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Mathematics Higher Prelim Eamination 04/05 Paper Assessing Units & + Vectors NATIONAL QUALIFICATIONS Time allowed - hour 0 minutes Read carefully Calculators may NOT be used in this paper. Section A - Questions - 0 (40 marks) Instructions for the completion of Section A are given on the net page. For this section of the eamination you should use an HB pencil. Section B (0 marks). Full credit will be given only where the solution contains appropriate working.. Answers obtained by readings from scale drawings will not receive any credit. Pegasys 04 Current Higher Units and + Vectors

Read carefully Check that the answer sheet provided is for Mathematics Higher Prelim 04/05 (Section A). For this section of the eamination you must use an HB pencil and, where necessary, an eraser. Make sure you write your name, class and teacher on the answer sheet provided. 4 The answer to each question is either A, B, C or D. Decide what your answer is, then, using your pencil, put a horizontal line in the space below your chosen letter (see the sample question below). 5 There is only one correct answer to each question. 6 Rough working should not be done on your answer sheet. 7 Make sure at the end of the eam that you hand in your answer sheet for Section A with the rest of your written answers. Sample Question A line has equation y 4. If the point (k,7) lies on this line, the value of k is A B 7 C 5 D The correct answer is A. The answer A should then be clearly marked in pencil with a horizontal line (see below). Changing an answer If you decide to change an answer, carefully erase your first answer and using your pencil, fill in the answer you want. The answer below has been changed to D. Pegasys 04 Current Higher Units and + Vectors

FORMULAE LIST Circle: The equation y g fy c 0 represents a circle centre ( g, f ) and radius g f c. The equation ( a) ( y b) r represents a circle centre ( a, b ) and radius r. Trigonometric formulae: sin A B cos A B sina cosa sin Acos B cos Asin B cos Acos B sin Asin B sin Acos A cos A sin A cos A sin A Scalar Product: a. b a b cosθ, where θ is the angle between a and b. or a.b a b a b a b where a a a a and b b b b Table of standard derivatives: f () f () sin a cos a acos a asin a Table of standard integrals: f () f () d sin a cos a C a cos a sin a C a Pegasys 04 Current Higher Units and + Vectors

SECTION A ALL questions should be attempted. The remainder when is divided by ( ) is A 9 B 7 C D. A sequence is defined by the recurrence relation U n au n b with U 0 4. An epression in terms of a and b for U is A B 4 a ab 6a 5b C a b 4 D 4 a ab b. Any line perpendicular to the line with equation y 6 5, has as its gradient A B 6 C 5 D 4. The rate of change of the function f ( ) when is A B C D Pegasys 04 Current Higher Units and + Vectors

5. Part of the graph of the function y f () is shown opposite. y o Which of the following graphs represents the related function y f ( ) 4? -4 y f () A y B y 8 o (,4) (, 4) o C y o (, 4) D (,4) y o 6. Given that sin A, the eact value of A cos is A B 7 9 C D 9 Pegasys 04 Current Higher Units and + Vectors

7. The minimum turning value of the function f ( ) is A 6 B 0 C 6 D - ( ) 8. A vector v is given by 6 What is the length, in units, of v? A 7 B C 5 D 49 9. 4 ( ) d is 0 A B C 8 D 0 0. For the equation k 0 to have real roots, k must take the values A 6 B 6 k 6 C k 6 or k 6 D k 6 or k 6 Pegasys 04 Current Higher Units and + Vectors

. Part of the graph of the curve y f () is shown below. y (, 4) O Which of the following is/are true for this function f?. f ( 0) 0. f ( ) 0. f ( ) 0 4. f ( 4) 0 A B C D only, and 4 only and only and 4 only. The equation of a circle, centre (, ), with the -ais as a tangent is A ( ) ( y ) 9 B ( ) ( y ) 4 C ( ) ( y ) 4 D ( ) ( y ) 9. If cos( ) sin 40 for 0 80 then has the value A 0 B 60 C 0 D 50 Pegasys 04 Current Higher Units and + Vectors

4. A recurrence relationship is defined as U 0 U b n 4 n. If the limit of the recurrence relationship is 50, the value of b is A 0 B 0 C D 5 5. Given that the points S(-4, 5, ), T(-6, -4, 6) and U(-4, -0, 6) are collinear, calculate the ratio in which T divides SU. A : B : C : 5 D : 5 4 6. The function f is defined as f ( ),. The value of f ( f ( )) equals A 8 B 8 C D 7. f ( ) a has a stationary value when. The value of a is A 4 B C 0 D 5 Pegasys 04 Current Higher Units and + Vectors

8. O is the centre of a circle of radius 4cm. Angle POQ is 0 5 radians. Q The area of the shaded sector POQ in square centimetres is O 4cm P A B C 4 D 8 9. If f ( ) where 0, then the composite function f ( f ( )) is equal to A B C D 0 0. The diagram opposite is made from 4 semi-circular arcs each of radius centimetre. The resulting shape has two aes of symmetry as shown by the dotted lines. The area of the shaded shape in square centimetres is A B 4 C D 4 [ END OF SECTION A ] Pegasys 04 Current Higher Units and + Vectors

SECTION B ALL questions should be attempted. Triangle ABC has B(8,) as one of its vertices. The line through A which meets side BC at D has as its equation y. y B(8,) A D O C (a) If side BC has a gradient of, find the equation of BC. (b) Establish the coordinates of D. (c) If D is in fact the mid-point of side BC, write down the coordinates of C. (d) Show clearly that this triangle is right-angled at B. (e) Hence find the equation of the circle passing through the points A, B and C.. Angle is acute and is such that tan. 4 (a) Show clearly that the eact value of sin is. (b) Hence show that sin 9 4. 4 Pegasys 04 Current Higher Units and + Vectors

. A sequence is defined by the recurrence relation U n pu n q, where p and q are constants. (a) Given that U 0 0, U and q 7, find the value of p. (b) Given that S U U U, calculate the value of S. 4. The small bo below is in the shape of a cuboid. k k The bo has dimensions, k and as shown. All the lengths are in centimetres. The volume of the bo is 60 cubic centimetres. (a) By forming an equation for the volume of the bo, and simplifying it, show that the following equation can be formed ( k k 0) 0. (b) Hence find the value of k, given that the above equation has equal roots. 4 [ END OF SECTION B ] [ END OF QUESTION PAPER ] Pegasys 04 Current Higher Units and + Vectors

Mathematics Higher Prelim Eamination 04/05 Paper Assessing Units & + Vectors NATIONAL QUALIFICATIONS Time allowed - hour 0 minutes Read carefully. Calculators may be used in this paper.. Full credit will be given only where the solution contains appropriate working.. Answers obtained from readings from scale drawings will not receive any credit. Pegasys 04 Current Higher Units and + Vectors

ALL questions should be attempted. The diagram below, which is not drawn to scale, shows part of the graph of the curve y 6. y A B O y 6 (a) (b) Find the coordinates of the point A, the maimum turning point of this curve. 4 The line through A, with gradient 5, intersects the curve at a further two points, one of which is B. Find algebraically the coordinates of B. Your answer must be accompanied with the appropriate working. 5. Solve algebraically the equation cos sin 0, for 0 60. 5 Pegasys 04 Current Higher Units and + Vectors

dy 9. A curve has as its derivative. d (a) Given that the point (, 6) lies on this curve, epress y in terms of. 4 (b) Hence find p if the point T(, p) also lies on this curve. (c) Find the equation of the tangent to this curve at T. 4. The circle below has as its equation y 8 6y 4 0. QR is a tangent to the circle. y R Q o y 8 6y 4 0 P(,-8) (a) Prove that the point P(,-8) lies on the circumference of the circle. (b) (c) Hence find the equation of the tangent to the circle at the point Q, where PQ is a diameter of the circle.. 5 Find the length of the line QR. Give your answer correct to - decimal place. Pegasys 04 Current Higher Units and + Vectors

5. A is the point (, -, 0), B is (, -, ) and C is (4, k, 0). (a) (i) Epress BA and BC in component form. (ii) Show that cos ABC = ( k 6k 4) 7 (b) If angle ABC = 0, find the possible values of k. 5 6. The diagram below shows a sketch of part of the graphs of y 4 8. The curves intersect at the point (0,) and at A. y 6 8 and The dotted line shown is parallel to the -ais and passes through (0,) and A. y A o (a) Establish the coordinates of the point A. (b) Hence calculate the area enclosed between the two curves. 5 Pegasys 04 Current Higher Units and + Vectors

7. Two functions in are defined on suitable domains as f ( ) ( a) and g ( ) a, where a is a constant. Given that f ( g( a )) 9, find the value of a where a 0. 4 k 8. (a) Given that 4 d show clearly that the eact value of k is 8 4. 4 0 (b) Hence show that sin k cos k can be written in the form p and state the value of p. [ END OF QUESTION PAPER ] Pegasys 04 Current Higher Units and + Vectors

Mathematics Higher Prelim Eamination 04/05 Paper - Section A - Answer Sheet NATIONAL QUALIFICATIONS NAME : CLASS : TEACHER : You should use an HB pencil. Erase all incorrect answers thoroughly. Indicate your choice of answer with a single mark as in this eample A B C D A B C D Section A 40 4 5 6 Section B 0 7 8 9 0 Total (P) 70 4 5 6 7 Total (P) 60 8 9 0 Overall Total 0 % Please make sure you have filled in all your details above before handing in this answer sheet. Pegasys 04 Current Higher Units and + Vectors

Higher Grade - Paper 04/05 ANSWERS - Section A C D A 4 D 5 B 6 A 7 C 8 B 9 B 0 D D C A 4 B 5 B 6 A 7 A 8 C 9 D 0 B 4 5 6 7 8 9 0 4 5 6 7 8 9 0 A B C D Pegasys 04 Current Higher Units and + Vectors

Higher Prelim Paper 04/05 Give mark for each Marking Scheme Illustration(s) for awarding each mark (a) ans: y + = 7 ( marks) substitutes info into equation y = ( 8) rearranges y + = 7 (b) ans: D(, ) ( marks) knows to use system of equations evidence solves for one variable y = solves for other variable and states point D = ; D(, ) (c) ans: C(6, 5) ( mark) establishes coordinates of C C(6, 5) (d) ans: proof ( marks) establishes coordinates of A when = 0; y = 7 finds gradients of AB and BC 7 5 m AB ; mbc 8 0 8 6 reason since m AB m BC ; AB is perp to BC so triangle is right angled at B (e) ans: ( 8)² + (y )² = 0 ( marks) knows AC is diameter and finds midpoint midpoint AC = (8, ) [centre] finds radius (8, ) to (0, 7) = (8² + 6²) = 0 subs into equation of circle ( 8)² + (y )² = 00 (a) ans: proof ( marks) interprets information triangle drawn or any acceptable method finds hypotenuse ( ) 4 8 finds sin and simplifies sin (b) ans: proof (4 marks) finds cos 4 cos substitutes values in epansion of sin 4 sin starts to simplify 8 sin 9 4 rationalises denominator to answer 4 8 8 4 9 8 9 Pegasys 04 Current Higher Units and + Vectors

Give mark for each Illustration(s) for awarding each mark (a) ans: p = 0 5 ( marks) substitutes values 0p 7 solves for p p 0 5 (b) ans: 5 ( marks) finds U U = 0 finds U U = 9 5 totals 5 4(a) ans: proof ( marks) subs values 60 k( ) rearranges to answer k 4k + 60 = 0. (b) ans: 5 k (4 marks) 6 knows condition for equal roots b 4ac 0 for equal roors identifies a, b and c a k; b k; c 0 substitutes values and simplifies 44k 4 k 0 44k 0k 0 4 solves and discards 4 5 4k (6k 5) 0; k 6 Total: 70 marks Pegasys 04 Current Higher Units and + Vectors

Higher Prelim Paper 04/05 Give mark for each Marking Scheme Illustration(s) for awarding each mark (a) ans: A(4, ) (4 marks) finds derivative and equates to zero dy 0 d solves for (4 ) 0; 0; 4 chooses appropriate value and subs when = 4, y = 6(4²) 4³ = 4 states coordinates of A 4 A(4, ) (b) ans: (, 7) (5 marks) establishes equation of AB y = 5( 4); y = 5 + equates equations of line and curve 5 + = 6² ³ ; ³ 6² + 5 + = 4 6 5 knows to use synthetic division 4 8 0 4 solves and chooses solution 4 ( 4)( )( + ) = 0; = 5 subs and states coordinates of B 5 y = 5( ) + = 7 ans: 0 o ; 0 o (5 marks) subs for cos o ( sin ) sin 0 simplifies and arranges to quadratic sin sin 0 factorises ( sin )( sin ) 0 4 knows to discard and solve 4 sin 5 finds angles 5 0 o ; 0 o Pegasys 04 Current Higher Units and + Vectors

Give mark for each Illustration(s) for awarding each mark (a) ans: 9 y 4 (4 marks) knows to integrate function y 9 integrates including C 9 y C subs and solves for C 6 9 C ; C 4 4 states y in terms of 4 9 y 4 (b) ans: p = 4 ( mark) subs values 9 p 4 4 (c) ans: y + 7 = ( marks) dy subs = into d m 9 7 subs into equation y 4 7( ) rearranges y 7 4(a) ans: proof ( mark) subs point to make true statement ² + ( 8)² 8() + 6( 8) 4 = 0 and conclusion so point lies on circle (b) ans: 5y + = (5 marks) uses acceptable method to find Q stepping out method or any acceptable establishes coordinates of Q Q(6, ) finds gradient of PQ 8 5 m PQ 6 4 finds perpendicular gradient 4 m perp = 5 5 subs in straight line equation and simplifies 5 y ( 6); 5y + = 5 (c) ans: 6 5 units ( marks) establishes coordinates of R 5y + 0 = ; y = 4 4; R(0, 4 4) uses distance formula QR² = 6² + 4² finds QR correctly rounded QR = 6 5 units Pegasys 04 Current Higher Units and + Vectors

Give mark for each Illustration(s) for awarding each mark 5(a) ans: proof (7 marks) (b) ans: k = -, k = -4 (5 marks) Pegasys 04 Current Higher Units and + Vectors

Give mark for each Illustration(s) for awarding each mark 6(a) ans: A(, ) ( marks) equates equations 6 8² = ; 8( + ) = 0 solves and states coordinates = 0 or ; A(, ) (b) ans: 0⅔ units² (5 marks) sets up limits 0... upper lower 6 8 (4 8 ) d integrates 0 4 4 4 0 4 0 4 0 4 0 4 4 substitutes values 4 4 0( ) [(0) [( ) ( ) )]] 5 evaluates 5 60 [( 0) [6 48)] ( 0 ) 0 7 ans: A(, 9) (4 marks) finds g(a + ) g(a + ) = a(a + ) = a² + a finds f(g(a + ) f(g(a + ) = (a² + a a) = (a² ) equates f(g(a + ) to 9 (a² ) = 9 4 solves for a 4 a = 8 (a) ans: proof (4 marks) integrates k 0 subs values and equates to k 8 8 starts to simplify 6k 4 solves for k 4 4k ; k 4 0 0 d = (b) ans: p = ( marks) uses eact values rationalises denominator and states p ; p = Pegasys 04 Current Higher Units and + Vectors Total: 60 marks