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Section A (36 marks) Q1 differentiation

Section A (36 marks) Q1 differentiation Q2 sequence (periodic): find 48 th term; find sum of 48 terms Q3 given tan ϑ find cos 2 ϑ Q3 alternative method Q4 transformations of graph of y = f(x) Q5 indefinite integration Q6 sketch sine graph; solve sine equation

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Section A (36 marks) Q1 differentiation

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  1. Section A (36 marks) Q1 differentiation Q2 sequence (periodic): find 48th term; find sum of 48 terms Q3 given tan ϑ find cos2 ϑ Q3 alternative method Q4 transformations of graph of y = f(x) Q5 indefinite integration Q6 sketch sine graph; solve sine equation Q7 sequence from Σ notation; powers of 2; sketch a related graph Q8 GP given second and fourth terms; find the sum to infinity Q9 solve equations involving logs Section B (36marks) Q10 find an expression for an area; first and second derivatives; find minimum area Q11 find lengths in a non-RAT (non-right-angled-triangle); bearings; perimeter of a sector of a circle Q12 find area between curve and line graph; demonstrate steps in an algebraic proof; identify the result grade boundaries

  2. 1 differentiate 10x4 + 12 [2] answer = 40x3

  3. 2 A sequence begins 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 … and continues in this pattern (i) Find the 48th term of this sequence [1] (ii) Find the sum of the first 48 terms of this sequence [2] this is a periodic sequence, repeating every five terms … 5th, 10th, 15th, … are all 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 1 … … 40th 41st 42nd 43rd 44th 45th 46th 47th48th … answer = 3 every five terms adds up to 15 by the 45th term there have been 9 lots of 5 terms adding up 9 x 15 = 135 adding on the 46th, 47th and 48th terms 135 + 6 answer = 141

  4. 3 You are given that tanϑ = ½ and the angle ϑ is acute Show without using a calculator, that cos2 ϑ = ⅘ [3] draw a RAT with angle ϑ tanϑ= ½  opposite = 1 and adjacent = 2 √5 1 ϑ 2 by Pythagoras rule, hypotenuse = √5 QED

  5. alternative method for … 3 You are given that tanϑ = ½ and the angle ϑ is acute Show without using a calculator, that cos2 ϑ = ⅘ [3] using identity sin2ϑ + cos2ϑ = 1 divide each term by cos2ϑ QED

  6. 4 The diagram shows a sketch of the graph of y = f(x) On separate diagrams, sketch the graphs of the following, showing clearly the coordinates of the points corresponding to A, B and C (i) y = 2f(x) [2] (ii) y = f(x + 3) [2] translation stretch; factor 2; y-direction (1,6) (3,6) (-2,3) (0,3) (2,2) (-1,1)

  7. [5] write as powers of x tidy up coefficients

  8. 6 (i) Sketch the graph of y = sin ϑ for 0 ≤ ϑ ≤ 2π [2](ii) Solve the equation 2 sin ϑ = -1 for 0 ≤ ϑ ≤ 2π Give your answer in the form kπ [3] 2 sin ϑ = -1 sin ϑ = -½ ϑ = sin-1 (-½)

  9. 7 (i) Find [2] (ii) Find the value of n for which [1] (iii) Sketch the curve with equation y = 2x [2] (i) … add the 4 terms made by substituting,in turn, k = 2 3,4 and 5 into expression 22 + 23 + 24 + 25 = 4 + 8 + 16 + 32 = 60 (ii) 64 = 26 answer n = -6

  10. 8 The second term of a geometric progression is 18 and the fourth term is 2 The common ratio is positive Find the sum to infinity of this progression [3] GP: a = ? r = ? GP: kth term = ark-1 2nd term = ar = 18 4th term = ar3 = 2 4th ÷ 2nd  r2 = 1/9  r = ±⅓ in this case r = ⅓ from 2nd term = 18 and r = ⅓ a = 54 GP: sum to infinity = = 54 ÷ (1-⅓) = 54 ÷ ⅔ sum to ∞ = 81

  11. 9 You are given that log10 y = 3x + 2 (i) Find the value of x when y = 500 Give your answer correct to 2 decimal places [1] (ii) Find the value of y when x = -1 [1] (iii) Express log10 (y4) in terms of x [1] (iv) Find an expression for y in terms of x [1] y = 500  3x + 2 = log10 500 = 2.698970004 3x = 0.698970004 x = 0.2329900014 x = 0.23 to 2 decimal places x = -1  log10 y = -3 + 2 = -1  y = 10-1 y = 0.1 or 1/10 log10 (y4) = 4 log10 y  log10 (y4) = 4(3x + 2) = 12x + 8 log10 y = 3x + 2  y = 103x + 2

  12. 10 This is a solid cuboid with square base of side x cm and height h cm Its volume is 120 cm3 (i) Find h in terms of x Hence show that the surface area, A cm2, of the cuboid is given by [3] Volume = x2h = 120 Surface Area = 2x2 + 4xh

  13. 10 This is a solid cuboid with square base of side x cm and height h cm Its volume is 120 cm3 (i) Find h in terms of x Hence show that the surface area, A cm2, of the cuboid is given by [3] [4] write A in powers of x A = 2x2 + 480x-1 differentiate … axn naxn-1 differentiate again … axn naxn-1

  14. (iii) Hence find the value of x which gives the minimum surface area Find also the value of the surface in this case [5] A is a minimum when x3 = 120 x= 3√120= 4.932424149 x= 4.93 cm ( to 3 s f) when x= 3√120 which is positive, so A is a minimum A = 146 cm2 A = 145.9728469

  15. 11 (i) The course for a yacht race is a triangle, as shown in the diagram The yachts start at A, then travel to B, then to C and finally back to A A Calculate the total length of the course for this race [4] B Given that the bearing of the first stage, AB, is 175°, calculate the bearing of the second stage, BC [4] given two sides and the angle between them  use the cosine rule to find the third side a2 = b2 + c2 - 2bc cos A a2 = 3022 + 3482 – 2 x 302 x 348 x cos 72 a2 = 147355.0999 a = 384 m (to 3 sig figs) length of course = 302 + 348 + 384 = 1034 metres

  16. B Given that the bearing of the first stage, AB, is 175°, calculate the bearing of the second stage, BC [4] 175° the back bearing BA is (180 + 175) = 355°  bearing BC is (355 – angle ABC) 384m 355° use the sine rule to find angle ABC bearing of C from B B = sin-1 (0.7479663227) = 48.4° sin B = 302 × sin (72) ÷ 384 355 - 48.4 = 306.6° bearing of BC is 307° (to nearest degree)

  17. (ii) The diagram shows the course of another yacht race The course follows the arc of a circle from P to Q, then a straight line back to Q The circle has a radius of 120m and centre O; angle POQ = 136° Calculate the total length of the course for this race [4] whole circumference = π × diameter = 240π = 469.1445029 arc PQ = × 240π 224° 120m x = 120 × sin 68 x 68° x = 111.2620625 course length = arc PQ + 2x = 692 metres (to 3 sig figs)

  18. 12 The diagram shows part of the curve y = x4 and the line y = 8x, which intersect at the origin and the point P A Find the coordinates of P, and show that the area of the triangle OPQ is 16 square units [3] B Find the area of the region bounded by the line and the curve [3] make the y expressions equal x4 = 8x rearrange to equal 0 , then factorise x4 - 8x = 0 x(x3 – 8) = 0 either x = 0, (at the origin) or x3 = 8 at P P is ( 2, 16) area of triangle = ½ base × height = ½ × 2× 16 = 16 square units

  19. 12 The diagram shows part of the curve y = x4 and the line y = 8x, which intersect at the origin and the point P A Find the coordinates of P, and show that the area of the triangle OPQ is 16 square units [3] B Find the area of the region bounded by the line and the curve [3] (2, 16) shaded area = area triangle − area under curve area under curve = = 32 ÷ 5 = 6.4 shaded area = 16 − 6.4 = 9.6 square units

  20. 12 (ii) You are given that f(x) = x4 A Complete this identity for f(x + h) f(x + h) = (x + h)4 = x4 + 4x3h + … [2] B Simplify [2] C Find [1] D State what this limit represents [1] Recall the features of the binomial expansion (a + b)4 = 1a4 + 4a3b1 + 6a2b2 + 4ab3 + 1b4 A (x + h)4 = x4 + 4x3h + 6x2h2 + 4xh3 + h4 B = 4x3 + 6x2h + 4xh2 + h3

  21. 12 (ii) You are given that f(x) = x4 A Complete this identity for f(x + h) f(x + h) = (x + h)4 = x4 + 4x3h + … [2] B Simplify [2] C Find [1] D State what this limit represents [1] 4x3 + 6x2h + 4xh2 + h3 C substitute h = 0 in simplified answer to B = 4x3 D this is the gradient function of the curve y = x4

  22. 72 marks in total 22

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