Cambridge IGCSE Mathematics 0580 — 2014 Oct/Nov Paper 4 · Variant 1

3 y 8 7 6 5 4 3 2 A 1 x –8 –7 –6 –5 –4 –3 –2 –1 0 1 2 3 4 5 6 7 8 –1 –2 –3 B –4 –5 –6 –7 –8 (a) Draw the image when triangle A is refl ected in the line x = 0. [1] (b) Draw the image when triangle A is rotated through 90° anticlockwise about (–4, 0). [2] (c) (i) Describe fully the single transformation that maps triangle A onto triangle B. Answer(c)(i) ................................................................................................................................ ..................................................................................................................................................... [3] (ii) Complete the following statement. Area of triangle A : Area of triangle B = .................... : .................... [2] (d) Write down the matrix that represents a stretch, factor 4 with the y-axis invariant. Answer(d) [2] f p (e) (i) On the grid, draw the image of triangle A after the transformation represented by the 1 0 matrix . e 2 1 o [3] (ii) Describe fully this single transformation. Answer(e)(ii) ............................................................................................................................... ..................................................................................................................................................... [3] 1 0 (iii) Find the inverse of the matrix . e 2 1 o Answer(e)(iii) [2] f p __________________________________________________________________________________________

Mark scheme: 3 (a) Correct reflection 1 (0, 1) (3, 1) (3, 3) (b) Correct rotation 2 SC1 for rotation of 90º anticlockwise about the wrong (–5, 1) (–7, 1) (–5, 4) centre or 90º clockwise about (–4, 0) or for 3 correct points plotted but not joined (c) (i) Enlargement 3 B1 for each [scale factor] 2 [centre] (–7, 7) (ii) 1 : 4 or 3 : 12 or ¼ : 1 2 M1 for 1 : 22 oe, e.g. (3 × 2)/2 : (6 × 4)/2 or SC1 for 4 : 1 or 12 : 3 or 1 : ¼  4 0   k 0  (d)   2 B1 for   , k may be algebraic or numeric  0 1   0 1  but ≠ 0 or 1  1 0  or SC1 for    0 4  (e) (i) Correct shear drawn 3 B2 for two correct points plotted or if not plotted (0, 1) (–3, –5) (–3, –3) correctly shown in working or  1 0 − 3   1 0 − 3   1 0  0  B1 for    or    or    or  2 1  3   2 1  1   2 1  1  better (ii) Shear 3 B1 for each y-axis or x = 0 invariant [factor] 2 (iii)  1 0  2 B1 for [determinant =] 1 shown or stated   oe  − 2 1   1 0  or k   soi, k ≠ 0  − 2 1  Qu Answers Mark Part Marks

More questions on Transformations

4 (a) Expand and simplify. (i) 4(2x – 1) – 3(3x – 5) Answer(a)(i) ................................................ [2] (ii) (2x – 3y)(3x + 4y) Answer(a)(ii) ................................................ [3] (b) Factorise. x3 – 5x Answer(b) ................................................ [1] (c) Solve the inequality. 2x + 1 5x - 8 Y= 3 4 Answer(c) ................................................ [3] (d) (i) x2 – 9x + 12 = (x – p)2 – q Find the value of p and the value of q. Answer(d)(i) p = ................................................ q = ................................................ [3] (ii) Write down the minimum value of x2 – 9x + 12. Answer(d)(ii) ................................................ [1] (iii) Write down the equation of the line of symmetry of the graph of y = x2 – 9x + 12. Answer(d)(iii) ................................................ [1] __________________________________________________________________________________________

Mark scheme: 4 (a) (i) 11 – x final answer 2 M1 for 8x – 4 – 9x + 15 or B1 for final answer 11 – kx or k – x (ii) 6x2 – xy – 12y2 final answer 3 M2 for 6x2 + 8xy – 9xy – 12y2 [= 0] or for final answer with one error in a coefficient (includes sign) but otherwise correct or M1 for any two of 6x2, 8xy, –9xy, –12y2 (b) x(x2 – 5) final answer 1 Condone x(x – 5 ) (x + 5 ) as final answer (c) x [ 4 or 4 Y x final answer 3 B2 for 4 with no/incorrect inequality or equals sign as nfww answer or M2 for 8x + 4 Y 15x – 24 or better or M1 for 4(2x + 1) Y 3(5x – 8) (d) (i) p = 4.5 oe 3 B2 for one correct answer q = 8.25 oe or for (x – 4.5)2 – 8.25 oe seen or M1 for (x – 4.5)2 oe seen or x2 – px – px + p2 seen and M1 for p2 – q = 12 or 2p = 9 (ii) –8.25 oe 1FT FT – their q (iii) x = 4.5 oe 1FT FT x = their p

More questions on Algebraic manipulation

35 (a) Complete the table of values for y = x2 + , x ≠ 0. x x –3 –2 –1 –0.5 0.4 0.6 1 1.5 2 3 y 8 2.5 –5.8 7.7 5.4 4 4.3 10 [2] 3 (b) Draw the graph of y = x2 + for –3 Y x Y –0.5 and 0.4 Y x Y 3. x y 10 8 6 4 2 x –3 –2 –1 0 1 2 3 –2 –4 –6 [5] 3 (c) Use your graph to solve the equation x2 + = 5. x Answer(c) x = .................... or x = .................... or x = .................... [3] 3 (d) By drawing a suitable straight line, solve the equation x2 + = x + 5. x Answer(d) x = .................... or x = .................... or x = .................... [4] __________________________________________________________________________________________

Mark scheme: 5 (a) –2, 5.5 2 B1 for each value (b) Correct curve 5 B5 for correct curve over full domain 10 y or B3FT for 9 or 10 points or B2FT for 7 or 8 points 5 or B1FT for 5 or 6 points Point must touch line if exact or be in correct square if not exact (including boundaries) and x −3 −2 −1 1 2 3 B1 independent for one branch on each side of the y-axis and not touching or crossing the y-axis −5 SC4 for correct curve with branches joined (c) –2.6 Y x Y –2.4 3 B1 for each value 0.6 Y x Y 0.7 1.8 Y x Y 1.9 If B0 then SC1 for y = 5 used Qu Answers Mark Part Marks (d) y = x + 5 ruled correctly 4 B1 for y = x + 5 ruled correctly and –2.2 Y x Y –2.0 B1indep for each value 0.5 Y x Y 0.6 2.4 Y x Y 2.6

More questions on Equations

6 A company tested 200 light bulbs to fi nd the lifetime, T hours, of each bulb. The results are shown in the table. Lifetime Number (T hours) of bulbs 0 < T Y 1000 10 1000 < T Y 1500 30 1500 < T Y 2000 55 2000 < T Y 2500 72 2500 < T Y 3500 33 (a) Calculate an estimate of the mean lifetime for the 200 light bulbs. Answer(a) ...................................... hours [4] (b) (i) Complete the cumulative frequency table. Lifetime (T hours) T Y 1000 T Y 1500 T Y 2000 T Y 2500 T Y 3500 Number of bulbs [2] (ii) On the grid, draw a cumulative frequency diagram to show this information. 200 150 Cumulative frequency 100 50 T 0 500 1000 1500 2000 2500 3000 3500 Lifetime (hours) [3] (iii) The company says that the average lifetime of a bulb is 2200 hours. Estimate the number of bulbs that lasted longer than 2200 hours. Answer(b)(iii) ................................................ [2] (c) Robert buys one energy saving bulb and one halogen bulb. 9 The probability that the energy saving bulb lasts longer than 3500 hours is . 10 3 The probability that the halogen bulb lasts longer than 3500 hours is . 5 Work out the probability that exactly one of the bulbs will last longer than 3500 hours. Answer(c) ................................................ [4] __________________________________________________________________________________________

Mark scheme: 6 (a) 2000 or 1998.75 or 1998.8 or 4 M1 for midpoints soi (condone 1 error or omission) 1999 nfww (500, 1250, 1750, 2250, 3000) and with x in correct interval M1 for use of ∑fx including both boundaries (condone 1 further error or omission) (5000, 37500, 96250, 162000, 99000) and M1 (dep on 2nd M1) for ∑fx ÷ 200 (b) (i) 10, 40, 95, 167, 200 2 B1 for 2 correct (ii) Correct curve or ruled polygon 3 B1FT their (b)(i) for 5 correct heights within 1mm vertically and B1 for 5 points at upper ends of intervals on correct vertical line and B1FT (dep on at least B1) for increasing curve or polygon through 5 points After 0 scored, SC1FT for 4 correct points plotted (iii) 68 to 80 2 M1 for 120 to 132 seen 21 9 2 1 3 (c) oe 4 M3 for × + × oe or better 50 10 5 10 5 or 9 2 1 3 18 3 M2 for × or × or oe or oe 10 5 10 5 50 50 or 1 2 M1 for sight of and 10 5 Qu Answers Mark Part Marks

More questions on Statistical charts and diagrams

7 (a) The diagram shows a circle with two chords, AB and CD, intersecting at X. B NOT TO C SCALE X A D (i) Show that triangles ACX and DBX are similar. Answer(a)(i) [2] (ii) AX = 3.2 cm, BX = 12.5 cm, CX = 4 cm and angle AXC = 110°. (a) Find DX. Answer(a)(ii)(a) DX = .......................................... cm [2] (b) Use the cosine rule to fi nd AC. Answer(a)(ii)(b) AC = .......................................... cm [4] (c) Find the area of triangle BXD. Answer(a)(ii)(c) ......................................... cm2 [2] (b) D NOT TO C SCALE 30 m 37° 31° A B In the diagram, BC represents a building 30 m tall. A fl agpole, DC, stands on top of the building. From a point, A, the angle of elevation of the top of the building is 31°. The angle of elevation of the top of the fl agpole is 37°. Calculate the height, DC, of the fl agpole. Answer(b) ............................................ m [5] __________________________________________________________________________________________

Mark scheme: 7 (a) (i) Any two of with conclusion 2 B1 for two pairs without a conclusion Angle ACD = angle ABD Angle CAB = angle CDB Angle AXC = angle DXB AND ‘triangles have equal angles’ oe e.g. similar and AA or AAA OR All three of without conclusion Angle ACD = angle ABD Angle CAB = angle CDB Angle AXC = angle DXB DX 2.3 (ii) (a) 10 2 M1 for = oe 125. 4 (b) 42 + 3.22 – 2 × 4 × M2 or M1 for implicit version 3.2cos110 34.9 to 35 A1 Implied by answer 5.92 or 5.915 to 5.916 after M2 5.92 or 5.915 to 5.916 B1 (c) 58.7 or 58.73[…] 2FT FT for ½ × 12.5 × their 10 × sin110 oe correctly evaluated to 3 or more sig figs M1 for ½ × 12.5 × their 10 × sin110 oe or ½ × 4 × 3.2 × sin110 × (12.5/4)2 After 0 scored and 15.6... in (a)(ii)(a), allow SC1 for ½ × 4 × 3.2 × sin110 × (12.5/3.2)2 (b) 7.62 or 7.623 to 7.624 5 B4 for 37.6[2…] or 37.63 or 30 M2 for [AB =] or 30 × tan59 oe tan 31 30 AB or M1 for tan31 = or tan59 = oe AB 30 And M2 for [BD =] their AB × tan37 oe or BD M1 for tan37 = oe their AB Qu Answers Mark Part Marks 1 f OQ i d

More questions on Similarity

8 B P A NOT TO SCALE Q 9b 6a C 3c O In the diagram, O is the origin and = 6a, = 9b and = 3c. The point P lies on AB such that = 3b – 2a. The point Q lies on BC such that = 2c – 6b. (a) Find, in terms of b and c, the position vector of Q. Give your answer in its simplest form. Answer(a) ................................................ [2] (b) Find, in terms of a and c, in its simplest form (i) , Answer(b)(i) = ................................................ [1] (ii) . Answer(b)(ii) = ................................................ [2] (c) Explain what your answers in part (b) tell you about PQ and AC. Answer(c) ............................................................................................................................................ ............................................................................................................................................................. [2] __________________________________________________________________________________________

Mark scheme: 8 (a) 2c + 3b 2 M1 for OQ recognised as pos vector. (b) (i) 3c – 6a or 3(c – 2a) 1 (ii) 2c – 4a or 2(c – 2a) 2 M1 for any valid route from P to Q e.g. − (3b − 2 a ) − 6 a + their OQ or PQ = PA + AO + OQ or PQ = PB + BQ 2 (c) PQ = AC oe 2FT STRICT FT dep on PQ = k AC from (b)(i) and 3 (b)(ii) B1FT for each statement and PQ is parallel to AC After 0 scored and PQ = k AC in (b)(i) and (ii), allow SC1FT for correct statement, e.g. PQ is not parallel to AC

More questions on Vectors in two dimensions

9 Diagram 1 Diagram 2 Diagram 3 Diagram 4 The fi rst four diagrams in a sequence are shown above. The diagrams are drawn using white squares and grey squares . (a) Complete the columns in the table for Diagram 4 and Diagram n. Diagram 1 2 3 4 n Number of white squares 12 20 28 Number of grey squares 0 1 4 Total number of squares 12 21 32 (n + 1)(n + 5) [6] (b) Work out the number of the diagram which has a total of 480 squares. Answer(b) ................................................ [2] (c) The total number of squares in the fi rst n diagrams is 1 3 n3 + pn2 + qn. 2 (i) Use n = 1 in this expression to show that p + q = 11 3 . Answer(c)(i) [1] 1 (ii) Use n = 2 in the expression to show that 4p + 2q = 30 3 . Answer(c)(ii) [2] (iii) Find the values of p and q. Answer(c)(iii) p = ................................................ q = ................................................ [3] __________________________________________________________________________________________

Mark scheme: 9 (a) 36, 9, 45 2 B1 for two correct values 8n + 4 oe 2 M1 for 8n + k, for any k (n – 1)2 oe 2 M1 for a quadratic expression of form n2 [+ an + b ] oe (b) 19 2 M1 for (n + 1)(n + 5) = 480 or better or 20 × 24 seen 1 1 1 3 2 (c) (i) + p + q = 12 and no errors 1 Accept p + q = 12 – after [1 ] + p [1 ] + q []1 shown 3 3 3 seen 1 (ii) × 8 + 4p + 2q = 12 + 21 2 M1 for 12 + 21 seen or 33 seen 3 7 (iii) [p =] oe 3 M1 for correct multiplication and subtraction or 2 substitution using the correct given equations 49 7 49 [q =] oe B1 for [p =] or [q =] 6 2 6 After 0 scored, SC1 for 2 values satisfying one of the original correct given equations

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