2011 Maths. Advanced Higher. Finalised Marking Instructions

Size: px

Start display at page:

Download "2011 Maths. Advanced Higher. Finalised Marking Instructions"

Arleen Sanders
5 years ago
Views:

1 0 Maths Advanced Higher Finalised Marking Instructions Scottish Qualifications Authority 0 The information in this publication may be reproduced to support SQA qualifications only on a noncommercial basis. If it is to be used for any other purposes written permission must be obtained from SQA s NQ Delivery: Exam Operations Team. Where the publication includes materials from sources other than SQA (secondary copyright), this material should only be reproduced for the purposes of examination or assessment. If it needs to be reproduced for any other purpose it is the centre s responsibility to obtain the necessary copyright clearance. SQA s NQ Delivery: Exam Operations Team may be able to direct you to the secondary sources. These Marking Instructions have been prepared by Examination Teams for use by SQA Appointed Markers when marking External Course Assessments. This publication must not be reproduced for commercial or trade purposes.

2 General Marking Principles These principles describe the approach taken when marking Advanced Higher Mathematics papers. For more detailed guidance please refer to the detailed Marking Instructions. The main principle is to give credit for the skills demonstrated and the criteria met. Failure to have a correct method may not preclude a candidate gaining credit for their solution. The answer to one part of a question, even if incorrect, can be accepted as a basis for subsequent dependent parts of the question. 3 The following are not penalised: working subsequent to a correct answer (unless it provides firm evidence that the requirements of the question have not been met) legitimate variation in numerical values / algebraic expressions. 4 Full credit will only be given where the solution contains appropriate working. Where the correct answer might be obtained by inspection or mentally, credit may be given. 5 Sometimes the method to be used in a particular question is explicitly stated; no credit will be given where a candidate obtains the correct answer by an alternative method. 6 Where the method to be used in a particular question is not explicitly stated in the question paper, full credit is available for an alternative valid method. (Some likely alternatives are included but these should not be assumed to be the only acceptable ones.) In the detailed Marking Instructions which follow, marks are shown alongside the line for which they are awarded. There is one code used M. This indicates a method mark, so in question (a), M means a method mark for the product rule.

3 Advanced Higher Mathematics 0 Marks awarded for. (5) 3 x x + 4x 5 = 3 x (x ) (x + 5) = A x + B x x = A(x + 5) + B(x ) x = = 6A A = for first value x = 5 8 = 6B B = 3 for second value 3 x Hence x + 4x 5 = x 3 x x x + 4x 5 = x 3 x + 5 = ln x 3 ln x c for logs for moduli. (3) ( x 3) 4 = 4 C 0 ( ) x C ( ) x 3 ( 3) + 4 C ( ) x ( 3) + 4 C 3 ( ) x ( 3) C 4 ( 3) 4 = ( ) x ( ) x 3 ( 3) + 6 ( ) x ( 3) +4 ( ) x ( 3) 3 + ( 3) 4 for powers for coefficients = x4 6 3x3 + 7x 54x + 8. for simplifying 3. (6) (a) Method y + e y = x + ey = x ( + ) ey = x = Method x ( + e y ) M ln (y + e y ) = ln x ( + e y ) y + e y = x = (y + ey ) x ( + e y ) Method 3 y + e y = x e y = x y y = ln(x y) = x x y (x y) = x = (x y + ) = x x x y + for applying implicit differentiation for accuracy

4 3. (b) Method f (x) = sin x cos 3 x f (x) = cos 4 x + sin x ( 3 cos x sin x) = cos 4 x 3 cos x sin x M for using the product rule for first term for second term Method f (x) = sin x cos 3 x ln (f (x)) = ln sin x + ln (cos 3 x) f (x) = cos x f (x) sin x 3 cos x sin x cos 3 x = cos x sin x 3 sin x cos x M f (x) = ( cos x sin x 3 sin x cos x ) sin x cos3 x = cos 4 x 3 sin x cos x 4. (6) (a) Singular when the determinant is 0. 0 λ λ det ( ) det ( ) + ( ) det ( ) = 0 M λ (8 + ) (3λ 0) = 0 5λ 40 = 0 when λ = 8 ) ( 3α + β (b) From A, A = α β 4 3. for transpose 3 Hence α β = and 3α + β = 5. 4α β = 3α + β = 5 7α = 7 α = and β =. for both values

5 5. (6) Let f (x) = ( + x), then f (x) = ( + x) f (0) = f (x) = ( + x) f (0) = f (x) = 4 ( + x) 3 f (0) = 4 f (x) = 8 3 ( + x) 5 f (0) = 8 3 Hence { for derivatives for values + x = + x 4 x x3 6 = + x x 8 + x3 6 and replacing x by x gives Thus ( + + x = + x x4 8 + x6 6 ( + x) ( + x ) = x x 8 + x3 6 )( + x x4 8 + x6 6 ) M for multiplying = + x + x 8 x + 4 x3 + 6 x3 + = + x x x (4) Reflect in the line y = x to get y (a, 0) (0, ) x for position for coordinates Now apply the modulus function y ( 0, ) x for shape for coordinates ( a, 0)

6 7. (4) Method y = esin x ( + x) 3 x ln y = ln (e sin x ( + x) 3 ) ln ( x) M for use of logs = sin x + 3 ln ( + x) ln ( x) for preparing to differentiate y = cos x x + ( x) ( = y cos x x + ( x)) When x = 0, y = 8 gradient = 8 ( ) = 4. for final value Method y = esin x ( + x) 3 x = d (e sin x ( + x) 3 ) x e sin x ( + x) 3 ( ( x) x) = [cosx esin x ( + x) 3 + 3e sin x ( + x) ] ( x) ( x) 3/ M + esin x ( + x) 3 ( x) 3/ When x = 0, gradient = (3 + 3 ) + 3 = = 4 Method 3 y = esin x ( + x) 3 x y x = e sin x ( + x) 3 x y( x) / = cosx e sin x ( + x) 3 + 3e sin x ( + x), when x = 0, y = e0 3 = 8. This leads to = 4 M for use of quotient rule

7 8. (4) n r = r 3 ( n r = r) = n (n + ) 4 n r 3 + ( n r) = n (n + ) r = r = 4 = n (n + ) 4 = n (n + ) + n (n + ) 4 n(n + ) ( ) = 0 n(n + ) + ( ) Marks awarded for 9. (5) Method = 3 ( + y) + x + y = 3 ( + x) M separating variables ln ( + y) = ( + x) 3 + c for LHS for term in x + y = exp ( ( + x) 3 + c) for the constant y = exp ( ( + x) 3 + c). = A exp ( ( + x) 3 ). Method 3 + x y = 3 + x Integrating Factor exp ( 3 + x ) = exp ( ( + x) 3/ ) d (y exp ( ( + x) 3/ )) = y (exp ( ( + x) 3/ )) = 3 + x (exp ( ( + x) 3/ )) ( 3 + x) exp ( ( + x) 3/ )) = exp ( ( + x) 3/ ) + c y = + c exp ( ( + x) 3/ ) 0. (5) Let z = x + iy, so z = (x ) + iy. z = (x ) + y = 9. The locus is the circle with centre (, 0) Can subsume the first two and radius 3. marks. y x for circle for shading or other indication

8 . (7) (a) π/4 0 (secx x)(secx + x) = π/4 0 (sec x x ) = tan x x3 3 0 = π [0] π 4 = π3. Exact value only 9 (b) Method Let u = 7x, M then du = 4x. x 49x = 4 4 du u = 4 sin u + c = 4 sin 7x + c must be in terms of x Method x 49x = 4 4 4x (7x ) for fraction for numerator for (7x ) = 4 sin 7x + c must be in terms of x. (5) For n =, = 64 + = 65. True when n =. Assume true for k, i.e. that 8 k + 3 k is divisible by 5, i.e. can be expressed as 5p for an integer p. for the inductive hypothesis Now consider 8 k k = 8 8 k + 3 k = 8 (5p 3 k ) + 3 k for replacing 8 k = 40p 3 k (8 3) = 5 (8p 3 k ) which is divisible by 5. So, since it is true for n =, it is true for all n.

9 3. (9) Method Let d be the common difference. Then u 3 = = a + d and u = a = a + d = a + ( a a ) a = a + a a + a = 0 (a + )(a ) = 0 a = sincea < 0. a = gives d = 3 and hence d = 3. Method u = a, u =, a u 3 = M a a = a a = a a + a = 0 (a + )(a ) = 0 a = sincea < 0. d = u 3 u = a = 3 S n = n [a + (n ) d] = n n 3 = 4 [3n n] 3n n > 4000 n 3 n > ( n 6 ) > n 6 > 48 6 n > = So the least value of n is 39. for value for suitable justification

10 4. (0) Auxiliary equation m m = 0 (m ) (m + ) = 0 m = or Complementary function is: y = Ae x + Be x The particular integral has the form y = Ce x + D y = Ce x + D = Cex Hence we need: d y d y = Cex y = ex + [Ce x ] [Ce x ] [Ce x + D] = e x + Ce x D = e x + Hence C = and D = 6. So the General Solution is y = Ae x + Be x e x 6. x = 0 and y = 3 A + B 6 = 3 x = 0 and = Setting up the equations A + B = 3B 7 = B = A = 3 So the particular solution is y = 3e x + e x e x 6.

11 5. (0) (a) In terms of a parameter s, L is given by x = + ks, y = s, z = 3 + s In terms of a parameter t, L is given by x = 4 + t, y = 3 + t, z = 3 + t Equating the y coordinates and equating the z coordinates: s = 3 + t 3 + s = 3 + t} Adding these 3 = 6 + 3t t = s =. From the x coordinates + ks = 4 + t Using the values of s and t + k = 5 k = The point of intersection is: (5,, ). (b) L has direction i j + k. L has direction i + j + k. For both directions. Let the angle between L and L be θ, then (i j + k). (i + j + k) cos θ = i j + k i + j + k = + = = θ = 60 The angle between L and L is 60.

12 6. () (a) I n = 0 ( + x ) n = 0 ( + x ) n for showing that is integrated = [( + x ) n ] 0 + 0(nx( + x ) n ) = [x( + x ) n ] 0 + 0nx ( + x ) n = 0 + n x ( + x ) n n 0 = n + n 0 x ( + x ) n +. (b) A ( + x ) + B n ( + x ) = x n + ( + x ) n + A( + x ) + B = x A =, B = ( + x ) + n ( + x ) = x n+ ( + x ) n+ (*) I n = + n x n 0 ( + x ). n + = + n n 0 ( + x ) + n n 0 ( + x n + ) Using (*) (c) 0 = + ni n n ni n + Recognising I n and I n + ni n + = n + (n ) I n I n + = n ( + n n + n ) I n. ( + x ) 3 = I 3 = I = ( 4 + I ) = x = [tan x] 0 = π 8 4 = 4 + 3π 3. END OF SOLUTIONS

2010 Maths. Advanced Higher. Finalised Marking Instructions

2010 Maths. Advanced Higher. Finalised Marking Instructions 00 Maths Advanced Higher Finalised Marking Instructions Scottish Qualifications Authority 00 The information in this publication may be reproduced to support SQA qualifications only on a noncommercial