MATHEMATICS Higher Grade - Paper I (Non~calculator)

Transcription

1 Prelim Eamination 006 / 007 (Assessing Units & ) MATHEMATICS Higher Grade - Paper I (Non~calculator) Time allowed - hour 0 minutes Read Carefully. Calculators may not be used in this paper.. Full credit will be given only where the solution contains appropriate working.. Answers obtained by readings from scale drawings will not receive any credit.. This eamination paper contains questions graded at all levels. Pegasys 006

2 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: ( A ± ) ( A ± ) sin = cos = sina = cosa = = = sin Acos ± cos Asin cos Acos m sin Asin sin Acos A cos A sin A cos A sin A Pegasys 006

3 All questions should be attempted. Two functions, defined on suitable domains, are given as f ( ) = ( ) and g ( ) =. (a) Show that the composite function, ( ) f ( g ( ) ) h =, can be written in the form h ( ) = a + b + c, where a, b and c are constants, and state the value(s) of a, b and c. (b) Hence solve the equation h ( ) = 6, for, showing clearly that there is only one solution.. Part of the line, L, with equation y = +, is shown in the diagram. The line L is parallel to L and passes through the point (0,-). Point A lies on the -ais. y y = + L L O - A (a) Establish the equation of line L and write down the coordinates of the point A. (b) Given that the line A is perpendicular to both lines, find, algebraically, the coordinates of point. 5 (c) Hence calculate the eact shortest distance between the lines L and L.. For what value of p, where p > 0, does the equation ( p + ) p + p = 0 have equal roots? 6 Pegasys 006

4 . Given that clearly that sin A = and cos =, with angles A and both being acute, show 6 cos ( A ) = The diagram shows part of the graph of y = f (). y A(,) O 5 Sketch the graph of = [ f ( + ) ] y marking clearly the new positions of the highlighted points and stating their new coordinates. 6. A function, f, is defined on a suitable domain as () = f ( ) (a) Differentiate f with respect to, epressing your answer with positive indices. (b) Hence find when f ( ) = A circle, centre C(8, k), has the points P(,-) and Q on its circumference as shown. y M(0,) is the mid-point of the chord PQ. Q C(8, k) (a) Find the coordinates of Q. (b) Given that radius CQ is horizontal, M(0, ) write down the value of k, the O P(, -) y-coordinate of C. (c) Hence establish the equation of the circle. Pegasys 006

5 8. A sequence is defined by the recurrence relation U = au 0, where a is a constant. n+ n + (a) Given that U 0 and U 6, find a. 0 = = (b) Find the value of S, if S = U + U. dy 9. A curve has as its derivative =. d Given that the point (, 7) lies on this curve, epress y in terms of. 0. A function is given as f ( θ) = cos θ + 8cosθ + 6 for 0 θ π. (a) Epress the function in the form f ( θ) = a (cos θ + b) + c and write down the values of a, b and c. (b) Hence state the minimum value of this function and the corresponding replacement for θ. [ END OF QUESTION PAPER ] Pegasys 006

6 Higher Grade Prelim 006/007 Give mark for each Marking Scheme - Paper I Illustration(s) for awarding each mark. (a) ans: a =, b =, c = marks for knowing g through f for correct substitution (algebra) epanding and simplifying for a, b and c (b) ans: = marks solving to zero strategy - synthetic division finding root showing only one solution (a) strategy f ( g( )) = ( ) ( ) (or equivalent) f ( g( )) = + a =, b =, c = ( ) (b) + 6 = = + = 0 =.. no solution (or equivalent). (a) ans: y = ; A(,0) marks for gradient writing down equation of line establishing the coordinates of A (b) ans: (,) for perpendicular gradient equation of line A for strategy of a system for first coordinate 5 second coordinate 5 marks (c) ans: 7 units marks strategy + lengths to use in Pyth. calculation to answer. ans: p = 5 6 marks for discriminant statement for a, b and c substituting correctly simplifying 5 factorising and answer(s) 6 discarding (a) y = = + m y = = 0 = = 0 (b) m = y 0 = ( ) y = + 6 y + = 6 y =.. or equiv. y = 5 = Pupils may attempt to step out using - gradient and then check if point satisfies equation - award marks on your discretion. (c) Pyth + using and d = + = 7 d = 7 for equal roots b ac = 0 a = p +, b = p, c = p ( p ) ( p + ) p = 0 (or equivalent) 00 p p = 0 5 p (5 p ) = 0 p = 0 or ± 5 6 p = 5 discard 0 and -5 Pegasys 006

7 Give mark for each. ans: Proof 6 marks strategy of drawing R.A. triangles calculating missing sides by Pyth. correct epansion substitution 5 calculation and simplifying surd 6 rationalising denom. to answer Illustration(s) for awarding each mark and cos( A ) = cos Acos + sin Asin 5 A 6... = = = = = = = (or equivalent) 5. ans: sketch as opposite marks for units to the left for reflection in the -ais new positions of S.P.'s (-,0) and (0,-) and for new position of root (,0) - A (0,-) 6. (a) ans: f ( ) = + marks for preparing to differentiate epanding brackets differentiating epress with positive indices (b) ans: = marks for forming equation for to subject answer various ways to solve use own discretion (a) f ( ) = ( ) f ( ) = f ( ) = + f ( ) = + (b) + = 5 mult. by then + = 0 = 8 then = 8 square both sides = = 6 alternative realise that = then 8 = Pegasys 006

8 Give mark for each 7. (a) ans: Q(-,6) mark answer (b) ans: k = 6 mark answer (c) ans: ( 8) + ( y 6) = 00 marks for strategy. radius and centre finding radius answer 8. (a) ans: a = 0 6 marks forming equation solving (b) ans: S = 6 6 marks finding U finding S Illustration(s) for awarding each mark (a) stepping out to answer (b) answer (c) strategy r can be found from horiz. line but some pupils will use points P and C. r = = 00 ( 8) + ( y 6) = 00 (a) 6 = a (0) a = 6 a = 6 (b) U = 0 6(6) + 0 = 5 6 S = = ans: y = 6 marks knows to integrate integrates correctly (with c) substitutes and finds c writes down answer y = ( ) d y = + c 7 = ( ) + c c = 6 y = 6 0. (a) ans: f ( θ) = (cos θ + ) + a =, b =, c = marks common factor completing the square multiplying back through answer (b) ans: min of at = π θ marks knowing minimum of knowing to solve to zero establishing answer no marks off if given in degrees (a) f ( θ) = (cos θ + cos θ) = ((cos θ + ) ) = (cos θ + ) + 6 f ( θ) = (cos θ + ) + a =, b =, c = (b) min cos θ + = 0 cos θ = θ = π (80) θ = π (90) Total 60 marks Pegasys 006

9 Prelim Eamination 006 / 007 (Assessing Units & ) MATHEMATICS Higher Grade - Paper II 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 by readings from scale drawings will not receive any credit.. This eamination paper contains questions graded at all levels. Pegasys 006

10 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: ( A ± ) ( A ± ) sin = cos = sina = cosa = = sin Acos ± cos Asin cos Acos m sin Asin sin Acos A cos A sin A cos A = sin A Pegasys 006

11 All questions should be attempted. Triangle PQR has as its vertices P(-7,-), Q(,8) and R(9,-0) as shown. y Q(,8) P(-7,-) O S R(9,-0) (a) Find the equation of side PR. (b) Find the equation of the altitude QS. (c) Hence find the coordinates of S, the point where the altitude QS meets side PR. (d) Establish the equation of the circle which passes through the points Q, S and R.. A recurrence relation is defined as u = 0 75u. n+ n + Given that U, find the difference between the limit of the sequence and the 0 = third term, U. 5. A curve has as its equation y = ( 6) + 8. Given that the line with equation y = 5 is a tangent to this curve, establish the coordinates of the point T, the point of contact between the curve and the line. Pegasys 006

12 . Part of the graph of the curve with equation y = is shown below. The curve passes throught the point (-,0). y y = - A O Find, algebraically, the coordinates of the points A and A circle has as its equation ( 9) + ( y + ) = 7. (a) Given that the point A(, k) lies on this circle, find k where k > 0. (b) Find the equation of the tangent to this circle at the point A. (c) Show clearly that this tangent passes through the centre of the circle with equation + y + 6 8y + = The functions f and g, defined on suitable domains, are given as ( ) = 5a f and g( ) = a, where a is a constant. (a) Given that f ( a) = g(), find the value of a, where a < 0. (b) With a taking this value, find the rate of change of g. Pegasys 006

13 7. A small feeding trough is shown opposite. The end face has an ais of symmetry A. 60cm Edge CD is perpendicular to the ais of symmetry. D C A When the end face is rotated through 90 o and then halved along the ais of symmetry, shape AC can be placed on a coordinate diagram as shown below. A lies along the -ais with the curved edge CA being part of the curve y = with equation ( ) y C y = ( 60 + ) o A (a) Establish the coordinates of A and. (b) (c) Hence calculate the area of shape AC given that all the units are in centimetres. Given that the trough is a prism and measures 60cm from back to front, calculate the volume of feed the trough can hold when full, giving your answer correct to the nearest litre. 8. Solve algebraically the equation sin = o o 5cos, where 0 < 60 <. 6 Pegasys 006

14 9. The captain of a small pleasure boat wishes to take a group of passengers from one island to the net, a journey of 00 kilometres. The amount of fuel used is dependent upon the speed, v kilometres per hour, of the boat. (a) Given that the rate of fuel used is ( v ) gallons per hour, show clearly that the total fuel used, F, for this 00 kilometre journey is given by F 00 v = v gallons. (b) Hence find the speed which keeps the amount of fuel used to a minimum and the amount of fuel needed, at this speed, for the voyage. 5 [ END OF QUESTION PAPER ] Pegasys 006

15 Higher Grade Prelim 006/007 Marking Scheme - Paper Give mark for each Illustration(s) for awarding each mark. (a) ans: y = (or equiv.) marks for gradient for equation of line (b) ans: y = + marks knowing gradients mult. to - for gradient equation of altitude (c) ans: S(-,-) marks knowing to solve as a system system strategy. (subst. or elimin.) first coordinate second coordinate (d) ans: ( 6) + ( y + ) = 90 marks realising strategy of R.A. QR = diam. finding centre calculating value of r equation of circle 0 + (a) m = = y + 0 = ( 9) (or equivalent) (b) if perpen. m m = stated or implied m = y 8 = ( ) (or equiv.) (c) y = y = + attempts to substitute or eliminate 5y = 0 y = = + = (d) strategy ( 0) C, ) C(6, ) ( r = 9 + = 90 ( 6) + ( y + ) = 90. ans: marks for finding U for U and U knowing how to find limit finding limit 5 calculating difference U = 0 75() + = 6 U = 0 75(6) + = 9 U = 0 75(9) + = 5 b L = (or equivalent) a L = = diff. = 8 5 = ans: T(7,9) marks knows to solve a system and sub. for y simplifies first coordinate second coordinate 5 = ( 6) = = + 9 ( 7)( 7) = 0 = 7 y = (7) 5 = 9 Pegasys 006

16 Give mark for each. ans: A(,0), (,-) 7 marks to find A set up synth. division use -.. or other find coordinate of A and hence A for know to diff. and solve to 0 5 differentiate correctly 6 find coordinate of 7 find y coordinate of Illustration(s) for awarding each mark set up synth. division for root = 0 ( )( + ) = 0 =, = A(,0) dy know S.P. = 0 d dy 5 = = 0 d 6 ( ) = 0 = ( discard ) 7 y = () = (, ) 5. (a) ans: k = 8 marks for subst. point in equation for simplifying to quadratic factorise and solve for k discarding (or implied) and answer (b) ans: y = + 8 marks for coordinates of centre gradient of radius gradient of tangent equation of tangent (a) ans: proof marks drawing out centre show point satisfies equation of tangent 6. (a) ans: a = marks for writing down f(a) for g() for equating and factorising answer (b) ans: 5 marks for g when a = for differentiation to answer (a) ( 9) + ( k + ) = 7 k + k 80 = 0 ( k + 0)( k 8) = 0, k = 0 or 8 k = 8 (b) C(9,-) 8 m r = = 9 m = T 8 = ( y ) (or equivalent) (c) C(-,) () = ( ) + 8 = proved (a) f ( a) = a 5a g( ) = a a = 5a a a a = 0 ( a )( a + ) = 0 a = (other root discard, implied) (b) 5 g ( ) = ( ) (or equiv.) 5 g ( ) = Pegasys 006

17 7. (a) ans: A(0,0), (,0) marks for solving to zero factorising and roots stating A finding (b) ans: Give mark for each 85 cm marks (for.. pupils may diff. find -coordinate of S.P.) for setting up integral for integration substitution correct calculation to answer (c) ans: 0 litres marks Illustration(s) for awarding each mark (a) (60 + ) = 0 (0 )(6 + ) = 0 = 0 or = 6 A(0,0) half way between roots. ( 0 + ( 6)) = (,0) 0 (b) A = ( 5 + ) d = [ 5 + ] 0 8 = ( ) (0 + ) = ( 6 ) ( ) = 85 (or equiv.) knows to double area finds volume answers to nearest litre (c) A face = 85 = 70 cm V = = 00 cm V = 0 litres (to nearest litre) 8. ans: { 5, 6 9, 70} 6 marks correct substitution simplifying and rearranging to zero factorising correct roots 5 for first angle 6 for remaining two 9. (a) ans: proof marks for epression for time of journey realising multiplication to find fuel used simplifying to answer (b) ans: v = 0 km/h, 7 5 gallons 5 marks knows to diff and solve to zero differentiates correctly strategy for solving equation solves equation 5 substitutes 0 into F and calc. fuel used sin = 5( sin ) sin = 5 0sin 0sin + sin 8 = 0 (5sin )(sin + ) = 0 (or equiv.) sin = 5 or sin = 5 o = 5 6 o = 6 9, o = 70 D 00 (a) T = = S v F = T rate of fuel used 00 = v v 00 = v v ( ) (b) at min F ( v) = 0 (stated or implied) 00 F ( v) = v v v = 0 ( v ) v v = 0 v = 8000 = 0 5 F = (0 ) = Pegasys 006 Total 70 marks