2010 Maths. Advanced Higher. Finalised Marking Instructions

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1 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 basis. If it is to be used for any other purposes written permission must be obtained from the External Print Team, Centre Services, Dalkeith. 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 External Print Team, Centre Services, at Dalkeith 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 00. (6 (a For f (x e x sin x, f (x e x sin x + e x (x cos x. M using the Product Rule, one for each correct term (b Method x 3 For g (x, ( + tan x Method g (x g (x 3x ( + tanx x 3 sec x. ( + tanx M using the Quotient Rule first term and denominator second term g (x x 3 ( + tan x for correct rewrite 3x ( + tanx + x 3 ( ( + tanx sec x, x ( + tan x (3 + 3 tan x x sec x for accuracy. (5 Let the first term be a and the common ratio be r. Then ar 6 and ar 3 {both terms needed} Hence r ar 3 ar 6. evaluating r So, since r <, the sum to infinity exists. justification S a correct formula r ( 3 8. the sum to infinity

4 3. (7 (a t x 4 dt 4x 3 dx correct differential x 3 + x 8dx 4 4x 3 + (x 4 dx 4 + t dt 4 tan t + c 4 tan x 4 + c correct integral in t correct answer (b x ln x dx (ln x x dx ln x x dx x x3 3 dx 3x 3 ln x 3 x dx 3x 3 ln x 9x 3 + c M, for using integration by parts for differentiating ln x 4. (4 0 The matrix 0 scale factor. The matrix 3 gives an enlargement, 3 gives a clockwise rotation of 60 about the origin. M ( correct matrix correct matrix correct order 5. n + (4 ( 3 n 3 (n +! 3!(n! n! 3!(n 3! (n +! n!(n 3!(n! 3!(n! (n +! n!(n 3!(n! n! [(n + (n ] 3! (n! n! 3 3!(n! n!!(n! n both terms correct {alternative methods will appear} correct numerator correct denominator for knowing (anywhere (n! (n (n 3!

5 6. (4 i j k v w 3 4 i j 3 + k i j + 5k u. (v w ( i + 0j + 5k.(9i j + 5k M a valid approach 7. (6 3x + 5 (x + (x + (x + 3dx 3x + 5 (x + (x + (x + 3 A x + + B x + + C x + 3 3x + 5 A(x + (x B(x + (x C(x + (x + M x A A x B B x 3 4 C C Hence 3x + 5 (x + (x + (x + 3 x + + x + x + 3 3x + 5 (x + (x + (x + 3dx ( x + + x + x +3dx [ln (x + + ln (x + ln (x + 3] ln3 + ln4 ln5 ln ln3 + ln4 ln ln 3 5 for first correct coefficient for second correct coefficient for last coefficient and applying them for correct integration and substitution 8. (6 (a Write the odd integers as: n + and m + where n and m are integers. M for unconnected odd integers Then (n + (m + 4nm + n + m + (nm + n + m + demonstrating clearly which is odd. (b Let n, p p which is given as odd. Assume p k is odd and consider p k +. M p k + p k p Since p k is assumed to be odd and p is odd, p k + is the product of two odd integers is therefore odd. Thus p n + is an odd integer for all n if p is an odd integer. for a valid explanation from a previous correct argument

6 9. (4 Let f (x ( + sin x. Then f (0 f (x sin x cos x f (0 0 sin x f (x cos x f (0 f (x 4 sin x f (0 0 f (x 8 cos x f (0 8 one for each non-zero term f (x + x! 8x4 4! + + x 3 x4 + Alternative f (0 f (x sin x cos x f (0 0 f (x cos x sin x f (0 f (x 4( sinx cosx f (0 0 4 cos x sin x f (x 8 cos x + 8 sin x f (0 8 etc Alternative f (x ( + sin x + cos x (3 cos x (3 ( (x! + (x4 4! (3 + x 3x 4 + x 3x 4 introducing cosx expanding cosx simplifying finishing 0. (3 The graph is not symmetrical about the y-axis (or f (x f ( x so it is not an even function. The graph does not have half-turn rotational symmetry (or f (x f ( x so it is not an odd function. The function is neither even nor odd. {apply follow through}

7 . (7. (4 d y dx + 4dy dx + 5y 0 m + 4m (m + m ± i The general solution is y e x (A cos x + B sin x x 0, y 3 3 A x π, y e π e π e π (3 cos π + Bsin π B The particular solution is: y e x (3 cosx + sinx. Assume + x is rational and let + x p where p, q are integers. q M appropriate CF for accuracy So x p q p q q as a single fraction Since p q and q are integers, it follows that x is rational. This is a contradiction. 3. (0 y t 3 5 t dy dt 3t 5t x t t / dx dt t / dy dx 3t 5t dx d d y dx dt ( dy dx dt t / 6t 5/ 0t 3/ 6 t 5 3/ 0 t 3 / t / 30t 30t i.e. a 30, b 30 for eliminating fractions M d At a point of inflexion, y dx 0 t 0 or But t > 0 t dy dx 4 the value of the gradient and the point of contact is (, 3 Hence the tangent is y (x i.e. y + 8x 5

8 4. (0 y + 0 a 0 0 a a 5 z for a structured approach for triangular form ; one correct variable a 5 a + 5 y a 5 a 5 y a a 5 ; x a a 5 + a 5 x a 5 a 5 + a a 5 a 4 a 5. for the two other variables {other justifications for uniqueness are possible} which exist when a 5 0. From the third row of the final tableau, when a 5, there are no solutions When a 3, x, y, z. ( ( C( ( 5 3 AB 0 3 From above, we have 0 and A( also 0 which suggests AC I and this can be verified directly. Hence A is the inverse of C (or vice versa. A candidate who obtains AC I directly may be awarded full marks.

9 5. (x 8x x 4 8x x 0, values of x Area 4 0 ( 8x x dx 4 8 ( 3 x3/ 3 x M Volume of revolution about the y -axis π x dy. M So in this case, we need to calculate two volumes and subtract: V π [ 4 0 y dy] π [ dy] y4 4 π y π y π π 5 ( 4π 5 ( 5 4, 4 0 the rest each term 6. (0 z 3 r 3 (cos 3θ + i sin 3θ (cos π 3 + i sin π 3 3 cos π + i sin π necessary Method a ; b 0 r 3 (cos 3θ + i sin 3θ 8 r 3 cos 3θ 8 and r 3 sin 3θ 0 r ; 3θ 0, π, 4π Roots are, (cos π 3 + i sin π 3, (cos 4π 3 + i sin 4π 3. In cartesian form:, ( + i 3, ( i 3 Method z (z (z + z (z ((z + + ( 3 0 so the roots are:, ( + i 3, ( i 3 or by using quadratic formula (a z + z + z 3 0 (b Since z 3 z 3 z it follows that z 6 + z 6 + z 6 3 (z 3 + (z 3 + (z END OF SOLUTIONS

2011 Maths. Advanced Higher. Finalised Marking Instructions

2011 Maths. Advanced Higher. Finalised Marking Instructions 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.