Differentiation — 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

(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

11 (a) OA = AB = AC = 3 - 7 6 Find (i) OB , OB = … [3] (ii) BC. BC = [2] f p (b) S R NOT TO SCALE b X P Q a PQRS is a parallelogram with diagonals PR and SQ intersecting at X. PQ = a and PS = b . Find QX in terms of a and b. Give your answer in its simplest form. QX = … [2] 2 5 (c) M = c 1 8 m Calculate (i) M2, M2 = [2] f p (ii) M -1 . M -1 = [2] f p

11 marks

Mark scheme: 11(a)(i) 12.6 or 12.64 to 12.65 3 2 2 M2 for 12 + ( − 4 ) OR  12  B1 for    − 4  M1 for (their12)2 + ( their − 4 ) 2 11(a)(ii)  −11  2  −11   k    B1 for   or   or for  13   k   13  JJJG  − 8  [ BA = ]    7  11(b) 1 2 M1 for correct route or correct (b − a) oe unsimplified answer 2 JJJG or B1 for QS = b – a oe 11(c)(i)  9 50  2 B1 for 2 correct elements    10 69  11(c)(ii) 1  8 − 5  2  8 −5  1  a b    oe isw B1 for k   or   11  − 1 2   − 1 2  11  c d  or det = 11 soi

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

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

(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

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

11 A curve has equation y = x 3 - 3x + 4 . (a) Work out the coordinates of the two stationary points. ( … , … ) ( … , … ) [5] (b) Determine whether each stationary point is a maximum or a minimum. Give reasons for your answers. [3]

8 marks

Mark scheme: 11(a) (1, 2) 5 B2 for [derivative oe = ] 3 x 2 − 3 ( −,1 6) 2 or B1 for [derivative oe = ] 3x or f( x ) − 3 M1 for their derivative = 0 d y or recognition of = 0 oe dx B1 for [x =] −,1 1 or for one coordinate pair 11(b) (1, 2) minimum with reason 3 Reasons could be e.g. a reasonable sketch ( −,1 6) maximum with reason correct use of 2nd derivative = 6x = 6 , 6 > 0, so (1, 2) minimum oe 2nd derivative = 6x = –6 , –6 < 0 so (–1, 6) maximum oe, or finds gradient on each side of both correct stationary points with correct conclusion B2 for 1 correct with reason or M1 for showing [2nd derivative =] 6x or gradients for one value on either side of one correct stationary point or for reasonable sketch of cubic

This question in 0580/42 Feb/March 2020

10 (a) A rhombus ABCD has a diagonal AC where A is the point (-3, 10) and C is the point (4, -4). (i) Calculate the length AC. … [3] (ii) Show that the equation of the line AC is y =- 2x + 4 . [2] (iii) Find the equation of the line BD. … [4] (b) A curve has the equation y = x 3 + 8x 2 + 5x . (i) Work out the coordinates of the two turning points. ( … , … ) and ( … , … ) [6] (ii) Determine whether each of the turning points is a maximum or a minimum. Give reasons for your answers. [3]

18 marks

Mark scheme: 10(a)(i) 15.7 or 15.65... 3 2 2 M2 for ( 4 – 10) + (4 – –3) oe or M1 for (–4 –10)2 + (4 – – 3)2 oe 10(a)(ii) –10 – 4 M1 [= –2] oe 4 – –3 10 = –2(–3) + c A1 Or –4 = –2(4) + c and correct completion to y = –2x + 4 10(a)(iii) 1 11 4 M1 for grad = ½ soi y = x + oe M1 for [midpoint =] (½, 3) 2 4 M1 for substitution of (1/2, 3) into their y = mx + c oe 10(b)(i)  1 22  6 B2 for 3x2 + 16x + 5  − , −  oe and (–5, 50) Or B1 for one correct  3 27  M1 for derivative = 0 or their derivative = 0 1 M1 for [x =] – and [ x =] –5 3 22 B1 for – and 50 27 10(b)(ii)  1 22  3 B2 for one correct with reason  − , −  minimum or M1 for correct attempt e.g. 2nd derivatives,  3 27  gradients or sketching (–5, 50) maximum with correct reasons

This question in 0580/41 May/June 2020

10 (a) y = x 4 - 4x 3 (i) Find the value of y when x =- 1. y = … [2] (ii) Find the two stationary points on the graph of y = x 4 - 4x 3 . ( … , … ) ( … , … ) [6] (b) y = x p + 2x q d y 10 4 d y = 11x + 10x , where is the derived function. d x d x Find the value of p and the value of q. p = … q = … [2]

10 marks

Mark scheme: 10(a)(i) 5 2 M1 for (–1)4 – 4(–1)3 10(a)(ii) (0, 0) and (3, –27) 6 B2 for 4x3 – 12x2 [ = 0] or B1 for 4x3 or 12x2 AND M1 for derivative = 0 or their derivative = 0 M1 for 4x2(x – 3)[= 0] B1 for [x =] 0 and [ x =] 3 or [y =] 0 and [y =] –27 or for one correct coordinate pair 10(b) [p =] 11 2 B1 for each [q =] 5 dy p −1 q −1 or M1 for = px + 2qx dx

This question in 0580/42 May/June 2020

10 (a) The diagrams show the graphs of two functions. Write down each function. (i) f(x) 5 – 5 0 x f(x) = … [2] (ii) f(x) 2 0 x 180° 360° – 2 f(x) = … [2] (b) f(x) 4 3 P 2 1 – 0.5 0 0.5 1 1.5 2 2.5 x – 1 The diagram shows the graph of another function. By drawing a suitable tangent, find an estimate for the gradient of the function at the point P. … [3]

7 marks

Mark scheme: 10(a)(i) x + 5 2 B1 for linear equation with positive gradient or intercept 5 10(a)(ii) 2 sin x oe 2 B1 for recognition of sin or cos(x – 90) 10(b) tangent ruled at P B1 1.3 to 1.4 B2 dep on tangent drawn M1 for rise/run

This question in 0580/43 May/June 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

12 (a) Find the gradient of the curve y = 2x 3 - 7x + 4 when x =- 2 . … [3] (b) A is the point (7, 2) and B is the point (−5, 8). (i) Calculate the length of AB. … [3] (ii) Find the equation of the line that is perpendicular to AB and that passes through the point (−1, 3). Give your answer in the form y = mx + c . y = … [4] (iii) AB is one side of the parallelogram ABCD and - a • BC = where a 2 0 and b 2 0 e- bo • the gradient of BC is 1 • BC = 8 . Find the coordinates of D. ( … , … ) [4]

14 marks

Mark scheme: 12(a) 17 3 M2 for 3 × 2 x 2 − 7 or better isw or M1 for 3 × 2 x 2 oe or kx2 – 7 seen 12(b)(i) 13.4 or 13.41 to 13.42 3 2 2 M2 for ( −−5 7 ) + ( 8 − 2 ) oe 2 2 or M1 for ( −−5 7 ) + ( 8 − 2 ) oe 12(b)(ii) [ y = ] 2 x + 5 final answer 4 8 − 2 M1 for [gradient of AB =] oe −−5 7 1 M1dep for gradient p = −÷1 their − oe 2 M1dep on previous M1 for substituting (−1, 3) into y = their px + c oe where their p ≠ 0 12(b)(iii) ( 5, 0) 4   − 2   2 B3 for AD =   or DA =  − 2  2   12  or coordinates of C (−7, 6) and  CD =      oe  − 6  seen or B2 for a = b = 2 soi or coordinates of C (−7, 6) 2 8 oe or M1 for a = b oe soi or for a 2 + b 2 = ( ) a or cos 45 = oe 8   −12    12  =  DC =  or for      or  CD    seen  6   − 6  y − 8 y − 2 or = 1 oe or = 1 x −−5 x − 7

This question in 0580/42 Feb/March 2021

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

9 (a) (i) The equation y = x 3 - 4x 2 + 4x can be written as y = x ( x - a) 2 . Find the value of a. a = … [2] (ii) On the axes, sketch the graph of y = x 3 - 4x 2 + 4x , indicating the values where the graph meets the axes. y O x [4] (b) Find the equation of the tangent to the graph of y = x 3 - 4x 2 + 4x at x = 4. Give your answer in the form y = mx + c . y = … [7] Question 10 is printed on the next page.

13 marks

Mark scheme: 9(a)(i) 2 2 M1 for x(x2 – 4x + 4) or x (x – 2)2 or (x2 – 2x) (x – 2) or x3 – 2ax2 + a2x 9(a)(ii) Correct sketch with curve passing through 4 B1 for any positive cubic O and touching (2, 0) B1 for sketch through or touching O B1 for sketch with min or max touching x-axis once only but not at (0, 0) B1FT their (a)(i) for sketch with min or max touching x-axis at (their 2, 0) and their 2 is labelled or clearly indicated 9(b) y = 20x – 64 final answer nfww 7 B6 for equivalent correct equation OR B2 for 3x2 – 8x + 4 isw or B1 for 3x2 or –8x seen M2dep for [grad =] 20 soi nfww or M1dep for substituting 4 into their derivative isw B1 for (4, 16) soi M1dep for 16 = their 20 × 4 + c oe

This question in 0580/42 May/June 2021

10 f ( )x = 3 x - 2 g ( )x = 5 x - 7 h ( )x = x 2 + x j ( )x = 3 x (a) Find (i) f(2), … [1] (ii) g(2), … [1] (iii) gf(2). … [1] (b) Find f -1 ( )x . f -1 ( )x = … [2] (c) Find hf(x), giving your answer in the form ax 2 + bx + c . … [3] (d) Find the derivative of h(x). … [1] (e) (i) Find x when j -1 ( )x = 4 . x = … [1] (ii) Simplify j -1 j ( )x . … [1]

11 marks

Mark scheme: 10(a)(i) 4 1 10(a)(ii) 3 1 10(a)(iii) 13 1 FT 5 × their (a)(i) – 7 10(b) x + 2 2 y 2 final answer M1 for y + 2 = 3x or for = x − 3 3 3 or for x = 3y – 2 10(c) 9x2 – 9x + 2 final answer 3 2 M1 for ( 3 x − 2 ) + 3 x − 2 2 2 B1 for ( 3 x − 2 ) = 9 x − 6 x − 6 x + 4 10(d) 2x + 1 1 10(e)(i) 81 1 10(e)(ii) x 1 Not y = x

This question in 0580/43 May/June 2021

10 (a) Find the coordinates of the turning points of the graph of y = x 3 - 12x + 6 . You must show all your working. ( … , … ) and ( … , … ) [5] (b) Determine whether each turning point is a maximum or a minimum. Show how you decide. [3]

8 marks

Mark scheme: 10(a) (2, –10) and (–2, 22) 5 B2 for 3x2 – 12 isw or B1 for 3x2 + k or px2 – 12 (p ≠ 0) or for 3x2 – 12 + 6 isw M1 for setting their derivative = 0 dy or = 0 dx B1 for x = ± 2 or for one correct coordinate pair 10(b) (2, –10) minimum with correct reason 3 B2 for 1 correct with correct reasoning or sketch or B2FT for correct evaluation with correct 2nd derivative for both of their different x (–2, 22) maximum with correct reason values or sketch or M1 for showing [2nd derivative =] 6x or gradients for one value on either side of one correct stationary point or for reasonable sketch of cubic

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

12 (a) Solve the equation tan x = 11.43 for 0° G x G 360 ° . x = … or x = … [2] (b) Sketch the curve y = x 3 - 4x . y x O [3] (c) A curve has equation y = x 3 + ax + b . The stationary points of the curve have coordinates (2, k) and (-2, 10 - k). Work out the value of a, the value of b and the value of k. a = … , b = … , k = … [6]

11 marks

Mark scheme: 12(a) 85[.0], 265[.0] and no others 2 B1 for each If 0 scored SC1 for two values in the range with a difference of 180 but not multiples of 90 12(b) correct shape and passes through 3 B1 for any positive cubic shape origin B1 for sketch with one max and one min and with 3 roots including zero If 0 scored, SC1 for x(x + 2)(x – 2) soi 12(c) a = –12 6 B5 for 2 correct b = 5 OR k = –11 B2 for 3x2 + a or B1 for 3x2 isw d y M1dep on at least B1 for their = 0 d x M1dep on at least B1M1 for x = 2 or x = – 2 d y substituted in their = 0 equation d x M1 for k = 23 + 2 × their a + b and 3 10 − k = ( −2 ) + ( −2 ) × their a + b

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

12 A curve has equation y = x 3 - kx 2 + 1. When x = 2 , the gradient of the curve is 6. (a) Show that k = 1.5 . [5] (b) Find the coordinates of the two stationary points of y = x 3 - 1.5x 2 + 1. You must show all your working. ( … , … ) and ( … , … ) [4] (c) Sketch the curve y = x 3 - 1.5x 2 + 1. y O x [2]

11 marks

Mark scheme: 12(a) 3 x 2  2 kx M2 M1 for 3x 2 or kx2 dy M1 Dep on at least M1 for derivative their = 6 dx dy M1 Dep on at least M1 for derivative x = 2 substituted in their dx Correct working leading to 1.5 oe A1 A0 if any errors in working leading to 1.5 12(b) ( 0, 1) ( 1, 0.5) 4 B3 for x = 0 and x = 1 or for (1, 0.5) OR dy M1 for their = 0 dx B1 for 3 x 2  3 x oe or better 12(c) correct sketch 2 with max on positive y-axis and min in 1st quadrant B1 for positive cubic or for graph with one max which is on pos y-axis and one min which is in 1st quadrant

This question in 0580/42 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

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

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

9 f ( x) = x 3 - 3x 2 - 4 (a) Find the gradient of the graph of y = f ( x) where x = 1. … [3] (b) Find the coordinates of the turning points of the graph of y = f ( x) . ( … , … ) , ( … , … ) [4] (c) Sketch the graph of y = f ( x) . y O x [2]

9 marks

Mark scheme: 9(a) –3 3 B2 for 3x2 – 6x or B1 for 3x2 – kx or for kx2 – 6x or for 3x2 – 6x + c 9(b) (0, –4) and (2, –8) 4 B3 for x = 0 and 2 or for (2, –8) OR d y M1 for their 3x2 – 6x = 0 or stating = 0 oe d x M1 for correct method to solve their 3x2 – 6x = 0 9(c) Correct sketch10101010 2 Max on negative y-axis and min in correct quadrant and extends into first quadrant 5555 -4-4-4-4 -2-2-2-2 0000 0000 2222 4444 B1 for positive cubic graph and two turning points -5-5-5-5 -10-10-10-10

This question in 0580/42 Feb/March 2023

8 (a) (i) Show that the equation y = ( x - 4)( x + 1)( x - 2) can be written as y = x 3 - 5x 2 + 2x + 8 . [2] (ii) On the diagram, sketch the graph of y = x 3 - 5x 2 + 2x + 8 , indicating the values where the graph crosses the axes. y O x [4] (b) The graph of y = x 3 - 5x 2 + 2x + 8 has two tangents with a gradient of 10. Find the equations of these two tangents. You must show all your working and give your answers in the form y = mx + c . y = … y = … [7]

13 marks

Mark scheme: 8(a)(i) Correct expansion of a pair of brackets M1 accept x2 – 4x + [1]x – 4 x2 – 3x – 4 or x2 – 4x – 2x + 8 or x2 – 6x + 8 or x2 + [1]x – 2x – 2 or x2 – [1]x – 2 x3 – 4x2 + x2 – 4x – 2x2 + 8x – 2x + 8 A1 Accept leading to and stating x3 – 3x2 – 4x – 2x2 + 6x + 8 [y = ] x3 – 5x2 + 2x + 8 or x3 – 6x2 +[1] x2 + 8x – 6x + 8 or x3 –[1] x2 – 2x – 4x2 + 4x + 8 leading to and stating [y = ] x3 – 5x2 + 2x + 8 8(a)(ii) Correct labelled sketch 4 positive cubic Crossing x-axis at –1, 2 and 4 only Crossing y – axis at 8 only B1 for positive cubic B2 for three intercepts only with x -axis labelled at – 1, 2 and 4 or B1 for 1 or 2 correctly labelled x – intercepts B1 for a single intercept on y-axis labelled at 8 but not if line y = 8 8(b) 3x2 – 10x – 8 [= 0] M3 B2 for derivative = 3x2 – 10x + 2 isw OR B1 for derivative with 3x2 or –10x given in expression isw M1dep on B1 for their first derivative = 10 2 B1 x = 4 and x =  3  2 112  B1 (4, 0) and   ,  oe  3 27  [y =] 10x – 40 B2 B1 for each and or for two different equations of the form 292 [y = ] 10x + c (c must be numeric) [y =] 10 x  27 292 or for c = –40 and 27

This question in 0580/42 May/June 2023

12 The equation of a curve is y = x 4 - 8x 2 + 5 . d y 4 2 (a) Find the derivative, of y = x - 8x + 5 . e d x o, … [2] (b) Find the coordinates of the three turning points. You must show all your working. ( … , … ) and ( … , … ) and ( … , … ) [4] (c) Determine which one of these turning points is a maximum. Justify your answer. [2]

8 marks

Mark scheme: 12(a) 4 x 3  16 x cao 2 M1 for 4x 3 +kx or kx 3  16 x or 4 x 3  16 x +k or 4 x 3  16 as final answers 12b dy dy B1 Their = 0 or stating = 0 dx dx Correct method to solve their 4x3 – 16x = 0 M1 e.g. 4x(x2 – 4) or 4x(x – 2)(x + 2) oe [x =] 0, –2, 2 A1 Or B1 for (–2, –11) and (2, –11) (0, 5) (–2, –11) (2, –11) A1 12(c) (0, 5) with correct reasoning 2 M1 for any of  correct use of 2nd derivative 12x2 −16  evaluates correctly both values of y on either side  evaluates correctly the gradient on either side  reasonable correct sketch

This question in 0580/43 May/June 2023

11 (a) Differentiate x 3 - 4 x 2 - 3x . … [2] (b) A curve has equation y = x 3 - 4x 2 - 3x . Work out the coordinates of the two stationary points. Show all your working. ( … , … ) ( … , … ) [5] (c) Determine whether each stationary point is a maximum or a minimum. Show all your working. [3]

10 marks

Mark scheme: 11(a) 3x2 – 8x – 3 2 B1 for two terms correct or correct answer seen 11(b) 3x2 – 8x – 3 = 0 M1 FT their part (a) Correct method to solve their 3- M2 term quadratic (3x + 1)(x – 3) [=0] 2 M1 for (3x + a)(x + b) [=0] −−( 8)  ( −8) − 4(3)( −3) where ab = –3 or 3b + a = –8 2(3) or for ( −8) 2 − 4(3)( −3) p  q or for where p = –(–8) and r = 2(3) seen r or for a correct method for solving a 2-term quadratic (3, –18) B2 B1 for one correct point or for two correct x-  1 14  values, − ,   or M1 for substitution of their x-values into  3 27  y = x 3 − 4 x 2 − 3 x shown 11(c) (3, –18) minimum with reason 3 Reasons could be e.g. 1. A reasonable sketch of a positive cubic  1 14  − ,   maximum with 2. Correct use of 2nd derivative = 6x – 8 = 10,  3 27  10 > 0, so (3, –18) is a minimum oe. reason 2nd derivative = 6x – 8 = –10, –10 < 0  1 14  so  − ,  is a maximum oe.  3 27  3. Evaluates correctly values of y on both sides of both correct stationary points 4. Finds gradient on each side of both correct stationary points. B2 for 1 correct with a reason for that stationary point or for both x-values correct with correct conclusions and reasonable sketch of a positive cubic, or for correct substitution of both of their x-values into their second derivative shown, or substitution shown for one x-value either side of both of their stationary points to find the gradients. Or M1 for showing [2nd derivative =] 6x – 8 or substitution shown for one x-value either side of one of their stationary points to find the gradients. or for reasonable sketch of positive cubic.

This question in 0580/41 Oct/Nov 2023

9 y B NOT TO SCALE A x O The diagram shows a sketch of the graph of y = 4x 3 - x 4 . The graph crosses the x-axis at the origin O and at the point A. The point B is a maximum point. (a) Differentiate 4x 3 - x 4 . … [2] (b) Find the coordinates of B. ( … , … ) [3] (c) Find the gradient of the graph at the point A. … [3]

8 marks

Mark scheme: 9(a) 12 x 2 − 4 x 3 oe final answer 2 B1 for 12x 2 or –4x3 in final answer or for correct answer seen 9(b) (3, 27) 3 B2 for x = 3 OR M1 for their 12 x 2 − 4 x 3 = 0 or better dy or states = 0 dx M1dep for substituting their x into y = 4 x 3 − x 4 shown 9(c) –64 3 M1 for 4 x 3 − x 4 = 0 B1 for x = 4

This question in 0580/42 Oct/Nov 2023

11 y NOT TO SCALE B A O C x The diagram shows a sketch of y = 18 + 5 x - 2x 2 . (a) Find the coordinates of the points A, B and C. A ( … , … ) B ( … , … ) C ( … , … ) [4] (b) Differentiate 18 + 5x - 2x 2 . … [2] (c) Find the coordinates of the point on y = 18 + 5x - 2x 2 where the gradient is 17. ( … , … ) [3]

9 marks

Mark scheme: 11(a) (–2, 0) 4 B1 for B = (0, 18) (0, 18) (4.5, 0) oe B3 for A = (–2, 0) and C = (4.5, 0) oe or B2 for x = –2 and x = 4.5 oe or B1 for (9 – 2x)(2 + x) oe or either A or C correct 11(b) 5 – 4x final answer 2 B1 for one correct term when simplified 11(c) (– 3, –15) 3 B2FT for x = – 3 OR M1 for their (b) = 17 M1 dep for correct substitution of their x into 18 + 5x – 2x2 shown

This question in 0580/43 Oct/Nov 2023

8 (a) On the axes, sketch the graph of y = 4 - 3 x . y O x [2] (b) On the axes, sketch the graph of y =- x2 . y O x [2] (c) (i) Find the coordinates of the turning points of the graph of y = 10 + 9 x 2 - 2x 3 . You must show all your working. ( … , … ) and ( … , … ) [5] (ii) Determine whether each turning point is a maximum or a minimum. Show how you decide. [3]

12 marks

Mark scheme: 8(a) Ruled line with negative gradient and 2 positive y-intercept B1 for ruled line with negative gradient or for ruled line with positive y-intercept or straight line with negative gradient and positive y-intercept 8(b) Negative quadratic, with vertex at origin 2 B1 for negative quadratic in other position or for sketch in 3rd and 4th quadrants only with single maximum at (0, 0) and no other turning point or for positive quadratic, with vertex at origin 8(c)(i) 18x – 6x2 isw B2 B1 for one correct term 18x or –6x2 seen d y M1 Dep on at least B1 earned setting their derivative = 0 or = 0 or their derivative = ±18x ± 6x2 d x (0, 10) and (3, 37) B2 B1 for x = 0 and x = 3 or for (0, 10) or (3, 37) 8(c)(ii) (0, 10) minimum with correct reason 3 Reasons could be e.g. 1 A reasonable sketch of a negative cubic AND 2 Correct use of 2nd derivative = –12(0) + 18 (3, 37) maximum with correct reason = 18, 18 > 0, so (0, 10) is a minimum oe. 2nd derivative = –12(3) + 18 = –18, –18 < 0 so (3, 37) is a maximum oe. 3 Evaluates correctly values of y on both sides of both correct stationary points 4 Finds gradient on each side of both correct stationary points. B2 for 1 correct with correct reason for that stationary point or for both x-values correct and reasonable sketch of a negative cubic, or for correct substitution and evaluation of both of their x-values into their second derivative or substitution and evaluation for one x-value on both sides of both of their stationary points to find the gradients soi or M1 for showing [2nd derivative =] –12x + 18 or correct FT their 2nd derivative or substitution and evaluation shown for one x-value on both sides of one of their stationary points to find the gradients soi or for sketch of any negative cubic. 9(a)(i) 5 3 (12800 − 8000 )  100 M2 for 8000  12 8000  12  r or M1 for [12800 − 8000 =] 100 or 400 seen If 0 scored, SC1 for answer 13.3 or 13.33…

This question in 0580/42 Feb/March 2024

11 (a) The point ( - 1, 6) lies on a curve. dy 3 2 This curve has the derived function =- 4 x - 9 x + 5 . dx Show that ( - 1, 6) is a stationary point of the curve. [2] (b) A different curve has equation y = 2x 3 - 6x + 8 . (i) Calculate the gradient of the tangent to this curve at the point ( - 2, 2) . … [3] (ii) Find the x-coordinates of the stationary points of this curve. x = … and x = … [2]

7 marks

Mark scheme: 11(a) –4 (–1)3 – 9 (–1)2 + 5 or better M1 = 0 [so stationary point] A1 with no errors 11(b)(i) 18 3 B2 for 6 x 2  6 isw OR B1 for 6 x 2  k (any k) isw or px 2  6 isw (p ≠ 0) or 6 x 2  6 + 8 M1dep on B1 for x = −2 substituted into d y their d x 11(b)(ii) 1 and –1 2 M1 for 6 x 2  6 = 0 oe seen d y or for their = 0 if B1 scored in part d x (b)(i)

This question in 0580/42 May/June 2024

10 y = x 7 - 7x 6 (a) Find the derivative of y with respect to x. … [2] (b) Find the equation of the tangent to the graph of y = x 7 - 7x 6 at the point where x =- 1. Give your answer in the form y = mx + c . y = … [4] (c) The graph of y = x 7 - 7x 6 has two turning points. Find the coordinates of these points. You must show all your working. ( … , … ) ( … , … ) [5]

11 marks

Mark scheme: 10(a) 6 5 2 B1 for one correct term 7x6 or 42x5 or for 7 x  42 x final answer 7 x 6  42 x 5 seen and spoiled 10(b) 49x + 41 4 M1 for substituting x = – 1 into [y = ] x7 – 7x6 M1 for x = – 1 substituted in their (a) or the correct derivative to give their m M1 for their –8 = (their m)(–1) + c oe 10(c) (0, 0) 5 B4 for (6, –46 656) (6, –46 656) or B3 for x = 0 and 6 OR d y d y M1 for their = 0 or stating = 0 dx dx and M1 for a correct method to solve their 7x6 – 42x5

This question in 0580/43 May/June 2024

10 A curve has the equation y = x 3 - 9x 2 - 48 x . (a) Differentiate x 3 - 9x 2 - 48x . … [2] (b) Find the coordinates of the turning points of the graph of y = x 3 - 9x 2 - 48 x . You must show all your working. ( … , … ) and ( … , … ) [4] (c) Determine whether each of the turning points is a maximum or a minimum. Give reasons for your answers. [3]

9 marks

Mark scheme: 10(a) 3x2 – 18x – 48 final answer 2 B1 for two correct terms or for correct answer seen then spoiled 10(b) d y M1 their = 0 soi d x [3](x – 8)(x + 2) oe M1 −−( 18)  ( −18) 2 − 4(3)( −48) or 2  3 oe 18  900 oe or 6 3 ± 16 + 32 (–2, 52) B2 B1 for one correct pair of coordinates or for two (8, –448) correct values of x 10(c) (–2, 52) maximum with reason 3 Reasons could be e.g. 1. A reasonable sketch of a positive cubic and 2. Correct evaluation and use of 2nd derivative 6x – 18 = –30, –30 < 0, so (–2, 52) is a ( 8, − 448 ) minimum with reason maximum oe. 6x – 18 = 30, 30 > 0 , so ( 8, − 448 ) is a and no incorrect statement minimum oe. 3. Evaluates correctly values of y on both sides of both correct stationary points 4. Finds gradient on each side of both correct stationary points. Any incorrect statement MAX B2 B2 for 1 correct with correct reason for that stationary point or for both x-values correct and reasonable sketch of a positive cubic or for correct substitution and evaluation of both of their x-values into their second derivative or substitution and evaluation for one x-value on both sides of both of their stationary points to find the gradients soi or M1 for showing [2nd derivative =] 6x – 18 or correct FT their 2nd derivative from part (a) or substitution and evaluation shown for one x- value on both sides of one of their stationary points to find the gradients soi or for sketch of any positive cubic.

This question in 0580/42 Oct/Nov 2024

10 y NOT TO B SCALE A O x The diagram shows a sketch of the graph of y = 3 + 2x - x 2 . A is the point (-1, 0) and B is the point (2, 3). (a) Find the derivative of 3 + 2x - x 2 . … [2] (b) (i) Show that the equation of the tangent at A is y = 4x + 4 . [3] (ii) The line L is perpendicular to the line y = 4x + 4 . The line L passes through the point B. Find the equation of the line L. Give your answer in the form y = mx + c . y = … [3] (c) Find the coordinates of the maximum point on the graph of y = 3 + 2x - x 2 . ( … , … ) [3]

11 marks

Mark scheme: 10(a) 2 – 2x 2 B1 for k – 2x or 2 – kx or 3 + 2 – 2x 10(b)(i) Gradient (m) = correct substitution of –1 M1 into their (a) 2 – 2(–1) 0 = their m × –1 + c M1 Dep on previous M1 or y [– 0] = their m(x – –1) oe c = 4 and leading to y = 4x + 4 A1 10(b)(ii) 1 7 3 1 − x + oe M1 for − 4 2 4 M1 for 3 = their m  2 + c or better or y – 3 = their m(x – 2) or better 10(c) (1, 4) 3 B2 for x = 1 or M1 for their (a) = 0 M1 for substituting their 1 into y = 3 + 2x – x2 OR B2 for x = 1 or M2 for 4 – (x – 1)2 or M1 for (x – 1)2

This question in 0580/43 Oct/Nov 2024

26 y = ax 11 + 3 x b d y 10 c = 44 x + 18 x d x Find the values of a, b and c. a = … b = … c = … [2]

2 marks

Mark scheme: 26 [a =] 4 2 B1 for one correct [b =] 6 [c =] 5

This question in 0580/42 May/June 2025

20 y = x 3 + 3x 2 - 13 x dy (a) Find . dx … [2] (b) Find the gradient of the curve y = x 3 + 3x 2 - 13 x at the point where x = 3 . … [2]

4 marks

Mark scheme: 20(a) 3x2 + 6x – 13 final answer 2 B1 for two terms correct or correct answer seen then spoilt 20(b) 32 2 d y FT their dep on B1 earned in (a) d x M1 for correct substitution of x = 3 into d y their d x

This question in 0580/43 Oct/Nov 2025