Chapter 2 Relations And Functions

Given:

Set A = $\{1, 2, 3, 4\}$

Set B = $\{5, 7, 9\}$

To Find:

(i) A × B

(ii) B × A

(iii) Whether A × B = B × A

(iv) Whether n(A × B) = n(B × A)

Solution:

(i) A × B is the Cartesian product of set A and set B, which is the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$.

A × B = $\{(1, 5), (1, 7), (1, 9), (2, 5), (2, 7), (2, 9), (3, 5), (3, 7), (3, 9), (4, 5), (4, 7), (4, 9)\}$

(ii) B × A is the Cartesian product of set B and set A, which is the set of all ordered pairs $(b, a)$ where $b \in B$ and $a \in A$.

B × A = $\{(5, 1), (5, 2), (5, 3), (5, 4), (7, 1), (7, 2), (7, 3), (7, 4), (9, 1), (9, 2), (9, 3), (9, 4)\}$

(iii) To check if A × B = B × A, we compare the elements (ordered pairs) of both sets.

The elements of A × B are ordered pairs of the form $(a, b)$ where $a \in A$ and $b \in B$.

The elements of B × A are ordered pairs of the form $(b, a)$ where $b \in B$ and $a \in A$.

For the sets to be equal, they must contain exactly the same elements.

For example, $(1, 5) \in A \times B$ but $(1, 5) \notin B \times A$ because the order of elements in an ordered pair matters.

Therefore, A × B is not equal to B × A.

Answer to (iii): No, A × B is not equal to B × A.

(iv) To check if n(A × B) = n(B × A), we count the number of elements in each set.

n(A) is the number of elements in set A. $n(A) = 4$.

n(B) is the number of elements in set B. $n(B) = 3$.

The number of elements in A × B is given by the formula $n(A \times B) = n(A) \times n(B)$.

$n(A \times B) = 4 \times 3 = 12$.

From part (i), counting the elements in A × B confirms there are 12 ordered pairs.

The number of elements in B × A is given by the formula $n(B \times A) = n(B) \times n(A)$.

$n(B \times A) = 3 \times 4 = 12$.

From part (ii), counting the elements in B × A confirms there are 12 ordered pairs.

Since $12 = 12$, we have $n(A \times B) = n(B \times A)$.

Answer to (iv): Yes, n(A × B) is equal to n(B × A).

Principle of Equality of Ordered Pairs:

Two ordered pairs $(a, b)$ and $(c, d)$ are equal if and only if their corresponding components are equal, i.e., $a = c$ and $b = d$.

Solution:

(i) Given the equation of ordered pairs: $(4x + 3, y) = (3x + 5, -2)$

Using the principle of equality of ordered pairs, we equate the corresponding components:

From equation (1):

$4x - 3x = 5 - 3$

$x = 2$

From equation (2), we directly get:

$y = -2$

So, the values are $x = 2$ and $y = -2$.

(ii) Given the equation of ordered pairs: $(x - y, x + y) = (6, 10)$

Using the principle of equality of ordered pairs, we equate the corresponding components:

We now have a system of two linear equations in two variables:

Equation (3): $x - y = 6$

Equation (4): $x + y = 10$

We can solve this system using the elimination method. Add Equation (3) and Equation (4):

Now substitute the value of $x$ from Equation (5) into Equation (4):

So, the values are $x = 8$ and $y = 2$.

Given:

Set A = $\{2, 4, 6, 9\}$

Set B = $\{4, 6, 18, 27, 54\}$

Conditions for ordered pair $(a, b)$: $a \in A$, $b \in B$, $a$ is a factor of $b$, and $a < b$.

To Find:

The set of ordered pairs $(a, b)$ that satisfy the given conditions.

Solution:

We need to find ordered pairs $(a, b)$ where $a$ is an element from set A and $b$ is an element from set B, such that $a$ divides $b$ evenly (i.e., $a$ is a factor of $b$) and $a$ is strictly less than $b$.

Let's examine each element $a \in A$ and find elements $b \in B$ that satisfy both conditions:

For $a = 2 \in A$:

- $b = 4 \in B$: Is 2 a factor of 4? Yes ($4 = 2 \times 2$). Is $2 < 4$? Yes. Include $(2, 4)$.

- $b = 6 \in B$: Is 2 a factor of 6? Yes ($6 = 2 \times 3$). Is $2 < 6$? Yes. Include $(2, 6)$.

- $b = 18 \in B$: Is 2 a factor of 18? Yes ($18 = 2 \times 9$). Is $2 < 18$? Yes. Include $(2, 18)$.

- $b = 27 \in B$: Is 2 a factor of 27? No.

- $b = 54 \in B$: Is 2 a factor of 54? Yes ($54 = 2 \times 27$). Is $2 < 54$? Yes. Include $(2, 54)$.

For $a = 4 \in A$:

- $b = 4 \in B$: Is 4 a factor of 4? Yes ($4 = 4 \times 1$). Is $4 < 4$? No.

- $b = 6 \in B$: Is 4 a factor of 6? No.

- $b = 18 \in B$: Is 4 a factor of 18? No.

- $b = 27 \in B$: Is 4 a factor of 27? No.

- $b = 54 \in B$: Is 4 a factor of 54? No.

For $a = 6 \in A$:

- $b = 4 \in B$: Is 6 a factor of 4? No.

- $b = 6 \in B$: Is 6 a factor of 6? Yes. Is $6 < 6$? No.

- $b = 18 \in B$: Is 6 a factor of 18? Yes ($18 = 6 \times 3$). Is $6 < 18$? Yes. Include $(6, 18)$.

- $b = 27 \in B$: Is 6 a factor of 27? No.

- $b = 54 \in B$: Is 6 a factor of 54? Yes ($54 = 6 \times 9$). Is $6 < 54$? Yes. Include $(6, 54)$.

For $a = 9 \in A$:

- $b = 4 \in B$: Is 9 a factor of 4? No.

- $b = 6 \in B$: Is 9 a factor of 6? No.

- $b = 18 \in B$: Is 9 a factor of 18? Yes ($18 = 9 \times 2$). Is $9 < 18$? Yes. Include $(9, 18)$.

- $b = 27 \in B$: Is 9 a factor of 27? Yes ($27 = 9 \times 3$). Is $9 < 27$? Yes. Include $(9, 27)$.

- $b = 54 \in B$: Is 9 a factor of 54? Yes ($54 = 9 \times 6$). Is $9 < 54$? Yes. Include $(9, 54)$.

Combining the pairs that satisfy both conditions:

The set of ordered pairs is $\{(2, 4), (2, 6), (2, 18), (2, 54), (6, 18), (6, 54), (9, 18), (9, 27), (9, 54)\}$.

Given:

The relation R is defined as $R = \{(x, y) : y = x + \frac{6}{x} ; \text{where } x, y \in N \text{ and } x < 6\}$.

Here, N is the set of natural numbers $\{1, 2, 3, ...\}$.

To Find:

The domain and range of the relation R.

Solution:

The conditions for an ordered pair $(x, y)$ to be in the relation R are:

1. $x \in N$

2. $y \in N$

3. $x < 6$

4. $y = x + \frac{6}{x}$

From conditions 1 and 3, the possible values for $x$ are the natural numbers less than 6.

Possible values for $x$: $\{1, 2, 3, 4, 5\}$.

Now, we calculate the corresponding value of $y$ for each possible $x$ using the equation $y = x + \frac{6}{x}$ and check if $y$ is a natural number (condition 2).

When $x = 1$:

$y = 1 + \frac{6}{1} = 1 + 6 = 7$.

Since $7 \in N$, the ordered pair $(1, 7)$ is in R.

When $x = 2$:

$y = 2 + \frac{6}{2} = 2 + 3 = 5$.

Since $5 \in N$, the ordered pair $(2, 5)$ is in R.

When $x = 3$:

$y = 3 + \frac{6}{3} = 3 + 2 = 5$.

Since $5 \in N$, the ordered pair $(3, 5)$ is in R.

When $x = 4$:

$y = 4 + \frac{6}{4} = 4 + \frac{3}{2} = 4 + 1.5 = 5.5$.

Since $5.5 \notin N$, the ordered pair $(4, 5.5)$ is not in R.

When $x = 5$:

$y = 5 + \frac{6}{5} = 5 + 1.2 = 6.2$.

Since $6.2 \notin N$, the ordered pair $(5, 6.2)$ is not in R.

Thus, the relation R is the set of ordered pairs that satisfy all conditions:

$R = \{(1, 7), (2, 5), (3, 5)\}$.

The domain of R is the set of all first components of the ordered pairs in R.

Domain(R) = $\{1, 2, 3\}$.

The range of R is the set of all second components of the ordered pairs in R.

Range(R) = $\{7, 5, 5\} = \{5, 7\}$.

Definition of a Function:

A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B.

In other words, for a relation to be a function, no two distinct ordered pairs in the relation can have the same first element.

Given:

(i) Relation R$_1$ = $\{(2, 3), (\frac{1}{2}, 0), (2, 7), (-4, 6)\}$.

(ii) Relation R$_2$ = $\{(x, |x|) \text{ | } x \text{ is a real number}\}$.

To Determine:

Whether R$_1$ and R$_2$ are functions and justify the answer.

Solution:

(i) Consider the relation R$_1$ = $\{(2, 3), (\frac{1}{2}, 0), (2, 7), (-4, 6)\}$.

The domain of R$_1$ is the set of first elements of the ordered pairs: $\{2, \frac{1}{2}, -4\}$.

Observe the ordered pairs in R$_1$. The element $2$ from the domain is associated with two different elements in the set of second components: $3$ (in $(2, 3)$) and $7$ (in $(2, 7)$).

Since the element $2$ in the domain is related to more than one element in the set of second components, the relation R$_1$ does not satisfy the definition of a function.

Conclusion for (i): R$_1$ is not a function.

Justification: The first element $2$ is paired with two different second elements, $3$ and $7$.

(ii) Consider the relation R$_2$ = $\{(x, |x|) \text{ | } x \text{ is a real number}\}$.

The domain of R$_2$ is the set of all real numbers, denoted by $\mathbb{R}$.

For any real number $x$, its absolute value $|x|$ is uniquely defined. For instance, if $x = 5$, $|x| = 5$; if $x = -5$, $|x| = 5$; if $x = 0$, $|x| = 0$.

For every specific real number $x$, there is only one value for $|x|$. Therefore, each element in the domain ($\mathbb{R}$) is associated with exactly one element in the set of second components (which is the set of non-negative real numbers, $[0, \infty)$).

This satisfies the definition of a function.

Conclusion for (ii): R$_2$ is a function.

Justification: For every real number $x$, there is a unique value of $|x|$.

Given:

Function $f(x) = 2x^2 - 1$

Function $g(x) = 1 - 3x$

To Find:

The domain (set of x-values) for which $f(x) = g(x)$.

Solution:

For the functions $f(x)$ and $g(x)$ to be equal for a specific value of $x$, their values must be the same at that point.

So, we set $f(x)$ equal to $g(x)$:

Now, we rearrange the equation to form a standard quadratic equation by moving all terms to one side:

This is a quadratic equation in the form $ax^2 + bx + c = 0$. We can solve this by factoring.

We look for two numbers that multiply to $a \times c = 2 \times (-2) = -4$ and add up to $b = 3$. These numbers are $4$ and $-1$.

We rewrite the middle term ($3x$) using these two numbers:

Now, factor by grouping the terms:

Factor out the common binomial factor $(x + 2)$:

For the product of two factors to be zero, at least one of the factors must be zero.

Case 1:

Case 2:

The values of $x$ for which $f(x) = g(x)$ are $x = -2$ and $x = \frac{1}{2}$.

The domain for which the functions are equal is the set containing these values.

Domain = $\{-2, \frac{1}{2}\}$.

Concept:

The domain of a function is the set of all possible input values (x-values) for which the function is defined.

Solution:

(i) Given function: $f(x) = \frac{x}{x^2 + 3x + 2}$

This is a rational function (a fraction where the numerator and denominator are polynomials).

A rational function is defined for all real numbers except those values of $x$ that make the denominator equal to zero, because division by zero is undefined.

To find the values of $x$ that are excluded from the domain, we set the denominator equal to zero:

We can factor the quadratic expression:

For the product of two factors to be zero, at least one of the factors must be zero:

So, the function $f(x)$ is undefined when $x = -1$ or $x = -2$.

The domain of $f(x)$ is all real numbers except $-1$ and $-2$.

Using set notation, the domain is $\mathbb{R} \setminus \{-1, -2\}$, where $\mathbb{R}$ represents the set of all real numbers.

Domain of f(x): $\mathbb{R} \setminus \{-1, -2\}$.

(ii) Given function: $f(x) = [x] + x$

Here, $[x]$ denotes the greatest integer function (also known as the floor function), which gives the greatest integer less than or equal to $x$.

The greatest integer function $[x]$ is defined for all real numbers.

The function $x$ (the identity function) is also defined for all real numbers.

The sum of two functions is defined for all values of $x$ for which both functions are defined.

Since both $[x]$ and $x$ are defined for all real numbers, their sum, $f(x) = [x] + x$, is also defined for all real numbers.

The domain of $f(x)$ is the set of all real numbers.

Using set notation, the domain is $\mathbb{R}$.

Domain of f(x): $\mathbb{R}$.

Concept:

The range of a function is the set of all possible output values (y-values or function values) that the function can produce.

Solution:

(i) Given function: $f(x) = \frac{|x - 4|}{x - 4}$

First, let's determine the domain of the function. The function is defined when the denominator is not zero, i.e., $x - 4 \neq 0$, which means $x \neq 4$.

The domain of $f(x)$ is all real numbers except 4, i.e., $\mathbb{R} \setminus \{4\}$.

Now, let's analyze the function based on the definition of the absolute value function $|u|$.

$|u| = u$ if $u \geq 0$

$|u| = -u$ if $u < 0$

In our case, $u = x - 4$.

Case 1: $x - 4 > 0$, which means $x > 4$.

In this case, $|x - 4| = x - 4$.

So, $f(x) = \frac{x - 4}{x - 4}$. Since $x - 4 \neq 0$, we can cancel the terms.

$f(x) = 1$ for all $x > 4$.

Case 2: $x - 4 < 0$, which means $x < 4$.

In this case, $|x - 4| = -(x - 4)$.

So, $f(x) = \frac{-(x - 4)}{x - 4}$. Since $x - 4 \neq 0$, we can cancel the terms.

$f(x) = -1$ for all $x < 4$.

The function $f(x)$ can only take the values $1$ or $-1$.

The range of $f(x)$ is the set $\{-1, 1\}$.

Range of $\frac{|x - 4|}{x - 4}$: $\{-1, 1\}$.

(ii) Given function: $f(x) = \sqrt{16 - x^2}$

First, let's determine the domain of the function. For the square root of a real number to be a real number, the expression under the square root must be non-negative.

We can rewrite this as:

Taking the square root of both sides (and remembering both positive and negative roots for the variable):

This inequality means that $x$ must be between $-4$ and $4$, inclusive.

The domain of $f(x)$ is the closed interval $[-4, 4]$.

For any $x$ in the domain $[-4, 4]$, the value of $16 - x^2$ will be between $16 - (4)^2 = 16 - 16 = 0$ (when $x = -4$ or $x = 4$) and $16 - (0)^2 = 16$ (when $x = 0$).

So, $0 \leq 16 - x^2 \leq 16$ for all $x \in [-4, 4]$.

Now, consider the function value $f(x) = \sqrt{16 - x^2}$.

Since the square root symbol $\sqrt{\phantom{x}}$ by convention represents the principal (non-negative) square root, the value of $f(x)$ will always be non-negative.

As $16 - x^2$ ranges from $0$ to $16$ for $x \in [-4, 4]$, the value of $\sqrt{16 - x^2}$ will range from $\sqrt{0} = 0$ to $\sqrt{16} = 4$.

So, $0 \leq \sqrt{16 - x^2} \leq 4$ for all $x \in [-4, 4]$.

The range of $f(x)$ is the closed interval $[0, 4]$.

Range of $\sqrt{16-x^2}$: $[0, 4]$.

Given:

The function $f(x) = |x - 1| + |1 + x|$ with the domain $-2 \leq x \leq 2$.

To Redefine:

Rewrite the function $f(x)$ without using the absolute value signs, considering the given domain.

Solution:

The function involves the sum of two absolute value expressions: $|x - 1|$ and $|1 + x|$.

We need to determine the sign of the expressions inside the absolute value ($x - 1$ and $1 + x$) within the given domain $-2 \leq x \leq 2$.

The critical points where the expressions inside the absolute value change sign are when $x - 1 = 0$ (i.e., $x = 1$) and when $1 + x = 0$ (i.e., $x = -1$).

These critical points divide the given domain $[-2, 2]$ into three intervals:

1. $-2 \leq x < -1$

2. $-1 \leq x < 1$

3. $1 \leq x \leq 2$

Let's analyze the sign of $(x - 1)$ and $(1 + x)$ in each interval:

Interval 1: $-2 \leq x < -1$

- For $x < -1$, $x - 1$ is negative (e.g., if $x = -1.5$, $x - 1 = -2.5$). So, $|x - 1| = -(x - 1) = 1 - x$.

- For $x < -1$, $1 + x$ is negative (e.g., if $x = -1.5$, $1 + x = -0.5$). So, $|1 + x| = -(1 + x) = -1 - x$.

In this interval, $f(x) = (1 - x) + (-1 - x) = 1 - x - 1 - x = -2x$.

Interval 2: $-1 \leq x < 1$

- For $x < 1$, $x - 1$ is negative (e.g., if $x = 0$, $x - 1 = -1$). So, $|x - 1| = -(x - 1) = 1 - x$.

- For $x \geq -1$, $1 + x$ is non-negative (e.g., if $x = 0$, $1 + x = 1$). So, $|1 + x| = 1 + x$.

In this interval, $f(x) = (1 - x) + (1 + x) = 1 - x + 1 + x = 2$.

Interval 3: $1 \leq x \leq 2$

- For $x \geq 1$, $x - 1$ is non-negative (e.g., if $x = 1.5$, $x - 1 = 0.5$). So, $|x - 1| = x - 1$.

- For $x \geq 1$ (which is also $x > -1$), $1 + x$ is positive (e.g., if $x = 1.5$, $1 + x = 2.5$). So, $|1 + x| = 1 + x$.

In this interval, $f(x) = (x - 1) + (1 + x) = x - 1 + 1 + x = 2x$.

We can summarize the function definition in these intervals:

This is the redefined function without absolute values over the given domain.

Given:

The function $f(x) = \frac{1}{\sqrt{[x]^{2}\;-\;[x]\;-\;6}}$.

Here, $[x]$ denotes the greatest integer function, which gives the greatest integer less than or equal to $x$.

To Find:

The domain of the function $f(x)$.

Solution:

For the function $f(x)$ to be defined in the real numbers, two conditions must be met:

1. The expression under the square root must be non-negative: $[x]^2 - [x] - 6 \geq 0$.

2. The denominator cannot be zero, which means the expression under the square root cannot be zero: $[x]^2 - [x] - 6 \neq 0$.

Combining these two conditions, the expression under the square root must be strictly positive:

$[x]^2 - [x] - 6 > 0$

Let $y = [x]$. The inequality becomes a quadratic inequality in $y$:

$y^2 - y - 6 > 0$

We factor the quadratic expression:

$(y - 3)(y + 2) > 0$

This inequality holds true when both factors have the same sign (both positive or both negative).

Case 1: Both factors are positive.

$y - 3 > 0$ and $y + 2 > 0$

$y > 3$ and $y > -2$

The intersection of these conditions is $y > 3$.

Case 2: Both factors are negative.

$y - 3 < 0$ and $y + 2 < 0$

$y < 3$ and $y < -2$

The intersection of these conditions is $y < -2$.

So, the inequality $(y - 3)(y + 2) > 0$ is satisfied when $y < -2$ or $y > 3$.

Now, substitute back $y = [x]$:

$[x] < -2$ or $[x] > 3$

We need to find the values of $x$ that satisfy these conditions on $[x]$.

Condition A: $[x] < -2$

The greatest integer less than or equal to $x$ is an integer strictly less than $-2$. This means $[x]$ can be $-3, -4, -5, \dots$.

If $[x]$ is any integer less than or equal to $-3$, then by definition of $[x]$, we have $[x] \leq x < [x] + 1$.

For example, if $[x] = -3$, then $-3 \leq x < -2$. All values in this interval satisfy $[x] = -3$, which is $<-2$.

In general, $[x] < -2$ is equivalent to saying that $x$ is strictly less than $-2$.

$[x] < -2 \iff x < -2$.

This corresponds to the interval $(-\infty, -2)$.

Condition B: $[x] > 3$

The greatest integer less than or equal to $x$ is an integer strictly greater than $3$. This means $[x]$ can be $4, 5, 6, \dots$.

If $[x]$ is any integer greater than or equal to $4$, then by definition of $[x]$, we have $[x] \leq x < [x] + 1$.

For example, if $[x] = 4$, then $4 \leq x < 5$. All values in this interval satisfy $[x] = 4$, which is $>3$.

In general, $[x] > 3$ is equivalent to saying that $x$ is greater than or equal to $4$.

$[x] > 3 \iff x \geq 4$.

This corresponds to the interval $[4, \infty)$.

The domain of $f(x)$ is the set of all $x$ that satisfy either Condition A or Condition B.

Domain = $\{x \in \mathbb{R} \text{ | } x < -2 \text{ or } x \geq 4\}$.

In interval notation, the domain is $(-\infty, -2) \cup [4, \infty)$.

Given:

The function $f(x) = \frac{1}{\sqrt{x - |x|}}$.

To Find:

The domain of the function $f(x)$.

Solution:

For the function $f(x)$ to be defined in the real numbers, the expression under the square root must be non-negative, and the denominator cannot be zero.

This means the expression under the square root must be strictly positive:

We need to analyze the expression $x - |x|$ based on the sign of $x$.

Case 1: $x \geq 0$

By the definition of absolute value, if $x \geq 0$, then $|x| = x$.

So, the expression becomes $x - x = 0$.

The condition $x - |x| > 0$ becomes $0 > 0$, which is false.

Thus, no non-negative value of $x$ is in the domain.

Case 2: $x < 0$

By the definition of absolute value, if $x < 0$, then $|x| = -x$.

So, the expression becomes $x - (-x) = x + x = 2x$.

The condition $x - |x| > 0$ becomes $2x > 0$.

To solve $2x > 0$ for $x$, we divide by $2$ (a positive number):

In Case 2, we assumed $x < 0$. The condition derived is $x > 0$. These two conditions ($x < 0$ and $x > 0$) are contradictory.

Thus, no value of $x$ less than 0 satisfies the condition $x - |x| > 0$.

Combining both cases, there is no real number $x$ for which $x - |x| > 0$.

Therefore, the function $f(x)$ is not defined for any real number $x$.

The domain of the function is the empty set, denoted by $\emptyset$ or \{\}.

Comparing this result with the given options:

(A) R (Set of all real numbers) - Incorrect.

(B) R$^+$ (Set of positive real numbers) - Incorrect.

(C) R$^-$ (Set of negative real numbers) - Incorrect.

(D) None of these - Correct, as the domain is the empty set.

The correct option is (D).

Given:

The function $f(x) = x^3 + \frac{1}{x^3}$.

To Find:

The value of $f(x) + f(\frac{1}{x})$.

Solution:

We are given the function $f(x) = x^3 + \frac{1}{x^3}$.

To find $f(\frac{1}{x})$, we substitute $\frac{1}{x}$ for $x$ in the expression for $f(x)$:

Using the properties of exponents and fractions, $(\frac{a}{b})^n = \frac{a^n}{b^n}$ and $\frac{1}{(\frac{a}{b})} = \frac{b}{a}$:

Now we compute the sum $f(x) + f(\frac{1}{x})$:

Combine like terms:

The result of $f(x) + f(\frac{1}{x})$ for the given function $f(x) = x^3 + \frac{1}{x^3}$ is $2x^3 + \frac{2}{x^3}$.

Comparing this result with the given options:

(A) $2x^3$

(B) $2\frac{1}{x^{3}}$

(C) 0

(D) None of these

The calculated expression $2x^3 + \frac{2}{x^3}$ does not match options (A), (B), or (C).

Therefore, based on the given function definition, none of the provided options are correct.

The most appropriate answer among the choices, given the mismatch between the calculated result and options (A), (B), and (C), is (D) None of these.

Given:

Set A with number of elements $n(A) = q$.

Set B with number of elements $n(B) = p$.

To Find:

The total number of functions $f : A \to B$.

Concept:

A function $f : A \to B$ assigns to each element in set A exactly one element in set B.

If $A = \{a_1, a_2, \dots, a_q\}$ and $B = \{b_1, b_2, \dots, b_p\}$, then for each element $a_i \in A$, there are $n(B) = p$ possible choices for its image $f(a_i)$ in set B.

Since there are $q$ elements in set A, and the choice of image for each element is independent, the total number of functions is the product of the number of choices for each element.

Calculation:

Number of choices for $f(a_1)$ = $p$

Number of choices for $f(a_2)$ = $p$

...

Number of choices for $f(a_q)$ = $p$

Total number of functions = $p \times p \times \dots \times p$ (q times)

Total number of functions = $p^q$

The total number of functions f : A $\to$ B is equal to $p^q$.

Let A and B be any two sets such that n(B) = p, n(A) = q then the total number of functions f : A $\to$ B is equal to $p^q$.

Given:

Function $f = \{(2, 4), (5, 6), (8, – 1), (10, – 3)\}$

Function $g = \{(2, 5), (7, 1), (8, 4), (10, 13), (11, – 5)\}$

To Find:

The domain of the function $(f + g)$.

Concept:

If f and g are two functions, the sum function $(f + g)$ is defined by $(f + g)(x) = f(x) + g(x)$.

The domain of $(f + g)$ is the set of all $x$ in the real numbers such that $x$ is in the domain of f AND $x$ is in the domain of g.

In set notation, Domain$(f + g)$ = Domain$(f) \cap$ Domain$(g)$.

Solution:

The domain of a function represented by a set of ordered pairs is the set of all the first components of the ordered pairs.

Domain$(f) = \{2, 5, 8, 10\}$

Domain$(g) = \{2, 7, 8, 10, 11\}$

Now, we find the intersection of Domain$(f)$ and Domain$(g)$:

Domain$(f + g)$ = Domain$(f) \cap$ Domain$(g)$

Domain$(f + g)$ = $\{2, 5, 8, 10\} \cap \{2, 7, 8, 10, 11\}$

The elements common to both sets are 2, 8, and 10.

Domain$(f + g) = \{2, 8, 10\}$

The domain of f + g is $\{2, 8, 10\}$.

Given sets are $A = \{-1, 2, 3\}$ and $B = \{1, 3\}$.

(i) A × B:

The Cartesian product $A \times B$ is the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$.

$A \times B = \{(-1, 1), (-1, 3), (2, 1), (2, 3), (3, 1), (3, 3)\}$.

(ii) B × A:

The Cartesian product $B \times A$ is the set of all ordered pairs $(b, a)$ where $b \in B$ and $a \in A$.

$B \times A = \{(1, -1), (1, 2), (1, 3), (3, -1), (3, 2), (3, 3)\}$.

(iii) B × B:

The Cartesian product $B \times B$ is the set of all ordered pairs $(b_1, b_2)$ where $b_1 \in B$ and $b_2 \in B$.

$B \times B = \{(1, 1), (1, 3), (3, 1), (3, 3)\}$.

(iv) A × A:

The Cartesian product $A \times A$ is the set of all ordered pairs $(a_1, a_2)$ where $a_1 \in A$ and $a_2 \in A$.

$A \times A = \{(-1, -1), (-1, 2), (-1, 3), (2, -1), (2, 2), (2, 3), (3, -1), (3, 2), (3, 3)\}$.

Given:

Set P = $\{x : x < 3, x \in N\}$, where N is the set of natural numbers {1, 2, 3, ...}.

Set Q = $\{x : x \leq 2, x \in W\}$, where W is the set of whole numbers {0, 1, 2, 3, ...}.

Determine the elements of sets P and Q:

For set P, $x$ is a natural number less than 3. Thus, $P = \{1, 2\}$.

For set Q, $x$ is a whole number less than or equal to 2. Thus, $Q = \{0, 1, 2\}$.

Find P ∪ Q:

$P \cup Q$ is the set of elements that are in P or Q or both.

$P \cup Q = \{1, 2\} \cup \{0, 1, 2\} = \{0, 1, 2\}$.

Find P ∩ Q:

$P \cap Q$ is the set of elements that are common to both P and Q.

$P \cap Q = \{1, 2\} \cap \{0, 1, 2\} = \{1, 2\}$.

Find (P ∪ Q) × (P ∩ Q):

This is the Cartesian product of the set $(P \cup Q)$ and the set $(P \cap Q)$. It is the set of all ordered pairs $(a, b)$ where $a \in (P \cup Q)$ and $b \in (P \cap Q)$.

Let $A = P \cup Q = \{0, 1, 2\}$ and $B = P \cap Q = \{1, 2\}$.

$(P \cup Q) \times (P \cap Q) = A \times B = \{(a, b) \mid a \in \{0, 1, 2\}, b \in \{1, 2\}\}$.

$(P \cup Q) \times (P \cap Q) = \{(0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2)\}$.

The final answer is: $(P \cup Q) \times (P \cap Q) = \{(0, 1), (0, 2), (1, 1), (1, 2), (2, 1), (2, 2)\}$.

Given:

Set A = $\{x : x \in W, x < 2\}$, where W is the set of whole numbers {0, 1, 2, ...}.

Set B = $\{x : x \in N, 1 < x < 5\}$, where N is the set of natural numbers {1, 2, 3, ...}.

Set C = $\{3, 5\}$.

Determine the elements of sets A and B:

For set A, $x$ is a whole number less than 2. Thus, $A = \{0, 1\}$.

For set B, $x$ is a natural number strictly between 1 and 5. Thus, $B = \{2, 3, 4\}$.

(i) A × (B ∩ C):

First, find $B \cap C$. This is the set of elements common to both B and C.

$B \cap C = \{2, 3, 4\} \cap \{3, 5\} = \{3\}$.

Now, find the Cartesian product $A \times (B \cap C)$. This is the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in (B \cap C)$.

$A \times (B \cap C) = \{0, 1\} \times \{3\} = \{(0, 3), (1, 3)\}$.

(ii) A × (B ∪ C):

First, find $B \cup C$. This is the set of elements that are in B or C or both.

$B \cup C = \{2, 3, 4\} \cup \{3, 5\} = \{2, 3, 4, 5\}$.

Now, find the Cartesian product $A \times (B \cup C)$. This is the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in (B \cup C)$.

$A \times (B \cup C) = \{0, 1\} \times \{2, 3, 4, 5\} = \{(0, 2), (0, 3), (0, 4), (0, 5), (1, 2), (1, 3), (1, 4), (1, 5)\}$.

When two ordered pairs are equal, their corresponding components are equal.

(i) (2a + b, a – b) = (8, 3)

Equating the first components:

$2a + b = 8$

Equating the second components:

$a – b = 3$

We have a system of two linear equations with two variables 'a' and 'b'. We can solve this system using the elimination method.

Adding equation (1) and equation (2):

$(2a + b) + (a - b) = 8 + 3$

$3a = 11$

Solving for 'a':

$a = \frac{11}{3}$

Substitute the value of 'a' into equation (2):

$\frac{11}{3} - b = 3$

Solving for 'b':

$-b = 3 - \frac{11}{3}$

$-b = \frac{9 - 11}{3}$

$-b = -\frac{2}{3}$

$b = \frac{2}{3}$

Therefore, $a = \frac{11}{3}$ and $b = \frac{2}{3}$.

(ii) $\left( \frac{a}{4} ,a-2b\right)$ = (0, 6 + b)

Equating the first components:

$\frac{a}{4} = 0$

Equating the second components:

$a - 2b = 6 + b$

From equation (1):

$\frac{a}{4} = 0$

$a = 0 \times 4$

$a = 0$

Substitute the value of 'a' into equation (2):

$0 - 2b = 6 + b$

Collect terms involving 'b' on one side:

$-2b - b = 6$

$-3b = 6$

Solving for 'b':

$b = \frac{6}{-3}$

$b = -2$

Therefore, $a = 0$ and $b = -2$.

Given:

Set $A = \{1, 2, 3, 4, 5\}$.

Set $S = \{(x, y) : x \in A, y \in A\}$. This means S is the set of all ordered pairs where both the first and second components are elements of A. In other words, $S = A \times A$.

We need to find the ordered pairs in $A \times A$ that satisfy the given conditions.

(i) Condition: $x + y = 5$

We need to find pairs $(x, y)$ from $A \times A$ such that the sum of the components is 5.

Possible pairs are:

If $x = 1$, then $1 + y = 5 \implies y = 4$. Since $4 \in A$, $(1, 4)$ is a valid pair.

If $x = 2$, then $2 + y = 5 \implies y = 3$. Since $3 \in A$, $(2, 3)$ is a valid pair.

If $x = 3$, then $3 + y = 5 \implies y = 2$. Since $2 \in A$, $(3, 2)$ is a valid pair.

If $x = 4$, then $4 + y = 5 \implies y = 1$. Since $1 \in A$, $(4, 1)$ is a valid pair.

If $x = 5$, then $5 + y = 5 \implies y = 0$. Since $0 \notin A$, $(5, 0)$ is not a valid pair from $A \times A$.

The ordered pairs satisfying $x + y = 5$ are: $\{(1, 4), (2, 3), (3, 2), (4, 1)\}$.

(ii) Condition: $x + y < 5$

We need to find pairs $(x, y)$ from $A \times A$ such that the sum of the components is less than 5.

Possible pairs are:

If $x = 1$: $1 + y < 5 \implies y < 4$. Since $y \in A = \{1, 2, 3, 4, 5\}$, possible values for y are 1, 2, 3. This gives pairs $(1, 1), (1, 2), (1, 3)$.

If $x = 2$: $2 + y < 5 \implies y < 3$. Since $y \in A$, possible values for y are 1, 2. This gives pairs $(2, 1), (2, 2)$.

If $x = 3$: $3 + y < 5 \implies y < 2$. Since $y \in A$, the only possible value for y is 1. This gives the pair $(3, 1)$.

If $x = 4$: $4 + y < 5 \implies y < 1$. Since $y \in A$ contains no values less than 1, there are no pairs starting with 4.

If $x = 5$: $5 + y < 5 \implies y < 0$. Since $y \in A$ contains no values less than 0, there are no pairs starting with 5.

The ordered pairs satisfying $x + y < 5$ are: $\{(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)\}$.

(iii) Condition: $x + y > 8$

We need to find pairs $(x, y)$ from $A \times A$ such that the sum of the components is greater than 8.

Possible pairs are:

If $x = 1$: $1 + y > 8 \implies y > 7$. Since the maximum value in A is 5, there are no pairs starting with 1.

If $x = 2$: $2 + y > 8 \implies y > 6$. Since the maximum value in A is 5, there are no pairs starting with 2.

If $x = 3$: $3 + y > 8 \implies y > 5$. Since the maximum value in A is 5, there are no pairs starting with 3.

If $x = 4$: $4 + y > 8 \implies y > 4$. Since $y \in A$, the only possible value for y is 5. This gives the pair $(4, 5)$.

If $x = 5$: $5 + y > 8 \implies y > 3$. Since $y \in A$, possible values for y are 4, 5. This gives pairs $(5, 4), (5, 5)$.

The ordered pairs satisfying $x + y > 8$ are: $\{(4, 5), (5, 4), (5, 5)\}$.

Given:

Relation $R = \{(x, y) : x \in W, y \in W, x^2 + y^2 = 25\}$, where W is the set of whole numbers {0, 1, 2, 3, ...}.

We need to find ordered pairs $(x, y)$ such that both x and y are whole numbers and satisfy the equation $x^2 + y^2 = 25$.

Find the ordered pairs in R:

We need to find pairs of whole numbers $(x, y)$ that satisfy the equation $x^2 + y^2 = 25$.

Consider possible whole number values for x and check if the corresponding y value is also a whole number.

If $x = 0$, $0^2 + y^2 = 25 \implies y^2 = 25$. Since $y \in W$, $y = 5$. The pair is $(0, 5)$.

If $x = 1$, $1^2 + y^2 = 25 \implies 1 + y^2 = 25 \implies y^2 = 24$. 24 is not a perfect square, so $y \notin W$.

If $x = 2$, $2^2 + y^2 = 25 \implies 4 + y^2 = 25 \implies y^2 = 21$. 21 is not a perfect square, so $y \notin W$.

If $x = 3$, $3^2 + y^2 = 25 \implies 9 + y^2 = 25 \implies y^2 = 16$. Since $y \in W$, $y = 4$. The pair is $(3, 4)$.

If $x = 4$, $4^2 + y^2 = 25 \implies 16 + y^2 = 25 \implies y^2 = 9$. Since $y \in W$, $y = 3$. The pair is $(4, 3)$.

If $x = 5$, $5^2 + y^2 = 25 \implies 25 + y^2 = 25 \implies y^2 = 0$. Since $y \in W$, $y = 0$. The pair is $(5, 0)$.

If $x > 5$, then $x^2 > 25$, so $x^2 + y^2 > 25$ for any $y \in W, y \ge 0$. No further pairs exist.

The set of ordered pairs in the relation R is $\{(0, 5), (3, 4), (4, 3), (5, 0)\}$.

Domain of R:

The domain of R is the set of all first components of the ordered pairs in R.

Domain(R) = $\{0, 3, 4, 5\}$.

Range of R:

The range of R is the set of all second components of the ordered pairs in R.

Range(R) = $\{5, 4, 3, 0\}$.

Given:

Relation $R_1 = \{(x, y) \mid y = 2x + 7, x \in R, -5 \leq x \leq 5\}$.

Here, x is a real number and is restricted to the interval $[-5, 5]$.

Domain of R₁:

The domain of a relation is the set of all possible values for the first component (x) of the ordered pairs.

From the definition of the relation $R_1$, the possible values of x are given by the condition $x \in R$ and $-5 \leq x \leq 5$.

Therefore, the domain of $R_1$ is $\{x \in R \mid -5 \leq x \leq 5\}$.

In interval notation, the domain is $[-5, 5]$.

Range of R₁:

The range of a relation is the set of all possible values for the second component (y) of the ordered pairs.

The relation is given by $y = 2x + 7$, and the domain for x is $-5 \leq x \leq 5$.

To find the range, we can determine the minimum and maximum values of y as x varies within its domain.

Since the function $y = 2x + 7$ is a linear function with a positive slope (2), it is strictly increasing. This means the minimum value of y occurs at the minimum value of x, and the maximum value of y occurs at the maximum value of x.

When $x = -5$:

$y = 2(-5) + 7 = -10 + 7 = -3$.

When $x = 5$:

$y = 2(5) + 7 = 10 + 7 = 17$.

As x takes all real values from -5 to 5, y takes all real values from -3 to 17.

Therefore, the range of $R_1$ is $\{y \in R \mid -3 \leq y \leq 17\}$.

In interval notation, the range is $[-3, 17]$.

Given:

Relation $R_2 = \{(x, y) \mid x \in \mathbb{Z}, y \in \mathbb{Z}, x^2 + y^2 = 64\}$.

We need to find all ordered pairs of integers $(x, y)$ that satisfy the equation $x^2 + y^2 = 64$.

Find the ordered pairs in R₂:

We are looking for integer solutions to the equation $x^2 + y^2 = 64$.

Since $x$ and $y$ are integers, $x^2$ and $y^2$ must be perfect squares that are non-negative.

The possible perfect squares less than or equal to 64 are: $0^2=0, 1^2=1, 2^2=4, 3^2=9, 4^2=16, 5^2=25, 6^2=36, 7^2=49, 8^2=64$.

We need to find pairs of these perfect squares that sum up to 64.

Let $x^2$ and $y^2$ be two such perfect squares. Their sum must be 64.

If $x^2 = 0$, then $y^2 = 64 - 0 = 64$. This gives $x = 0$ and $y = \pm \sqrt{64} = \pm 8$. The pairs are $(0, 8)$ and $(0, -8)$.

If $x^2 = 1$, then $y^2 = 64 - 1 = 63$. 63 is not a perfect square, so no integer solution for y.

If $x^2 = 4$, then $y^2 = 64 - 4 = 60$. 60 is not a perfect square.

If $x^2 = 9$, then $y^2 = 64 - 9 = 55$. 55 is not a perfect square.

If $x^2 = 16$, then $y^2 = 64 - 16 = 48$. 48 is not a perfect square.

If $x^2 = 25$, then $y^2 = 64 - 25 = 39$. 39 is not a perfect square.

If $x^2 = 36$, then $y^2 = 64 - 36 = 28$. 28 is not a perfect square.

If $x^2 = 49$, then $y^2 = 64 - 49 = 15$. 15 is not a perfect square.

If $x^2 = 64$, then $y^2 = 64 - 64 = 0$. This gives $x = \pm \sqrt{64} = \pm 8$ and $y = 0$. The pairs are $(8, 0)$ and $(-8, 0)$.

Any integer $|x| > 8$ would result in $x^2 > 64$, making $y^2 = 64 - x^2$ negative, which has no real (and thus no integer) solutions for y.

The integer pairs $(x, y)$ that satisfy $x^2 + y^2 = 64$ are $(0, 8), (0, -8), (8, 0), (-8, 0)$.

The relation $R_2$ is the set of these ordered pairs.

$R_2 = \{(0, 8), (0, -8), (8, 0), (-8, 0)\}$.

Given:

Relation $R_3 = \{(x, y) \mid y = |x|, x \in R\}$.

Here, x is any real number, and y is the absolute value of x.

Domain of R₃:

The domain of a relation is the set of all possible values for the first component (x) of the ordered pairs.

From the definition of the relation $R_3$, the possible values of x are given by the condition $x \in R$. There are no other restrictions on x.

Therefore, the domain of $R_3$ is the set of all real numbers.

Domain($R_3$) = $R$ or $(-\infty, \infty)$.

Range of R₃:

The range of a relation is the set of all possible values for the second component (y) of the ordered pairs.

The relation is given by $y = |x|$.

The absolute value of any real number is always non-negative.

For any real number x, $|x| \geq 0$.

Also, for any non-negative real number $y \geq 0$, we can find a real number x such that $|x| = y$ (for example, $x = y$ or $x = -y$).

Thus, the possible values for y are all non-negative real numbers.

Therefore, the range of $R_3$ is the set of all non-negative real numbers.

Range($R_3$) = $\{y \in R \mid y \geq 0\}$ or $[0, \infty)$.

A relation R from set A to set B is a function if every element in set A has one and only one image in set B.

(i) h = {(4, 6), (3, 9), (– 11, 6), (3, 11)}

The domain of this relation is the set of first components: $\{4, 3, -11\}$.

In this relation, the element $3$ from the domain is associated with two different elements in the range: $9$ and $11$. This is seen in the ordered pairs $(3, 9)$ and $(3, 11)$.

Therefore, the relation h is not a function because an element in the domain (3) has more than one image.

(ii) f = {(x, x) | x is a real number}

The domain of this relation is the set of all real numbers, $R$. The relation is defined by $y = x$.

For every real number $x$, there is exactly one corresponding real number $y$ such that $y = x$. For example, if $x=5$, $y=5$; if $x=-2$, $y=-2$.

Every element in the domain (any real number) is mapped to a unique element in the range (itself).

Therefore, the relation f is a function.

(iii) g = { (n, $\frac{1}{n}$) | n is a positive integer}

The domain of this relation is the set of positive integers, $\{1, 2, 3, ...\}$. The relation is defined by $y = \frac{1}{n}$.

For every positive integer $n$, there is exactly one corresponding value $\frac{1}{n}$. For example, if $n=1$, $y=\frac{1}{1}=1$; if $n=2$, $y=\frac{1}{2}$.

Every element in the domain (a positive integer) is mapped to a unique element in the range (its reciprocal).

Therefore, the relation g is a function.

(iv) s = {(n, n²) | n is a positive integer}

The domain of this relation is the set of positive integers, $\{1, 2, 3, ...\}$. The relation is defined by $y = n^2$.

For every positive integer $n$, there is exactly one corresponding value $n^2$. For example, if $n=1$, $y=1^2=1$; if $n=2$, $y=2^2=4$; if $n=3$, $y=3^2=9$.

Every element in the domain (a positive integer) is mapped to a unique element in the range ( its square).

Therefore, the relation s is a function.

(v) t = {(x, 3) | x is a real number}

The domain of this relation is the set of all real numbers, $R$. The relation is defined by $y = 3$. This is a constant function.

For every real number $x$, the corresponding value of $y$ is always $3$. For example, if $x=5$, $y=3$; if $x=-2$, $y=3$.

Every element in the domain (any real number) is mapped to a unique element in the range (the value 3).

Therefore, the relation t is a function.

Given the real functions:

$f(x) = x^2 + 7$

$g(x) = 3x + 5$

(a) f (3) + g (– 5)

First, evaluate $f(3)$ by substituting $x=3$ into the expression for $f(x)$.

$f(3) = (3)^2 + 7 = 9 + 7 = 16$.

Next, evaluate $g(-5)$ by substituting $x=-5$ into the expression for $g(x)$.

$g(-5) = 3(-5) + 5 = -15 + 5 = -10$.

Now, find the sum $f(3) + g(-5)$.

$f(3) + g(-5) = 16 + (-10) = 16 - 10 = 6$.

(b) f $\left( \frac{1}{2} \right)$ × g (14)

First, evaluate $f\left(\frac{1}{2}\right)$ by substituting $x=\frac{1}{2}$ into the expression for $f(x)$.

$f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 + 7 = \frac{1}{4} + 7 = \frac{1}{4} + \frac{28}{4} = \frac{1+28}{4} = \frac{29}{4}$.

Next, evaluate $g(14)$ by substituting $x=14$ into the expression for $g(x)$.

$g(14) = 3(14) + 5 = 42 + 5 = 47$.

Now, find the product $f\left(\frac{1}{2}\right) \times g(14)$.

$f\left(\frac{1}{2}\right) \times g(14) = \frac{29}{4} \times 47 = \frac{29 \times 47}{4} = \frac{1363}{4}$.

(c) f (– 2) + g (– 1)

First, evaluate $f(-2)$ by substituting $x=-2$ into the expression for $f(x)$.

$f(-2) = (-2)^2 + 7 = 4 + 7 = 11$.

Next, evaluate $g(-1)$ by substituting $x=-1$ into the expression for $g(x)$.

$g(-1) = 3(-1) + 5 = -3 + 5 = 2$.

Now, find the sum $f(-2) + g(-1)$.

$f(-2) + g(-1) = 11 + 2 = 13$.

(d) f (t) – f (– 2)

The expression for $f(t)$ is obtained by substituting $x=t$ into the expression for $f(x)$.

$f(t) = t^2 + 7$.

From part (c), we already evaluated $f(-2)$.

$f(-2) = 11$.

Now, find the difference $f(t) - f(-2)$.

$f(t) - f(-2) = (t^2 + 7) - 11 = t^2 + 7 - 11 = t^2 - 4$.

(e) $\frac{f(t)\;-\;f(5)}{t\;-\;5}$ , if t ≠ 5

First, evaluate $f(t)$ which is $t^2 + 7$.

Next, evaluate $f(5)$ by substituting $x=5$ into the expression for $f(x)$.

$f(5) = (5)^2 + 7 = 25 + 7 = 32$.

Now, substitute $f(t)$ and $f(5)$ into the given expression.

$\frac{f(t)\;-\;f(5)}{t\;-\;5} = \frac{(t^2 + 7) - 32}{t - 5} = \frac{t^2 + 7 - 32}{t - 5} = \frac{t^2 - 25}{t - 5}$.

The numerator is a difference of squares, which can be factored as $t^2 - 25 = (t - 5)(t + 5)$.

So, the expression becomes:

$\frac{(t - 5)(t + 5)}{t - 5}$.

Since it is given that $t \neq 5$, the term $(t - 5)$ is non-zero, and we can cancel it from the numerator and the denominator.

$\frac{\cancel{(t - 5)} (t + 5)}{\cancel{t - 5}} = t + 5$.

The expression simplifies to $t + 5$ for $t \neq 5$.

Given the real functions:

$f(x) = 2x + 1$

$g(x) = 4x - 7$

(a) For what real numbers x, f (x) = g (x)?

We need to find the value(s) of x for which $f(x)$ is equal to $g(x)$.

Set the expressions for $f(x)$ and $g(x)$ equal to each other:

$f(x) = g(x)$

$2x + 1 = 4x - 7$

Now, solve this linear equation for x.

Subtract $2x$ from both sides:

$1 = 4x - 2x - 7$

$1 = 2x - 7$

Add 7 to both sides:

$1 + 7 = 2x$

$8 = 2x$

Divide both sides by 2:

$\frac{8}{2} = x$

$4 = x$

So, $x = 4$.

Thus, $f(x) = g(x)$ for the real number $x = 4$.

(b) For what real numbers x, f (x) < g (x)?

We need to find the value(s) of x for which $f(x)$ is less than $g(x)$.

Set the inequality:

$f(x) < g(x)$

$2x + 1 < 4x - 7$

Now, solve this linear inequality for x. We can manipulate inequalities much like equations, keeping in mind to reverse the inequality sign if multiplying or dividing by a negative number.

Subtract $2x$ from both sides:

$1 < 4x - 2x - 7$

$1 < 2x - 7$

Add 7 to both sides:

$1 + 7 < 2x$

$8 < 2x$

Divide both sides by 2 (which is positive, so the inequality sign does not reverse):

$\frac{8}{2} < x$

$4 < x$

So, $x > 4$.

Thus, $f(x) < g(x)$ for all real numbers x such that $x > 4$.

Given the real valued functions:

$f(x) = 2x + 1$

$g(x) = x^2 + 1$

(i) f + g:

The sum of two functions $f$ and $g$, denoted by $f + g$, is defined as $(f + g)(x) = f(x) + g(x)$.

$(f + g)(x) = (2x + 1) + (x^2 + 1)$

$(f + g)(x) = x^2 + 2x + 1 + 1$

$(f + g)(x) = x^2 + 2x + 2$.

The domain of both $f$ and $g$ is the set of all real numbers. The domain of $f+g$ is the intersection of the domains of $f$ and $g$, which is also the set of all real numbers.

(ii) f – g:

The difference of two functions $f$ and $g$, denoted by $f - g$, is defined as $(f - g)(x) = f(x) - g(x)$.

$(f - g)(x) = (2x + 1) - (x^2 + 1)$

$(f - g)(x) = 2x + 1 - x^2 - 1$

$(f - g)(x) = -x^2 + 2x$.

The domain of $f-g$ is the intersection of the domains of $f$ and $g$, which is the set of all real numbers.

(iii) f g:

The product of two functions $f$ and $g$, denoted by $fg$, is defined as $(fg)(x) = f(x) \times g(x)$.

$(fg)(x) = (2x + 1)(x^2 + 1)$

$(fg)(x) = 2x(x^2) + 2x(1) + 1(x^2) + 1(1)$

$(fg)(x) = 2x^3 + 2x + x^2 + 1$

$(fg)(x) = 2x^3 + x^2 + 2x + 1$.

The domain of $fg$ is the intersection of the domains of $f$ and $g$, which is the set of all real numbers.

(iv) $\frac{f}{g}$:

The quotient of two functions $f$ and $g$, denoted by $\frac{f}{g}$, is defined as $\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$, provided that $g(x) \neq 0$.

$\left(\frac{f}{g}\right)(x) = \frac{2x + 1}{x^2 + 1}$.

We need to check when the denominator $g(x) = x^2 + 1$ is equal to zero.

$x^2 + 1 = 0$

$x^2 = -1$

For real numbers x, $x^2$ is always non-negative ($x^2 \geq 0$). Therefore, $x^2 = -1$ has no real solutions.

This means that $g(x) = x^2 + 1$ is never zero for any real number x.

The domain of $\frac{f}{g}$ is the set of all real numbers for which $g(x) \neq 0$. Since $g(x)$ is never zero for real x, the domain of $\frac{f}{g}$ is the set of all real numbers.

$\left(\frac{f}{g}\right)(x) = \frac{2x + 1}{x^2 + 1}$ for all $x \in R$.

Given:

Function $f : X \to R$ defined by $f(x) = x^3 + 1$.

Domain $X = \{-1, 0, 3, 9, 7\}$.

To express the function as a set of ordered pairs, we need to find the value of $f(x)$ for each element $x$ in the domain X.

For $x = -1$:

$f(-1) = (-1)^3 + 1 = -1 + 1 = 0$. The ordered pair is $(-1, 0)$.

For $x = 0$:

$f(0) = (0)^3 + 1 = 0 + 1 = 1$. The ordered pair is $(0, 1)$.

For $x = 3$:

$f(3) = (3)^3 + 1 = 27 + 1 = 28$. The ordered pair is $(3, 28)$.

For $x = 9$:

$f(9) = (9)^3 + 1 = 729 + 1 = 730$. The ordered pair is $(9, 730)$.

For $x = 7$:

$f(7) = (7)^3 + 1 = 343 + 1 = 344$. The ordered pair is $(7, 344)$.

The function f as a set of ordered pairs is:

$f = \{(-1, 0), (0, 1), (3, 28), (9, 730), (7, 344)\}$.

The range of the function is the set of all second components (y-values) of the ordered pairs.

Range of f = $\{0, 1, 28, 730, 344\}$.

Listing the elements in increasing order:

Range of f = $\{0, 1, 28, 344, 730\}$.

Given:

Functions $f(x) = 3x^2 - 1$ and $g(x) = 3 + x$.

We need to find the values of x for which $f(x) = g(x)$.

Set the expressions for $f(x)$ and $g(x)$ equal to each other:

$f(x) = g(x)$

$3x^2 - 1 = 3 + x$

Rearrange the equation to bring all terms to one side, resulting in a quadratic equation:

$3x^2 - 1 - (3 + x) = 0$

$3x^2 - 1 - 3 - x = 0$

$3x^2 - x - 4 = 0$

We now have a quadratic equation in the form $ax^2 + bx + c = 0$, where $a=3$, $b=-1$, and $c=-4$. We can solve this by factoring or using the quadratic formula.

Using factoring, we look for two numbers that multiply to $a \times c = 3 \times (-4) = -12$ and add up to $b = -1$. The numbers are -4 and 3.

Rewrite the middle term ($-x$) using these numbers:

$3x^2 - 4x + 3x - 4 = 0$

Group the terms and factor by grouping:

$(3x^2 - 4x) + (3x - 4) = 0$

Factor out the common factor from each group:

$x(3x - 4) + 1(3x - 4) = 0$

Factor out the common binomial $(3x - 4)$:

$(3x - 4)(x + 1) = 0$

Set each factor equal to zero and solve for x:

Case 1: $3x - 4 = 0$

$3x = 4$

$x = \frac{4}{3}$

Case 2: $x + 1 = 0$

$x = -1$

Thus, the values of x for which the functions $f(x)$ and $g(x)$ are equal are $x = \frac{4}{3}$ and $x = -1$.

Part 1: Is g a function?

A relation is considered a function if every element in its domain is associated with exactly one element in its codomain (or range).

The given relation is $g = \{(1, 1), (2, 3), (3, 5), (4, 7)\}$.

The domain of the relation g is the set of all first components of the ordered pairs: $\{1, 2, 3, 4\}$.

Let's examine the image of each element in the domain:

The element 1 is associated only with 1.

The element 2 is associated only with 3.

The element 3 is associated only with 5.

The element 4 is associated only with 7.

Since each element in the domain is mapped to a unique element in the range, the relation g is a function.

Part 2: Finding the values of $\alpha$ and $\beta$ if $g(x) = \alpha x + \beta$.

Given that the function is described by $g(x) = \alpha x + \beta$, we can use the ordered pairs from the set g to form equations.

Using the ordered pair $(1, 1)$, where $x=1$ and $g(x)=1$:

$g(1) = \alpha(1) + \beta$

Using the ordered pair $(2, 3)$, where $x=2$ and $g(x)=3$:

$g(2) = \alpha(2) + \beta$

We now have a system of two linear equations with two variables $\alpha$ and $\beta$. We can solve this system using the elimination method.

Subtract equation (1) from equation (2):

$(2\alpha + \beta) - (\alpha + \beta) = 3 - 1$

$2\alpha + \beta - \alpha - \beta = 2$

$\alpha = 2$

Substitute the value of $\alpha = 2$ into equation (1):

$2 + \beta = 1$

$\beta = 1 - 2$

$\beta = -1$

So, the function is $g(x) = 2x - 1$. Let's verify with the remaining points in g:

For $(3, 5)$: $g(3) = 2(3) - 1 = 6 - 1 = 5$. Correct.

For $(4, 7)$: $g(4) = 2(4) - 1 = 8 - 1 = 7$. Correct.

The values that should be assigned to $\alpha$ and $\beta$ are $\alpha = 2$ and $\beta = -1$.

The domain of a real-valued function is the set of all real numbers for which the function is defined and produces a real output.

(i) f(x) = $\frac{1}{\sqrt{1\;-\;\cos x}}$

For the function to be defined, two conditions must be met:

1. The expression under the square root must be non-negative: $1 - \cos x \geq 0$. This implies $\cos x \leq 1$. This inequality is true for all real values of x, as the maximum value of $\cos x$ is 1.

2. The denominator cannot be zero: $\sqrt{1 - \cos x} \neq 0$. This implies $1 - \cos x \neq 0$, so $\cos x \neq 1$.

We know that $\cos x = 1$ when $x$ is an integer multiple of $2\pi$. That is, $x = 2n\pi$ for $n \in \mathbb{Z}$.

Combining the conditions, we need $\cos x \leq 1$ and $\cos x \neq 1$, which means $\cos x < 1$. This inequality holds for all real numbers x except those where $\cos x = 1$.

Therefore, the values of x for which the function is undefined are $x = 2n\pi$, where $n$ is any integer.

The domain of f(x) is the set of all real numbers except $2n\pi$, where $n \in \mathbb{Z}$.

Domain = $R \setminus \{2n\pi \mid n \in \mathbb{Z}\}$.

(ii) f(x) = $\frac{1}{\sqrt{x\;+\;|x|}}$

For the function to be defined, the expression under the square root must be non-negative and the denominator must be non-zero. This means the expression under the square root must be strictly positive: $x + |x| > 0$.

Consider two cases for the absolute value:

Case 1: If $x \geq 0$, then $|x| = x$. The inequality becomes $x + x > 0$, which is $2x > 0$. Dividing by 2 (a positive number) gives $x > 0$. So, all $x > 0$ satisfy the condition.

Case 2: If $x < 0$, then $|x| = -x$. The inequality becomes $x + (-x) > 0$, which simplifies to $0 > 0$. This is a false statement, so no values of x less than 0 satisfy the condition.

The only real numbers x that satisfy $x + |x| > 0$ are those where $x > 0$.

The domain of f(x) is the set of all positive real numbers.

Domain = $\{x \in R \mid x > 0\}$ or $(0, \infty)$.

(iii) f(x) = x |x|

The function is defined as the product of x and the absolute value of x. The absolute value $|x|$ is defined for all real numbers x. The product of any two real numbers is a real number.

There are no restrictions such as division by zero or taking the square root of a negative number.

The function $f(x) = x|x|$ is defined for all real numbers x.

The domain of f(x) is the set of all real numbers.

Domain = $R$ or $(-\infty, \infty)$.

(iv) f(x) = $\frac{x^{3}\;-\;x\;+\;3}{x^{2}\;-\;1}$

This is a rational function. A rational function is defined for all real numbers except those values of x that make the denominator zero.

Set the denominator equal to zero:

$x^2 - 1 = 0$

$(x - 1)(x + 1) = 0$

This equation is true if $x - 1 = 0$ or $x + 1 = 0$.

$x = 1$ or $x = -1$.

The function is undefined when $x = 1$ or $x = -1$.

The domain of f(x) is the set of all real numbers except 1 and -1.

Domain = $R \setminus \{-1, 1\}$.

(v) f(x) = $\frac{3x}{2x\;-\;8}$

This is also a rational function. It is defined for all real numbers except those values of x that make the denominator zero.

Set the denominator equal to zero:

$2x - 8 = 0$

$2x = 8$

$x = \frac{8}{2}$

$x = 4$

The function is undefined when $x = 4$.

The domain of f(x) is the set of all real numbers except 4.

Domain = $R \setminus \{4\}$.

The range of a function is the set of all possible output values (y-values).

(i) f(x) = $\frac{3}{2\;-\;x^{2}}$

Let $y = f(x) = \frac{3}{2-x^2}$. To find the range, we can determine the possible values of y.

The domain of the function is all real numbers except where the denominator is zero, i.e., $2 - x^2 \neq 0$, which means $x \neq \pm \sqrt{2}$.

Consider the term $x^2$. For any real number x, $x^2 \geq 0$.

So, $-x^2 \leq 0$.

Adding 2, we get $2 - x^2 \leq 2$.

Thus, the denominator $2 - x^2$ can be any value less than or equal to 2, except 0.

This means $2 - x^2 \in (-\infty, 0) \cup (0, 2]$.

Now consider the reciprocal $\frac{1}{2-x^2}$.

If $2 - x^2 \in (-\infty, 0)$, then $\frac{1}{2-x^2} \in (-\infty, 0)$.

If $2 - x^2 \in (0, 2]$, then $\frac{1}{2-x^2} \in [\frac{1}{2}, \infty)$. (Note that as the positive denominator approaches 0, the reciprocal approaches $\infty$; as the denominator increases to 2, the reciprocal decreases to $\frac{1}{2}$).

So, $\frac{1}{2-x^2} \in (-\infty, 0) \cup [\frac{1}{2}, \infty)$.

Finally, consider $f(x) = 3 \times \frac{1}{2-x^2}$. Multiply the set by 3.

If $\frac{1}{2-x^2} \in (-\infty, 0)$, then $3 \times \frac{1}{2-x^2} \in (-\infty, 0)$.

If $\frac{1}{2-x^2} \in [\frac{1}{2}, \infty)$, then $3 \times \frac{1}{2-x^2} \in [3 \times \frac{1}{2}, 3 \times \infty) = [\frac{3}{2}, \infty)$.

Combining these, the range of f(x) is $(-\infty, 0) \cup [\frac{3}{2}, \infty)$.

(ii) f(x) = 1 - |x - 2|

The domain of the function is all real numbers, since $|x-2|$ is defined for all real x.

Consider the term $|x - 2|$. The absolute value of any real number is always non-negative.

So, $|x - 2| \geq 0$ for all real x.

Multiply the inequality by -1, which reverses the inequality sign:

$-|x - 2| \leq 0$.

Add 1 to both sides of the inequality:

$1 - |x - 2| \leq 1 + 0$

$1 - |x - 2| \leq 1$.

Thus, the maximum value of $f(x)$ is 1 (which occurs when $x - 2 = 0 \implies x = 2$). As $|x-2|$ increases, $1 - |x-2|$ decreases without bound.

The range of f(x) is $(-\infty, 1]$.

(iii) f(x) = |x - 3|

The domain of the function is all real numbers, since $|x-3|$ is defined for all real x.

Consider the term $|x - 3|$. The absolute value of any real number is always non-negative.

So, $|x - 3| \geq 0$ for all real x.

The minimum value of $|x - 3|$ is 0 (which occurs when $x - 3 = 0 \implies x = 3$). As x moves away from 3, $|x-3|$ increases and can take any positive value.

Thus, the value of $f(x) = |x - 3|$ can be any non-negative real number.

The range of f(x) is $[0, \infty)$.

(iv) f(x) = 1 + 3 cos 2x

The domain of the cosine function, and thus of $\cos(2x)$, is all real numbers.

We know that the range of the cosine function is $[-1, 1]$.

So, for any real number x, we have:

$-1 \leq \cos (2x) \leq 1$

Now, we transform this inequality to match the expression for f(x).

Multiply the inequality by 3 (a positive number, so inequality signs remain the same):

$-1 \times 3 \leq 3 \cos (2x) \leq 1 \times 3$

$-3 \leq 3 \cos (2x) \leq 3$

Add 1 to all parts of the inequality:

$-3 + 1 \leq 1 + 3 \cos (2x) \leq 3 + 1$

$-2 \leq 1 + 3 \cos (2x) \leq 4$

So, the value of $f(x)$ is between -2 and 4, inclusive. Since $\cos(2x)$ attains all values in $[-1, 1]$, $1 + 3\cos(2x)$ attains all values in $[-2, 4]$.

The range of f(x) is $[-2, 4]$.

Given:

The function $f(x) = |x - 2| + |2 + x|$ for $-3 \leq x \leq 3$.

We need to redefine the function by removing the absolute value signs over the given interval. The definition of $|a|$ depends on the sign of $a$. We consider the expressions inside the absolute values: $(x - 2)$ and $(2 + x)$.

The expression $(x - 2)$ changes sign at $x - 2 = 0$, i.e., $x = 2$.

• If $x \geq 2$, then $x - 2 \geq 0$, so $|x - 2| = x - 2$.
• If $x < 2$, then $x - 2 < 0$, so $|x - 2| = -(x - 2) = -x + 2$.

The expression $(2 + x)$ changes sign at $2 + x = 0$, i.e., $x = -2$.

• If $x \geq -2$, then $2 + x \geq 0$, so $|2 + x| = 2 + x$.
• If $x < -2$, then $2 + x < 0$, so $|2 + x| = -(2 + x) = -2 - x$.

The given interval is $-3 \leq x \leq 3$. The critical points within this interval are $x = -2$ and $x = 2$. These points divide the interval $[-3, 3]$ into three subintervals:

Case 1: $-3 \leq x < -2$

In this interval, $x < 2$ and $x < -2$.

$|x - 2| = -(x - 2) = -x + 2$

$|2 + x| = -(2 + x) = -2 - x$

$f(x) = (-x + 2) + (-2 - x) = -x + 2 - 2 - x = -2x$

Case 2: $-2 \leq x < 2$

In this interval, $x < 2$ and $x \geq -2$.

$|x - 2| = -(x - 2) = -x + 2$

$|2 + x| = 2 + x$

$f(x) = (-x + 2) + (2 + x) = -x + 2 + 2 + x = 4$

Case 3: $2 \leq x \leq 3$

In this interval, $x \geq 2$ and $x > -2$ (since $x \geq 2$).

$|x - 2| = x - 2$

$|2 + x| = 2 + x$

$f(x) = (x - 2) + (2 + x) = x - 2 + 2 + x = 2x$

Combining the cases, the redefined function is:

$f(x) = \begin{cases} -2x & , & -3 \leq x < -2 \\ 4 & , & -2 \leq x < 2 \\ 2x & , & 2 \leq x \leq 3 \end{cases}$

Given:

The function $f(x) = \frac{x - 1}{x + 1}$.

The domain of f(x) is all real numbers $x$ such that $x+1 \neq 0$, i.e., $x \neq -1$.

(i) Show that $f\left( \frac{1}{x} \right) = -f(x)$

We need to evaluate both sides of the equation and show they are equal.

Consider the Left Hand Side (LHS): $f\left( \frac{1}{x} \right)$.

Substitute $x$ with $\frac{1}{x}$ in the expression for $f(x)$:

$f\left( \frac{1}{x} \right) = \frac{\frac{1}{x} - 1}{\frac{1}{x} + 1}$

To simplify the complex fraction, multiply the numerator and the denominator by $x$ (assuming $x \neq 0$):

$f\left( \frac{1}{x} \right) = \frac{x \left( \frac{1}{x} - 1 \right)}{x \left( \frac{1}{x} + 1 \right)} = \frac{x \cdot \frac{1}{x} - x \cdot 1}{x \cdot \frac{1}{x} + x \cdot 1} = \frac{1 - x}{1 + x}$

So, LHS $= \frac{1 - x}{1 + x}$.

Now, consider the Right Hand Side (RHS): $-f(x)$.

$-f(x) = - \left( \frac{x - 1}{x + 1} \right)$

Apply the negative sign to the numerator:

$-f(x) = \frac{-(x - 1)}{x + 1} = \frac{-x + 1}{x + 1} = \frac{1 - x}{x + 1}$

So, RHS $= \frac{1 - x}{1 + x}$.

Since LHS = RHS, we have shown that $f\left( \frac{1}{x} \right) = -f(x)$ for $x \in R \setminus \{0, -1\}$.

(ii) Show that $f\left( -\frac{1}{x} \right) = \frac{-1}{f(x)}$

We need to evaluate both sides of the equation and show they are equal.

Consider the Left Hand Side (LHS): $f\left( -\frac{1}{x} \right)$.

Substitute $x$ with $-\frac{1}{x}$ in the expression for $f(x)$:

$f\left( -\frac{1}{x} \right) = \frac{-\frac{1}{x} - 1}{-\frac{1}{x} + 1}$

To simplify the complex fraction, multiply the numerator and the denominator by $x$ (assuming $x \neq 0$):

$f\left( -\frac{1}{x} \right) = \frac{x \left( -\frac{1}{x} - 1 \right)}{x \left( -\frac{1}{x} + 1 \right)} = \frac{x \cdot \left(-\frac{1}{x}\right) - x \cdot 1}{x \cdot \left(-\frac{1}{x}\right) + x \cdot 1} = \frac{-1 - x}{-1 + x}$

So, LHS $= \frac{-1 - x}{-1 + x}$. This can also be written as $\frac{-(1 + x)}{-(1 - x)} = \frac{1 + x}{1 - x}$ if we factor out -1 from both numerator and denominator. However, let's keep the form $\frac{-1 - x}{-1 + x}$ for comparison initially.

Now, consider the Right Hand Side (RHS): $\frac{-1}{f(x)}$.

$\frac{-1}{f(x)} = \frac{-1}{\frac{x - 1}{x + 1}}$

To divide by a fraction, multiply by its reciprocal (assuming $f(x) \neq 0$, i.e., $\frac{x-1}{x+1} \neq 0$, which means $x-1 \neq 0 \implies x \neq 1$):

$\frac{-1}{f(x)} = -1 \times \frac{x + 1}{x - 1} = \frac{-(x + 1)}{x - 1} = \frac{-x - 1}{x - 1}$

So, RHS $= \frac{-x - 1}{x - 1}$.

Let's compare LHS and RHS:

LHS $= \frac{-1 - x}{-1 + x}$

RHS $= \frac{-x - 1}{x - 1}$

Notice that the numerator of LHS is $-1 - x$ and the numerator of RHS is $-x - 1$, which are the same. The denominator of LHS is $-1 + x$ and the denominator of RHS is $x - 1$, which are also the same.

Thus, LHS = RHS.

Alternatively, from LHS: $\frac{-1 - x}{-1 + x} = \frac{-(1 + x)}{-(1 - x)}$. This does not simplify to $\frac{1+x}{1-x}$ unless $x \neq 1$. Let's recheck the RHS calculation.

RHS $= \frac{-1}{f(x)} = \frac{-1}{\frac{x-1}{x+1}} = -1 \times \frac{x+1}{x-1} = \frac{-(x+1)}{x-1} = \frac{-x-1}{x-1}$.

LHS $= f(-\frac{1}{x}) = \frac{-\frac{1}{x} - 1}{-\frac{1}{x} + 1} = \frac{\frac{-1-x}{x}}{\frac{-1+x}{x}} = \frac{-1-x}{x} \times \frac{x}{-1+x} = \frac{-1-x}{x-1}$.

Yes, the calculations are correct. LHS = $\frac{-1-x}{x-1}$ and RHS = $\frac{-x-1}{x-1}$. They are indeed equal.

We have shown that $f\left( -\frac{1}{x} \right) = \frac{-1}{f(x)}$ for $x \in R \setminus \{-1, 0, 1\}$.

Given:

Functions $f(x) = \sqrt{x}$ and $g(x) = x$.

Both functions are defined on the domain $R^+ \cup \{0\}$, which is the set of non-negative real numbers, i.e., $[0, \infty)$.

(i) (f + g) (x):

The sum of two functions $f$ and $g$ is defined as $(f + g)(x) = f(x) + g(x)$.

$(f + g)(x) = \sqrt{x} + x$.

The domain of $f+g$ is the intersection of the domains of $f$ and $g$. Both are defined on $[0, \infty)$, so the domain of $(f+g)(x)$ is $[0, \infty)$.

$(f + g)(x) = \sqrt{x} + x$ for $x \geq 0$.

(ii) (f – g) (x):

The difference of two functions $f$ and $g$ is defined as $(f - g)(x) = f(x) - g(x)$.

$(f - g)(x) = \sqrt{x} - x$.

The domain of $f-g$ is the intersection of the domains of $f$ and $g$, which is $[0, \infty)$.

$(f - g)(x) = \sqrt{x} - x$ for $x \geq 0$.

(iii) (fg) (x):

The product of two functions $f$ and $g$ is defined as $(fg)(x) = f(x) \times g(x)$.

$(fg)(x) = \sqrt{x} \times x$.

Using exponent rules, $\sqrt{x} = x^{1/2}$ and $x = x^1$.

$(fg)(x) = x^{1/2} \times x^1 = x^{1/2 + 1} = x^{3/2}$.

The domain of $fg$ is the intersection of the domains of $f$ and $g$, which is $[0, \infty)$.

$(fg)(x) = x^{3/2}$ for $x \geq 0$.

(iv) $\left( \frac{f}{g} \right)(x)$:

The quotient of two functions $f$ and $g$ is defined as $\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$, provided that $g(x) \neq 0$.

$\left(\frac{f}{g}\right)(x) = \frac{\sqrt{x}}{x}$.

The domain of $\frac{f}{g}$ is the intersection of the domains of $f$ and $g$, excluding values of x where $g(x) = 0$.

The domain of $f$ and $g$ is $[0, \infty)$.

We need to exclude values where $g(x) = x = 0$. So, $x \neq 0$.

The domain of $\left(\frac{f}{g}\right)(x)$ is $\{x \in [0, \infty) \mid x \neq 0\}$, which is the set of positive real numbers, $(0, \infty)$.

Simplify the expression for $x > 0$:

$\frac{\sqrt{x}}{x} = \frac{x^{1/2}}{x^1} = x^{1/2 - 1} = x^{-1/2} = \frac{1}{x^{1/2}} = \frac{1}{\sqrt{x}}$.

$\left(\frac{f}{g}\right)(x) = \frac{1}{\sqrt{x}}$ for $x > 0$.

Given:

The function $f(x) = \frac{1}{\sqrt{x\;-\;5}}$.

Domain of f(x):

For the function $f(x) = \frac{1}{\sqrt{x\;-\;5}}$ to be defined in the set of real numbers, two conditions must be satisfied:

1. The expression under the square root must be non-negative: $x - 5 \geq 0$.

2. The denominator cannot be zero, which means $\sqrt{x - 5} \neq 0$. This implies $x - 5 \neq 0$.

Combining these two conditions, we require that the expression under the square root is strictly positive:

$x - 5 > 0$

Solving for x:

$x > 5$

The domain of the function is the set of all real numbers x that are strictly greater than 5.

Domain = $\{x \in R \mid x > 5\}$ or $(5, \infty)$.

Range of f(x):

Let $y = f(x) = \frac{1}{\sqrt{x\;-\;5}}$. We want to find the set of all possible values that y can take.

From the domain, we know that $x > 5$.

If $x > 5$, then $x - 5 > 0$.

Taking the square root of a positive number results in a positive number:

$\sqrt{x - 5} > 0$.

Now, consider the reciprocal $\frac{1}{\sqrt{x - 5}}$. The reciprocal of a positive number is always a positive number.

So, $y = \frac{1}{\sqrt{x - 5}} > 0$.

As x approaches 5 from the right side ($x \to 5^+$), $\sqrt{x - 5}$ approaches 0 from the positive side ($\sqrt{x - 5} \to 0^+$). In this case, $y = \frac{1}{\sqrt{x - 5}}$ approaches $\frac{1}{0^+}$, which tends to positive infinity ($y \to \infty$).

As x increases towards infinity ($x \to \infty$), $\sqrt{x - 5}$ also increases towards infinity ($\sqrt{x - 5} \to \infty$). In this case, $y = \frac{1}{\sqrt{x - 5}}$ approaches $\frac{1}{\infty}$, which tends to 0 ($y \to 0$).

Since y can take any value strictly greater than 0, the range of the function is the set of all positive real numbers.

Range = $\{y \in R \mid y > 0\}$ or $(0, \infty)$.

Given:

The function $f(x) = y = \frac{ax - b}{cx - a}$.

The domain of $f(x)$ requires $cx - a \neq 0$, so $x \neq \frac{a}{c}$ (if $c \neq 0$).

To Prove:

$f(y) = x$.

Proof:

To find $f(y)$, we substitute $y$ for $x$ in the expression for $f(x)$.

$f(y) = \frac{ay - b}{cy - a}$

Now, substitute the given expression for y, which is $y = \frac{ax - b}{cx - a}$.

$f(y) = \frac{a \left( \frac{ax - b}{cx - a} \right) - b}{c \left( \frac{ax - b}{cx - a} \right) - a}$

To simplify this complex fraction, multiply both the numerator and the denominator by $(cx - a)$, assuming $cx - a \neq 0$.

Numerator: $a \left( \frac{ax - b}{cx - a} \right) - b = \frac{a(ax - b)}{cx - a} - b = \frac{a(ax - b) - b(cx - a)}{cx - a} = \frac{a^2x - ab - bcx + ab}{cx - a} = \frac{a^2x - bcx}{cx - a} = \frac{x(a^2 - bc)}{cx - a}$

Denominator: $c \left( \frac{ax - b}{cx - a} \right) - a = \frac{c(ax - b)}{cx - a} - a = \frac{c(ax - b) - a(cx - a)}{cx - a} = \frac{acx - bc - acx + a^2}{cx - a} = \frac{a^2 - bc}{cx - a}$

Now, substitute these simplified expressions back into the fraction for $f(y)$:

$f(y) = \frac{\frac{x(a^2 - bc)}{cx - a}}{\frac{a^2 - bc}{cx - a}}$

To divide by a fraction, multiply by its reciprocal. We assume $a^2 - bc \neq 0$ for the function to be non-trivial.

$f(y) = \frac{x(a^2 - bc)}{cx - a} \times \frac{cx - a}{a^2 - bc}$

Since $cx - a \neq 0$ and $a^2 - bc \neq 0$, we can cancel the common terms:

$f(y) = x \times \frac{\cancel{a^2 - bc}}{\cancel{cx - a}} \times \frac{\cancel{cx - a}}{\cancel{a^2 - bc}}$

$f(y) = x$

Thus, we have proved that $f(y) = x$.

Given:

n(A) = m (the number of elements in set A)

n(B) = n (the number of elements in set B)

A relation R from set A to set B is a subset of the Cartesian product $A \times B$.

The number of elements in the Cartesian product $A \times B$ is given by n($A \times B$) = n(A) $\times$ n(B) = $m \times n = mn$.

The total number of possible relations from A to B is the total number of subsets of $A \times B$.

The total number of subsets of a set with k elements is $2^k$.

In this case, the number of elements in $A \times B$ is $mn$.

So, the total number of relations from A to B is $2^{mn}$.

The question asks for the total number of non-empty relations.

The total number of relations includes the empty relation (which is the empty set, $\emptyset$).

To find the number of non-empty relations, we subtract the empty relation from the total number of relations.

Number of non-empty relations = (Total number of relations) - (Number of empty relations)

Number of non-empty relations = $2^{mn} - 1$.

Comparing this result with the given options, we see that option (D) matches our result.

The correct answer is (D) $2^{mn} – 1$.

Given:

The equation $[x]^2 - 5 [x] + 6 = 0$, where $[x]$ denotes the greatest integer function.

Let $y = [x]$. Substituting y into the equation, we get a quadratic equation in terms of y:

$y^2 - 5y + 6 = 0$

We can solve this quadratic equation by factoring. We look for two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3.

$(y - 2)(y - 3) = 0$

This equation is satisfied if either factor is equal to zero.

Case 1: $y - 2 = 0 \implies y = 2$

Case 2: $y - 3 = 0 \implies y = 3$

Since $y = [x]$, we have two possibilities for the value of $[x]$.

Case 1: $[x] = 2$

By the definition of the greatest integer function, $[x] = n$ if and only if $n \leq x < n + 1$, where n is an integer.

So, $[x] = 2$ implies $2 \leq x < 2 + 1$, which means $2 \leq x < 3$.

Case 2: $[x] = 3$

Similarly, $[x] = 3$ implies $3 \leq x < 3 + 1$, which means $3 \leq x < 4$.

The values of x that satisfy the original equation are those in the union of the intervals from Case 1 and Case 2.

The solution set for x is $\{x \mid 2 \leq x < 3\} \cup \{x \mid 3 \leq x < 4\}$.

This union represents all real numbers x such that $2 \leq x < 4$.

In interval notation, the solution is $[2, 4)$.

Now, compare this result with the given options:

(A) x ∈ [3, 4]

(B) x ∈ (2, 3]

(C) x ∈ [2, 3]

(D) x ∈ [2, 4)

The interval $[2, 4)$ corresponds to option (D).

The correct answer is (D) x ∈ [2, 4).

Given:

The function $f(x) = \frac{1}{1-2 \cos x}$.

To find the range of the function, we analyse the possible values of $f(x)$.

The function is defined for real numbers x where the denominator is non-zero, i.e., $1 - 2 \cos x \neq 0$, which implies $\cos x \neq \frac{1}{2}$.

We know that for any real number x, the range of the cosine function is the closed interval $[-1, 1]$.

$-1 \leq \cos x \leq 1$

Multiply the inequality by -2. Since -2 is a negative number, the inequality signs are reversed.

$(-1) \times (-2) \geq -2 \cos x \geq 1 \times (-2)$

$2 \geq -2 \cos x \geq -2$

Rearranging the inequality:

$-2 \leq -2 \cos x \leq 2$

Now, add 1 to all parts of the inequality:

$1 - 2 \leq 1 - 2 \cos x \leq 1 + 2$

$-1 \leq 1 - 2 \cos x \leq 3$

So, the denominator $1 - 2 \cos x$ can take any value in the interval $[-1, 3]$.

However, the denominator cannot be zero. The denominator is zero when $1 - 2 \cos x = 0$, which means $\cos x = \frac{1}{2}$. Since $\frac{1}{2}$ is within the interval $[-1, 1]$, there are values of x for which $\cos x = \frac{1}{2}$, and thus $1 - 2 \cos x = 0$.

Therefore, the denominator $1 - 2 \cos x$ can take any value in the interval $[-1, 3]$ except for 0.

The set of possible values for the denominator is $[-1, 0) \cup (0, 3]$.

Now, consider $f(x) = \frac{1}{1 - 2 \cos x}$. Let $y = f(x)$. We need to find the set of values that $y = \frac{1}{\text{denominator}}$ can take.

Case 1: The denominator $D = 1 - 2 \cos x$ is in the interval $[-1, 0)$.

If $D \in [-1, 0)$, then $y = \frac{1}{D}$ is in the interval $(-\infty, \frac{1}{-1}]$, which is $(-\infty, -1]$.

Case 2: The denominator $D = 1 - 2 \cos x$ is in the interval $(0, 3]$.

If $D \in (0, 3]$, then $y = \frac{1}{D}$ is in the interval $[\frac{1}{3}, \infty)$.

The range of $f(x)$ is the union of the possible values from these two cases.

Range of $f(x) = (-\infty, -1] \cup [\frac{1}{3}, \infty)$.

Comparing this result with the given options, option (C) matches our result.

The correct answer is (C) $\left(-\infty,-1 \right]\cup \left[ \frac{1}{3},\infty \right)$.

Given:

The function $f(x) = \sqrt{1+x^{2}}$. The domain of this function is all real numbers, as $1+x^2 \geq 1$ for any real $x$, so the term under the square root is always non-negative.

We need to compare $f(xy)$ and $f(x) \cdot f(y)$.

Evaluate $f(xy)$ by substituting $xy$ for $x$ in the expression for $f(x)$:

$f(xy) = \sqrt{1 + (xy)^2} = \sqrt{1 + x^2y^2}$.

Evaluate $f(x) \cdot f(y)$ by multiplying the expressions for $f(x)$ and $f(y)$:

$f(x) \cdot f(y) = \sqrt{1+x^2} \cdot \sqrt{1+y^2}$.

Since $1+x^2 \geq 0$ and $1+y^2 \geq 0$ for all real $x, y$, we can use the property $\sqrt{a}\sqrt{b} = \sqrt{ab}$.

$f(x) \cdot f(y) = \sqrt{(1+x^2)(1+y^2)}$.

Expand the product inside the square root:

$(1+x^2)(1+y^2) = 1 \cdot 1 + 1 \cdot y^2 + x^2 \cdot 1 + x^2 \cdot y^2 = 1 + y^2 + x^2 + x^2y^2$.

So, $f(x) \cdot f(y) = \sqrt{1 + x^2 + y^2 + x^2y^2}$.

Now we need to compare $f(xy) = \sqrt{1 + x^2y^2}$ and $f(x) \cdot f(y) = \sqrt{1 + x^2 + y^2 + x^2y^2}$.

Comparing $\sqrt{A}$ and $\sqrt{B}$ for non-negative values is equivalent to comparing $A$ and $B$. We compare $1 + x^2y^2$ and $1 + x^2 + y^2 + x^2y^2$.

Let's look at the difference between the terms under the square root:

$(1 + x^2 + y^2 + x^2y^2) - (1 + x^2y^2) = 1 + x^2 + y^2 + x^2y^2 - 1 - x^2y^2 = x^2 + y^2$.

Since $x$ and $y$ are real numbers, $x^2 \geq 0$ and $y^2 \geq 0$. Therefore, $x^2 + y^2 \geq 0$.

This implies that $1 + x^2 + y^2 + x^2y^2 \geq 1 + x^2y^2$.

Taking the square root of both sides (since both expressions are non-negative):

$\sqrt{1 + x^2 + y^2 + x^2y^2} \geq \sqrt{1 + x^2y^2}$.

Substituting back the function definitions:

$f(x) \cdot f(y) \geq f(xy)$.

This is equivalent to $f(xy) \leq f(x) \cdot f(y)$.

Comparing this result with the given options:

(A) $f(xy) = f(x) \cdot f(y)$

(B) $f(xy) \geq f(x) \cdot f(y)$

(C) $f(xy) \leq f(x) \cdot f(y)$

(D) None of these

The inequality $f(xy) \leq f(x) \cdot f(y)$ matches option (C).

The correct answer is (C) f (xy) ≤ f (x) . f (y).

Given:

The function is $f(x) = \sqrt{a^2 - x^2}$, where $a > 0$.

For a real-valued function involving a square root, the expression under the square root must be non-negative.

So, we require $a^2 - x^2 \geq 0$.

Add $x^2$ to both sides of the inequality:

$a^2 \geq x^2$

This inequality can be rewritten as $x^2 \leq a^2$.

To solve $x^2 \leq a^2$, we can take the square root of both sides. Remember that $\sqrt{x^2} = |x|$ and $\sqrt{a^2} = |a|$. Since $a > 0$, $|a| = a$.

$\sqrt{x^2} \leq \sqrt{a^2}$

$|x| \leq a$

The inequality $|x| \leq a$ means that x is between -a and a, inclusive.

$-a \leq x \leq a$

The domain of the function is the closed interval $[-a, a]$.

Comparing this result with the given options:

(A) (– a, a)

(B) [– a, a]

(C) [0, a]

(D) (– a, 0]

The interval $[-a, a]$ corresponds to option (B).

The correct answer is (B) [– a, a].

Given:

The function $f(x) = ax + b$, where a and b are integers.

We are given two conditions:

$f(-1) = -5$

$f(3) = 3$

Use the given conditions to form equations involving a and b.

Substitute $x = -1$ and $f(x) = -5$ into the function definition $f(x) = ax + b$:

$f(-1) = a(-1) + b$

$-5 = -a + b$

Substitute $x = 3$ and $f(x) = 3$ into the function definition $f(x) = ax + b$:

$f(3) = a(3) + b$

$3 = 3a + b$

We have a system of two linear equations with two variables a and b.

Equation (1): $-a + b = -5$

Equation (2): $3a + b = 3$

We can solve this system using the elimination method. Subtract equation (1) from equation (2):

$(3a + b) - (-a + b) = 3 - (-5)$

$3a + b + a - b = 3 + 5$

$4a = 8$

Divide by 4:

$a = \frac{8}{4}$

$a = 2$

Now, substitute the value of $a = 2$ into either equation (1) or (2) to find b. Using equation (1):

$-a + b = -5$

$-(2) + b = -5$

$-2 + b = -5$

Add 2 to both sides:

$b = -5 + 2$

$b = -3$

The values of a and b are 2 and -3 respectively. Both are integers, as required.

So, $a = 2$ and $b = -3$.

Comparing these values with the given options:

(A) a = – 3, b = –1

(B) a = 2, b = – 3

(C) a = 0, b = 2

(D) a = 2, b = 3

The values $a = 2, b = -3$ match option (B).

The correct answer is (B) a = 2, b = – 3.

Given:

The function $f(x) = \sqrt{4-x} + \frac{1}{\sqrt{x^{2}-1}}$.

The domain of the sum of two functions is the intersection of their individual domains. Let $f_1(x) = \sqrt{4-x}$ and $f_2(x) = \frac{1}{\sqrt{x^{2}-1}}$. We need to find the domain of $f_1(x)$ and the domain of $f_2(x)$ and then find their intersection.

Domain of $f_1(x) = \sqrt{4-x}$:

For $\sqrt{4-x}$ to be a real number, the expression under the square root must be non-negative.

$4 - x \geq 0$

Adding x to both sides:

$4 \geq x$

So, $x \leq 4$.

The domain of $f_1(x)$ is the interval $(-\infty, 4]$.

Domain of $f_2(x) = \frac{1}{\sqrt{x^{2}-1}}$:

For $\frac{1}{\sqrt{x^{2}-1}}$ to be a real number, two conditions must be met:

1. The expression under the square root must be non-negative: $x^2 - 1 \geq 0$.

2. The denominator must not be zero: $\sqrt{x^2 - 1} \neq 0$, which means $x^2 - 1 \neq 0$.

Combining these two conditions, we require that the expression under the square root is strictly positive:

$x^2 - 1 > 0$

Add 1 to both sides:

$x^2 > 1$

This inequality holds if $x > \sqrt{1}$ or $x < -\sqrt{1}$.

$x > 1$ or $x < -1$.

The domain of $f_2(x)$ is the union of the intervals $(-\infty, -1)$ and $(1, \infty)$.

Domain of $f_2(x) = (-\infty, -1) \cup (1, \infty)$.

Domain of $f(x) = f_1(x) + f_2(x)$:

The domain of $f(x)$ is the intersection of the domain of $f_1(x)$ and the domain of $f_2(x)$.

Domain(f) = $(-\infty, 4] \cap ((-\infty, -1) \cup (1, \infty))$

We find the intersection by considering the overlap of the intervals:

The interval $(-\infty, 4]$ overlaps with $(-\infty, -1)$ in the region $(-\infty, -1)$.

The interval $(-\infty, 4]$ overlaps with $(1, \infty)$ in the region $(1, 4]$.

The intersection is the union of these overlapping regions.

Domain(f) = $(-\infty, -1) \cup (1, 4]$.

Comparing this result with the given options, option (A) matches our result.

The correct answer is (A) (– ∞, – 1) ∪ (1, 4].

Given:

The real function $f(x) = \frac{4 \;-\; x}{x \;-\; 4}$.

Domain of f(x):

For a rational function, the domain is all real numbers except for the values of x that make the denominator zero.

Set the denominator equal to zero:

$x - 4 = 0$

$x = 4$

The function is undefined when $x = 4$.

The domain of f(x) is the set of all real numbers except 4.

Domain = $R \setminus \{4\}$.

Range of f(x):

Consider the expression for $f(x)$:

$f(x) = \frac{4 - x}{x - 4}$

Notice that the numerator $(4 - x)$ is the negative of the denominator $(x - 4)$, i.e., $4 - x = -(x - 4)$.

So, for any value of x where the denominator is not zero (i.e., $x \neq 4$), we can simplify the expression:

$f(x) = \frac{-(x - 4)}{x - 4}$

Since $x \neq 4$, $x - 4 \neq 0$. We can cancel the term $(x - 4)$ from the numerator and the denominator.

$f(x) = \frac{-1 \cancel{(x - 4)}}{\cancel{x - 4}} = -1$.

Thus, for all values of x in the domain ($x \neq 4$), the value of the function $f(x)$ is always -1.

The range of the function is the set containing only the value -1.

Range = $\{-1\}$.

Comparing the domain and range with the given options:

(A) Domain = R, Range = {–1, 1} (Incorrect Domain and Range)

(B) Domain = R – {1}, Range = R (Incorrect Domain and Range)

(C) Domain = R – {4}, Range = {– 1} (Correct Domain and Range)

(D) Domain = R – {– 4}, Range = {–1, 1} (Incorrect Domain and Range)

The correct answer is (C) Domain = R – {4}, Range = {– 1}.

Given:

The real function $f(x) = \sqrt{x - 1}$.

Domain of f(x):

For the function $f(x) = \sqrt{x - 1}$ to be defined in the set of real numbers, the expression under the square root must be non-negative.

$x - 1 \geq 0$

Adding 1 to both sides:

$x \geq 1$

The domain of the function is the set of all real numbers x that are greater than or equal to 1.

Domain = $[1, \infty)$.

Range of f(x):

Let $y = f(x) = \sqrt{x - 1}$. We want to find the set of all possible values that y can take.

From the domain, we know that $x \geq 1$.

If $x \geq 1$, then $x - 1 \geq 0$.

The square root of a non-negative number is a non-negative real number.

So, $y = \sqrt{x - 1} \geq 0$.

When $x = 1$, $y = \sqrt{1 - 1} = \sqrt{0} = 0$. This is the minimum value of y.

As x increases, $x - 1$ increases, and $\sqrt{x - 1}$ also increases. As x approaches infinity, $\sqrt{x - 1}$ approaches infinity.

Thus, the value of y can be any non-negative real number.

Range = $[0, \infty)$.

Comparing the domain and range with the given options:

(A) Domain = (1, ∞), Range = (0, ∞) (Incorrect Domain and Range includes 0)

(B) Domain = [1, ∞), Range = (0, ∞) (Incorrect Range does not include 0)

(C) Domain = [1, ∞), Range = [0, ∞) (Correct Domain and Range)

(D) Domain = [1, ∞), Range = [0, ∞) (Same as C, correct Domain and Range)

Options C and D are identical. Let's assume one was intended to be different. Based on our calculation, option (C) or (D) is the correct answer.

The correct answer is (C) or (D) Domain = [1, ∞), Range = [0, ∞).

Given:

The function $f(x) = \frac{x^{2}\;+\;2x\;+\;1}{x^{2}\;-\;x\;-\;6}$.

This is a rational function. The domain of a rational function is the set of all real numbers except for the values of x that make the denominator zero.

Set the denominator equal to zero:

$x^2 - x - 6 = 0$

We need to solve this quadratic equation for x. We can factor the quadratic expression. We look for two numbers that multiply to -6 and add up to -1. These numbers are -3 and 2.

Rewrite the middle term ($-x$) using these numbers:

$x^2 - 3x + 2x - 6 = 0$

Group the terms and factor by grouping:

$(x^2 - 3x) + (2x - 6) = 0$

$x(x - 3) + 2(x - 3) = 0$

Factor out the common binomial $(x - 3)$:

$(x - 3)(x + 2) = 0$

Set each factor equal to zero and solve for x:

Case 1: $x - 3 = 0 \implies x = 3$

Case 2: $x + 2 = 0 \implies x = -2$

The denominator is zero when $x = 3$ or $x = -2$. These values must be excluded from the domain.

The domain of the function is the set of all real numbers except 3 and -2.

Domain = $R \setminus \{-2, 3\}$.

Comparing this result with the given options:

(A) R – {3, – 2} (Correct)

(B) R – {–3, 2} (Incorrect values)

(C) R – [3, – 2] (Incorrect notation and order)

(D) R – (3, – 2) (Incorrect notation and order)

The correct answer is (A) R – {3, – 2}.

Given:

The function $f(x) = 2 - |x - 5|$.

Domain of f(x):

The function involves the absolute value of a linear expression. The absolute value function $|a|$ is defined for all real numbers $a$. The term $|x - 5|$ is defined for all real numbers x. Subtracting $|x-5|$ from 2 is also defined for all real numbers x.

Therefore, the domain of $f(x)$ is the set of all real numbers.

Domain = $R$ or $(-\infty, \infty)$.

Range of f(x):

Let $y = f(x) = 2 - |x - 5|$. We want to find the set of all possible values that y can take.

We know that for any real number $a$, $|a| \geq 0$.

So, for any real number x, $|x - 5| \geq 0$.

Multiply the inequality by -1, which reverses the inequality sign:

$-|x - 5| \leq 0$.

Add 2 to both sides of the inequality:

$2 - |x - 5| \leq 2 + 0$

$2 - |x - 5| \leq 2$.

Thus, the maximum value of $f(x)$ is 2 (which occurs when $x - 5 = 0 \implies x = 5$). As $|x-5|$ increases (as x moves away from 5), $2 - |x-5|$ decreases without bound.

The range of f(x) is $(-\infty, 2]$.

Comparing the domain and range with the given options:

(A) Domain = R+, Range = ( – ∞, 1] (Incorrect Domain and Range)

(B) Domain = R, Range = ( – ∞, 2] (Correct Domain and Range)

(C) Domain = R, Range = (– ∞, 2) (Incorrect Range - does not include 2)

(D) Domain = R+ , Range = (– ∞, 2] (Incorrect Domain)

The correct answer is (B) Domain = R, Range = ( – ∞, 2].

Given:

The functions $f(x) = 3x^2 - 1$ and $g(x) = 3 + x$.

We need to find the values of x for which $f(x) = g(x)$.

Set the expressions for $f(x)$ and $g(x)$ equal to each other:

$f(x) = g(x)$

$3x^2 - 1 = 3 + x$

Rearrange the equation to form a quadratic equation by moving all terms to one side:

$3x^2 - x - 1 - 3 = 0$

$3x^2 - x - 4 = 0$

We need to solve this quadratic equation for x. We can factor the quadratic expression $3x^2 - x - 4$. We look for two numbers that multiply to the product of the leading coefficient (3) and the constant term (-4), which is $3 \times (-4) = -12$, and add up to the coefficient of the x term (-1). The numbers that satisfy these conditions are -4 and 3.

Rewrite the middle term ($-x$) using -4x and 3x:

$3x^2 - 4x + 3x - 4 = 0$

Group the terms and factor by grouping:

$(3x^2 - 4x) + (3x - 4) = 0$

Factor out the common factor from each group:

$x(3x - 4) + 1(3x - 4) = 0$

Factor out the common binomial term $(3x - 4)$:

$(3x - 4)(x + 1) = 0$

Set each factor equal to zero and solve for x:

Case 1: $3x - 4 = 0$

$3x = 4$

$x = \frac{4}{3}$

Case 2: $x + 1 = 0$

$x = -1$

The values of x for which the functions $f(x)$ and $g(x)$ are equal are the solutions to the equation $3x^2 - x - 4 = 0$. These values are $x = -1$ and $x = \frac{4}{3}$.

The question asks for the "domain" for which the functions are equal, which refers to the specific set of x-values where the equality holds.

The set of these values is $\{-1, \frac{4}{3}\}$.

Comparing this result with the given options:

(A) $\left\{ -1,\;\frac{4}{3}\right\}$ - This is the set containing the two values we found.

(B) $\left[ -1,\;\frac{4}{3} \right]$ - This is an interval, not a set of discrete points.

(C) $\left(-1,\frac{4}{3} \right)$ - This is an interval, not a set of discrete points.

(D) $\left[ -1,\frac{4}{3} \right)$ - This is an interval, not a set of discrete points.

The correct answer represents the set of specific x values where the equality $f(x) = g(x)$ is true.

The correct answer is (A) $\left\{ -1,\;\frac{4}{3}\right\}$.

Given:

Function f = {(0, 1), (2, 0), (3, – 4), (4, 2), (5, 1)}

Function g = {(1, 0), (2, 2), (3, – 1), (4, 4), (5, 3)}

The domain of a function given as a set of ordered pairs is the set of the first components of the ordered pairs.

The domain of f is the set of all x-values in the ordered pairs of f.

Domain(f) = $\{0, 2, 3, 4, 5\}$.

The domain of g is the set of all x-values in the ordered pairs of g.

Domain(g) = $\{1, 2, 3, 4, 5\}$.

For the product function $(f \cdot g)(x) = f(x) \cdot g(x)$ to be defined, both $f(x)$ and $g(x)$ must be defined.

The domain of $(f \cdot g)$ is the intersection of the domains of f and g.

Domain$(f \cdot g)$ = Domain(f) $\cap$ Domain(g)

Domain$(f \cdot g)$ = $\{0, 2, 3, 4, 5\} \cap \{1, 2, 3, 4, 5\}$

The elements common to both sets are 2, 3, 4, and 5.

Domain$(f \cdot g)$ = $\{2, 3, 4, 5\}$.

The domain of f . g is given by $\{2, 3, 4, 5\}$.

Given:

Function f = {(2, 4), (5, 6), (8, – 1), (10, – 3)}

Function g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, 5)}

To perform operations on functions defined as sets of ordered pairs, the operations are performed on the elements in the intersection of their domains.

Domain(f) = $\{2, 5, 8, 10\}$

Domain(g) = $\{2, 7, 8, 10, 11\}$

The intersection of the domains is Domain(f) $\cap$ Domain(g) = $\{2, 8, 10\}$.

All the operations $(f-g)$, $(f+g)$, $(f \cdot g)$, and $(\frac{f}{g})$ will have this intersection as their domain (with the additional condition that the denominator is non-zero for division).

(a) $f - g$: $(f - g)(x) = f(x) - g(x)$

For $x=2$: $(f-g)(2) = f(2) - g(2) = 4 - 5 = -1$. Ordered pair: $(2, -1)$.

For $x=8$: $(f-g)(8) = f(8) - g(8) = -1 - 4 = -5$. Ordered pair: $(8, -5)$.

For $x=10$: $(f-g)(10) = f(10) - g(10) = -3 - 13 = -16$. Ordered pair: $(10, -16)$.

$f - g = \{(2, -1), (8, -5), (10, -16)\}$.

This matches option (iii).

(b) $f + g$: $(f + g)(x) = f(x) + g(x)$

For $x=2$: $(f+g)(2) = f(2) + g(2) = 4 + 5 = 9$. Ordered pair: $(2, 9)$.

For $x=8$: $(f+g)(8) = f(8) + g(8) = -1 + 4 = 3$. Ordered pair: $(8, 3)$.

For $x=10$: $(f+g)(10) = f(10) + g(10) = -3 + 13 = 10$. Ordered pair: $(10, 10)$.

$f + g = \{(2, 9), (8, 3), (10, 10)\}$.

This matches option (iv).

(c) $f \cdot g$: $(f \cdot g)(x) = f(x) \cdot g(x)$

For $x=2$: $(f \cdot g)(2) = f(2) \cdot g(2) = 4 \cdot 5 = 20$. Ordered pair: $(2, 20)$.

For $x=8$: $(f \cdot g)(8) = f(8) \cdot g(8) = -1 \cdot 4 = -4$. Ordered pair: $(8, -4)$.

For $x=10$: $(f \cdot g)(10) = f(10) \cdot g(10) = -3 \cdot 13 = -39$. Ordered pair: $(10, -39)$.

$f \cdot g = \{(2, 20), (8, -4), (10, -39)\}$.

This matches option (ii).

(d) $\frac{f}{g}$: $\left(\frac{f}{g}\right)(x) = \frac{f(x)}{g(x)}$, where $g(x) \neq 0$.

The common domain is $\{2, 8, 10\}$. The values of $g(x)$ for these x are $g(2)=5$, $g(8)=4$, $g(10)=13$. None of these are zero, so the function is defined for all x in the common domain.

For $x=2$: $\left(\frac{f}{g}\right)(2) = \frac{f(2)}{g(2)} = \frac{4}{5}$. Ordered pair: $(2, \frac{4}{5})$.

For $x=8$: $\left(\frac{f}{g}\right)(8) = \frac{f(8)}{g(8)} = \frac{-1}{4}$. Ordered pair: $(8, \frac{-1}{4})$.

For $x=10$: $\left(\frac{f}{g}\right)(10) = \frac{f(10)}{g(10)} = \frac{-3}{13}$. Ordered pair: $(10, \frac{-3}{13})$.

$\frac{f}{g} = \{(2, \frac{4}{5}), (8, \frac{-1}{4}), (10, \frac{-3}{13})\}$.

This matches option (i).

Matching Results:

(a) $f - g \quad \to \quad$ (iii)

(b) $f + g \quad \to \quad$ (iv)

(c) $f \cdot g \quad \to \quad$ (ii)

(d) $\frac{f}{g} \quad \to \quad$ (i)

Given:

The relation $R = \{(x, y) : y = x - 5, x \in \mathbb{Z}, y \in \mathbb{Z}\}$.

We need to determine if the ordered pair $(5, 2)$ belongs to this relation.

For the ordered pair $(5, 2)$ to belong to the relation R, the components must satisfy the conditions of the relation.

Here, $x = 5$ and $y = 2$.

Both x and y must be integers. $5 \in \mathbb{Z}$ and $2 \in \mathbb{Z}$, so this condition is satisfied.

The components must satisfy the equation $y = x - 5$.

Substitute $x = 5$ and $y = 2$ into the equation:

LHS = $y = 2$

RHS = $x - 5 = 5 - 5 = 0$

Since $2 \neq 0$, the equation $y = x - 5$ is not satisfied for the ordered pair $(5, 2)$.

Therefore, the ordered pair $(5, 2)$ does not belong to the relation R.

The statement "The ordered pair (5, 2) belongs to the relation R = {(x, y) : y = x – 5, x, y ∈ Z}" is False.

The correct answer is False.

Given:

Set P = {1, 2}.

The Cartesian product $P \times P \times P$ is the set of all ordered triplets $(a, b, c)$ where $a \in P$, $b \in P$, and $c \in P$.

The number of elements in set P is n(P) = 2.

The number of elements in $P \times P \times P$ is n($P \times P \times P$) = n(P) $\times$ n(P) $\times$ n(P) = $2 \times 2 \times 2 = 8$.

Let's list the elements of $P \times P \times P$ systematically.

The first element can be 1 or 2.

The second element can be 1 or 2.

The third element can be 1 or 2.

Possible ordered triplets:

• Starting with 1: (1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2)
• Starting with 2: (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)

So, $P \times P \times P = \{(1, 1, 1), (1, 1, 2), (1, 2, 1), (1, 2, 2), (2, 1, 1), (2, 1, 2), (2, 2, 1), (2, 2, 2)\}$.

The statement claims that $P \times P \times P = \{(1, 1, 1), (2, 2, 2), (1, 2, 2), (2, 1, 1)\}$.

The claimed set has only 4 elements, whereas $P \times P \times P$ has 8 elements.

Moreover, the claimed set is a subset of the actual $P \times P \times P$, but it is not equal to it.

For the statement to be true, the two sets must be exactly the same, containing all and only the same elements.

The statement "If P = {1, 2}, then P × P × P = {(1, 1, 1), (2, 2, 2), (1, 2, 2), (2, 1, 1)}" is False.

The correct answer is False.

Given:

Set A = {1, 2, 3}

Set B = {3, 4}

Set C = {4, 5, 6}

We need to verify if $(A \times B) \cup (A \times C)$ is equal to the given set.

First, calculate the Cartesian product $A \times B$. This is the set of all ordered pairs $(a, b)$ where $a \in A$ and $b \in B$.

$A \times B = \{(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)\}$.

Next, calculate the Cartesian product $A \times C$. This is the set of all ordered pairs $(a, c)$ where $a \in A$ and $c \in C$.

$A \times C = \{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\}$.

Now, find the union of $(A \times B)$ and $(A \times C)$. The union contains all elements that are in either set (or both), without duplication.

$(A \times B) \cup (A \times C) = \{(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)\} \cup \{(1, 4), (1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 4), (3, 5), (3, 6)\}$

The elements in the union are:

$\{(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)\}$.

The set calculated above is exactly the same as the set given in the statement.

The statement is equivalent to the property $A \times (B \cup C) = (A \times B) \cup (A \times C)$.

Let's calculate $A \times (B \cup C)$ as an alternative check.

$B \cup C = \{3, 4\} \cup \{4, 5, 6\} = \{3, 4, 5, 6\}$.

$A \times (B \cup C) = \{1, 2, 3\} \times \{3, 4, 5, 6\}$

$A \times (B \cup C) = \{(1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6)\}$.

This matches the given set.

The statement is True.

The correct answer is True.

Given:

Two equal ordered pairs $(x - 2, y + 5) = \left(-2, \frac{1}{3}\right)$.

If two ordered pairs are equal, their corresponding components must be equal.

Equating the first components:

$x - 2 = -2$

Equating the second components:

$y + 5 = \frac{1}{3}$

Solve the first equation for x:

$x - 2 = -2$

Add 2 to both sides:

$x = -2 + 2$

$x = 0$

The statement claims that $x = 4$. Our calculation shows $x = 0$.

Solve the second equation for y:

$y + 5 = \frac{1}{3}$

Subtract 5 from both sides:

$y = \frac{1}{3} - 5$

To subtract, find a common denominator, which is 3.

$y = \frac{1}{3} - \frac{5 \times 3}{1 \times 3} = \frac{1}{3} - \frac{15}{3}$

$y = \frac{1 - 15}{3} = \frac{-14}{3}$

The statement claims that $y = \frac{-14}{3}$. Our calculation confirms this.

However, for the entire statement to be true, both claims about x and y must be correct based on the given information.

Since the claim $x = 4$ is incorrect (we found $x = 0$), the entire statement is false.

The statement "If (x - 2, y + 5) = $\left(-2,\frac{1}{3} \right)$ are two equal ordered pairs, then x = 4 , y = $\frac{-14}{3}$" is False.

The correct answer is False.

Given:

The Cartesian product $A \times B = \{(a, x), (a, y), (b, x), (b, y)\}$.

The Cartesian product $A \times B$ is defined as the set of all ordered pairs $(u, v)$ where the first element $u$ belongs to set A, and the second element $v$ belongs to set B.

So, the elements of set A are the unique first components of the ordered pairs in $A \times B$.

From the given ordered pairs, the first components are a, a, b, b.

The set of unique first components is $\{a, b\}$.

Therefore, $A = \{a, b\}$.

The elements of set B are the unique second components of the ordered pairs in $A \times B$.

From the given ordered pairs, the second components are x, y, x, y.

The set of unique second components is $\{x, y\}$.

Therefore, $B = \{x, y\}$.

The statement claims that if $A \times B = \{(a, x), (a, y), (b, x), (b, y)\}$, then $A = \{a, b\}$ and $B = \{x, y\}$.

Our derivation shows that $A = \{a, b\}$ and $B = \{x, y\}$.

The derived sets match the sets claimed in the statement.

The statement is True.

The correct answer is True.