Trigonometric functions — Paper 4 · IGCSE Mathematics - International 0607

10 y 3 x 0 90 180 270 360 –3 f x = 2 sin x + cos x for 0° G x G 360 ° ^ h g x = 2 - log x for 0° G x G 360° ^ h (a) On the diagram, sketch the graph of y = f x [3] ^ h. (b) On the same diagram, sketch the graph of y = g x [2] ^ h. (c) Solve the equation. 2 sin x + cos x = 2 - log x … [3]

8 marks

Mark scheme: 10(a) Correct Graph 3 M1 for sine graph with one max and one min A1 for x-intercepts at 150 and 330 (approx.) y f(x)=2sin(x)+cos(x) 3 f(x)=2-log(x) A1 for positive y-intercept x 90 180 270 360 3 10(b) Correct Graph with second 2 M1 for correct shape intersection with other graph (if correct) below x-axis 10(c) 6.18 or 6.175... 3 B1 for each 159 or 158.5 to 158.6 320 or 320.3 to 320.4

This question in 0607/41 May/June 2017

8 y 5 x –400 0 600 f(x) = 3sinx (a) Sketch the graph of y = f(x) for - 400 ° G x G 600 ° . [3] (b) Find the x co-ordinates of the local maximum points of f(x) for - 400° G x G 600° . x = … or x = … or x = … [3] (c) The point (30, 3) is on the graph. The point (a, 3) is also on the graph where 600° 1 a 1 900 ° . Find the two possible values of a. a = … or a = … [2] x (d) g(x) = 3 − 100 Solve the inequality g x 2 f x . ^ h ^ h … [3]

11 marks

Mark scheme: 8(a) 3 WithW correct shape with ttwo max on right of y-axxis anda one on leeft, all abovee x-axis and reasonabler qualityq oro B2 for corrrect shape anand all above x-axis oro B1 for corrrect shape Correct skettch 8(b) –270, 90, 4550 3 B1B for each SC2S for all correctc but wwith y co-ordss oro SC1 for twwo correct wwith y co-ordss 8(c) 750, 870 2 B1B for each 8(d) x < 54.7 1 54.745 to 54.775 164 < x < 2667 2 163.51 to 163.6 , 266.6... B1B for one innequality oro B1 for botth values seeen IfI 0 scored, B1B for straighht line with negativen gradientg crosssing curve thhree times beetween xx = 0 and x == 400. May bbe freehand.

This question in 0607/42 May/June 2017

9 C x cm NOT TO SCALE 60° A B (x + 2) cm In the diagram AC = x cm, AB = (x + 2) cm and angle A = 60°. (a) (i) Find an expression, in terms of x, for the area of triangle ABC. Give your answer in surd form. … cm2 [2] (ii) The area of triangle ABC = 18 3 cm2. Show that x2 + 2x – 72 = 0. [2] (b) (i) Solve the equation x2 + 2x – 72 = 0. x = … or x = … [2] (ii) Find the shortest distance between the line AB and the point C. … cm [2]

8 marks

Mark scheme: 9(a)(i) 1 3 2 1 x + 2 ) × sin60 × x × ( x + 2 ) × oe or better M1 for × x × ( 2 2 2 final answer 9(a)(ii) equating to 18 3 and correct M1 Dependent on correct answer used from (a)(i) or answer to (a)(i) contains sin60 but is elimination of 3 otherwise correct. Completion with at least one step A1 No errors or omissions 9(b)(i) 7.54 or 7.544... , –9.54 or –9.544... 2 B1 for each If 0 scored, M1 for substitution in formula or sketch or (x + 1)2 – 73 or better 9(b)(ii) 6.53 or 6.54 or 6.529 to 6.536... 2 [ ] M1 for sin 60 = oe their 7.54

This question in 0607/42 May/June 2017

6 V NOT TO SCALE 8 cm B C √72 cm P A D √72 cm The diagram shows a pyramid with a square base ABCD of side 72 cm. The diagonals of the base, AC and BD, meet at P. The vertex, V, is vertically above P and VP = 8 cm. (a) Find the volume of the pyramid. Give the units of your answer. … … [3] (b) Find the length AC. AC = … cm [2] (c) Find the length DV. DV = … cm [3] (d) Find angle VDP. Angle VDP = … [2] (e) X is the midpoint of the side CD. (i) Find the length VX. VX = … cm [3] (ii) Find angle VXP. Angle VXP = … [2] (f) The pyramid is cut parallel to ABCD to form a smaller pyramid VEFGH. The volume of VEFGH is 24 cm3. Find the vertical height of this pyramid. … cm [3]

18 marks

Mark scheme: 6(a) 192 2 1 2 M1 for × 72 × 8 oe ( ) 3 cm3 1 6(b) 12 2 M1 for ( 72) 2 + ( 72) 2 oe 6(c) 10 3 2 2 M2 for 8 + ( 0.5 their (b) ) or M1 for [PD oe =] 0.5 × their (b) (d) 53.1 or 53.13 2 8 8 M1 for tan = or sin = 0.5 × their (b) their (c) 0.5 × their (b) or cos = their (c) 6(e) (i) 82 or 9.06 or 9.055... 3 M2 for 8 2 + (0.5 × 72) 2 or (their (c))2 – (0.5 × 72) 2 or M1 for (0.5 × 72) 2 6(e)(ii) 62.1 or 62[.0] or 62.00 to 62.10 2 8 M1 for tan = oe 0.5 × 72 6(f) 4 cao 3 24 their (a) M2 for 3 or 3 their (a) 24 1 soi by 2 or 2 24 their (a) or M1 for or their (a) 24 1 soi by 8 or 8

This question in 0607/43 Oct/Nov 2017

5 y 2 1 x 0 180 360 540 −1 −2 J N 1 f (x) = sin x° g (x ) = log KK OO sin x° L P (a) (i) On the diagram, sketch the graph of y = f (x) for 0 G x G 540 . [2] (ii) Write down the range of f ()x for 0 G x G 540 . … [1] (b) (i) On the same diagram, sketch the graph of y = g (x) for values of x between 0 and 540. [2] (ii) Give a reason why there are no values of g ()x for 180 G x G 360 . … [1] (iii) Write down the co-ordinates of the minimum points on the graph of y = g (x) . ( … , … ) and ( … , … ) [2] (iv) Write down the equations of the four asymptotes to the graph of y = g (x) . … , … , … , … [2] (c) (i) f (k) = g (k) and 0 G k G 90 . Find the value of k. k = … [1] (ii) Solve the inequality f (x) 2 g (x) for values of x between 0 and 540. … [2] (iii) j is an integer. The equation f ()x = j has no solutions. The equation g ()x = j has no solutions. Write down a possible value of j. j = … [1]

14 marks

Mark scheme: 5(a)(i) Correct2222 sketch 2 B1 for sine graph with different amplitude and/or period but must go through (0, 0) 1111 or for correct sine graph but only one cycle 0000 0000 100100100100 200200200200 300300300300 400400400400 500500500500 -1-1-1-1 -2-2-2-2 5(a)(ii) −-1 f ( x ) - 1 1 5(b)(i) Correct2222 sketch 2 i.e. Correct shape with 2 branches above x- axis and gap of at least 120 between the 1111 branches and only slightly crossing either 0000 0000 100100100100 200200200200 300300300300 400400400400 500500500500 axis. -1-1-1-1 B1 for 2 branches above x-axis but gap less -2-2-2-2 than 120 between the branches and only slightly crossing either axis or one branch correct 5(b)(ii) logarithms of negative numbers do 1 not exist oe 5(b)(iii) (90, 0), (450, 0) 2 B1 for each 5(b)(iv) x = 0, x = 180, x = 360, x = 540 2 B1 for 2 or 3 correct 5(c)(i) 23.5 or 23.51 to 23.52 1 5(c)(ii) 23.5 < x < 156.5 2 B1 for each 383.5< x < 516.5 Allow 23.51 to 23.52, 156.48 to 156.49 Allow 383.51 to 383.52, 516.48 to 516.49 5(c)(iii) Any integer less than – 1 1

This question in 0607/41 May/June 2018

11 f ()x = 10 x g ()x = 2x - 1 (a) Find the value of g(3). … [1] (b) Find the range of f (x) for the domain {-1, 0, 1, 2}. { … } [2] (c) Find x when g ()x = 12 . x = … [2] 2 (d) The graph of y = g (x) is translated by the vector onto the graph of h(x). e3o Find h(x). Give your answer in its simplest form. h(x) = … [3] (e) Find f -1 ()x . f -1 ()x = … [2] (f) tan (g (x)) = 1 and 0° G x G 180° . Find the two values of x. x = … or x = … [3]

13 marks

Mark scheme: 11(a) 5 1 11(b) 0.1oe , 1, 10, 100 2 B1 for 3 correct or all correct seen and spoilt. 11(c) 6.5 oe 2 M1 for 2x – 1 = 12 11(d) 2x – 2 or 2(x – 1) 3 B2 for correct unsimplified answer OR M1 for substituting x − 2 for x M1 for adding 3 to a function in x oe OR M1 for y = 2x + c (c ≠ – 1) leading to answer with gradient 2 M1 for substituting coords of valid point into y = 2x + c 11(e) log x 2 M1 for log y = x or x = 10y 11(f) 23, 113 3 B2 for 23 or B1 for [g(x) =] 45 soi

This question in 0607/41 May/June 2018

5 16 m A P D NOT TO SCALE 30 m S Q 18 m 12 m B R C 24 m 40 m In the diagram, ABCD is a rectangle. (a) Find PS. PS = … m [2] (b) Find angle BRS. Angle BRS = … [2] (c) Find the perimeter of PQRS. … m [3] (d) Find the shaded area. … m2 [3] (e) Explain why triangle ASP is similar to triangle BSR. … … [2]

12 marks

Mark scheme: 5(a) 20 2 2 2 M1 for 16 + ( 30 − 18 ) 5(b) 36.9 or 36.86 to 36.87 2 18 M1 for tan[ ] = oe 24 5(c) 100 3 2 2 M2 for 2 × (their (a) + 18 + 24 ) oe or M1 for 182 + 24 2 or RS = 30 or PQ = 30 seen 5(d) 576 3 M2 for (40 × 30) − 2 × (0.5 × 18 × 24) − 2 × (0.5 × 16 × 12) oe or M1 for any correct and relevant area 5(e) Correct explanation 2 B1 for partial explanation e.g. ratio of two sides the same, with names or numbers given.

This question in 0607/42 May/June 2018

1 (a) Solve the following equations. (i) 12 - x = 4 x = … [1] (ii) 9x - 4 = 6x + 8 x = … [2] 12 (iii) + 5 = 9 x x = … [2] (b) (i) Solve 6x 2 - 5x + 1 = 0 . x = … or x = … [3] (ii) Use your answer to part (b)(i) to solve 6 sin 2 x - 5 sin x + 1 = 0 for 0° G x G 90 ° . x = … or x = … [3]

11 marks

Mark scheme: Question Answer Marks Partial Marks 1(a)(i) 8 1 1(a)(ii) 4 2 M1 for correct 1st step 1(a)(iii) 3 2 M1 for correct 1st step 1(b)(i) 1 1 3 M2 for (3 x − 1)(2 x − 1) [ = 0] [ x = ] , 2 3 or M1 for ( ax ± 1)(bx ± 1) where ab = 6 or a + b = –5 or 3 x (2 x − 1) − 1(2 x − 1) or 2 x (3 x − 1) − 1(3 x − 1) OR M2 for correct sketch or M1 for any U-shaped parabola crossing x-axis twice OR 5 ± ( −5) 2 − 4 × 6[× 1] M2 for 2 × 6 b 2 or M1 for or b − 4 ac correct 2 a 1(b)(ii) 30, 19.5 or 19.47… 3 B2 FT for one correct answer 1  or M1 for sin x = their   2  1  sin x = their    3 

This question in 0607/41 Oct/Nov 2018

3 y 4 x 0 –90 90 –4 f (x) = 1 - 2 sin (2x - 10) ° (a) On the diagram sketch the graph of y = f (x) , for - 90 G x G 90 . [3] (b) Write down the co-ordinates of the x-intercepts. ( … , … ) ( … , … ) [2] (c) Write down the co-ordinates of the local maximum. ( … , … ) [1] x ° (d) The graph of y =- intersects the graph of y = 1 - 2 sin (2x - 10 ) three times. 60 Find the value of the x co-ordinate at each point of intersection. x = … or x = … or x = … [3]

9 marks

Mark scheme: 3(a) Correct sketch 3 Intersections with x-axis both positive and not 90 and maximum below 4 B1 correct sine graph shape B1 max and min in correct quadrant 3(b) (20, 0) (80, 0) 2 B1 for each 3(c) (–40, 3) 1 3(d) –80.2 or –80.16… 3 B1 for each 28.9 or 28.90… 56.7 or 56.71 to 56.72

This question in 0607/41 Oct/Nov 2018

11 B NOT TO 8.6 cm SCALE A 9.3 cm C The area of triangle ABC = 23.5 cm2. (a) Show that angle BAC = 36.0°, correct to 1 decimal place. [2] (b) Use the cosine rule to find BC. BC = … cm [3] (c) All the angles in triangle ABC are acute. Use the sine rule to find the largest angle in the triangle ABC. … [3]

8 marks

Mark scheme: 11(a) 0.5 × 8.6 × 9.3 × sin A = 23.5 M1 35.99...[ = 36.0] A1 11(b) 2 2 or M1 for [ x = ] 8.6 + 9.3 −×2 8.6 × 9.3 × cos36 M2 [ x 2 = ] 8.6 2 + 9.32 − 2 × 8.6 × 9.3 × cos36 5.57 or 5.569 to 5.571... A1 M1 with correct answer scores full marks 11(c) 9.3 × sin36 M2 9.3 their ( b ) sin B = or M1 for = their ( b ) sin B sin36 78.8 or 78.9 or 79.[0] A1 M1 with correct answer scores full marks or 78.77 to 78.98...

This question in 0607/41 Oct/Nov 2018

7 North North 10 km B 150° NOT TO A SCALE 12 km 18 km C 21 km D The diagram shows four villages A, B, C and D and five straight roads connecting them. B is 10 km due east of A. C is 12 km from B on a bearing of 150°. D is 21 km from C and 18 km from A. (a) Calculate the distance AC and show that your answer rounds to 19.08 km, correct to 2 decimal places. [4] (b) Using the sine rule, calculate angle ACB and show that your answer rounds to 27.0°, correct to 1 decimal place. [3] (c) Calculate the bearing of D from C. … [4] (d) A straight path, BP, connects B to the closest point, P, on AC. Calculate the length of this path. … km [2] (e) The area within triangle ABC is grassland. Calculate the area of this grassland. … km2 [2]

15 marks

Mark scheme: 7(a) [Angle ABC = ] 120 B1 10 2 + 12 2 − 2 × 10 × 12cos(their ABC ) M1 19.078 to 19.079 A2 A1 for 364 7(b) 10sin120 M2 19.08 10 M1 for = oe 19.08 sin120 sin ACB 26.99... A1 M1 only and A1 imply M2 A1 7(c) 249.8 to 250[.0] 4 212 + 19.082 − 182 M2 for [cos ACD = ] oe 2 × 21 × 19.08 or M1 for 18 2 = 212 + 19.08 2 −×2 21 × 19.08 × cos( ACD ) M1 for 360 – (30 + 27 + their ACD) oe 7(d) 5.45 or 5.446 to 5.448 2 M1 for 12sin27.0 oe 7(e) 52[.0] or 51.94 to 51.99... 2 1 M1 for × 19.08 × their (d) 2 1 or for × 10 × 12 × sin120 oe 2 1 or for × 19.08 × 12 × sin 27 2

This question in 0607/41 Oct/Nov 2019

11 y 4 x 0 180 –4 f (x) = 3 sin (3 x°) (a) On the diagram, sketch the graph of y = f ( x) for 0 G x G 180. [2] (b) Write down the amplitude and the period of f (x). Amplitude = … Period = … [2] (c) Solve the inequality f (x) 1- 1.5 for 0 G x G 180. … [2] (d) g (x) = 3 sin ( x°) (i) On the same diagram, sketch the graph of y = g (x) for 0 G x G 180. [1] (ii) On the diagram, shade the regions where f (x) H g (x ). [1] (iii) Describe fully the single transformation that maps the graph of y = g (x) onto the graph of y = f (x). … … [3]

11 marks

Mark scheme: 11(a) Correct4444 sketch 2 B1 for sine graph with incorrect amplitude and/or incorrect period, passing through (0, 0). 2222 0000 0000 50505050 100100100100 150150150150 -2-2-2-2 -4-4-4-4 11(b) 3 2 B1 for each 120 11(c) 70 < x < 110 2 B1 for 70 and 110 seen 11(d)(i) Correct4444 sketch 1 2222 0000 0000 50505050 100100100100 150150150150 -2-2-2-2 -4-4-4-4 11(d)(ii) Two areas shaded, which are below 1 graph of y = f(x) and above graph of y = g(x) 11(d)(iii) Stretch 3 B1 for each 1 [factor] 3 invariant line y-axis oe

This question in 0607/41 Oct/Nov 2019

10 D NOT TO SCALE 20° 50° A 12 m B C The diagram shows a vertical pole CD. ABC is a straight line on level ground. Find DC. DC = … m [6]

6 marks

Mark scheme: 10 6.29 or 6.288 to 6.293 6 12 × tan 20 × tan50 M5 for tan50 − tan 20 12tan20 or M4 for BC = tan50 − tan20 or M3 for BC tan50 = (12 + BC )tan 20 DC or M2 for tan20 = 12 + BC DC and tan50 = BC DC or M1 for tan20 = 12 + BC DC or tan50 = BC

This question in 0607/42 May/June 2020

10 In this question, all lengths are in centimetres. NOT TO SCALE 2x + 4 2x + 1 30° 4x 4x + 5 The areas of the two triangles are equal. (a) Show that 8x 2 + 18 x - 5 = 0 . [5] (b) Solve 8x 2 + 18 x - 5 = 0 . You must show all your working. x = … or x = … [3] (c) Find the area of each of the triangles. … cm2 [2]

10 marks

Mark scheme: 10(a) 1 M2 M1 for either area × 4 x ( 2 x + 4 ) = 2 1 ( 2 x + 1)( 4 x + 5 ) sin30 2 1 M1 sin30 = and eliminating fractions 2 Expanding brackets M1 FT Completion to 8x2 + 18x – 5 = 0 A1 with no errors 10(b) (4x – 1)(2x + 5) = 0 M1 2 − 18 ± 18 − 4 × 8 × ( − 5) or x = 2 × 8 or sketch of parabola (U shaped) with one +ve and one –ve zero. 1 1 A2 A1 for each. , – 2 oe 1 1 4 2 If 0 scored, SC1 for , – 2 4 2 10(c) 2.25 2 M1 for substituting their positive solution in either area formula.

This question in 0607/43 May/June 2020

8 E F NOT TO SCALE 12 cm B A 8 cm D C 20 cm ABCDEF is a triangular prism. ABCD is a rectangle. Find (a) AC, AC = … cm [2] (b) ED, ED = … cm [2] (c) angle EAD, Angle EAD = … [2] (d) angle FAC. Angle FAC = … [2]

8 marks

Mark scheme: 8(a) 21.5 or 21.54… 2 M1 for 20 2 + 8 2 8(b) 8.94 or 8.944… 2 M1 for 12 2 − 8 2 8(c) 48.2 or 48.15 to 48.19... 2 8 M1 for cos[ x ] = oe 12 8(d) 22.5 or 22.6 or 22.54 to 22.59 2 2 2 12 − 8 M1 for tan[ x ] = oe 20 2 + 8 2 their (b ) or tan[ x ] = oe their ( a )

This question in 0607/41 Oct/Nov 2020

9 F NOT TO SCALE B A E 20 cm D C 12 cm The diagram shows rectangle ABCD and two right-angled isosceles triangles, ABF and BCE. (a) Find the perimeter of the quadrilateral CDFE. … cm [3] (b) (i) Find the area of the quadrilateral CDFE. … cm2 [3] (ii) Quadrilateral Q is similar to quadrilateral CDFE. The area of quadrilateral Q is 158 cm2. Find the length of the shortest side of quadrilateral Q. … cm [2] (c) Calculate angle AFE. Angle AFE = … [2]

10 marks

Mark scheme: 9(a) 106 or 106.4 to 106.5 3 2 2 2 2 M2 for 32 + 12 and 20 + 20 oe or M1 for 322 + 122 or 202 + 202 oe 9(b)(i) 632 3 M2 for 0.5 × 20 × 20 and 0.5 × 32 × 12 and 20 × 12 oe or M1 for 0.5 × 20 × 20 or 0.5 × 32 × 12 or 0.5 × 12 × 12 or 0.5 × 12 × 12 9(b)(ii) 6 2 their 632 158 M1 for or 158 their 632 9(c) 69.4 or 69.42 to 69.45 2 32 M1 for tan [ x ] = oe 12

This question in 0607/41 May/June 2021

7 In this question all lengths are in centimetres. (a) C B 8x° ( x + 5 )° NOT TO SCALE A In triangle ABC, AC = BC, angle ABC = ( x + 5)° and angle ACB = 8x° . Find the value of x. x = … [3] (b) NOT TO ( p - 2) SCALE ( p + 1) The diagram shows a rectangle with sides of length ( p + 1) and ( p - 2) . The area of the rectangle is 90 cm2 . Find the value of p. p = … [4] (c) ( y - 1) ( y - 4) NOT TO SCALE 30° The diagram shows a right-angled triangle. Find the value of y. y = … [3] (d) 13 ( w + 1 ) NOT TO SCALE ( 2w + 3) The diagram shows a right-angled triangle with sides of length ( w + 1) , ( 2w + 3) and 13. Work out the area of the triangle. … cm2 [6]

16 marks

Mark scheme: 7(a) 17 3 M2 for x + 5 + 8 x + x + 5 = 180 oe or M1 for angle A = x + 5 7(b) 10.1 or 10.10... 4 B3 for correct sketch indicating roots −−( 1) ± ( − 1) 2 − 4(1)( − 92) or for oe 2(1) or B2 for p 2 − 2 p + p − 2 [ = 90] or better or M1 for ( p + 1)( p − 2) [ = 90] 7(c) 7 3 M2 for 2(y – 4) = y – 1 or better y − 4 or M1 for = sin30 y − 1 If 0 scored SC1 for sin 30 = 0.5 7(d) 2.04 oe 6 B4 for (5 w − 1)( w + 3) or correct sketch indicating roots − 14 ± 14 2 − 4(5)( −3) or 2(5) or B3 for 5 w 2 + 14 w − 3 = 0 and M1 for correct calculation of area of triangle with their positive w OR 2 2 M1 for ( w + 1) + ( 2 w + 3 ) = 13 B1 for w 2 + w + w + 1 oe or 4 w 2 + 6 w + 6 w + 9 oe and M1 for correct calculation of area of triangle with their positive w

This question in 0607/43 May/June 2021

8 D 7 cm NOT TO C SCALE 18 cm A 13 cm 16 cm B (a) Calculate angle BCA and show that it rounds to 59.57°, correct to 2 decimal places. [3] (b) Find the area of quadrilateral ABCD. … cm2 [3] (c) Find the shortest distance from A to BC. … cm [2] (d) D is due north of B. Find the bearing of B from C. … [6]

14 marks

Mark scheme: 8(a) 18 2 + 132 − 16 2 M2 M1 for 16 2 = 182 + 132 −×2 18 × 13cos(...) 2 × 18 × 13 59.574 to 59.575 A1 8(b) 164 or 163.8 to 163.9 3 1 M1 for × 18 × 7 oe 2 1 M1 for × 18 × 13 × sin59.57 oe 2 8(c) 15.5 or 15.52... 2 distance M1 for sin 59.57 = oe 18 8(d) 191 or 190.5 to 190.6 6 M2 for 7 2 + 132 −×2 7 × 13cos(90 + 59.57) or B1 for [angle BCD =] 149.57 7sin(90 + 59.57) M2 for theirBD theirBD 7 or M1 for = sin(90 + 59.57) sin DBC M1 for 180 + their DBC oe

This question in 0607/43 May/June 2021

9 (a) x cm 4 cm NOT TO SCALE 40° Calculate the value of x. x = … [3] (b) C 8 cm 9 cm NOT TO SCALE A B 10 cm (i) Calculate angle ABC. Angle ABC = … [3] (ii) T is the point on AB that is the shortest distance from C. Calculate BT. BT = … cm [3] (c) Another triangle PQR has QR = 12 cm, PR = 7 cm and angle PQR = 35°. Calculate the difference between the two possible values of angle QPR. … [5]

14 marks

Mark scheme: 9(a) 6.22 or 6.222 to 6.223 3 4 M2 for oe sin40 4 or M1 for sin 40 = oe x 9(b)(i) 49.5 or 49.45 to 49.46 3 9 2 + 10 2 − 8 2 M2 for [cos=] oe 2.9.10 or M1 for 82 = 92 + 102 – 2 × 9 × 10 cos(...) 9(b)(ii) 5.85 or 5.845 to 5.851... 3 BT M2 for = cos(their(b)(i)) oe or better 9 or M1 for CT drawn and right angle at T 9(c) 21[.0] or 20.98 to 21.00 5 12sin35 M2 for 7 7 12 or M1 for = oe sin35 sin P A1 for 79.5 or 79.50 to 79.51 M1 for 180 – their 79.5 If 0 scored, SC1 for diagram showing the two angles

This question in 0607/42 Feb/March 2022

11 (a) 2r NOT TO r SCALE y° The diagram shows a sector of a circle with radius r and angle y°. The length of the arc of the sector is 2r. Calculate the value of y. y = … [3] (b) NOT TO 8 cm SCALE x° The diagram shows a sector of a circle with radius 8 cm and angle x°. The area of the shaded segment is A cm 2. 8x (i) Show that A = r - 32 sin x . 45 [2] (ii) Find the value of A when x = 90. … [1] 8x (iii) By sketching the graph of A = r - 32 sin x , find the value of x when A = 5.5 . 45 A 20 0 90 x x = … [3]

9 marks

Mark scheme: 11(a) 114.6 or 114.5 to 114.6 3 y M2 for × 2πr = 2 r oe 360 y or M1 for × 2 πr oe 360 11(b)(i) x 2 1 2 M2 x 2 1 2 × π× 8 − × 8 × sin x = A M1 for × π× 8 or × 8 × sin x 360 2 360 2 11(b)(ii) 18.3 or 18.26 to 18.27... 1 11(b)(iii) Correct sketch of curve and line B2 B1 for correct shape of curve 6666 4444 2222 0000 0000 20202020 40404040 60606060 80808080 58.9 or 58.90 to 58.92 1

This question in 0607/42 Feb/March 2022

12 B NOT TO A SCALE 80° 9 m 16 m D 115° 22 m C (a) Calculate the area of triangle BCD. … m2 [2] (b) Calculate angle ADB. Angle ADB = … [6]

8 marks

Mark scheme: 12(a) 160 or 159.5… 2 1 M1 for  16  22  sin115 oe 2 12(b) 84[.0] or 84.02 to 84.03 6 B3 for 32.2 or 32.21.. soi OR M2 for 16 2  22 2 2 16  22  cos115 or M1 for 16 2  22 2  2  16  22  cos115 AND 9sin80 M2 for dependent on their (32.21) cosine rule used sin ABD sin80 or M1 for  9 their (32.21)

This question in 0607/41 May/June 2022

11 F X E NOT TO SCALE B A C D The diagram shows a triangular prism ABCDEF. X is a point on FE. AB = 8 m , AD = 15 m , AF = 10 m , EC = 6 m and FX = 5 m . Angle ABF = 90° and angle DCE = 90° . (a) Calculate angle CDE. Angle CDE = … [2] (b) Calculate AC. AC = … m [2] (c) Calculate angle CXA. Angle CXA = … [5] (d) Calculate the area of triangle CXA. … m2 [2] Question 12 is printed on the next page.

11 marks

Mark scheme: 11(a) 36.9 or 36.86 to 36.87 2 6 6 M1 for tan[CDE ] = or sin[CDE ] = 8 10 8 or cos[CDE ] = 10 11(b) 17 2 M1 for [ AC 2 ] = 15 2 + 8 2 11(c) 96.2 or 96.16… 5 M1 for AX 2 = 52 + 10 2 M1 for CX 2 = 10 2 + 6 2 M2 dep for their AX 2 + theirCX 2 − their AC 2 [cos CDX =] 2  their AX  theirCX their AC 2 = their AX 2 + theirCX 2 or M1dep for −2 their AX  theirCX  cos CDX both dependent on Pythagoras or trigonometry used for AX and CX 11(d) 64.8 or 64.81 to 64.82 2 M1 dep for area CXA = 0.5  theirCX  their AX  sin(theirCXA) dependent on Pythagoras or trigonometry used for AX and CX

This question in 0607/41 Oct/Nov 2022

3 y 1 0 x 360 –1 f ( x) = cos x° for 0 G x G 360 (a) On the diagram, sketch the graph of y = f ( x) . [2] (b) Find the zeros of f ( )x . … [2] (c) (i) Solve the equation f ( x) = 0.5 . … [2] (ii) Solve the inequality f ( x) 1 0 .5 . … [2] (iii) On the diagram, shade the regions that satisfy the inequalities y 1 0.5 and y 2 f ( x) . [1] (d) The equation f ( )x = k has four solutions. Complete the statement to show the range of possible values of k. … 1 k 1 … [1]

10 marks

Mark scheme: 3(a) Correct sketch 2 B1 for correct shape but inaccurate or different domain 3(b) 90, 270 2 B1 for each –1 if y cords (0) included 3(c)(i) 60, 120, 240, 300 2 B1 for two or three correct with no extras or four correct with extras –1 if y cords (0.5) included 3(c)(ii) 60 < x < 120, 240 < x < 300 2 B1 for each 3(c)(iii) Correct areas shaded, 1 i.e. below y = 0.5 and above y = f(x) 3(d) 0 [ < k < ] 1 1

This question in 0607/41 May/June 2023

9 C D NOT TO 118° SCALE 5 m 12 m 35° A B 16 m (a) B is due east of A. Find the bearing of A from C. … [2] (b) Calculate the area of triangle ABC. … m2 [2] (c) Calculate angle CAD. Angle CAD = … [4] (d) Calculate the length of the straight line BD. … m [3]

11 marks

Mark scheme: 9(a) 235 2 M1 for 180  55 or for 360 – 125 or for 270 – 35 or for 35 or 55 or 125 or 145 correctly indicated at C. 9(b) 55.1 or 55.06... 2 1 M1 for  12  16 sin 35 oe 2 9(c) 40.4 or 40.41... 4 5sin118 M2 for [sinC = ] 12 12 5 or M1 for  oe sin118 sinC M1 dep for 180 – 118 – their C dependent on sine rule used to find angle. 9(d) 15.5 or 15.51... nfww 3 M2 for 5 2  16 2  2 5 16cos(35 theirA) or M1 for 52 + 162 – 2 516cos(35 + theirA) A1 for 241 or 240.6 to 240.7...

This question in 0607/41 May/June 2023

5 (a) The diagram shows a regular pentagon with sides of 10 cm and centre O. B 10 cm A C NOT TO O SCALE E D (i) Find angle AOB. Angle AOB = … [1] (ii) Show that OA = 8. 51 cm correct to 3 significant figures. [3] (iii) Find the area of the pentagon. … cm2 [2] (b) V NOT TO 18 cm SCALE B A C O E 10 cm D The regular pentagon in part (a) is the base of a pyramid. The sloping edges, VA, VB, VC, VD, and VE, are each of length 18 cm. (i) Calculate the perpendicular height, VO, of the pyramid. VO = … cm [3] (ii) Calculate the volume of the pyramid. … cm3 [2] (iii) A geometrically similar pyramid has volume 1500 cm 3. Calculate the length of a side of the base of this pyramid. … cm [3]

14 marks

Mark scheme: 5(a)(i) 72 1 5(a)(ii) 5 M2  1 5 oe oe M1 for sin  their 72   1   2  OD sin  their 72   2  8.506 to 8.507 A1 5(a)(iii) 172 or 172.0 to 172.2 2 1 M1 for × 8.512 × sin(their 72) oe 2 1 or  8.51  (10or5)sin54 oe 2 1 or  (10or5)  5tan54 oe 2 5(b)(i) 15.9 or 15.86... 3 M2 for 182 – 8.512 or M1 for VO2 + 8.512 = 182 5(b)(ii) 909 to 913 2 1 M1 for  (their172)  (their15.9) 3 5(b)(iii) 11.8 or 11.79 to 11.82 3 1500 M2 for 10 × 3 oe their(b)(ii) 1500 their(b)(ii) or M1 for 3 or 3 their(b)(ii) 1500 their (b)(ii)  10  3 or   oe 1500  x 

This question in 0607/42 May/June 2023

5 B NOT TO 50 cm SCALE 55 cm A 64 cm C D E In the diagram, AB is parallel to ED. ACD and BCE are straight lines. AB = 50 cm, BC = 55 cm and AC = 64 cm . (a) Show that angle ACB = 49.0° correct to one decimal place. [3] (b) Use the sine rule to calculate angle CAB. Angle CAB = … [3] (c) Calculate the area of triangle ABC. … cm2 [2] 2(d) AC = AD 3 Calculate the area of triangle CDE. … cm2 [2]

10 marks

Mark scheme: 5(a) 55 2  64 2  50 2 M2 M1 for cos ACB  2 2 2 2  55  64 50  55  64 2 55  64  cos ACB  ACB   48.97...   49.0  A1 5(b) 55sin49 55 50 sin CAB  M2 M1 for  oe 50 sin CAB sin 49 56.1 or 56.07 to 56.12 B1 5(c) 1330 or 1327 to 1328… 2 M1 for 0.5  64  55  sin49 oe 5(d) 331.9 to 333 2 FT their 5(c) ÷ 4 M1 for 0.5 2 or 2 2 1 or  32  27.5sin 49.0 oe 2

This question in 0607/43 May/June 2023

2 y 10 0 x 360 -10 1 f ( x) = for 0 G x G 360 sin x° (a) On the diagram, sketch the graph of y = f ( x) . [3] (b) Find the coordinates of the local minimum point. ( … , … ) [1] (c) Write down the equations of the three asymptotes of the graph of y = f ( x) . … , … , … [2] (d) The equation f ( x) = k has no solutions. Write down the range of values of k. … [2] 1 x ° (e) By sketching another graph on the diagram, solve the equation = 5 sin for 0 G x G 360 . sin x° b 2 l … [3]

11 marks

Mark scheme: 2(a) 10 Correct sketch 3 Two branches with small gap at approx x = 180 5 B2 for two correct shaped branches but with large gap or too much overlap 0 0 50 100 150 200 250 300 350 or B1 for one correct branch -5 -10 2(b) (90, 1) 1 2(c) x = 180 2 B1 x = 0 and x = 360 B1 If 0 scored, SC1 for all 3 values seen 2(d) –1 < k < 1 2 B1 for each If 0 scored, SC1 for –1 ⩽ k ⩽ 1 2(e) 38[.0] or 37.95... AND 3 B1 for either solution correct 168 or 168.4... or B1 for both solutions expressed in coordinate form 10101010 AND 5555 0000 0000 50505050 100100100100 150150150150 200200200200 250250250250 300300300300 350350350350 B1 Correct sine curve through (0, 0), (360, 0) and with amplitude -5-5-5-5 approximately 5 -10-10-10-10

This question in 0607/41 Oct/Nov 2023

11 C NOT TO SCALE 8 cm 47° A B 10 cm (a) Calculate the area of triangle ABC. … cm2 [2] (b) Calculate the shortest distance from C to AB. … cm [3] (c) Show that BC = 7.41 cm correct to 2 decimal places. [3] (d) C NOT TO SCALE 8 cm O 47° A B 10 cm In triangle ABC, O is the centre of the circle that passes through A, B and C. Calculate the radius of this circle. … cm [4]

12 marks

Mark scheme: 11(a) 29.3 or 29.25... 2 1 M1 for 8 10sin47 2 11(b) 5.85 to 5.86 3 distance M2 for sin 47 = oe or for 8 2 their(a) oe 10 or M1 for recognition of shortest distance 11(c) 2 2 M2 M1 for 82 + 102 – 2  8  10cos47 oe 8 + 10 −2 8 10cos47 A1 for 54.9 or 54.88... 7.408... A1 11(d) 5.06 or 5.07 or 5.064 to 5.066 4 1 2  7.41 M3 for oe sin 47 or 1 2  7.41 M2 for sin47 = oe radius or M1 for angle BOC = 94 soi

This question in 0607/41 Oct/Nov 2023

7 A NOT TO SCALE 5 cm 7.63 cm B 7 cm 12° D C In triangle ACD, AB = 5 cm , AD = 7. 63 cm and BD = 7 cm . Angle BDC = 12° . (a) Show that angle ABD = 77.0° correct to 1 decimal place. [3] (b) Calculate the area of triangle ABD. … cm2 [2] (c) Calculate BC. … cm [4]

9 marks

Mark scheme: 7(a) 7 2 + 5 2 − 7.632 M2 M1 for 7.632 = 7 2 + 52 −2 7 5 cos ABD cos ABD = 2 7 5 ABD = 76.96... A1 no errors or omissions 7(b) 17.1 or 17.[0] or 17.04 to 17.05… 2 M1 for 0.5 5 7 sin77 7(c) 1.61 or 1.605 to 1.606… 4 7sin12 M3 for oe sin (77 − 12) BC 7 or M2 for = sin12 sin(77 − 12) or B1 for [ ACD =]65

This question in 0607/42 Oct/Nov 2023

11 (a) NOT TO SCALE 12 cm p° 18 cm Calculate the value of p. p = … [2] (b) B NOT TO SCALE 230 m 190 m C A 150 m 180 m D (i) Show that angle ABC = 67.0° correct to 1 decimal place. [4] (ii) Calculate the shortest distance from A to the side BC. … m [3] Question 12 is printed on the next page.

9 marks

Mark scheme: 11(a) 33.7 or 33.69... 2 12 M1 for tan[p] = 18 11(b)(i) 1502 + 1802 M1 192 2 + 230 2 − their (180 2 + 150 2 ) M2 M1 for their (1802 + 1502) = 1902 + 2302 [cos=] – 2 × 190 × 230 cosB 2  190  230 67.03 to 67.04 A1 11(b)(ii) 175 or 174.8 to 174.9... 3 x M2 for sin67[.0] = oe 190 or M1 for distance required is perpendicular to BC oe

This question in 0607/43 Oct/Nov 2023

1 y 1.5 x -180 0 180 -1.5 f ( x) = ( sin x°) 2 (a) On the diagram, sketch the graph of y = f ( x) for - 180 G x G 180 . [2] (b) Write down the amplitude and period of f(x). Amplitude … Period … [2] (c) g ( x) = 0 .002 x + 0 .5 (i) On the diagram, sketch the graph of y = g ( x) for - 180 G x G 180 . [2] (ii) Solve g ( x) = f ( x) for - 180 G x G 180 . … [4] (iii) Solve g ( x) 1 f ( x) for - 180 G x G 180 . … [2]

12 marks

Mark scheme: Question Answer Marks Partial Marks 1(a) Correct curve 2 B1 if cusp at (0, 0) or ‘correct curve’ but height clearly incorrect. 1(b) 0.5 2 B1 for each 180 1(c)(i) Correct sketch 2 maximum of 1 mark if it does not intersect curve 4 times B1 for positive gradient and positive y- intercept 1(c)(ii) –154 or –154.0... 4 B1 for each –40.4 or –40.36 to –40.35 Max 3 if y coordinates included. 50.9 or 50.87... 121 or 120.5 to 120.6 1(c)(iii) –154 < x < –40.4 2 FT from (c)(ii) 50.9 < x < 121 B1 for each Same accuracy as (ii)

This question in 0607/42 Feb/March 2024

8 8 m A B 73.4° NOT TO 36° SCALE 7 m C D The diagram shows a shape ABDC formed from triangle ABC and a sector of a circle BCD, centre B. (a) Show that BC = 9.0 m , correct to 1 decimal place. [3] (b) Use the sine rule to find angle BCA. Angle BCA = … [3] (c) Find the area of triangle ABC. … m2 [2] (d) Find the area of the shaded region. … m2 [3] (e) Find the perimeter of the shape ABDC. … m [2]

13 marks

Mark scheme: 8(a) M2 or M1 for [ BC 2 ]  82  7 2 2 8 7 cos73.4 [ BC  ] 8 2  7 2  2 8 7 cos73.4 M1 for 81.[00…] 9.00[0…] A1 8(b) 8  sin73.4 M2 8 9 [sin...]  M1 for  oe 9 sin BCA sin73.4 58.4 or 58.41… B1 8(c) 26.8 or 26.83… 2 1 M1 for 7 8 sin73.4 oe 2 8(d) 1.64 or 1.641 to 1.642 3 36 M1 for areasector BCD  π 9 9 360 M1 for area triangle BCD  0.5 9 9 sin36 8(e) 29.7 or 29.65 to 29.66 NFWW 2 36 M1 for  2π  9 oe 360

This question in 0607/41 May/June 2024

9 B 78.2° NOT TO A SCALE 43.2° 110.9° D C 9.9 cm Triangle ABC is isosceles with AB = BC . (a) Show that AC = 13.5 cm correct to 3 significant figures. [3] (b) Calculate the length AB. … cm [3] (c) Find the area of ABCD. … cm2 [3]

9 marks

Mark scheme: 9(a) 9.9  sin110.9 M2 9.9 AC [ AC ] = M1 for = sin43.2 sin43.2 sin110.9 13.51… seen A1 9(b) 10.7 or 10.70 to 10.71… 3 0.5  13.5 M2 for [ AB ] = sin(0.5  78.2) 0.5  13.5 or M1 for sin(0.5  78.2) = AB OR 13.5  sin 12 (180 − 78.2) M2 for [ AB ] = sin78.2 13.5 AB or M1 for = sin78.2 sin 12 (180 − 78.2) OR 2 13.52 M2 for x = 2  (1 − cos78.2) or M1 for 13.52 = x 2 + x 2 −2 x x cos78.2 9(c) 85.2 to 85.4 3 M1 for 1 2  (their AB) 2  sin78.2 M1 for 12  13.5  9.9  sin(180 − (110.9 + 43.2))

This question in 0607/43 Oct/Nov 2024

14 C B 75° NOT TO SCALE 10 cm 65° A D 14 cm Triangle BCD is isosceles. (a) Find the area of triangle ABD. … cm2 [2] (b) Find the shortest distance from D to AB. … cm [3] (c) Find the perimeter of ABCD. … cm [6]

11 marks

Mark scheme: 14(a) 63.4 or 63.44… 2 1 M1 for  10  14  sin 65 2 14(b) 12.7 or 12.68 to 12.69 nfww 3 x M2 for sin65 = oe 14 1 or their area =  10  x 2 or M1 for recognising shortest distance 14(c) 44.2 or 44.18… to 44.23 6 2 2 M2 for  BD =  10 + 14 −2 10  14cos65 or M1 for 102 + 142 – 2 × 10 × 14 × cos65 their BD sin(theirBDC ) M2 for  BC =  oe sin75 BC their BD or M1 for = oe sin(theirBDC ) sin 75 OR M2 for (theirBD ) 2 + (theirBD ) 2 − 2(theirBD )(theirBD )cos30 or M1 for (theirBD)2 + (theirBD)2 – 2(theirBD)(theirBD)cos30 OR M2 for 2 × theirBD × cos 75 1 2 BC or M1 for cos75 = theirBD M1 for 10 + 14 + their BC + their CD dependent on at least trigonometry used for BC and CD

This question in 0607/42 Feb/March 2025

15 B 30° NOT TO SCALE 12.4 cm A C The area of triangle ABC is 74.4 cm2. AB = 12.4 cm and angle ABC = 30°. (a) Show that BC = 24 cm. [2] (b) Find AC. AC = … cm [3] (c) Find obtuse angle CAB. Angle CAB = … [3] (d) NOT TO Y SCALE X Z Triangle XYZ is similar to triangle ABC. The area of triangle XYZ is 62 cm2. Find YZ. YZ = … cm [3]

11 marks

Mark scheme: 15(a) 74.4 = 0.5  12.4  BC  sin30 oe M1 Use of 24 scores M0 74.4 A1 74.4 [=24] oe oe e.g. 0.5  12.4  sin30 3.1 15(b) 14.6 or 14.63 to 14.64 3 2 2 M2 for 12.4 + 24 −2 12.4  24  cos30 or M1 for [ AC 2 ] = 12.4 2 + 24 2 −2 12.4  24  cos30 15(c) 124.7 to 125.3 3 their 14.6 must come from trig 12.4 2 + (their14.6) 2 − 24 2 M2 for [cos A = ] 2  12.4  their14.6 or M1 for 24 2 = (their14.6) 2 + 12.4 2 −2 their14.6  12.4cos A OR 24  sin30 M2 for [sin A] = oe their 14.6 sin A sin30 or M1 for = oe 24 their 14.6 OR 74.4 M2 for [sin A] = 1  12.4  their 14.6 2 1 or M1 for  12.4  their 14.6sin A = 74.4 2 15(d) 21.9 or 21.90 to 21.92 or 4 30 oe 3 62 74.4 M2 for 24  or 24  74.4 62 62 74.4  YZ 2 62 oe or M1 for or or for  = 74.4 62  24  74.4

This question in 0607/42 May/June 2025