G r a d e 1 1 P r e - C a l c u l u s M a t h e m a t i c s ( 3 0 S ) Final Practice Exam Answer Key

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1 G r a d e P r e - C a l c u l u s M a t h e m a t i c s ( 3 0 S ) Final Practice Eam Answer Key

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3 G r a d e P r e - C a l c u l u s M a t h e m a t i c s Final Practice Eam Answer Key Name: Student Number: Attending q Non-Attending q Phone Number: Address: For Marker s Use Only Date: Final Mark: /00 = % Comments: Answer Key Instructions The final eamination will be weighted as follows: Modules 8 00% The format of the eamination will be as follows: Module : 5 marks Module 5: 9 marks Module : 5 marks Module 6: 9 marks Module 3: 5 marks Module 7: 3 marks Module 4: 5 marks Module 8: 9 marks Time allowed:.5 hours Note: You are allowed to bring the following to the eam: pencils ( or 3 of each), blank paper, a ruler, a scientific or graphing calculator, and your Final Eam Resource Sheet. Your Final Eam Resource Sheet must be handed in with the eam. Show all calculations and formulas used. Include units where appropriate. Clearly state your final answer. Diagrams may not be drawn to scale. Final Practice Eam Answer Key 3 of 34

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5 Name: Answer all questions to the best of your ability. Show all your work. Module : Sequences and Series (5 marks). Write the defining linear function of the following arithmetic sequence. ( marks) (Lesson ) 99, 03, 07,... Points on line: (, 99), (, 03), (3, 07) Slope: d = 4 Defining Linear Function: y y = m( ) y 99 = 4( ) y = y = k k 8. Use a formula to find the value of 8 3 = Notice k s initial value is. Thus, n is only 7, not 8. The first few terms are 8 3 Thus, t = 7 and r = 3. S S n 7 t = n ( r ) r 3 8 8, 3, 3, = = = 7 86 = (3 marks) (Lesson 4) Final Practice Eam Answer Key 5 of 34

6 Module : Factoring and Rational Epressions (5 marks). Simplify. Identify all non-permissible values. (5 marks) (Lessons 4 and 5) ( 5 + 9) ( ) ( + )( + ) 3( 4) 3 ( 4) + ( 5+ 9) ( ) = ( + )( + ) ( 5+ 9) = + 3 = ( ) + 5 ( ) ( ) + ( ) ( ) + ( + 5) ( ) 3( 5 ( )( + ) + ) ( ( + ) ( + 5) + + ) ( + 5) ( + ) ( 5+ 9) ( ) ( + 3) ( + ) + ( + ) ( + 5) + + ( + 5) ( + ) ( 5+ 9) = + 3 ( ) ( )( + ) ( + ) ( + 5) ( 5+ 9) = + 3 ( ) ( ( )( + ) + ) ( + 5) ( ) = ( + 3) + ( ( ) + ) ( + 5) ( )( + )( ) ( )( + ) = ,, 4,, 5,, 5 6 of 34 Grade Pre-Calculus Mathematics

7 Name: Module 3: Quadratic Functions (5 marks). Given the following parabola in verte form, complete the following questions. ( mark each, for a total of 5 marks) y= ( 3) + a) Identify the range. R: {y y ³, y Î Â} (Lessons to 4) b) Identify the direction of the opening. The parabola opens upward c) Identify the ais of symmetry. = 3 d) Identify the y-intercept. y= ( 3) + let = 0 y = ( ) y = ( )+ 9 3 y = Final Practice Eam Answer Key 7 of 34

8 e) Sketch the graph of the parabola. The verte is (3, ), the curve opens up, and its size is wide. The y-values are of the normal y-values. y y = ( 3) of 34 Grade Pre-Calculus Mathematics

9 Name: Module 4: Solving Rational and Quadratic Equations (5 marks). Solve the following quadratic equation using any method you wish. Eplain why you chose the method you used. ( marks) (Lessons to 4) + 5 = 5 + 5= = = 0 ( 5) 5( 5)= 0 ( 5) ( 5)= 0 5 = or = 5 You can use any method. As long as you complete the question correctly and eplain why you chose the method you did, you will get full marks. A possible eplanation is: I chose to solve this quadratic equation by factoring because it is easily factorable. Final Practice Eam Answer Key 9 of 34

10 . Solve the following rational equation. Identify any non-permissible values. (3 marks) 3 + = + (Lesson 6) Non-permissible values: = 0 or = LCD = ( + ) 3 + ( ) ( + )= + ( ) + ( ) + 3+ ( )= ( ) = ( ) = = 0 ( + )+ ( + )= 0 ( + ) = 0 = 0 of 34 Grade Pre-Calculus Mathematics

11 Name: Module 5: Radicals (9 marks). Order the following radical epressions from least to greatest. Do not use a calculator. ( marks) (Lesson ) 6 3, 7, 9 5, 5 6 3= 36 3 = 08 7 = 7 = = 8 5 = 405 5= 4 5 = 60 Thus, 60 < 08 < 405 < 847 and 5 < 6 3 < 9 5 < 7.. Simplify each of the following. All answers must have rationalized denominators. Assume all variables are non-negative. (Lesson ) a) = = ( ) 9 = = 7 9 ( mark) Final Practice Eam Answer Key of 34

12 3 ( )( ) b) 8 y 4 y ( marks) (Lesson 3) 3 ( 8y )( 4y ) = 4 y 4 y y 4 y y = 4y y y c) = (3 marks) (Lesson 4) = = ( 3) = = of 34 Grade Pre-Calculus Mathematics

13 Name: 3. Identify the values for each of the variables for which each radical epression is defined. ( mark each, for a total of marks) a) 6 3 (Lesson ) In order for this radical epression to be defined, the radicand has to be greater than or equal to zero. 6 3 ³ 0 6 ³ 3 ³ Thus, in order for this radical epression to be defined. b) 3 (Lesson ) Because the inde is two, the radicand has to be greater than or equal to zero for this epression to be defined. As any value squared will always be greater than or equal to zero, Î Â for this epression to be defined. Final Practice Eam Answer Key 3 of 34

14 4. Find the solutions for each of the following equations. Check your solutions for etraneous roots. Determine any restrictions on the variable. a) 3 = 3 (4 marks) (Lesson 5) Restrictions on the variable: ( 3 ) = = 3 8+ = 0 ( 6) ( )= 0 = 6 or = Check = Check = 6 LHS RHS LHS RHS ( ) ( 6) 3 LHS = RHS 9 3 LHS ¹ RHS Therefore, the only solution is =. 4 of 34 Grade Pre-Calculus Mathematics

15 Name: b) = 9 (3 marks) (Lesson 5) Restrictions on the variable: + 4 ³ 0 ³ 4 ³ = 9 + 4= 6 + 4= 6 + 4= 36 = 3 = 6 Check = 6 LHS RHS 9 ( 6) LHS = RHS \ The solution is = 6. Final Practice Eam Answer Key 5 of 34

16 5. The period, P, measured in seconds, of a pendulum is the time it takes to complete one L full swing. The period can be found using the formula P = π, where L measures 98. the length of the pendulum in metres. How long should a pendulum be to complete one full swing in 4.3 seconds? Round your answer to four decimal places. Check your answer for etraneous solutions. ( marks) (Lesson 5) P = π L 98. Restrictions: L ³ 0 and P ³ 0 Note: P is positive or zero because it is equal to a positive number, p, multiplied by the principal square root. Also, it would not make sense for either variable to be negative. P = π 43. = π L 98. L 98. L =( ) 43. π 98. 4π L 8. 49= L = 4π = L Check: L ³ 0 LHS P 4.3 π π LHS = RHS RHS L The pendulum should be m long. 6 of 34 Grade Pre-Calculus Mathematics

17 Name: Module 6: Systems of Equations and Inequalities (9 marks). Solve the following system of equations algebraically and graphically. (6 marks) y= ( 3) ( ) + y= 3 + Algebraic Solution ( ) + y= ( 3) y= 3 + (Lesson ) y= y= 3( + + )+ y= Substitute the epression for y in Equation () into the y-variable of Equation () = = 0 = 0 Solve for y: ( ) y = 3 y = 0 3 y = 9 ( ) Thus, the solution to this system of equations is (0, 9). Final Practice Eam Answer Key 7 of 34

18 Graphing Solution The parabola, y = ( 3), has its verte at (3, 0), it opens up, and has normal shape. The parabola, y = 3( + ) +, has its verte at (, ), opens down, and its shape is narrower. The y-coordinates are multiplied by 3. y y = ( 3) 6 y = 3( ) The solution to this system of equations is the point at which the two functions intersect, or (0, 9). 8 of 34 Grade Pre-Calculus Mathematics

19 Name:. The sum of two numbers is. Their product is 7. a) Write a system of equations to represent this problem. ( mark) (Lesson ) Let the first number be represented by m and the second number be represented by n. The system of equations is: m+ n= mn= 7 b) Solve the system of equations to find the two numbers. (3 marks) Substitute m = n into nm = 7. ( n)(n) = 7 n n 7 = 0 n n + 7 = 0 (n 3)(n 9) = 0 n = 3 or n = 9 Substituting into the equation m = n, you get: n = 3 n = 9 m = n m = n m = 3 m = 9 m = 9 m = 3 Therefore, the two numbers are 9 and 3. Final Practice Eam Answer Key 9 of 34

20 3. Solve the following inequalities by graphing. ( marks each for a total of 6 marks) a) y < 3 8 (Lesson 3) The slope of the line is 3. The y-intercept is 8. The boundary line is dotted and the shading is below the line. y y < b) 0 4 (Lesson 4) You are interested in the domain where the epression is greater than or equal to 0. Write 4 ³ 0 and write the corresponding function as y = 4. The parabola has its verte at (0, 4), opens up, and has normal shape. y y = 4 4 The parabola crosses the -ais at and. The parabola lies on or above the -ais when and when ³. Therefore, the solution set is { - or ³, Î Â}. 0 of 34 Grade Pre-Calculus Mathematics

21 Name: c) y > ( + ) 3 (Lesson 5) The parabola has its verte at (, 3), opens up, and has normal shape. The boundary line is dotted and the region above the parabola is shaded. y y > ( + ) 3 4 Final Practice Eam Answer Key of 34

22 4. Use the sign analysis method to solve the following quadratic inequality. (3 marks) (Lesson 4) Write as 5 3 ³ 0. You want to consider all values of the epression that are greater than or equal to zero. Factoring the corresponding quadratic equation: 5 3 = = 0 ( 3) + ( 3) = 0 ( + )( 3) = 0 + = 0, 3 = 0 = or = 3 The critical points are and 3. Factor + 3 Sign Diagram ( + )( 3) The solution set contains the points where 5 3 ³ 0 and is or 3, R. of 34 Grade Pre-Calculus Mathematics

23 Name: Module 7: Trigonometry (3 marks). P(, 3) is a point on the terminal side of angle q in standard position. Determine sin q, cos q, and tan q for the following point. Also, determine the distance from the origin to the point P(, 3). Rationalize any denominators. (5 marks) (Lesson ) r = + ( 3) = 3 y y = = 3 = 3 3 sin θ r 3 3 cos θ = = = r r 3 y = = 3 3 tan θ = P(, 3) Final Practice Eam Answer Key 3 of 34

24 . Given the following angles in standard position, determine the eact values of the sine, cosine, and tangent ratios. Show how you determined the eact values. ( marks each, for a total of 4 marks) (Lesson 3) a) 40 q r = = 60 3 y 60 3 sin 40 = cos 40 = 3 tan 40 = = 3 b) 45 q r = 45 y 45 sin 45 = cos 45 = tan 45 = 4 of 34 Grade Pre-Calculus Mathematics

25 Name: 3. Sketch the angle 34 in standard position and find its reference angle. Determine the other angles that have the same reference angle as the given angle for q in the interval [0, 360 ]. (3 marks) (Lesson ) y = 34 The reference angle for 34 is = 8. The other angles that have this same reference angle are: 8 in Quadrant I 80 8 = 6 in Quadrant II = 98 in Quadrant III Final Practice Eam Answer Key 5 of 34

26 4. Determine the solution set for each of the following trigonometric equations over the interval [0, 360 ]. Round to the nearest degree where necessary. ( marks each, for a total of 4 marks) (Lesson ) a) cos θ= 5 θ θ r r = cos 5 = 78 As cos q > 0, q must be in Quadrants I or IV. In Quadrant I, q = 78. In Quadrant IV, q = = 8. b) sin q = 0 sin q = This is a quadrantal angle through the point (0, ) and only occurs when q is of 34 Grade Pre-Calculus Mathematics

27 Name: 5. In DABC, ÐC = 4, c = 6, and a = 8. Find all possible values for b and ÐB. Draw a diagram and round off answers to two decimal places. (5 marks) (Lessons 4 and 6) Since AB < BC, find h: h = 8 sin 4 = 5.5 Since h < AB, there are two solutions. Draw two triangles and solve. C 4 B 8 6 h A B B C A b 8 4 C A b sin 4 = sin A 6 sin A = 8 sin 4 8 sin 4 sin A = 6 A = sin 8 sin 4 6 Arelated = 6. 0, with A in Quadrants Iand II. A = 6. 0 ( ) B = = = b sin 4 sin b 6 sin = sin 4 b = 895. units A = = ( ) B = = sin 4 = b sin 0. 0 b b 6 sin 00. = sin 4 = 33. units or 3. 4 units Final Practice Eam Answer Key 7 of 34

28 6. Find the measure of ÐA in DABC (to the nearest degree). ( marks) (Lesson 5) C 4 6 A 7 B a = b + c bc cos A 6 = ( 4)( 7) cos A = ( 4)( 7) cos A 9 = 56 cos A = cos A A = cos ( ) A = 59 8 of 34 Grade Pre-Calculus Mathematics

29 Name: Module 8: Absolute Value and Reciprocal Functions (9 marks). Evaluate. ( mark each, for a total of marks) (Lesson ) a) 4 3( 7) ( 7) = = 4(7) 3 = 65 b) (3) (3) = = (5) + 5 = 5. Solve the following absolute value equation algebraically. (4 marks) (Lesson 3) = 6 Case : = 6 and ³ 0 6 = 0 ( ) = 0 ( 3)( + ) = 0 = 0, = 3, + + Î (, 0] È [, ) Since = 3 is in the interval [, ) and = is in the interval (, 0], both are solutions. Case : ( ) = 6 and < 0 + = = = 0 b 4ac = ( ) 4()(6) = 4 = 3 There is no solution. 0 Therefore, the solutions to this absolute value equation are = 3 and = Final Practice Eam Answer Key 9 of 34

30 3. Solve the following absolute value equation graphically. (4 marks) + 4 = 5 (Lesson 3) From the left side of the equation, graph y = + 4. From the right side, graph y = 5. The two graphs intersect at the points ( 9, 5) and (, 5). The solutions for the equation are = 9 and =. y y = y = of 34 Grade Pre-Calculus Mathematics

31 Name: 4. Sketch the graph of the reciprocal of the given function. State the equation(s) of the horizontal and vertical asymptote(s). (4 marks) (Lesson 4) y = 6 The reciprocal function is: f ( )= 6 y-intercept: Find f(0): f ( 0 )= 0 6 = ( ) 6 = 6 Vertical asymptote: 6 = 0 = 6 = 3 Horizontal asymptote: The horizontal asymptote is the line y = 0. Invariant points: Let 6= = 7 = 7 or 6= = 5 5 = The invariant points are 7 5, and,. Final Practice Eam Answer Key 3 of 34

32 Draw the curve: y y = 6 4 = 3 3 of 34 Grade Pre-Calculus Mathematics

33 Name: 5. Given the graph of y =, f ( ) sketch the graph of y = f(). (5 marks) (Lesson 5) y 3, 9 = = 4 The vertical asymptotes of this reciprocal graph are at = and = 4, and are thus the -intercepts in the graph of f(). In other words, f() is of the form f() = a( + )( 4). The key point is,. The corresponding point in the graph of the function, f(), is 9 (, 9). The invariant points can be read from the graph and are approimately (., ), (4., ), (.8, ), and (3.8, ). Final Practice Eam Answer Key 33 of 34

34 The parabola has its verte at (, 9) and opens up. y y = f of 34 Grade Pre-Calculus Mathematics