Grade 11 Pre-Calculus Math Exam Answer Key, Schemes and Mind Maps of Pre-Calculus

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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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Answer all questions to the best of your ability. Show all your work.

Module 1: Sequences and Series (5 marks)

1. Write the defining linear function of the following arithmetic sequence. (2 marks) (Lesson 1) 99, 103, 107,... Answer: Points on line: (1, 99), (2, 103), (3, 107) Slope: d = 4 Defining Linear Function: y – y 1 = m ( x – x 1 ) y – 99 = 4( x – 1) y = 4 x – 4 + 99 y = 4 x + 95
2. Use a formula to find the value of 81

8 ^1

^

− ∑ k

k

. (^) (3 marks) (Lesson 4)

Answer: Notice k ’s initial value is 2. Thus, n is only 7, not 8.

The first few terms are 81 1 3

^1 2

^

^

^

^

^

Thus, t 1 = 27 and r = 1 3

S

t r r

S

n

n

( − ) −

^

^

1

7

7

^

^

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Module 3: Quadratic Functions (5 marks)

1. Given the following parabola in vertex form, complete the following questions. (1 mark each, for a total of 5 marks)

y = 1 (^ x − )^ +

3 2 2 (Lessons 1 to 4)

a) Identify the range. Answer: R: { y | y ³ 2, y Î Â}

b) Identify the direction of the opening. Answer: The parabola opens upward

c) Identify the axis of symmetry. Answer: x = 3

d) Identify the y -intercept. Answer: y x x

y

y

y

= (^ − )^ +

= (^ − )^ +

2

2

let

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e) Sketch the graph of the parabola. Answer: The vertex is (3, 2), the curve opens up, and its size is wide. The y -values are 1 2 of the normal y -values.

x

y



y = 12 ( x  3)^2 + 2

2 

3



13 2

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1. Solve the following rational equation. Identify any non-permissible values. (3 marks) 3 1 1

x x

x x

• (Lesson 6)

Answer: Non-permissible values: x = 0 or x = – LCD = x ( x + 1) 3 1 1 1 1

3 1 1 3 4 1 4

2 2

x x x^ x^

x x x^ x x x x x x x x x

( (^) + )=

( ) ( (^) + )

( + ) ( + )=− ( )

• • =− 22 2

2

( (^) + ) + ( (^) + )=

( + ) =

=−

x x x x x x x x

x

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Module 5: Radicals (19 marks)

1. Order the following radical expressions from least to greatest. Do not use a calculator. (2 marks) (Lesson 1) 6 3 , 11 7 , 9 5 , 2 15

Answer: 6 3 36 3 108 11 7 121 7 847 9 5 81 5 405 2 15 4 15 60 60 108 405

Thus, < < < 8847 and 2 15 < 6 3 < 9 5 < 11 7.

1. Simplify each of the following. All answers must have rationalized denominators. Assume all variables are non-negative. (Lesson 2) a) 3 19  2 76 (1 mark) Answer: 3 19 2 76 3 19 2 4 19 3 19 2 2 19 3 19 4 19 7 19

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1. Identify the values for each of the variables for which each radical expression is defined. (1 mark each, for a total of 2 marks) a) 6  3 x (Lesson 1) Answer: In order for this radical expression to be defined, the radicand has to be greater than or equal to zero. 6 – 3 x ³ 0 6 ³ 3 x 2 ³ x x £ 2 Thus, x £ 2 in order for this radical expression to be defined.

b) 32 x^2 (Lesson 1) Answer: Because the index is two, the radicand has to be greater than or equal to zero for this expression to be defined. As any value squared will always be greater than or equal to zero, x Î Â for this expression to be defined.

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1. Find the solutions for each of the following equations. Check your solutions for extraneous roots. Determine any restrictions on the variable. a) 3 − x = 2 x − 3 (4 marks) (Lesson 5) Answer: Restrictions on the variable: 2 3 0 2 3 3 2

x x

x

2 2 2

x x x x x x x x x x or x Check x = 2 Check x = 6 LHS RHS LHS RHS 3 – x 3 – 2 1

x −

3 – x 3 – 6

x −

LHS = RHS LHS ¹ RHS

Therefore, the only solution is x = 2.

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1. The period, P , measured in seconds, of a pendulum is the time it takes to complete one

full swing. The period can be found using the formula P = 2 L 9 8 π . , where L measures 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 extraneous solutions. (2 marks) (Lesson 5) Answer:

P = 2 L 9 8 π . Restrictions: L ³ 0 and P ³ 0 Note: P is positive or zero because it is equal to a positive number, 2p, multiplied by the principal square root. Also, it would not make sense for either variable to be negative.

P L

L

L

L

L

=(^ )

^

2 2

2

π

π

π

π

. ××

2

π L Check: L ³ 0 LHS RHS P 4.3 (^2) 9 8

π

π

L

LHS = RHS

The pendulum should be 4.5899 m long.

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Module 6: Systems of Equations and Inequalities (19 marks)

1. Solve the following system of equations algebraically and graphically. (6 marks) y x y x

= (^ − )

=− ( + ) +

2 2 (Lesson 2)

Answer: Algebraic Solution y x y x y x x y x x y x x

= (^ − )^ =− ( + ) + = − + =− ( + + ) + =− − +

2 2 2 2 2

Substitute the expression for y in Equation (1) into the y -variable of Equation (2). x x x x x x

2 2 2

Solve for y : y x y y

= (^) ( − ) = (^) ( − ) =

2 2

Thus, the solution to this system of equations is (0, 9).

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1. The sum of two numbers is 22. Their product is 117. a) Write a system of equations to represent this problem. (1 mark) (Lesson 1) Answer: Let the first number be represented by m and the second number be represented by n. The system of equations is: m n mn

b) Solve the system of equations to find the two numbers. (3 marks) Answer: Substitute m = 22 – n into nm = 117. (22 – n )( n ) = 117 22 n – n^2 – 117 = 0 n^2 – 22 n + 117 = 0 ( n – 13)( n – 9) = 0 n = 13 or n = 9 Substituting into the equation m = 22 – n , you get: n = 13 n = 9 m = 22 – n m = 22 – n m = 22 – 13 m = 22 – 9 m = 9 m = 13 Therefore, the two numbers are 9 and 13.

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1. Solve the following inequalities by graphing. (2 marks each for a total of 6 marks) a) y < 3 x – 8 (Lesson 3) Answer: The slope of the line is 3. The y -intercept is –8. The boundary line is dotted and the shading is below the line.

x

y

 8







y < 3 x  8



 2

b) 0 £ x^2 – 4 (Lesson 4) Answer: You are interested in the domain where the expression is greater than or equal to 0. Write x^2 – 4 ³ 0 and write the corresponding function as y = x^2 – 4. The parabola has its vertex at (0, –4), opens up, and has normal shape.

x

y

 4

 2   2



y = x^2  4

The parabola crosses the x -axis at – and 2. The parabola lies on or above the x -axis when x £ –2 and when x ³ 2. Therefore, the solution set is { x | x £ -2 or x ³ 2, x Î Â}.