Principles of Mathematics pril 999 Provincial Examination NSWER KEY / SCORING GUIDE Organizers CURRICULUM: Sub-Organizers. Problem Solving Problem Set. Patterns and Relations B C D E F Sequences and Series Polynomials Logarithms and Exponents Quadratic Relations Quadratic Systems 3. Shape and Space G H Trigonometry Geometry Part : Multiple Choice Q K C CO PLO Q K C CO PLO. B K C 4. C U B. C U C3 5. D U B4 3. C U C6 6. B U B4 4. H C4 7. D H B4 5. C H C 8. B H B6, 6. D K E5 9. C U 3 G 7. B U E 30. B K 3 G5 8. B U F5 3. D U 3 G 9. C U E5 3. D U 3 G3 0. C U E6 33. B U 3 G. D U F 34. C U 3 G5. U F 35. U 3 G8 3. H E7 36. B U 3 G7 4. D U D5 37. U 3 G7 5. U D5 38. H 3, G9, 7 6. D U D 39. B U 3 H 7. K D 40. U 3 H 8. C U D6 4. D U 3 H 9. U D5 4. C H 3 H4 0. H D5 43. D H 3. C K B 44. B U 3. B U B4 45. D U 3 3. C U B5 Multiple Choice = 45 s 994pmak - - May 4, 999
Part B: Written Response Q B C S CO PLO. U 3 F. U 3 C7 3. 3 U 3 E4 4. 4 U 3 3 G9, 7 5. 5 U 3 D5, 7 6. 6 U 3 3 H4 7. 7 U 3 3 8. 8 U 4 3 H Written Response = 5 s Multiple Choice = 45 (45 questions) Written Response = 5 (8 questions) EXMINTION TOTL = 70 s LEGEND: Q = Question Number K = Keyed Response C = Cognitive Level B = Score Box Number S = Score CO = Curriculum Organizer PLO = Prescribed Learning Outcome 994pmak - - May 4, 999
PRT B: WRITTEN RESPONSE Value: 5 s INSTRUCTIONS: Suggested Time: 45 minutes Rough-work space has been incorporated into the space allowed for answering each question. You may not need all the space provided to answer each question. Where required, place the final answer for each question in the space provided. If, in a justification, you refer to information produced by the calculator, this information must be presented clearly in the response. For example, if a graph is used in the solution of the problem, it is important to sketch the graph, showing its general shape and indicating the appropriate window dimensions. When using the calculator, you should provide a decimal answer that is correct to at least two decimal places (unless otherwise indicated). Such rounding should occur only in the final step of the solution. Full s will NOT be given for the final answer only.. Solve the following system algebraically. Express all solutions as ordered pairs. (3 s) x + y = 5 x = y 5 x + y = 5 x y = 5 x + x = 0 x + x 0 = 0 ( x + 5) ( x 4) = 0 x = 5 or x = 4 x + y = 5 x + y = 5 5 + y = 5 6 + y = 5 y = 0 y = 9 y = 0 y =±3 ( 5, 0) ( 4, 3) ( 4, 3) deduction if answers not listed as ordered pairs 994pmak - 3 - May 4, 999
. Solve the following system algebraically. Express all solutions as ordered pairs. (3 s) x + y = 5 x = y 5 ( y 5) + y = 5 y 4 0y + 5 + y = 5 y 4 9y = 0 y ( y 9) = 0 y = 0, x =±3 x = y 5 x = 9 5 = 0 5 = 4 = 5 ( 5, 0) ( 4, 3) ( 4, 3) deduction if answers not listed as ordered pairs 994pmak - 4 - May 4, 999
. polynomial function of degree 3 has a zero of and a double zero of 4. Determine this function if it passes through the point (, 0). nswer may be left in factored form. (3 s) ( )( x 4) ( x 4) y = k x+ { 444 4443 ( for y = 0) k( + ) ( 4) ( 4) = 0 for x = and k( ) ( 3) ( 3) = 0 8k = 0 k = 0 8 = 5 9 y = 5 9 ( x + ) ( x 4) ( x 4) 994pmak - 5 - May 4, 999
( ) moves such that it is always equidistant from the point,3 ( ) and the 3. point P x, y line y =. Determine the equation of this locus, in standard form. (3 s) y P (x, y) d (, 3) d x (x, ) d = d ( x ) + ( y 3) = ( x x) + ( y + ) ( x ) + ( y 3) = ( y + ) ( x ) + y 6y + 9 = y + y + ( x ) = 8y 8 y = 8 ( x ) or y = 8 ( x ) + 994pmak - 6 - May 4, 999
4. Use a graphing calculator to solve the following equation for x where 0 x < π. (3 s) cos x = sin 3x Sketch the graph in the viewing window below and indicate appropriate window dimensions. State the function(s) used in your graph. Ensure that the relative maximum and relative minimum points of the function(s) are visible within the viewing window. Y = cos x + sin 3x for equation for graph x [ 0, π] y [ 3, 3] for window dimensions x =.6, 4. 4 s 994pmak - 7 - May 4, 999
4. Use a graphing calculator to solve the following equation for x where 0 x < π. (3 s) cos x = sin 3x Sketch the graph in the viewing window below and indicate appropriate window dimensions. State the function(s) used in your graph. Ensure that the relative maximum and relative minimum points of the function(s) are visible within the viewing window. Y = cos x Y sin 3x for graph of each curve x [ 0, π] y [ 3, 3] for window dimensions x =.6, 4. 4 s 994pmak - 8 - May 4, 999
5. Solve the following system using a graphing calculator. Express all solutions as ordered pairs. (3 s) y = 3log( x + )+ y = x Sketch the graph in the viewing window below. State the function(s) that you entered to obtain your graph and your solution. Indicate the dimensions of the viewing window that will show enough of the graph so that recognizable characteristics of the function(s) and all intersection points are visible. Y = 3log( x + )+ Y = x for each graph x [ 5, 5] y [ 3, 4] for window dimensions Therefore the solutions are: (.9,.78), (.73, 0.70) s for answers 994pmak - 9 - May 4, 999
5. Solve the following system using a graphing calculator. Express all solutions as ordered pairs. (3 s) y = 3log( x + )+ y = x Sketch the graph in the viewing window below. State the function(s) that you entered to obtain your graph and your solution. Indicate the dimensions of the viewing window that will show enough of the graph so that recognizable characteristics of the function(s) and all intersection points are visible. Y = 3log( x + )+ x + for graph x [ 5, 5] y [ 3, 4] for window dimensions The zeros of the function are: x =.9, x =. 73 Therefore, by substitution, the solution to the system is: (.9,.78), (.73, 0.70) for obtaining both y-values 994pmak - 0 - May 4, 999
6. In the diagram below, B and CB are tangents to a circle with radius 0. If BC = 80, find x, the shortest distance from B to the circle. (3 s) C x B C sin 40 = 0 OB OB = 0 sin 40 = 5.557 x = 5.557 0 0 O = 5.56 40 x B 994pmak - - May 4, 999
7. Pam and Donna are playing snooker on a 6 snooker table. To establish the position of the balls, consider a coordinate system with ( 0, 0) at the bottom left corner and the other corners at ( 6, 0), ( 6, ) and ( 0, ). The cue ball is at (, 0) and the ball that must be hit is at B( 4, ). If the cue ball at must bounce off side CD and then hit the ball at B, determine the coordinates of the point on side CD that the cue ball must hit. (3 s) (0, ) C (6, ) (0, ) C (6, ) (, 0) (, 0) B (4, ) B (4, ) (0, 0) D (6, 0) (0, 0) D (6, 0) 994pmak - - May 4, 999
4 8 x = x 4x = 6 x 6x = 6 (0, ) (, 0) 4 C (6, ) x 6 8 = = = 6. 6 3 x + =. 6 + = 46. 8 x for s the coordinates of the point on CD are x for other numbers with variables ( 6 4 6) 6 4,. or, 3 B (4, ) each (0, 0) D (6, 0) 4 0 y = y (0, ) C (6, ) 4y = 8 0 y 6y = 8 (, 0) 4 y = 8 6 = 4 3 0 y for s the point is 6, 4 3 or ( 6, 4. 6 ) each B (4, ) (6, y) y for other numbers with variables (0, 0) D (6, 0) 994pmak - 3 - May 4, 999
4 8 x = x (0, ) C (6, ) 4x = 6 x 6x = 6 (, 0) 4 x = 6 6 = 8 3 =. 6 8 x the coordinates of the point on CD are ( 6, 4. 6 ) or 6, 4 3 each B (4, ) x for s for other numbers with variables (0, 0) D (6, 0) 8 6 = k 4 (0, ) C (6, ) 3 = 6k k = 6 3 (, 0) 4 0 6 3 = 4 3 k 8 the coordinates of the point on CD are 6, 4 3 each B (4, ) for s for other numbers with variables (0, 0) D (6, 0) 994pmak - 4 - May 4, 999
8. Complete the proof. (4 s) Given: Circle with centre C = 6 34 Prove: B = D B 5 C 6 D Method : Statement PROOF Reason C is the centre given C = BC radii = = 3 s opposite = sides are = C = DC radii = 4 = 6 = 6 3= 4 s opposite = sides are = given substitution = 5 3rd s of s are = B = D chords opposite = central s are = 994pmak - 5 - May 4, 999
8. Complete the proof. (4 s) Given: Circle with centre C = 6 34 Prove: B = D B 5 C 6 D Method : Statement C is the centre C = BC = 3 C = DC 4 = 6 = 6 3= 4 C = C BC DC B = D PROOF given Reason radii = s opposite = sides are = radii = s opposite = sides are = given substitution same side S CPCTC 994pmak - 6 - May 4, 999
8. Complete the proof. (4 s) Given: Circle with centre C = 6 34 Prove: B = D B 5 C 6 D Method : Statement C is the centre C = BC = 3 C = DC 4 = 6 = 6 3= 4 BC = DC BC DC B = D PROOF given Reason radii = s opposite = sides are = radii = s opposite = sides are = given substitution radii = S CPCTC 994pmak - 7 - May 4, 999
8. Complete the proof. (4 s) Given: Circle with centre C = 6 34 Prove: B = D B 5 C 6 D Method : Statement C is the centre C = BC = 3 C = DC 4 = 6 = 6 3= 4 BC = DC C = C = 5 BC DC B = D PROOF given Reason radii = s opposite = sides are = radii = s opposite = sides are = given substitution radii = same side 3rd s of s are = SS CPCTC 994pmak - 8 - May 4, 999
8. Complete the proof. (4 s) Given: Circle with centre C = 6 34 Prove: B = D B 5 C 6 D Method : Since all radii are = ( ), the two triangles are isosceles ( ) = 3, 4 = 6 because s opposite = sides are = Since = 6 3= 4 = 5( ) because 3rd angles of s are = ( ) So B = D since chords opposite = central s are = 994pmak - 9 - May 4, 999
8. Complete the proof. (4 s) Given: Circle with centre C = 6 34 Prove: B = D B 5 C 6 D Method : Since all radii are = ( ), the two triangles are isosceles ( ) = 3, 4 = 6 because s opposite = sides are = Since = 6 3= 4 since C = C by same side ( ) ( ) BC DC by S B = D by CPCTC END OF KEY 994pmak - 0 - May 4, 999