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 ) Midterm Practice Exam Answer Key

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1 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 ) Midterm Practice Eam Answer Key

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3 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 Midterm Practice Eam Answer Key Name: Student Number: Attending q Non-Attending q Phone Number: Address: For Marker s Use Only Date: Final Mark: /100 = % Comments: Answer Key Instructions The midterm eamination will be weighted as follows: Modules % The format of the eamination will be as follows: Module 1: Sequences and Series 4 marks Module : Factoring and Rational Epressions 3 marks Module 3: Quadratic Functions 5 marks Module 4: Solving Rational and Quadratic Equations 8 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 Midterm Eam Resource Sheet. Your Midterm Eam Resource Sheet must be handed in with the eam. You will receive your Midterm Eam Resource Sheet back from your tutor/ marker with the net module work that is submitted for marking. Show all calculations and formulas used. Include units where appropriate. Clearly state your final answer. Midterm Practice Eam Answer Key 3 of 6

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5 Name: Answer all questions to the best of your ability. Show all your work. Module 1: Sequences and Series (4 marks) 1. For each sequence below, indicate whether it is geometric, arithmetic, or neither. If it is geometric, state the value of r (the common ratio). If it is arithmetic, state the value of d (the common difference). (1 mark each, for a total of 3 marks) (Lesson 3) a) 81, 9, 1, 1,... 9 Geometric sequence with a common ratio of 1 9. b) 1, 6, 9, 10,... Neither. c) 1, 4, 7, 10,... Arithmetic sequence with a common difference of 3.. Eplain the relationship between arithmetic sequences and linear functions. ( marks) (Lesson 1) If the domain of a linear function is the set of natural numbers (or positive integers), the range is called an arithmetic sequence. Also, the common difference of an arithmetic sequence is the same as the slope in the corresponding linear function. Midterm Practice Eam Answer Key 5 of 6

6 3. Write the defining linear function of the following arithmetic sequence. ( marks) (Lesson 1) 3, 8, 13,... One point on line: (1, 3) Slope: d = 8 3 = 5 Defining linear function: y y 1 = m( 1 ) y 3 = 5( 1) y = y = 5 4. Which term of the arithmetic sequence -4,, 8,... is the number 170? (3 marks) (Lesson 1) Method 1: Using the general term formula t n = t 1 + (n 1)d: 170 = 4 + (n 1)(6) = 6n = 6n 180 = 6n 30 = n Therefore, the 30th term is 170. Method : Using the defining linear equation: m = 6 and points (1, 4) and (n, 170) y y 1 = m( 1 ) 170 ( 4) = 6(n 1) 174 = 6(n 1) 9 = n 1 30 = n Therefore, the 30th term is of 6 Grade 11 Pre-Calculus Mathematics

7 Name: 5. Find the sum of the first 15 terms defined by t 1 = 4, t n = t n (3 marks) (Lesson ) The arithmetic sequence is 4, 11, 18,... The common difference = 7 To find t 15, use the formula t n = t 1 + (n 1)d Or t 15 = 4 + (15 1)7 t 15 = t 15 = Sum of the series: S 15 = + ( 15)= S n Sn = t + ( n ) d = ( 4)+ ( 15 1)( 7) 15 S15 = 8 + ( 14)( 7) S = Find the total distance travelled by a ball in coming to rest, if it is dropped from a height of 1 m and it rebounds 3 of its previous height every time it hits the ground (4 marks) (Lesson 5) down series: ( )+ 1 3 ( )+... down distance: 1 = 36 m up series: 3 1 ( ) ( )+ 3 ( )+... up distance = down distance 1 = 4 m Total distance travelled = up distance + down distance = 4 m + 36 m = 60 m. Midterm Practice Eam Answer Key 7 of 6

8 7. Consider the geometric sequence 1,, 4,... (Lesson 3) a) Write the function for the geometric sequence. ( marks) t r = = 1 t t 1 n n = 1 = t r n 1 1 = 1( ) n 1 b) Using this function, find t 14 of the sequence. (1 mark) t 14 = 1( ) 14 1 t 14 = 1( ) 13 t 14 = 1( 819) t 14 = 819 c) Which term of the geometric sequence is the number 4096? ( marks) t n = 1( ) n = 1( ) n = ( ) n 1 ( ) 1 = ( ) n 1 1 = n 1 13 = n Therefore, 4096 is the 13th term in this geometric sequence. 8 of 6 Grade 11 Pre-Calculus Mathematics

9 Name: k 1 k 8. If possible, find the value of = 3 r= and r satisfies the condition: r < 1 4 t1 = 64 t S = 1 if r < 1 1 r S = 64 = ( marks) (Lesson 5) Midterm Practice Eam Answer Key 9 of 6

10 Module : Factoring and Rational Epressions (3 marks) 1. Factor the following epressions completely. (Lessons 1 and ) a) y 7y + 10y ( marks) = y( ) = y( )( 5) b) ( marks) = = ( + 3) + 3( + 3) = ( + 3)( + 3) = ( + 3) c) 36 64y (1 mark) = 4(9 16y ) = 4(3 4y)(3 + 4y) d) 5( 4) 49(y 5) (3 marks) 5( 4) 49(y 5) Let w = 4 t = y 5 5w 49t (5w 7t)(5w + 7t) [5( 4) 7(y 5)][5( 4) + 7(y 5)] [5 0 7y + 35][ y 35] (5 7y + 15)(5 + 7y 55) 10 of 6 Grade 11 Pre-Calculus Mathematics

11 Name:. Without factoring, determine whether the binomial ( 1) is a factor of the following polynomial. Eplain. ( marks) (Lesson ) 1 If ( 1) were a factor of this polynomial, then when = 1 is substituted into the polynomial, the polynomial would equal zero. (1) 1 1 = 1 1 = 1 1 = 0 This polynomial equals zero when = 1 is substituted in for. Therefore, ( 1) is a factor of this polynomial. 3. Create an equivalent rational epression for the following rational epression. Eplain how you know the rational epression you created is equivalent to the original rational epression. State the non-permissible values of your rational epression. Do not simplify your answer. ( marks) (Lesson 3) Answers may vary. A sample answer is: ( )( ) + + = ( ) The numerator and denominator are both multiplied by the same non-zero polynomial. Because any polynomial divided by itself is one, you are essentially multiplying the epression by 1. Anything multiplied by 1 is equal to itself. The non-permissible values are = 0 and =. Midterm Practice Eam Answer Key 11 of 6

12 4. Perform the indicated operation and simplify your answer. State the non-permissible values. a) (4 marks) (Lesson 5) = ( ) ( + 3) + ( ) + 1 ( )( + LCD = 3) + 1 = = = ( ) ( ) ( + 1) ( + 1) ( ( )( + )( + ) + + 3) 3 1 ( ) ( + 1) ( + 3) ( ) ( + 3) ( + 1) ( ) ( + 3) ( + 1) Non-permissible values: =, 3, 1 b) (3 marks) (Lesson 4) ( 1 4 = + )( ) ( ( 3) ( + ) + ) = + 3 ( ) ( ) ( )( ) ( 4 3 ( )( + ) + )( ) ( 5) ( + 4) 1 = 5 ( ) Non-permissible values: = 3,, 5, 4 1 of 6 Grade 11 Pre-Calculus Mathematics

13 Name: c) (3 marks) (Lesson 4) ( ) = 4 6 ( ) ( + ) ( + 1) = ( )( + ) ( 3) + 4 = 3 Non-permissible values: = 0, 1,, 3, ( ) 5. Create a rational epression with the following non-permissible values. Do not simplify your answer. (1 mark) (Lesson 3) 3 and 5 If 3 and 5 are non-permissible values, then ( + 3) and ( + 5) are factors of the denominator of the rational epression. Answers may vary. However, ( + 3) and ( + 5) have to be factors of the denominator. A sample answer is: 1 ( + 3) ( + 5) Midterm Practice Eam Answer Key 13 of 6

14 Module 3: Quadratic Functions (5 marks) 1. Given the graph below, answer the following questions. (Lessons 1 to 4) y a) What are the coordinates of the verte? (1 mark) (1, 5) b) What is the range? (1 mark) R: {y y ³ 5} or [ 5, ) or {y y ³ 5, y Î Â} c) What is the equation of the ais of symmetry? (1 mark) = 1 d) State whether the graph has a maimum or a minimum value and what that value is. (1 mark) It has a minimum value when y = 5. e) Write a quadratic function in verte form to represent this graph. ( marks) Since the graph is normal width, the value of a is 1. y = ( 1) 5 14 of 6 Grade 11 Pre-Calculus Mathematics

15 Name:. Match each equation to its corresponding graph. Place (a), (b), (c), and (d) net to its corresponding graph below. (1 mark each, for a total of 4 marks) (Lessons 1 to 4) a) y = ( ) 1 b) y= ( ) + 1 c) y= ( + ) + d) y = ( + ) y y 0 0 (b) (c) y y 0 0 (a) (d) 3. For what value of k is the epression 11 + k a perfect square trinomial? (1 mark) (Lesson 5) 11 = 11 4 Midterm Practice Eam Answer Key 15 of 6

16 4. Consider the parabola y = (Lesson 5) a) Write the function in verte form by completing the square. ( marks) y = 3( + 4) 36 y = 3( ) 36 4(3) y = 3( + ) 48 b) Find the coordinates of the verte. (1 mark) (, 48) c) Find the y-intercept. (1 mark) Let = 0 y = y = 3(0) + 1(0) 36 y = 36 d) State the domain. (1 mark) D: { Î Â} or (, ) e) Write an equation for the ais of symmetry. (1 mark) = 16 of 6 Grade 11 Pre-Calculus Mathematics

17 Name: f) Sketch the graph. ( marks) The verte is (, 48). The curve opens up and has a narrow width. The y-values are three times the normal ones. y g) Find the -intercepts from the graph. (1 mark) The -intercepts are located at = 6 and =. Midterm Practice Eam Answer Key 17 of 6

18 5. A projectile is shot straight up from a height of 40 m with an initial velocity of 30 m/s. Its height in metres after t seconds is given by h(t) = t 5t. (1.5 marks each, for a total of 3 marks) (Lesson 7) a) After how many seconds does the projectile reach its maimum height? h(t) = 5t + 30t + 40 b 30 Time at maimum height = = = 3seconds a ( 5 ) Note: This question could also be solved by completing the square to find the verte at (3, 85), where 3 seconds is the answer to (a) and 85 metres is the answer to (b). b) Find the maimum height above the ground that the ball reaches. Maimum height = h(3) = 5(3) + 30(3) + 40 = 85 metres 6. Consider the following quadratic functions. Determine how many -intercepts the corresponding graph has by considering the values of a and q. (1 mark each, for a total of marks) (Lesson 6) a) y = ( + 3) 15 In this equation, a = < 0, and q = -15 < 0. Therefore, as both a and q are less than zero, the corresponding graph has no -intercepts. Or The graph opens down and the verte is below the -ais. There are no -intercepts. b) y = 9( + ) In this equation, a = 9 > 0, and q = 0. Therefore, the corresponding graph has one (1) -intercept, as the verte is on the -ais. Or The graph opens up and the verte is on the -ais. There is one -intercept. 18 of 6 Grade 11 Pre-Calculus Mathematics

19 Name: Module 4: Solving Rational and Quadratic Equations (8 marks) 1. Find the roots of the following equations. Eplain which method you used to solve each equation and why. (Lessons 1 to 4) a) = 0 ( marks) This equation is easily factorable. ( 6)( 3) = 0 = 6, 3 b) 5 1 = 0 (3 marks) This equation is not easily factorable. Also, if you were to complete the square, there would be fractions. Therefore, the quadratic formula should be used. Note: If you use the completing-the-square method, you should still arrive at the correct answer. b b ac = ± 4 a 5 1= 0 a= 1, b= 5, c = 1 = ± ( ) ( 1 )( 1 ) ( 1) 5 = ± = ± 9 5 = , Note: Unless specifically mentioned, you may use any method to solve quadratic equations. Midterm Practice Eam Answer Key 19 of 6

20 . Use the quadratic formula to find the roots of = 0. ( marks) (Lesson 4) b b ac = ± 4 a a=, b= 6, c = 1 = ± ( )( 1 ) ( ) = 6 ± = 6± = ± = ± = ± = +, 0 of 6 Grade 11 Pre-Calculus Mathematics

21 Name: 3. Solve 3 = 0 by completing the square. (3 marks) (Lesson 4) 3= 0 = = = =± 1 = ± = , 4. Solve ( 7) = 36 by taking square roots. ( marks) (Lesson ) ( 7) = 36 7 = ±6 = 7 ± 6 = 7 + 6, 7 6 = 13, 1 Midterm Practice Eam Answer Key 1 of 6

22 5. Solve ( ) 1 = 0 by graphing. ( marks) (Lesson 1) Let y = ( ) 1 Verte is (, 1); the curve opens up and its width is normal. y 4 4 The -intercepts (or zeros, or roots) of this quadratic equation are 3 and 1. of 6 Grade 11 Pre-Calculus Mathematics

23 Name: 6. Solve for. State the non-permissible values of the rational equation. (4 marks) (Lesson 6) ( ) + = ( ) + = ( )( ) LCD = + 3 Non-permissible values are = 3, ( + 3) ( ) 3( + 3) ( )= + 3 ( ( ) ( + 3) ( ) ) ( ) 3( + 3 6)= 4( + 3) + 3 ( ) ( 4+ 4) 3( + 6)= ( ) = = = = = 3( + 1) ( )( + ) 0= = 14 1, 3 ( ) Midterm Practice Eam Answer Key 3 of 6

24 7. The discriminant of the quadratic equation 4 + k = 0 is 84. (1 mark each, for a total of marks) (Lesson 5) a) Find the value of k. a = 4, b =, c = k The discriminant = b 4ac 84 = () 4(4)( k) 84 = k 80 = 16k 5 = k b) State the nature of the roots without solving for the roots. As the discriminant is positive and not a perfect square, there are real, irrational, unequal roots. 4 of 6 Grade 11 Pre-Calculus Mathematics

25 Name: 8. Tia can ride her bicycle 1 km/h faster than Denzel. They begin at the same spot and at the same time, travelling in opposite directions. After travelling for the same length of time, they stop and realize that Tia has travelled 1 km and Denzel has travelled 19 km. How fast was Denzel travelling? (4 marks) (Lesson 6) Recall: Distance = Rate Time, or Time = Distance Rate Let r = Denzel s rate. Rate Time = Distance Tia r r 1 1 Denzel r 19 r 19 Tia s time of travel = Denzel s time of travel 1 19 r + 1 = r LCD = ( r + 1)( r) 1 19 ( r + 1)( r)= ( r + 1)( r) r + 1 r ( ) 1r= 19 r + 1 1r= 19r + 19 r = 19 r = 95. Therefore, Denzel was travelling at 9.5 km/h. Midterm Practice Eam Answer Key 5 of 6

26 9. Andrea can mow a lawn by herself in 4 hours. Samantha can mow the same lawn in 3 hours. If the girls work together, how long will it take to mow this lawn? (4 marks) (Lesson 6) The following equation can be used to solve this problem: (per hour rate) (hours worked) = portion of lawn mowed In one hour, the Andrea can mow 1 4 of the lawn. Therefore, Andrea has a per-hour rate of 1 4. In one hour, Samantha can mow 1 3 of the lawn. Therefore, Samantha has a per-hour rate of 1 3. Andrea Samantha Per Hour Rate Hours Worked Portion of Lawn Mowed When both girls are working together, the portion of the lawn mowed in hours is the whole lawn, and is represented in the equation as 1. The equation becomes: + = LCD = = = = 1 1 = 17. hours 7 Therefore, both of these girls working together can mow the lawn in approimately 1.7 hours of 6 Grade 11 Pre-Calculus Mathematics