BAULKHAM HILLS HIGH SCHOOL 03 YEAR YEARLY EXAMINATIONS Mathematics General Instructions Reading time 5 minutes Working time hours Write using black or blue pen Black pen is preferred Board-approved calculators may be used A table of standard integrals is provided at the back of this paper Show all necessary working in Questions 4 Marks may be deducted for careless or badly arranged work Total marks 70 Section I Pages 4 0 marks Attempt Questions 0 Allow about 5 minutes for this section Section II Pages 5 9 60 marks Attempt Questions 4 Allow about hour 45 minutes for this section
Section I 0 marks Attempt Questions 0 Allow about 5 minutes for this section Use the multiple-choice answer sheet for Questions 0 If x x + is expressed in the form (x b) + c, what is the value of c? (A) 5 (B) 7 (C) (D) 9 For what values of k does the equation x 3x + k =0 have equal roots? (A) 9 4 (B) (C) 0 (D) 9 4 3 What is the derivative of x 3 +4 3? (A) 3 x 3 +4 (B) 3 3x +4 (C) 9x x 3 +4 (D) 9x 3x +4 - -
4 A function is given by f(x) = 9 x. What is a suitable domain of f(x)? (A) x 3 (B) x 3 (C) 3 x 3 (D) 9 x 9 5 The line through P(7,p) and Q(4, 5) has a gradient of 3. What is the value of p? (A) 4 (B) 4 (C) 6 (D) 8 6 What is the gradient of the perpendicular to the line with equation 3y = x +? (A) (B) 3 (C) (D) 3 7 Find all the values of x in the interval 0 x 360 for which tanx = 3. (A) 30, 330 (B) 0, 40 (C) 0, 300 (D) 50, 330-3 -
8 Which of the following is true for the function g(x) where g (x) =x +x +? (A) g(x) is never decreasing. (B) g(x) is increasing then decreasing. (C) g(x) is decreasing then increasing. (D) g(x) is never increasing. 9 If a > b, which of the following is always true? (A) a > b (B) a > b (C) a > b (D) a > b 0 If ABC has area 36 cm, then the area of ADE is A 5 cm 3 cm D 4 cm E cm C B (A) 4 cm (B) 8 cm (C) cm (D) 6 cm END OF SECTION I - 4 -
Section II 90 marks Attempt Questions 4 Allow about hour 45 minutes for this section Answer each question on the appropriate answer sheet. Each answer sheet must show your name. Extra paper is available. All necessary working should be shown in every question. Question (5 marks) Use a separate answer sheet Marks (a) Express 0.6. 0. 4 as a fraction in its simplest form. (b) Factorise x 3 x 5x (c) Solve 5x 9 > (d) Write 3 + with a rational denominator (e) Find the coordinates of the focus of the parabola x =(y 4) (f) Simplify x + x x x (g) Solve the following pair of simultaneous equations 3 xy = x y =3-5 -
Question (5 marks) Use a separate answer sheet Marks (a) Find the derivative of (i) 5x 3 +4x 7+ 3 x (ii) x 5 3x +7 (iii) 4 x (b) If and are the roots of x x +5=0, find (i) + (ii) + (c) If is a reflex angle such that tan = 3, find the value of cos. (d) Use the cosine rule to find the largest angle in this triangle, correct to the 3 nearest minute. Y 0 cm 8 cm X 5 cm Z - 6 -
Question 3 (5 marks) Use a separate answer sheet (a) In the diagram, the line x + y = 4 cuts the x-axis at A and the y-axis at B Marks y B A x x + y = 4 Copy the diagram onto your answer sheet. (i) Find the coordinates of A and B. (ii) Find the perpendicular distance of the point C(5,) from the line x + y = 4. (iii) Show that the gradient of the line AC is 3. (iv) Hence, or otherwise, show that the equation of the line AC is x 3y = 4. (v) Find the distance AB. (vi) Find the exact area of ABC. (vii) On your diagram, sketch the region described by the inequalities x + y 4 and x 3y 4. (b) The function y = x 3 3x 9x + is defined in the domain x 5. (i) Find the coordinates of any turning points and determine their nature. 3 (ii) Find the coordinates of any points of inflection. (iii) Draw a neat sketch of the curve, clearly labelling the important features. - 7 -
Question 4 (5 marks) Use a separate answer sheet Marks (a) A square garage door of length x metres is surrounded on three sides, by brickwork. The brickwork is.5 metres wide on each side of the door and of height metres above the door, as shown below. In order to meet building regulations, the area of brickwork must be more than twice the area of the garage door. (i) Show that x 5x 6<0 (ii) Hence find the range of values of x, in order to satisfy building regulations. (b) WXYZ is a parallelogram. XP bisects WXY and ZQ bisects WZY. Copy or trace the diagram onto your answer sheet. (i) Explain why WXY = WZY (ii) Prove WXP YZQ (iii) Hence find the length of PQ, given WY = 0 cm and QY= 8 cm Question 4 continues on page 9-8 -
Question 4 (continued) Marks (c) A rectangular hot water tank x metres wide, y metres high and.5 metres long, which fits exactly into the roof of the house. The cross-section of the roof is an isosceles triangle with base 8 metres and equal sides 5 metres in length, as shown below. x m TANK y m 8 m (i) Explain why the roof of the house is 3 metres high. (ii) Show that y = 3 (8 x). 8 (iii) Show that the volume of the tank is given by V = 9 x(8 x). 6 (iv) Calculate the maximum volume of the tank. 3 End of paper - 9 -
BAULKHAM HILLS HIGH SCHOOL YEAR MATHEMATICS YEARLY EXAMINATION 03 SOLUTIONS Solution Marks Comments SECTION I. B - x x += x 6x + = x 6x +9 8+ =(x 3) 7 c = 7. D equal roots occur when =0 i.e. b 4ac =0 ( 3) 4()(k) =0 9 4k =0 4k =9 k = 9 4 3. C - f(x) = x 3 +4 3 f (x) =3 x 3 +4 3x =9x x 3 +4 4. C - domain:9 x 0 x 9 3 x 3 5. B - m PQ =3 p +5 7 4 =3 p +5 =3 3 p +5=9 p =4 6. D - 3y = x + y = 3 x + 3 gradient = 3 gradient of perpendicular= 3 7. C - tanx = 3 0 x 360 Q & Q4 tan = 3 =60 x = 80, 360 x = 0, 300 8. A g (x) =x +x + =(x +) 0 g (x) is never negative Thus g(x) is never decreasing 9. D (A) > however (-) >( ) is FALSE (B) 3 > however 3 > is FALSE (C) 3 > however 3> is FALSE (D) as x increases as x increases then if a > b then a > b
Solution Marks Comments 0. A - DAE = CAB (common ) AD AC = 3 9 = 3 AE AB = 5 5 = 3 AD AC = AE AB thus DAE CAB (SAS - in ratio :3) If sides are in the ratio :3 then the areas are in the ratio :3 =:9 area A =4 cm
(a) Let x =0.6 0. 4. x = 0.604040404... 00x = 60.4040404... 99x = 59.8 x = 59.8 99 = 598 990 = 99 495 (b) x 3 x 5x = x x x 5 =x(x 5)(x +3) (c) 5x 9 > 5x 9 > or 5x +9> 5x > 30 5x > x > 6 x < 5 x < 5 or x >6 (d) 3 + 3 = 6 3 3 4 (e) x =(y 4) 4a = a =3 = 6 = 6 (f) x + x x x = x(x +) x(x ) (g) xy = y = x = (x ) (x +) x(x +)(x ) = x(x +)(x ) Solution Marks Comments SECTION II QUESTION OR 0.6. 0. 4= 604 6 990 = 598 990 = 99 495 Vertex = (0, 4) Focus = (0, 4 + a) = (0, 7) x x =3 x 4=3x x 3x 4=0 (x +)(x 4)=0 x = or x =4 y = or y = x =, y = or x =4, y = 3 marks Unsimplified correct answer Correctly uses a valid technique Bald simplified answer marks Fully factorised answer Partially factorised answer marks Finds two possible critical points via a valid method 0 marks Solves without considering the absolute value marks. Attempts to multiply by the conjugate marks Finds focal length Finds vertex marks Creates a common denominator 3 marks marks Correctly finds one answer Finds two correct x- values or y-values Creates a quadratic equation in an attempt to find a solution Note: do not penalise for writing answer as coordinates.
Solution Marks Comments QUESTION (a) (i) f(x) =5x 3 +4x 7+ 3 marks x =5x 3 +4x 7+3x f (x) =5x +4 6x 3 Correctly differentiates at least two terms =5x +4 6 Note: may be left in index form x 3 (a) (ii) f(x) = x 5 3x +7 (3x +7)() (x 5)(3) f (x) = (3x +7) (a) (iii) f(x) = 4 x 6x +4 6x +5 = (3x +7) 9 = (3x +7) = 4 x (b) (i) + = =5 + = + (b) (ii) = 5 + =( + ) =() (5) =4 0 = 6 f (x) = 4 x ( x) = x 4 x = x 4 x (c) is reflex and tan >0, 3rd quadrant and thus cos <0 3 3 θ cos = 3 3 marks Correct use of the quotient rule or equivalent merit marks Correct use of the chain rule or equivalent merit Note: may be left in index form marks. Correctly finds + and Rewrites + in terms of sum and product of roots marks. Rewrites + in terms of sum and product of roots marks Establishes the sign of the answer Finds the correct ratio, disregarding sign (d) y = x + z xzcosy cosy = x + z y xz = 8 +0 5 8 0 Y =cos - 6 60 = 5 3 3 marks marks Correctly finds one of the other angles Uses the cosine rule in an attempt to find the largest angle Correctly identifies Y as the largest angle Uses the cosine rule in an attempt to find one of the other angles Note: do not penalise for the incorrect rounding of minutes
Solution Marks Comments QUESTION 3 3(a) (i) x-intercept occurs when y = 0 y-intercept occurs when x = 0 x =4 y =4 for both A x = B is (0, 4) and B A is (, 0) (5) + () 4 marks 3(a) (ii) d = + = 8 units Correctly substitutes into 5 the perpendicular distance 3(a) (iii) m AB = 0 5 = 3 3(a) (iv) y 0= (x ) 3 3y =x 4 x 3y =4 3(a) (v) d AB = ( 0) +(0 4) = 4+6 = 0 = 5 units OR By Pythagoras AB = +4 =0 AB = 0 = 5 units 3(a) (vi) Area ABC = 5 8 5 = 8 units 3(a) (vii) y formula marks Correctly substitutes into the distance formula or equivalent merit Note: Need to use answers found in part (i) Note: May use answers found in (ii) and (v) B C(5,) x 3y = 4 A x x + y = 4
3(b) (i) y = x 3 3x 9x + dy dx =3x 6x 9 d y =6x 6 dx d 3 y dx 3 =6 Stationary points occur when dy dx =0 i.e. 3x 6x +9=0 x x +3=0 (x 3)(x +)=0 x = 3 or x = when x =, d y dx =6( ) 6 = < 0 (, 6) is a maximum turning point 3(b) (ii) possible inflection points occur when d y dx =0 i.e. 6x 6=0 x = OR x (0) + () d y 6 0 6 dx Solution Marks Comments 3 marks marks Finds both stationary points Finds one stationary point and determines its nature Acknowledges the 3 condition for stationary points when x =3, d y dx =6(3) 6 = > 0 (3, 6) is a minimum turning point when x =, d3 y dx 3 =6 0 Finds the possible inflection point and explains the change in concavity 3(b) (iii) (, ) (,6) there is a change in concavity thus (, 0) is a point of inflection y (5,6) 3 9 x marks Neat sketch with key points labelled A complete sketch with a poor scale Neat sketch with correct basic shape, with only some key features labelled A correct sketch that extends outside the given domain (, 0) (3, 6)
QUESTION 4 4(a) (i) The total face of the garage must be >3x (x +3)(x +)>3x x +5x +6>3x 0>x 5x 6 x 5x 6<0 4(a) (ii) x 5x 6<0 5 5 4 ( 6) 5 73 4 Solution Marks Comments < x < 5 5 +4 ( 6) < x < 5+ 73 4 However x > 0, 0<x < 5+ 73 4 4(b) (i) Opposite 's in a parallelogram are = 4(b) (ii) WXP = WXY (XP bisects WXY) YZQ = WZY (ZQ bisects WZY) WXY = WZY (proven in (i) ) WXP = YZQ (A) PWX = QYZ (alternate 's are =, WX ZY) (A) WX = YZ (opposite sides in gram are = ) (S) WXP YZQ (AAS) 4(b) (iii) WP = YQ (matching sides in 's ) WY = WP + PQ + QY (common side) WY=QY+PQ PQ = WY QY =0 8 =4 cm marks Obtains desired inequality using a valid argument Makes some logical statement that would lead to a correct solution marks Finds the two roots of x 5x 6 Discounts the negative solutions of their inequality Correct explanation marks Significant progress towards a correct solution
4(c) (i) Using Pythagoras h =5 4 =9 h =3 m 4(c) (ii) 4 y 3 = x 4 4y = 3 x ratio of sides in 's ) Solution Marks Comments Correct explanation y =3 3 8 x 4 y = 3 8 (8 x) y 3 marks Shows without using similar figures Uses similar triangles to establish a relationship between x and y 4 4(c) (iii) V =.5 xy = 3 x 3 (8 x) 8 = 9 x(8 x) 6 4(c) (iv) V = 9 x 9 6 x dv dx = 9 9 8 x d V dx = 9 8 d V when x =4, dx = 9 8 <0 Volume is a maximum when x = 4 maximum V = 9 (4) 9 6 (4) =9 m 3 stationary points occur when dv dx =0 OR i.e. 9 9 8 x =0 8 x = x =4 Function is a concave doen parabola, thus the maximum is the y-value of the vertex 9 AOS = 9 maximum V = 9 (4) 9 6 (4) 6 = 9 8 =9 m 3 9 3 3 marks marks Proves that x = 4 gives the maximum volume Finds the volume without proving it is a maximum Finds the axis of symmetry Finds the discriminant of the function Obtains the value of x that will produce a maximum volume Attempts to show that a maximum is obtained Recognises the significance of the concavity of the function. =4 OR maximum V = 4a 9 = 0 4 9 6 = 8 4 4 9 =9 m 3