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Columbia College Mathematics 11 Final Examination December 2010 (3 hours) 1. Find the slope-intercept equation of the line that contains the points (3,2) and (5,18) and prove that the point (10,28) is also on this line. (4) 2. Find the equation of the perpendicular bisector of the line segment that has the endpoints (1,3) and (7,1). (4) 3. Use elimination to solve the system of equations: 1. 2 x 3y 30 2. 4x 5y 16 (4) 4. Solve the system of equations: 1. 2x y z 18 2. 3x y 2z 18 3. 4x y z 0 (6) 5. Using set notation, describe the shaded region shown at the left. (6) 6. By completing the square, find the vertex of the parabola y 2x 2 12x 16 and then accurately draw the graph indicating the vertex and two points on the parabola that are to the right of the vertex and two points on the parabola that are to the left of the vertex. (6) 7. In the same coordinate plane draw the graphs of y x 2 6x 6 and y x 4 and find their points of intersection by solving: x 2 6x 6 x 4 . (6) 8. Solve the equation 9 x 2 54 x 56 0 by completing the square. (4) 9. Solve the equation 15x 2 32x 7 0 by factoring. (4) 10. What is a function? (2) 11. Find the domain of f ( x ) 2x 10 x 2 8x 7 (4) 12. If g ( x ) 3x 2 and f ( x ) 2x 2 5x 2 , express the composite function f(g(x)) in standard form. (4) 13. If g ( x ) x2 3x 2 and g (f ( x )) , find f(x). 2x 1 6x 9 (4) 14. Find all the solutions of 9 3x x 15 . (4) 15. Find all the solutions of 2x 1 x 3 x 4 x 5 (4) 16. Let P( x ) 3x 4 5x3 2x 2 3x 4 and D( x ) x 2 2x 3 . Divide P(x) by D(x) and then check. (4) 17. Given that 1 and 2 are the two real zeros of P( x) x 4 2x3 x 2 4x 4 , find the other two zeros. Express these zeros in standard form. (4) 18. Use the Rational Zero theorem to solve 2x 3 3x 2 7 x 3 0 . (4) 2 19. Let f ( x ) x 3 be restricted to the domain [6,3] 3 a) In the same coordinate plane, draw the graphs of f and f 1 . (4) b) State the domain and the range of f and the domain and the range of f 1 . (2) c) Find the formula for f 1 . (2) 20. Let f ( x ) 3x 2 36x 46 be restricted to the domain (,6] . Find the formula for f 1 . (4) 21. Solve the following: a) 23x 15 77 (2) b) x 4 2x 11 17 (4) c) 5 3x 7 (2) 22. Given that O is the centre of the circle and BDA = 40o , find BAC (4) 23. The smaller circle with centre O1 has radius = 1 and the bigger circle with centre O2 has radius = 2. O1 is on the bigger circle and the two circles intersect at points A and C. The line AC intersects the line O1O2 at point B. a) Find the length of the line segment (2) O1O2 . b) Prove that ABO 2 CBO2 (5) 24. Given that AB and CD are chords of equal length in a circle with centre O and that the lines AC and BD intersect at E, prove that BEA CED . (4) Total = 113 Answers. 18 2 2 . Therefore the equation of the line is of the form 5 ( 3) y = 2x+C for some constant C. Since (5,18) is a solution: 18 2 5 C C 8 . Therefore the equation of the line is y = 2x + 8. Since 28 2 10 8 , (10,28) is a solution of the equation and therefore a point on the line. (4) 1. The slope of the line is 2. The slope of the given line segment is 1 3 1 . Therefore the slope of its 7 1 3 1 perpendicular bisector is 3 (because 3 1 ) The midpoint of the given line 3 1 7 3 1 , segment is (4,2) . Therefore the equation of the perpendicular bisector is 2 2 of the form y = 3x + C where (4,2) is a solution: 2 3 4 C C 10 . Therefore the equation of the perpendicular bisector is y 3x 10 . (4) 3. Therefore the system has the unique solution (9,-4). (4) 4. Adding equation 1. to equation 3. yields 6x = 18 and therefore x = 3. Adding equation 1. to equation 3. yields 5x + 3z = 36 and therefore 15 + 3z = 36 and z = 7. Substituting x = 3 and z = 7 into all three equations yields y = 5. Therefore the system has the unique solution (3,5,7). (6) 5. The line containing (-2,1) and (2,1) is the horizontal line y = 1. The line containing (-2,1) and (1,-2) is y x 1 and the line containing (2,1) and (1,-2) is y 3x 5 . Therefore the shaded area is the set {( x, y) : y 1 and y x 1 and y 3x 5} . (6) 6. y 2x 2 12x 16 2( x 2 6x) 16 2 ( x 3)2 9 16 2( x 3)2 2 . Therefore the vertex of the parabola is (3,-2). The graph is on the next page. (6) 7. a) y ( x 2 6x) 6 (x 3)2 9 6 ( x 3)2 3 is a parabola with vertex (3,3). The line y x 4 has intercepts (0,4) and (4,0). The graph is on the next page. (3) b) x 4 x 2 6x 6 x 2 7x 10 0 (x 2)( x 5) 0 which means that the two intersections are (2,2) and (5,-1). (3) 8. 9x 2 54x 56 0 9(x 2 6x) 56 0 9 (x 3)2 9 56 0 9( x 3)2 25 (x 3)2 25 5 14 4 x 3 x or . (4) 9 3 3 3 9. 15x 2 32x 7 0 (5x 1)(3x 7) 0 5x 1 0 or 3x 7 0 5x 1 or 3x 7 x 1 7 or x . (4) 5 3 10. Definition: A function is a rule or a set of rules that pairs each number in a set called the domain with exactly one number in a set called the range. (2) 11. Domain of f = {x : 2x 10 0 and x 2 8x 7 ( x 7)( x 1) 0} = {x : x 5 and x 7} . (4) 12. f (g(x )) f (3x 2) 2(3x 2)2 5(3x 2) 2 2(9x 2 12x 4) 15x 10 2 = 18x 2 39x 20 . (4) 13. g(f ( x )) 14. f ( x ) 2 3x 2 f ( x ) 3x 4 . (4) 2f ( x ) 1 6x 9 9 3x x 15 9 3x x 2 30x 225 x 2 33x 216 0 ( x 9)( x 24) 0 x 9 or x 24 but only x 9 is a solution of the given equation. ( 24 is extraneous). (4) 15. After multiplying both sides of the equation by the denominators: (2x 1)( x 5) (x 3)( x 4) 2x 2 9x 5 x 2 x 12 x 2 8x 7 0 ( x 1)( x 7) 0 x 1 or x 7 . (4) 16. (4) 17. Since 1 and 2 are zeros of P(x), x 1 and x 2 are factors of P(x) by the Factor theorem and therefore ( x 1)9x 2) x 2 3x 2 is also a factor of P(x): Therefore P( x ) ( x 1)( x 2)( x 2 x 2) so that the other zeros of P(x) are the zeros of 1 8 1 7 i and x2 x 2 : 2 2 2 1 7 i . (4) 2 2 18. Since the coefficients of the polynomial P( x ) 2x3 3x 2 7x 3 are integers, we have by the Rational zero theorem that if r is a rational zero of P then it must be a number of the form p/q where p is a factor of the constant term, 3, and q is a factor of the leading coefficient, 2. Since p = 1 or 3 and q = 1 or 2 , r must be in the set: 1 3 1 1 {1, ,3, } . P(1) 3 , P( ) 0 . Therefore x is a factor of P(x) and by 2 2 2 2 1 dividing P(x) by this factor we have: P( x ) ( x )( 2 x 2 2 x 6) = (2x 1)( x 2 x 3) 2 1 11 1 11 i and Therefore the other zeros of P are the zeros of x 2 x 3 : = 2 2 2 1 11 i . (2) 2 2 19. b) Domain of f = [-6,3], Range of f = [1,7], Domain of f 1( x ) = [1,7], Range of f 1( x ) = [-6,3]. (2) 2 2 c) y x 3 (where y = f(x)) Interchange x and y to get x y 3 (where 3 3 2 3 9 3 9 y f 1( x) ), x 3 y y x . Therefore f 1( x ) x . (2) 3 2 2 2 2 20. f (x ) 3( x 2 12x) 46 3 (x 6)2 36 46 3( x 6)2 62 . Therefore f is one-toone on the interval (,6] (the graph of f is the left arm of a parabola with its vertex at (6,61) ). Setting y 3( x 6)2 61 where y = f(x) and then interchanging x and y to x 61 x 61 but ( y 6)2 y 6 3 3 since the values of y must be 6 (the domain of f) it follows that x 61 . (4) f 1( x) 6 3 get x 3( y 6)2 61 where y f 1( x) ): 21. a) 23x 15 77 23x 15 77 or 23x 15 77 23x 92 or 23x 62 x 4 or 62 . (2) 23 b) There are three cases to consider: i) When x 4 then x 4 x 4 and 2x 11 2x 11 and the equation becomes: 3x 10 x 10 which is not less than -4. Therefore there are no solutions from 3 case 1. 11 then x 4 x 4 and 2x 11 2x 11 and the equation 2 becomes x 2 or x 2 which is in the interval and therefore is a solution. 11 iii) When x then both expressions in the absolute values are positive and the 2 equation becomes: 3x = 24 or x = 8 which is a solution. Therefore the equation has two solutions: x 2 and x = 8. (4) ii) When 4 x c) 5 3x 7 5 3x 7 or 5 3x 7 3x 2 or 3x 12 x or x 4 . (2) 2 3 22. BDA 40o (Given) BCA 40o (Both angles in the same arc AB) CBA 90o (Angle subtended by a diameter) BAC 180o CBA BCA 50o (Sum of the angles in a triangle = 180 o . (4) 23. a) O1O2 2 (the radius of the big circle) (2) b) First prove: O1AO2 O1CO2 : (radii of the small circle) O1A O1C O1O2 O1O2 (common side) AO2 CO2 A (radii of large circle) Therefore O1AO2 O1CO2 by the Side-side-side theorem and therefore: (corresponding angles of congruent triangles) BO2A BO2C To prove ABO 2 CBO2 (common side) BO 2 BO 2 BO2A BO2C (proved) (radii of large circle) O2A O2C Therefore ABO 2 CBO2 by the Side-angle-side theorem. (5) 24. BEA CED Proof: ABE DCE (both angles subtended by the chord AD) AB = BC (given) BEA CED (both angles subtended by the chord BC) Therefore BEA CED by the Angle-side-angle theorem. (4)