High School Pre-Calculus Tutor

By The Editors of REA

Specifically designed to meet the needs of high school students, REA’s High School Pre-Calculus Tutor presents hundreds of solved problems with step-by-step and detailed solutions. Almost any imaginable problem that might be assigned for homework or given on an exam is covered. Topics include algebraic laws and operations, coordinate system relations, linear functions, sequences, series, graphing, limits, and applications. A valuable study aid for students taking upper-level mathematics courses. Fully indexed for locating specific problems rapidly.

• Language: English
• Publisher: Research & Education Association
• Release date: Jan 1, 2013
• ISBN: 9780738668420

Book preview

Preparation

SECTION 1

ALGEBRA

CHAPTER 1

NUMBERS AND NOTATIONS

PROBLEM 1-1

Given these two number lines, explain how the space between the points (numbers) on each line differs.

SOLUTION:

The spacing between any two points on line 1 represents an equal difference. The numbers on this line are spaced in an arithmetic progression. (For more on progression see Chapter 14.) The spacing on line 2 is based on a geometric progression. The space between any two points represents a progressively larger difference as one moves from left to right on the number line.

PROBLEM 1-2

Is it possible to establish a one-to-one correspondence between the set of real numbers and the points on a number line?

SOLUTION:

It is not possible since there are an infinity of real numbers. However, by the Number Line Postulate, one must assume that such a correspondence exists. The Number Line Postulate is also known as the axiom of continuity.

PROBLEM 1-3

Use the field properties to derive the equation x = 5 from the equation 5x − 3 = 2(x + 6).

SOLUTION:

We could also derive 5x − 3 = 2(x + 6) from x = 5 by reversing the steps in the solution. Let us see if 5 will make the equation 5x − 3 = 2 (x + 6) true.

Two equations are equivalent if and only if they have the same solution set. Since 5x – 3 = 2(x + 6) and x = 5 have the same solution set, {5}, the two equations are equivalent.

PROBLEM 1.4

Is zero a natural number? Is zero an integer?

Is zero a positive number or a negative number?

Is zero an odd number or an even number?

Give three examples of each of the following:

Integers

Rational numbers

Irrational numbers

Natural numbers

Prime numbers

Complex numbers

What is a real number?

SOLUTION:

(a) Zero is not a natural number. The natural numbers begin with 1 and continue up through positive infinity, i.e., 1, 2, 3, 4, 5, 6, 7, .... Zero is an integer. Zero divides the positive integers from the negative integers on a number line. An example of a set of integers is ..., −5, −4, −3, −2, −1, 0, 1, 2, 3, 4, 5, .... Zero is an even number as 2, 4, 6, and 8. Zero is neither a positive number nor a negative number.

(b)

An integer is a whole number (not a fraction or a decimal) which can be either positive or negative. Examples of integers are 1, 5, and −9.

where p and q are integers, and q ≠ 0.

.

. Examples of irrational numbers are −√2, π.

Examples of natural numbers are 2, 7, and 9.

A prime number is a number which is divisible by only ±1, and plus or minus itself. Examples of prime numbers are 2, 3, and 13.

A complex number is a number of the form a + bi, where a and b . The real part of a complex number is a, and the imaginary part of a complex number is bi. Examples of complex numbers are 6 + √3i, 4.5 − 2i, and −9 + 16i.

(c) The real numbers consist of all positive and negative rational and irrational numbers, and zero.

PROBLEM 1-5

Classify each of the following numbers into as many different sets as possible. Example: real, integer, rational ...

0

9

√6

1.5

SOLUTION:

Zero is a real number and an integer.

9 is real, rational, natural number, and an integer.

√6 is an irrational, real number.

is a rational, real number.

is a rational, real number.

1.5 is a rational, real number, and a decimal.

PROBLEM 1-6

Find the absolute value for each of the following:

0

4

– π

a, where a is a real number

SOLUTION:

The absolute value of a number is represented by 2 vertical lines around the number, and is equal to the given number, regardless of sign.

= 0

= 4

= π

for a a = a for a a = 0 for a a = – a,

PROBLEM 1-7

as a repeating decimal.

SOLUTION:

To write a fraction as a repeating decimal, divide the numerator by the denominator, until a pattern of repeated digits appears.

2 ÷ 7 = .285714285714...

Identify the entire portion of the decimal which is repeated. The repeating decimal can then be written in the shortened form:

PROBLEM 1.8

Find the common fraction form of the repeating decimal 0.4242....

SOLUTION:

Let x represent the repeating decimal.

(1)

Divide both sides of equation (1) by 99.

The repeating decimal of this example had 2 digits that repeated. The first step in the solution was to multiply both sides of the original equation by the 2nd power of 10 or 10² or 100. If there were 3 digits that repeated, the first step in the solution would be to multiply both sides of the original equation by the 3rd power of 10 or 10³ or 1,000.

PROBLEM 1-9

Use scientific notation to express each number.

4,375

186,000

0.00012

4,005

SOLUTION:

A number expressed in scientific notation is written as a product of a number between 1 and 10 and a power of 10. The number between 1 and 10 is obtained by moving the decimal point of the number (actual or implied) the required number of digits. The power of 10, for a number greater than 1, is positive and is one less than the number of digits before the decimal point in the original number. The power of 10, for a number less than 1, is negative and one more than the number of zeros immediately following the decimal point in the original number. Hence,

4,375 = 4.375 x 10³

186,000 = 1.86 x 10⁵

0.00012 = 1.2 × 10 – 4

4,005 = 4.005 × 10³

CHAPTER 2

DEFINITIONS AND NOTATIONS OF SETS AND SET OPERATIONS

PROBLEM 2-1

List all the subsets of C = {1, 2}.

SOLUTION:

{1}, {2}, {1,2}, ø, where ø is the empty set. Each set listed in the solution contains at least one element of the set C. The set {2, 1} is identical to {1, 2} and therefore is not listed. ø is included in the solution because ø is a subset of every set.

PROBLEM 2-2

Given the set S = {1, 2, 3, 4, 5, 6}, find a partition of S.

SOLUTION:

A partition of a set S is a subdivision of the set into subsets that are disjoint and exhaustive, i.e., every element of S must belong to one and only one of the subsets. Each subset in the partition is also called a cell.

Therefore, (S1, S2, ..., Sn) is a partition of S if

(a) Si ∩ Sj = ø (where ø is the empty set) for all i ≠ j (the cells are disjoint), and

(b) S1 ∪S2 ∪S3∪...∪Sn = S (the cells are exhaustive).

Hence, one of the partitions of S is

{ {1, 2, 3},{ +4},{ +5, 6} }. The partition { {1}, {2}, {3}, {4}, {5}, {6} } is a partition

into unit sets.

PROBLEM 2-3

U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}, P = {2, 4, 6, 8, 10}, Q .

SOLUTION:

are the complements of P and Q, respectively.

is the set of all elements in the universal set, U, that are not elements of Pis the set of elements in U that are not in Q, Therefore,

(a)

; (b)

.

PROBLEM 2-4

Show that the complement of the complement of a set is the set itself.

SOLUTION:

The complement of set A is given by A’. Therefore, the complement of the complement of a set is given by (A’)’. This set, (A’)’, must be shown to be the set A; that is, that (A’)’ = A In the figure, the complement of the set A, or A’, is the set of all points not in set A; that is, all points in the rectangle that are not in the circle. This is the shaded area in the figure. Therefore, this shaded area is A’. The complement of this set, or (A’)’, is the set of all points of the rectangle that are not in the shaded area; that is, all points in the circle, which is the set A. Therefore, the set (A’)’ is the same as set A; that is (A’)’ = A.

PROBLEM 2-5

State the laws of set operations.

SOLUTION:

If S is an algebra of sets and if A, B, C, ... , ø, U,... are elements of S, then the following hold for ∩, U, and ’.

IDENTITY LAWS

1a. A ∪ ø = A

2a. A ∪ U = U

1b. A ∩ ø = ø

2b. A ∩ U = A

IDEMPOTENT LAWS

3a. A ∪ A = A

3b. A ∩ A = A

COMPLEMENT LAWS

4a. A ∪ A’ = U

5a. (A’)’ = A

4b. A ∩ A’ = ø

5b. ø’ = U; U’ = ø

COMMUTATIVE LAWS

6a. A ∪ B = B ∪ A

6b. A ∩ B = B ∩ A

ASSOCIATIVE LAWS

7a. (A ∪ B) U C = A ∪ (B ∪ C)

7b. (A ∩ B) ∩ C = A ∩ (B ∩ C)

DISTRIBUTIVE LAWS

8a. A ∪ (B ∩ C) = (A ∪ B) n (A ∪ C)

8b. A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)

DE MORGAN’S LAWS

9a. (A ∪ B)’ = A’ ∩ B’

9b. (A ∩ B)’ = A’ ∪ B’

PROBLEM 2-6

If A = {2, 3, 5, 7) and B = {1, −2, 3, 4, −5, √6}, find (a) A U B and (b) A ∩ B.

SOLUTION:

(a) A U B is the set of all elements in A or in B or in both A and B, with no element included twice in the union set.

A U B = {1, 2, −2, 3, 4, 5, −5, √6, 7}

(b) A ∩ B is the set of all elements in both A and B.

A ∩ B = {3}

Sometimes two sets have no elements in common. Let S = {3, 4, 7} and T = {2, −4, 6}. What is the intersection of S and T? In this case S ∩ T has no elements. Hence S ∩ T = ø, the empty set. In that case, the sets are said to be disjoint.

The set of all elements entering a discussion is called the universal set, U. When the universal set is not given, we assume it to be the set of real numbers. The set of all elements in the universal set that are not elements of A is called the complement of A.

PROBLEM 2-7

Find A − B and A − (A ∩ B) for

A = {1, 2, 3, 4} and

B = {2, 4, 6, 8, 10}.

SOLUTION:

The relative complement of subsets A and B of the universal set U is defined as the set

A − B = {xx ∈ A and x ∉ B}. Note that it is not assumed that B ⊆ A.

Hence, A − B = {1, 3}.

To find A − (A ∩ B), first find A ∩ B. The set A ∩ B is a set of elements that are common to both A and B, so A ∩ B = {2, 4}.

Therefore, A − (A ∩ B) = {1, 3}.

CHAPTER 3

FUNDAMENTAL ALGEBRAIC LAWS AND OPERATIONS WITH NUMBERS

PROBLEM 3-1

Evaluate 2 − {5 + (2 − 3) + [2 − (3 − 4)]}.

SOLUTION:

When working with a group of nested parentheses, we evaluate the innermost parenthesis first.

Thus, 2 − {5 + (2 − 3) + [2 − (3 − 4)]}

= 2 − {5 + (2 − 3) + [2 − (−1)]}

= 2 − {5 + (− 1) + [2 + 1]}

= 2 − {5 + (− 1) + 3}

= 2 − {4 + 3}

= 2 − 7

= −5

PROBLEM 3-2

Simplify 4[−2(3 + 9) ÷ 3] + 5.

SOLUTION:

To simplify means to find the simplest expression. We perform the operations within the innermost grouping symbols first. That is 3 + 9 = 12.

Thus, 4[−2(3 + 9) ÷ 3] + 5 = 4[−2(12) ÷ 3] + 5

Next we simplify within the brackets:

= 4 [−24 ÷ 3] + 5

= 4 • (−8) + 5

We now perform the multiplication, since multiplication is done before addition:

= −32 + 5

= −27

Hence, 4 [−2(3 + 9) ÷ 3] + 5 = −27.

PROBLEM 3-3

Calculate the value of

to the fifth decimal place.

SOLUTION:

The rule for solving this type of problem is to work from the innermost parenthesis out. The order of operations are multiplication and division, then subtraction and addition.

The first step in this problem is to change everything to decimals.

+ 4 ÷ (1.44444)]

The second step is to reduce the parentheses.

5 − [6 . [1.5 • 0.1 + (4.05) + 0.5] - 2 ÷ 3.7 + 4 ÷ (1.44444)]

= 5 − [6 • [0.15 + 8.1] − 0.540541 + 2.769231]

= 5 − [6 • (8.25) − 0.540541 + 2.769231]

= 5 − [49.5 + 2.228690]

= 5 − 51.72869

= −46.72869

PROBLEM 3-4

Calculate the value of each of the following expressions:

)

SOLUTION:

Before solving this problem, one must remember the order of operations: parenthesis, multiplication and division, addition and subtraction.

= 5

0.016 + 0.18 + 12(26 − (−2) ÷ 2) 0.016 + 0.18 + 12(26 − (−1)) 0.196 + 12(27) = 0.196 + 324 = 324.196

PROBLEM 3-5

.

SOLUTION:

, we must convert 2 into thirds.

Since division by a fraction is equivalent to multiplication by that fraction’s reciprocal

Therefore,

.

PROBLEM 3-6

Simplify

SOLUTION:

into sixths:

into twelfths:

Thus,

Division by a fraction is equivalent to multiplication by the reciprocal; hence,

Cancelling 6 from the numerator and denominator:

A second method is to multiply both the numerator and the denominator by the least common denominator of the entire fraction. Since we have already seen that LCD of the numerator is 6 and the LCD of the denominator is 12, and 12 is divisible by 6, we use 12 as the LCD of the entire fraction. Thus,

PROBLEM 3-7

when a , and c =