , 0.4 to 0.5, A2 A1 for two correct values dep on M2 1.25 to 1.4 If M0 or M1 scored, SC1 for three correct values

This question in 0580/42 Oct/Nov 2025