Graphs of functions — Paper 2 · IGCSE Mathematics 0580

14 y 5 4 3 2 1 _3 _2 _1 0 1 2 3 x _1 _2 _3 _4 _5 (a) Write down the coordinates of the points where the gradient of the curve is zero. Answer(a) ( , ) and ( , ) [2] (b) Write down the range of values of x when the gradient of the curve is negative. Answer(b) [1]

3 marks

Mark scheme: 14 (a) (-1, 0) (1, -4) 1, 1 (b) -1 < x < 1 1 Allow in words provided ± 1 clearly excluded

This question in 0580/21 Oct/Nov 2004

19 The braking distance, d metres, for Alex’s car travelling at v km/h is given by the formula Examiner's Use 200d = v(v + 40). (a) Calculate the missing values in the table. v 0 20 40 60 80 100 120 (km/h) d 0 16 48 96 (metres) [2] (b) On the grid below, draw the graph of 200d = v(v + 40) for 0 Y v Y 120. d 100 90 80 70 60 Distance 50 (metres) 40 30 20 10 v 0 10 20 30 40 50 60 70 80 90 100 110 120 Speed (km / h) [3] (c) Find the braking distance when the car is travelling at 110 km/h. Answer(c) m [1] (d) Find the speed of the car when the braking distance is 80 m. Answer(d) km/h [1]

7 marks

Mark scheme: 19 (a) 6, 30, 70 2 B1 for 2 correct (b) graph 3 P2 7 plots correct from table P1 5 or 6 plots correct from table C1 smooth curve through the points in the given range within one small square of the plots or the correct position (c) 82.5 or ft ±1 1ft (d) 108 or ft ±1 1ft

This question in 0580/22 May/June 2010

18 For 60 Examiner's Use 50 40 Speed 30 (m / s) 20 10 t 0 5 10 15 20 25 30 35 Time (seconds) The graph shows the speed of a sports car after t seconds. It starts from rest and accelerates to its maximum speed in 12 seconds. (a) (i) Draw a tangent to the graph at t = 7. [1] (ii) Find the acceleration of the car at t = 7. Answer(a)(ii) m/s2 [2] (b) The car travels at its maximum speed for 13 seconds. Find the distance travelled by the car at its maximum speed. Answer(b) m [2]

5 marks

Mark scheme: 18 (a) (i) Tangent 1 Correct tangent drawn (ii) 4.4 to 6 2 dep M1 attempting to find gradient of their tangent (b) 780 2 M1 evidence of finding the area under the graph ONLY from t = 12 to t = 25

This question in 0580/21 Oct/Nov 2010

5 y x 0 NOT TO SCALE The sketch shows the graph of y = axn where a and n are integers. Write down a possible value for a and a possible value for n. Answer a = n = [2]

2 marks

Mark scheme: 5 a any negative integer 2 B1 for one correct n any even (positive) integer

This question in 0580/23 May/June 2011

20 A car passes through a checkpoint at time t = 0 seconds, travelling at 8 m/s. It travels at this speed for 10 seconds. The car then decelerates at a constant rate until it stops when t = 55 seconds. (a) On the grid, draw the speed-time graph. 10 8 6 Speed (m/s) 4 2 t 0 10 20 30 40 50 60 Time (seconds) [2] (b) Calculate the total distance travelled by the car after passing through the checkpoint. Answer(b) … m [3]

5 marks

Mark scheme: 20 (a) B1 line from (0, 8) to (10, 8) 8 B1 line from their (10, 8) to (55, 0) 10 55 (b) 260 3FT M2FT for 8 × 10 + 0.5 × 8 × 45 oe or for a fully correct area calculation for their graph or M1FT for 8 × 10 or 0.5 × 8 × 45 or for one correct area calculation for their graph 7 2 15 25

This question in 0580/22 Oct/Nov 2015

19 The curve y = x3 + 2x2 – 4x is shown on the grid. y 9 8 7 6 5 4 3 2 1 x –3 –2 –1 0 1 2 –1 –2 –3 –4 –5 (a) By drawing a suitable tangent, find an estimate of the gradient of the curve when x = 1. … [3] (b) A point D lies on the curve. The x co-ordinate of D is negative. The gradient of the tangent at D is 0. Write down the co-ordinates of D. ( … , … ) [1]

4 marks

Mark scheme: 19 (a) Correct tangent B1 No daylight between tangent and curve at point of contact. Consider point of contact as midpoint between two vertices of daylight, the midpoint must be between x = 0.8 and x = 1.2 2.1 ⩽ grad ⩽ 3.9 2 dep on B1 rise M1 for also dep on any tangent drawn or run close attempt at tangent at any point Must see correct or implied calculation from a drawn tangent (b) (–2, 8) 1

This question in 0580/23 Oct/Nov 2016

16 4 3.5 3 2.5 2 1.5 P 1 0.5 0 0.5 1 1.5 2 2.5 3 3.5 4 By drawing a suitable tangent, estimate the gradient of the curve at the point P. … [3]

3 marks

Mark scheme: 16 tangent ruled at x = 2 B1 −0.7 to –0.3 B2 dep on B1 or a close attempt at tangent at x = 2 or M1 for rise/run for their tangent at x = 2 must see correct or implied calculation from a drawn tangent

This question in 0580/22 Feb/March 2019

10 Sketch the graph of each function. (a) y = x - 3 y O x [1] 1 (b) y = x y O x [2]

3 marks

Mark scheme: 10(a) Correct sketch 1 Line with positive gradient and negative y intercept 10(b) Correct sketch 2 B1 for only one branch or attempt at correct shape

This question in 0580/22 Feb/March 2020

19 y 1 0 x 90° 180° 270° 360° – 1 (a) On the diagram, sketch the graph of y = cos x for 0° G x G 360° . [2] (b) Solve the equation 4 cosx + 2 = 3 for 0° G x G 360 ° . x = … and x = … [3] Questions 20 and 21 are printed on the next page.

5 marks

Mark scheme: 19(a) Correct sketch 2 Needs all three features for 2 marks: • Correct curve shape • Maximum at (0, 1) and at (360, 1) and minimum at (180, −1) • Passing through (90, 0) and (270, 0) only B1 for two correct features 19(b) 75.5 or 75.52… 3 B2 for one correct and 1 or M1 for cos x = oe 284.4 to 284.5 4 If 0 scored, SC1 for two answers with a sum of 360

This question in 0580/22 Feb/March 2020

21 (a) Differentiate 6 + 4x - x 2 . … [2] (b) Find the coordinates of the turning point of the graph of y = 6 + 4x - x 2 . ( … , … ) [2]

4 marks

Mark scheme: 21(a) 4 – 2x 2 B1 for 4 or – 2x 21(b) (2, 10) 2 B1 for x-coordinate of 2 or M1 for their 4 – 2x = 0

This question in 0580/22 Oct/Nov 2020

16 y 4 3 2 1 -1 0 1 2 3 4 x -1 The region R satisfies these three inequalities. y 2 1 y 1 2x + 2 x + y G 3 By drawing three suitable lines, and shading unwanted regions, find and label the region R. [5]

5 marks

Mark scheme: 16 3 correct ruled lines 5 B1 for each line and R clearly indicated y = 1 dashed y = 2 x + 2 dashed x + y = 3 solid B2 for correct region R or B1 for region satisfying 2 inequalities 0 1 1 2 0 1 0 or SC1 for shading of the wanted region only

This question in 0580/22 Feb/March 2021

21 On the axes, sketch the graph of each of these functions. 1 (a) y = x y x O [2] (b) y = 4x y x O [2]

4 marks

Mark scheme: 21(a) Correct sketch 2 B1 for one correct branch or attempt at correct shape 21(b) Correct sketch 2 B1 for correct shape but crossing x-axis or correct shape but just in one quadrant

This question in 0580/22 Feb/March 2021

24 A curve has equation y = x 3 - 2x 2 + 5 . Find the coordinates of its two stationary points. ( … , … ) and ( … , … ) [5]

5 marks

Mark scheme: 24 (0, 5) 5 2 B2 for 3x − 4 x or B1 for 3x 2 or − 4x  4 103   ,  oe dy  3 27  M1 for their derivative = 0 oe or = 0 dx 4 B1 for [x =] 0 and 3 or for 1 correct coordinate pair

This question in 0580/22 Feb/March 2021

19 (a) Sketch the graph of y = tan x for 0 ° G x G 360° . y 0 90° 180° 270° 360° x [2] (b) Solve the equation 5 tanx = 1 for 0° G x G 360 ° . x = … or x = … [2]

4 marks

Mark scheme: 19(a) Correct sketch 2 1 for one correct branch or correct sketch but with branches joined 19(b) 11.3 or 11.30 to 11.31 2 B1 for each and If 0 scored SC1 for two answers with a difference of 180° 191.3 or 191.30 to 191.31

This question in 0580/21 May/June 2021

24 On the axes, sketch the graph of each of these functions. 2 (a) y = x y O x [2] (b) y = 2 -x y O x [2] Questions 25 and 26 are printed on the next page.

4 marks

Mark scheme: 24(a) Correct sketch 2 B1 for one correct branch or attempt at 4444 correct shape 2222 -10-10-10-10 -5-5-5-5 0000 0000 5555 10101010 -2-2-2-2 -4-4-4-4 24(b) Correct sketch 2 B1 for correct shape but crossing x-axis or for correct shape but just drawn in one quadrant

This question in 0580/23 May/June 2021

25 Find the x-coordinates of the points on the graph of y = x 5 - 5x 4 where the gradient is 0. … [4]

4 marks

Mark scheme: 25 0 and 4 final answer 4 B3 for 5x3(x – 4) or better or B2 for 5 x 4 − 20 x 3 or B1 for 5x 4 or − 20 x 3

This question in 0580/23 May/June 2021

17 (a) Sketch the graph of y = sin x for 0° G x G 360° . y 1 O x 360 – 1 [2] (b) Solve the equation 3 sinx + 1 = 0 for 0° G x G 360° . x = … or x = … [3]

5 marks

Mark scheme: 17(a) 2 Correct sketch to go through B1 for correct sine curve shape through the origin (0, 0), (180, 0) and (360, 0) 17(b) 199.5 or 199.47… 3 B2 for one correct and 340.5 or 340.52 to 340.53… 1 or M1 for sin x =  oe 3 If 0 scored SC1 for two reflex angles with sum of 540 or two non-reflex angles with sum of 180

This question in 0580/22 May/June 2022

21 The graph of a cubic function has two turning points. When x 1 0 and when x 2 4 the gradient of the graph is positive. When 0 1 x 1 4 the gradient of the graph is negative. The graph passes through the origin. Sketch the graph. y O x [2]

2 marks

Mark scheme: 21 Correct sketch with maximum at origin 2 B1 for any cubic with exactly 2 distinct and minimum in fourth quadrant turning points 2222 0000 0000 2222 4444 6666 -1-1-1-1 -2-2-2-2 -3-3-3-3 -4-4-4-4 -5-5-5-5

This question in 0580/21 Oct/Nov 2022

22 y 1 0 x 360° – 1 (a) On the diagram, sketch the graph of y = cos x for 0° G x G 360° . [2] 1 (b) Solve the equation cosx =- for 0° G x G 360° . 2 x = … or x = … [2]

4 marks

Mark scheme: 22(a) Correct sketch 2 To go through (0, 1) and close to (360, 1) and reasonably close to (180, –1) 1111 B1 for correct cosine curve shape through 0.50.50.50.5 (0, 1) 0000 0000 50505050 100100100100 150150150150 200200200200 250250250250 300300300300 350350350350 -0.5-0.5-0.5-0.5 -1-1-1-1 Correct sketch to go through (0, 1), (360, 1) and (180, –1) 22(b) 120, 240 2 B1 for each or for two values with sum of 360

This question in 0580/21 Oct/Nov 2022

20 (a) y 1 0 x 360° – 1 Sketch the graph of y = sin x for 0° G x G 360° . [2] 13 (b) Solve 3 - 2 sinx = for 0° G x G 360° . 4 x = … or x = … [3]

5 marks

Mark scheme: 20(a) 2 B1 for correct sine curve shape through the origin Correct sketch to go through (0, 0), (180, 0) and (360, 0) 20(b) 187.2 3 B2 for one correct value, if more than two and 352.8 answers given award B2 if any of the correct answers found and may be in the working 1 or M1 for sin x = − oe soi 8 If 0 scored, SC1 for two reflex angles with a sum of 540 or two non-reflex angles with a sum of 180

This question in 0580/23 Oct/Nov 2022

19 (a) On the diagram, sketch the graph of y = cos x for 0° G x G 360° . y 1 0 180° 360° x – 1 [2] (b) Solve the equation 5 cosx + 3 = 0 for 0° G x G 360° . x = … or x = … [3]

5 marks

Mark scheme: 19(a) correct sketch 2 B1 for correct cosine curve shape through (0, 1) Correct sketch to go through (0, 1), (360, 1) and (180, –1) 19(b) 126.9 or 126.86 to 126.87 3 B2 for 1 correct angle 233.1 or 233.13 to 233.14 3 or M1 for cos x =  oe 5 If M1 or 0 scored SC1 for two angles with a sum of 360

This question in 0580/21 May/June 2023

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

7 marks

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

This question in 0580/21 May/June 2023

15 y 8 7 6 5 4 3 2 1 x 0 - 5 - 4 - 3 - 2 - 1 1 2 3 4 5 6 7 - 1 - 2 - 3 By shading the unwanted regions of the grid, draw and label the region R which satisfies these inequalities. y 2 1 x G 2 y H x + 2 [5]

5 marks

Mark scheme: 15 5 B1 for y = 1 dashed line B1 for x = 2 solid line B1 B1 for y = x + 2 solid line R B2 for region identified satisfying all 3 inequalities B1 or B1 for region identified satisfying only 2 of these inequalities with y = 1, x = 2 and y = x + B1 2 all drawn

This question in 0580/21 Oct/Nov 2023

19 (a) y 1 x 0 360° – 1 Sketch the graph of y = cos x for 0° G x G 360 ° . [2] (b) When cosx = 0 .21 , find the reflex angle x. … [2]

4 marks

Mark scheme: 19(a) 2 B1 for correct cosine curve shape through (0,1) Correct sketch to go through (0, 1), close to (360, 1) and reasonably close to (180, –1) 19(b) 282.1 or 282.12… 2 B1 implied by 77.9 or 77.87 to 77.88 or 282.13 or M1 for 360 – their acute angle

This question in 0580/21 Oct/Nov 2023

23 (a) On the axes, sketch the graph of y = cos x , for 0° G x G 360° . y 1 0 x 180° 360° – 1 [2] (b) Solve the equation cosx = 0 .294 for 0° G x G 360° . x = … or x = … [2]

4 marks

Mark scheme: 23(a) 2 M1 for correct cosine curve shape through (0, 1) Correct sketch to go through (0, 1), close to (360, 1) and reasonably close to (180, –1) 23(b) 72.9 and 287.1 2 B1 for one correct If 0 scored, SC1 for two angles with a sum of 360

This question in 0580/22 Feb/March 2024

18 y 18 16 14 12 10 8 6 4 2 0 x 0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 The graph of y = f ( x) is drawn on the grid. (a) Draw the tangent to the graph at the point x = 3 . [1] (b) Use your tangent to find an estimate for the gradient of the curve at the point x = 3 . … [2]

3 marks

Mark scheme: 18(a) tangent ruled at x = 3 1 18(b) 4.8 to 5.8 2 dep on a close attempt at a tangent rise M1 for also dep on close attempt at tangent run

This question in 0580/22 May/June 2024

22 (a) For each sketch, put a ring around the correct type of function shown. (i) y x O linear cubic quadratic reciprocal exponential [1] (ii) y O x linear cubic quadratic reciprocal exponential [1] (b) (i) On the grid, sketch the curve y = sin x for 0° G x G 360 ° . y 1 0 x 180° 360° – 1 [2] (ii) Solve the equation sinx + 0.4 = 0 for 0° G x G 360° . x = … or x = … [3]

7 marks

Mark scheme: 22(a)(i) cubic 1 22(a)(ii) reciprocal 1 22(b)(i) correct sine curve sketch through 2 (0, 0), (180, 0) and (360, 0) M1 for correct sine curve shape through the origin 22(b)(ii) 203.6 and 336.4 3 B2 for one correct or M1 for sin x = −0.4 oe If 0 or M1 scored, SC1 for two reflex angles with a sum of 540 or two non-reflex angles with a sum of 180

This question in 0580/22 May/June 2024

22 A curve has equation y = x 3 + x 2 - x . 1 5 The curve has a stationary point at e ,3 - 27 o. (a) Find the coordinates of the other stationary point. ( … , … ) [5] (b) By sketching the graph of y = x 3 + x 2 - x , determine whether the stationary point 1 5 e ,3 - 27 o is a maximum or a minimum. y x O 1 5 e ,3 - 27 o is a … [2] (c) The equation x 3 + x 2 - x = k has fewer than 3 solutions. Find the range of possible values for k. … [2] Question 23 is printed on the next page.

9 marks

Mark scheme: 22(a) (–1, 1) nfww 5 B4 for x = – 1 nfww or answer (–1, k) nfww OR B2 for 3x2 + 2x – 1 or B1 for two terms correct dy M1 for setting their = 0 or dx dy stating = 0 dx M1 for correct method to solve their 3-term quadratic e.g. (3x – 1)(x + 1) 22(b) Correct sketch of positive cubic 2 B1 for correct shape of positive with minimum in correct quadrant cubic and minimum 22(c) 2 B1 strict FT for each If their y coordinate Strict FT: from (a) is: or SC1FT for non-inclusive versions of both correct strict FT 5 k their yin ( a ) − inequalities 27 5 k − 27 5 k their yin ( a ) − 27 5 k − 27

This question in 0580/22 Feb/March 2025

11 The diagram shows the graph of y = f ( x) and the point P ( - 2 , 11) . y 15 P 10 5 – 2 – 1 0 1 2 3 x – 5 – 10 The tangent from P touches the graph of y = f ( x) at the point (a, b). The values of a and b are integers. (a) By drawing this tangent, find the value of a and the value of b. a = … , b = … [2] (b) Find the equation of the tangent. Give your answer in the form y = mx + c . y = … [3]

5 marks

Mark scheme: 11(a) For correct ruled tangent and 2 B1 for correct ruled tangent or both values [a =] 1, [b =] 2 correct without a correct tangent 11(b) [y =] 5 – 3x 3 3FT their (a) provided m < 0, c ≠ 0 B1 for (their –3)x + c rise or M1 for correct for their line run B1 for mx + c where c is the correct intercept for their graph, m ≠ 0

This question in 0580/21 May/June 2025

22 A curve has equation y = x n + qx 2 + 9x . dy 2 = 3 x - 12 x + 9 dx (a) Find the value of n, and the value of q. n = … q = … [2] (b) Work out the coordinates of the turning points of the curve. ( … , … ) and ( … , … ) [4]

6 marks

Mark scheme: 22(a) [n =] 3, [q =] – 6 2 B1 for each correct value 22(b) (1, 4) and (3, 0) 4 B3 for (1, 4) or (3, 0) or for two correct values of x or M2 for [3](x – 1)(x – 3) [ = 0] oe −−( 12 )  ( −12 ) 2 −4 3 9 or x = oe 2  3 d y or M1 for writing = 0 d x or for 3 x 2 − 12 x + 9 = 0

This question in 0580/21 May/June 2025

15 y 5 4 3 2 1 – 8 – 7 – 6 – 5 – 4 – 3 – 2 – 1 0 1 2 3 4 5 6 7 8 x – 1 – 2 – 3 – 4 – 5 2 The diagram shows the graph of y = - 1. x (a) Write down the coordinates of the point where the graph crosses the x-axis. ( … , … ) [1] (b) Write down the equation of each asymptote. … … [2] 2 (c) By drawing a suitable straight line on the grid, solve - x - 1 = 0 . x x = … or x = … [3]

6 marks

Mark scheme: 15(a) (2, 0) 1 15(b) x = 0, y = –1 2 B1 for each 15(c) y = x ruled B1 x = –2 and x = 1 B2 B1 for one correct or for two correct answers FT from their line

This question in 0580/22 May/June 2025

21 y Q NOT TO A O B x SCALE P The diagram shows the graph of y = 3 x - x 3 . The graph crosses the x-axis at A, at O and at B. The turning points of the graph are at P and at Q. (a) Find the x-coordinate of A and the x-coordinate of B. Give your answers as exact values. x-coordinate of A … x-coordinate of B … [3] (b) (i) Differentiate 3x - x 3 . … [2] (ii) Find the coordinates of P and Q. P ( … , … ) Q ( … , … ) [4]

9 marks

Mark scheme: 21(a) [A =] − 3 oe 3 B2 for – 3 oe or 3 oe [B =] 3 oe or M1 for x 3 − x 2 = 0  or better ( ) 0  0 2 −−4 1 3 or oe 2 −1 21(b)(i) 3 – 3x2 final answer 2 B1 for 3 or – 3x2 correct in an expression or for correct answer spoilt 21(b)(ii) [P =] (–1, –2) 4 B3 for (–1, –2) or (1, 2) and or for two correct values of x [Q =] (1, 2) or M2 for x2 = 1 or for [3](1 – x)(1 + x) [= 0] oe factorised 0  0 2 −−4 3 3 or oe 2 − 3 OR M1 for [3](1 – x2) [= 0] or their (b)(i) = 0 d y or for stating = 0 d x M1 for correct method to solve their quadratic

This question in 0580/22 May/June 2025

23 Find the coordinates of the turning point on the graph of y = 7 - 2x - x 2 . ( … , … ) [4]

4 marks

Mark scheme: 23 (–1, 8) 4 B3 for x = –1 OR M1 for –2 – 2x M1 for their derivative = 0 OR −−( 2 ) M2 for [x =] or better 2 ( −1) −b or B1 for 2 a OR M2 for – ((x + 1)2 – 8) [= 0] oe or M1 for [±](x + 1)2 + k

This question in 0580/21 Oct/Nov 2025

21 (a) y 1 O 180° 360° x – 1 On the diagram, sketch the graph of y = cos x for 0° G x G 360° . [2] (b) Solve the equation 2 cosx + 3 = 0 for 0° G x G 360° . x = … or x = … [3]

5 marks

Mark scheme: 21(a) Correct sketch to go through (0, 1), 2 M1 for correct cos curve shape through (180, –1) and (360, 1) (0, 1) or for almost correct sketch within 1 tolerance but with an omission at either end 0 or for almost correct sketch within 360 tolerance but with incorrect curvature in one place only –1 21(b) 150 and 210 3 B2 for one correct answer 3 or M1 for cos x = − or better 2 or B1 for 30 or –30 If M1, B1 or 0 scored, award SC1 for two answers in range with a sum of 360

This question in 0580/22 Oct/Nov 2025

22 A graph with equation y = x 2 + bx + c has a minimum point at (-5, 12). Find the value of b and the value of c. b = … c = … [3]

3 marks

Mark scheme: 22 [b =] 10 3 B2 for b = 10 or c = 37 [c =] 37 OR M1 for correct method for b e.g. • (x + 5)2 + 12 • 2x + b = 0 oe −b • = −5 2 1 M1 for correct method for c e.g. • x2 + 5x + 5x + 25 [+ 12] oe • 12 = (–5)2 + their b × –5 + c oe ( their b ) 2 • c − = 12 oe 4 1

This question in 0580/22 Oct/Nov 2025

14 The minimum point on a quadratic curve is (–3, –5). (a) Find the equation of the line of symmetry of the curve. … [1] (b) Write the equation of the curve in the form y = ( x + a) 2 + b . y = … [1]

2 marks

Mark scheme: 14(a) x = –3 1 14(b) [y =] (x + 3)2 – 5 1

This question in 0580/23 Oct/Nov 2025

17 (a) Sketch the graph of y = cos x for 0 ° G x G 360 ° . y 1 0 x 180° 360° –1 [2] 1 (b) cos x° = and x is a reflex angle. 2 Find the value of x. x = … [2]

4 marks

Mark scheme: 17(a) Correct sketch to go through (0, 1), 2 M1 for correct cos curve shape through (180, –1) and (360, 1) (0, 1) or for almost correct sketch within 1 tolerance but with an omission at either end 0 or for almost correct sketch within 360 tolerance but with incorrect curvature in one place only –1 17(b) 315 2 B1 for 45

This question in 0580/23 Oct/Nov 2025