2003 Mathematics. Advanced Higher. Finalised Marking Instructions

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1 2003 Mathematics Advanced Higher Finalised Marking Instructions

2 2003 Mathematics Advanced Higher Section A Finalised Marking Instructions

3 Advanced Higher 2003: Section A Solutions and marks A. (a) Given f (x) = x ( + x) 0, then (b) Given f (x) = ( + x) 0 + x.0 ( + x) 9, = ( + x)( + x) 9 * y = 3 x, then ln y = x ln 3 y dy dx = ln 3 dy dx = ln 3 y = ln 3. 3x. A2. S n = n u k k = = n ( 2k) k = = n 2 n k k = k = = n 2 2n (n + ), = n 2 + 0n. n 2 + 0n = 2 (n 3)(n 7) = 0 The sum is 2 when there are 3 terms and when there are 7 terms. Alternative for first 2/3 marks Using results for Arithmetic Series. a = 9, d = 2 S n = n [8 +(n )( 2)] 2 A3. y 3 + 3xy = 3x 2 5 3y 2 dy dx + 3xdy + 3y = 6x dx, dy dx = 6x 3y 3y 2 + 3x = 2x y y 2 + x Thus at (2, ), the gradient is and an equation is (y ) = (x 2). i.e. x = y + or y = x. * The ( + x) 9 must be pulled out. If trial and error is used, both values are needed for the last mark. Page 2

4 A4. Let z = x + iy. z + i = (x + iy) +i = x +( + y) i z + i = x 2 +( + y) 2 x 2 +( + y) 2 = 4 which is a circle, centre (0, ) radius 2. (The centre could be given as i.) A5. x = + sin θ dx = cos θ dθ θ = 0 x = ; θ = π /2 x = 2 cos θ ( + sin θ) dθ = 2 3 x dx 3 π/2 0 = 2 x 3 dx = x = 8 2 ( 8) = 3 * A6. x + y + 3z = 3x + ay + z = x + y + z =. Hence x + y + 3z = (a 3) y 8z = 2 2z = 2 2 When a 3, we can solve to give a unique solution. 6 z = ; y = a 3 ; x = a. 2E When a = 3, we get z = from the second equation but from the third, i.e. inconsistent 4 z =. 2 * optional off for lower triangular form for identifying the two values for z for conclusion Page 3

5 A7. f (x) = x + x 2 f (x) = ( + x2 ) 2x 2 ( + x 2 ) 2, = x2 ( + x 2 ) 2 f (x) = 0 x 2 = 0 x = ±. The graph of f (x) has two stationary values: (, 2) and (, 2) and passes through (0, 0). Thus g has two turning points (, 2) and (, 2) and its third critical value is (0, 0) {as its gradient is discontinuous}. A8. Statement A is true: p(n) =n(n + ) and one of n and (n + ) must be even. 3 Or: n 2 and n are either both odd or both even. In either case n 2 + n is even. * Statement B is false: when n =, n 2 + n = 2. A9. w = cosθ + i sinθ = cosθ i sinθ cosθ + i sinθ cosθ i sinθ = w k + w k = cos θ i sin θ = cos θ i sin θ cosθ i sinθ cos 2 θ i 2 sin 2 θ = w k + (w k ) = (cos θ + i sin θ) k + (cos θ + i sin θ) k = cos kθ + i sin kθ + cos kθ + i sin kθ = cos kθ + i sin kθ + cos kθ i sin kθ = 2 cos kθ (w + w ) 4 = w 4 + 4w w 2 + w 4 2E (2 cos θ) 4 = (w 4 + w 4 ) + 4 (w 2 + w 2 ) + 6, 6 cos 4 θ = 2 cos 4θ + 8 cos 2θ + 6 so cos 4 θ = 8 cos 4θ + 2 cos 2θ * First mark needs attempt at justification; alternative proofs (e.g. induction) are acceptable needs justifying. Page 4

6 A0. (a) I = 0 xe x dx = x e x dx. e x dx. dx 0 2E (b) 0 = [ xe x e x ] 0 = e e (0 ) ( = 2 = e 0.264).* x n e x dx = x n e x dx (nx n e x dx) dx = [ x n e x ] 0 + n xn e x dx = e ( 0)+n x n e x dx 0 = ni n e (c) I 3 = 3I 2 e = 3 (2I e ) e = 3 (2 4e e ) e = 6 6e E A. V (0) = 5, so dv dt = V (0 V) dv V (0 V) = dt 0 V + 0 V dv = dt 2 (ln V ln (0 V)) = t + C 0 0 ln V ln (0 V) = t + C 0 0 ln 5 0 ln 5 = 0 + C C = 0 ln V ln (0 V) = 0t V ln ( 0 V ) = 0t V 0 V = e0t V = 0e 0t Ve 0t V ( + e 0t ) = 0e 0t 0e0t 2E V = + e 0t V = 0e0t 0 = + e 0t e 0t + 0 as t. * optional [END OF MARKING INSTRUCTIONS] Page 5

7 2003 Mathematics Advanced Higher Section B Finalised Marking Instructions

8 B. Let then Advanced Higher 2003: Section B Solutions and marks x 3 4 = y 2 x = 3 + 4t = z + 2 = t y = 2 t 2E Thus z = + 2t. 2 (3 + 4t) +(2 t) ( + 2t) = t = 4 t = so the point is (, 3, 3). B2. A 2 = 4A 3I A 3 = 4A 2 3A = 6A 2I 3A = 3A 2I A 4 = 3A 2 2A = 52A 39I 2A i.e. p = 40, q = 39. Alternative = 40A 39I B3. Let λ represent a fixed point, then A 4 = (A 2 ) 2 = 6A 2 24A + 9I = 64A 48I 24A + 9I = 40A 39I* λ = 2 { λ + 7 λ} 2λ = λ + 7 λ λ = 7 λ λ 2 = 7. The fixed points are 7 and 7. * Using I = at any stage costs a mark. For full marks, exact values are needed. Iterative method giving ±2.645 gets 2 marks. Page 2

9 B4. either or f (x) = sin 2 x f (0) = 0 f (x) = 2 sin x cos x = sin 2x f (0) = 0 f (x) = 2 cos 2 x 2 sin 2 x = 2 cos 2x f (0) = 2 f (x) = 4 cos x sin x 4 sin x cos x = 4 sin 2x f (0) = 0 f (x) = 8 cos 2 x + 8 sin 2 x = 8 cos 2x f (0) = 8 2E f (x) = x + 2. x x x4 24 = x 2 3 x4 Since cos 2 x + sin 2 x =, cos 2 x = x 2 + 3x 4 OR sin x = x 3! x3 + (sin x) 2 = ( x 6 x3 + )( x 6 x3 + ) = x x4 + =, B5. (a) When n =, LHS = 0, RHS = 0 2 = 0. Thus true when n =. (b) k r = 3(r 2 r) =(k )k(k + ) k + Assume and consider the sum to. k + 3 (r 2 r) = 3 ((k + ) 2 (k + )) + k 3 (r 2 r) r = r = = 3 (k + ) 2 3 (k + ) +(k ) k (k + ) = (k + ) [3k k 2 k] = (k + ) (k 2 + 2k) = k (k + )(k + 2) ((k + ) ) (k + ) ((k + ) +). Thus true for k +. Since true for, true for all n (r 2 r) = 40 3 (r 2 r) 0 3 (r 2 r) r = r = r = = = = Page 3

10 B6. Consider first Auxiliary equations is d 2 y dx 2 4dy dx + 4y = 0. m 2 4m + 4 = 0 (m 2) 2 = 0 Thus the complementary function is y = (A + Bx) e 2x. To find the particular integral, let f (x) = ae x. Then f (x) = ae x and f (x) = ae x. ae x 4ae x + 4ae x = ae x so a =. Therefore the general solution is y = (A + Bx) e 2x + e x Initial conditions give dy dx = Be2x + 2 (A + Bx) e 2x + e x 2 = A + = B + 2A + i.e. A = and B = 2. The required solution is y = ( 2x) e 2x + e x. [END OF MARKING INSTRUCTIONS] Page 4

11 2003 Mathematics Advanced Higher Section C Finalised Marking Instructions

12 Advanced Higher 2003: Section C Solutions and marks C. P(Breast cancer Mammogram positive) = P(Breast cancer and Mammogram positive) P(Mammogram positive) = P(Mammogram positive Breast cancer)p(breast cancer) P(M+ BC).P(BC) + P(M+ BC ). P( BC ) = = = (= 0.083) * 2 C2. (a) X Bin (20, 0 25), (b) P (X 3) = (c) The hypothesis p = 0.25 cannot be rejected at the 5% significance level since the probability calculated in (b) exceeds Thus there is no evidence from the data to support the manager's belief. C3. (a) Y = 5 (X 32) 9 E (Y) = 5 (04 32) = 40 M, 9 V (Y) = ( 5 9) σ = 2 3 (b) For central 95% probability, z = 96. limits are 40 ± i.e. (38 7, 4 3) * optional for conclusion not sufficient, limits have to be evaluated Page 2

13 C4. (a) (b)(i) (ii) We require ˆ p 2 96 ± 96 ˆ p ( ˆ p) n, n ( =.8 *, n).8 n 0. n (.8 0.) 2 n 324 Thus a sample size of 324 or greater is required. C5. (a) z = x σ / n = 63 / 25 = 2 22 H 0 : µ = 502 H : µ > 502 The critical region is z > 2 33 Since 2 22 lies outside in the critical region, the null hypothesis is accepted i.e. there is no evidence of an increase (b) p-value = P (Z > 2 22) = which is greater that 0 0 confirming the conclusion (c) The central limit theorem guarantees that sample means will be approximately normally distributed so the test will still be valid [END OF MARKING INSTRUCTIONS] * optional Page 3

14 2003 Mathematics Advanced Higher Section D Finalised Marking Instructions

15 D. L(2 5) = Advanced Higher 2003: Section D Solutions and marks (2 5 4)(2 5 6) ( 3)( 5) (2 5 )(2 5 6) (3)( 2) (2 5 )(2 5 4) (5)(2) = = D2. f (x) = ln (3 + 2x) f (x) = Taylor polynomial is 2 (3 + 2x) f (x) = 4 (3 + 2x) 2 p (x) = p ( + h) = ln 5 2h 5 2h For ln 5 4, take f ( 2), h = 0 2; p( 2) = = Coefficient of h in Taylor polynomial is substantially smaller than. Hence f (x) is likely to be very insensitive to small changes in x near x =. D3. 2 f 0 = f f 0 = (f 2 f ) (f f 0 ) = f 2 2f + f 0 3 f 0 = (f 3 2f 2 + f ) (f 2 2f + f 0 ) = f 3 3f 2 + 3f f 0 2 Maximum error is ε + 3ε + 3ε + ε = 8ε. This occurs when f and f 3 have been rounded up and f 0 and f 2 rounded down by the maximum amount, or vice versa. D4. (a) Difference table is: i x f (x) diff diff2 diff (b) 2 f 3 = (c) Third degree polynomial would be suitable. (Differences are approximately constant (well within rounding error).) (d) p = 0 f (0 63) = (0 28)+ (0 )( 0 9) 2 (0 249)+ (0 )( 0 9)( 9) (0 020) 6 = = 96 3 (from f (0 3) with p =, f (0 63) = = 98). Page 2

16 D5. (a) x x 0 f (x)dx = 0 f (x 0 + ph)h dp = h 0 [f (x 0 )+f (x 0 )ph + 2f (x 0 )p 2 h 2 ] dp = h f (x 0 )p + f (x 0)hp 2 2 = h f (x 0 )+ f (x 0)h 2 + f (x 0)h 2 p 3 6 = h f (x 0 )+ 2 f (x ) 2 f (x 0) f (x 0)h 2 = h(f 0 + f ) 2 h3 f (x 0 ) 2 + f (x 0)h f (x 0)h 2 6 (using f (x )=f(x 0 )+hf (x 0 )+ 2h 2 f (x 0 )+ ) First term is trapezium rule; second term is principal truncation error. 5 (b) Trapezium rule calculation is: x f (x) m m f (x) π / π / π / π / π / Hence I = π / (c) f (x) =2 cosx x sinx whose magnitude has maximum on [π /4, π /2] at x = π /2since f (π/2) = π /2 =.57 and f (π/4) = and f (x) 0 on the interval. maximum truncation error = (π /6) 2 π /4 57 / Hence estimate for I is I = [END OF MARKING INSTRUCTIONS] Page 3

17 2003 Mathematics Advanced Higher Section E Finalised Marking Instructions

18 Advanced Higher 2003: Section E Solutions and marks E. E2. d 2 s (a) Given that = a, ds = at + c. dt 2 dt ds ds Since = U when t = 0, c = U. Thus = U + at. dt dt s = (U + at) dt = Ut + 2at 2 + c. When t = 0, s = 0 so c = 0, hence s = Ut + 2at 2. (b) Use s = 2gt 2. When s = H, t = 6 so H = 8g. Hence, when s = 2H 2gt 2 = 2H t 2 = 8 t = seconds B q A By the sine rule 40 f 30 sin θ 30 = sin sin θ = θ 5 3 Then φ = 80 ( ) = 34 7 and the required bearing is = E3. braking force (a) By Newton II so m d2 s dt 2 = 2m ( 4) + t d 2 s dt 2 = 2 ( 4) + t Page 2

19 ds Integrating gives = 2t t2 dt 4 + c. ds Since dt = 2 when t = 0, c = 2. So ds = 2 2t t2 dt 4. ( ) ds The car is stationary when dt = 0, i.e. when t 2 2t 2 = 0 4 (b) t 2 + 8t 48 = 0 (t + 2)(t 4) = 0 t = 4 (As t 0, the root 2 is ignored.) Integrating (*) gives s = 2t t 2 t3 2 The stopping distance is (as s (0) = 0) s (4) = ( ) = m. E4. R F (a) P 5 3 mg a a 4 Resolving perpendicular to the plane R = mg cos α = 4 5mg and hence F = µr = 4 5 µmg. Resolving parallel to the plane mg sin α = P + F = P µmg P = mg ( µ) = mg (3 4µ). 5 (b) As in (a) F = 4 5 µmg Resolving parallel to the plane 2P = 3 5 mg + 4 µmg, 5 P = mg (3 + 4µ). 0 To find µ, equate these expressions for P. 0 (3 + 4µ) = 5 (3 4µ) 3 + 4µ = 6 8µ µ = 4. Page 3

20 E5. (a) V = V (cos 45, sin 45 ) = V (, ). 2 From the equations of motion x = 0 x = V 2 t y = g y = V 2 t 2 gt2. 2x Substituting t =, gives V y = V 2 2x V 2 ( g 2x V ) 2 = x gx2 V. 2 (b) To hit A, we require y = h when x = 0h. 0h g (0h)2 V 2 = h 9h = g V 2 (0h)2 V 2 gh = 02 9 V = 0 3 gh. (b) At By< : h when x = h So h g (h)2 V 2 < h 0 < 0 3 < V gh < 0. gh 2 V 2 V 2 gh < 2 0 V gh < 0. [END OF MARKING INSTRUCTIONS] Page 4