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The Further Mathematis Network

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Information about The Further Mathematis Network
Education

Published on September 30, 2008

Author: aSGuest301

Source: authorstream.com

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Slide 1: Worked Solutions to Additional Mathematics : June 2003 The Further Mathematics Network Slide 2: 1 Solve simultaneously y = x + 6 (1) y = x2 – x + 3 (2) Equate RHS: x + 6 = x2 – x + 3 Rearrange: 0 = x2 – 2x – 3 Factorise: 0 = (x + 1)(x – 3) Solve: x = –1 or x = 3 Sub. in (1) to find y: y = –1 + 6 = 5 or y = 3 + 6 = 9 Points of intersection: (–1, 5) and (3, 9) Slide 3: Stationary points for: 2 y = x3 – 6x2 + 9x + 5 (1) Differentiate: 3x2 – 12x + 9 = 0 Factorise: (x – 1)(x – 3) = 0 Solve: x = 1 or x = 3 Sub. in (1) to find y: y = 13 – 6×12 + 9×1 + 5 = 9 or y = 33 – 6×32 + 9×3 + 5 = 5 Stationary points are: (1, 9) and (3, 5) x2 – 4x + 3 = 0 Simplify: Slide 4: 2 (1, 9) & (3, 5) Examine gradients either side of stationary points: Maximum Stationary points for y = x3 – 6x2 + 9x + 5: Slide 5: 2 Maximum Minimum (1, 9) & (3, 5) Examine gradients either side of stationary points: Stationary points for y = x3 – 6x2 + 9x + 5: Slide 6: Gradient function of curve: 3 = 2 + 2x – x2 Integrate: y = 2x + x2 – ⅓x3 + c (1) Find constant: x = 3  y = 10: Sub. in (1) y = 2x + x2 – ⅓x3 + 4  10 = 2×3 + 32 – ⅓×33 + c  c = 10 – 6 – 9 + 9  c = 4 Slide 7: 4 Solve the equation: sin 2θ = 0.5 for 0 ≤ θ ≤ 360 sin 2θ = 0.5 2θ = 30 or 2θ = 180 – 30 = 150 or 2θ = 30 + 360  = 390 or 2θ = 150 + 360  = 510 Solution set: θ = 15, 75, 195 or 255 Remember: 0 ≤ θ ≤ 360  0 ≤ 2θ ≤ 720 Slide 8: 5 3x + 4y = 24 Region for 3x + 4y ≤ 24 Slide 9: 5 3x + 4y = 24 3x + y = 15 Region for 3x + 4y ≤ 24 and 3x + y ≤ 15 Slide 10: 5 3x + 4y = 24 3x + y = 15 Region for 3x + 4y ≤ 24 and 3x + y ≤ 15 Maximum value of 2x + y is 11, when x = 4 and y = 3 (4, 3) Slide 11: (2 + x)7 = 27 + 7C1×26×x + 7C2×25×x2 + 7C3×24×x3 + …. 6 = 128 + 7×64×x + 21×32×x2 + 35×16×x3 + …. = 128 + 448x + 672x2 + 560x3 + …. To estimate the value of 1.997, let x = –0.01: (2 + (–0.01))7  128 + 448×(–0.01) + 672×(–0.01)2 + 560×(–0.01)3 = 128 – 4.48 + 0.0672 – 0.00056 = 123.5866 (to 4 decimal places) (a + b)7 = a7 + 7C1×a6×b + 7C2×a5×b2 + 7C3×a4×b3 + …. Slide 12: 7 a c b θ Pythagoras: a2 + b2 = c2 Divide through by a2: Trig. ratios: Simplify: Simplify further: Express in trig. ratios: Slide 13: 8 A B C D V O 6 6 7 Identify right-angled triangle AOV and angle VAO = θ – required angle θ First find AC, then AO, then θ : From right-angled triangle ABC: AC2 = AB2 + BC2 = 62 + 62 = 72  AC = √72  AO = ½√72 Slide 14: 9 (i) When f(x) is divided by (x – 3) remainder is Function f(x) = x3 + 2x2 – 5x – 6 f(3) f(3) = 33 + 2×32 – 5×3 – 6 = 27 + 18 – 15 – 6 = 24 (ii) (x – 2) is a factor of f(x)  f(2) = 0 f(2) = 23 + 2×22 – 5×2 – 6 = 8 + 8 – 10 – 6 = 0  (x – 2) is a factor of f(x) (iii) f(x) = 0  x3 + 2x2 – 5x – 6 = 0  (x – 2)(px2 + qx + r) = 0  (x – 2)(x2 + qx + 3) = 0  (x – 2)(x2 + 4x + 3) = 0  (x – 2)(x + 1)(x + 3) = 0  x = 2, x = –1 or x = –3 Slide 15: 10 (i) To find speed after 5 seconds, i.e. value of v when t = 5: Constant acceleration: s = ?, u = 20, v = ?, a = 1.2, t = ? Use: v = u + at  v = 20 + 1.2×5 = 26 ms-1 (ii) To find distance in 5 seconds, i.e. value of s when t = 5: Use: s = ut + ½at2  s = 20×5 + ½×1.2×52  s = 100 + 15 = 115 m Slide 16: 11 (i) y = 3x – x2  = 3 – 2x (ii) Gradient of tangent at B = 3 – 2×1 = 1 Equation of tangent at B is y = mx + c, where m = 1  y = x + c Since tangent passes through B (1, 2), y = 2 when x = 1  2 = 1 + c  c = 1 Equation of tangent at B is y = x + 1 Slide 17: 11 (iii) At A, y = 0  0 = x + 1  x = –1  A is (–1, 0) (iv) Area of sail = area Δ ABC – area under curve OBC Area Δ ABC = ½×2×2 = 2 units2 Area of sail = units2 Area under curve OBC = Slide 18: 12 (i) P(no faulty bulbs) = P(X = 0) = 0.98 = 0.430 (to 3 s.f.) Binomial Distribution: Let X represent number of faulty bulbs: X ~ B(8, 0.1) (ii) P(two or more faulty bulbs) = P(X ≥ 2) = 1 – [P(X = 0) + P(X = 1)] = 1 – P(X ≤ 1) = 1 – [0.98 + 8C1×0.1×0.97] = 1 – 0.81310473 = 1 – [0.43046721 + 0.38263752] = 0.187 (to 3 s.f.) n = 8, p = 0.1  q = 1 – p = 0.9 (a) P(X = r) = 8Cr×0.1r×0.98–r Slide 19: 12 P(batch accepted) Binomial Distribution: Let X represent number of faulty bulbs: X ~ B(8, 0.1) = P(X = 0) + [P(X = 1) × P(X = 0)] = 0.98 + [(8C1×0.1×0.97) × 0.98] = 0.595 (to 3 s.f.) = 0.43046721 + (0.38263752× 0.43046721) n = 8, p = 0.1  q = 1 – p = 0.9 (b) P(X = r) = 8Cr×0.1r×0.98–r = P(1st pack has 0 faulty bulbs or 1st pack has 1 faulty bulb and 2nd pack has 0 faulty bulbs) Slide 20: 13 (i) 10 tonnes ≡ 10×1000 = 10000 kg (ii) If bags too light by x kg, then mass of bag =  Number of bags = 10000  5 = 2000 5 – x kg If bags too heavy by x kg, then mass of bag = 5 + x kg  Largest number of bags = 10000  (5 – x) =  Smallest number of bags = 10000  (5 + x) = Slide 21: 13 (iii) Largest number of bags – smallest number = 100 Common denominator is (5 – x)(5 + x) Multiply equation (1) throughout by (5 – x)(5 + x):  10000(5 + x) – 10000(5 – x) = 100(5 – x)(5 + x)  100(5 + x) – 100(5 – x) = (5 – x)(5 + x)  500 + 100x – 500 + 100x = 25 – 5x + 5x – x2  x2 + 200x – 25 = 0 Slide 22: 13 (iv) Solve x2 + 200x – 25 = 0 Equation does not factorise so use quadratic formula.  Largest mass = 5 + 0.125 = 5.125 kg  x = 0.125 (to 3 s.f.) – ignore negative root  Smallest mass = 5 – 0.125 = 4.875 kg Slide 23: 14 N N 40 170 L A C 6 50 (i) Bearing of C from A is 170 Bearing of L from A is 40 + 180 = 220 Angle LAC is 220 – 170 = 50 Slide 24: 14 N N 40 L A C 6 50 40 LC = 6 cos 40 or 6 sin 50 = 4.596 nm AC = 6 sin 40 or 6 cos 50 = 3.857 nm Speed of ship = 21 knots  Time taken from A to C = 3.857  21 = 0.1837 hr = 11.02 min  Time when ship is at C is 12.11 (ii) Slide 25: 14 N N 40 L A C 6 50 40 Let ship be at point D when on a bearing of 110 from lighthouse.   CLD = 30 and  CDL = 60 Using the sine rule for ΔALD: 30 D 60  Time taken from A to C = 6.510  21 = 0.3100 hr = 18.6 min  Time when ship is at D is 12.19 (to nearest minute) (iii)

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