Cambridge IGCSE Mathematics 0580 — 2012 May/June Paper 4 · Variant 1

6 (a) A parallelogram has base (2x O 1) metres and height (4x O7) metres. For The area of the parallelogram is 1 m2. Examiner's Use (i) Show that 4x2 O 9x + 3 = 0 . Answer (a)(i) [3] (ii) Solve the equation 4x2 O 9x + 3 = 0 . Show all your working and give your answers correct to 2 decimal places. Answer(a)(ii) x = or x = [4] (iii) Calculate the height of the parallelogram. Answer(a)(iii) m [1] (b) (i) Factorise x2 O 16. For Examiner's Use Answer(b)(i) [1] 2 x + 3 x + 40 (ii) Solve the equation + = 2 . x − 4 x 2 − 16 Answer(b)(ii) x = [4]

Mark scheme: 6 (a) (i) ( 4 x − 7 )( 2 x − )1 = 1 M1 or ( 4 x − 7 )( 2 x − )1 − 1 = 0 only 8 x 2 − 14 x − 4 x + 7 B1 allow − 18x and/or + 6 = 0 or = − 6 4 x 2 −x9 + 3 = 0 E1 at least one more line e.g. 8 x 2 − 18 x + 6 = 0 with no errors or omissions seen (ii) 2 B2 B1 for ( −9 ) 2 − 4( 4)(3) or better seen ( 33 ) − ( − 9) ± ( − 9) − 4( 4)(3) ( x = ) 2 × 4 B1 for p = − (− 9) and r = 2×4 or better as long as p + or − q in the form r B1,B1 After B0B0, SC1 for 0.4 or 0.406(929...) (x = ) 0.41, 1.84 cao and 1.8 or 1.843(070...) (iii) 0.36 or 0.3720 to 0.3724 or 0.37 B1ft ft their value to give positive ( 4 x − 7 ) (b) (i) ( x − 4 )( x + 4 ) B1 (ii) ( 2 x + 3)( x + 4) + ( x + 40 ) = 2( x 2 − 16 ) M2 fractions cleared or could all still be over ( x 2 − 16) oe or ( 2 x + 3)( x 2 − 16 ) + ( x + 40)( x − 4) = 2( x − 4)( x 2 − 16 ) 2 x 2 + 8 x + 3 x + 12 or B1 Condone sign slips 2 x 3 + 3 x 2 − 32 x − 48 x = −7 www 4 A1 IGCSE – May/June 2012 0580 41

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7 For y Examiner's Use 10 9 8 7 6 5 4 P 3 2 1 x –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 9 –1 R –2 –3 Q –4 –5 –6 (a) Describe fully (i) the single transformation which maps triangle P onto triangle Q, Answer(a)(i) [3] (ii) the single transformation which maps triangle Q onto triangle R, Answer(a)(ii) [3] (iii) the single transformation which maps triangle R onto triangle P. Answer(a)(iii) [3] (b) On the grid, draw the image of For Examiner's Use  − 4  (i) triangle P after translation by   , [2]  − 5  (ii) triangle P after reflection in the line x = −1 . [2] (c) (i) On the grid, draw the image of triangle P after a stretch, scale factor 2 and the y-axis as the invariant line. [2] (ii) Find the matrix which represents this stretch.     Answer(c)(ii)   [2]    

Mark scheme: 7 In any part of part (a) all marks are independent but mention of a second transformation scores 0 out of 3 (a) (i) Rotation 1 accept R (centre/about) origin (O) (0,0) 1 180° 1 SC3 for all of enlargement, sf – 1, (0, 0) (ii) Enlargement 1 accept E (centre/about) (0,− 3) 1 SF − 3 1 (iii) Enlargement 1 accept E (centre/about) (0, 6) 1 1 1 SF 3 − 4   k  (b) (i) image at (−4, −2) (−2, −2) and (−1, 0) 2 SC1 for translation by   or   , k ≠ 0  k   −5  (ii) image at (− 2, 3) (− 4, 3) and (− 5, 5) 2 SC1 for reflection in y = − 1 (c) (i) image at (0, 3) (4, 3) and (6, 5) 2 SC1 for stretch sf 2 with x-axis invariant ie at (0,6) (2,6) (3,10) (ii)  2 0  2 ft ft their stretch factor only   ft  1 0   0 1  SC1 for correct left hand column ft or ft    0 2 

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8 = {1, 2, 3, 4, 5, 6, 7, 8, 9} For Examiner's E = {x : x is an even number} Use F = {2, 5, 7} G = {x : x2 O 13x + 36 = 0} (a) List the elements of set E. Answer(a) E = { } [1] (b) Write down n(F ). Answer(b) n(F ) = [1] (c) (i) Factorise x2 O 13x + 36. Answer(c)(i) [2] (ii) Using your answer to part (c)(i), solve x2 O 13x + 36 = 0 to find the two elements of G. Answer(c)(ii) x = or x = [1] (d) Write all the elements of in their correct place in the Venn diagram. E F G [2] (e) Use set notation to complete the following statements. (i) F ∩ G = [1] (ii) 7 E [1] (iii) n(E F ) = 6 [1]

Mark scheme: 8 (a) 2 4 6 8 1 (b) 3 1 (c) (i) ( x − 4 )( x − 9 ) 2 SC1 any other ( x + a )( x + b ) where a × b = 36 or a + b = − 13 (ii) 4 9 B1 ft ft or can recover (d) E E 2 Must have all 9 numbers on diagram and no extras 8 5 F 6 2 7 1 SC1 for 5 or more correct elements 4 3 9 G (e) (i) ∅ or { } cao 1 (ii) ∉ cao 1 (iii) ∪ cao 1 IGCSE – May/June 2012 0580 41

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9 f(x) = 3x + 5 g(x) = 7 O 2x h(x) = x2 O 8 For Examiner's (a) Find Use (i) f(3), Answer(a)(i) [1] (ii) g(x O 3) in terms of x in its simplest form, Answer(a)(ii) [2] (iii) h(5x) in terms of x in its simplest form. Answer(a)(iii) [1] (b) Find the inverse function g –1(x). Answer(b) g –1(x) = [2] (c) Find hf(x) in the form ax2 + bx + c . Answer(c) hf(x) = [3] (d) Solve the equation ff(x) = 83. Answer(d) x = [3] (e) Solve the inequality 2f(x) I g(x). Answer(e) [3] Question 10 is printed on the next page.

Mark scheme: 9 (a) (i) 14 1 (ii) 13 − 2 x 2 M1 for 7 − 2( x − 3) (iii) 25 x 2 − 8 final answer 1 7 − y (b) 7 − x 2 M1 for 2 x = 7 − y , x = oe oe 2 2 or x = 7 − 2 y , 2 y = 7 − x oe i.e one step from answer 2 (c) 9 x 2 + 30 x + 17 3 M1 for (3 x + 5 ) − 8 seen B1 for 9 x 2 + 30 x + 25 (d) 7 cao 3 M2 for 3(3 x + 5) + 5 = 83 or better or B1 for 3(3 x + 5) + 5 oe (e) 3 3 M1 for 2 (3 x + 5) < 7 − 2 x oe x < − oe cao 8 B1 for 8x * – 3 or – 8x * 3 3 Do not accept − 8

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10 For Examiner's Use NOT TO 24 cm SCALE 9 cm A solid metal cone has base radius 9 cm and vertical height 24 cm. (a) Calculate the volume of the cone. 1 [The volume, V, of a cone with radius r and height h is V = πr2h.] 3 Answer(a) cm3 [2] (b) NOT TO 16 cm SCALE 9 cm A cone of height 8 cm is removed by cutting parallel to the base, leaving the solid shown above. Show that the volume of this solid rounds to 1960 cm3, correct to 3 significant figures. Answer (b) [4] (c) The 1960 cm3 of metal in the solid in part (b) is melted and made into 5 identical cylinders, each of length 15 cm. Show that the radius of each cylinder rounds to 2.9 cm, correct to 1 decimal place. Answer (c) [4]

Mark scheme: 10 (a) 2030 or 2040 or 2034 to 2036. (…) 2 (V = ) 13 × π × 9 2 × 24 Accept 648π for 2 marks if final answer 8 (b) (upper radius =) 3 B1 accept 9× oe 24 (vol cut off =) 13 × π × their 32 × 8 M1 (= 75.36 to 75.41) their r must be less than 9 their (a) − their 75.39 M1 [ alternate method M1 for ratio sides 1:3 dep M1 ratio vols 1 : 27 M1 their (a) × 26 ÷ 27 ] 624π implies B1 M2 or M3 1958 to 1964.(...) E1 must see a figure after decimal point if 1960 (c) 1960 = 5 × π × r 2 × 15 soi M1 r 2 = 1960 ÷ π ÷ 15 ÷ 5 M1 implied by 8.318... √ their 8.318 M1 dep on M1 M1 SC2 for 5 × π × 9.2 2 × 15 = 1980 to 1982 2.88 to 2.89 E1

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