X100/701 MATHEMATICS ADVANCED HIGHER. Read carefully

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1 X/7 N A T I O N A L Q U A L I F I C A T I O N S 9 T H U R S D A Y, M A Y. P M. P M MATHEMATICS ADVANCED HIGHER Read carefully. Calculators may be used in this paper.. Candidates should answer all questions. 3. Full credit will be given only where the solution contains appropriate working. L I X / 7 6 / 9 *X/7*

2 Answer all the questions. Marks. (a) Given f(x) = (x + )(x ) 3, obtain the values of x for which f (x) =. (b) Calculate the gradient of the curve defined by x x y 5 at the point y + = (3, ). 3 ( ) t + 3t. Given the matrix A =. 3 5 (a) Find A in terms of t when A is non-singular. (b) Write down the value of t such that A is singular. ( ) 6 3 (c) Given that the transpose of A is, find t Given that y dy x e = and y = when x =, find y in terms of x.. Prove by induction that, for all positive integers n, n =. r( r + ) n + r = 5 5. Show that ln ln 3 e e x x + e e x x = 9 ln Express z = ( + i) in the form a + ib where a and b are real numbers. 7 i Show z on an Argand diagram and evaluate z and arg (z). 6 [X/7] Page two

3 7. Use the substitution x = sin θ to obtain the exact value of (Note that cos A = sin A.) x x. Marks 6 8. (a) Write down the binomial expansion of ( + x) 5. (b) Hence show that. 9 5 is Use integration by parts to obtain the exact value of x tan x. 5. Use the Euclidean algorithm to obtain the greatest common divisor of 36 and 65, expressing it in the form 36a + 65b, where a and b are integers.. The curve y = x x + dy is defined for x >. Obtain the values of y and at the point where x =. 5. The first two terms of a geometric sequence are a = p and a = p. Obtain expressions for S n and S n in terms of p, where Given that S n = 65S n show that p n = 6. = aj. Given also that a 3 = p and that p >, obtain the exact value of p and hence the value of n. S k k j=,, 3. The function f(x) is defined by f( x) = x + x ( x ± ). x Obtain equations for the asymptotes of the graph of f(x). Show that f(x) is a strictly decreasing function. Find the coordinates of the points where the graph of f(x) crosses (i) the x-axis and (ii) the horizontal asymptote. Sketch the graph of f(x), showing clearly all relevant features. 3 3 [Turn over for Questions to 6 on Page four [X/7] Page three

4 . Express x + 6x in partial fractions. ( x + ) ( x ) Hence, or otherwise, obtain the first three non-zero terms in the Maclaurin expansion of x + 6x. ( x + ) ( x ) Marks 5 5. (a) Solve the differential equation dy ( x + ) 3 y = ( x + ) (b) given that y = 6 when x =, expressing the answer in the form y = f(x). Hence find the area enclosed by the graphs of y = f(x), y = ( x) and the x-axis (a) Use Gaussian elimination to solve the following system of equations (b) x + y z = 6 x 3y + z = 5x + y z =. Show that the line of intersection, L, of the planes x + y z = 6 and x 3y + z = has parametric equations 5 x = λ y = λ z = 5λ. (c) Find the acute angle between line L and the plane 5x + y z =. [END OF QUESTION PAPER] [X/7] Page four

5 9 Mathematics Advanced Higher Finalised Marking Instructions! Scottish Qualifications Authority 9 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 Question Paper Operations Team, 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 Question Paper Operations Team 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.

6 Solutions for AH maths 9. (a) f (x) = (x + ) (x ) 3 f (x) = (x ) (x + ) (x ) = (x ) ((x ) + 3 (x + )) = (x ) (x + ) = when x = and when x =. (b) Method Note: a candidate may obtain Method Mathod 3 x y + x = y 5 x + xy = y 5y x + x dy + y = ydy 5dy dy = dy 5dy 5 = dy dy = dy = x + y y x 5 x y + x = y 5 xy x dy y 6 9 dy + = dy + = dy 6 9 dy + = dy dy = x y ( + x = y 5 x x y ( + x dy y) + = dy and then substitute. y) + x = y dy + = dy dy = dy xy + y Note: a candidate may obtain = y (in and 3) and then + x substitute. E E E E E

7 . (a) t + 3t det ( 3 5 ) = 5 (t + ) 9t = t A 5 3t = t ( 3 t + ), (b) t = t = 5 (c) t + 3 ( ) ( ) = 6 3 t = 3t e y x dy = e y dy = x y = when x = so ey dy = x e y = x + c = + c c = e y = x y = ln ( x). When n =, LHS =, RHS. So true when = = = n =. Assume true for n = k, k. r = r (r + ) = k + Consider n = k + k + r = r (r + ) = k r = = = r (r + ) + (k + ) (k + ) k + + (k + ) (k + ) k + (k + ) (k + ) = k + (k + ) ((k + ) + ) = ((k + ) + ) Thus, if true for n = k, statement is true for n = k +, and, since true for n =, true for all n.

8 5. Method ln e x + e x ln 3 e x e x Let u = e x e x, then du = (e x + e x ). When x = ln 3, u = 3 3 = 6 5 and when x = ln, u = = 3. ln ln 3 e x + e x e x e = 3/ du x 5/6 u = [ln u] 3/ 5/6 = ln 3 ln 5 6 = ln 9 5 Method ln ln 3 e x + e x e x e x = [ln (ex e x )] ln ln 3, = ln ( ) ln ( 3 3) = ln 3 ln 5 6 = ln ( + i) 7 i = + i 7 i 3 + i = 7 + i 7 i 7 + i ( 3 + i) (7 + i) = 5 = + i z = + = arg z = tan = tan ( ) = 3π (or 35 ).

9 7. x = sin θ = cos θ dθ x = θ = ; x = sin θ = θ = π x x = π/ = π/ sin θ ( cos θ) dθ sin θ sin θ cos θ = π/ ( sin θ) dθ ( cos θ) dθ = π/ ( cos θ) dθ = θ sin θ π/ = { π } = π 8. (a) ( + x) 5 = + 5x + x + x 3 + 5x + x 5 (b) Let x =, then 9 5 = ( + ( )) 5 = = = x tan x = tan x x = x tan x = x tan x x x, + x x 3 + x ln ( + ) x = tan ln ln = π 8 ln,

10 . 65 = = = 3 3 = = 36 9 (65 36) = When x =, y =. y = x x + ln y = ln (x x + ) = (x + ) ln x dy y = x + x + x ln x, Hence, when x =, y = and dy = 3 + = 3.. a j = p j S k = p + p + + p k = p (pk ) p S n = p (pn ) p S n = p (pn ) p p (p n ) p = 65p (pn ) p (p n + ) (p n ) = 65 (p n ) p n + = 65 p n = 6 a = p a 3 = p 3 but a 3 = p so p 3 = p p = p = since p >. p n = 6 = 6 = ( ) n =

11 3. f (x) = x + x x = x + x (x ) (x + ) Hence there are vertical asymptotes at x = and x =. f (x) = x + x x = + x x x = + x x as x. So y = is a horizontal asymptote. Hence f (x) f (x) = x + x x f (x) = (x + ) (x ) (x + x) x (x ) = x3 x + x x 3 x (x ) = (x + x + ) (x ) = ((x + ) + 3 ) < (x ) is a strictly decreasing function. f (x) = x + x x = x = or x = f (x) = x + x x = x + x = x x = for shape for position Alternatively for the horizontal asymptote: x x + x x x + f (x) = + x + x as x

12 . x + 6x (x + ) (x ) = A (x + ) + B x + + C x x + 6x = A(x ) + B(x + ) (x ) + C (x + ) Let x = then = 6A A =. Let x = then 6 + = 36C C =. Let x = then = A 8B + C = 8 8B + B =. Thus x + 6x (x + ) (x ) = (x + ) + x. Let f (x) = (x + ) + (x ) then f (x) = (x + ) + (x ) f () = = f (x) = (x + ) 3 (x ) f () = 6 = 6 9 f (x) = (x + ) + (x ) 3 f () = 3 3 = 3 M 3 Thus x + 6x (x + ) (x ) = 9x 6 + 3x 6 + E

13 5. (a) (x + ) dy 3y = (x + ) dy 3 x + y = (x + )3 Integrating factor: since 3 = 3 ln (x + ). x + Hence the integrating factor is (x + ) 3. dy (x + ) 3 3 (x + ) y = d ((x + ) 3 y) = y (x + ) 3 = = x + c y = 6 when x =, so = + c c =. Hence y = (x + ) (b) (x + ) = ( x) x + = x x = or x + = + x which has no solutions. y = f (x) y = ( x) Area = (x + ) + ( x) M = (x + ) = 5 [(x + )5 ] = 5 = 5

14 6. (a) x + y z = 6 x 3y + z = x + y z = ,, z = 7 ( 7 5 ) = 5 5y = y = x + 5 = 6 x = 3 (b) Let x = λ. Method In first plane: x + y z = 6. λ + (λ ) (5λ ) = 5λ 5λ + 6 = 6. In the second plane: x 3y + z = λ 3(λ ) + (5λ ) = 5λ 5λ + =. Method Method y z = 6 λ y = 6 + z λ 3y + z = λ ( 8 3z + 3λ) + z = λ z = 5λ z = 5λ and y = λ x + y z = 6 () x 3y + z = () 5x z = () + 3 () x y = () + () y = x z = 5x x = λ, y = λ, z = 5λ (c) Direction of L is i + j + 5k, direction of normal to the plane is 5i + j k. Letting θ be the angle between these then cos θ = M, 5 = 7 3 This gives a value of 3. which leads to the angle 3. 9 = 3..,

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.